structural analysis

STABILITY, DETERMINACY OF STRUCTURES AND CONSISTENT DEFORMATIONS METHOD 1

CHAPTER ONE

STABILITY, DETERMINACY OF STRUCTURES AND

CONSISTENT DEFORMATIONS METHOD 1.1. STABILITY OF STRUCTURES: Before deciding the determinacy or indeterminacy of a structure we should first of all have a structure which is stable. The question of determinacy or indeterminacy comes next. We shall now discuss 2-D or single plane structures. (Defined and accommodated in a single plane). 1.1.1. STABLE STRUCTURE: A stable structure is the one, which remains stable for any conceivable (imaginable) system of loads. Therefore, we do not consider the types of loads, their number and their points of application for deciding the stability or determinacy of the structure. Normally internal and external stability of a structure should be checked separately and if its overall stable then total degree of indeterminacy should be checked. 1.2. ARTICULATED STRUCTURES:

This may be defined as “A truss, or an articulated structure, composed of links or bars, assumed to be connected by frictionless pins at the joints, and arranged so that the area enclosed within the boundaries of the structure is subdivided by the bars into geometrical figures which are usually triangles.”

1.3. CONTINUOUS FRAME:

“ A continuous frame is a structure which is dependent, in part, for its stability and load carrying capacity upon the ability of one or more of its joints to resist moment.” In other words, one or more joints are more or less rigid. 1.4. DETERMINACY:

A statically indeterminate structure is the one in which all the reactive components plus the internal forces cannot be calculated only from the equations of equilibrium available for a given force system.These equations, of course, are

∑ H = 0, ∑ V = 0 and ∑ M = 0 The degree of indeterminacy for a given structure is, in fact , the excess of total number of reactive components or excess of members over the equations of equilibrium available.

It is convenient to consider stability and determinacy as follows.

a) With respect to reactions, i.e. external stability and determinacy.

b) With respect to members, i.e. internal stability and determinacy.

c) A combination of external and internal conditions, i.e. total stability and determinacy.

2 THEORY OF INDETERMINATE STRUCTURES

1.4.1. EXTERNAL INDETERMINACY: A stable structure should have at least three reactive components, (which may not always be sufficient) for external stability of a 2-D structure, which are non-concurrent and non-parallel.

Fig. 1.1. Stable & determinate.

000

Fig. 1.2. Stable & determinate.

External indeterminacy is, in fact, the excess of total number of reactive components over the equations of equilibrium available.

3 + 2 = 5 Fig. 1.3.

No. of reactions possible = 5 No. of Equations of equilibrium available = 3 Degree of External indeterminacy = 5 − 3 = 2

3 + 3 = 6

Fig. 1.4 Stable & Indeterminate to 2nd degree. (Fig. 1.3)

Fig. 1.4. Stable & externally indeterminate to 3rd degree.

2 2 = 4 Fig. 1.5.


Stable & Indeterminate to Ist degree. (Fig. 1.5)

3 + 1 + 2 + 2 = 8 Fig. 1.6.

Stable & externally indeterminate to 5th degree. (Fig. 1.6) Remove any five suitable redundant reactions to make it statically determinate. 1.4.2. INTERNAL INDETERMINACY:

This question can be decided only if the minimum number of reactive components necessary for external stability and determinacy are known and are acting on the structure. This type of indeterminacy is normally associated with articulated structures like trusses. We assume that the structure whose internal indeterminacy is being checked is under the action of minimum reactive components required for external stability at the supports.

The basic form of the truss is a triangle.

To make the truss, add two members and one joint and repeat.

Fig 1.7

Let us assume that

j = Total number of joints.

b = Total number of bars.

r = Minimum number of reactive components required for external stability/determinacy. b + r = 2j total number of total number of unknowns. equations available (at joints).


1. If b + r = 2 j Stable & internally determinate. Check the arrangement of members also. 2. If b + r > 2 j Stable & internally indeterminate. (degree of indeterminacy would be decided by the difference of these two quantities).

3. If b + r < 2 j Unstable. A structure is said to have determinacy or indeterminacy only if it is stable. Now we consider some examples.

2 3 5 7 9 11

4 8 y

x1 6 10

Fig. 1.8. b = 11

r = 3 (Minimum external reactions required for external stability/determinacy)

j = 7

b + r = 2 j

11 + 3 = 2 × 7

14 = 14

This truss of fig. 1.8 is stable and internally determinate.

1 3 5 9 15

4 8 12

11 13

2 6 10 14

7

Fig. 1.9. b = 15 r = 3 j = 9 b + r = 2 j 15 + 3 = 2 × 9 18 = 18


The truss of fig. 1.9 is stable and internally determinate.

4 6 12 15

2 8 13 17

181 3 5 7 910

11

14 16

Fig. 1.10.

b = 18 r = 3 j = 10 b + r = 2 j 18 + 3 = 2 × 10 21 > 20 This truss of fig. 1.10 is stable & internally indeterminate to 1st degree.

2 6 10 13

4 8 11 15

161 3 5 79 12 14

17

Fig. 1.11.

b = 16 r = 3 j = 10 b + r = 2 j 17 + 3 = 2 × 10 20 = 20 This truss is Unstable by inspection although the criterion equation is satisfied. The members in

indicated square may get displaced and rotated due to gravity loads.

Always inspect member positions. Insert one member in the encircled box or manage prevention of sliding by external supports to make it stable.

NOTE:- The difference between the internal and the external indeterminacy is only in the definition of ‘r’

1.4.3. TOTAL INDETERMINACY The question of total indeterminacy is of little interest and we have got different equations for different types of structures. For example, the previous equation, i.e., b + r = 2 j can be used to check the total degree of indeterminacy of an articulated structure like truss by slightly modifying the definition of “ r ” which should now be considered as the “total number of reactive components available”.


b + r = 2 j where b = Total number of bars.

r = Total number of reactive components available.

j = Total number of joints

Example No. 1: Determine the external and internal conditions of stability and determinateness for the

following structures:-

3 2

14

9

7 8

65

Fig. 1.12

(i) External Stability And Determinacy:- Number of reactive components available = 2

Number of equations of equilibrium available = 3

∴ Unstable. (Visible also)

(ii) Internal Stability And Determinacy

b = 9 r = 3 j = 6 b + r = 2 j 9 + 3 = 2 × 6 12 = 12 Degree of Indeterminacy = D = 12 − 12 = 0

∴ Stable and Internally Determinate, if arrangement is improved to have Σ = 3.

Example No. 2:

Link

Fig. 1.13.


* In this case the presence of a pin at each end of the link makes one additional type of movement possible if reaction components are removed. Two condition equations are therefore provided by the link in terms of algebraic sum of moments equal to zero at the joints of link.

External Stability and Determinacy. Number of reactive components = 5 Number of equations of equilibrium available = 3 + 2* = 5 Degree of indeterminacy = 5 − 5 = 0 ∴ Stable and Externally Determinate. (Structure of fig. 1.13.) Example No. 3:

0 0

1 2 3

4 5 6

7 8 9 1011

1213

1415

1617

18 19

2021

22

Fig. 1.14.

(i) External Stability and Determinacy:− Number of reactions = 3 Number of equations = 3 D = 3 − 3 = 0 ∴ Externally Stable and Determinate (ii) Internal Stability and Determinacy:- b = 22 r = 3 j = 11 b + r = 2 j D = ( b + r ) − 2 j = ( 22 + 3 ) − ( 2 × 11 ) = 25 − 22 D = 3 where D = Degree of indeterminacy. ∴ Stable and indeterminate to 3rd degree. Example No. 4:

Continuous frame

Fig. 1.15. External Stability and Determinacy:- Number of reactions = 9 Number of equations = 3 D = 9 − 3 = 6 ∴ Stable and Indeterminate to 6th degree. (fig. 1.15).


Example No. 5:

1

2

6

3

4

5

Fig 1.16

(i) External Stability And Determinacy :- Number of reactions = 6 Number of equations = 3 Degree of indeterminacy = 6 − 3 = 3 ∴ Stable and externally Indeterminate to 3rd degree. (ii) Internal Stability and Determinacy :- b = 6 r = 3, where r is the minimum reactive components required for external j = 6 stability and determinacy. Degree of indeterminacy of rigid jointed structure. (Fig. 1.16) D = (3b + r ) − 3 j D = ( 3 × 6 + 3 ) − ( 3 × 6 ) D = 21 − 18 D = 3 ∴ Stable and indeterminate to 3rd degree. Example No. 6:

(i) External Stability and Determinacy :-

3

20 214

185

1917

16 15

1412

13

11 10

6

7

2

18

9

Fig. 1.17.


Number of reactions = 4 Number of equations = 3 D = 4 − 3 = 1 ∴ Stable and indeterminate to Ist degree.

(ii) Internal Stability and Determinacy :- b = 21 r = 3 j = 11 D = ( b + r ) − 2 j = ( 21 + 3 ) − 2 × 11 D = 24 − 22 = 2 ∴ Stable and indeterminate to 2nd degree. Note: In case of a pin jointed structure, there is one unknown per member and in case of rigid jointed

structure there are three unknowns at a joint. Example No. 7:

o o

Fig. 1.18. (i) External Stability and Determinacy :- Number of reactions = 3 Number of equations = 3 D = 3 − 3 = 0 ∴ Stable and Determinate. (ii) Internal Stability and Determinacy :- b = 38 r = 3 j = 20 D = ( b + r ) − 2 j = (38 + 3) − 2 × 20 = 41 − 40 D = 1 ∴ Stable and indeterminate to Ist degree. (Fig. 1.18) Example No. 8:

o o

Fig. 1.19.


(i) External Stability and Determinacy :- Number of reactions = 3 Number of equations = 3 D = 3 − 3 = 0 ∴ Stable and Determinate. (ii) Internal Stability and Determinacy :- b = 54

r = 3

j = 25

b + r = 2 j

54 + 3 > 2 × 25

57 > 50 D = 57 − 50 = 7 ∴ Stable and indeterminate to 9th degree. (Fig. 1.19) Example No. 9:

1 2

3

7

8

9

5

10

11

12 16

15

14 17

18

19

13

4

6

Fig. 1.20.

(i) External Stability and Determinacy :- Number of reactions = 12 Number of equations = 3 D = 12 − 3 = 9 ∴ Stable and indeterminate to 9th degree.

(ii) Internal Stability and Determinacy :- b = 19 r = 3 j = 16 D = ( 3 b + r ) = 3 j

= ( 3 × 19 + 3 ) = 3 × 16

= 60 > 48

D = 60 − 48 = 12

∴ Stable and Internally Indeterminate to twelfth degree. (Fig. 1.20)


Example No. 10:

5 3

2

8

9

10 1

4

6

7

11

Fig. 1.21.

(i) External Stability and Determinacy :- Number of reactions = 6 Number of equations = 3 D = 6 − 3 = 3 ∴ Stable and Indeterminate to 3rd degree. (ii) Internal Stability and Determinacy : - b = 11 r = 3 j = 9 D = ( 3 b + r ) − 3 j = ( 3 × 11 + 3 ) − 3 × 9 = 36 − 27 D = 9 ∴ Stable and indeterminate to 9th degree. (Fig. 1.21)

Example No. 11:

2

1

4

3

7 8

9 10

5

6

Fig. 1.22.


(i) External Stability and Determinacy :- Number of reactions = 6 Number of equations = 3 D = 6 − 3 = 3 ∴ Stable and indeterminate to 3rd degree.

(ii) Internal Stability and Determinacy :- b = 10 r = 3 j = 9 D = ( 3 b + r ) − 3 j = ( 3 × 10 + 3 ) − 3 × 9 D = 33 − 27 D = 6 ∴ Stable and indeterminate to 6th degree. (Fig. 1.22)

Example No. 12:

12

11

10

98

7 2

3

4

6

513

14 1

o o o o Fig. 1.23.

(i) External Stability and Determinacy :- Number of reactions = 2 Number of equations = 3 ∴ Unstable Externally. (Visible also) (ii) Internal Stability and Determinacy :-

b = 14 r = 3 j = 8 D = ( b + r ) − 2 j = ( 14 + 3 ) − 2 × 8 D = 1 ∴ Stable and Internal Indeterminacy to Ist degree.


Example No. 13:

1

3 5

2

8

6

74 10 12 15

19

2018

1716149

13

M=0

W

11

Fig. 1.24.

(i) External Stability and Determinacy :- Number of reactions = 4 Number of equations = 3 + 1 = 4 D = 4 − 4 = 0 ∴ Stable and Determinate. (ii) Internal Stability and Determinacy :- b = 20 r = 4 (Note this. A roller at either support will create instability) j = 12 ( b + r ) = 2 j ( 20 + 4) = 2 × 12 24 = 24 D = 24 − 24 = 0 (Here minimum r is 4 for internal stability and determinacy.) ∴ Stable and determinate.

Example No. 14:

1 5 9 13 22 26 30

2

3

46

78

11

10 1412

15

16

18

17 1921

20

23

24

2725 29

28

31

3332

34

38

3537

36

403941

42 43

M=0r = 1

M=0r = 1

W W

Fig. 1.25.



Number of reactions = 6 Number of equations = 3 + 2 = 5 D = 6 − 5 = 1 ∴ Stable and Indeterminate to Ist degree.

(ii) Internal Stability and Determinacy :- b = 43 r = 3 + 2 = 5 (take notice of it). Two pins where ΣM = 0 j = 24 b + r = 2 j 43 + 5 = 2 × 24 48 = 48 D = 48 − 48 = 0 ∴ Stable and Determinate. (Fig. 1.25)

Example No. 15:

M=0

M=0

M=0

M=0

M=0 Fig. 1.26.


Number of reactions = 8 Number of equations = 8 = (3 + 5) D = 8 − 8 = 0 ∴ Stable and Determinate.

(ii) Internal Stability and Determinacy :- b = 42 r = 3 + 5 = 8. There are 5 joints where ΣM = 0 j = 25 b + r = 2 j 42 + 8 = 2 × 25 50 = 50 D = 50 − 50 = 0 ∴ Stable and Determinate.


Example No. 16:

o o o o

2

1

8

7

11

12

14

16

15

3

4

9

10

6 5

13


Number of reactions = 4 Number of equations = 3 D = 4 − 3 = 1 ∴ Stable and Indeterminate to Ist degree. (ii) Internal Stability and Determinacy :-

b = 16 r = 3 j = 9 D = ( b + r ) − 2 j = ( 16 + 3 ) − 2 × 9 = 19 − 18 D = 1 ∴ Stable and Indeterminate to Ist degree. In the analysis of statically determinate structures, all external as well as internal forces are completely known by the application of laws of statics.Member sizes do not come into the picture as no compatibility requirements are to be satisfied. However, in the analysis of indeterminate structures we should have member sizes, sectional and material properties before doing the analysis as member sizes would be involved in the determination of deflections or rotations which are to be put in compatibility equations afterwards. Now we discuss methods for finding deflection and rotations. 1.5. METHODS FOR FINDING DEFLECTION AND ROTATION;-

Usually following methods are used in this classical analysis of structures.. --- Unit - load method. (Strain energy method). --- Moment - area method. --- Conjugate beam method (a special case of moment - area method). 1.5.1. MOMENT AREA THEOREM (1) ;-

The change of slope between tangents drawn at any two points on the elastic curve of an originally straight beam is equal to the area of the B.M.D between these two points when multiplied by 1/EI (reciprocal of flexural stiffness),


A B

Tangent at A

Tangent at B

AB

= --- (Area of B.M.D.1EI between A & B)

AB

= --- (AREA)1EIAB AB

Signs of Change of Slope:-

AB Tangent at A

Tangent at B(a)

A BTangent at A

Tangent at B

(b)

A B

Elastic curve

Elastic curve

Elastic curve

Fig 2.1Fig 2.1

θ

θ

θ

θ

ABθ

(a) Positive change of slope, θAB is counterclockwise from the left tangent. (Fig. 2.1a)

(b) Negative change of slope, θAB is clockwise from the left tangent. (Fig. 2.1b)

1.5.2. MOMENT AREA THEOREM (2) :-

“The deviation of any point on elastic curve from the tangent drawn at some other point on the

elastic curve is equal to 1EI multiplied by the moment of the area of the bending moment diagram

between these two points”. The moment may generally be taken through a point where deviation is being measured.

A B

Elastic curve.

BA = Deviation of point B w.r.t tangent at A

t

BAt = (Area)AB

XB1EI

Fig 2.2

AB

tangent at A

tangent at B


1.5.3. SIGN CONVENTION FOR DEVIATIONS:-

BA=Deviation of point B on elastic curve w.r.t. tangent at point A on elastic curve

t

AB

(a) Positive deviationFig 2.2

BA

Fig 2.2 (b) Negative deviation

tangent at A

Elastic curve

BA=Deviation of point B on elastic curve w.r.t. tangent at point A on elastic curve

t

(a) Positive Deviation:- B located above the reference tangent. (Tangent at A; Fig. 2.2a)

(b) Negative Deviation:- B located below the reference tangent. (Tangent at A; Fig. 2.2b) 1.5.4. INEQUALITY OF tBA AND tAB

Depending upon loading, these two deviations tab and tba may not be equal if loading is unsymmetrical about mid span of the member.

tABtBA

Reference tangent at B Reference tangent at A

Fig. 2.3

Elastic curveA B

tAB tBA

1.6. BENDING MOMENT DIAGRAM BY PARTS:

In order to compute deviations and change of slope by moment area method, bending moment diagram may be drawn in parts i.e. one diagram for a particular load starting from left to right. Same sign convention would be followed for bending moment and shear force as have been followed in subjects done earlier. Bending moment would be positive if elastic curve resembles sagging i.e. compression at top fibers and tension at the bottom fibers while shear force would be


positive at a section of a portion being considered as a free body when left resultant force acts upwards and right resultant force acts downwards. Negative bending moment and shear force would be just opposite to this.

1.6.1. SIGN CONVENTIONS FOR SHEAR FORCE AND BENDING MOMENT

Compression

Compression

Tension

Tension

Positive B.M. Positive Shear Force

Negative Shear Force

Negative B.M.

L

L

R

R

L

L

R

R

Fig 2.4 Consider the following loaded beam. Start from faces on LHS and move towards RHS. Construct BMS due to all forces encountered treating one force at a time only.

P P

DC

1 2

BA

o o

+

-

-

-

Ra

B.M.D. due to Ra = Ra x L

B.M.D. due to P

B.M.D. due to P

1

2

2

1

2

P x --- L

P x ---

34

L2

B.M.D. due to U.D.L.W x (L/2) x ___ = ___(L/2)

2WL8

L/4 L/2L/4Rb

We observe that the moment effect of any single specified loading is always some variation of the general equation. Like

y = kXn (1)


This Relationship has been plotted below. While drawing bending moment diagrams by parts and starting from left, for example, Ra is acting at A. Imagine that Ra is acting while support at A has been removed and beam is fixed adequately at B ( just like a cantilever support), the deflected shape whether sagging or hogging will determine the sign of B.M.D. Similar procedure is adopted for other loads.

y

xx

y=kXn

n=2

dxX

where k = constant n = degree of

curve ofB.M.D

i.e. y=PX k=P, n=1y = k = w/2,wx

22

b

h

y

A

Fig. 2.6Generalized variation of B.M. w.r.t. x

X

In general X = ∫ Xd AA

Area of the strip = ydX = kXn dX by putting value of y.

Total area = A = b

∫o kXn dX

A =

kXn+1

n + 1

b o

A = Kb(n+1)

(n + 1)

We want to find the total area under the curve in terms of ‘b’ and ‘h’ and for that the constant ‘k’ has to be evaluated from the given boundary conditions. At X = b , y = h Put this in (1) , y = kX n we get h = kb n

or k = hbn Put this in equation for A above.


A = h bn+1

bn (n+1) Simplifying

= h bn . bbn (n+1)

So A = bh

(n+1) (2)

Now its centroid would be determined with reference to fig. 2.6..

X−

= ∫ X d AA

= ∫ X (ydX)A Put y= kXn

= ∫ X kXn dXA

= b

∫o k Xn+1 dX

A Now put k= hbn and A=

bh(n+1) we have

= b

∫o h/bn (X)n+1 dX

bh/(n+1)

= b

∫o h (Xn+1) dX(n+1)

hbn+1 simplifying step by step

= (n + 1)b n+1

b

∫o Xn+1dX

= (n + 1)b n+1

Xn+2

(n+2)

b o

= (n + 1)b n+1

bn+2

(n+2)

= (n + 1)b (n+1) .

bn+1 . b(n+2)

__X =

b (n+1)(n+2) (3)

X−

is the location of centroid from zero bending moment From above figure 2.6, we have

X−

+ X/ = b

∴ X/ = b − X−

= b − b (n + 1)(n + 2) Simplify


= b (n + 2) − b (n + 1)

(n + 2)

= bn + 2b − bn − b

(n + 2)

X/ = b

(n+2) (4)

This gives us the location of centroid from the ordinate of B.M.D

A= bh

(n+1) (2)

Note:- While applying these two formulae to calculate the deflection and the rotation by moment area

method and with diagrams by parts, it must be kept in mind that these two relationship assume zero

slope of the B.M. Diagram at a suitable point. It may not be applied to calculate A & X−

within various segments of the B.M.D where this condition is not satisfied. Apply the above equations for area and centroid to the following example.

L

x = ---L4

Tangent of elastic curve at A.

Elastic curve

0

WL2

WL6

2

3A =

( - )

2nd degree curve

ab Cantilever under u.d.l

B.M.D

Fig 2.7

Aθ

B


Tangent at A onElastic curve.

L

ab

P

A B

PL

Eleastic curve

A

a= t AB

B.M.D

Fig. 2.8

X = 2/3L

X = L/3

(−ve) sign in the deflection of diagram below does not mean that area is (−ve) but ordinate of BMD is (−ve). For loads the fig. 2.7.

∆a = 1EI

A ×

3L4

= 1EI

-WL3

6 × 3L4

= −WL4

8EI

1.7. FIRST THEOREM OF CONJUGATE BEAM METHOD :−

In simple words the absolute slope at any point in the actual beam is equal to the shear force at the

corresponding point on the conjugate beam which is loaded by MEI diagram due to loads on actual

beam. 1.7.1. SECOND THEOREM OF CONJUGATE BEAM METHOD :-

The absolute deflection at any point in the actual beam is equal to the B.M at the corresponding

point on the conjugate beam which is loaded by MEI diagram.

The reader is reminded to draw conjugate beams for actual beams under loads very carefully by giving due consideration to support conditions of actual beam. In general for a fixed and free end of actual beam, the corresponding supports would be free and fixed in conjugate beam respectively. Deflection ∆ at any point on actual beam is associated with the bending moment at corresponding point on conjugate beam while rotation θ at any point on actual beam is associated with shear force at corresponding point on conjugate beam. At an actual hinge support ∆ is equal to zero and θ is there indicating non development of moment at the support (Shear force present,


bending moment zero). The corresponding support conditions in conjugate beam would be such where bending moment is zero and shear force may be there i.e., a hinge is indicated. See the following example.

EXAMPLE :- Calculate the central deflection by the conjugate beam method:

EI=Constt.

L/2

P

C

PL16EI

2PL

16EI2 PL

16EI2

PL16EI

2

L/6C/A B

//

A B

PL/4P/2

+

PL/8EI

P/2

B.M.D/EI

+

A = --- x L x ---12

PL4EI

= ----PL8EI

2

2 a = b = -----PL

16EI

Actual beamunder load

Fig. 2.9

Conjugate beamunder M/EI diagramas a load

θ θ

∆C = Mc′ = PL2

16EI × L2 −

PL2

16EI × L6 (considering forces on LHS of

= PL3

32EI − PL3

96EI = 3PL3 − PL3

96EI = 2EPL3

96EI point C of shaded area)

∆C = PL3

48EI

1.8. STRAIN ENERGY :-

“The energy stored in a body when it undergoes any type of deformation (twisting, elongation, shortening & deflection etc.) under the action of any external force is called the strain energy.” If this strain energy is stored in elastic range it is termed as elastic strain energy. All rules relating to strain energy apply. The units of strain energy are the same as that of the work i.e., joule (N − mm, N − m).


1.8.1. TYPES OF STRAIN ENERGY :- 1.8.1.1 STRAIN ENERGY DUE TO DIRECT FORCE :-

PL

P

AE = Axial Stiffness

Fig. 2.10

Work done by a gradually increased force ‘P’ is equal to area of load − deflection diagram = P/2 ∆. (From graph) … Stress ∝ Strain (Hooke’s Law)

So f ∝ ∈

f = Constt . ∈

f = E . ∈

PA = E ×

∆L

so ∆ = PLAE Strain energy will be

12 P∆ from above. So putting it we have.

⇒ U = P2

PL

AE , where U is the internal strain energy stored.

U = P2L2AE (for single member)

U = Σ P2L2AE (for several members subjected to axial forces)

1.8.1.2. STRAIN ENERGY DUE TO BENDING, SHEAR FORCE AND TORSION :-

(1) U = L

∫O M2 dX2 EI . This is elastic strain energy stored due to bending.

(2) Strain Energy Due to shear force:- U = L

∫O Q2 ds2AG where Q is shear force and G is shear modulus


(3) Strain Energy Due to Torsion:- U = L

∫O T2 ds2GJ (Consult a book on strength of Materials). Where

T is Torque and J is polar moment of inertia. 1.9. CASTIGLIANO’S THEOREM :-

In 1879, Castigliano published two theorems connecting the strain energy with the deformations and the applied loads.

1.9.1 CASTIGLIANO’S FIRST THEOREM :-

The partial derivative of the total strain energy stored with respect to a particular deformation gives the corresponding force acting at that point.

Mathematically

PM

∂U∂∆ = P Where U is strain energy stored in bending

and ∂U∂θ = M . Here ∆ is connected with loads and θ with moment.

1.9.2. CASTIGLIANO’S SECOND THEOREM :- The partial derivative of the total strain energy stored with respect to a particular force gives the

corresponding deformation at that point. Mathematically,

∂U∂P = ∆

and ∂U∂M = θ Here ∆ is connected with loads and θ with moment.

1.10. CONSISTENT DEFORMATION METHOD :-

This method may be termed as redundant force method or simply a force method. In this method, the statically indeterminate structure is idealized as a basic determinate structure under the action of applied loads plus the same structure under the action of redundant forces considered one by one. The deformations produced at the points of redundancy are calculated in the above-mentioned basic determinate structures and then these calculated deformations are put into compatibility requirement for the structure. Normally these are satisfied at a joint.


Now for a given beam, various possible Basic determinate structures (BDS) would be given. A clever choice of BDS for a given structure can reduce the amount of time and labour.

1. First alternative

θ θ

θ is presentis present

An indeterminate structure can be made determinate in several ways and the corresponding quantities may be calculated very easily. However, we will notice that a clever choice of making a basic determinate structure will reduce the time of our computations tremendously. In Figs. 2.11 and 2.12 various options regarding choice of BDS are given while Figs. 2.13 and 2.14 illustrate how to make conjugate beam for a given beam using the guidelines stated earlier. Consider another loaded beam in Fig. 2.15.


A BMa

Ra

P P

B

B

B

P P

A

/

Basic determinte structureunder applied loads only.

Rb has been chosenas redundant.

Fig. 2.15

Fig. 2.15 a where ∆B is the deflection at point B due to the applied loads.

BA

B.D.S. under unitredundant force at B.

1

bb

Fig. 2.15 b So compatibility of deformation at B requires that ∆B + Rb × δbb = 0 (Deflection Produced by loads Plus that by redundant should where ∆B = Deflection at B due to applied loads in a BDS. be equal to zero at point B) δbb = deflection at B due to redundant at B in a BDS.

or Rb = − ∆Bδbb (sign is self-adjusting)

A

P P

Ba Ma has beenconsidered asredundant force.

Fig. 2.16

θ

θa = Slope at point. A due to applied loads only in a BDS.

The other option of a simple beam as BDS is shown in fig. 2.16.


A Ba

B.D.S. under unit redundant moment at A.where aa = slope at A due to unit redundant moment at A.

M = 1

Fig 2.16a

∝

a

Compatibility equation θ a + Ma . ∝ aa = 0 (Slope created by loads + slope created by redundant moment should be zero)

or Ma = − θ a∝aa

“In consistent deformation method (force method ), there are always as many conditions of geometry as is the number of redundant forces.” 1.11. Example No. 1:- Analyze the following beam by the force method. Draw S.F. & B.M. diagrams. SOLUTION :-

P

Ma

A

Ra

EI = Constt.

BL/2L/2

RbFig2.17

Number of reactions = 3 Number of equations = 2 Degree of Indeterminacy = 3 − 2 = 1 Indeterminate to Ist degree. SOLUTION: (1) Chose cantilever as a basic determinate structure.

B

L/2 L/2

PB

B

+ L

B

B

bbEI = Constant

1

Fig 2.17a Fig 2.17b δbb=Deflection of point B due to unit load at B B.D.S. under applied loads. B.D.S. under unit redundant force at B.


Therefore, now compatibility requirement is

∆B + Rb × δbb = 0 ( Deflection created by actual loads + deflection created by redundant Rb should be equal to zero at support B)

or Rb = − ∆Bδbb → (1)

Therefore, determine these deflections ∆B and δbb in equation (1) either by moment area method or by unit load method.

1.11.1. DETERMINE ∆B AND δbb BY MOMENT - AREA METHOD :-

L/2 L/2

P

P

B

B/

BEI = Constant

PL2

12

PL8

PL2

L2 =x x

2

o BMD due to applied loads.

PL/82

L/6 L/3 L/2

PL/2

Area of BMD = BDS underapplied loads

A

∆B = I

EI

−

PL2

8

L

2 + L3

= I

EI

−

PL2

8 × 5L6

∆B = − 5PL3

48EI

L

I

bb

BA

I

L

2/3 L

L2

12 x L x L

2

=o o

L=Lx1

Fig 2.18 a

BMD dueUnit redundant

BDS under unitredundant at B

δbb = I

EI

−

L2

2 × 2L3


δbb = − L3

3EI , Putting ∆B and δbb in equation (1)

Rb = −

−5PL2

48EI / −L3

3EI By putting ∆B and δbb in compatibility equation

= − 5PL3

48EI × 3EIL3 = −

5P16

The (− ve) sign with Rb indicates that the direction of application of redundant force is actually

upwards and the magnitude of redundant force Rb is equal to 5P16 . Apply evaluated redundant at point B.

P

L/2L/2y

x11P16

5P16

Ma = 3PL16

Fig. 2.19

∑fy = 0 Ra + Rb = P

Ra = P − Rb = P − 5P16 =

11P16 . Now moment at A can be calculated.

Direction of applied moment at A = 5P16 × L − P .

L2 =

5PL16 −

PL2

= 5PL − 8PL

16

= − 3 PL16

The (−ve) sign with 3 PL16 indicates that the net applied moment about ‘A’ is clockwise. Therefore, the

reactive moment at the support should be counterclockwise (giving tension at top). Apply loads and evaluate redundant on the given structure.


L/2 L/2

Ma =3PL16

EI = Constant 5P1611P

16

Rb =

0 S.F.D

11P16

5P165P165PL

32

3PL16

+

(-ve) B.M0

X = L811

0 B.M.D

0+

P

Fig. 2.20 LOCATION OF POINT OF CONTRAFLEXURE :-

MX = 5 PX

16 − P

X −

L2 = 0

= 5 PX

16 − PX + PL2 = 0

= − 11PX

16 + PL2 = 0

= PL2 =

11PX16

X = 8L11

Note:- In case of cantilever, moment − area method is always preferred because slope is absolute everywhere.

L/2 L/2P

BEI = Constant

A

Elastic curveFig. 2.21


Solution: (2) As a second alternative, Chose Simply Supported Beam as a basic determinate structure.

+

Fig. 2.21a1

Fig. 2.21bB.M.D dueto unit redundantmoment at A

Fig 2.21d

L/3EI

Fig. 2.21c

BDS underloads

BDS underunit redundant

L

(by 1st momentarea theorem)

diagram onconjugate beam

2

2 2

22

6

∝ aa = L

3EI

θ a = PL2

16EI (by 1st moment area theorem)

For fixed end, there is no rotation. Therefore compatibility equation becomes θ a + Ma ∝ aa = 0 (slope at A created by loads + slope at A created

So Ma = − θ a∝aa by redundant should be zero).

θ a & ∝ aa are the flexibility co−efficients. Putting these in compatibility equation

we have, Ma = − PL2

16EI × 3EIL

Ma = − 3PL16

The (−ve) sign with Ma indicates that the net redundant moment is in opposite direction to that assumed. Once Ma is known, Ra and Rb can be calculated.

L/2 L/2P

BEI = Constant

A

3PL16

11P16

Ra= 5P16

Rb=

Fig. 2.22


To calculate Rb, ∑Ma = 0

Rb × L − P × L2 +

3PL16 = 0

Rb × L = PL2 −

3PL16

= 8 PL − 3 PL

16

Rb × L = 5PL16

Rb = 5P16

∑fy = 0 Ra + Rb = P so Ra = P − Rb

= P − 5P16

Ra = 11P16

Note:- In case of simply supported beam, conjugate beam method is preferred for calculating slopes and deflections.

1.12. Example No. 2:- Analyze the following beam by the force method. Draw S.F. and B.M. diagrams. SOLUTION :-

Ra

L

No. of reactions = 4No. of equations = 2Degree of Indeteminacy = 4 - 2 = 2Indeterminate to 2nd degree.

Rb

WKN/m

A

Ma Mb

EI = ConstantB

Fig. 2.23 Choosing cantilever with support at A as BDS. Vertical reaction at B and moment at B will be redundants. To develop compatibility equations at B regarding translation and rotation at B, we imagine the BDS under applied loads and then under various redundants separately.


L

WKN/m

AB

B

bB

tangent at BFig. 2.23a B.D.S under loads

L

A B

bb

B bb

+ 1

EI = constant

Fig. 2.23b B.D.S. under redundant unit vertical force at B

L

AB

bb

B bb

+1

Fig. 2.23c B.D.S. under unit redundant moment at B

EI=constant

Compatibility Equations

∆B + Vb × δbb + Mb × δ′bb = 0 → (1) For vertical displacement at B

θB + Vb × ∝′ bb + Mb × ∝bb = 0 → (2) For redundant moment at B

Notice that rotation produced by Unit load at B (α'bb) and deflection produced by unit moment of B (δ'bb) are denoted by dash as superscript to identify them appropriately.

In matrix form

δbb δ′bb

∝′bb ∝bb

Vb

Mb =

- ∆B

- θB

↑ ↑ ↑ Structure flexibility Column vector Column vector of matrix. of redundants. flexibility coefficients.

Vb

Mb =

δbb δ′bb

∝′bb ∝bb

- ∆B

- θB


Now we evaluate ∆B, θb, δbb, α'bb, δ'bb and ∝bb with the help of moment area theorems separately, where ∆ = Deflection at B in BDS due to applied loads

θb = Rotation at B in BDS due to applied loads.

L

WKN/m

AB

WL

WL2

2

WL2

2WL6

3

X = L/43L4

B.M.S. due toapplied loads.

0 0

B.M.D

B.D.S. under loadsFig. 2.24a

Calculate area of BMD and fix its centroid

A = bh

(n+1) = L × (− WL2)

(2+1) = − WL3

6 b = width of BMD.

h = ordinate of BMD.

X′ = b

n + 2 = L

(2 + 2) = L4 By applying second theorem of moment area, we have

∆B = 1EI

−

WL3

6 × 34 L = −

WL4

8EI

θb = 1EI

−

WL3

6 = − WL3

6EI


AB1

L

0 0

L2

212 x L x L =

L/3 2L/3L

L

Fig. 2.24b B.M.D. due to unit redundant force at B

B.D.S. under unit redundant force at B.

δbb = 1EI

−

L2

2 × 23 L = −

L3

3EI ; δbb = Deflection at B due to unit redundant at B

α′bb = 1EI

−

L2

2 = − L2

2EI ; α′bb = Rotation at B due to unit redundant at B

A B1 1

L

L/2

L x 1 = L

0

1

0

1B.M.D

Fig. 2.24c B.D.S under unit redundant moment at B

B/

α′bb

δbb

δ′bb = 1EI

− L ×

L2 = −

L2

2EI

∝bb = 1EI [ ]− L = −

LEI


Normally BMD’s are plotted on the compression side of beam. Putting values in first equation, we have

− WL4

8EI − Vb × L3

3EI − L2

2EI Mb = 0 (1) multiply by 24 and simplify to get

equation (3) Putting values in second equation, we have (2) multiply by 6 and simplify to get

− WL3

6EI − Vb x L2

2EI − L x Mb

EI = 0 equation (4)

− 3 WL4 − 8 L3 × Vb − 12 L2 × Mb = 0 (3) or 3 WL4 + 8 L3 Vb + 12 L2 Mb = 0 (3) − WL3 − 3 L2 Vb − 6 L Mb = 0 (4) or WL3 + 3 L2 Vb + 6 L Mb = 0 (4) Multiply (4) by 2 L & subtract (4) from (3) 3 WL4 + 8 L3 Vb + 12 L2 Mb = 0 (3) 2 WL4 + 6 L3 Vb + 12 L2 Mb = 0 (4) WL4 + 2 L3 Vb = 0 WL4 = − 2 L3 Vb

Vb = − WL4

2L3

Vb = − WL

2

The (−ve) sign with Vb shows that the unit redundant load at B is in upward direction.( Opposite to that assumed and applied) Putting the value of Vb in (3)

3 WL4 + 8 L3

−

WL2 + 12 L2 Mb = 0

or 3 WL4 − 4 WL4 + 12 L2 Mb = 0 WL4 = 12 L2 Mb

Mb = WL4

12L2

Mb = WL2

12

The ( +ve) sign with Mb indicates that the assumed direction of the unit redundant moment at B is correct. Now apply the computed redundants at B and evaluate and apply reactions at A.


L

WKN/m

A B

Va=WL/2 Vb=WL/2

WL WL2 211

2 2

Fig. 2.25

Ma= Mb=

0 0

B.M.D

0.789 L

0.211L 0.578L 0.211L

WL21

2WL21

2

WL42

2

Points of Contraflexure : -

B as origin :- write moment expression

Mx = WL

2 X − WL2

12 − WX2

2 = 0

Multiply by 12W and re-arrange.

6 X2 − 6 LX + L2 = 0

X = + 6L ± 36 L2 − 4 × 6 × L2

2 × 6

= 6L ± 36 L2 − 24 L2

12

= 6L ± 12 L2

12

= 6L ± 2 3 L2

12

= 6 L ± 3.464 L

12


= 9.464 L

12 , 2.536 L

12

X = 0.789 L , 0.211 L Location of point of contraflexure From both ends.

X = 0.211 L

Same can be done by taking A as origin and writing moment expression : −

Mx′ = WLX′

2 − WL2

12 − WX′2

2 = 0

6 WLX′ − WL2 − 6 WX′2 = 0 Simplify

LX′ − L2

6 − X′2 = 0

X′2 − LX′ + L2

6 = 0

X′ = L ± L2 − 4 × 1 ×

L2

62 × 1

= L ± L2 −

2 L2

32

= L ±

L2

32

= L ±

13 . L2

2

X′ = L ± 0.577 L

2

X′ = 0.789 L , 0.211 L Location of points of contraflexure.

X′ = 0.211 L We get the same answer as before.

This is a flexibility method and was written in matrix form earlier. The matrix inversion process is given now for reference and use.


1.13. MATRIX INVERSION : - These co-efficients may also be evaluated by matrix Inversion so basic procedures are given.

Inverse of matrix = Adjoint of matrix

Determinant of matrix

Adjoint a matrix = Transpose ( Interchanging rows & columns) of matrix of co-factors. Co-factors of an element = (− 1)i+j × minor of element.where i = Row number in which

that element is located and j = Column number in which that element is located.

Minor of element = Value obtained by deleting the row & the column in which that particular element is located and evaluating remaining determinant. Let us assume a matrix :

A =

1 3 7

4 5 98 10 11

Determinant of matrix A = 1 (5 × 11 − 10 × 9 )− 3 (44 − 72) + 7 ( 4 × 10 − 8 × 5 )

= − 35 + 84 + 0

= 47

MINORS OF MATRIX :- Find out the minors for all the elements of the matrix. Then establish matrix of co-factors.

Matrix of Minors =

-35 -28 0

-37 -45 -14-8 -19 -7

Matrix of co-factors =

-35 28 0

37 -45 14-8 19 -7

Adjoint of matrix A =

-35 37 -8

28 -45 190 14 -7

Inverse of matrix = 1

49

-35 37 -8

28 -45 190 14 -7

A-1 =

-0.71 0.755 -0.163

0.571 -0.918 0.3870 0.286 -0.143

A x A−1 = I =

1 0 0

0 1 00 0 1

Check for correct matrix inversion

Aij x Bjk = Cik


A A−1 =

1 3 7

4 5 98 10 11

−0.71 0.755 −0.163

0.571 −0.918 0.3870 0.286 −0.143

=

−1×0.71+3×0.571+7×0 1×0.755−3×0.918+7×0.286 −1×0.163+3×0.387 −7×0.143

0 1 0 0 0 1

AA−1 =

1 0 0

0 1 00 0 1

Proved.

1.14. 2ND DEGREE INDETERMINACY :- Example No. 3: Solve the following continuous beam by consistent deformation method.

A B C D

40 kN

3m 4m 5m

EI = constantFig. 2.26

In this case, we treat reaction at B and C as redundants and the basic determinate structure is a simply supported beam AD.

B C

40 kN

Fig. 2.26 a

A D

Bending under applied loads

B C1

Fig. 2.26 b

A D

Bending under unit redundant force at B

bb cb

B C

1

Fig. 2.26 c

A D

Bending under unit redundant force at C

bc cc


Compatibility equations are as follows:

∆B + δbb × Rb + δbc × Rc = 0 → (1) For compatibility at B

∆C + δcb × Rb + δcc × Rc = 0 → (2) For compatibility at C

Evaluate the flexibility co-efficients given in equation (1) and (2). Using Conjugate beam method.

A 5 m 7 m D

16.67 KN23.33 KN

∑

∑

23.33+

+

0 S.F.D.16.67

116.67 KN

70/EI 83.35/EI116.67

EI

B.M.D.

Wb

D’

6.335.67

MD=0RAx12 - 40x7=0RA=23.33 KN FY=0RA+RD=40RD=16.67 KN

a

40 KN

M=Wab

L

Fig. 2.27

A B C

369.455EI

291.675EI

700.02EI

408.345EI

330.565EI

3 5

L

L + a L + b3 3

In general for a simple beam loaded as below,the centroid is a shown

MEI

diagram

∑MD′ = 0, Calculate RA'

RA′ × 12 = 291.675

EI

7 +

13 5 +

408.345EI

2

3 x 7

= 2527.85

EI + 1905.61

EI

RA′ = 369.455

EI

∑Fy = 0


RA′ + RD′ = 369.455

EI + RD' = 700.02

EI

RD′ = 700.02

EI − 369.455

EI

RD′ = 330.565

EI . Now ordinates of MEI diagram are determined by comparing

Similar triangles.

116.67

5 EI = Y3 ⇒ Y =

70EI

Now by using conjugate beam method (theorem 2)

∆B = 1EI

369.455 × 3 −

1

2 × 3 × 70 × 33

∆B = 1003.365

EI KN − m3

Determine

116.67

7 = Y5

Y = 83.34

∆C = 1EI

330.565 × 5 −

1

2 × 5 × 83.34 × 53

∆C = 1305.575

EI KN − m3

Now apply unit redundant at B.

B

B

C1

Fig. 2.28

A

A

D

D

bb cb3m 4m 5m

2/3 1/3

2.25/EI1.25/EI

13.55 7

Conjugate beam under M/EI7.875/EI 5.625/EI

C


Computing Co-efficients by Conjugate beam method. (Theorem 2)

MB' = δbb = 1EI [ 7.875 × 3 − 3.375 × 1 ]

δbb = 20.25

EI KN − m3

Determine ordinate 2.25

9 = Y5

Y = 1.25EI

MC' = δcb = 1EI

5.625 × 5 − 3.125 ×

53

δcb = 22.92

EI KN − m3

Now apply unit redundant at C. I

1 x 7 x 512

=2.92

2.92EI

8.28EI

17.52EI

9.24

bc cc

bc cc

2.92

Fig. 2.29

DA

B C

Conjugate beam under M/EI

6.33 m 5.67 m

Moment at B’ in conjugate beam gives

MB' = δbc = 1EI

8.28 x 3 −

12 x 1.25 x 3 x 1

MC' = δbc = 22.965

EI KN − m3 (δbc = δcb ) PROVED.

δcc = 1EI

9.24 x 5 −

12 x 2.92 x 5 x

53

δcc = 34.03

EI KN − m3.


Inserting evaluated Co-efficients in equation (1) and (2)

1003.365

EI + 20.25

EI Rb + 22.965

EI Rc = 0 (1)

1003.365 + 20.25 Rb + 22.965 Rc = 0 (3) Canceling 1/EI throughout

1305.575

EI + 22.92

EI Rb + 34.03

EI Rc = 0 (4) Cancelling 1EI throughout

1305.575 + 22.92 Rb + 34.03 Rc = 0 (4) Multiply (3) by 22.92 and (4) by 20.25 & subtract (4) from (3) 22997.1258 + 464.13 Rb + 526.357 Rc = 0 (3)

26437.8938 + 464.13 Rb + 689.1075 Rc = 0

− 3460.768 − 162.75 Rc = 0 (4)

Rc = − 21.264 KN Putting this in equation (3) 1003.365 + 20.25 Rb − 22.963 × 21.264 = 0

Rb = − 25.434 KN The ( −ve) signs with the values of the redundants are suggestive of the fact that the directions of the actual redundants are in fact upwards. Now apply loads and evaluated redundants to original beam calculate remaining reaction.

A B C D3m 4m 5m

Fig. 2.30

∑Fy = 0 Considering all upwards at this stage as Ra and Rd are unknown. RA + RD + 25.434 + 21.264 − 40 = 0 RA + RD = −6.698 → (1) ∑MD = 0 Considering all upward reactions RA × 12 + 25.454 × 9 − 40 × 7 + 21.264 × 5 = 0

RA = − 4.602 KN . It actually acts downwards. RD = − RA − 6.698 = 4.602 − 6.698 RD = − 2.096 KN All determined reactions are shown in figure 2.30 above sketch SFD and BMD.


Fig. 2.31

Elastic curve

S.F.D.

B.M.D.

2 LOCATION OF POINTS OF CONTRAFLEXURE :- These are in Span BC. A as origin. Write moment expression and equate to zero.

MX1 = − 4.602 X1 + 25.434 ( X1 − 3 )

− 4.602 X1 + 25.434 X1 − 76.302 = 0

X1 = 3.663 m from A.

D as origin. Write moment expression and equate to zero.

MX2 = − 2.096 X2 + 21.264 ( X2 − 5 ) = 0

− 2.096 X2 + 21.264 X2 − 106.32 = 0

19.168 X2 − 106.32 = 0

X2 = 5.547 m.

These locations are marked above in BMD.


1.15. 3RD DEGREE INDETERMINACY :- Example No. 4: Solve the frame shown below by consistent deformation method.

Fig. 2.32 B.M is +ve forTension on inner sides

inner sides

outer sides

outer sidesouter sides

1.15.1. SOLUTION:

Sign convention for S.F. and B.M. remains the same and are shown above as well. In this case, any force or moment which creates tension on the inner side of a frame would be considered as a (+ve) B.M. Removing right hand support to get BDS. The loads create three defermations as shown.

Fig. 2.33 (a) Fig. 2.33 (b) Note: ∆DH = Deflection of point D in horizontal direction due to applied loads on BDS. ∆DV = Deflection of point D in vertical direction due to applied loads on BDS. θ D = Rotation of point D due to applied loads on BDS.

A

1

B

C

D

6m

4m4m

ddv

dd ddh

1

1

Fig. 2.33c B.D.S. under unit vertical redundant force at D

Fig. 2.33d B.D.S. under unit rotational redundant moment at D

m -Diagram


Where (See mH diagram Fig. 2.33b) δddh = Deflection of point D due to unit load at D in horizontal direction acting on BDS. δ'ddv = Deflection of point D, in vertical direction due to unit load at D in horizontal direction. α'ddh= Rotation of point D, due to unit load in horizontal direction at D acting on BDS. (See mV diagram Fig: 2.33c) δddv = Deflection of point D due to unit load at D in vertical direction. δ'ddh = Deflection of point D (in horizontal direction) due to unit vertical load at D. α'ddv = Rotation of point D due to unit vertical load at D.

(See mθ diagram Fig: 2.33d)) α'ddh = Horizontal deflection of point D due to unit moment at D. α'ddv = Vertical deflection of point D due to unit moment at D. αdd = Rotation of point D due to unit moment at D.

Compatibility equations :- ∆DH + HD × δddh + VD × δ′ddv + MD × ∝′ddh = 0 (1) Compatibility in horizontal direction at D. ∆DV + HD × δ′ddh + VD × δddV + MD × ∝′ddV = 0 (2) Compatibility in vertical direction at D. θD + HD × ∝′ddh + VD × ∝′ddv + MD × ∝dd = 0 (3) Compatibility of rotation at D Now evaluate flexibility co-efficients used in above three equations. We know that

∆ or θ = ∫ 1EI ( Mmdx )

There are 12 co-efficients to be evaluated in above three equations.

So ∆ DH = ∫ M × mHEI dx (1)

δddh = ∫ (mH)2 dxEI (2)

δ′ddh = ∫ mH mv dxEI (3)

∆ Dv = ∫ M × (mv ) dxEI (4)

δ′ddv = ∫ (mH × mv ) dxEI (5)

δ ddv = ∫ (mv)2 dxEI (6)

∝′ddv = ∫ mv × mθEI dx (7)


θD = 1EI ∫ ( M ) ( m θ ) dx (8)

∝′ddh = 1EI ∫ ( mH ) ( m θ ) dx (9)

∝′ddv = 1EI ∫ ( mv ) ( mθ ) dx (10)

∝dd = 1EI ∫ ( mθ )2 dx (11)

Multiplying the corresponding moment expressions in above equations, we can evaluate above deformations. Draw M-diagram.

A

80

KN-m

3m

E

20

KN

3m

B

2m

F

10KN

2m

C

4m

D

20KN

x

10KNM - Diagram

M = 10 x 2 + 20 x 3 = + 80KN-m Fig. 2.34 B.D.S under applied loads

M – Diagram by parts

A

x

80KN-m

3m

20KN

x

3m

20KN-m

B

10KN

20KN-m

x

B

2m

10KN

F

10KN

2m

x

CC

4m

x D

M=20

x

6-20

x

3

-

80

=

20KN-m

10KN

20KN

E


A

2

1

B

C

D

1

MH - Diagram

A

2

1

E

6m

B 4

1

4

B F

4m

4

C

1

+

C

1

4m

D

1

1

Fig. 2.34a Fig 2.34b

4

A

1

E

B

41

1

BF

C

1

1C

D

1

mv-diagram (by parts)

4

A1

E

B 1

1B

F

C 1

C1

1

D

m -diagram (by parts)

Fig 2.34c Fig 2.34d Moments expressions in various members can now be written in a tabular form.

Portion Origin Limits M mH mv MO AE A 0 − 3 20X − 80 X − 2 − 4 −1 BE B 0 − 3 − 20 − X + 4 − 4 −1 BF B 0 − 2 10X − 20 4 X − 4 −1 CF C 0 − 2 0 4 − X −1 CD D 0 − 4 0 X 0 −1

Put these moment expressions, integrate and evaluate co-efficients


∆DH = 1EI ∫ M ( mH ) dX

∆DH = 1EI

3

∫o (20X − 80) (X − 2) dX +

3

∫o (−X+4) (−20) dX +

2

∫o(10X − 20) 4 dX + 0 + 0

= 1EI

3

∫o (20X2 − 80X − 40X + 160 ) +

3

∫o

(20X − 80 )dX+2

∫o ( 40X − 80 ) dX

= 1EI

20X3

3 − 80X2

2 − 4X2

2 + 160X3 o +

20X2

2 − 80X 3|o +

40X2

2 − 80 X 4 o

= 1EI

20 × 33

3 − 40 × 32 − 20 × (3)2 + 160 × 3 + (10 × 9 − 80 × 3) + (20 × 4 − 80 × 2)

∆ DH = − 110EI

δ ddh = 1EI ∫ ( mH )2 dX

= 1EI

3

∫o(X − 2)2 dX +

3

∫o (−X + 0)2dX +

2

∫o 16 dX +

2

∫o16 dX +

4

∫o X2 dX

δddh = 1EI

3

∫o(X − 4X + 4) dX +

3

∫o (16 − 8X + X2) dX +

2

∫o 16 dX +

2

∫o 16 dX +

4

∫o X2 dX

= 1EI

X3

3 − 4X2

2 + 4X 3|o +16X −

8X2

2 + X3

3 3|o + 16X

2|o + 16X

2|o +

X3

3

4 o

= 1EI

33

3 − 2 (3)2 + 4 × 3 +

16 × 3 − 4 × 9 +

33

3 +

(16 × 2) + (16 × 2) +

(4)3

3 − 0

δddh = 109.33

EI

δ'ddV = 1EI ∫ ( mH ) ( mv ) dX

= 1EI

3

∫o (X − 2) ( − 4 ) dX +

3

∫o (−X + 4 ) (−4 ) dX +

2

∫o ( 4 ) (X−4 ) dX +

2

∫o 4 (−X ) dX + 0

= 1EI

3

∫o (− 4X+8 ) dX +

3

∫o (4X − 16 ) dX +

2

∫o (4X−16 ) dX +

2

∫o − 4XdX

= 1EI

−

4X2

2 + 8X 3|o +

4X2

2 −16X 3|o +

4X2

2 − 16X 2|o + −

4X2

2

2 o


= 1EI [ ] − 2 × (3)2 + 8 × 3 + (2 × 32 − 16 × 3 ) + ( 2 × 22 − 16 × 2 ) + ( − 2 × 22 )

δ′ddV = − 56EI

∝′ddh = 1EI ∫ ( mH ) ( mθ ) dX

= 1EI

3

∫o (−1 ) (X − 2) dX +

3

∫o (−1) (−x + 4) dX +

2

∫o − 4 dX +

2

∫o − 4 dX +

4

∫o − XdX

= 1EI

−

X2

2 + 2X 3|o +

X2

2 −4X 3|o + − 4X

2|o + − 4X

2|o + −

X2

2

4 o

= 1EI

−

92 + 2 × 3 +

9

2 − 4 × 3 + (− 4 × 2) + (− 4 × 2) +

−

42

2 − 0

α′ddh = −30EI

θD = 1EI ∫ M ( mθ ) dX

= 1EI

3

∫o − (20X − 80 ) dX +

3

∫o 20 dX +

2

∫o (−10X + 20 ) dX + 0 + 0

= 1EI

−

20X2

2 + 80X 3|o + 20X

3|o + −

10X2

2 + 20X 2|o

= 1EI [(−10 × 32 + 80 × 3) + (20 × 3) + (− 5 × 4 + 20 × 2)]

θD = 230EI

∆ Dv = 1EI ∫ M ( mv ) dX

= 1EI

3

∫o (20X − 80) (−4) dX +

3

∫o (−20) (−4) dX +

2

∫o (10X − 20) (X − 4) dX + 0 + 0

= 1EI

3

∫o (− 80X + 320) dX +

3

∫o 80 dX +

2

∫o (10X2 − 20X − 40X + 80) dX

= 1EI

− 80

X2

2 + 320X 3|o + 80X

3|o + 10

X3

3 − 60X2

2 + 80X 2|o


= 1EI

(−40 × 9 + 320 × 3) + (80 × 3) +

10

3 × 8 − 30 × 4 + 80 × 2

∆Dv = 906.67

EI

δ′ddh = 1EI ∫ ( mH ) ( mv ) dX

= 1EI

3

∫o (X−2) (−4) dX +

3

∫o (−X + 4) (−4) dX +

2

∫o 4 (X−4) dX +

2

∫o − 4XdX + 0

= 1EI

3

∫o (−4X + 8)dX +

3

∫o (4X − 16) dX +

2

∫o (4X − 16) dX +

2

∫o

− 4XdX

= 1EI

−

4X2

2 + 8X 3|o +

4X2

2 − 16X 3|o +

4X2

2 − 16X 2|o +

4X2

2 o

= 1EI [(−2 × 9 + 8 × 3) + (2 × 9 − 16 × 3) + (2 × 4 − 16 × 2) + (−2 × 4)]

δ′ddh = − 56EI

δddv = 1EI ∫ ( mv2 ) dX

= 1EI

3

∫o 16 dX +

3

∫o 16 dX +

2

∫o (X − 4)2 dX +

2

∫o (−X)2dX + 0

= 1EI

3

∫o16 dX+

3

∫o 16 dX+

2

∫o(X2 − 8X +16)dX+

2

∫o + X2 dX

= 1EI

16X

3|o + 16X

3|o +

X3

3 − 8X2

2 + 16X 2|o + | +

X3

3

2 o

= 1EI

(16 × 3 ) + ( 16 × 3 ) +

8

3 − 4 × 4 + 16 × 2 +

+

83

δddv = 117.33

EI


∝′ddv = 1EI ∫ mv × mθ dX

= 1EI

3

∫o + 4 dX +

3

∫o + 4 dX +

2

∫o (−X + 4) dX +

2

∫o XdX

= 1EI

4X

3|o + 4X

3|o + −

X2

2 + 4X 2|o + |

X2

2

2 o

= 1EI

(4 × 3) + (4 × 3) + (−2 + 4 × 2) +

22

2

∝′ddv = 32EI

∝dd = 1EI ∫ ( mθ )2 dX

= 1EI

3

∫o

(−1)2 dX + 3

∫o

(−1)2 dX + 2

∫o

(−1)2 dX + 2

∫o

(−1)2 dX + 4

∫o

(−1)2 dX

= 1EI

X

3|o + X

3|o + X

2|o + X

2|o + X

4|o

= 1EI [ 3 + 3 + 2 + 2 + 4 ]

∝dd = 14EI

Putting all values of evaluated co-efficients, equations 1,2 and 3 become

− 110EI +

109.33EI × HD −

56EI × VD −

30EI MD = 0 (1)

and 906.67

EI − 56EI × HD +

117.33EI × VD +

32EI MD = 0 (2)

and 230EI −

30EI × HD +

32EI × VD +

14EI MD = 0 (3) Simplifying

−110 + 109.33 HD − 56 VD − 30 MD = 0 → (1) 906.67 − 56 HD + 117.33 VD + 32 MD = 0 → (2) 230 − 30 HD + 32 VD + 14 MD = 0 → (3)


From Eq (1)

MD = −110 + 109.33 HD − 56 VD

30 = −3.67 + 3.64 HD − 1.86 VD → (4)

Putting in Eq (2) 906.67 − 56 HD + 117.33 VD + 32 (−3.67 + 3.64 HD − 1.86 VD) = 0 906.67 − 56 HD + 117.33 VD − 117.44 + 116.5 HD − 59.52 VD = 0 789.23 + 60.5 HD + 57.81 VD = 0 HD = −13.045 − 0.95 VD → (5) Putting the value of HD in Eq (4) MD = −3.67 + 3.64 (−13.045 − 0.95 VD) − 1.86 VD MD = −51.15 − 5.32 VD → (6) Putting the values of MD & HD in Eq (3) 230 − 30 (−13.045 − 0.95 VD) + 32 VD + 14 (−51.15 − 5.32 VD) = 0 230 + 391.35 + 28.5 VD + 32 VD − 716.1 − 74.5 VD = 0 −14 VD − 94.75 = 0 VD = −6.78 KN Putting in (5) & (6) HD = −6.61 KN, MD = −15.08 KN−m

From any equation above. We get

VD = − 12.478 KN

Apply the evaluated structural actions in correct sense on the frame. The correctness of solution can be checked afterwards by equilibrium conditions.

AMa=1.8 KN

20KN

3m

3m

2mB

2m

10KNC

D15.08KN=m

6.61KN

12.478 KNHa=13.39 KN

Va = 2.478 KN

4m

Fig. 2.35 shows all reactions after Evaluation


∑Fx = 0 20 − Ha − 6.61 = 0

Ha = 13.39 KN ∑Fy = 0 Va + 12.478 − 10 = 0 (asuming Va upwards)

Va = − 2.478 KN 0 Ma+ 20 × 3 + 10 × 2 − 12.478 × 4 − 6.61 × 2 − 15.08 = 0 (assuming Ma clockwise)

Ma = − 1.8 KN-m ΣMa = 0 12.478 × 4 + 15.08 + 6.61 × 2 + 1.8 − 20 × 3 − 10 × 2 = 0 Proved.

1.16. ANALYSIS OF STATICALLY EXTERNALLY INDETERMINATE TRUSSES :- A truss may be statically indeterminate if all external reactive components and internal member

forces may not be evaluated simply by the help of equations of equilibrium available. The indeterminacy of the trusses can be categorized as follows :-

(1) Trusses containing excessive external reactive components than those actually required

for external stability requirements. (2) Trusses containing excessive internal members than required for internal stability

requirements giving lesser the number of equations of equilibrium obtained from various joints.

(3) A combination of both of the above categories i.e. excessive external reactions plus

excessive internal members.

INTERNAL INDETERMINACY:-

b + r = 2j There are two equations of equilibrium per joint where b = number of bars or members. r = minimum number of external reactive components required for external stability (usually 3). j = number of joints.

The above formula can also be used to check the total indeterminacy of a truss if we define ‘r’ as the total number of reactive components which can be provided by a typical support system. 1.17. METHOD OF MOMENTS AND SHEARS :

A simple method is presented to evaluate axial member forces in parallel chord trusses. For other types of trusses method of joints, method of sections or Maxwell’s diagram may be used. For determining forces in members of trusses, this method has been used throughout this text. To develop the method, consider the truss loaded as shown below:


AB C

D

E F G H

2P 3P

h

3 a

RA = P73

@ RD = P83

Fig. 2.36 A typical Truss under loads

Consider the equilibrium of L.H.S. of the section. Take ‘D’ as the moment centre: we find Ra

Ra × 3a = 2P × 2a + 3 P × a

Ra = 7Pa3a =

7P3

∑ Mc = 0 and assuming all internal member forces to be tensile initially, we have Ra x 2a − 2P × a + SFG × h = 0 (considering forces on LHS of section)

or SFG = −

Ra × 2a − 2 Pa

h

The ( −ve ) sign indicates a compressive force. Or

SFG =

Ra × 2a − 2 Pa

h = Mch where numerator is Mc. Therefore

The force in any chord member is a function of bending moment. “To find out the axial force in any chord member, the moment centre will be that point where other two members completing the same triangle meet and the force will be obtained by taking moments about that point and dividing it by the height of truss. The signs of the chord members are established in the very beginning by using an analogy that the truss behaves as a deep beam. Under downward loads, all upper chord members are in compression while all lower chord members are in tension. Similarly, SBC =

MFh (using the guide line given in the above para)

Consider the equilibrium of left hand side of the section and ∑Fy = 0

Ra − 2P − SFC Cos θ = 0

SFC =

Ra − 2P

Cos θ where Ra − 2P is equal to shear force V due to applied loads at

the section. So in general the force in any inclined member is a function of shear force.

SFC = V

Cosθ

The general formula is : S =

± (V)± (Cos θ)


Where V is the S.F. at the section passing through the middle of inclined member and ‘θ‘ is the angle measured from “the inclined member to the vertical” at one of its ends. Use (+ve) sign as a pre-multiplier with the Cosθ if this angle is clockwise and (−ve) sign if θ is anticlockwise. Take appropriate sign with the S.F also. This treatment is only valid for parallel chord trusses. The force in the vertical members is determined by inspection or by considering the equilibrium of forces acting at the relevant joints. To illustrate the method follow the example below.

1.17.1: EXAMPLE :− Analyze the following truss by the method of moment & shear. SOLUTION:- Determine reactions and Draw SFD and BMD.

A

1.5P

I J KP

LP

MP

N O

Hh

GFEDCB P

1.5 P

8 @ a

0.5P

0.5P0

S.F.D.

1.5P

1.5Pa3 Pa

4.5

Pa 5 Pa 4.5 P

0

0

3 Pa1.5 Pa

B.M.D.

1.5PGiven Truss under loads

Fig. 2.37 TOP CHORD MEMBERS. Considering the beam analogy of truss, all top chord members are in compression. Picking bending

moment, at appropriate moment centers, from BMD and dividing by height of Truss.

Sij = −3 Pa

h

Sjk = −3 Pa

h

Stl = −5 Pa

h

Slm = −5 Pa

h

Smn = −3 Pa

h

Sno = −3 Pa

h Negative sign means compression.


BOTTOM CHORD MEMBERS.

All are in tension. Taking appropriate moment point and dividing by height of Truss.

Sap = Spb = + 1.5 Pa

h

Sbc = Scd = + 4.5 Pa

h

Sde = Sef = + 4.5 Pa

h

Sfg = Sgh = + 1.5 Pa

h

INCLINED MEMBERS.

The force in these members has been computed by the formula. ±V

±(Cosθ). Follow the guidelines.

Sai = 1.5 P− Cosθ

Sib = 1.5 P

+ Cosθ Length AI = a2 + h2

(if a and h are given, length and Cos θ will have also late values)

Sbk = 1.5 P− Cosθ Cos θ =

ha2 + h2

Skd = 0.5 P

+ Cosθ

Sdm = − 0.5 P− Cosθ =

0.5 PCosθ

Smf = − 1.5 P+ Cosθ

Sfo = − 1.5 P− Cosθ =

1.5 PCosθ

Soh = − 1.5 P+ Cosθ

VERTICAL MEMBERS. For all vertical members of trusses in this book, member forces have been determined by Inspection or by Equilibrium of joints. So Sip = Sbj = Sck = Sem = Sfn = Sgo = 0 Sld = − P ( If a and h values are given, all forces can be numerically evaluated)

1.18. EXTERNALLY REDUNDANT TRUSSES – FIRST DEGREE EXAMPLE 5 :- Analyze the following truss by the force method. (consistent deformation method). The following data is given. E =200 × 106 KN/m2 A=5x10−3m2 for inclineds and verticals,

A=4x10−3m2 for top chord members, A=6x10−3m2 for bottom chord members


SOLUTION:-

A

F

G

H

36KN 72KN

I J

E

1.8m

D CB

4

@

1.8m Fig. 2.38 Given Truss under loads

TOTAL INDETERMINACY :-

b + r = 2 j where r = total reactions which the supports are capable of providing.

17 + 4 ≠ 2 × 10

21 ≠ 20

D = 21 − 20 = 1

Indeterminate to Ist degree.

Apply check for Internal Indeterminacy :- b + r = 2 j where r = Minimum number of external reactions required for stability.

17 + 3 = 2 × 10

20 = 20

This truss is internally determinate and externally indeterminate to 1st degree, therefore, we select reaction at point “C” as the redundant force. Remove support at C, the Compatibility equation is :

∆ C + δcc × Rc = 0 (Deflection at C due to loads plus due to redundant

should be zero.)

or Rc = − ∆cδcc . Now we have to calculate ∆c and δcc to get Rc.

where ∆c = ∑ F′ UL

AE where F' = Force induced in members due to applied loads

acting on BDS.

δcc = ∑ U2 LAE U = Forces in members due to Unit load applied in direction

of applied loads, at external redundant support in BDS.


A

F G

36K

H

72K

I J

E

1.8m

DCCB4 @ 1.8m

.

A

F

G

H

J

E

D

1

C

cc

B

Fig 2.39b B.D.S under unit Vertical Redundant at C(U-Diagram)

(F-Diagram)

Analyze the given truss by the method of moments and shears as explained already for F' and U forces in members.

A

F G36KN 72KN

I J

E

Re = 45 KN4

@

1.8m

1

(F -Diagram)/ Ra

=

63

B C D

H

1.8m

0

0

63

27+

0

S.F.D.

45

B.M.D.

0

81

16245

113.4


Determine forces in all members of trusses loaded as shown in this question and enter the results in a tabular form. (using method of moments and shears, F' and U values for members have been obtained).

A

F G H I J

ED1CB

U=Diagram

S.F.D.

B.M.D.

0.91.8

0.9

+

+

½½

Fig 2.41 B.D.S under Unit redundant force at C

Member F′ (KN)

U Ax 10−3

(m)2

L (m)

F′ULAE × 10 −3

(m)

U2LAE × 10−3

(m)

Fi = Fi′ − Rc × U1 (KN)

FG 0 0 4 1.8 0 0 0 GH − 90 − 1 ″ ″ 0.2025 2.25 × 10−3 + 2.5 H I − 90 − 1 ″ ″ 0.2025 2.25 × 10−3 + 2.5 I J 0 0 4 ″ 0 0 0 AB +63 +0.5 6 1.8 0.04725 0.375 × 10−3 +16.75 BC +63 +0.5 ″ ″ 0.04725 0.375 × 10−3 +16.75 CD +45 +0.5 ″ ″ 0.03375 0.375 × 10−3 − 1.25 DE +45 +0.5 ″ ″ 0.03375 0.375 × 10−3 − 1.25 AG − 89.1 −0.707 ″ 2.55 0.16063 1.275 × 10−3 − 23.7 GC +38.2 GC 5 ″ 0.06887 1.275 × 10−3 − 27.2 C I +63.64 +0.707 ″ ″ 0.11473 1.275 × 10−3 − 1.76 I E −63.64 −0.707 ″ ″ 0.11473 1.275 × 10−3 +1.76 AF 0 0 ″ 1.8 0 0 0 BG 0 0 ″ ″ 0 0 0 HC − 72 0 ″ ″ 0 0 − 72 I D 0 0 ″ ″ 0 0 0 J E 0 0 ″ ″ 0 0 0

∑

F′ULAE = 1.02596

× 10−3

∑ U2LAE =11.1

× 10−6


∆ C = ∑ F′ UL

AE = 1.02596 × 10−3 = 1025.96 × 10−6 m

δ cc = ∑ U2 LAE = 11.1 × 10−6 m . Putting these two in original compatibility equation

Rc = − ∆ Cδcc =

−1025.96 × 10−6

11.1 × 10−6

Rc = − 92.5 KN. The (−ve) sign with Rc shows that the assumed direction of redundant is incorrect and Rc acts upward. If Fi is net internal force due to applied loading and the redundants, acting together, then member forces an calculated from

Fi = Fi′ − Rc × Ui The final axial force in any particular member can be obtained by applying the principle of superposition and is equal to the force in that particular member due to applied loading ( ± ) the force induced in the same member due to the redundant with actual signs. Apply the principle of superposition and insert the magnitude of redundant Rc with its sign which has been obtained by applying the compatibility condition to calculate member forces. 1.19. SOLUTION OF 2ND DEGREE EXTERNALLY INDETERMINATE TRUSSES:- Example-6 : Solve the following truss by consistent deformation method use previous

member properties.

A

F

36KN

G

72KN

H

I

J

E

1.8m

D

C

B

4

@

1.8m

Fig 2.42 Given Truss

0

113.40

63

63KN

36KN

DC45KN

S.F.D.

0

45

B.M.D.

0

81+

162

1.8m

Fig 2.42a B.D.S under loads

72KN

(F -diagram)/


0

0.9

dccc

S.F.D.

0 B.M.D.

0.9(+)

1.8

12

12

12

+

1

+

(U diagram)1

0

0

0.25

0.25cd dd

0.75

0S.F.D.

0.75

0 B.M.D.

(-)

1.35

(+)0.9

(+)

0.45

Fig 2.42 c B.D.S under unit redundant at D

(U diagram)21

Compatibility equations are: ∆C + Rc. δcc + Rd × δcd = 0 (1) Compatibility of deformations at C ∆D + Rc . δdc + Rd . δdd = 0 (2) Compatibility of deformations at D δcd = δdc by the law of reciprocal deflection. δcc = deflection of point C due to unit load at C. δdc = deflection of point D due to unit load at C. δdd = deflection of point D due to unit load at D. δcd = deflection of point C due to unit load at D. Flexibility coefficients of above two equations are evaluated in tabular form (Consult the attached table)

∆C = ∑ F′U1L

AE = 1026.2 × 10 −6 m

∆D = ∑ F′U2L

AE = 579.82 × 10−6 m

δcc = ∑ U1

2LAE = 11.1 × 10−6 m


δdd = ∑ U2

2LAE = 9.3565 × 10−6 m

δcd = ∑ U1U2L

AE = 6.291 × 10−6 m

δdc = ∑ U1U2L

AE = 6.291 × 10−6 m Put these in equations 1 and 2

1026.2 × 10−6 + 11.1 × 10−6 Rc + 6.291 × 10−6 Rd = 0 → (1) 579.82 × 10−6 + 6.291 × 10−6 Rc + 9.3565 × 10−6 Rd = 0 → (2)

Simplify 1026.2 + 11.1 Rc + 6.291 Rd = 0 → (3) 579.82 + 6.291 Rc + 9.3565 Rd = 0 → (4) From (3)

Rc =

−1026.2 − 6.291 Rd

11.1 → (5)

Put Rc in (4) & solve for Rd

579.82 + 6.291

−1026.2 − 6.291 Rd

11.1 + 9.3565 Rd = 0

− 1.786 + 5.791 Rd = 0

Rd = + 0.308 KN

So, from (5), ⇒ Rc =

−1026.2 − 6.291 × 0.308

11.1

Rc = − 92.625 KN

∴ Rc = − 92.625 KN Rd = + 0.308 KN These signs indicate that reaction at C is upwards and reaction at D is downwards. By superposition, the member forces will be calculated as follows Fi = Fi + Rc × U1 + Rd × U2 which becomes. Fi = Fi − Rc × U1 + Rd × U2. It takes care of (−ve) sign with Rc. Equilibrium checks:−

1.082

0.308

1.082

0.308

Joint D∑ Fx = 0

∑ Fy = 0

Equilibrium is satisfied. Only check at one joint has been applied. In fact this check should be satisfied at all joints.


Table 79−A


23.722 27.178 721.954 0.308

1.519

1.0821.082

16.76516.765

G 0 F

0

A

36KN2.471

72KNH 2.471 I 0 J

0

EDCB

0

16.965KN 92.625KN 0.308KN 1.082KN

Fig 2.43 Result of analyzed Truss

Now find remaining reactions Ra and Re. ∑Fy = 0 Ra + Re + 92.625 − 0.308 − 36 − 72 = 0 Ra + Re = 15.683 → (1) ∑MA = 0 Re × ∆ × 1.8 − 0.308 × 3 × 1.8 + 92.625 × 2 × 1.8 − 72 × 2 × 1.8 − 36 × 1.8 = 0

Re = − 1.082 KN

As Ra + Re = 15.863 So Ra = 15.863 + 1.082

Ra = 16.945 KN Now truss is determinate. Calculate member forces and apply checks in them. Joint (C) ∑Fx = 0

16.765

27.178 72 1.954

1.082

92.625

− 1.082 − 16.765 − 1.954 × 0.707 + 27.178 × 0.707 = 0 − 0.0136 = 0 0 ≅ 0 equilibrium is satisfied. ∑Fy = 0 − 72 + 92.625 − 1.954 × 0.707 − 27.178 × 0.707 = 0 0.0286 = 0 0 ≅ 0 equilibrium is satisfied


Joint (E) ∑Fy = 0

1.082

1.519

1.082

1.519 × 0.707 − 1.087 = 0 0 = 0 ∑Fx = 0 082 − 1.519 × 0.707 = 0 0 = 0 equilibrium is satisfied. 1.20. Example–7:- SOLUTION OF 3RD DEGREE EXTERNALLY INDETERMINATE TRUSSES:-

Now we solve the following truss by consistent deformation method. Choosing reaction of B, C and D as redundant.

SOLUTION:- First step. Choose BDS Draw BDS under loads and subsequently under applied unit loads at points

of redundancy also.

A

F G H I J

E

1.8m

DCB

4 @ 1.8m

36KN 72KN

Fig 2.44 Given 3rd degree externally indeterminate truss under loads

=

A

F G H I J

E

1.8m

DCB B CD

72 KN36 KN

Fig 2.44(a) B.D.S under loads

+


bb cbdb1

Fig 2.44(b) B.D.S under redundant unit load at B(U1 diagram)

+

A

F G H I J

EDCB bc cc dc

1

Fig 2.44(c) B.D.S under redundant unit load at C(U2 diagram)

1

+

A

F G H I J

EDCB bd cd dd

1

Fig 2.44(d) B.D.S under redundant unit load at D(U3 diagram)

1

Step No.2: Compatibility equations are:

∆B + Rb.δbb + Rc.δbc + Rd x δbd = 0 For joint B → (1)

∆C + Rb.δcb + Rc.δcc + Rd x δcd = 0 For joint C → (2)

∆D + Rb.δdb + Rc.δdc + Rd x δdd = 0 For joint D → (3)


Step No.3: Evaluation of Flexibility co-efficients

∆B = ∑ F′U1L

AE ∆C = ∑ F′U2L

AE ∆D = ∑ F′U3L

AE

δbb = ∑ U1

2LAE δbc = ∑

U1U2LAE δbd = ∑

U1U3LAE

δcb = ∑ U1U2L

AE δcc = ∑ U2

2LAE δcd = ∑

U2U3LAE

δdb = ∑ U1U3L

AE δdc = ∑ U2U3L

AE δdd = ∑ U3

2LAE

By law of reciprocal deflections :- We know that δbc = δcb

δbd = δdb

δcd = δdc

In order to find member forces due to applied forces in BDS, consider.

A

F G H I J

EDCB63

630

6327 27

45

173.4

162

+81

45

B.M.D.

0S.F.D.

45

72KN36 KN

B.D.S under loads (F’ diagram)

The above SFD and BMD are used to calculate member forces by method of moments and shears. Finally ∆B, ∆C and ∆D due to applied loads on BDS are calculated in a tabular form as given below:


Table 84−A


0.750.75

0.75(+)

0 0.25

1.3

(+)0.9

0.45

(-)00.25

S.F.D.

0.25

B.M.D.

1.0

δB.D.S under unit load at Bfor calculating bb, cb and dbδ δ

(U1 - diagram)

δB.D.S under unit load at Cfor calculating cc, bc and dcδ δ

10.5

0.50 +

0.5

0.5

0 S.F.D.

0.91.8

0.9+

B.M.D.

U2 - diagram

(+)

(-)(+)

0.25 0.751Same as above

0.25

0.75

BMD

1.3

δbd, δcd and δdd U3 diagram for

SDF

From the previous table we have the values of all flexibility co-efficients as given below:

∆B=391.65 × 10−6 m ∆C=1026.2 × 10−6 m ∆D=692.42 × 10−6 m δbb = 9.3616 × 10−6 m, and δcc = 11.1 × 10−6 m, δdd = 9.3565 × 10−6 m δbc = δcb = 6.417 × 10−6 m δbd = δdb = 3.517 × 10−6 m δcd = δdc = 6.291 × 10−6 m


Putting the values of flexibility co-efficients into compatibility equations we have.

391.65 × 10−6 +9.3616 × 10−6 Rb+6.292 × 10−6 Rc+3.517 × 10−6 Rd= 0 → (1)

1026.2 × 10−6 +6.292 × 10−6 Rb+11.1 × 10−6 Rc + 6.291 × 10−6 Rd = 0 → (2)

579.82 × 10−6 +3.517 × 10−6 Rb+6.291 × 10−6 Rc+9.3565 × 10−6Rd = 0 → (3)

Step No. 4

Simplify equation (1), (2) and (3), we have

391.65 +9.3620 Rb+6.292 Rc+3.517 Rd = 0 → (4)

1026.2 + 6.292 Rb + 11.1 Rc + 6.291 Rd = 0 → (5)

579.82 + 3.517 Rb + 6.291 Rc+9.357 Rd = 0 → (6)

Multiply (4) by 6.291 & (5) by 3.517 & subtract (5) from (4)

391.65 × 6.291+9.362 × 6.291Rb+6.292 × 6.291 Rc+3.517 × 6.291Rd=0

1026.2 × 3.517+6.292 × 3.517 Rb+11.1 × 3.517 Rc+3.517 × 6.291Rd=0 − 1145.275 + 36.767 Rb + 0.544 Rc = 0 → (7)

Multiply (5) by 9.357 & (6) by 6.291 & subtract (6) from (5) :-

1026.2 × 9.357+6.292 × 9.357 Rb+11.1 × 9.357 Rc+6.291 × 9.357Rd=0

579.82 × 6.291+3.517 × 6.291Rb+6.291 × 6.291 Rc+6.291 × 9.357Rd=0

5954.506 + 36.749 Rb + 64.286 Rc = 0 → (8)

From (7), Rb =

1145.275 − 0.544 Rc

36.767

Put Rb in (8) & solve for Rc

5954.506 + 36.749

1145.275 − 0.544 Rc

36.767 + 64.286 Rc = 0

5954.506 + 1144.71 − 0.544 Rc + 64.286 Rc = 0

7099.22 + 63.742 Rc = 0

Rc = − 111.374 KN

Put this value in equation (7) and solve for Rb

Rb =

1145.275 − 0.544 × 111.374

36.767

Rb = +32.797 KN

Put Rb and Rc values in equation (4) to get Rd. 391.65 + 9.362 × 32.797 + 6.292 × (111.374) +3.517 Rd = 0

Rd = + 0.588 KN


After reactions have been calculated, truss is statically determinate and member forces can be easily calculated by Fi = Fi/ + RbU1 + RcU2 + RdU3 as given in table. Apply checks on calculated member forces. Step No. 5: Equilibrium checks. Joint (C)

32.058

51.814 723.828

2.047

111.374 ∑ Fx = 0 − 2.047 − 32.058 − 3.828 × 0.707 + 51.814 × 0.707 = 0 − 0.179 ≅ 0 0 = 0 ∑ Fy = 0 111.374 − 72 − 3.828 × 0.707 − 51.814 × 0.707 = 0 0.035 ≅ 0 0 = 0 (satisfied) Solution is alright. 1.21: ANALYSIS OF 3-DEGREE REDUNDANT FRAMES Example No. 8: Analyze the following frame by consistent deformation method.

A3m

36KN

3m

B

3m 96KN

6m

C

32

7.5m

D

F

E

I

I

I

SOLUTION :- The given frame is statically indeterminate to the 3rd degree. So that three redundants have to be removed at support D or A. Consider HD, VD & MD as the redundants

A3m

36KN

3m

B

3m 96KN

6m

C

32

7.5m

D

F

E

I

I

I

=


96KN

396KN-mA

36KN2m

2 E36KN

3mB 3m

96KN6m C

7.5m

D

D

DH

Dv

I I

3I

Fig. 2.45 B.D.S under loads

A

1.53m

E

3m

B

3m

F

6m

C

7.5m

D

1

1

1

1.5m

dvdh

mH-Diagram

9

E

6m

B

F

9m

C

7.5m

1

dvdv

1

1

mV-Diagram

1

+

dhdv

A

3m

3m

B

3m

F

6m C

D

d d

m -diagramd dh

d dv

dhddvd

dh dh

(BDS under redundants) Compatibility Equations:- ∆DH + HD × δdh.dh + VD × δdhdv + MD × αdhdθ =0 (1) compatibility in horizontal direction at D.

∆DV + HD × δdv.dh + VD × δdvdv + MD × αdvdθ =0 (2) compatibility in vertical direction at D.

θD + HD × αdθ.dh +VD × αdθdv + MD × αdθdθ =0 (3) rotational compatibity at D.

We have to determine the following flexibility co-efficients.

∆DH = Horizontal deflection of point D due to applied loads.

∆DV = Vertical deflection of point D due to applied loads.

θD = Rotation of point D due to applied loads.

δdhdh = Horizontal deflection of point D due to unit horizontal redundant force at D


δdhdv = Horizontal deflection of point D due to unit vertical redundant force at D

αdθdh = angular deflection of point D due to unit angular redundant force at D

δdvdh = Vertical deflection of point D due to unit horizontal redundant force at D

δdvdv = Vertical deflection of point D due to unit vertical redundant force at D

αdθdv = Rotation deflection of point D due to unit vertical redundant force at D

αdhdθ = Horizontal rotation of point D due to unit rotation at pt D

αdvdθ = Vertical rotation of point D due to unit rotation at pt D

αdθdθ = Rotation rotation of point D due to unit rotation at pt D

δdvdh = δdhdv ( reciprocal deformations)

αdθdh = αdhdθ ( reciprocal deformations)

αdθdv = αdvdθ ( reciprocal deformations)

Now these flexibility co-efficients can be evaluated by following formulae.

∆DH = ∫ M × mHEI dX

∆DV = ∫ M × mVEI dX

θD = ∫ M x mθEI dX

δdhdh = ∫ (mH)2 dXEI

δdvdv = ∫ (mv)2 dXEI

αdθdh = αdhdθ = ∫ mH × mθ

EI dX

δdhdv = δdvdh = ∫ mv × mH

EI dX from law of reciprocals deformations

αdθdv = αdvdθ = ∫ mv × mθ

EI dX

αdθdθ = ∫ m2θ

EI dX


ESTABLISH MOMENT EXPRESSIONS BY FREE BODY DIAGRAMS: Note: Moments giving compression on outside and tension on inside of frame (sagging) will be positive.

96KN

396KN-mA

36KN3m

E

3m

36KN

B 288KN-m

96KN

288KN-m

3m

96KN

6m C

C

7.5m

D

B

96KNF

Fig 2.46 B.D.S under loads (M-diagram)

ΣMb = 0 Mb + 36 × 6 − 396 − 36 × 3 = 0

Mb = + 288 KN − m.

ΣMc = 0 Mc + 96 × 9 − 288 − 96 × 6 = 0

Mc + 0 = 0

Mc = 0

Free body m − Diagrams

A

1.53m

E

3m7.5

B1

7.5 B3m F 6m

C

7.5

1

C

7.5

7.5m

1

D

MFig. 2.46a mH-Diagam

1

9

3m

E

3m

9

1

B

9

3m

1

F

6m

C

1

1

C

7.5m

D1

MFig. 2.46b mv-diagram

1

1


1

1

3m

E3m

B

1

3m F 6m C

1

1C

7.5m

D1

Fig. 2.46 m diagram

B

Write moment expressions alongwith limits in a tabular form Portion Origin Limits M MH Mv Mθ I

AE A 0 − 3 36X−396 X + 1.5 − 9 − 1 2I

BE B 0 − 3 − 288 −X + 7.5 − 9 − 1 2I

BF B 0 − 3 96X−288 + 7.5 + X − 9 − 1 3I

CF C 0 − 6 0 + 7.5 − X − 1 3I

CD D 0 −7.5 0 + X 0 − 1 I It may be done in a tabular form or may be directly evaluated. CALCULATIONS OF FLEXIBILITY CO-EFFICIENTS:-

∆DH = 1EI ∫ M × mH dX

= 1

2EI 3

∫o (36X −396)(X+1.5 )dX+

12EI

3

∫o

(−288)(−X+7.5) dX + 1

3EI 3

∫o (96X−288)(7.5)dX +

6

∫o

0 + 7.5

∫o

0

= 1

2EI 3

∫o (36X2+54X −396X − 594) dX +

12EI

3

∫o (288X−2160) dX +

13EI

3

∫o (720X − 2160) dX

= 1

2EI 3

∫o (36X2 −54X−2754) dX +

13EI

3

∫o (720X − 2160)dX , (First two integrals have been combined)

= 1

2EI

36X3

3 − 54X2

2 − 2754 X3 o +

13EI

720X2

2 − 2160X3 o

= 1

2EI

12 × 33 −

542 × 32 − 2754 × 3 +

13EI

720

2 × 32 − 2160 × 3 − 4090.5

EI − 1080

EI

∆DH = 51.705

EI


δdhdh = 1EI ∫ mH2 dX

= 1

2EI 3

∫o

(X + 1.5)2dX + 1

2EI 3

∫o (−X+7.5)2dX +

13EI

3

∫o (7.5)2dX +

13EI

6

∫o (7.5)2dX +

1EI

7.5

∫o

X2dX

= 1

2EI 3

∫o(X2+3X+2.25)dX+

12EI

3

∫o(X2−15X+56.25)dX+

13EI

3

∫o56.25 dX+

13EI

6

∫o56.25 dX+

1EI

7.5

∫o

X2 dX

= 1

2EI

X3

3 + 3X2

2 + 2.25X3 o+

12EI

X3

3 − 15X2

2 +56.25X3 o+

13EI 56.25X

3|o+

13EI 56.25X

6|o+

1EI

X 3

3

7.5 o

= 1

2EI

33

3 ×32×32+2.25×3 +

12EI

33

3 −152 ×32+56.25×3 +

13EI (56.25×3) +

13EI(56.25×6) +

13EI

7.53

3

= 14.625

EI + 55.125

EI + 56.25

EI + 112.5

EI + 140.625

EI

δdhdh = + 379.125

EI

αdhdθ = 1EI ∫ (mH × mθ) dX

αdhdθ =1

2EI 3

∫o(X+1.5)(−1)dX+

12EI

3

∫o(−X+7.5)(−1)dX+

13EI

3

∫o(7.5)(−1)dX+

13EI

6

∫o(7.5)(−1)dX+

1 EI

7.5

∫o

(X)(−1)dX

= 1

2EI 3

∫o

(−X−1.5)dX + 1

2EI 3

∫o

(X−7.5)dX + 1

3EI 3

∫o

(−7.5)dX + 1

3EI 6

∫o

(−7.5)dX + 1EI

7.5

∫o

(−X)

= 1

2EI 3

∫o

(−9)dX + 1

2EI 3

∫o

(−7.5)dX + 1

3EI 6

∫o

(−7.5)dX + 1EI

7.5

∫o

(−X)dX

= 1

2EI −9X 3|o +

13EI −7.5X

3|o +

13EI −7.5X

6|o+

1EI

-

X2

2

7.5 o

= 1

2EI (−9 × 3) + 1

3EI (−7.5 × 3) + 1

3EI (−7.5 × 6) + 1EI

−

(7.5)2

2

αdhdθ = − 64.125

EI

∆Dv = 1EI ∫ (M × mv) dX

∆Dv = 1

2EI 3

∫o

(36X − 396 )(−9 ) dX + 1

2EI 3

∫o

(−288 )(−9 ) dX +1

3EI 3

∫o

(96X − 288) (X−9)dX + 0 + 0


= 1

2EI 3

∫o

(−324X+3564) dX + 1

2EI 3

∫o 2592 dX +

13EI

3

∫o(96X2−864X −288X + 2592) dX

= 1

2EI 3

∫o(−324X + 6156) dX +

13EI

3

∫o(96X2 − 1152X + 2592) dX

= 1

2EI

−324X2

2 + 6156X3 o +

13EI

96X3

3 − 1152X2

2 + 2592X3 o

= 1

2EI (−162 × 32 + 6156 × 3) + 1

3EI (32 × 33 − 576 × 32 + 2592 × 3)

= 8505

EI + 1152

EI

∆Dv = 9657

EI

δdvdv = 1EI ∫ (mv)2 dX

= 1

2EI 3

∫o(−9 )2 dX +

12EI

3

∫o(−9 )2 dX +

13EI

3

∫o(X−9 )2 dX +

13EI

6

∫o (−X)2 dX +

1EI

7.5

∫o

( 0 ) dX

= 1

2EI 3

∫o162 dX +

13EI

3

∫o(X2 −18X + 81) dX +

13EI

6

∫o X2 dX

= 1622EI X

3|o +

13EI

X2

3 − 18X2

2 + 81X3 o +

13EI

X3

3

6 o

= 81(3)

EI + 1

3EI

33

3 − 9 × 32 + 81 × 3 + 1

3EI 63

3

δdvdv = + 324EI

αdvdθ = 1EI ∫ (mv × mθ) dX

αdvdθ = 1

2EI 3

∫o

9 dX + 1

2EI 3

∫o 9 dX +

13EI

3

∫o

(−X + 9) dX + 1

3EI 6

∫o × dX + 0

= 1

2EI 9X 3|o +

12EI 9X

3|o +

13EI

−

X2

2 + 9X3|o+

13EI

X2

2

6 o

= 1

2EI (9 × 3) + 1

2EI (9 × 3) + 1

3EI

−9

2 + 9 × 3 + 1

3EI

36

2

αdvdθ = + 40.5EI


αdθdθ = 1EI ∫ (mθ )2 dX

αdθdθ = 1

2EI 3

∫o 1dX +

12EI

3

∫o 1dX +

13EI

3

∫o 1dX +

13EI

6

∫o 1dX +

1EI

7.5

∫o

1dX

= 1

2EI X 3|o +

13EI X

3|o +

13EI X

6|o +

1EI X

7.5|o

= 1EI (3) +

13EI (3) +

13EI (6) +

1EI (7.5)

αdθdθ = + 13.5EI

θD = 1EI ∫ (M x mθ) dX

= 1

2EI 3

∫o (−36X +396) dX +

12EI

3

∫o 288 dX +

13EI

3

∫o (−96X + 288) dX

= 1

2EI 3

∫o (−36X + 684) dX +

13EI

3

∫o (−96X + 288) dX

= 1

2EI

−36

X2

2 + 684X3 o +

13EI

−96

X2

2 + 288X3 o

= 1

2EI (−18 × 9 + 684 × 3) + 1

3EI (− 48 × 9 + 288 × 3)

θD = + 1089

EI

δdhdv = 1EI ∫ (mH × mv ) dX

δdhdv = 1

2EI 3

∫o (−9X − 13.5)dX +

12EI

3

∫o (+9X − 67.5)dX +

13EI

3

∫o (7.5x − 67.5)dX +

13EI

6

∫o (−7.5X) dX +0

= 1

2EI 3

∫o (− 81)dX +

13EI

3

∫o (7.5X − 67.5) dx +

13EI

6

∫o (− 7.5X) dX


= 1

2EI −81X 3|o +

13EI

7.5X2

2 − 67.5X3|o +

13EI

−

7.5X2

2

6 o

= 1

2EI (−81 × 3) + 1

3EI

7.5

2 × 9 − 67.5 × 3 + 1

3EI

−

−7.52 × 36

δdhdv = − 222.75

EI

Putting above evaluated flexibility co−efficients in compatibility equations , we have.

(1) ⇒ −5170.5 + 379.125 HD − 222.75 VD − 64.125 MD = 0 → (4)

(2) ⇒ +9657 − 222.75 HD + 324 VD + 40.5 MD = 0 → (5)

(3) ⇒ + 1089 − 64.125 HD + 40.5 VD + 13.5 MD = 0 → (6)

Multiply (4) by 222.75 & (5) by 379.125 Then add (4) & (5) to eliminate HD

− (5170.5 × 222.75) +(379.125 × 222.75)HD−(222.75)2VD−(64.125 × 222.75)MD =0

+(9657×379.125)− (379.125×222.75)HD+(324×379.125)VD+(40.5×379.125) MD=0

2509481.25 + 73218.9375 VD +1070.72 MD = 0 → (7)

Multiply (5) by 64.125 & (6) by 222.75 & subtract (6) from (5) to eliminate HD again

619255.125 − 14283.84 HD + 20776.5 VD + 2597.06 MD = 0

− 242574.75 − 14283.84 HD + 9021.375 VD + 3007.125 MD= 0

376680.375 + 11755.125 VD − 410.065 MD = 0 → (8)

Now equation (7) and (8) are in terms of VD and MD From (7), VD =

−1070.72 MD − 2509481.25

73218.9375 → (9)

Put VD in (8) to get MD

376680.375 + 11755.125

−1070.72 MD − 2509481.25

73218.9375 − 410.065MD = 0

376680.375 − 171.90 MD − 402891.20 − 410.065 MD = 0 − 26210.83 − 581.965 MD = 0 MD = − 45.04 KN−m, put this in (9) to get VD

VD =

−1070.72 × (45.04) − 2509481.25

73218.9375


VD = − 33.62 KN. Now put values of VD and MD in (4) to get HD − 5170.5+379.125 × HD+222.75 × 33.62 + 64.125 × 45.04 = 0 379.125 HD + 5205.44 = 0 HD = − 13.73 KN HD = − 13.73 KN

VD = − 33.62 KN MD = − 45.64 KN − m

These reactions are applied to frame which becomes statically determinate now and shear force and moment diagram can be sketched (by parts) now.

Ma=68.98Kn-mVA =62.38KN

A3m

HA=22.27KN

E

23

36KN

3m

B3m 6m C

7.5m

D 45.04KN-m

13.73KN

33.62KN

I

I

I

Fig. 2.47

96

Applying condition of equilibrium at A, reactions can be obtained.

∑ FX = 0

36 − HA − 13.73 = 0

HA = 22.27 KN

∑Fy = 0

VA + 33.62 − 96 = 0

VA = 62.38 KN


∑ M = 0

MA + 45.04 − 13.73 × 1.5 + 33.62 × 9 − 96 × 3− 36 × 3 = 0

MA − 68.98 = 0

MA = 68.98 KN-m Applying these reactions to frame, various free-body diagrams can be drawn and moments expressions can be set-up for

determining combined deflections of any point due to applied loads and reactions (at supports) acting simultaneously.

62.38Kn

22.27Kn68.98KN-m

3m

E36KN3m 13.73KN

43.36KN-m

62.38KN

43.36KN-m 96KN 57.94KN-m

57.94KN-m

13.73KN C

7.5m

D45.04KN-m

13.73Kn

33.62KN

E

M = 0 , Mb+22.27 x 6 68.98-36 x 3 = 0b -Mb = 43.36 KN-m

M+62.38 x 9-43.36-96x 6=0Mc=57.94 KN-m (for beam)

62.38KN

13.73KN 6m3mF

C 13.73KN

33.62KN33.62KN6m

B

B

A

Mc=0 ,

BENDING MOMENT AND SHEAR FORCE DIAGRAMS :−

For beam BC

43.36

0

-

x=0.695

33.62143.78

0+

62.3862.38KN

B43.36KN-m

3m96KN

6m57.94KN-mC

33.62KN

S.F.D.033.62

=1.723mx

0 B.M.D.

57.94

m

Mx = −45.04 + 13.73x = 0

x = 3.28 m


FOR COLUMN AB (Seen rotated at 90°)

68.98

0

0+22.27

22.27KN

A

68.98KN-m

3m

36KN

3m B

43.36KN-m

13.73KN

S.F.D.013.73

0

43.36

2.17 FOR COLUMN DC (Seen rotated at 90°)

45.04

0

X=3.28m

+0

+

13.73

13.73KND45.04KN-m

7.5m57.94KN-m

C13.73KN

13.73

0 S.F.D.

0

57.94

B.M.D.

+

Mx=-45.04+13.73x = 0x = 3.28m

68.98

2.17B.M.D.

43.36 4 .36

+

143.78

57.94

57.94+

45.04

3

22.27

+

13.73S.F.D.

62.38

13.73

33.62

+

13.73Composite S.F.D. for analysed frame

Fig. 2.48


Elastic Curve:-

1.22: Analysis of Continuous Beams Example No. 9: Analyze the following beam by consistent deformation method. Check the results by the method of

least work. SOLUTION:-

A 30m B 40m C 40m D 30m E

15m 10KN10m

5KN E1=Constt

Number of reactions=5number of equations=2

Fig. 2.56 Step No.1: In this structure, we treat reactions at B, C & D as redundants and the B.D.S. is a simply supported

beam AE.

140m

A B

B

C

C

D E

D

15m10KN

10m5KN

B.D.S. Under applied loads. Fig. 2.56a

A B C D E

bb x Vb cb x VbD

dbxVb

B.D.S. Under Unit redundant load at B. Fig. 2.56 b

1

A B C D E bc x Vc

cc

x

Vc

dc x

Vc

1

B.D.S. under Unit redundant load at C. Fig. 2.56c

U


A B C D E bd

x

Vd

cd

x

Vd

dd

x

Vd

1

B.D.S. under Unit redundant load at D. Fig. 2.56d

U

Step No.2: Compatibility Equations. ∆B + Vb × δbb + Vc × δbc + Vd × δbd = 0 → (1) Compatibility of deformations at B

∆C + Vb × δcb + Vc × δcc + Vd × δcd = 0 → (2) Compatibility of deformations at C

∆D + Vb × δdb + Vc × δdc + Vd × δdd = 0 → (3) Compatibility of deformation at D

Sketch BDS, Draw SFD, and MEI diagram for use in conjugate beam method.

A B C D E

A1 A2

A3

A4

9748.339/E111631.161/E1

0 03.93

S.F.D.++

1.07 1.0711.07

= 11.07KN

3.93KN = RE

140m80m 60m

AB C D E

15m10KN 5KN

RA =

10x 125140

5x60140

xFig. 2.57

65 m

M/EI diagram overconjugate beam

166.05/EI235.8/EI

Splitting above MEI in 4 parts as shown, calculate areas of these portions.

A1 = 12 × 15 ×

166.05EI =

1245.375EI

A2 = 166.05

EI × 65 = 10793.25

EI


A3 = 12 ×

69.75EI × 65 =

2266.875EI

A4 = 12 × 235.8 × 60 =

7074EI

A1+A2+A3+A4 = 21379.5

EI

∑ M′E = 0

RA′ x 140 = 1EI

1245.375

125+

153 +10793.25

60+

652 +2266.875

60+

653 +7074

2

3×60

RA′ = 11631.161

EI

RE′ = 21379.5

EI − 11631.161

EI

RE′ = 9748.339

EI

Isolating the upper part of MEI diagram between two loads.

15

6555

235.8

Y1 Yy2y166.05/EI

BC

y255 =

235.865 By conjugate beam method, ∆B would be moment at B' of conjugate beam

loaded with MEI diagram.

y2 = 199.52 y1 = 54.4

∆B = 1EI

11631.161×30−1245.375

15+

153 − (166.05×15) × 7.5 −

54.42×

152 ×

15

3

= 303080.955

EI KN−m3

Moment at C' of conjugate beam

∆C = 1EI

11631.161×70−(1245.375)

15

3 +55 −(166.05×55)

55

2 −

1

2×100.52×5.5 ×

1

3×55

= 387716.812

EI KN−m3


60m

Y y3

30235.8

Dy = 117.9/EI3

Isolating the portion of

MEI diagram between right support and 5 KN load.

Moment at D' of conjugate beam

∆D = 1EI

9748.339 × 30 −

1

2 × 117.9 × 30 × 303

∆D = 274765.17

EI KN−m3

If we construct MEI diagram for above figures 2.56b, 2.56c and 2.56d and place them over conjugate beam,

we have δcb= 34501.88, δcc= 57166.66, δcd= 34501.88 on similar lines as above. From conjugate beam for fig: 2.56b, you will have

δbb = 1EI

982.086 × 30 − (353.565)

30

30 = 25926.93

EI

δcb = 1EI

667.884 × 70 −

1

2 × 15 × 70

70

3 = 34501.88

EI

δdb = 1EI

667.884 × 30 −

1

2 × 6.423 × 30

30

3 = 19073.07

EI

We already know from law of reciprocal deflections that δcb = δbc δbd = δdb δcd = δdc From conjugate beam for fig: 2.5d, you will have

δcd = 1EI

667.884 × 70 −

15 × 70

2

70

3 = 34501.88

EI

δdd = 1EI

982.086 × 30 −

1

2 × 23.571 × 30

30

EI = 25926.93

EI

Putting above flexibility co-efficients in compatibility equations, we have

303080.955 + 25926.93 Vb + 34500 Vc + 19073.07 Vd = 0 → (1) 387716.812 + 34501.88 Vb + 57166.67 Vc + 34501.88 Vd = 0 → (2) 274765.17 + 1907307 Vb + 34500 Vc + 25926.93 Vd = 0 → (3)

Solving above three linear – simultaneous equations, we have

Vd = − 14.30 KN

Vc = 12.98 KN

Vb = 18.44 KN


Now the continuous beam has become determinate. Apply loads and redundants reactions, other support reactions can be determined.

A B C D E

Va 18.44KN 12.98KN 14.30KN Ve

10KN 5KN15m 10m

∑ME = 0 Va × 140 − 10 × 125 − 18.44 × 110 − 12.98 × 70 − 5 × 60 + 14.3 × 30 = 0

Va = 28.9 KN ∑ Fy = 0 gives Ve = 3.22 KN upwards Now shear force and BMD can be plotted as the beam is statically determinate now.

METHOD OF LEAST WORK

91

CHAPTER TWO


The method of least work is used for the analysis of statically indeterminate beams, frames and trusses. Indirect use of the Castigliano’s 2nd theorem is made and the following steps are taken. (1) The structure is considered under the action of applied loads and the redundants. The

redundants can be decided by choosing a particular basic determinate structure and the choice of redundants may vary within a problem.

(2) Moment expressions for the entire structure are established in terms of the applied loads

and the redundants, which are assumed to act simultaneously for beams and frames. (3) Strain energy stored due to direct forces and in bending etc. is calculated and is partially

differentiated with respect to the redundants. (4) A set of linear equations is obtained, the number of which is equal to that of the

redundants.Solution of these equations evaluates the redundants. NOTE:− Special care must be exercised while partially differentiating the strain energy expressions and

compatibility requirements of the chosen basic determinate structure should also be kept in mind. For the convenience of readers, Castigliano’s theorem are given below:

2.1. CASTIGLIANO’S FIRST THEOREM:−

“The partial derivative of the total strain energy stored with respect to a particular deformation gives the corresponding force acting at that point.”

Mathematically this theorem is stated as below:

∂U∂∆ = P

and

∂U∂θ = M

It suggests that displacements correspond to loads while rotations correspond to moments. 2.2. CASTIGLIANO’S SECOND THEOREM :− “The partial derivative of the total strain energy stored with respect to a particular force gives the corresponding deformation at that point.” Mathematically,

∂U∂P = ∆

and

∂U∂M = θ


2.3. STATEMENT OF THEOREM OF LEAST WORK. “In a statically indeterminate structure, the redundants are such that the internal strain energy stored is minimum.” This minima is achieved by partially differentiating strain energy and setting it to zero or to a known value. This forms the basis of structural stability and of Finite Element Method. 2.4. Example No.1: 1st Degree Indeterminacy of Beams. Analyze the following loaded beam by the method of least work.

LRb

Ma

A Bx

wKN/m

Number of reactions = 3nNumber of equations = 2

Ra

The beam is redundant to first degree. In case of cantilever, always take free end as the origin for establishing moment expressions. Choosing cantilever with support at A and Rb as redundant. Apply loads and redundant simultaneously to BDS.

LRa Rb

BA

MaWwKN/m

x

Taking B as origin (for variation of X)

MX =

RbX −

wX2

2 0 < X < L

U = 1

2EI L

∫o M2 dX. A generalized strain energy expression due to moments.

Therefore, partially differentiating the strain energy stored w.r.t. redundant, the generalized form is:

∂U∂R =

1EI

L

∫o M

∂M

∂R dX Where R is a typical redundant.

Putting moment expression alongwith its limits of validity in strain energy expression.

U = 1

2EI L

∫o

RbX −

wX2

2

2

dX

Partially differentiate strain energy U w.r.t. redundant Rb, and set equal to zero.

So ∂U∂Rb = ∆b = 0 =

1EI

L

∫o

RbX −

wX2

2 (X) dX, because at B, there should be no deflection.


93

0 = 1EI

L

∫o

RbX2 −

wX3

2 dX

0 = 1EI

RbX3

3 − wX4

8

L o

Or RbL3

3 = wL4

8

and

Rb = +38 wL

The (+ve) sign with Rb indicates that the assumed direction of redundant Rb is correct. Now calculate Ra. ∑ Fy = 0

Ra + Rb = wL

Ra = wL − Rb

= wL − 38 wL

= 8 wL − 3 wL

8

Ra = 58 wL

Put X = L and Rb = 38 wL in moment expression for MX already established before to get Ma.

Ma = 38 wL .L −

wL2

2

= 38 wL2 −

wL2

2

= 3 wL2 − 4 wL2

8

Ma = − wL2

8

The (−ve) sign with Ma indicates that this reactive moment should be applied such that it gives us tension at the top at point A.


Example No.2: Solve the following propped cantilever loaded at its centre as shown by method of least work.

Ra L Rb

BAMa x

xP

C

B.D.S. is a cantiever supported at A.I

Ra L

L/2

Rb

BAMa x

xPC

Rb is a redundant as shown.

BDS under loads and redundant. Taking point B as origin.

Mbc = RbX 0 < X < L2

and Mac = RbX − P

x −

L2

L2 < X < L. Now write strain energy expression.

U = 1

2EI L/2

∫o

(RbX)2 dX + 1

2EI L

∫ L/2

RbX − P

X −

L2

2 dX. Partially differentiate

w.r.t redundant Rb.

∂U∂Rb = ∆b = 0 =

1EI

L/2

∫o

[RbX] [X] dX + 1EI

L

∫ L/2

Rbx − P

X −

L2 [X] dX

0 = 1EI

L/2

∫o

RbX2 dX + 1EI

L

∫ L/2

RbX2 − PX2 + P

L2 X dX

0 = 1EI

Rb.

X3

3 L/2 o

+ 1EI

RbX3

3 − PX3

3 + PL4 X2

L L/2

. Put limits

0 = 1EI

RbL3

24 − 0 + 1EI

RbL3

3 − PL3

3 + PL3

4 − RbL3

24 + PL3

24 − PL3

16

0 = 1EI

RbL3

24 + RbL3

3 − RbL3

24 − PL3

3 + PL3

4 + PL3

24 − PL3

16

0 = 1EI

RbL3

3 +

− 16PL3 + 12PL3 + 2PL3 − 3PL3

48


95

0 = RbL3

3 − 5PL3

48

Or RbL3

3 = 5PL3

48

Rb = +5P16

The (+ve) sign with Rb indicates that the assumed direction of redundant Rb is correct. Now Ra can be calculated. ∑ Fy = 0 Ra + Rb = P Ra = P − Rb

Ra = P − 5P16 =

16P − 5P16

Ra = 11P16

Put X = L and Rb = 5P16 in expression for Mac to get Ma.

Ma = 5P16 L − P

L2

= 5 PL − 8 PL

16

Ma = − 3 PL

16

The (−ve) sign with Ma indicates that this reactive moment should be acting such that it gives us tension at the top. 2.5. 2ND DEGREE INDETERMINACY:−

EXAMPLE NO. 3: Analyze the following fixed ended beam loaded by Udl by least work method.

LRbRa

WwKN/m

BAMa Mb

B.D.S. is chosen as a cantilever supported at A. Rb and Mb are chosen as redundants.

L RbRa

WwKN/mx

BAMa

Mb

BDS UNDER LOADS AND REDUNDANTS


Mx = RbX − wX2

2 − Mb 0 < X < L Choosing B as origin.

Write strain energy expression.

U = 1

2EI L

∫o

RbX −

wX2

2 − Mb2 dX

Differentiate strain energy partially w.r.t. redundant Rb and use castigations theorem alongwith boundary condition.

∂U∂Rb = ∆b = 0 =

1EI

L

∫o

RbX −

wX2

2 − Mb [X] dX

0 = 1EI

L

∫o

RbX −

wX2

2 − Mb dX

0 = 1EI

Rb

X3

3 − wX4

8 − MbX2

2

L o

0 = 1EI

Rb

L3

3 − wL4

8 − MbL2

2

0 = Rb L3

3 − wL4

8 − MbL2

2 → (1)

As there are two redundants, so we require two equations. Now differentiate strain energy expression w.r.t. another redundants Mb. Use castigations theorem and boundary condition.

∂U

∂Mb = θb = 0 = 1EI

L

∫o

RbX −

wX2

2 − Mb ( −1) dX

0 = 1EI

L

∫o

− RbX +

wX2

2 + Mb dX

0 = 1EI

−

RbX2

2 + wX3

6 + MbX L o

0 = − Rb L2

2 + wL3

6 + MbL.

Rb L2

2 − wL3

6 = MbL

So Mb = RbL

2 − wL2

6 → (2) Put Mb in equation 1, we get

0 = RbL3

3 − wL4

8 −

RbL2 −

wL2

6 L2

2


97

0 = RbL3

3 − wL4

8 − RbL3

4 + wL4

12

0 = RbL3

12 − wL4

24

Rb = wL2

Put Rb value in equation 2, we have

Mb =

wL

2 L2 −

wL2

6

Mb = +wL2

12 The (+ve) value with Rb and Mb indicates that the assumed directions of these two redundants are correct. Now find other reactions Ra and Mb by using equations of static equilibrium.

∑ Fy = 0 Ra + Rb = wL Ra = wL − Rb

= wL − wL2

Ra = wL2

Put X = L , Rb = wL2 & Mb =

wL2

12 in MX expression to get Ma

Ma = wL2 . L −

wL2

2 − wL2

12

Ma = − wL2

12

The (−ve) sign with Ma indicates that this moment should be applied in such direction that it gives us tension at the top. Example No. 4: Solve the same previous fixed ended beam by taking a simple beam as B.D.S.:−

Choosing Ma and Mb as redundants.

L RbRa

WwKN/mx

BA

MaMb


B.D.S. is a simply supported beam , So Ma and Mb are redundants.


∑ Ma = 0

Rb × L + Ma = Mb + wL2

2

Rb × L = (Mb − Ma ) + wL2

2

Rb =

Mb − Ma

L + wL2 So taking B as origin. Write MX expression.

MX = RbX − Mb − wX2

2 0 < X < L

Put Rb value

MX =

Mb − Ma

L + wL2 X −

wX2

2 − Mb 0 < X < L. Set up strain energy

expression.

U = 1

2EI L

∫o

Mb − Ma

L + wL2 X −

wX2

2 − Mb2 dX. Differentiate w.r.t. Ma first.

Use castigations theorem and boundary conditions.

∂U

∂Ma = θa = 0 = 1EI

L

∫o

Mb − Ma

L + wL2 X −

wX2

2 − Mb

−

XL dX. In general R.H.S.

is 1EI ∫ N.m.dX.

0 = 1EI

L

∫o

MbX

L − MaX

L + wL2 X −

wX2

2 − Mb

−

XL dX

0 = 1EI

L

∫o

−

MbX2

L2 + MaX2

L2 − wX2

2 + wX3

2L + MbX

L dX . Integrate it.

0 = 1EI

−

MbL2

X2

3 + MaL2

X3

3 − wX3

6 + wX4

8L + MbX2

2L L o . Simplify it.

0 = MbL

6 + MaL

3 − wL3

24 → (1)

Now differentiate U Partially w.r.t. Mb. Use castiglianos theorem and boundary conditions.

∂U

∂Mb = θb = 0 = 1EI

L

∫o

Mb − Ma

L + wL2 X −

wX2

2 − Mb

X

L − 1 dX

0 = 1EI

L

∫o

MbX

L − MaX

L + wL2 X −

wX2

2 − Mb

X

L − 1 dX

0 = L

∫o

MbX2

L2 − MaX2

L2 + wLX2

2L − wX3

2L − MbX

L − MbX

L + MaX

L − wLX

2 + wX2

2 + Mb dX


99

0 =

MbX3

3L2 − MaX3

3L2 + wX3

6 − wX4

8L − MbX2

2L − MbX2

2L + MaX2

2L − wLX2

4 + wX3

6 + MbXL o

Put limits now.

0 =

MbL3

3L2 − MaL3

3L2 + wL3

6 − wL4

8L − MbL2

2L − MbL2

2L + MaL2

2L − wLL2

4 + wL3

6 + MbL

Simplifying we get.

0 = MbL

3 + MaL

6 − wL3

24

or MbL

3 = − MaL

6 + wL3

24

so Mb = wL2

8 − Ma2 (2), Put Mb in equation (1) we get.

0 =

wL2

8 − Ma2

L6 +

MaL3 −

wL3

24 Simplify to get Ma.

0 = wL3

48 − MaL12 +

MaL3 −

wL3

24

Ma = wL2

12

Put Ma in equation (2) , we have

Mb = wL2

8 − wL2

12 × 12

or Mb = wL2

12 ; Now Rb =

Ma + Mb

L + wL2 Putting Ma and Mb we have.

Rb =

wL2

12 − wL2

12L +

wL2

Rb = wL2 , Calculate Ra now.

∑ Fy = 0 Ra + Rb = wL Put value of Rb. Ra = wL − Rb

Ra = wL − wL2

Ra = wL2

We get same results even with a different BDS. The beam is now statically determinate. SFD and BMD can be drawn. Deflections at can be found by routine methods.


2.6. 2ND DEGREE INDETERMINACY OF BEAMS:− Exmaple No. 5: Solve the following loaded beam by the method of least work.

L/2 L/2

W W

AB

C

EI=Constant

B.D.S. is a cantilever supported at A. Rb & Rc are chosen as redundants.

L/2 L/2

Ra Rb Rc

A CB

xxW WMa


Choosing C as origin, Set-up moment expressions in different parts of this beam.

Mbc = Rc.X − wX2

2 0 < X < L2

Mab = Rc.X + Rb

X −

L2 −

wX2

2 L2 < X < L . Write strain energy expression for entire

structure.

U = 1

2EI L/2

∫o

Rc.X −

wX2

2 2

dX + 1

2EI L

∫L/2

Rc.X + Rb

X −

L2 −

wX2

2

2 dX

Partially differentiate it w.r.t. redundant Rc first. Use castiglianos theorem and boundary conditions.

∂U∂Rc = ∆c = 0 =

1EI

L/2

∫o

Rc.X −

wX2

2 [X]dX + 1EI

L

∫L/2

Rc.X + Rb

X −

L2 −

wX2

2 [X] dX

0 = 1EI

L/2

∫o

Rc.X2 −

wX3

2 dX + 1EI

L

∫L/2

Rc.X2 + Rb.X2 −

Rb.LX2 −

wX3

2 dX . Integrate it.

0 = 1EI

Rc.

X3

3 − wX 4

8 L/2 o

+ 1EI

Rc.

X3

3 + Rb. X3

3 − RbLX2

4 . − wX 4

8

L L/2

. Insert limits and

simplify.

0 = Rc.L3

3 + 5Rb.L3

48 − wL4

8 → (1)


101

Now partially differentiate strain energy w.r.t. Rb. Use Castiglianos theorem and boundary conditions.

∂U∂Rb = ∆b = 0 =

1EI

L/2

∫o

Rc.X −

wX2

2 (0) dX + 1EI

L

∫L/2

Rc.X + Rb

X −

L2 −

wX2

2

X −

L2 dX

0 = 0 + 1EI

L

∫L/2

Rc.X2 + RbX2 −

RbLX2 −

wX3 2 −

Rc.L.X2 −

RbL.X2 +

Rb.L2

4 + wL.X2

4 dX.

Integrate.

0 = 1EI

Rc.X3

3 + Rb.X3

3 − Rb.L.X2

4 − wX4

8 − Rc.L.X2

4 − Rb.LX2

4 + Rb.L2.X

4 + wL.X3

12

L L/2

.

Put limits

0 = Rc.L3

3 + Rb.L3

3 − Rb.L3

4 − wL4

8 − Rc.L3

4 − Rb.L3

4 + Rb.L3

4 + wL4

12 − Rc.L3

24 − Rb.L3

24

+ Rb.L3

16 + wL4

128 + Rc.L3

16 + Rb.L3

16 − Rb.L3

8 − wL4

96

Simplify to get

Rc. = − 25 Rb. +

1740 wL → (2) Put this value of Rc in equation ( 1), to get Rb

0 =

−

25 Rb. +

1740 wL

L3

3 + 5

48 Rb.L3 − wL4

8 (1)

0 = − 215 Rb.L3 +

17120 wL4 +

548 Rb.L3 −

wL4

8

Simplify to get

Rb. = 1221 wL

Put value of Rb in equation (2) and evaluate Rc,

Rc = − 25 ×

1221 wL +

1740 wL

Rc = 1156 wL

The (+ve) signs with Rb & Rc indicate that the assumed directions of these two redundants are correct. Now calculate Ra. ∑ Fy = 0 Ra + Rb + Rc = wL or Ra = wL − Rb − Rc . Put values of Rb and Rc from above and simplify.


= wL − 1221 wL −

11 wL56

Ra = 373

1176 wL

Ra = 91

392 wL

Putting the values of these reactions in Mx expression for span AB and set X = L, we have

Ma = Rc.L + Rb. L2 −

wL2

2 . Put values of Rb and Rc from above and simplify.

= 11 wL

56 .L + 1221 wL ×

L2 −

wL2

2

Ma = − 21

1176 wL2

Ma = − 7

392 wL2

The (−ve) sign with Ma indicates that this reactive moment should be applied in such a direction that gives us tension at the top. Now the beam has been analyzed and it is statically determinate now. 2.7. INTERNAL INDETERMINACY OF STRUCTURES BY FORCE METHOD :−

The question of internal indeterminacy relates to the skeletal structures like trusses which have discrete line members connected at the ends. The structures which fall in this category may include trusses and skeletal frames. For fixed ended portal frames, the question of internal indeterminacy is of theoretical interest only.

1 2

Relative displacementof horizontal number =

Consider he truss shown in the above diagram. If this truss is to be treated as internally indeterminate, more than one members can be considered as redundants. However, the following points should be considered for deciding the redundant members.

(1) The member which is chosen the redundant member is usually assumed to be removed or cut. The selection of redundant should be such that it should not effect the stability of the remaining structure.


103

(2) The skeletal redundant members will have unequal elongations at the two ends and in the direction in which the member is located. For example, if a horizontal member is chosen as redundant, then we will be concerned with the relative displacement of that member in the horizontal direction only.

(3) Unequal nodal deflection (∆1 − ∆2 ) of a typical member shown above which is often termed as relative displacement is responsible for the self elongation of the member and hence the internal force in that member.

2.7.1. FIRST APPROACH: WHEN THE MEMBER IS REMOVED :− With reference to the above diagram, we assume that the redundant member (sloping up to left) in the actual structure is in tension due to the combined effect of the applied loads and the redundant itself. Then the member is removed and now the structure will be under the action of applied loads only.

A D

CBB1

TogetherB2

Apart

Due to the applied loads, the distance between the points B and D will increase. Let us assume that point B is displaced to its position B2. This displacement is termed as ∆ apart. Now the same structure is considered under the action of redundant force only and let us assume that point B2 comes to its position B1 (some of the deflections have been recovered). This displacement is termed as ∆ together. The difference of these two displacements ( ∆apart − ∆together) is infact the self lengthening of the member BD and the compatibility equation is

∆apart − ∆together = self elongation. 2.7.2. 2ND APPROACH We assume that the member is infact cut and the distance between the cut ends has to vanish away when the structure is under the action of applied loads and the redundant. In other words, we can say that the deformation produced by the applied loads plus the deformation produced by the redundant should be equal to zero.

A D

B C

F-Diagram

A D

B C

U-Diagram

1

1


Total Deflection produced by redundants ∆ × R = n

∑i = 1

2

UiLiAiEi

× X

Total Deflection produced by loads ∆ × L = n

∑i = 1

FiUiLiAiEi

If deflection is (+ve), there is elongation. If deflection is (−ve), there is shortening.

Now U = P2L2AE Elastic strain energy stored due to axial forces

PL AE

P PROOF:− Work done = 1/2 P.∆ = shaded area of P − ∆ diagram. Now f α ∈ (Hooke’s Law)

or PA α

∆L (For direct stresses)

PA = E

∆L where E is Yung’s Modulus of elasticity.

∆ = PLAE

Therefore work done = P∆2 =

12 P.

PLAE ( Shaded area under P−∆ line __ By putting value of ∆)

Work done = P2L2AE (for single member)

Work done = ∑ P2L2AE (for several members)

We know that Work done is always equal to strain energy stored.


105

EXAMPLE NO 6:

Analyze the truss shown below by Method of Least work. Take

(1) Member U1L2 as redundant.

(2) Member U1U2 as redundant. Number in brackets ( ) are areas × 10−3 m2. E = 200 × 106 KN/m2

(2.4) L1 (2.4) L2 (2.4) L0 L3

3 @ 4,5m48KN

6m(3.0)

(24)U1 U2

(3.0)(1.2)

(1.8)

(1.8)

L0U1 = 7.5mCos = 0.8Sin = 0.6

Note: In case of internally redundant trusses, Unit load method (a special case of strain energy method) is preferred over direct strain energy computations followed by their partial differentiation.

SOLUTION: Case 1 – Member U1L2 as redundant

F-Diagram

[email protected]

L32.0 2.4 L1 2.4 L2

6m3.01.21.8

U2U1

1.2

L0

L0 U1=7.5mCos = 0.8Sin = 0.6

3.0

(1) U1L2 is redundant.: STEPS 1 − Remove this member. (See – diagram)

2 − Assume that tensile forces would be induced in this member.

3 − Analyze the structure without U1L2 (B.D.S.) or F' diagram.

4 − Displacement of members due to redundant + that due to loads should be equal to zero. OR

∆ × L + ∆ × R = 0

5 − Analyze the truss with unit tensile force representing U1L2 or U−diagram.


Condition: ∆ apart = ∑81

F′ULAE ∆ together= ∑8

1 U2LAE × PU1 L2

U1 2.4 U2

3.0

3.0 6m

L1 L2 L3

L0

1.2

1.8

1.2

2.4

[email protected]

16

0

+

0

32

144

0

72+

+0 B.M.D.

(BDS under loads) F - diagram/

SFD

48

We shall determine member forces for F/ - diagram by method of moments and shears as

explained earlier. These are shown in table given in pages to follow. Member forces in U-diagram are determined by the method of joints.

U1 U20.60

LoO L1 L2 o

L3Cos

Sin

+1.0

11

(BDS under) U-diagram redundant unit force.

JOINT (L2)

L1 L2

U2L21

∑FX = 0

1 × Sinθ + L1L2 = 0

L1L2 = − Sinθ = − 0.60

∑ Fy = 0

U2L2 + 1 × Cosθ = 0

U2L2 = − Cosθ = − 0.80


107

Joint (L1) U1

L1

L1U2

0.6L1L2

∑ FX =0 L1U2 Sinθ − 0.6 = 0

L1U2 = 0.60.6 = + 1

∑ Fy = 0 L1U2 × 0.80 + UL1 = 0 ⇒ U1L1 = −0.80 Now Book F/ forces induced in members as determined by moments and shears method and U forces as determined by method of joints in a tabular form.

Member A × 10-3

(m2)

L (m)

Fi′ (KN)

Ui

F′ULAE ×10-3

(m)

U2LAE × 10-3

(m)

Fi=Fi′ +UiX (KN)

U1U2 2.4 4.5 − 12 −0.6 +0.0675 3.375×10-3 − 25.15 LoL1 2.4 4.5 +12 0 0 0 +12 L1L2 2.4 4.5 +24 −0.6 − 0.135 3.375×10-3 +10.84 L2L3 2.4 4.5 +24 0 0 0 +24 LoU1 3.0 7.5 − 20 0 0 0 − 20 L1U2 4.8 7.5 − 20 +1.0 − 0.416 20.83×10-3 + 1.93 U2L3 3.0 7.5 − 40 0 0 0 − 40 U1L1 1.2 6.0 +16 −0.8 − 0.32 16×10-3 − 1.54 U2L2 1.2 6.0 +48 −0.8 − 0.96 16×10-3 +30.456 U1L2 1.8 7.5 0 +1.0 0 20.83×10-3 +21.96

∑−1.7635× 10−3

∑ 80.91 × 10−6

Compatibility equation is ∆ × L + ∆ × R = 0

∆ × L = n

∑1

F′ULAE

∆ × R = n

∑1

U2LAE . X Putting values from above table in compatibility equation. Where R = X = force

in redundant Member U1L2


− 1.7635 × 10-3 + 80.41 × 10-6. X = 0 or − 1.7635 × 10-3 + 0.08041 × 10-3. X = 0

− 1.7635 + 0.08041 × X = 0

0.08041 X = 1.7635

X = 1.7635

0.08041

X = + 21.93 KN (Force in members U1L2) Now final member forces will be obtained by formula Fi = Fi' + Ui X. These are also given in above table. Apply check on calculated forces. Check on forces Joint Lo

20

12

16 Note: Tensile forces in above table carry positive sign and are represented as acting away from joint.

Compressive forces carry negative sign and are represented in diagram as acting towards the joint. ∑ Fx = 0

12 − 20 Sin θ = 0

12 − 20 × 0.6 = 0

0 = 0 ∑ Fy = 0

16 − 20 Cos θ = 0

16 − 20 × 0.8 = 0

0 = 0 Checks have been satisfied showing correctness of solution. EXMAPLE NO. 7: CASE 2: Analyze previous loaded Truss by taking U1 U2 as Redundant

16 48 32L1 36 L2 24

L3L0

20

U1 U2

406420

40

32

F/ =Diagram

Cos = 0.8Sin = 0.6


109

In this case member forces in BDS (F/ diagram) have been computed by method of joints due to obvious reasons.) Joint Lo:-

16

LoU1

LoL1

∑ Fy = 0

16 + LoU1 × Cosθ = 0

LoU1 = − 160.8 = − 20

∑ FX = 0

LoL1 + LoU1 Sinθ = 0

LoL1 + LoU1 × 0.6 = 0

LoL1 − 20 × 0.6 = 0

LoL1 = + 12 Joint U1

U1L1

U1L2

20

∑ FX = 0

20 Sinθ+ U1L2 Sinθ = 0

20 × 0.6 + U1L2 × 0.6 = 0

U1L2 = − 20

∑ Fy = 0

20 × 0.8 − U1L1 − U1L2 × 0.8 = 0

20 × 0.8 − U1L1 + 20 × 0.8 = 0

U1L1 = 32

Joint L1:

12

32L1U2

L1 L2

∑ Fy = 0

L1U2 Cosθ + 32 = 0


L1U2 = − 320.8

L1U2 = − 40

∑ FX = 0

L1L2 + L1U2 Sinθ − 12 = 0

L1L2 − 40 × 0.6 − 12 = 0

L1L2 = 36 Joint U2

40 U2L2 U2L3

∑ FX = 0

40 Sinθ + U2L3 Sinθ = 0

40 × 0.6 + U2L3 × 0.6 = 0

U2L3 = − 40

∑ Fy = 0

40 Cosθ − U2L3 Cosθ − U2L2 = 0

40 × 0.8 − ( − 40) × 0.8 − U2L2 = 0

U2L2 = 64

Joint L2

36

20 64

48

L2 L3

∑ FX = 0

L2L3 + 20 Sinθ − 36 = 0

L2L3 + 20 × 0.6 − 36 = 0

L2L3 − 24 = 0

L2L3 = 24


111

Joint L3 (Checks)

2432

40

∑ FX = 0

40 Sinθ − 24 = 0

40 × 0.6 − 24 = 0

0 = 0

∑ Fy = 0

32 − 40 Cosθ = 0

32 − 40 × 0.8 = 0

0 = 0 Checks are satisfied. Results are OK and are given in table at page to follow:

Now determine member forces in U diagram.

0 L1 1 L2 0 L3

00

1 1

1.3281.661.328

U1 U2

1.66

L0

U-Diagram (BDS under unit redundant force)

Joint U1

1

U1 LU1 L21

∑ FX = 0

1 + U1L2 × Sinθ = 0

1 + U1L2 × 0.6 = 0


U1L2 = − 1.66

∑ Fy = 0 U1L1 +U1L2 × Cosθ = 0

U1L1 + ( − 1.66) × 0.8 = 0

U1L1 = 1.328

Joint L1 :-

1.328 L1 U2

L1 L2

∑ Fy = 0

1.328 + L1U2 × 0.8 = 0

L1U2 = − 1.3280.8 = − 1.66

∑ FX = 0

L1L2 + L1L2 × 0.6 = 0

L1L2 − 1.66 × 0.6 = 0

L1L2 = +1

Entering results of member forces pertaining to F/ diagram and U diagram alongwith member properties in a tabular form.

Mem-ber

A × 10-3 (m)

L (m)

Fi′ (KN)

U1 F′ULAE × 10-3

(m)

U2LAE × 10-3

(m)

Fi=Fi+UiX (KN)

U1U2 2.4 4.5 0 +1 0 9.375 × 10-3 −25.34 LoL1 2.4 4.5 +12 0 0 0 + 12 L1L2 2.4 4.5 + 36 + 1 +0.3375 9.375 × 10-3 +10.66 L2L3 2.4 4.5 +24 0 0 0 + 24 LoU1 3.0 7.5 − 20 0 0 0 − 20 L1U2 1.8 7.5 − 40 −1.66 +1.383 57.4 × 10-3 +2.06 U2L3 3.0 7.5 − 40 0 0 0 − 40 U1L1 1.2 6.0 + 32 1.328 1.0624 44.09 × 10-3 + 65.65 U2L2 1.2 6.0 + 64 1.328 2.1248 44.09 × 10-3 + 97.65 U1L2 1.8 7.5 − 20 −1.66 0.691 57.4 × 10-3 − 62.06 ∑ 5.6 × 10-3 ∑221.73 × 10-6


113

Compatibility equation is

∆ × L + ∆ × R = 0 Putting values of ∆ × L and ∆ × R due to redundant from above table.

56 × 10-3 + 221.73 × 10-6 X = 0 , where X is force in redundant member U1U2. or 5.6 × 10 -3 + 0.22173 × 10-3 X = 0

X = 5.6 × 10-3

0.22173 × 10-3

X = − 25.34 KN. Therefore forces in truss finally are as follows. (by using formula (Fi = Fi' + UiX and are given in the last column of above table)

FU1 U2 = 0 + Ui.x = 0 − 25.34 × 1 = − 25.34

FLoL1 = 12 − 25.34 × 0 = + 12

FL1L2 = 36 − 25.34 × 1 = + 10.66

FL2L3 = 24 − 0 = + 24

FLoU1 = − 20 − 0 × 25.34 = − 20

FL1U2 = − 40 + 1.66 × 25.34 = + 2.06

FU2L3 = − 40 + 0 × 25.34 = − 40

FU1L1 = + 32 + 1.328 × 25.34 = + 65.65

FU2L2 = + 64 + 1.328 × 25.34 = + 97.65

FU1L2 = − 20 − 1.66 × 25.34 = − 62.06. Now based on these values final check can be applied. Joint Lo.

20

12

16 ∑ FX = 0

12 − 20 Sinθ = 0

12 − 20 × 0.6 = 0

0 = 0 ∑ Fy = 0

16 − 20 Cosθ = 0

16 − 20 × 0.8 = 0

16 − 16 = 0

0 = 0 Results are OK.


2.8. STEPS FOR TRUSS SOLUTION BY METHOD OF LEAST WORK. Now instead of Unit load method, we shall solve the previous truss by direct use of method of least work. (1) Consider the given truss under the action of applied loads and redundant force X in member U1L2 (2) The forces in the relevant rectangle will be a function of applied load and redundant force X. (As was seen in previous unit load method solution) (3) Formulate the total strain energy expression due to direct forces for all the members in the truss. (4) Partially differentiate the above expressions with respect to X. (5) Sum up these expressions and set equal to zero. Solve for X. (6) With this value of X, find the member forces due to applied loads and redundant acting simultaneously (by applying the principle of super positions). EXAMPLE NO. 8 :-

Analyze the loaded truss shown below by least work by treating member U1L2 as redundant. Numbers in ( ) are areas × 10-3 m2 . E = 200 × 106 KN/m2. SOLUTION:-

48 x 4.5= 16KN 32

48

b = 10 r = 3 j = 6 b + r = 2 j 10 + 3 = 2 × 6 13 = 12 D = 13 − 12 = 1


115

Stable Indeterminate to 1st degree.

16 3248

X

X

F − Diagram (Truss under loads and redundant) NOTE: Only the rectangle of members containing redundant X contains forces in terms of X as has been

seen earlier. Now analyze the Truss by method of joints to get Fi forces. JOINT L0

16KN

L0L1

L0U1

∑ Fy = 0 LoU1 Cosθ + 16 = 0

LoU1 = − 16Cosθ

= − 160.8

LoU1 = − 20 KN ∑ FX = 0 LoL1 + LoU1 Sinθ = 0 LoL1 + (−20) × 0.6 = 0 LoL1 − 12 = 0

LoL1 = 12 KN Joint U1

20

X

U1U2

U1L1 ∑ FX = 0 U1 U2 + X Sinθ + 20 Sinθ = 0


U1 U2 + X × 0.6 + 20 × 0.6 = 0

U1 U2 = − (0.6 X +12) ∑ Fy = 0 − U1 L1 − X Cosθ + 20 Cosθ = 0 − U1 L1 − X × 0.8 + 20 × 0.8 = 0 U1 L1 = − 0.8 X + 16

U1L1 = − (0.8 X − 16)

Joint L1 :-

U2L10.8X - 16

L1L212

∑ Fy = 0 − (0.8X − 16) + L1 U2 Cosθ = 0 L1U2 × 0.8 = 0.8 X − 16

L1U2 = (X − 20) ∑ FX = 0 L1L2 + L1U2 Sinθ − 12 = 0 Put value of L1U2.

L1L2 + (X − 20 ) × 0.6 − 12 = 0

L1L2 + 0.6 X − 12 − 12 = 0

L1 L2 = − (0.6X − 24) Joint U2

(0.6X+12)

(X-20)U2L2

U2L3

∑ FX = 0

(0.6 X + 12) + U2L3 Sinθ − (X − 20) Sinθ = 0

0.6 X + 12 + U2L3 × 0.6 − (X − 20) × 0.6 = 0


117

0.6 X + 12 + 0.6U2L3 − 0.6 X + 12 = 0

U2L3 = − 240.6

U2L3 = − 40 KN

∑ Fy = 0

− U2L2 − (X − 20) Cosθ − U2L3 Cosθ = 0

− U2L2 − (X − 20) × 0.8 − (− 40) × 0.8 = 0

− U2L2 − 0.8 X + 16 + 32 = 0

− 0.8 X + 48 = U2L2

U2L2 = − (0.8X − 48) Joint L2:-

X0.8 X- 48

L2 L3

48

0.6 X-24

∑ FX = 0

L2L3 + 0.6 X − 24 − X Sinθ = 0

L2L3 = − 0.6 X + 24 + 0.6 X

L2L3 = 24 KN ∑ Fy = 0

− (0.8X − 48) − 48 + X Cosθ = 0

− 0.8X + 48 − 48 + 0.8X = 0

0 = 0 (Check) Joint L3 :- At this joint, all forces have already been calculated. Apply checks for corretness.

24

40

32


∑ FX = 0 40 Sinθ − 24 = 0 40 × 0.6 − 24 = 0 24 − 24 = 0 0 = 0 O.K. ∑ Fy = 0 − 40 Cosθ + 32 = 0 − 40 × 0.8 + 32 = 0 − 32 + 32 = 0 O.K. Checks have been satisfied. 0 = 0 This means forces have been calculated correctly. We know that strain energy stored in entire

Truss is U = ∑ Fi2L2AE

So, ∂U∂X = ∆ = 0 =

∑ Fi ∂Fi∂X . Li

AE

∑ Fi

∂Fi∂X . Li

AE = 0 = 80.41 × 10−6X − 1764.17 × 10−6 Values of Fi and Li for various

members have been picked up from table annexed. 0 = 80.41 X − 1764.17

or 80.41 X = 1764.17

X = 1764.1780.41

X = 21.94 KN Now putting this value of X in column S of annexed table will give us member forces. Now apply equilibrium check on member forces calculated. You may select any Joint say L1. Joint L1 :-

12

15.5 1.74

10.84 ∑ FX = 0,

10.84 − 12 + 1.94 Sinθ = 0 or 10.84 − 12 + 1.94 × 0.6 = 0 ,

or 0 = 0 (Check) It means that solution is correct.


119

Insert here Page No. 138−A


EXAMPLE NO. 9:- By the force method analyze the truss shown in fig. below. By using the forces in members L1U2 and L2U3 as the redundants. Check the solution by using two different members as the redundants. E = 200 × 10 6 KN/m2

SOLUTION:-

(1.5) (1.5) (1.5) (1.5) L4L0 L1 L2 L3

6m

U1 (1.8) U2 (1.8) U3

(2.4)(0.90) (1.2) (1.2) (0.60) (0.90)

(1.2)(1.2)0 0

48KN 96KN 72KN

L0U1 = 7.5mCos = 0.8Sin = 0.6

[email protected]+96+72-114 = 102KN

F - Diagram

48x4.518

96x918

+

72x13.518

+ = 114KN

(1.5) (1.5) (1.5) (1.5) L4(1.5) (1.5) (1.5) (1.5) L4L0

L0

L0 L4

L4L1 L2 L3

L1 L2 L3

6m

6m

6m

(1.2) (1.2)

(1.2)(1.2)

0

0 0

00

0

00

0

0

00

0 0

0 0

0

0

[email protected]

(0.90) (0.60) (0.90)(2.4)(2.4) loads only.

102KN

102KN

48KN 96KN 72KN 114KN

114KN

54KN

42KN

S.F.D.

F-Diagram

B.M.D.

+

+

-

459 KN-m 702KN-m 513KN-m

Or

0.6

0.6

0.6

0

U2-diagram for redundant X2

U1-Diagram for redundant X10.80.8

0.8 0.8

(1) (2)

L1 L2 L3

Compatibility equations are:

∆X1L + ∆X1R1 + ∆X1R2 = 0 → (1) Change in length in member 1 due to loads and two redundants should be zero.

∆X2L + ∆X2R1 + ∆X2R2 = 0 → (2) Change in length in member 2 due to loads and two redundants should be zero.

Here R1 = X1 R2 = X2


121

Where ∆X1L = ∑.F′U1 L

AE = Deflection produced in member (1) due to applied loads.

∆X1R1 = Deflection produced in member (1) due to redundant R1 = ∑

U1

2LAE . X1

∆x1R2 = Deflection produced in member (1) due to redundant R2 = ∑

U1U2L

AE . X2

∆x2L = Deflection produced in member (2) due to loads = ∑ F′U2L

AE

∆x2R1 = Deflection produced in member (2) due to redundant R1 = ∑

U1U2L

AE . X1

∆x2R2 = Deflection produced in member (2) due to redundant R 2 = ∑

U2

2LAE . X2

From table attached, the above evaluated summations are picked up and final member forces can be seen in the same table. All member forces due to applied loads (Fi' diagram) have been determined by the method of moments and shears and by method of joints for U1 and U2 diagrams. Evaluation of member forces in verticals of F′ − Diagram :- Forces in verticals are determined from mothod of joints for different trusses shown above.

(Joint L1)

76.5

48

76.5

U1 L1

∑ Fy = 0

U1L1 − 48 = 0

U1L1 = 48 (Joint U2)

117 85.5

52.5U2L2

∑ Fy = 0 − U2L2 + 52.5 Cosθ = 0 − U2L2 + 52.5 × 0.8 = 0 U2L2 = 52.5 × 0.8

U2L2 = + 42


(Joint U3) 85.5

U3L3142.5

∑ Fy = 0

− U3L3 + 142.5 Cosθ = 0

U3L3 = 142.5 × 0.8

U3L3 = + 114

Evaluation of forces in verticals of U1 − Diagram:- (Joint L1)

U1L11

L1L2

∑ FX = 0

L1L2 + 1 Sin θ = 0

L1L2 = − 0.6

∑ Fy = 0

U1L1 + 1 Cos θ = 0

U1L1 = − 0.8

(Joint U1 )

0.8U1 L2

U1U2

∑ FX = 0

U1U2 + U1L2 Sinθ = 0


123

∑ Fy = 0

+ 0.8 − U1L2 Cos θ = 0

0.8 = U1L2 × 0.8

U1L2 = 1

so U1U2 + 1 × 0.6 = 0 Putting value of U1L2 in ∑ FX.

U1 U2 = − 0.6

Now from the table, the following values are taken. ∆X1L = − 0.671 × 10 -3

∆X1R1 = 125.7 × 10−6X1 = 0.1257 × 10-3X1 ∆X1R2 = 32 × 10-6 X2 = 0.032 × 10-3X2 ∆X2L = − 6.77 × 10-3

∆X2R1 = 0.032 × 10-3 X1

∆X2R2 = 125.6 × 10-6X2 = 0.1256 × 10−3X2

Putting these in compatibility equations, we have.

− 0.671 × 10−3+0.1257 × 10−3X1+0.032 × 10−3X2 = 0 → (1)

− 6.77 × 10−3+0.032 × 10−3 X1+0.1256 × 10−3X2 = 0 → (2)

dividing by 10−3 − 0.671+0.1257X1 + 0.032X2 = 0 → (1)

− 6.77 + 0.032X1 + 0.1256X2 = 0 → (2)

From (1), X1 = 0.671 − 0.032X2

0.1257 → (3)

Put X1 in (2) & solve for X2

− 6.77 + 0.032

0.671 − 0.032X2

0.1257 + 0.1256X2 = 0

− 6.77 + 0.171 − 8.146 × 10-3X2 + 0.1256X2 = 0

− 6.599 + 0.1174X2 = 0

0.1174X2 = 6.599

X2 = 56.19 KN

From (3) X1 = 0.671 − 0.032 × 56.19

0.1257

X1 = − 8.96 KN After redundants have been evaluated, final member forces can be calculated by using the formula shown in last column of table. Apply checks on these member forces.


CHECKS:-

(Joint Lo) 127.5

76.5

102

∑ FX = 0

76.5 − 127.5 Sinθ = 0

76.5 − 127.5 × 0.6 = 0

0 = 0

∑ Fy = 0

102 − 127.5 Cosθ = 0

102 − 127.5 × 0.8 = 0

0 = 0

The results are O.K. Follow same procedure if some other two members are considered redundant. See example No. 12.


125

Insert Page No. 143−A


2.9. SIMULTANEIOUS INTERNAL AND EXTERNAL TRUSS REDUNDANCY EXAMPLE NO. 10: Determine all reactions and member forces of the following truss by using castiglianos theorem or method of least work. Consider it as: (i) internally redundant; (ii) internally and externally redundant.

Nos. in ( ) are areas in × 10-3m2. E = 200 × 106 KN/m2

A6m

F(2)

(5)(5)

(2)

(4) E6KN B

(3)(3)(2)

D(4)3KN C20KN 20KN

8m

8m

SOLUTION:

DEGREE OF INDETERMINACY :-

D = (m + r ) − 2 j = (10 + 4 ) − 2 × 6 = 2

Therefore, the truss is internally statically indeterminate to the 2nd degree. There can be two approaches, viz, considering two suitable members as redundants and secondly taking one member and one reaction as redundants for which the basic determinate structure can be obtained by cutting the diagonal CE and replacing it by a pair of forces X1 − X1 and replacing the hinge at F by a roller support with a horizontal redundant reaction HF = X2. Applying the first approach and treating inclineds of both storeys sloping down to right as redundants.

(I) WHEN THE TRUSS IS CONSIDERED AS INTERNALLY REDUNDANT :-

A6m

F(2)

(5)(5)

(2)

(4) E6KN B

(3)(3)(2)

D(4)3KN C20KN 20KN

8m

8m

X1

X1

X2

Applying method of joints for calculating member forces.


127

Consider Joint (C) and all unknown forces are assumed to be in tension to begin with , acting away from the joint. Length AE= 10 m , cos θ = 0.6 , sin θ = 0.8

Joint (C)

3KN

20KN

SCD

X1

SBC ∑ FX = 0 Scd + 3 + X1 Cos θ = 0 Scd = − (3 + 0.6 × X1) ∑ Fy = 0 − Sbc − X1 Sin θ − 20 = 0 Sbc = − ( 20 + 0.8 X1 ) Joint (D)

SBD SDE

(3+0.6X1)

20KN

∑ FX = 0 3 + 0.6X1 − SBD × 0.6 = 0 SBD = ( 5 + X1 ) ∑ Fy = 0 − SDE − 20 − SBD Sinθ = 0 − SDE − 20 − ( 5 + X1 ) × 0.80 = 0 SDE = − ( 24 + 0.8X1 ) Joint (B)

6KN

(20+0.8X1)

(5+X1)

SBE

X2SAB

∑ FX = 0 SBE + (5+X1) × 0.6 + X2 × 0.6 + 6 = 0 SBE = − ( 9 + 0.6 X1 + 0.6 X2) ∑ Fy = 0


− SAB − X2 Sinθ − (20 + 0.8 X1) + (5+X1) Sinθ = 0 − SAB − 0.8 X2 − 20 − 0.8 X1 + 4 + 0.8 X1 = 0 SAB = − (16 + 0.8 X2 ) Joint (E)

(24 + 0.8 x 1)

SAE SEF

X1

9+0.6X + 0.6X1 2

∑ FX = 0 9 + 0.6 X1 + 0.6 X2 − X1 x 0.6 − SAE × 0.6 = 0 9 + 0.6 X2 = SAE × 0.6 SAE = ( 15 + X2 )

∑ Fy = 0 − SEF − 24 − 0.8 X1 + X1 × 0.8 − (15 + X2 ) × 0.8 = 0 SEF = − 24 − 0.8 X1 + 0.8 X1 − 12 − 0.8 X2 = 0 SEF = − 36 − 0.8 X2 SEF = − (36 + 0.8 X2 )

Enter Forces in table. Now applying Catiglianos’ theorem and taking values from table attached.

∑ S . ∂S∂X1

. L

AE = 0 = 485.6 + 65.64X1 + 2.7X2 = 0 (1)

and

∑ S. ∂S∂X2

. L

AE = 0 = 748.3 + 2.7X1 + 62.94 X2 = 0 (2)

or 485.6 + 65.64 X1 + 2.7 X2 = 0 → (1)

748.3 + 2.7 X1 + 62.94 X2 = 0 → (2) From (1)

X2 = −

485.6 + 65.64 X1

2.7 putting in (2)

748.3 + 2.7 X1 − 62.94

485.6 + 65.64 X1

2.7 = 0 → (2)

748.3+2.7X1 −11319.875 − 1530.141X1 − 10571.575 − 1527.441 X1 = 0 → (3)

X1 = − 6.921 KN

From (3) X2 = −

485.6 − 65.64 × 6.921

2.7

X2 = − 11.592 KN Now put values of X1 and X2 in 5th column of S to get final number forces SF as given in last column of table. Apply equilibrium check to verify correctness of solution.


129



EQUILIBRIUM CHECKS :- Joint (A)

HA

6.726KN

3.408KN

4KN ∑ FX = 0 3.408 Cosθ − HA − 0 HA = 2.045 KN ∑ Fy = 0 −6.726 + 4 + 3.408 Sinθ = 0 0 = 0 Check is OK. Joint (F)

11.592KN26.726KN

HF

36KN

∑ FX = 0 − HF + 11.592 Cosθ = 0 HF = + 6.955 KN ∑ Fy = 0 36 − 27.726 − 11.592 × Sinθ = 0 0 = 0 (check) It means solution is correct. Now calculate vertical reactions and show forces in diagram.


131

VA=4KN6m

VF=+36KN

A FHA=2.045Kn 3.408HF=6.955KN

11.5926.726 26.726

BE6KN

4.4261.921

14.4636.921

18.463

1.153C D3KN

20KN 20KN

8m

8m

ANALYZED TRUSS

∑ MA = 0 VF × 6 − 20 × 6 − 3 × 16 − 6 × 8 = 0 VF = + 36 KN ∑ Fy = 0 VA + VF = 40 KN VA = + 4 KN EXAMPLE NO. 11: CASE II : When the Truss is considered as both externally & internally redundant. Taking SCE & HF as redundants. Now Truss is determinate and calculate vertical reactions.

6m4KN 36Kn

(9-HF) A F HF

8mSin

Cos =0.6 8m

6KNB E

3kn 20KN 20KN

0.8 =0.8

Fy = 0VA + VF = 40

MA = 0VFx6 - 3x16-20x6-6x8=0

VF = 36KNand

VA

=

4KN

C DX

X

Fig. 2.51


Compatibility Equations are:

∑ S.∂S

∂HF . L

AE = 0 (1) Partial differentiation of strain energy w.r.t. HF = ∆H = 0.

(Pin support)

∑ S. ∂S∂X .

LAE = 0 (2) Partial differentiation of strain energy w.r.t. X = elongation of

member CE due to X = 0. As before determine member forces Si in members by method of joints. Joint (A)

(9-HF)

SAB

SAE

4 ∑FX = 0 SAE Cosθ − (9 − HF) = 0 SAE × 0.6 − (9 − HF) = 0

SAE =

9 − HF

0.6

SAE = 15 − 1.67 HF ∑ Fy = 0 4 + SAB + SAE Sinθ = 0 4 + SAB + (15 − 1.670 HF ) × 0.8 = 0 4 + SAB + 12 − 1.33 HF = 0 SAB = − 16 + 1.33 HF SAB = − (16 − 1.33 HF ) Joint (F)

SBF SEF

HF

36


133

∑ FX = 0 − HF − SBF Cosθ = 0 − HF − 0.6 SBF = 0 − HF = 0.6 SBF

SBF = − 1.67 HF ∑ Fy = 0 36 + SEF + SBF Sinθ = 0 36 + SEF − 1.67 HF × 0.8 = 0 SEF = − (36 − 1.33 HF)

Joint (E)

SBE

XSDE

(36 1.33H )- F

(15-1.67HF) ∑ FX = 0 − SBE − X Cosθ − (15 − 1.67 HF) Cosθ = 0 − SBE − 0.6X − ( 15 − 1.67 HF ) × 0.6 = 0 − SBE − 0.6X − 9 + HF = 0 HF − 0.6X − 9 = SBE

SBE = (HF − 0.6 X − 9)

∑ Fy = 0 SDE +36 − 1.33 HF + X Sinθ − (15 − 1.67HF ) Sinθ = 0 by putting Sinθ = 0.08 SDE + 36 − 1.33 HF + 0.8X − 12 + 1.33 HF = 0 SDE = − 0.8X − 24 SDE = − ( 24 + 0.8X)

Joint (C) 20KN

3KNSCD

XSBC


∑ FX = 0 SCD + 3 + X Cosθ = 0 SCD = − ( 3 + 0.6X)

∑ Fy = 0 − 20 − SBC − X Sin θ = 0 − 20 − SBC − 0.8X = 0 SBC = − ( 20 + 0.8 X )

Joint (D)

SBD(24+ 0.8X)

20KN

(3+0.6X)

∑FX = 0

3 + 0.6X − SBD Cosθ = 0

3 + 0.6X − 0.6 SBD = 0

SBD = ( 5 + X)

∑ Fy = 0

− 20 + 24 + 0.8X − SBD Sinθ = 0

− 20 + 24 + 0.8X − ( 5 + X ) 0.8 = 0

− 20 + 24 + 0.8X − 4 − 0.8X = 0

0 = 0 (check)

Calculation of HF & X :−

From the attached table, picking up the values of summations, we have.

∑. S. ∂ S∂HF

. L

AE = 0 = (−1247.03 + 175.24 HF − 4.5 × X) 10−6


135

and ∑. S. ∂S∂X .

LAE = 0 = (460.6 − 4.5 HF + 65.64X) 10-6

−1247.03 + 175.24 HF − 4.5X = 0 → (1) + 460.6 − 4.5 HF + 65.64X = 0 → (2) From (1)

X =

− 1247.03 + 175.24 HF

4.5 → (3)

Put in (2) to get HF

460.6 − 4.5 HF + 65.64

− 1247.03 + 175.24 HF

4.5 = 0

460.6 − 4.5 HF − 18190.01 + 2556.17 HF = 0 −17729.41 + 2551.67 HF = 0 HF = 6.948 KN Put this value in 3 to get X.

X =

−1247.03 + 175.24 × 6.948

4.5 (3)

or X = − 6.541 KN Now calculate number Forces by putting the values of X and

HF in S expressions given in column 5 of the attached table. These final forces appear in last column for SF.

6m4kn 36KN

2.052Kn A F 6.948KN11.603

26.759 8m6.7593.392

B E1.8736KN

14.7626.641

18.7678m

1.541

0.925C D3KN

20KN 20KN

Fig 2.52 ANALYZED TRUSS


137

Equilibrium checks for the accuracy of calculated member Forces:- Joint (A)

2.052

6.7593.397

4 ∑ FX = 0 3.397 Cosθ − 2.052 = 0 0 = 0 Check ∑ Fy = 0 − 6.759 + 4 + 3.397 × 0.8 = 0 0 = 0 Check Joint (F)

11.603 26.759

6.948

36 ∑ FX = 0 − 6.948 + 11.603 × 0.6 = 0 0 ≅ 0 Check ∑ Fy = 0 36 − 26.759 − 11.603 × 0.8 = 0 0 ≅ 0 Check Joint (C)

20

0.925

6.54114.767

3

∑ FX = 0 0.925 − 6.541 × 0.6 + 3 = 0 0 = 0 Check ∑ Fy = 0 14.767 − 20 + 6.541 × 0.8 = 0 0 = 0 Check. This verifies correctness of solution.


EXAMPLE NO. 12:-

By the unit load−method analyze the internally indeterminate truss shown below. Take the forces

in members L1U2 and U2L3 as the redundants.

Note: The same truss has already been solved in Example No. 9, by taking L1U2 and L2U3 as redundants.

E = 200 × 106 KN/m2

SOLUTION:-

[email protected] 114KN

114KN

F-Diagram6m

48KN

48KN

96KN

96KN

72KN

72KN

102KN

102KN

L0

L0

L0

L0

L1 L2 L3

L1 L2 L3

L1 L2 L3

L30.6

0.6

0.6

0.6

L2 L1

L4

L4

L4

L4

U1 U2 U3

U1 U2 U3

U1 U2 U3

U1 U2 U3

(1.8) (1.8)

(2.4) (2.4)(1.2) (1.2)(1.2)

(1.2)0.90 (0.90)(1.5) (1.5) (1.5) (1.5)

LoU1 = 7.5 mCos = 0.8Sin = 0.6

B.D.S. Under appliedload only.

Or F -Diagram/

10254

42

702513459

114

0

00

0 00

0 0 0

0

0 0

00

0

0

B.M.D.+

U2-Diagam

U1 -Diagram

0.8

0.80.8

0.8

+

1

S.F.D.


139

Compatibility equations are : ∆X1 + ∆X1 R1 + ∆X1R2 = 0 → (1) Here X1 = R1 X2 = R2

Deflection created by applied loads and redundants shall be zero. ∆X2L + ∆X2R1 + ∆X2R2 = 0 → (2)

∆X1L = ∑. F′U1L

AE (Change in length of first redundant member by applied loads)

∆X1R1 = ∑

U1

2LAE X1 (Change in length in first redundant member due to first redundant force)

∆X1R2 = ∑

U1U2L

AE . X2 (Change in length in first redundant member due to second redundant force)

∆X2L = ∑ F′U2L

AE (Change in second redundant member due to applied load.)

∆X2R1 = ∑

U1U2L

AE . X1 (Change in length of second redundant member due to first redundant force.)

∆X2R2 = ∑

U2

2LAE . X2 (Change in length of second redundant member due to redundant force in it.)

Picking up the above deformations from the table (158−A) and calculate final member forces by following formula.

F = F' + U1X1 + U2X2 Forces in chord members and inclineds are determined by the method of moments and shears as explained already, while for verticals method of joints has been used. Evaluation of force in verticals of F′ − Diagram (Joint L2)

96

76.5

67.5

85.5

52.5U2L2

∑ FX = 0 85.5 − 76.5 + 52.5 Sinθ − 67.5 Sinθ = 0 85.5 − 76.5 + 52.5 × 0.6 − 67.5 × 0.6 = 0 0 = 0 (Check) ∑ Fy + 0 U2L2 + 52.5 Cosθ + 67.5 Cos θ − 96 = 0 U2L2 = − 52.5 × 0.8 − 67.5 × 0.8 + 96 = 0 U2L2 = 0


141

Picking the following values from attached table (Table for example No.12) ∆X1L = + 1.009 × 10−3

∆X1R1 = + 125.7 × 10−6 X1 = + 0.1257 × 10−3 X1 ∆X1R2 = + 32 × 10−6 X2 = + 0.032 × 10−3 X2 ∆X2L = − 0.171 × 10−3 ∆X2R1 = + 32 × 10−6 X1 = + 0.032 × 10−3 X1 ∆X2R2 = + 125.7 × 10−6 X2 = + 0.1257 × 10−3 X2 Putting these in compatibility equals. 1.009 × 10−3+0.1257 × 10−3 X1+0.032 × 10−3 X2 = 0 (1) − 0.171 × 10−3+0.032 × 10−3X1+0.1257 × 10−3X2 = 0 (2) Simplify 1.009 + 0.1257 X1 + 0.032 X2 = 0 → (1) − 0.171 + 0.032 X1 + 0.1257X2 = 0 → (2)

From (1) X1 =

−1.009 − 0.032 X2

0.1257 → (3)

Put in (2) & solve for X2

− 0.171 + 0.032

−1.009 − 0.032 X2

0.1257 + 0.1257 X2 = 0

− 0.171 − 0.257 − 8.146 × 10−3 X2 + 0.1257X2 = 0 − 0.428 + 0.1176 X2 = 0

X2 = 0.4280.1176

X2 = 3.64 KN Put this in equation (3) to get X1

(3) ⇒ X1 =

−1.009 − 0.032 × 3.64

0.1257

X1 = − 8.95 KN So final forces in members are calculated by the following given formula. F = F′+ U1 X1 + U2 X2 FLoL1 = 76.5 + 0 + 0 = + 76.5 KN FL1 L2 = 76.5 + ( − 0.6) ( − 8.95) + 0 = + 81.87 KN FL2 L3 = 85.5 + 0 + 3.64) (− 0.6) = + 83.32 KN FL3 L4 = 85.5 + 0 + 0 = + 85.5 KN FU1 U2 = −117 + (− 0.6) (− 8.95) + 0 = − 111.63 KN FU2 U3 = −117 + o +(− 0.6) (3.64) = − 119.18 KN FU1 L1 = + 48 + (− 0.8) (− 8.95) + 0 = + 55.16 KN FU2 L2 = 0 + (− 0.8) (− 8.95) + (− 0.8) (3.64) = + 4.25 KN


FU3 L3 = + 72 + 0 + (− 0.8) (3.64) = + 69.09 KN Flo U1 = − 127.5 + 0 + 0 = − 127.5 KN FU1 L2 = + 67.5 + (1) (− 8.95) + 0 = 58.55 KN FL1 U2 = 0 + (1) (− 8.95) + 0 = − 8.95 KN FU2 L3 = 0 + 0 + (1) (3.64) = + 3.64 KN FL2 U3 = 52.5 + 0 + (1) (3.64) = + 56.14 KN FU3 L4 = − 142.5 + 0 + 0 = − 142.5 KN

CHECK ON FORCE VALUES We may apply check at random at any joint. If solution is correct, equilibrium checks will be satisfied at all joint. Joint Lo.

102

76.5

127.5

∑ FX = 0 76.5 − 127.5 Sinθ = 0 76.5 − 127.5 × 0.6 = 0 0 = 0 ∑ Fy = 0 102 − 127.5 × 0.8 = 0 0 = 0 OK. Results seem to be correct. The credit for developing method of least work goes to Alberto Castiglianos who worked as an engineer in Italian Railways. This method was presented in a thesis in partial fulfillment of the requirement for the award of diploma engineering of associate engineer. He published a paper for finding deflections which is called Castiglianos first theorem and in consequence thereof, method of least work which is also known as Castiglianos second theorem. Method of least work also mentioned earlier in a paper by an Italian General Menabrea who was not able to give a satisfactory proof. Leonard Euler had also used the method about 50 years ago for derivation of equations for buckling of columns wherein, Daniel Bernolli gave valuable suggestion to him.

Method of least work or Castiglianos second theorem is a very versatile method for the analysis of indeterminate structures and specially to trussed type structures. The method does not however, accounts for erection stresses, temperature stresses or differential support sinking. The reader is advised to use some other method for the analysis of such indeterminate structures like frames and continuos beams.

It must be appreciated in general, for horizontal and vertical indeterminate structural systems, carrying various types of loads, there are generally more than one structural actions present at the same time including direct forces, shear forces, bending moments and twisting moments. In order to have a precise analysis all redundant structural actions and hence strain energies must be considered which would make the method laborious and cumbersome. Therefore, most of engineers think it sufficient to consider only the significant strain energy. The reader should know that most of structural analysis approaches whether classical or matrix methods consider equilibrium of forces and displacement/strain compatibility of members of a system.


143

The basis of the method of consistent deformation and method of least work are essentially the same. In consistent deformation method, the deformation produced by the applied loads are equated to these produced by the redundants. This process usually results in the evolution of redundants. However, in the method of least work, total strain energy expression of a structural system in terms of that due to known applied loads and due to redundants is established. Then the total strain energy is partially differentiated with respect to redundant which ultimately result in the evolution of the redundant. It must be appreciated that, for indeterminate structural system like trusses, the unknown redundants maybe external supports reaction or the internal forces or both. And it may not be very clear which type of redundants should be considered as the amount of work involved in terms of requisite calculation may vary. Therefore, a clever choice of redundants (or a basic determinate structure as was the case with consistent deformation method) may often greatly reduce the amount of work involved. There is often a debate going on these days regarding the utility or justification of classical structural analysis in comparison to the computer method of structural analysis. It is commented that in case of classical methods of structural analysis the student comes across basic and finer points of structural engineering after which a computer analysis of a complex structure maybe undertaken. In the absence of basic knowledge of classical structural analysis, the engineer maybe in a difficult position to justify to computer results which are again to be checked against equilibrium and deformation compatibility only. EXAMPLE NO. 13:

The procedure for analysis has already been given. Utilizing that procedure, analyze the following truss by the method of least work. Areas in ( ) carry the units of 10−3 m2 while the value of E can be taken as 200 × 106 KN/m2.

A B

C D

E F

4

44 22 2

2 2 22

4.5m

[email protected] 15 kN

where i = total degree of indeterminacy b = number of bars. r = total number of reactive components which the support can provide. b + r = 2j

10 + 3 > 2 × 6 13 > 12 so i= 1 . First degree internal indeterminancy.

U = F2L

2 AE Strain energy due to direct forces induced due to applied loads in a BDS Truss.

∂U∂X = F.

∂F∂X .

LAE = 0


Note:− We select the redundants in such a way that the stability of the structure is not effected. Selecting member EC as redundant.

5 +

+22.5

45 10

5KN 10KN

4.5m

A B C D

E F

15KN

xx

F-diagram B.D.S. under the action of applied loads & redundant.

S.F.D. due to applied load only.

load only.

Method of moments and shears has been used to find forces in BDS due to applied loads. A table has been made. Forces vertical in members in terms of redundant X may be determined by the method of joints as before. From table.

∑ F. ∂F∂x .

LAE = 0 = − 331.22 × 10−6 + 51.49 × 10−6X

or − 331.22 + 51.49X = 0

X = + 6.433 KN The final member forces are obtained as below by putting value of X in column 5 of the table. Member Force (KN) AB + 5

BC +5.45

CD + 10

EF − 9.55

BE + 0.45

CF + 10.45

CE + 6.43

BF − 0.64

AE − 7.07

DF − 14.14


145

Insert page No. 164−A


CHECK. Joint A.

5

5 ∑ FX = 0 5 − 7.07 Cosθ = 0 5 − 7.07 × 0.707 = 0 0 = 0 ∑ Fy = 0 − 7.07 × 0.707 + 5 = 0 0 = 0 Check is OK.

EXAMPLE NO. 14:− Analyze the following symmetrically loaded second degree internally

indeterminate truss by the method of least work. Areas in ( ) are 10−3m2 . The value of E can be taken as 200 × 106 KN/m2

AB

C

FED

3 332 2 2 2

4

4 4

4

3m

2@3m15KN

AB

C

FE

X2X1

X2

15KN

7.5KN 7.5KN

3m

2@3m

D

X1

Selecting member BD and Before as redundants.

BDS under loadsand redundants.


147

SOLUTION: Note :− By virtue of symmetry, we can expect to have same values for X1 and X2. It is known before hand.

+

+

7.5

7.5

22.5

S.F.D.

B.M.D.

SFD and BMD in BDS due to applied loads are shown above. As in previous case determine member forces in BDS due to applied loads by the method of moments shears while method of joints may be used to determine member forces due to redundants acting separately. Apply super position principal. Then these are entered in a table given. Summation of relavant columns due to X1 and X2 gives two equations from which these can be calculated. Putting values from table and solving for X1 and X2.

[−2.65 × 10−3 (7.5 − 0.707X1 ) − 2.65 × 103 (− 0.707X1 ) −3.53 × 10−3 (− 0.707X1 ) −3.53 × 10−3(15 − 0.707X1 − 0.707X2 ) +10.6 × 10−3 (−10.6+X1 ) + 10.6 × 10−3 (X2 ) ]10−3 = 0

− 19.875 + 1.874X1 + 1.874 X1 + 2.450 X1 − 52.45 + 2.50 X1 + 2.5 X2 − 112.36 + 10.6 X1 + 10.6 X1 = 0 29.898 X1 + 2.50 X2 − 185.185 = 0 → (1) ( ∑ col 8 ) − 2.65 × 10−3(7.5−0.707 X2) − 2.65 × 10−3 (− 0.707 X2) − 3.53 × 10−3 (15−0.707 X1 − 0.707 X2) − 3.53 × 10−3 (− 0.707 X2 ) + 10.6 × 10−3 (−10.6+X2) + 10.6 × 10−3 X2 = 0 − 19.875+1.874 X2+1.874 X2−52.95+2.50 X1+2.50X2+2.450 X2−112.36+10.6X2+10.6 X2 = 0 2.50 X1 + 29.898 X2 − 185.185 = 0 → (2) ( ∑ col 9 )

From (1), X1 =

185.185 − 2.50 X2

29.898 → (3) Put in 2 above

(2) ⇒ 2.50

185.185 − 2.50 X2

29.898 + 29.898X2 − 185.185 = 0

15.465 − 0.21 X2 + 29.898 X2 − 185.185 = 0 29.689 X2 − 169.7 = 0

X2 = + 5.716 KN Put X2 in equation 3 to get X1. The final member forces are given in last column. These are obtained by putting values of X1 and X2, whichever is applicable, in column 5 of the table.


149

Then X1 =

185.185 − 2.50 × 5.716

29.898

X1 = + 5.716 KN

Equilibrium Check. 4.04

4.884

3.459

7.5

∑ FX = 0 3.459 − 4.884 × Cosθ = 0 3.459 − 4.884 × 0.707 = 0 0 = 0 ∑ Fy = 0 7.5 − 4.04 − 4.884 × 0.707 = 0 0 = 0 Checks are satisfied. Results are OK. EXAMPLE NO. 15:− Analyze the following internally indeterminate truss by the method of least work. Areas in ( ) are 10−3m2 . The value of E can be taken as 200 × 10 6 KN/m2. SOLUTION:−

b = 13 , r = 3 , j = 7 so degree of indeterminacy I =( b + r ) –2j =2 Choosing members EB and BG as redundants, forces due to loads have been determined by the method of moments and shears for the BDS and are entered in a table. While forces due to redundants X1 and X2.

A B C

D

GFE

3m

3@3m

10KN 5KN

15KN

10

0 05

30

15

0

0

+

+S.F.D

B.M.D

X1 X1 X2X2


A B C

D

GFE

3m

3@3m

X1 X2

X1 X2

10KN 5KN

15KN

10

0 05

30

15

0

0

+

+S.F.D

B.M.D

Member Forces Due to Redundants Only. Please number that due to separate action of redundants X1 and X2 member forces will be induced only in the square whose inclineds are X1 and X2. There will be no reaction at supports.

Joint D:− DG

CD ∑ Fy = 0 DG Sinθ − 0 DG = 0 ∑ FX = 0 DG Cos θ + CD = 0 CD = 0 Joint G :−

FG

CGX2 ∑ FX = 0 − FG − X2 Cos θ = 0

FG = − 0.707 X2

∑ Fy = 0 − CG − X2 Sin θ = 0

CG = − 0.707 X2


151

Joint C :−

CF

BC

0.707X2

∑ Fy = 0 CF Sin θ − 0.707 X2 = 0

CF = 0.707 X2

0.707

CF = + X2 ∑ FX = 0 − BC − CF Cos θ = 0 BC = −0.707 X2 Joint B.

AB 0.707X2

X2BFX1

∑ FX = 0

− 0.707 X2 − AB + X2 Cos θ − X1 Cos θ = 0

AB = − 0.707 X1 ∑ Fy = 0

X1 Sin θ + X2 Sin θ + BF = 0

BF = − 0.707X1 − 0.707X2 Joint A.

AEAF

0.707X1 ∑ FX = 0 − 0.707 X1 + AF Cos θ = 0

AF = X1


∑ Fy = 0

AE + AF Sin θ = 0

AE = − 0.707X1 Joint E.

EF

X10.707 X1

∑ FX = 0

EF + X1 Cos θ = 0 EF = − 0.707 X1

∑ Fy = 0

0.707 X1 − 0.707 X1 = 0

0 = 0 (Check)

Entering the values of summations from attached table, we have.

∑ F. ∂F∂X1

. L

AE = 0 = − 229.443 × 10−6 +29.848 × 10−6 X1+2.45 × 10−6X2

∑ F. ∂F∂X2

. L

AE = 0 = −168.9 × 10−6 +2.45 × 10−6 X1+29.848 × 10−6 X2

Simplifying

− 229.443 + 29.848 X1 + 2.45 X2 = 0 → (1)

− 168.9 + 2.45 X1 + 29.848 X2 = 0 → (2)

From (1)

X1 =

− 2.45 X2 + 229.443

29.848 → (3)

Put in (2) & solve for X2

− 168.9 + 2.45

− 2.45 X2 + 229.443

29.848 + 29.848 X2 = 0

− 168.9 − 0.201 X2 + 18.833 + 29.848 X2 = 0 − 150.067 + 29.647 X2 = 0

X2 = 150.06729.647

X2 = + 5.062 KN


153

Insert page No. 170−A


So X1 = − 2.45 × 5.062 + 229.443

29.848 by putting value of X2 in (3)

X1 = + 7.272 KN EQUILIBRIUM CHECKS :−

A4.859 6.421 5

7.07

G8.579F5.141E

5.141 6.286.87 2.000

5.6621.421

B CD7.272

10KN 15KN5KN

Joint B:−

4.859

15

6.421

5.0626.287.272

∑ FX = 0 6.421 + 5.062 Cosθ − 7.272 Cosθ − 4.859 = 0 0 = 0

∑ Fy = 0 6.28 − 15 + 5.062 Sinθ + 7.272 Sinθ = 0 0 = 0 The results are OK.

Joint C:− 1.421

5

2.008

6.421

∑ FX = 0 5 + 2.008 Cosθ − 6.421 = 0 0 = 0 ∑ Fy = 0 1.421 − 2.008 Sinθ = 0 0 = 0 Results are OK.

INTRODUCTION TO TWO-HINGED ARCHES 155

CHAPTER THREE

INTRODUCTION TO TWO-HINGED ARCHES

3.0. TWO-HINGED ARCHES:- The following issues should be settled first. − Definition. − Types. − Basic Principle and B.M. − Linear Arch. − Mathematical Generalized Expressions. − Segmental Arches. Some information is contained elsewhere where determinate arches have been dealt. 3.1. DEFINITION OF AN ARCH. “An arch can be defined as a humped or curved beam subjected to transverse and other loads as well as the horizontal thrust at the supports.” An efficient use of an arch can be made only if full horizontal restraint is developed at the supports. If either of the support allows some movement in the horizontal direction, it will tend to increase the B.M. to which an arch is subjected and arch would become simply a curved beam. The B.M., in arches due to the applied loads is reduced due to the inward thrust. Analysis is carried out to find the horizontal thrust and also to find the B.M., to which an arch is subjected. Beam action Vs arch action :

A

A

B

BH

Va

Va

Vb

Vb

P

P

P

P

P

Simple beam subjecte to applied transverse loads.

Arch carrying vertical loads & horizontal thrust

Support, abutments orspringing.

P

Mo

y M=Mo-HyH

x

One reaction at support only

Two reactions at supports

The above beam and arch carry similar loadings. If Mo = B.M. due to applied loads at a distance X on the simple span of a simple beam where rise is y. then bending moment in the arch is, MX = Mo ± Hy where MX is the B.M., in the arch at a distance x . H is the horizontal thrust at the springings & y is the rize of the arch at a distance. ‘x’ as shown in the diagram. The ( ± ) sign is to be used with care and a (−) sign will be used if the horizontal thrust is inwards or vice versa. In later case it will behave as a beam.


P P P

A B HH

Va Vb Under transverse loads, the horizontal thrust at either of the springings abutments is equal. In the arch shown above, the degree of indeterminacy is one and let us consider the horizontal thrust at support B as the redundant. The above loaded arch can be considered equal to the following two diagrams wherein a BDS arch is under the action of loads plus the same BDS arch under the action of inward unit horizontal load at the springings. =

P P P

A

Va Vb

BB

BL B.D.S. under applied loads (loads try to flatten the arch) ∆BL stands for displacement of point B due to applied loads in a BDS arch..

A B

BR

1

+

(Flattened arch recovers some of horizontal displacement at B due to unit horizontal loads and will recover fully if full horizontal thirst is applied at B.) (Arch flattens out under the action of applied loads because freedom in the horizontal direction has been provided at point B.) and all due to full redundant value. This forces the basis of compatibility.

∆BR stands for displacement of point. B (in the direction of force) due to unit horizontal redundant force at B. Remember that a horizontal reactive component cannot be realized at the roller support. However, we can always apply a horizontal force at the roller.

3.2. Compatibility equation.

∆BL − ( ∆BR ) H = 0 ( If unit load is applied in opposite sense so that it also produces flettening, +ve sign may be used in the equation and the final sign with H will be self adjusting.)


or H = ∆BL∆BR =

displacement at B due to loadsdisplacement at B due to unit horizontal redundant

We will be considering strain energy stored in bending only.The modified expression for that for curved structural members is as follows.

U = ∫ M2ds

2EI

Where ds is the elemental length along the centre line of the arch and U is the strain energy stored in bending along centre-line of arch. The bending moment at a distance x from support is MX = Mo − Hy (Horizontal thrust is inwards). (1) Where Mo = Simple span bending moment ( S.S.B.M.) in a similar loaded simple beam.

U = ∫ M2 ds2EI

If H is chosen as redundant, then differentiating U w.r.t. H , we have

∂U∂H = ∆BH = 0 =∫ 1

EI . M.

∂M

∂H ds Put M= Mo – Hy and then differentiate.

∂U∂H = ∆BH = 0 = ∫ 1

EI . (Mo − Hy)(−y) ds by putting M from (1)

0 = ∫ (Hy2 − Mo y) ds

EI Simlifying

∫ H y2 ds

EI − ∫ Mo y ds

EI = 0

∫ H y2 ds

EI = ∫ Mo y ds

EI

or

H = ∫

Moy.dsEI

∫ y2dsEI

Applying Castigliano’s 2nd theorem, ∆BL becomes = ∫ Mo y ds

EI

and ∆BR = ∫ y2 dsEI


The algebraic integration of the above integrals can also be performed in limited number of cases when EI is a suitable function of S ( total curved arch length), otherwise, go for numerical integration. For prismatic (same cross section) members which normally have EI constant, the above expression can be written as follows:

H = ∫ Mo y ds

∫ y2ds

3.3. TYPES OF ARCHES :− The arches can be classified into a variety of ways depending mainly upon the material of construction and the end conditions. (1) Classification Of Arches Based On Material of Construction :− The following arches fall in this particular category: a) Brick masonary arches. b) Reinforced concrete arches. c) Steel arches. The span of the arches which can be permitted increases as we approach steel arches from the brick masonary arches. (2) Classification Of Arches Based On End Conditions :− The following arches fall in this particular category: a) Three hinged arches. b) Two hinged arches. c) Fixed arches. In the ancient times, three hinged arches have been used to support wide spans roofs. However, their use is very rare in bridge construction since the discontinuity at the crown hinge is communicated to the main deck of the bridge. In three hinged arches, all reactive components are found by statical considerations without considering the deformations of the arch rib. Therefore, they are insensitive to foundation movements and temperature changes etc., and are statically determinate. These are covered as a separate chapter in this book. The Romans exploited the potential of arches to a great extent. However, their emperical analysis approach became available in the early 18th century. 3.4. LINEAR ARCH :− This is just a theoretical arch at every X−section of which the B.M. is zero.

M = Mo − Hy = 0

or Mo = Hy (The B.M. due to applied loads is balanced by Hy).

therefore, y = MoH


This is the equation for the centre line of a linear arch. With the change in position and the number of loads on the arch, the corresponding linear arch would also change as Mo keeps on changing. Therefore, there are infinite number of such arches for every load pattern and position on the actual arch.

EXAMPLE NO. 1: 3.5. ANALYSIS OF TWO – HINGED SEGMENTAL ARCHES We develop the method for indeterminate arches starting with the simplest cases of segmental arches. Solve the following segmental arch by using the basic principles of consistent deformation method and by treating horizontal thrust at support D as the redundant. The segmental arches could be used in tunnels and in water ways.

20KN/m

A

B C

D

4m EI=Constant

2m4m2m

40 kN 40 kN8m

Ha=20 kN

20KN/m

A

B C

D

4m EI=Constant

2m4m2m

40 kN 40 kN8m

Ha=20 kN

(Ha will occur only point D is a hinge support)

M − Diagram. Due to applied loads. Similarly reactions due to supermetrical loading.

A D

CB

11

DR m − Diagram. Due to unit redundant at D. (X is varied along length of members). Find Cosθ and Sinθ. cos θ = 0.4472 , sin θ = 0.8944.


Sab sin θ + 40 =0 so Sab= −40

0.8944 = − 44.722. Consider equilibrium of joint A and project forces

in y-direction. (M-diagram)

Consider same diagram with roller at D. Now consider joint A and Project forces in X direction to evaluate Ha. Sab Cosθ + Ha = 0 or –44.722 x 0.4472 + Ha = 0

or Ha= 20KN Compatibility equation ∆DL − ∆DR. H = 0

Or H = ∆DL∆DR =

Horizontal displacement of D due to loadsHorizontal displacement of D due to redundants

∆DL = ∫ Mmdx

EI

Applying Unit load method concepts,

∆DR = ∫ m2 dx

EI

Now we attempt the evaluation of these integrals in a tabular form. X is measured along member axis.

Mem ber

Origin. Limits. M m

AB A 0 − 4.472 40 X Cosθ =40X0.477= 17.88X

+1.XSinθ=+0.894X

BC B 0 − 4 40(2+X)−10X2= 80 + 40 X−10 X2

+ 4

CD D 0 − 4.472 17.88 X + 0.894 X

∆DL = ∫ MmdX

EI = 1EI

4.472

∫o

(17.88X)(+0.894X)dX + 1EI

4

∫o 80+40X −10X)(+4) dX

+ 1EI

4.472

∫o

(17.88 X)(+0.894 X) dX

= 2EI

4.472

∫o

(+15.985 X2)dX + 1EI

4

∫o(+320+160X − 40X2) dX Integrate and put limits

= +31.969

EI

X3

3

4.472 o

+ 1EI

+320X +

160X2

2 − 40X2

3 4 o

= +10.656

EI ( 4.4723 − 0) + 1EI

+320 × 4 +80 × 16 −

403 × 16


∆DL = + 2659.72

EI

∆DR = ∫ m2dX

EI = 1EI

4.472

∫o

(+ 0.894X)2 dX + 1EI

4

∫o 16 dX +

1EI

4.472

∫o

(+ 0.894X)2 dX

= 2EI

4.472

∫o

0.799 X2 dX + 16EI

4

∫o dX

= 1.598

EI

X3

3 4.472 o

+ 16EI X

4 o

= 0.533

EI [(4.472)3 − 0] + 16EI (4 − 0)

∆DR = 111.653

EI

H = ∆DL∆DR

= 2659.72/EI111.653/EI

H = 23.82 KN EXAMPLE NO. 2:− Solve the following arch by using consistent deformation method.

2m4m2m

A D

CB

4m EI-Constt

20KN/m40KN

The above redundant / segmental arch can be replaced by the following similar arches carrying loads and redundant unit load.

A

Bx

C

D

4m

4m2m

Ra=20KN

M-DiagramRd=60KN

DL

40KN

20KN/m

2m

40KN

X is varied alongmember lengths.

BDS UNDER LOADS


ΣMa = 0 ; Rd × 8 = 20 × 4 × 4 + 40× 4 ∴ Rd = 60 KN so Ra = 20 KN

X is varied alongmember lengths.

A

B C

D

+

x

x

1 1

DR m − Diagram

BDS UNDER UNIT REDUNDANT AT D

Compatibility equation is ∆DL − ∆DR.H = 0 Where ∆DL = Horizontal deflection of D in BDS due to applied loads. ∆DR = Horizontal deflection at D due to Unit redundant. H = Total Horizontal redundant.

Or H = ∆DL∆DR

and ∆DL = ∫ MmdX

EI

∆DR = ∫ m2 dX

EI

Member Origin Limits M m EI

AB A 0−4.472 20X Cosθ+40X Sinθ 20X ×0.447+40X × 0.894 = 44.72X

XSinθ=0.894X Constt.

BC B 0 − 4 20(2+X)+40 × 4 −10X2 40+20X +160 − 10X2 = −10X2 + 20X + 200

+ 4

Constt.

CD D 0−4.472 60X Cosθ=60X × 0.447 = 26.82 X

0.894X Constt.

∆DL = ∫ MmdX

EI = 1EI

4.472

∫o

(+44.72X)(0.894X) dX + 1EI

4

∫o (−10X2 +20X − 200 ) 4 dX


+ 1EI

4.472

∫o

(26.82X ) (0.894X)dX

1.33X = 2 × 23.977

EI

4.472

o

X3

3 + 4EI

−10X3

3 + 20X2

2 + 200X 4 o

∆DL = 63.97

EI

4.4723

3 + 4EI

−103 × 43 + 10 × 42 + 200 × 4 =

+4893.8EI

∆DR = ∫ m2 dX

EI = 1EI

4.472

∫o

(0.894X)2 + 1EI

4

∫o 16dX +

1EI

4.472

∫o

(0.894X)2

= 2EI

4.472

∫o

0.799X2 dX + 16EI

4

∫o dX

= 1.598

EI

X3

3 4.472 o

+ 16EI X

4 o

= 0.533

EI [ (4.472)3 − 0 ] + 16EI ( 4 − 0)

∆DR = 111.653

EI

∴ H = ∆DL∆DR

= + 4893.8/EI111.653/EI

So H = + 43.83 KN EXAMPLE NO. 3:- Determine the horizontal thrust for the for following loaded segmental arch. Take EI equal to constant.

A

B

C G

P P

D

E

F3m 5m 4m 3m4m

3m

4m


SOLUTION :−

A

X

B

C

P P

D

E

F

G

P P

X

X is varied alongmember length

Now consider a BDS under Loads and redundant separately for the same arch and evaluate integrals.

An inspection of the arch indicates that it is symmetrical about point G and is indeterminate to the first degree choosing horizontal reaction at F as the redundant, we draw two basic determinate structures under the action of applied loads and the redundant horizontal thirst at support F.

PX

P

4m 5m 4m 3m

PPM-Diagram (BDS under loads)

3m

G

B

A

C D

E

F

A

B

C E

B.D.S. under unit horizontal redundant load at F.

E

F1 1

m-Diagram


Because of symmetry, Moments and hence strain energy is computed for half frame.

Portion Origin Limits M m

AB A 0 − 5 PX Cosθ = X0.6 PX 0.8 X

BC B 0 − 5 P (3+0.8X) 4 + 0.6X

CG C 0 − 2.5 P (7+X) − PX = 7 P 7

∆FL = 2 5

∫o (0.6 PX)(0.8X)

EI dX + 2 5

∫o P(3+0.8X)(4+0.6X)

EI dX+ 2 2.5

∫o

49 PEI dX

= 2 PEI [

5

∫o 0.48 X2 dX +

5

∫o (0.48 X2+5X+12)dX +

2.5

∫o

49 dX]

= 2 PEI

0.48X3

3

5 o+

0.48 X3

3 + 5 X2

2 + 12X 5|o+49X

2.5|o

= 2 PEI

0.483 × 53 +

0.48 × 53

3 + 5 × 52

2 + 12 × 5 + 49 × 2.5

∆FL = 570 P

EI (deflection of point F due to loads)

∆FR = 2EI

5

∫o (0.8X)2dX +

2EI

5

∫o (16 + 0.36X2 + 4.8X) dX +

2EI

2.5

∫o

49dX

= 2EI

0.64X3

3

5 o+

16X +

0.36X3

3 + 4.8X2

2 5 o+ 49X

2.5|o

= 2EI

0.64 × 53

3 + 16 × 5 + 0.36

3 × 53 + 4.8 × 52

2 + 49 × 2.5

∆FR = 608.33

EI , H = ∆FL∆FR

H = 570 P608.32

So H = 0.937 P


NOTE :− Compatibility equation is ∆FL − ∆FR × H = 0 ∆FL = ∆FR × H

H = ∆FL∆FR

We take compression on outer side & tension on inner side +ve in case of M and m-diagram. EXAMPLE NO. 4 :− Determine the horizontal thrust provided that EI = Constt for the following loaded segmental arch.:

SOLUTION:

X

X

PX

P

4m 5m 4m 3m

RRa

3mA F

B

C D

EP

1

3m

4m

f Taking horizontal reaction at F as redundant. ΣMa=0 Rf. 19 = P . 12 + P .7 + 4. P , So

Rf = 1.211 P and therefore Ra is, Ra = 2P − 1.211 P

Ra = + 0.789 P

X

X

PX

P

4m 5m 4m 3mP

M-Diagram

3m

0.789 P 1.211 P

P 1


4m 5m 4m 3m3m1 1

A

B

C D

E

F

m-diagram (Unit redundant at F)

3m

4m

Portion Origin Limits M m AB A 0 − 5 0.789 PX Cosθ+PX Sinθ

= 0.4734 PX + 0.8 PX = 1.2734 PX

1 × X Cos θ = 0.8X

BC B 0 − 5 0.789 P(3 + XCosθ1) +P(4 + XSinθ1) − PX Sinθ1 = 0.6312 PX+6.367 P

1(4 + X Sinθ1) = 4 + 0.6X

CD C 0 − 5 0.789P (7+X)+P×7−P×3−PX = − 0.211 PX + 9.523 P

+ 7

DE E 0 − 5 1.211 P(3+X Cosθ1) = 3.633 P + 0.9688 PX

1(4 + X Sinθ1) = 4 + 0.6X

EF F 0 − 5 1.211 PX Cos θ= 0.7266 PX X Sin θ = 0.8X Determine Sines and Cosines of θ and θ1.

∆FL = 1EI [

5

∫o (1.2734 PX)(0.8 X)dX +

5

∫o (0.6312 PX + 6.367 P)

(4 + 0.6X) dX + 5

∫o (− 0.211 PX + 9.523 P)(7)dX

+ 5

∫o (3.633 P + 0.9688 PX)(4 + 0.6X) +

5

∫o (0.7266PX (0.8X) dX]

= PEI [

5

∫o 1.01872X2 dX +

5

∫o (2.5248X + 0.37872X2 + 25.468 + 3.8202 X) dX

+ 5

∫o (− 1.477X + 66.661) dX +

5

∫o (14.532 + 2.1798X

+ 3.8752X + 0.58128X2 ) dX + 5

∫o 0.58128X dX . Simplifying we get.

= PEI

5

∫o (1.97872X2 + 11.50428X + 106.661) dX

∆FL = PEI

1.97972

X3

3 + 11.50428 X2

2 + 106.661X 5 o


= PEI

1.97872 ×

53

3 + 11.50428 × 52

2 + 106.661 × 5

∆FL = 759.56 P

EI

∆FR = 1EI [

5

∫o(0.8X)2dX +

5

∫o (16+0.36X2+ 4.8X) dX

+ 5

∫o 49 dX +

5

∫o (16+0.36X2+4.8X) dX +

5

∫o 0.64 X2 dX]

= 1EI

0.64X3

3 + 16X + 0.36 X3

3 + 4.8 X2

2 + 49X + 16X + 0.36 X3

3 + 4.8X2

2 + 0.64X3

3 5 o

= 1EI [

0.643 × 53 + 16 × 5 +

0.36 × 53

3 + 4.8 × 52

2 + 49 × 5

+ 16 × 5 + 0.36

3 × 53 + 4.82 × 52 +

0.643 × 53 ] . Simplifying

∆FR = 608.33

EI . Compatibility equation remains the same. Putting values of integrals, we have

H = ∆FL∆FR

= 759.56 P

EI 608.33

EI

H = 1.2486 P Now all reactions are shown.

A

P B

C

P PD

E

F

0.789P 1.211P

0.2486P 1.2486P

ANALYZED SEGMENTAL ARCH

Check : ∑ Mc = 0 0.789P × 7 − 0.2486 P × 7 − P × 3 + P × 5 + 1.2486 P × 7 − 1.211P × 12 = 0 0 = 0 O.K.


3.6. ANALYSIS OF TWO HINGED CIRCULAR ARCHES :−

AF

X CE

B

R

0

R

L

ycy

D

P P

The circular arches are infact a portion of the circle and are commonly used in bridge construction. From the knowledge of determinate circular arches, it is known that the maximum thrust and the vertical reactions occur at the springings. Therefore, logically there should be a greater moment of inertia near the springings rather than that near the mid−span of the arch. The approach is called the secant variation of inertia and is most economical. However, to establish the basic principles, we will first of all consider arches with constant EI. The following points are normally required to be calculated in the analysis. (1) Horizontal thrust at the springings.

(2) B.M. & the normal S.F. at any section of the arch.

Usually, the span and the central rise is given and we have to determine;

(i) the radius of the arch;

(ii) the equation of centre line of the circular arch.

Two possible analysis are performed.

(1) Algebraic integration.

(2) Numerical integration.

After solving some problems, it will be amply demonstrated that algebraic integration is very laborious and time consuming for most of the cases. Therefore, more emphasis will be placed on numerical integration which is not as exact but gives sufficiently reliable results. Some researches have shown that if arch is divided in sixteen portions, the results obtained are sufficiently accurate. In general, the accuracy increases with the increase or more in number of sub−divisions of the arch. We will be considering two triangles. 1 − ∆ ADO 2 − ∆EFO By considering ∆ ADO OB2 = OD2 + BD2


R2 = (R−yc)2 + (L /2 )2 R2 = R2 − 2Ryc + yc2 + L2/4 0 = yc ( yc − 2 R) + L2/4 or yc ( yc − 2 R) = − L2/4 − yc ( yc − 2 R ) = L2/4

yc (2R − yc) = L2

4 (1)

By considering ∆ EFO OF2 = OE2 + EF2 R2 = ( R − yc + y )2 + X2 R2 − X2 = ( R − yc + y )2

R − yc + y = R2 − x2

S

R2

y = R2 − X2 − (R − yc) (2)

The detailed derivation of this equation can be found in some other Chapter of this book. In this case, S = R ( 2 θ ) where θ is in radiains. S is the total length along centre line of the arch.

H = ∫ Myds ∫y2ds as before obtained By eliminating EI as we are considering EI = Constt

EXAMPLE NO. 5:− A two− hinged circular arch carries a concentrated force of 50 KN at the centre. The span & the rise of the arch are 60m & 10m respectively. Find the horizontal thrust at the abutments.

SOLUTION :− The arch span is divided in ten equal segments and ordinates are considered at the centre of each segment.

AD

0

B

50KN30m

60m

R=50m

(1) (2) (3) (4)(5) (6) (7)(8)(9) (10)10


R = L2

8yc + yc2 , where R = Redious, yc = Central rise and L = Span of arch.

= (60)2

8 × 10 + 102

R = 50 m

Sinα = 3050 = 0.6 . Now compute angle α is radians.

α = 36.87° , we know π rad = 180° 180° = π rad

1° = π

180 rad

So 36.87° = π

180 × 36.87 radians

36.87° = 0.6435 rad = α

α = 0.6435 rad

S = R (2 α) = 50 (2 × 0.6435) , Where S is length of arch along its centre-line For circular arches. X is varied from centre to abutments. S = 64.35 m

A y x E yc=10m

C

B

0

R=50m

R

D 25KN25KN30-x

50KN

H = ∫ Myds∫y2ds

where M = Simple span ( S.S ) B.M. in the arch due to applied loads only. Mbc = Mac = 25 ( 30 − X ) in two portions at a distance X from mid span.

OE = R Cos θ


OD = R − yc = 50 − 10 = 40 m y = OE – OD [Since OC = OD + CD = 50 and CD = 10 = Yc] y = R Cos θ − 40

and ds = Rdθ

X = R Sin θ

Evaluation of Numerator :− Mx = 25 (30 − X), ds = Rdθ, y = RCosθ − 40

∫ Myds = 2 α

∫o[25 (30 − R Sinθ)] [R Cosθ−40] [Rdθ], By putting X, y and ds from above. Also put

value of α which is in radians.

= 50 R 0.6435

∫o

(30 − R Sinθ)(R Cosθ − 40) dθ, we know, 2Sinθ Cosθ = Sin 2θ.

= 50R 0.6435

∫o

(30R Cosθ Cosθ − R2Sinθ Cosθ − 1200 + 40R Sinθ) dθ

= 50R

30R Sinθ +

R2

2 . Cos 2θ

2 − 1200 θ − 40R Cosθ 0.6435 o

Put limits now

= 50 × 50

30×50×0.6+

25004 × 0.28−1200×0.6435−40×50×0.8 −

502

4 ×1+ 40 × 50 ×1

= + 194500 ∫ Myds = 194.5 × 103

Evaluation of Denominator :−

We know Cos2θ = 12 (1 + Cos2θ)

and Sin2θ = 12 (1 − Cos2θ)

∫y2ds = 2 0.6435

∫o

(RCosθ − 40)2 (Rdθ)

= 2R 0.6435

∫o

(R2 Cos2 θ − 80R Cosθ + 1600) dθ

= 2R 0.6435

∫o

R2

2 (1 + Cos 2θ) − 80 R Cos θ + 1600 dθ Integrate

= 2R

R2

2

θ +

Sin 2θ2 − 80R Sin θ + 1600 θ

0.6435 o

Put limits now


= 2 × 50

502

2

0.6435 +

0.962 − 80 × 50 × 0.6 + 1600 × 0.6435

= 3397.5 ∫y2ds = + 3.3975 × 103

H = 194.5 × 103

3.3975 × 10-3

H = 57.2 KN EXAMPLE NO. 5: BY NUMERICAL INTEGRATION :− The values of X, y and M are determined at the mid ordinates of the segments. The basic philosophy is that if we consider a very small arc length that would be regarded as a straight line and therefore we tend to average out these values. y = R2 − X2 − (R − yc) or y = 502 − X2 − (50 − 10 ) or y = 502 − X2 − (40) (1) See segments of Example 5 about 4 page before. For section (1) X1 = 27, from (1), y1 = 502 − 272 − (40) = 2.08 m For section (2) X2 = 21 from (1), y2 = 502 − 212 − (40) = 5.738 m and so on. M = 25 ( 30 − X ) = (750 − 25X) 0 < X < 30 as before Now do numerical integration in a tabular form as under.

Section. X y. M My y2 1 27 2.08 75 156.00 4.33 2 21 5.380 225 1210.50 28.94 3 15 7.69 375 3883.75 59.14 4 9 9.18 525 4819.50 84.27 5 3 9.91 675 6689.25 98.21 6 3 9.91 675 6689.25 98.21 7 9 9.18 525 4819.50 84.27 8 15 7.69 375 2883.75 59.14 9 21 5.380 225 1210.50 28.94

10 27 2.08 75 156.00 4.33 ∑31518 ∑549.78

S = 64.35 m

and ds = 64.35

10

ds = 6.435 m


H = ∫ Myds∫ y2ds =

∑ Myds∑ y2ds

= 31518 × 6.435549.78 × 6.435 ( Note:- ds cancels out )

H = 57.33 KN A result similar to that already obtained from algebraic solution 3.7. ARCHES WITH SECANT VARIATION OF INERTIA :− If Io is the second moment of area of arch rib at the crown: Then secant variation of inertia means. I = Io sec. α and ds Cos α =dX

dy ds

dx Or ds = dX Sec α

H = ∫

MydsEI

∫ y2dsEI

If it is built of the same material, then E would cancel out:

H = ∫

MydsI

∫ y2ds

I

Put I= Io sec α

H = ∫

My dX Sec αIo Secα

∫ y2dX Sec α

Io Sec α

H = ∫ MydX∫ y2 dX


If we utilize the above expression for horizontal thrust, it may be kept in mind that integration can now take place in the Cartesian coordinate system instead of the polar coordinate system. 3.8. BY SECANT VARIATION USING ALGEBRAIC INTEGRATION :− EXAMPLE NO. 6: Analyze the arch in Example No. 5: We know, y = R2 − X2 − (R − yc) y = 502 − X2 − 40 Mac = Mbc = 25 ( 30 − X ) 0 < X < 30

∫ MydX = 2 30

∫o

25 (30 − X)[ 502 − X2 − 40 ] dX

= 50 [ 30 30

∫o

502 − X2 . dX − 30

∫o

1200dX − 30

∫o

502 − X2 . XdX + 40 30

∫o

XdX ]

= 1500 30

∫o

502 − X2 dX − 1200 × 50 30

∫o

dX − 50 30

∫o

502 − X2 XdX + 2000 30

∫o

XdX

Put X = 50 Sin θ= R sinθ dX = 50 Cosθ dθ At X = 0 θ = 0

At X = 30 θ = 0.6435 Now Evaluate integrals Substitutions

Cos2θ = 1 + Cos2θ

2

∫Cos2θ = θ2 +

Sin2θ4

∫Cos2θ Sinθdθ = − Cos3θ

3

by letting X = Cosθ dX = −Sin θdθ

∫ MydX = 1500 0.6435

∫o

502 (1 − Sin2θ) (50 Cosθdθ ) − 60000 X 30 o

+ 25

(502 − X2 )3/2

3/2

30 o

+ 2000

X2

2 30 o

= 1500 × 502 0.6435

∫o

(1 + Cos2θ)

2 dθ − 6 × 104 (30)


+ 503 [(502 − 302)3/2 − (302) + 1000 (302 )]

= 187.5 × 104

θ +

Sin 2θ2

0.6435 o

− 180 × 104 − 1016666.666 + 90 × 104

∫ MydX = 187.5 × 104

0.6435 +

Sin(2 × 0.6435)2 − 1916666.666

= 2106561.918 − 1916666.666 ∫ MydX = 189895.252

∫ y2dX = 2 30

∫o

(502 − X2 +402 − 80 502 − X2 ) dX

= 2 30

∫o

(4100 − X2 − 80 502 − X2 ) dX

Substitutions: X = 50 Sin θ dX = 50 Cosθdθ 1 − Sin2θ = Cos2θ

= 8200 30

∫o

dX − 2 30

∫o

X2 dX − 160 0.6435

∫o

502 Cos2 θ dθ

= 8200 X 30|o

− 2

X3

3

30 o

− 160 × 502

2 0.6435

∫o

(1 + Cos 2θ)dθ

= 8200 (30) − 23 (303) −

160 × 502

2

θ +

Sin 2θ2

0.6435 o

= 228000 − 160 × 502

2

0.6435 +

Sin(2 × 0.6435)2

= 228000 − 224699.938 ∫y2dX = 3300.062

H = ∫ MydX∫ y2dX

= 189895.252

3300.062

H = 57.543 KN


EXAMPLE NO. 7:- A circular arch carries a uniformly distributed load on its left half, calculate the horizontal thrust.

A B

10KN/m

60m

0

yc=10mC

D

SOLUTION :− Determine Vertical Support reactions as usual and write moment expressions due to applied loads only without considering horizontal thrust.

0

A B

10KN/m

D

Cx xE

60m225KN

75KN

yc=10my

From diagram, X = R Sinθ

Mac = 225 (30 − R Sinθ) − 5 (30 − R Sin θ)2, in other words. Mac = Va ( 30 – X ) – w X2/2

where X = R sinθ

and Mbc = 75 (30 − R Sinθ) OD = OC − CD = 50 − 10 = 40 m

y = OE − OD = R Cosθ − 40

so H = ∫ Myds∫ y2ds


Evaluation of Numerator.

∫ Myds = 0.6435

∫o

[225 (30 − R Sinθ) − 5(30 − R Sinθ)2 ] [ R Cosθ − 40] (Rdθ)

+ 0.6435

∫o

[75(30 − R Sinθ)] [R Cosθ − 40 ] [Rdθ ]. This consists of two integrals.

Evaluate First Integral

= I1 = R 0.6435

∫o

[6750 − 225 R Sinθ − 4500 − 5 R2 Sin2 θ + 300 R Sinθ] [R Cos θ − 40]

I 1 = R 0.6435

∫o

[2250 + 75 R Sinθ − 5 R2 Sin2 θ][R Cos θ − 40] dθ

= R 0.6435

∫o

[2250 R Cos θ + 75 R2 Sin θ Cosθ −5 R3 Sin2 θ Cosθ

− 90000 − 3000 R Sin θ + 200 R2 Sin2 θ ] dθ

= R 0.6435

∫o

[2250 R Cos θ + 75 R2 Sin θ Cos θ − 5 R3Sin2θ Cos θ

Let X = Sinθ dX = cosθ dθ

So ∫Sin2θ Cosθ dθ = ∫X2 dX = X3

3 = Sin3θ

3

− 90000 − 3000 R Sinθ + 200 R2

1 − Cos2θ

2 ] dθ

= R 2250 R Sin θ − 752 R2

Cos2 θ 2 − 5

R3 Sin3θ 3 − 90000 θ

+ 3000 R Cosθ + 200

2 R2

θ −

Sin2 θ2

0.6435|o

= 50 [ 2250 × 50 × 0.6 − 754 × 2500 × 0.28 − 5 × 503 ×

0.2163

− 90000 × 0.6435 + 3000 × 50 × 0.8 + 200

2 × 502

0.6435 −

0.962

+ 754 × 2500 × 1 − 3000 × 50 × 1]


= 50 [ 67500 − 13125 − 45000 − 57915 + 120000 + 160875

− 120000 + 46875 − 150000 ] = 50 ( 9210 )

I1 = 460.5 × 103

Now Evaluate

2nd Integral = I2 = R 0.6435

∫o

(2250 − 75 R Sin θ)(R Cos θ − 40) ( dθ ) multiply two expressions.

I2 = R 0.6435

∫o

2250 R Cos θ − 75 R2 Sin θ Cos θ − 90000 + 3000 R Sin θ) dθ Integrate now.

= R 2250R Sin θ + 752 R2 Cos 2θ

2 − 90000 θ − 3000R Cosθ 0.6435|o

= 50 (2250 × 50 × 0.6 + 754 × 2500 × 0.28 − 90000 × 0.6435

− 3000 × 50 × 0.8 − 754 × 2500 × 1 + 3000 × 50 × 1)

I2 = 291.75 × 103

Add these two integrals (I1 and I2) of ∫Myds.

∫ Myds = I1 + I2

= 460.5 × 103 + 291.75 × 103

or ∫ Myds = 752.25 × 103

Now Evaluate

∫ y2ds = 2 0.6435

∫o

(R Cosθ − 40)2 ( R dθ )

= 2 R 0.6435

∫o

(R2Cos2θ + 1600 − 80 R Cosθ)dθ ; We know that Cos2θ = 1+Cos2θ

2

= 2 R 0.6435

∫o

R2

2 (1 + Cos2θ) + 1600 − 80 R Cosθ dθ

= 2 R R2

2

θ + Sin 2θ

2 + 1600 θ − 80 R Sinθ 0.6435|o

= 2 × 50 [502

2

0.6435+

0.962 +1600 × 0.6435 − 80 × 50 × 0.6], So ∫ y2ds=3.3975×103


H = ∫ Myds∫ y2ds

H = 752.25 × 103

3.3975 × 103

H = 221.42 KN

EXAMPLE NO. 8: Analyze the same problem by numerical Integration. Write moment expression for segments in portions AC and BC due to applied loading only for a simple span. For segments 1 – 5, Mac = 225 (30 − X) − 5 (30 − X)2 as before but in Cartesian co-ordinate system. For segments 6 – 10, Mbc = 75 (30 − X)

A

225KN 60m 75KN

0

10KN/m

BD

R=50m

(1) (2) (3) (4)(5) (6) (7)(8)(9) (10)

C

Note: X is measured for mid span and y is corresponding rise. Now attempt in a tabular form.

Section X y M My y2 1. 27 2.08 630 1310.4 4.33 2 21 5.38 1620 8715.6 28.94 3 15 7.69 2250 17302.5 59.14 4 9 9.18 2520 23133.6 84.27 5 3 9.91 2430 24081.3 98.21 6 3 9.91 2025 20067.75 98.21 7 9 9.18 1575 14458.5 84.27 8 15 7.69 1125 8651.25 59.14 9 21 5.38 675 3624.75 28.94 10 27 2.08 225 468 4.33 ∑121813.65 ∑549.78

S = R (2 α )

= 50 × 2 × 0.6435

S = 64.35 m


so ds = 64.35

10 = 6.435 m , (Because S has been divided in Ten Segments)

H = ∫ Myds∫ y2ds

= ∑ Myds∑ y2ds

= 121813.65 × 6.435

549.78 × 6.435 ( Note: ds cancels out )

H = 221.57 KN Same answer as obtained by algebraic integration.

EXAMPLE NO. 9: Analyze the previous arch for by assuming secant variation of inertia. Integrate along the x − axis by considering arch to be a beam. Mac = 225 (30 − X) − 5 (30 − X)2 0 < X < 30

Mbc = 75 (30 − X) 0 < X < 30

y = 502 − X2 − 40

∫ MydX = 30

∫o

[225 (30 − X) − 5 (30 − X)2 ] [ 502 − X2 − 40] dX

+ 30

∫o

[75 (30 − X)] [ 502 − X2 − 40] dX , By taking y expression common, we have

∫ MydX = 30

∫o

[6750 − 225X − 5 (900 − 60X + X2 ) +2250 − 75X] [ 502 − X2 − 40)] dX

= 30

∫o

(− 5X2 + 4500)[ 502 − X2 − 40] dX X terms cancel out

Let X = 50 sinθ, then dX = 50 cosθ dθ, So (502 – X2 ) = 50 Cosθ. Putting these we have.

= 0.6435

∫o

( 4500 – 12500 sin2θ ) ( 50 cosθ − 40 ) ( 50 cosθ) dθ

Note : In solving the above expression , the following trignometrical relationships are used. 1. Sin2θ = 1− cos2θ and ∫ cos2θ = θ/2 + sin 2θ/4 2. ∫cos3θ = sinθ − sin3θ/3 3. ∫cos4θ = 3θ/8 + sin2θ/4 + sin4θ/32


By using the above formulas and solving the integral, we get the value as follows. ∫MydX = 730607.23 . Now evaluate ∫y2dX.

∫y2dX = 2 30

∫o

[ (502 − X2) − 40]2 dX. By evaluating on similar lines as stated above; we have.

= 3322.0

H = ∫ MydX∫ y2 dX

= 730607.23

3322.0

H = 220.0 KN The same may be solved by numerical integration 3.9. TWO HINGED PARABOLIC ARCHES

A

L

C

Byc

Equation of the centre line of a parabolic arch with either abutment as origin is y = CX (L − X) → (1)

At X = L2 y = yc Putting

yc = C × L2

L −

L2

yc = C. L2

L

2

yc = C. L2

4

C = 4 ycL2

Putting the value of ‘C’ in equation (1), we have.

y = 4 ycL2 X (L − X)

y = 4 yc X

L2 (L − X), rated for 0 < X < L


and dydX =

4 ycL2 (L − 2X) 0 < X < L

So H = ∫ MydX∫ y2dX

In parabolic arches, origin for X is usually their supports. EXAMPLE NO. 10:− A two−hinged parabolic arch with secant variation of inertia is subjected to the loads at 3rd points as shown in the diagram. Determine the horizontal thrust at abutments & plot the B.M.D. Verify your answer by numerical integration. SOLUTION:−

AD

40KN 40KN20m 20m

CD

B

60m40KN X 40KN

yc

It is a symmetrically loaded arch. So moment expression on simple span in portions AC and CD may be found and corresponding integrals may be evaluated and multiplied by 2. Mac = 40 X 0 < X < 20 Mcd = 40 X − 40 (X − 20) = 800 20 < X < 30

y = 4 yc X

L2 (L − X) , Put value of yc and L for simplification purpose.

=

4 . 10 . X602 (60 − X)

or y = 0.011 X (60 − X) = 0.011 × 60 X − 0.011 X2

∫ MydX = 2 20

∫o

(40 X)(0.011 × 60 X − 0.011 X2)dX

+2 30

∫20

800(0.66 X − 0.011X2 ) dX

Simplifying

= 2 20

∫o

(26.4 X2 − 0.44 X3 ) dX + 2 30

∫o

(528X − 8.8X2)dX

= 2

26.4 X3

3 − 0.44 X4

4

20 o

+ 2

528 X2

2 − 8.8 X3

3

30 20


= 2

26.4

3 × 20 3 − 0.44

4 × 204 + 2

528

2 × 302 − 8.83 × 303 −

5282 × 202 +

8.83 × 203

= 105600 + 152533.33

= 258133.33

∫ MydX = 258.133 × 103 . Now evaluate ∫ y2dX.

∫ y2dX = 60

∫o

(0.011 × 60 X − 0.011 X2 )2 dX

= 60

∫o

[(0.66)2 X2 + (0.011)2 X4 − 2 × 0.66 × 0.011 X3]dX

= 60

∫o

(0.4356 X2 + 1.21 × 10−4 X4 − 0.01452 X3) dX

=

0.4356 X3

3 + 1.21 × 10-4 X5

5 − 0.01452 X4

4 60 o

= 0.4356

3 × 603 + 1.21 × 10-4

5 × 605 − 0.01452

4 × 604

= 3136.32

∫ y2dX = 3.136 ×103


= 258.133 × 103

3.136 × 103

H = 82.3 KN M = Mo − Hy , y = 0.001 X (60 − X) , at X = 20, y = yE

yc = 0.011 × 20 (60 − 20 ) = 8.8 m = yE

Mc = 40 × 20 − 82.3 × 8.8 = 75.76 KN−m

MD = (40 × 30 − 40 × 10) − 82.3 × 10 = − 23 KN

ME = 40 × 20 − 82.3 × 8.8 = 75.76 KN


Now BMD can be plotted.

A

40KN 40KN60m

20m 20m40KN 40KN

C D

B8.8m8.8m 10m

823800

800

0 0

linear archparablic (2nd degree)8.0

724.2924.24

Note:− The length of the segment should be even multiple of span. More than 5 or 6 segments will give slightly improved answer. 3.10. EDDY’S THEOREM:− The difference between the linear arch and the actual arch is the BMD at that point. EXAMPLE NO. 11:- Analyze the following loaded two hinged arch by numerical integration method.

A

C

B

40KNL=60m40KN

y=10mc

D

1 2 3 4 5 6

E

40kN40kN20m 20m

Mac = 40 X 0 < X < 20

Mcd = 40 X − 40(X − 20) = 800 20 < X < 40

Meb = 40 X − 40(X − 20) − 40(X − 40) = 2400 − 40X 40 < X < 60

and y = 0.011 × (60 − X) = 0.66X − 0.011 X2 ( As before ) solving in a tabular forces.


Section X y M My y2 1 5 3.025 200 605 9.15 2 15 7.425 600 4455 55.13 3 25 9.625 800 7700 92.64 4 35 9.625 800 7700 92.64 5 45 7.425 600 4455 55.13 6 55 3.025 200 605 9.15

∑25520 ∑313.84

L = 60 m , dX = 606 = 10 m

H = ∑ MydX∑y2dX

= 25520 × 10313.84 × 10

H = 81.31 KN

Almost similar result was obtained by algebraic integration earlier.

EXAMPLE NO. 12:- A two−hinged parabolic arch with secant variation of inertia is subjected to a uniformly distributed load on its left half. Determine the horizontal thrust at abutments and plot the B.M.D. Verify your answer by numerical integration. SOLUTION :−

A B

C

L=60m225KN 75KN

yc=10m

10KN/m

Mac = 225X − 5 X2 0 < X < 30

Mbc = 75X 0 < X < 30

y = 4yc X

L2 (L − X)

= 4 . 10 . X

602 (L − X)

= 0.011 X (60 − X)


y = 0.66 X − 0.011 X2 and dydX = 0.66 − 0.022X = Tanθ

∫ MydX = 30

∫o

(225X − 5 X2) (0.66 X − 0.011 X2) dX + 30

∫o

75 X (0.66 X − 0.011 X2) dX

= 30

∫o

(148.5 X2 − 2.475 X3 − 3.3 X3 + 0.055 X4) dX + 30

∫o

(49.5 X2 − 0.825 X3) dX

=

148.5 X3

3 − 2.475 X4

4 − 3.3 X4

4 + 0.055 X5

5

30 o

+

49.5 X3

3 − 0.825 X4

4

30 o

=

148.5

3 × 303 − 2.475 × 304

4 − 3.3 × 304

4 + 0.055 × 305

5 +

49.5 × 303

3 − 0.825 × 304

4

= 712800.0174 ∫ MydX = 712.8 x 103

∫ y2dX = 60

∫o

(0.66 X − 0.011 X2)2 dX

= 60

∫o

[(0.662) X2 + (0.011)2 X4 − 2 . 0.66 . 0.011 X3] dX

=

(0.66)2

X3

3 + (0.011)2 X5

5 − 2 . 0.66 . 0.011X4

4

60 o

= 3.136 × 10−3

H = 712.8 103

3.136 103

H = 227.30 KN

EXAMPLE NO. 13:- Now Analyze the previous example. BY NUMERICAL INTEGRATION :−

A B

10KN

60m 75KN225KN

(1)(2)

(3) C (4)(5)

(6)

Mac = 225X − 5 X2 0 < X < 30 Mcb = 225X − 300 (X − 15) 30 < X < 60 y = 0.66 X − 0.011 X2 (same as before). Attempt in a tabular form.


Section X y M My y2 1 5 3.025 1000 3025 9.15 2 15 7.425 2250 16706.25 55.13 3 25 9.625 2500 24062.5 92.64 4 35 9.625 1875 18046.875 92.64 5 45 7.425 1125 8353.125 55.13 6 55 3.05 375 1134.375 9.15 ∑71328.125 ∑313.84

H = 71328.125 . 10

313.84 . 10

H = 227.28 KN WE GET THE SAME ANSWER AS WAS OBTAINED BY ALGEBRAIC INTEGRATION. y15 = 0.66 X 15 − 0.011 (15)2 = 2.425 m y45 = 7.425 m

A

10KN/m

7.425m 10m 7.425m B

C

60m 75KN225KN

2250 22501125+

22731687.71687.7

Mo-diagram

Hy-diagram

M-diagram

29.99m

1687.7

1125

22502273

Point of contraflexure. Write a generalized Mx expression and set that to zero. Mx = 225X − 5X2 − 227.30 + [0.011 X (60 − X)] = 0 225X − 5X2 − 150.02X + 2.50X2 = 0 − 2.5X2 + 74.98X = 0 − 2.5X + 74.98 = 0

X = 29.99 m Insert this value back in Mx expression to find M max in the arch. EXAMPLE NO. 14:- Analyze the following arch by algebraic and numerical integration. Consider : A. the arch to be parabolic and then circular. B. moment of inertia constant and then with secant variation.


2 KN/m

5KN20m

6m

70m Generally arches have been used by the engineers and architects dating back to old roman buildings, Mughal and Muslim architecture. Main applications are in bridges, churches, mosques and other buildings. Arch behaviour is dependent upon stiffness of supports, commonly called abutments or springings so that horizontal reaction develops. SOLUTION :− A. PARABOLIC ARCH AND ALGEBRAIC INTEGRATION

2KN/m

5KN20mD

Cx

B70m

21.07KN53.93KN

70x52.570

5x2070

+ = 53.93 A

Determine simple span bending moments. Mac = 53.93 X − X2 0 < X < 35 Mcd = 53.93X − 70(X − 17.5) 35< X < 50

= 53.93X − 70X + 1225

= − 16.07X + 1225

Mdb = 53.93X − 70(X−17.5) −5 (X−50) 0 < X < 70

= 53.93X − 70X + 1225 − 5X + 250

= −21.07X + 1475

Y = 4YcX

L2 (L − X)

= 4 . 6 . X

702 ( 70 − X)

= 4.898 . 10−3 X ( 70 − X ) Y = 0.343X − 4.898 . 10−3 X2

∫ MydX = 35

∫o

(53.93X − X2 ) (0.343X −4.898 × 10−3 X2 ) dX


+ 50

∫35

(−16.07X + 1225) (0.343X − 4.898 × 10−3 X2 ) dX

+ 70

∫50

( −21.07X + 1475 ) (0.343X −4.898 × 10−3 X2 ) dX Multiply the expressions

= 35

∫o

(18.498X2 − 0.264X3 − 0.343X3 + 4.898 × 10−3 X4 ) dX

+ 50

∫35

(−5.512X2 + 0.079X3 + 420.175X − 6X2 ) dX

+ 70

∫50

(−7.227X2 + 0.103X3 + 505.925X − 7.225X2 ) dX re-arranging we get

= 35

∫o

(4.898 × 10−3 X4 − 0.607X3 + 18.498X2 ) dX

+ 50

∫35

(0.079X3 − 11.512 X2 + 420.175 X) dX

+ 70

∫50

(0.103X3 − 14.452X2 + 505.925X) dX

=

4.898 × 10-3

X5

5 − 0.607 X4

4 + 18.498 X3

3

35 o

+

0.079

X4

4 − 11.512 X3

3 + 420.175 X2

2

50 35

+

0.103

X4

4 − 14.452 X3

3 + 505.925 X2

2

70 50

. Insert limits and simplify

= 88097.835 + 46520.7188 + 14251.3336

∫ MydX = 148869.8874 . Now calculate ∫y2dX

∫ y2dX = 70

∫o

(0.343X − 4.898 × 10−3 X2 )2 dX

= 70

∫o

( 0.118X2 + 2.399 × 10−5 X4 − 3.360 × 10−3 X3) dX

∫ y2dX =

0.118X3

3 + 2.399 × 10-5 X5

5 − 3.360 × 10-3 X4

4

70 o

= 1386.932

H = ∫ MydX∫y2dX

= 148869.8874

1386.932

H = 107.34 KN


B. SOLUTION OF SAME PARABOLIC ARCH BY NUMERICAL INTEGRATION:− We know Mac = 53.93X − X2 0 < X < 35

Mcd = 53.93X − 70 (X − 17.5) 35 < X < 50

Mdb = 53.93X − 70 (X − 17.5) −5 (X − 50) 50 < X < 70

y = 0.343X − 4.898 . 10−3 X2 . Solve in a tabular form.

SECTION X Y M MY Y2

1 3.5 1.14 176.51 201.22 1.30 2 10.5 3.06 456.02 1395.42 9.36 3 17.5 4.50 637.53 2868.89 20.27 4 24.5 5.46 721.04 3936.88 29.35 5 31.5 5.94 706.55 4196.91 35.34 6 38.5 5.94 606.31 3601.48 35.34 7 45.5 5.46 493.82 2696.26 29.85 8 52.5 4.50 368.83 1659.74 20.27 9 59.5 3.06 221.34 677.29 9.36 10 66.5 1.14 73.85 84.18 1.30 ∑ 21318.27 ∑ 192.24


= 21318.27 × 7

192.24 × 7

H = 110.89 KN Accuracy can be increased by increasing the number of segments. Now BMD is drawn.

637.53 706,5 706.8 676.74

499.00

368.83

499.00

M-Diagram

0

499.00

676.74

499.00

0

0

0

Hy-Diagram

110.98KN

21.07 KN

20m

4.506m4.5070m

110.98KNA

53.93KN

C

706.55421.4

368.83634.53

D

Mx-Diagram

2Kn/m 5kN


C. CONSIDERING IT TO BE A CIRCULAR ARCH WITH ALGEBRAIC INTEGRATION

0

C

BA

x x

2KN/m 5KN20mD

21.07KN

70M

6m

R=105.08m

53.93KNR

D

y

R = L2

8yc + yc2

R = 702

8X6 + 6y

R = 105.08 m

y = R2 − X2 − (h − yc) and dydX = tan θ =

−X105.082 − X2

y = 105.082 − X2 − (105.08 − 6 ) y = 105.082 − X2 − 99.08 . Establishment expressions.

Mac = 53.93 ( 35 − X) − ( 35 − X)2 0 < X < 35

Mbd = 21.07 (35 − X) 0 < X < 20

Mdc = 21.07 ( 35 − X) − 5 (15 − X) 20 < X < 35

∫ My dX = 35

∫o[53.93 (35 − X) − (35 − X)2][ 105.082 − X2 −99.08] dX

+ 20

∫o

21.07 (35 − X) [ 105.082 − X2 − 99.08] dX

+35

∫o[ 21.07( 35 − X) −5 (15 − X)] [ 105.082 − X2 −99.08] dX

∫ My dX = I1 + I2 + I3


( Where I1 , I2 and I3 are 1st , 2nd and 3rd integrals of above expression respectively). These are evaluated separately to avoid lengthy simultaneous evaluation of above ∫ My dX expression.

Evaluation of I1 =

35

∫o[53.93 × 35 − 53.93X − (352 + X2 −70X)] [ 105.082 − X2 − 99.08]dX

= 35

∫o

(662.55 + 16.07X − X2) [ 105.082 − X2 − 99.08] dX

= 35

∫o[662.55 105.082 − X2 + 16.07 X 105.082 − X2

− X2 105.082 − X2 − 65645.454 − 1592.216X + 99.08X2] dX

= 662.55 35

∫o

105.082 − X2 dX −16.07

2 35

∫o

105.082 − X2 (−2X)dX .

Taking constants out.

12

35

∫o

X 105.082 − X2 (− 2X)dX − 65645.454 35

∫o

dX −1592.216

35

∫o

XdX + 99.08 35

∫o

X2dX

Put X = 105.08 Sinθ

and dX = 105.08 Cosθ dθ

At X = 0 , θ = 0

At X = 35 , θ = 0.3396 radians = 19.4°

I1 = 662.55

0.3396

∫o

105.082 − 105.082 × Sin2 θ(105.08)Cosθdθ

− 16.07

2

105.082 − X2

3/2

3/2 35 o

+ 12

X

105.082 − X2

3/2

3/2 35 o

− 35

∫o

105.082 − X2

3/2

3/2

.dX

− 65645.454 X 35|o

− 1592.216

X 2

2

35 o

+ 99.08

X 3

3

35 o

= 662.55 × 105.082 0.3396

∫o

Cos2 θ dθ − 16.07

3 [(105.082 − 352 )3/2

− (105.082 )3/2 ] + 13 [35 (105.082 − 352 )3/2 −

35

∫o

(105.082 − X2 )3/2 dX]


− 65645.454 (35 − 0) − 1592.216

2 (352 ) + 99.08

353

3

I1 = 7315748.83 0.3396

∫o

1+Cos2 θ

2 dθ +1005048.922 + 11347550.55

− 13

0.3396

∫o

105.084 Cos4 θdθ − 1856804.857

= 7315748.83

2

θ +

Sin2 θ2

0.3396 o

− 13 (105.08)4

0.3396

∫o

Cos2 θ (1− Sin2 θ)dθ + 10495794.62

I1 = 7315748.83

2

0.3396 +

Sin (2 × 0.3396)2 + 10495794.62

− 13 × (105.08)4

0.3396

∫o

1 + Cos 2θ

2

1−Cos 2 θ

2 dθ

= 12886893.66 − 13 ×

105.08

2

4

θ+

Sin 2θ2

0.3396 o

+ 1

12 × (105.08)4 0.3396 o

(1 − Cos2 2θ) dθ

= 12886893.66 − 16 × (105.8)4

0.3396 +

Sin (2 × 0.3396)2 ]

+ 1

12 × (105.08)4 0.3396

∫o

1 −

1 + Cos 4 θ

2 dθ

= 12886893.66 − 13283049.35 + 1

12 × (105.08)4 0.3396

∫o

1

2 − 12 Cos 4 θ dθ

= − 396155.69 + 1

24 (105.08)4

θ −

Sin 4θ4

0.3396 o

= − 396155.69 + 1

24 (105.08)4

0.3396 −

Sin (4 × 0.3396)4

0.3396 o

= − 396155.69 + 483712.6275 = 87556.9375


I2 = 20

∫o

21.07 (35 − X) [ 105.082 − X2 − 99.08] dX

= 20

∫o

(737.45 − 21.07X) [ 105.082 − X2 − 99.08] dX

= 20

∫o

[737.45 105.082 − X2 − 73066.546

− 21.07X (105.08)2 − X2 + 2087.6162 ] dX

Put X = 105.08 Sin θ

dX = 105.08 Cos θ dθ

At X = 0 θ = 0 At X = 20 θ = 0.1915

I2 = 737.45 0.1915

∫o

(105.08)2 Cos2 θ dθ + 21.07

2 20

∫o

105.082 − X2

(−2X) dX − 73066.546 26

∫o

dX + 2087.616 20

∫o

XdX

= 8.143 × 106 0.1915

∫o

1+Cos 2θ

2 dθ + 21.07

2

(105.08)2 − X2

3/2

3/2 20 o

− 73066.546 | X 26|o

+ 2087.616

X 2

2

20 o

= 8.143 × 106

2

θ +

Sin 2θ2

0.1915 o

+ 21.07

3 [{(105.08)2 − (20)2}3/2 − (105.08)2x3/2 ]

− 73066.546 (20) + 2087.616

2 (400 )

= 8.143 × 106

2

0.1916 +

Sin (2 × 0.1915)2 − 438772.215

I2 = 58247.385

I3 = 35

∫20

(662.45 − 16.07X) [ 105.082 − X2 − 99.08 ] dX

= 35

∫20

[662.45 105.082 − X2 − 65635.546

− 16.07 × 105.082 − X2 + 1592.216X] dX


= 662.45 0.3396

∫0.1915

105.082 Cos2 θ dθ − 65635.546 | X 35|20

+ 1592.216

X2

2

35 20

+ 16.07

2 35

∫20

105.082 − X2 (−2X) dX

= 662.45 × 105.082 0.3396

∫0.1915

1 + Cos 2 θ

2 dθ − 65635.546 × 15

+ 1592.216

2 (352 − 202 ) + 16.07

2

(105.082 − X2 )3/2

3/2

35 20

= 662.45 × 105.082

2

θ +

Sin 2 θ2

0.3396 0.1915

− 65635.546 × 15

+ 1592.216

2 (352 − 202) + 16.07

3 [(105.082 − 352 )3/2 − (105.082 − 202 )3/2 ]

= 662.45 × 105.082

2

0.3396 − 0.1915 +

Sin (2 × 0.3396)2 −

Sin (2 × 0.1915)2

− 65635.546 × 15 + 1592.216

2 (352 −202 ) + 16.07

3 [(105.082 −352 )3/2 − (105.082 − 202)3/2 ]

I3 = 8838.028 . Adding values of three integrals. We have MydX = 87556.9375 + 58247.385 + 8838.028 = 154642.3505 . Now calculate ∫y2dX

∫y2 dX = 2 35

∫o

[ 105.082 − X2 − 99.08 ]2 dX

= 2 35

∫o

[105.082 − X2 + 99.082 − 2X99.08 105.082 − X2 ] dX

= 2 35

∫o

(20858.653 − X2 − 198.16 105.082 − X2 ) dX

= 2 × 20858.653 | X 35|o

− 23 | X3 | 35 − 198.16 × 2

0.3396

∫o

105.082 Cos2 θ dθ

= 2 × 20858.653 (35) − 23 (353 ) − 198.16 × 2 × 105.082

0.3396

∫o

1+Cos2 θ

2 dθ

= 2 × 20858.653 × 35 − 23 + 353 −

198.16 × 2 × 105.082

2

θ +

Sin 2θ2

0.3396 o


∫y2d X = 1229.761

H = ∫MydX∫y2dX

= 154642.3505

1239.761

H = 125.75 KN

D. CIRCULAR ARCH BY NUMERICAL INTEGRATION:- As you have seen algebraic integration is lengthy, laborious and time consuming. so it is better to store such question by numerical integration.

0

21.07KN53.93KN

A B

20m5KN

70m(1)

(2) (3) (4) (5) (6) (7) (8) (9) (10)

y = 105.082 − X2 − 99.08

Mac = 53.93 (35 − X) − (35 − X) 20 < X < 35

Mbd = 21.07 (35 − X) 0 < X < 20

Mdc = 21.07 (35 − X) − 5 (15 − X) 29 < X < 35 Attempting in a tabular form

Section X Y M MY Y2 1 31.5 1.167 176.505 205.981 1.362 2 24.5 3.104 456.015 1415.47 9.635 3 17.5 0.533 637.525 2889.901 20.548 4 10.5 5.474 721.035 3946.446 29.965 5 3.5 5.942 760.545 4198.29 35.307 6 3.5 5.942 606.205 3602.07 35.307 7 10.5 5.474 493.715 2702.596 29.965 8 17.5 4.533 368.725 1671.430 20.548 9 24.5 3.104 221.235 686.713 9.635 10 31.5 1.167 73.745 86.060 1.362 ∑21405.157 ∑193.634


S = 105.08 (2 × 0.3396) = 71.370 m

dS = 71.37

10 = 7.137 m

H = ∑Myds∑y2ds =

21405.157 × 7.137193.634 × 7.137

H = 110.54 KN , Accuracy can be increased by taking more segments. For secant variation of inertia follow the same procedures established already in this Chapter. Space for taking Notes:

SLOPE __ DEFLECTION METHOD 199

CHAPTER FOUR

4. SLOPE – DEFLECTION METHOD

This method is applicable to all types of statically indeterminate beams & frames and in this method, we solve for unknown joint rotations, which are expressed in terms of the applied loads and the bending moments. By inspection, the degree of indeterminacy is checked and the corresponding number of unknown joint rotations are calculated from the slope – deflections equations.

4.1. SIGN CONVENTION:–

(1) ROTATIONS:– Clockwise joint rotations are considered as (+ve).

(2) END MOMENTS:– Counterclockwise end moments are considered as (+ve).

4.2. PROCEDURE:–

The procedure is as follows: (1) Determine the fixed end moments at the end of each span due to applied loads acting on span by

considering each span as fixed ended. Assign ± Signs w.r.t. above sign convention.

+WL2____12

-WL2____12

A

w (u.d.l)

BL

P

baBA

-WL2____12MFba =+WL2____

12Mfab =

MFba = ____- Pa bL

2

2Mfab = + Pa b

L

22

L

(2) Express all end moments in terms of fixed end moments and the joint rotations by using slope –

deflection equations.

(3) Establish simultaneous equations with the joint rotations as the unknowns by applying the condition that sum of the end moments acting on the ends of the two members meeting at a joint should be equal to zero.

(4) Solve for unknown joint rotations.

(5) Substitute back the end rotations in slope – deflection equations and compute the end moments.

(6) Determine all reactions and draw S.F. and B.M. diagrams and also sketch the elastic curve


4.3. DERIVATION OF SLOPE – DEFLECTION EQUATION:– Consider a generalized beam under the action of applied loads and end moments as shown at (i).

(i) (ii) Fig: (i) can be equated to a fixed ended beam carrying applied loads which produce fixing moments plus two simple beams carrying end moments [figs (iii) and (iv)]

(iii) (iv) Draw moment diagrams. Determine their areas and centroid locations.

(Assuming these MEI diagrams are placed on conjugate beams)

Equating relevant rotations in above four diagrams according to sign conventions θa = 0 − θa1 + θa2 = − θa1 + θa2 and θb = 0 + θb1 − θb2 = θb1 − θb2 (1) Compatibility on rotations

During the same for moments.

So Mab = Mfab + Ma′ Mba = Mfba + Mb′ (2) Compatibility on moments


Where Ma′ and Mb′ are the additional moments required to produce the joint rotations at ends A and B respectively and Mfab & Mfba are the fixed ended moments which hold the tangents at points A and B straight.Conjugate beam theorem states that “ rotation at a point in actual beam is equal to the shear force at the corresponding point in the conjugate beam ). Applying it we have.

θa1 = 23

LMa′

2EI = LMa′3EI

θb1 = 13

LMa′

2EI = LMa′6EI

θa2 = 13

LMb′

2EI = LMb′6EI

θb2 = 23

LMb′

2EI = LMb′3EI

Putting the values of θa1, θa2, θb1 & θb2 in equation (1) and solve for Ma′ & Mb′.

θa = − LMa′3EI +

LMb′6EI = −

L3

Ma'EI +

LMb'6EI → (3)

and θb = Ma'L6EI −

LMb'3EI =

L6

Ma/

EI − L3

Mb/

EI → (4)

Equation (3) becomes θa + LMa'3EI =

LMb'6EI OR

6EIθa + 2LMa′

6EI = LMb′6EI

6EIθa + 2 LMa′ = LMb′

Mb′ = 6EIL θa + 2 Ma′ → (5)

From (4), θb = Ma ′L6EI −

L3EI

6EIθa

L + 2Ma' by putting Mb/ from (5)

θb = Ma ′L6EI − 2 θa −

2LMa′3EI

θb + 2 θa = Ma ′L6EI −

2LMa′3EI

θb + 2θa = Ma′L − 4 LMa′

6EI


θb + 2θa = − 3LMa′

6EI

So θb + 2θa = − LMa′

2EI From here Ma' is

Ma′ = − 2EI

L (2θa + θb )

or Ma′ = 2EIL ( − 2θa − θb) → (6)

From(5) Mb′ = 6EI θa

L + 4EIL (− 2 θa − θb ) By putting value of Ma’ from 6 in 5 and simplifying

Mb′ = 6EI θa

L − 8EI θa

L − 4EIL θb

Mb′ = − 2EI θa

L − 4EIL θb

or Mb′ = 2EIL (− θa − 2 θb ) → (7)

Putting the values of Ma′ and Mb′ from equations 6 and 7 in equation (2), we have.

Mab = Mfab + 2EIL (− 2θa−θb)

Mba = Mfba + 2EIL (−θa− 2θb)

Absolute values of 2EIL are not required in general except for special cases and we use relative

values of 2EIL in cases without settlement..

Where, K = IL if absolute stiffness (rotation) is not required.

Where K = relative stiffness Slope − deflection equation for members without settlement.

Mab = Mfab + 2EIL (− 2θa − θb )

Mba = Mfba + 2EIL (− 2 θb − θa )


without absolute value of 2EIL , above equations become

Mab = Mfab + Kab (− 2θa − θb ) Mba = Mfba + Kab (− 2 θb − θa ) Where Kab = relative stiffness of member ab

Kab =

2EI

L

ab

Now we apply the method to various indeterminate structures. EXAMPLE NO.1::− Analyze the continuous beam shown by slope − deflection method. Draw shear & moment diagram and sketch the elastic curve. SOLUTION :−

2KN 2KN/m 4KN1m 2m

A B C D2I 3I4m 6m 4m

4I

Step 1: Calculation of Relative Stiffness :−

Member. I L IL Krel.

AB 2 4 24 × 12 6

BC 4 6 46 × 12 8

CD 3 4 34 × 12 9

Step 2: Calculation of Fixed End Moments :− Treat each span as fixed ended.

a b

P

Pb aL2

2P b aL2

2

a bL

P

Pb aL2

2P b aL2

2

- (any generalized span carrying a single load)

Mfab = Mfba = 0 (there is no load acting on span AB)

Mfbc = 2 × 62

12 = + 6 KN−m (According to our sign convention)


Mfcb = − 6 KN−m (According to our sign convention)

Mfcd = 4 × 22 × 2

42 = + 2 KN−m

Mfdc = − 2 KN−m Step 3: Establish simultaneous equations :−

Mab = Mfab + Kab (−2 θa − θb) (General form–Put values of FEMs & relative stiffnesses)

Mab = 0 + 6 ( − 2θa − θb) = − 12 θa − 6 θb

Mba = 6 (− 2θb − θa) = − 12 θb − 6 θa

Mbc = 6 + 8 (− 2 θb − θc) = 6 − 16 θb − 8 θc

Mcb = − 6 + 8 (− 2 θc − θb) = − 6 − 16 θc − 8 θb

Mcd = 2 + 9 (− 2θc − θd) = 2 − 18 θc − 9 θd

Mdc = − 2 + 9 (–2θd – θc) = – 2 – 18 θd – 9 θc

Step 4: Joint Conditions :– at A : Mab – 2=0 or Mab = 2 KN–m

B Mba + Mbc = 0

C: Mcb + Mcd = 0

D: Mdc = 0

Put these joint conditions in the linear simultaneous equations set up in step No. (3).

Mab = 2, so – 12 θa – 6 θb = 2

– 12 θa – 6 θb – 2 = 0 → (1)

Mba + Mbc = 0

so – 12 θb – 6 θa + 6 – 16 θb – 8 θc = 0

– 6 θa – 28 θb – 8 θc + 6 = 0 → (2)

Mcb + Mcd = 0 so – 6 – 16θc – 8 θb + 2 – 18 θc – 9 θd = 0

– 8θb – 34 θc – 9 θd – 4 = 0 → (3)

Mdc = 0

– 2 – 18 θd – 9 θc = 0

– 9 θc – 18 θd – 2 = 0 → (4)

– 12 θa – 6 θb – 2 = 0 – 6 θa – 28 θb – 8 θc + 6 = 0 ( Symmetrical about θa and θd diagonal ) 0 – 8 θb – 34 θc – 9 θd – 4 = 0 0 − 0 – 9 θc – 18 θd– 2 = 0


If the linear simultaneous equations are established and are arranged in a sequence of joint conditions, we will find that the quantities on the leading diagonal are dominant in that particular equation and off diagonal quantities are symmetrical as far as the magnitude of rotations is concerned. This is a typical property of the stiffness method, which you will study later in matrix methods of structural analysis.

From (1) θa =

–2 – 6θb

12 → (5)

From (4) θd =

– 2 – 9 θc

18 → (6)

Putting these values in equations (2) & (3), all deformations are expressed in terms of θb & θc. Therefore, we get two linear simultaneous equations in terms of θb & θc. Hence, their values can be calculated. Put θa from equations (5) in equation (2)

– 6

–2 – 6θb

12 – 28 θb – 8 θc + 6 = 0

+ 1 + 3 θb – 28 θb – 8 θc + 6 = 0 or – 25 θb – 8 θc + 7 = 0 → (7) Put θd from equation (6) in equation (3)

– 8 θb – 34 θc – 9

– 2 – 9 θc

18 – 4 =0 Simplifying

– 8 θb – 34 θc + 1 + 4.5 θc – 4 = 0 – 8 θb – 29.5 θc – 3 = 0 → (8)

From (7) θb =

– 8 θc + 7

25 → (9)

Put in (8) – 8

– 8 θc + 7

25 – 29.5 θc – 3 = 0

or 2.56 θc – 2.24 – 29.5 θc – 3 = 0 – 26.94 θc – 5.24 = 0

θc = –5.2426.94

θc = – 0.1945 Radians


Put value of θc in equation (9) , we get

θb =

–8 ( – 0.1945) + 7

25

θb = + 0.3422 radians. Put θb in equation (5)

θa =

–2 – 6 × 0.3422

12

θa = – 0.3378 radians.

Put θc in equation (6)

θd = –2 – 9 . (– 0.1945)

18

θd = – 0.0139 radians.

Putting these values of rotations in simultaneous equations set up in step (3) & simplifying we get the values of end moments as under: Mab = 2 KN–m These two values should be the Mba = – 2.08 KN–m the same but with opposite signs to satisfy equilibrium at that

Mbc = + 2.08 KN–m joint. Mcb = – 5.63 KN–m Mcd = + 5.63 KN–m (Same comment) Mdc = 0 As the end moments have been calculated and they also satisfy the joint conditions, therefore, the structure is statically determinate at this stage. Reactions, shear force diagrams, B.M. diagrams & elastic curves can now be sketched. NOTE:– In slope – deflection method, the actual deformations are the redundants and stiffness matrix is symmetrical. In force – method, we can chose any redundant and therefore flexibility matrix is not generally symmetrical about leading diagonal. Now we can draw shear force and bending moment diagrams and sketch elastic curve. Free body diagrams of various spans are drawn.


2KN 2KN/m 4KN1m 2

A A 4m B B 6m C C 4m D

2.08 2.08 5.63 5.63 2m

+2 0 0 +6 +6 +2 +2 reactions due to applied loads

-0.02 +0.02 -0.592 +0.592 +1.408 -1.408 reactions due to end moments+2 -0.02 +0.02 +5.408 +6.592 + 3.408 +0.592

+1.98 +5.428 +10

2KN Elastic curve 2KN/m 4KN2m1m

A B D4m6m4m C

1.98KN 5.428KN 10KN 0.592KN

5.408 3.408 3.408

0 0

2.0 0.02 0.02

=2.704m6-X

6.592X

S.F.D.0.5920.592

=0.417mXX

2.082

0 0 B.M.D.++

++

1.184=1.008m

5.63a=1.652m

2.08

/

adding values on both sides of a support

Find the location of points of contraflexure & find the maximum +ve B. M. in portion BC by setting the relevant moment expression equal to zero and by setting the concerned S.F. expression equal to zero respectively. To Find Max B.M. in Portion BC :–

X

5.408 = 6 – X6.592

6.592 X = 6 × 5.408 – 5.408 X X = 2.704m


So Mbc = – 2.08 + 5.408 × 2.704 – 22 × (2.704)

Mbc = 5.237 KN–m Points of Contraflexure :– Near B:–

– 2.08 + 5.408 X – X2 = 0

X2 – 5.408 X + 2.08 = 0

X = 5.408 ± (5.408)2 – 4 × 1 × 2.08

2 × 1

X = 0.417 m, 4.991 m

X = 0.417 m

Near C :– In span CB

– 5.63 + 6.592 X' – X'2 = 0

X′2 – 6.592 X′ + 5.63 = 0

X′ = 6.592 ± (6.592)2 – 4 × 1 × 5.63

21

X′ = 6.592 ± 4.575

2

X′ = 5.584 , 1.008

X′ = 1.008 m

1.1842 − a =

5.63a in span CD.

1.184 a = 5.63 × 2 – 5.63 a

a = 1.652 m

These can be put in bending moment diagram and sketch elastic curve.


EXAMPLE NO. 2:– Analyse the continuous beam shown by slope –deflection method. Draw S.F.D. & B.M.D. Also sketch the elastic curve. SOLUTION :–

A CB

4KN

EI = Constt.6m

EI = Constt.

2m2m

Step 1: Calculation of Relative Stiffness :–

Member I L IL Krel.

AB 1 4 14 × 12 3

BC 1 6 16 × 12 2

Step 2: Calculation of Fixed End Moments :– Mab = Mfab + Krel (−2ba − θb)

Mfab = 4 × 22 × 2

42 = + 2 KN–m

Mfba = – 2 KN–m

Mfbc = 0

Mfcb = 0 ( As there is no load in portion BC ) Step 3: Establish Simultaneous Equations :– Mab = 2 + 3 ( – 2 θa – θb )

Mba = – 2 + 3 ( – 2 θb – θa )

Mbc = 0 + 2 ( – 2 θb – θc)

Mcb = 0 + 2 ( – 2 θc – θb ) Step 4: Joint Conditions :– A : θa = 0 ( Being a fixed joint )

B : Mba + Mbc = 0

C: θc = 0 (Being a fixed end) Putting these joint conditions in the linear simultaneous equations set up in step No. (3) Put θa = θc = 0 in above equations. The only equation is obtained from joint B. That becomes. – 2 – 6 θb – 3 θa – 4 θb – 2 θc = 0

– 2 – 6 θb – 0 – 4 θb – 0 = 0

– 2 – 10 θb = 0

θb = – 0.2 radians.


Put these values of rotations i.e., θa = θc = 0 and θb = −0.2 in simultaneous equations set up in step (3) & get the values of end moments. Mab = 2 + 3 ( – 2 × 0 + 0.2) = 2.6 KN–m

Mba = – 2 + 3 (– 2 × (– 0.2) – 0) = – 0.8 KN–m.

Mbc = 0 + 2 [ – 2 × (–0.2) – 0] = + 0.8 KN–m

Mcb = 0 + 2 ( 0 + 0.2) = + 0.4 KN–m . Now Draw SFD and BMD.

A

2.6KN-m

2.45KN

2m4KN

2mB 6m C

0.4KN-m

0.2KN1.75KN

2.452.45

0.2

1.551.55

++

+

0.20

0

S.F.D.0

2.3

+X=1.061m

0.4

B.M.D.

a=2m(6-a)

0.8X(2-X)

(2-X)

=0.516m2.6 As the end moments have been calculated and they satisfy the joint conditions, therefore, the structure is statically determinate at this stage. Reactions, S.F. diagram, B.M. diagram & elastic curve have now been sketched. LOCATION OF POINTS OF CONTRAFLEXURE :– Near A :–

2.6X =

2.32 − X

2.6 × 2 – 2.6 X = 2.3 X

X = 1.061 m

Near B :– X′0.8 =

2– X′2.3

2.3 X′ = 2 × 0.8 – 0.8 X′ X′ = 0.516 m


Near C :− a

0.4 =

6 – a

0.8

0.8 a = 6 × 0.4 – 0.4 a a = 2 m There have been shown on BMD.

EXAMPLE NO. 3:– Analyze the continuous beam shown by slope – deflection method. Draw S.F.D & B.M.D. Also sketch the elastic curve. SOLUTION:–

2KN 2KN/m 4KN

A B C D

2m

1m 4m 6m 4m

2 4 3I I I Step 1: Calculation of relative stiffness :–


AB 2 4 24 × 12 6

BC 4 6 46 × 12 8

CD 3 4 34 × 12 9

Step 2: Calculation of Fixed End Moments :–

Mfab = Mfba = 0 (no load over span AB)

Mfbc = 2 × 62

12 = + 6 KN–m

Mfcb = – 6 KN–m

Mfcd = 4 × 22 × 2

42 = + 2 KN–m

Mfdc = –2 KN–m


Step 3: Establish simultaneous equations :– Put values of fixing moments and Krel.

Mab = 0 + 6 (–2 θa – θb) = – 12 θa – 6 θb

Mba = 0 + 6 ( – 2 θb – θa ) = – 12 θb – 6 θa

Mbc = 6 + 8 ( – 2 θb – θc) = 6 – 16 θb – 8 θc

Mcb = – 6 + 8 ( – 2 θc – θb) = – 6 – 16 θc – 8 θb

Mcd = 2 + 9 ( – 2 θc – θd) 2 – 18 θc – 9 θd

Mdc = – 2 + 9 (– 2 θd – θc) = – 2 – 18 θd – 9 θc Step 4: Joint Conditions :– A: : Mab – 2 = 0 or Mab = 2 KN–m

B : Mba + Mbc = 0

C : Mcb + Mcd = 0

D : θd = 0 Putting these joint conditions in the linear simultaneous equations set up in step No. (3) – 12 θa – 6 θb = 2 ∴ Mab = 2

– 12 θa – 6 θb – 2 = 0 → (1)

Mba + Mbc = 0

– 12 θb – 6 θa + 6 – 16 θb – 8 θc = 0

– 6 θa – 28 θb – 8 θc + 6 = 0 → (2)

Mcb + Mcd = 0 – 6 – 16 θc – 8 θb + 2 – 18 θc – 9 θd = 0

– 8 θb – 34 θc – 9 θd – 4 = 0 → (3)

θd = 0 (4) Simplifying we get.

– 12 θa – 6 θb – 2 = 0 → (1)

– 6 θa – 28 θb – 8 θc + 6 = 0 → (2)

– 8 θb – 34 θc – 9 θd – 4 = 0 → (3)

θd = 0 → (4)

Putting the value of θd in equation (3) – 8θb – 34 θc – 0 – 4 = 0 – 8 θb – 34 θc – 4 = 0 → (5)


From (1) θa =

– 6 θb – 2

12 → (6)

Put in (2) – 6

– 6 θb – 2

12 – 28 θb – 8 θc + 6 = 0

+ 3 θb + 1 – 28 θb – 8 θc + 6 = 0

– 25 θb – 8 θc + 7 = 0 → (7)

From (5) θb =

–34 θc – 4

8 → (8)

Put in (7) – 25

–34 θc – 4

8 – 8 θc + 7 = 0

or 106.25 θc + 12.5 – 8 θc + 7 = 0

98.25 θc + 19.5 = 0

θc = – 0.1985 radians.

From (8) θb =

–34 (– 0.1985) – 4

8 by putting value of θc

From (6) θa =

– 6 x 0.3435 – 2

12

θb = + 0.3435 radians. θa = – 0.3384 radians.

Finally θa = – 0.3384 θb = + 0.3435 θc = – 0.1985 θd = 0 Putting these values of rotations in simultaneous equations set up in step # (3) & getting the values of end moments as follows. Mab = –12x (–0.3384) – 6 × 0..3435 = 1.9918 = + 2 KN–m Mba = – 12x (+0.3435)– 6x(– 0.3384) = – 2.092 KN–m Mbc = 6 – 16(+0.3435)–8 (– 0.1985) = + 2.092 KN–m Mcb = – 6 – 16(– 0.1985) – 8(+0.3435) = – 5.572 KN–m Mcd = 2 – 18 (– 0.1985) – 9 × 0 = + 5.573 KN–m Mdc = – 2 – 18 x 0 – 9 (– 0.1985) = – 0.214 KN–m.


As the end moments have been calculated and they satisfy the joint conditions. Therefore, the structure is statically determinate at this stage. Reactions, S.F.D., B.M.D. & elastic curve can now be sketched.

2KN 2KN/m 4KN

1m

22

2m2m+2 0 0 +6 +6 +2 +2 reactions due to

applied loads

4m 6m

2.092 2.092 5.572 5.573 0.214

0 -0.023 +0.023 -0.58 +0.58 +1.34 -1.34 reactions due to end moments +2 -0.023 +0.023 +5.42 +6.58 +3.34 +0.66 final reactions

+1.977 +5.443 +9.92

2KN1m

Elastic curve 2KN/m 4KN

A

A

B

B

C

C

D

D

4m 6M 4m+1.977KN +5.443KN 9.92KN 0.66KN

5.42 3.34+

-

(6-a)02 0.023 0.023

a

0 S.F.D.0.66

2

0-

- -

+

+

X XX X

0 B.M.D.

5.5722.092

-+

1.106

0.214

2KN 2KN/m 4KN

1m

22

2m2m+2 0 0 +6 +6 +2 +2 reactions due to

applied loads

4m 6m

2.092 2.092 5.572 5.573 0.214

0 -0.023 +0.023 -0.58 +0.58 +1.34 -1.34 reactions due to end moments +2 -0.023 +0.023 +5.42 +6.58 +3.34 +0.66 final reactions

+1.977 +5.443 +9.92

2KN1m

Elastic curve 2KN/m 4KN

A

A

B

B

C

C

D

D

4m 6M 4m+1.977KN +5.443KN 9.92KN 0.66KN

5.42 3.34+(6-a)

02 0.023 0.023

a

0 S.F.D.0.66

2

0+

+

X XX X

0 B.M.D.

5.5722.092

-+

1.106

0.214

5.25

TO LOCATE THE MAX. B.M. IN PORTION BC :–

5.42

a = 6.58

(6 − a)

5.42 × 6 – 5.42.a = 6.58 a a = 2.71 m

Mbc = – 2.092 +

5.42 × 2.71 –

22 × 2.712 = + 5.252 KN–m

= 5.25 KN−m


LOCATION OF POINTS OF CONTRAFLEXURE :– Near B :– (Span BC ) – 2.092 + 5.42 X – X2 = 0 X2 – 5.42 X + 2.092 = 0

X = 5.42 ± (5.42)2 – 4 × 1 × 2.092

2

X = 5.42 ± 4.583

2

= 0.418 , 5.002 , So X = 0.418 m Near C :– Span BC

5.572 + 6.58 X′ – X′2 = 0 X′2 – 6.58 X′ + 5.572 = 0

X′ = + 6.58 ± (6.58)2 – 4 × 1 × 5.572

2

= 6.58 ± 4.583

2

X′ = 0.998 , 5.582 X′ = 0.998 m Near C : ( Span CD ) 5.573 + 3.34 X"= 0 X" = 1.669 m Near D :– ( Span CD )

0.214 + 0.66 X = 0 X = 0.324 m These have been shown on BMD.


4.4. ANALYSIS OF INDETERMINATE BEAMS DUE TO MEMBER AXIS ROTATION (SETTLEMENT OF SUPPORTS) :–

L

B

B

RR = L

L

B

B

RR = L

Consider a generalized fixed ended beam settling differentially at B. The angle R is measured from the original members axis to the displaced member axis and will be +ve if it is clockwise. The

absolute values of 2EIL with consistent units are to be used in the settlement problem and the final slope –

deflection equation to be used for settlement problems is as follows:–

Mab = Mfab + 2EIL (– 2 θa – θb + 3 R)

Mba = Mfba + 2EIL (– 2 θb – θa + 3 R).

The above equation is general and can be used to find the end moments due to applied loading and

due to sinking of supports simultaneously. However, it is a common practice to consider end moments induced due to applied loading separately from those induced due to settlement. The superposition principle can then be applied afterwards and the final end moments can be obtained. If all supports of a continuous structure like beams and frames settle by the same amount, no additional end moments will be induced due to sinking. These will be induced only whenever there is a differential sinking of supports like the following case. Where support C sinks by ∆ w.r.t supports B and D.

A B C D

R R

RR

L1 L2A B C D

R R

RR

L1 L2

C/

(Sign of R is the same if determined at the two ends of a span ). So

Rab = 0 ( Both supports of span AB are at the same level )

Rbc = ∆L1

( Clock–wise angle is positive )

Rcd = – ∆L2

( Counterclock–wise angle is negative )


The following points are to be strictly followed : (1) Consideration and computation of values of ‘R’ in the span effected by the settlement. (2) Use proper sign for R keeping in view the corresponding sign convention.

(3) The units of the R.H.S. of the slope–deflection equation should be those of the B.M. (KN–m). EXAMPLE NO. 4:– Analyze the continuous beam shown due to the settlement of support B by slope- deflection method. Draw shear and moment diagrams and sketch the elastic curve.

A B CD

15mm

1m 4m 5m 4m2I 4I 3I

E=200X10 KN/m6 2

I=400X10 m-4 4

A B CD

15mm

1m 4m 5m 4m2I 4I 3I

E=200X10 KN/m6 2

I=400X10 m-4 4

B/

SOLUTION:– Step 1: Calculation of F.E.M :–

Mab = Mfab + 2EIL (– 2 θa – θb + 3 R). where R is in radians

As there is no applied loading on the beam, therefore all fixed end moments terms in the slope – deflection equation will be equal to zero.

Step 2: Calculation of R and 2EIL terms for various spans :–

Span AB.

R = + 0.015

4 = + 3.75 × 10–3 rad

2EIL =

2x(200 × 106 ) × (2 × 400 × 10–6 )4

KN/m2xm4

m

= 80,000 KN–m Span BC :–

R = – 0.015

5 = – 3 × 10–3 rad

2EIL =

2 (200 × 106 ) (4 × 400 × 10–6)5

= 128,000 KN–m


Span CD :– R = 0

2EIL =

2x (200 × 106) × (3 × 400 × 10–6)4

= 120,000 KN–m Step 3: Write Slope–deflection Equation in terms of Joint Rotations & R. Mab = 0 + 80,000 (– 2 θa – θb + 11.25 × 10–3)

Mba = 0 + 80,000 (– 2 θb – θa + 11.25 × 10–3)

Mbc = 128,000 (– 2 θb – θc – 9 × 10–3)

Mcb = 128,000 (– 2 θc – θb – 9 × 10–3)

Mcd = 120,000 (– 2 θc – θd)

Mdc = 120,000 (– 2θd – θc) Step 4: Joint Conditions (Conditions of Equilibrium + geometry) :– Joint A:– Mab = 0 (Pin support) → (1)

Joint B :– Mba+Mbc=0 (Continuous support) → (2)

Joint C :– Mcb + Mcd=0 (Continuous support) → (3)

Joint D :– θd = 0 (Fixed support) → (4) Step 5: Simultaneous Equations :– Putting joint conditions in slope – deflection equations

– 160,000 θa – 80,000 θb + 0 + 900 = 0 ∴ Mab = 0 → (1)

– 160,000 θb – 80,000 θa + 900 – 256,000 θb –128,000 θc – 1152 = 0 – 80,000 θa – 416,000 θb –128,000 θc–252=0 ∴ Mba + Mbc = 0 → (2)

– 256,000 θc – 128,000 θb – 1152 – 240,000 θc–0=0 Mcb + Mcd = 0 → (3)

– 128,000 θb – 496,000 θc – 1152 = 0 Simplifying, finally

– 160,000 θa – 80,000 θb + 0 + 900 = 0 → (1)

– 80,000 θa – 416,000 θb – 128,000 θc – 252=0 → (2)

– 128,000θb – 496,000θc–1152=0 → (3)

Solve the above three linear simultaneous equations to get the values of θa, θb & θc which will be put in the original slope–deflection equations to determine the final end moments.

From (1) θa =

900 – 80000 θb

160000

or θa = 5.625 × 10–3 – 0.5 θb → (4)


From (3) θc =

–128000 θb – 1152

496000

so θc = – 0.258 θb – 2.32 × 10–3 → (5) Put (4) and (5) in (2) , we have. – 80,000 [ 5.625 × 10–3 – 0.5 θb] – 416,000 θb – 128,000 [– 0.258 θb – 2.32 × 10–3] – 252 = 0 – 450 + 40,000 θb – 416,000 θb+33,024 θb+296.96–252=0

θb = –405.04342976

θb = – 1.181 × 10–3 radians.

Put θb in (1) because θa is dominant there. – 160,000 θa – 80,000 (–1.181 × 10–3) + 900 = 0

θa =

900 – 80000 (– 1.181 × 10–3)

160000

θa = + 6.215 × 10–3 radians. Put θb in (3) because θc is dominant there, we get.

θc = –128000 (– 1.181 × 10–3 ) – 1152

496000

θc = – 2.018 × 10–3 red.

θa = + 6.215 × 10–3 red.

θb = – 1.181 × 10–3 red.

θc = – 2.018 × 10–3 red.

θd = 0 red.

Step 6: End Moments :– Putting values of rotations in generalized slope – deflection equation. Mab = 80,000 (–2 × 6.215 × 10–3+1.181 × 10–3 + 11.25 × 10–3) = 0 KN−m (Check) Mba = 80,000 (+2 × 1.181 × 10–3 – 6.215 × 10–3 + 11.25 × 10–3) = + 592 KN–m Mbc = 128,000 (+ 2 × 1.181 × 10–3 +2.018 × 10–3 – 9 × 10–3 ) = – 592 KN–m

( Note: Mba = Mbc Check is OK )


Mcb = 128,000 (+2 × 2.018 × 10–3 + 1.181 × 10–3 – 9 × 10–3) = – 485 KN–m Mcd = 120,000 (+2 × 2.018 × 10–3 – 0 ) = + 485 KN–m Mdc = 120,000 (0+2.018 × 10–3 ) = + 242 KN–m Note:- A great care should be exercised while putting the direction of end moments in the free body

diagrams and then drawing the composite B.M.D. e.g., a (+ve) end moment would mean that it is counterclockwise at that particular joint or vice versa. After putting the correct directions according to the sign convention, we will decide by the nature of B.M. strictly by keeping in view the sign convention for B.M. (tension at a bottom means +ve B.M.).

A 4m B B 5m C C 4m D

592 592 485 485 242

+148 148 215.4 +215.4 +181.75 181.75- - -148KN 363.4KN 397.15KN 181.75KN

Reactions due toand moments at supportsFinal reaction

A B C D242KN-m

397.15KN15mm

1m 4m 5m 4m

363.4KN

148

0

0

+

++

148181.75

+

`81.75

-

-

215.4592(tension at the

bottom).X

242

B.M.D. (KN-m)

485 (Tension at the top)

X=2.75m

A B C D242KN-m

181.75KN397.15KN

15mm

148KN1m 4m 5m 4m

363.4KN

148

0

0

+

++

148181.75

+

`81.75

S.F.D. (KN)

215.4592(tension at the

bottom).X

242

485 (Tension at the top)

X=2.75m

0+

0+

Elastic curve


POINTS OF CONTRAFLEXURES:– Near B. Span BC

Let it be X.

MX = 592 – 215.4 X = 0 X = 2.75 m Near D. Span DC Let it be X′

MX′ = 242 – 181.75 X′ = 0 X′= 1.33 m

EXAMPLE NO. 5:- Analyze the following beam by slope – deflection method. Draw shear and moment diagrams. Sketch elastic curve. Take I = 400 × 10–6m4

and E = 200 × 106 KN/m2.

SOLUTION :– Consider each span fixed end and compute fixed ended moments. This is a case of continuous beam carrying loads and subjected to settlements.

A

3KN/m

Rab

B

20mm

Rbc

4m

10KN

C

10mm

Rcd

2m

5KN

D

3

I 10

I 2

I8m

6m 8m

6m

2m

5KN

4m

4m

10KN

6m

3KN/m

A B

B C

C D


Step 1: FIXED END MOMENTS

Mfab = 3 × 62 / 12 = 9 KN–m , Mfba = – 9 KN−m

Mfbc = 10 × 42 × 4 / 82 = 10 , Mfcb = –10 KN−m

Mfcd = 5 × 22 × 6/ 82 = 1.875 , Mfdc = –5 × 62 × 2/82 = –5.625 KN−m

Step 2: CALCULATION OF R & 2EI/L TERMS FOR VARIOUS SPANS :– SPAN AB :–

R = + 0.020

6 = + 3.33 × 10–3 rad.

2EIL =

2 × 200 × 106 × (3 × 400 × 10–6 )6 = 80,000 KN–m

SPAN BC :–

R = – 0.02

8 + 0.01

8 = – 1.25 × 10–3 rad

2EIL =

2 × 200 × 106 × (10 × 400 × 10–6 )8 = 200,000 KN–m

SPAN CD:–

R = – 0.01

8 = –1.25 × 10–3 rad

2EIL =

2 × 200 × 106 × (2 × 400 × 10–6)8 = 40,000 KN–m

Step 3: SLOPE – DEFLECTION EQUATIONS:–

Put values of fixed ended moments, Krel and R, we get. Mab = 9 + 80,000 (–2θa – θb + 10 × 10–3).

Mba = –9 + 80,000 (–2θb – θa + 10 × 10–3 )

Mbc = 10 + 200,000 (–2θb – θc – 3.75 × 10–3 ).

Mcb = –10 + 200,000 (–2θc – θb – 3.75 × 10–3 ).

Mcd = 1.875 + 40,000 (–2θc – θd – 3.75 × 10–3 ).

Mdc = –5.625 + 40,000 (–2θd – θc – 3.75 × 10–3 ).

Step 4: JOINT CONDITIONS :–

Joint A ⇒ θa = 0 (Fixed support)

Joint B ⇒ Mba + Mbc = 0 (Continuous support)

Joint C ⇒ Mcb + Mcd = 0 (Continuous support)

Joint D ⇒ Mdc = 0 (Pin support)


Step 5: SIMULTANEOUS EQUATIONS :–

Putting values of Mba, Mbc, Mcb , Mcd and Mdc in terms of θ

– 9 –160,000 θb+800+ 10 – 400,000 θb –200,000 θc – 750 = 0 Mba + Mbc = 0 and θa = 0

– 560,000 θb – 200,000 θc + 51 = 0 → (1)

– 10 – 400,000 θc – 200,000 θb –750 + 1.875 – 80,000 θc – 40,000 θd – 150 = 0

– 200,000 θb – 480,000 θc – 40,000 θd – 908.125=0 Mcb + Mcd = 0→ (2)

– 5.625 – 80,000 θd – 40,000 θc – 150 = 0 Mdc = 0

– 40,000 θc – 80,000 θd – 155.625 = 0 → (3) Writing again

– 560,000 θb – 200,000 θc + 51 = 0 → (1)

– 200,000 θb – 480,000 θc – 40,000 θd – 908.125 = 0 → (2)

– 40,000 θc – 80,000 θd – 155.625 = 0 → (3)

From (1) θb =

51 – 200000 θc

560000 → (4)

From (3) θd =

–155.625 – 40000 θc

80000 → (5)

Put θb and θd in equ. (2) – 200,000

51 – 200000 θc

560000 – 480,000 θc

− 40,000

–155.625 – 40000 θc

80000 – 908.125 = 0 Simplifying

– 18.2143 + 71428.5714 θc – 480,000 θc + 77.8125 + 200000 θc – 908.125 = 0

– 388571.4286 θc – 848.5268 = 0 we get θc = −21.8371 rad. From (4) and (5) θb and θd are calculated.

θc = – 21.8371 × 10–4 rad.θb = + 8.7097 × 10–4 rad.θd = – 8.5346 × 10–4 rad.

Step 6: END MOMENTS :–

Mab = 9+80,000 (–8.7097 × 10–4+10 × 10–3 ) = +739.32 KN–m

Mba =–9+80,000 (–2 × 8.7097 × 10–4 +10 × 10–3 ) = +651.64 KN–m

Mbc = 10+200,000 (–2 × 8.7097 × 10–4+21.8371 × 10–4–3.75 × 10–3) = –651.64 KN–m Mcb = –10+200,000 (+2 × 21.8381 × 10–4–8.7097 × 10–4–3.75 × 10–3) = –60.71 KN–m Mcd = 1.875+40,000 (+2 × 21.8371 × 10–4+8.5346 × 10–4 –3.75 × 10–3) = + 60.71 KN–m Mdc = –5.625+40,000 (+2 × 8.5346 × 10–4+21.8371 × 10–4 –3.75 × 10–3 ) = 0 KN−m


Step 7: SUPPORT REACTIONS:– By applying loads and end moments on free-body diagrams.

A

739.32

3KN/m

6m

651.64KN-m

B

B

651.64KN

-m

8m

4m

10KN

60.71KN-m

C

C

60.71KN-m

8m

2m

5KN

D

240=+9+231.83

222.8384.04

=+9-231.83=+5-89.04 94.04

=89.04+5

8.84=+1.25+7.59

3.84=+3.75-7.59

Net reactions, shear force and bending moment diagrams can now be plotted

Step 8: S.F & B.M. DIAGRAMS & ELASTIC CURVE :–

739.32KN-m

240.83KN

31

-222.83

3KN/m

651.64KN-m

10KN

4m

60.71KN-m

5KN

2m

D

3.84KN

+8.84

C

94.04

20mm

-84.04

101

102.88Kn

21

8m

8m

6m

306.87KN

739.32

0

-

X=3.13m

0

84.04

240.83

+

222.83

-

651.64

+

+

+

-

60.71

94.04

8.84

3.84

S.F.D. (KN).

B.M.D. (KN-m)

X = 0.646 m/

A

B

0+

0+

Elastic curve

Step 9: POINTS OF CONTRAFLEXURE :–

NEAR A: Let it be at X from A in Span AB MX = –739.32 + 240.83X – 1.5X2 = 0 1.5X2 – 240.83X + 739.32 = 0


X = + 240.83 ± (–240.83)2 –4 × 1.5 × 739.32

2 × 1.5

= 240.83 ± 231.44

3

= 3.13 , 157.42

X = 3.13 m

NEAR C: Let it be at X′ from C in Span BC − 60.71 + 94.04 X' = 0, X' = 0.646 m EXAMPLE NO.6:– Analyze the continuous beam shown due to settlement of support B by slope–deflection method. Draw S.F. & B.M. diagrams & sketch the elastic curve. SOLUTION –

A B CD

3KN/m 24KN 12KN

1m 4m 5mm 5m 4m2 4 3I I I

B

E=200X106

KN/m2

I = 400X10-6m4

2.5 m 1m

3KN/m

4m

24KN

2.5m 2.5m

12KN1m3m

A B

B C

C D

Step 1: FIXED END MOMENTS

Mfab = 3 × 42/12 = 4 KN–m , Mfba = – 4 KN−m Mfbc = 24 × 2.52 × 2.5/52 = 15 , Mfcb = – 15 KN−m Mfcd = 12 × 12 × 3/42 = 2.25 , Mfdc = – 12 × 32 × 1/42 = – 6.75 KN−m

Step2: CALCULATION OF R & 2EIL TERMS FOR VARIOUS SPANS:–

Span AB :–

R = + 0.015

4 = 3.75 × 10–3 rad

2EIL =

2(200 × 106) (2 × 400 × 10–6)4 = 80,000 KN–m


Span BC :–

R = – 0.015

5 = – 3 × 10–3 rad.

2EIL =

2 (200 × 106) (4 × 400 × 10–6)5

= 128,000 KN–m Span CD :– R = 0

2EIL =

2 × (200 × 106) (3 × 400 × 10–6)4

= 120,000 KN–m Step 3: SLOPE – DEFLECTION EQUATIONS.

Putting values of fixed end moments, 2EIL and 3R we have.

Mab = 4 + 80,000 (– 2 θa – θb + 11.25 × 10–3)

Mba = – 4 + 80,000 (– 2 θb – θa + 11.25 × 10–3)

Mbc = 15 + 128,000 ( – 2 θb – θc – 9 × 10–3)

Mcb = – 15 + 128,000 (– 2θc – θb – 9 × 10–3)

Mcd = 2.25 +120,000 (– 2θc – θd)

Mdc = – 6.75 + 120,000 (– 2 θd – θc)

Step 4: JOINT CONDITIONS :– Joint A ; Mab = 0 (Pin support) → (1)

Joint B ; Mba + Mbc = 0 (Continuous support) → (2)

Joint C ; Mcb + Mcd = 0 (Continuous support) → (3)

Joint D ; θd = 0 (Fixed end)

Step 5: SIMULTANEOUS EQUATIONS :– 4 – 160,000 θa – 80,000 θb + 900 = 0 ∴ Mab = 0

– 160,000 θa – 80,000 θb + 904 = 0 → (1)

– 4 – 160,000 θb – 80,000 θa + 900 + 15–256,000 θb– 128,000 θc – 1152 = 0 Mba + Mbc = 0

– 80,000 θa – 416,000 θb– 128,000 θc – 241=0 → (2)

– 15 –256,000 θc –128,000 θb –1152 + 2.25 –240,000 θc– 120,000 θd = 0 → (3) Mcb + Mcd = 0

– 128,000 θb – 496,000 θc – 120,000 × 0 – 1164.75 = 0

or – 128,000 θb –496,000 θc – 1164.75 = 0 Putting θd = 0 → (3)


Finally the equations become

– 160,000 θa – 80,000 θb + 904 = 0 → (1)

– 80,000 θa – 416,000 θb – 128,000 θc – 241 = 0 → (2)

– 128,000 θb – 496,000 θc – 1164.75 = 0 → (3)

From (1) θa =

904 – 80000 θb

160000 → (4)

From (3) θc =

–1164.75–128000 θb

496000 → (5)

Put θa & θc from (4) and (5) in (2)

–80,000

904 – 80000 θb

160000 – 416,000 θb – 128,000

–1164.75 – 128000 θb

496000 – 241 = 0

– 452 + 40,000 θb – 416,000 θb + 300.58 + 33032.26θb – 241=0

– 342967.74 θb – 392.42 = 0

θb = – 1.144 × 10–3 radians

From (4) θa =

904 + 80000 × 1.144 × 10–3

160000

θa = + 6.222 × 10–3 rad.

From (5) θc =

– 1164.75 + 128000 × 1.144 × 10–3

496000 = − 2.053 × 10−3 radians.

θc = – 2.053 × 10–3 rad. θa = +6.222 × 10–3 rad. θb = – 1.144 × 10–3 rad. θc = – 2.053 × 10–3 rad. θd = 0 rad. Step 6: END MOMENTS –

Putting the values of Fixed end moments, relative stiffness, and end rotations (θ values) in slope-deflection equations, we have.

Mab = 4 + 80,000 (– 2 x 6.222 × 10–3 + 1.144 × 10–3+ 11.25 × 10–3) = 0 KN-m Mba = – 4 + 80,000 (+ 2 × 1.144 × 10–3 – 6,222 × 10–3+ 11.25 × 10–3) = + 581 KN-m Mbc = 15 + 128,000 (+2 × 1.144 × 10–3 + 2.053 × 10–3– 9 × 10–3 ) = –581 KN-m Mcb = – 15 + 128,000 (+ 2 × 2.053 × 10–3 + 1.144 × 10–3– 9 × 10–3) = – 495 KN-m


Mcd = 2.25 + 120,000 (+ 2 × 2.053 × 10–3)= + 495 KN-m Mdc = – 6.75 + 120,000 (+2.053 × 10–3) = −495 KN-m Now plot SFD, BMD and sketch elastic curve by applying loads and end moments to

free-body diagram.


+145.25 -145.25 -215.2 +215.2 +183.75 -183.75 reaction due to end moments (KN)

+6 +6 +12 +12 +3 +9

+151.25 =139.25 -203.2 +227.2 +186.75 -174.75 final reactions (KN)

-342.45 +413.95

3KN/m 24KN 12KN581 581 495 495 2401m2.5m

reaction due to applied loads (KN)

Note: Reactions due to loads and end moments have been calculated separately and then added up appropriately.

A B CD

3KN/m 24KN 12KN2.5m 1m

151.25KN1m 4m 5m 4m

413.95KN 174.75KN

151.25 139.25 186.75 174.75174.75KN

0+

+

+

+

1.37m

240

995=2.86mX

203.2227.2 227.2

581

342.45KN

0+

0+

S.F.D. (KN)

B.M.D. (KN-M)

Elastic curve.

X = 1.37 m/

POINTS OF CONTRAFLEXURES :– Near B :– Span AB Let it be ‘X’ MX = 581 – 203.2 X = 0 X = 2.86 m


Near D :– Span CD Let it be X′ Mx′ = 240 – 174.75 X′ = 0 X′ = 1.37 m These have been shown on BMD. EXAMPLE NO. 7:– Analyze the continuous beam shown due to the settlement of support B alone by slope–deflection method. Draw S.F. & B.M. diagrams & sketch the elastic curve. SOLUTION :–

A B C D

1m 4m 5m 4m2I B 4 3I I

E=200X10 KN/m6 2

I=400X10 m-6 4

15mm

Step 1: FIXED END MOMENTS :–

Mab = Mfab + 2EIL (– 2 θa – θb + 3 R) __ A generalized slope–deflection equation.

As there is no applied loading on the beam, therefore, all fixed end moment terms in the slope–deflection equation will be equal to zero.

Step 2: CALCULATION OF R AND 2EIL TERMS FOR VARIOUS SPANS

Span AB :–

R = + 0.015

4 = + 3.75 × 10–3 rad.

2EIL =

2(200 × 106 ) (2 × 400 × 10–6 )4 = 80,000 KN–m

Span BC :–

R = – 0.015

5 = – 3 × 10–3 rad.

2EIL =

2(200 × 106 )(4 × 400 × 10–6)5 = 128,000 KN–m

Span CD :– R = 0 rad. (Both points C and D are at the same level)

2EIL =

2(200 × 106 )(3 × 400 × 10–6 )4 = 120,000 KN–m


Step 3: SLOPE–DEFLECTION EQUATIONS :–

Putting 2EIL and 3R values, we have.

Mab = 80,000 (– 2θa – θb – 11.25 × 10–3 )

Mba = 80,000 (– 2 θb – θa + 11.25 × 10–3 )

Mbc = 128,000 ( – 2 θb – θc – 9 × 10–3 )

Mcb = 128,000 (– 2 θc – θb – 9 × 10–3 )

Mcd = 120,000 (– 2 θc – θd)

Mdc = 120,000 (– 2 θd – θc )


Joint A ; Mab = 0 (Pin support)

Joint B ; Mba + Mbc = 0 (Continuous support)

Joint C ; Mcb + Mcd = 0 (Continuous support)

Joint D ; Mdc = 0 (Pin support)


Putting joint conditions in Slope – deflection equation, we have (Mab = 0)

– 160,000 θa – 80,000 θb + 900 = 0 → (1)

(Mba + Mbc = 0) – 160,000 θb – 80,000 θa + 900 – 256,000 θb –128,000 θc – 1152 = 0 – 80,000 θa – 416,000 θb – 128,000 θc–252=0 → (2)

(Mcb + Mcd = 0)

– 256,000 θc – 128,000 θb – 1152 – 240,000 θc – 120,000 θd = 0 – 128,000 θb – 496,000 θc – 120,000 θd – 1152=0 → (3)

(Mdc = 0)

– 240,000 θd – 120,000 θc = 0 – 120,000 θc – 240,000 θd = 0 → (4)

Re-writing – 160,000 θa – 80,000 θb + 0 + 0 + 900 = 0 → (1) – 80,000 θa – 416.000 θb – 128,000 θc + 0 – 252 = 0 → (2) 0 – 128,000 θb – 496,000 θc–120,000 θd–1152 = 0 → (3) 0 + 0 – 120,000 θc – 240,000 θd + 0 = 0 → (4)

From (1) θa =

900 – 80000 θb

160000 → (5)

From (4) θd = – 120000 θc

240000

θd = – 0.5 θc → (6)


Put (5) in (2)

– 80,000

900 – 80000 θb

160000 – 416,000 θb – 128,000 θc –252 =0

– 50 + 40,000 θb – 416,000 θb – 128,000 θc–252 = 0 – 376,000 θb – 128,000 θc – 702 = 0 → (7) Put (6) in (3) – 128,000 θb – 496,000 θc – 120,000 ( – 0.5 θc) – 1152 = 0 – 128,000 θb – 436,000 θc – 1152 = 0 → (8) From (7)

θb =

–702 – 128000 θc

376000 → (9)

Put θb from equation (9) in (8), we have.

– 128,000

– 702 – 128000 θc

376000 – 436,000 θc –1152 = 0

238.98 + 43574.47 θc – 436,000 θc – 1152 = 0

– 392,425.53 θc – 913.02 = 0

θc = – 2.327 × 10–3 radians.

from (9) θb =

– 702 + 128000 × 2.327 × 10–3

376000

θb = – 1.075 × 10–3 rad. Now calculate other rotations from equations.

from (5) θa =

900 + 80000 × 1.075 × 10–3

160000

θa = + 6.162 × 10–3 rad.

from (6) θd = – 0.5 (– 2.327 × 10–3)

θd = + 1.164 × 10–3 rad.

Final values of end rotations are:

θa = + 6.162 × 10–3 rad.

θb = – 1.075 × 10–3 rad.

θc = – 2.327 × 10–3 rad.

θd = + 1.164 × 10–3 rad.


Step 6: END MOMENTS :– Putting values of rotations in slope-deflection equations.

Mab = 80,000 (– 2 × 6.162 × 10–3 +1.075 × 10–3+11.25 × 10–3 = 0 KN-m

Mba = 80,000(+2 × 1.075 × 10–3 – 6.162 × 10–3+11.25 × 10–3 ) = +579 KN-m

Mbc = 128,000 (+2 × 1.075 × 10–3 +2.327 × 10–3 – 9 × 10–3 ) = –579 KN-m

Mcb = 128,000 (+2 × 2.327 × 10–3 +1.075 × 10–3 – 9 × 10–3 ) = – 419 KN-m

Mcd = 120,000 (+2 × 2.327 × 10–3 – 1.164 × 10–3 ) = + 419 KN-m

Mdc = 120,000 (–2 × 1.164 × 10–3 + 2.327 × 10–3 ) = 0 KN-m


579 579 419 419

+144.75 144.75 199.6 +199.6 +104.75 104.75- - -

144.75KN 344.35 KN 304.35 KN 104.75KN

Reaction due toend moments

Final reaction

A B C D

104.75KN304.35KN

1m 4m 5m 4m

144.75KN

144.75 144.75

344.35KN

104.75 104.75S.F.D. (KN)

++

199.6579 199.6

+

X=2.9m419

0+

0+ B.M.D. (KN-m)

Near B :– Span BC Let it be at ‘X’ from B.

MX = 579 – 199.6 X = 0

X = 2.9 m


4.5. APPLICATION TO FRAMES (WITHOUT SIDE SWAY):–

Lateral Loads UnsymmetericalLoad

(Side sway Present)

2I 2I

(i) side sway present (ii) I is same but support conditionsare different (side sway present)

Centre line

Load is symmetrical and

The side sway (relative displacement of two ends of a column) or the horizontal movement of the structure may become obvious once the structure and the loading is inspected in terms of inertia, E values and support conditions etc. However, following are the rules and guide lines which may be followed for deciding whether side sway is present or not. (1) In case of symmetrical frames subjected to symmetrical loading, the side sway may be neglected for columns having equal inertia values if support conditions are same.

(2) If a force is applied in horizontal direction to a symmetrical frame where no arrangement exists for preventing horizontal movement, the side sway must be considered.(with reference to all these diagrams).

(3) An unsymmetrical frame subjected to symmetrical loading might be considered to have side sway.

4.6. UNSYMMETRICAL FRAME :– “An unsymmetrical frame is that which has columns of unequal lengths and different end conditions and moment of inertia the load may be symmetrical or unsymmetrical.” 4.7. STIFFNESS :– “Stiffness can be defined as the resistance towards deformation which is a material, sectional and support parameter.” More is the stiffness, less is the deformation & vice versa. Stiffness attracts loads / stresses. The stiffness is of various types :

(1) Axial stiffness (AE).

(2) Flexural stiffness (EI).

(3) Shear stiffness (AG).

(4) Torsional stiffness (GJ).


EXAMPLE NO. 8:– Analyze the rigid frame shown by slope–deflection method.

A2m 3m

10KN 2KN/m

CB

D

2 3mI4m2I3I

SOLUTION :– Examining loads and support conditions, horizontal moment is not possible.

Step 1: Relative Stiffness :–

Member I L IL Krel.

AB 3 5 35 × 30 18

BC 2 4 24 × 30 15

BD 2 3 23 × 30 20

Step 2: Fixed End Moments :–

2m 3m

4m

2KN/m

10KNA B

B C

Mfab = 10 × 32 × 2/52 = 7.2 KN–m , Mfba = – 10 × 22 × 3/52 = –4.8 KN-m

Mfbc = 2 × 42/12 = 2.67 KN-m, Mfcb = – 2.67 KN-m

Mfdb = Mfdb = 0 (There is no load acting within member BD)

Step 3: Generalized Slope – deflection Equation :– Put values of fixed end moments.

Mab = 7.2 + 18(–2 θa – θb) = 7.2 – 36 θa – 18 θb

Mba = – 4.8 + 18 (– 2 θb – θa)= – 4.8 – 36 θb – 18 θa.

Mbc = 2.67 + 15 (– 2 θb – θc) = 2.67 – 30 θb – 15 θc.

Mcb = – 2.67 + 15 (–2 θc – θb) = – 2.67 – 30 θc – 15 θb

Mbd = 0 + 20 (– 2 θb – θd) = – 40 θb – 20 θd

Mdb = 0 + 20 (– 2 θd – θb) = – 40 θd – 20 θb.


Step 4: Joint Conditions:– Joint A : θa = 0 (Being fixed end) Joint B : Mba + Mbc + Mbd = 0 → (1) Continuous joint Joint C : Mcb = 0 (Pin end) → (2) Joint D : θd = 0 (Fixed end)

Step 5: Simultaneous equations Putting above joint conditions in slope deflection equations, we have. – 4.8–36 θb –18 θa+2.67 30 θb –15 θc – 40θb–20 θd = 0 → (1)

Mba + Mbc + Mbd = 0 Put θd = 0 and θa = 0.

– 4.8 – 36 θb – 0 + 2.67 – 30 θb – 15 θc – 40θb – 0 = 0 –106 θb – 15 θc – 2.13 = 0 → (1)

(Mcb = 0) – 2.67 – 30 θc – 15 θb = 0 → (2)

– 15 θb – 30 θc – 2.67 = 0 → (2) – 106 θb – 15 θc – 2.13 = 0 → (1)

– 15 θb – 30 θc – 2.67 → (2) Multiply (1) by 2 and subtract (2) from (1) – 212 θb – 30 θc – 4..26 = 0

15 θb 30 θc 2.67 = 0

– 197 θb – 1.59 = 0

θb = – 8.07 × 10–3 rad.

From (1) ⇒ – 106 (– 8.07 × 10–3 ) – 15 θc – 2.13 = 0

θc = – 84.96 × 10–3 rad.

θa = 0 rad.

θb = – 8.07 × 10–3 rad.

θc = – 84.96 × 10–3 rad.

θd = 0 rad.

Step 6: End moments. Putting values of FEM and rotations in slope-deflection equations.

Mab = 7.2 – 36 (0) – 18(–8.07 × 10–3 )= + 7.345 KN–m

Mba = – 4.8 – 36(– 8.07 × 10–3 )– 18 (0) = – 4.509 KN–m

Mbc = 2.67–30(–8.07×10–3 )–15 (–84.96 × 10–3 ) = + 4.187 KN–m

Mcb =– 2.67 – 30 (–84.96 × 10–3 ) – 15 (– 8.07 × 10–3 ) = 0

Mbd = – 40(–8.07 × 10–3 ) – 20 (0) = + 0.323 KN–m

Mdb = – 40(0) – 20 (–8.07 × 10–3 ) = + 0.161 KN–m


Draw SFD , BMD and sketch elastic curve.

0.167.345

2m 3m

10KN 2 KN/m

CA B B 4m8.48

4.509 4.187

+6 +4 +4 +4

+0.57 -0.57 +1.05 -1.056.57 3.44 0.16 5.05 2.95

3m0.16D

0.161

8.48

0.33B

2m 3m

6.57KN3.43KN4.509 KN-m

4.187KN-m2KN/m

4m 2.96KN5.05KN

5.04 1.48S.F.D. (KN)X=

++- -

6.57S.F.D. (KN)

03.433.43

5.795X=1.12m

6.57

Vx=2.96-2X=0x=2.96-2x=0Mx=2.96x1.48

-1.48 =2.190W-m

0

2.96

2.190B.M.D. (KN-m)

X=1.31m

B.M.D. (KN-m)

7.34 4.5094.187X=1.31m

Point of contraflexureX=1.12M Mx=2.96x-x=0

X(2.96-X)=0

-0X=2.96m

Mx=4.509x3.43Either x=0++

7.345KN 10KN

0+0

+

0+

0+

A B B C

0 00.16

1KN

-m3m

0.32

3KN

-m

0.16

KN

0.16

0.16 S.

F.D

. (KN

)

0.32

3 B.M

.D. (

KN

-m)

0.16

1

+X=

1m

0.16

KN

0+ 0+

BD


A

D

CB

+ 0.165.04

0.43

S.F.D.2.963.43

A

D

C

B.M.D.

0.523 B

4.187

2.195.795

+ +-

4.5

7.345

1

ELASTIC CURVE

6.57

EXAMPLE NO. 9:– Analyze the rigid frame shown by slope–deflection method

5KN

A

B C2m 2m

10KN

1.5m

1.5m

2I

2I

3I


SOLUTION:– Inspecting loads and support conditions, horizontal displacement is not possible. Step 1: Relative Stiffness :–

Member I L IL Krel.

AB 2 3 23 × 12 8

BC 3 4 34 × 12 9

Step 2: Fixed End Moments :–

Mfab =

5 × 1.52 × 1.532 = + 1.875 KN–m

Mfba = – 1.875 KN–m

Mfbc = 10 × 22 × 2

42 = + 5 KN–m

Mfcb = – 5 KN–m Step 3: Generalized Slope–deflection Equations :–

Put values of fixed end moments and Krel.

Mab = 1.875 + 8 (– 2 θa – θb)

Mba = – 1.875 + 8 (– 2 θb – θa )

Mbc = 5 + 9 (– 2 θb – θc)

Mcb = – 5 + 9 (– 2 θc – θb)

Step 4: Joint Conditions :–

Joint A : θa = 0 Being fixed End.

Joint B : Mba + Mbc = 0 Continuous end.

Joint C : θc = 0 Being fixed End. Step 5: Simultaneous Equations :– Put θa = θc = 0 in the joint condition at B. Mba + Mbc = 0 – 1.875 – 16 θb – 0 + 5 – 18 θb – 0 = 0

3.125 – 34 θb = 0

θb = + 0.092 radians.

θa = 0

θc = 0


Step 6: End moments. Put values of rotations in slope-deflection equations. Mab = 1.875 + 8 (0 –0.092) = + 1.140 KN–m Mba = – 1.875 + 8 (– 2 x 0.092 – 0 ) = – 3.346 KN–m Mbc = 5 + 9 (– 2 x 0.092 – 0 ) = + 3.346 KN–m Mcb = – 5 + 9 ( 0 – 0.092) = – 5.827 KN–m

Now draw SFD , BMD and sketch elastic curve. Doing it by-parts for each member.

3.235 2m 2m 3.23510KN 5.8273.346

-0.6204.380 KN

+0.6205.62 KN

+5 +5

5KN

1.5m

1.5m

3.346

4.38

+2.5+0.7353.235 KN

-0.7351.765 KN

+2.5

4.38

1.140

SHEAR FORCE AND B.M. DIAGRAMS

B B

0 0 0

C A2m 2m10KN 5.827KN-m3.346KN-m 3.346KN-m

4.380KN

4.384.38+

+ +

+

5.625.62S.F.D.

5KN

1.765KN

1.14KN.m1.5m 1.5m

5.62KN 3.235KN

1.765

0S.F.D.

3.235

1.508

0B.M.D.3.3461.14

0 00

3.346

Mx=3.346 +4.38X=0X=0.764m

Mx=5.62X -5.827=0X=1.037m

X=0.764m X =1.037m/

5.414


S.F.D BMD

A

B C

1

+

5.623.235

4.38

+

1.765

5.414

+

5.827

Elastic Curve+1.508

1 3.346

3.346

1.14

4.8. FRAMES WITH SIDE SWAY – SINGLE STOREY FRAMES :–

For columns of unequal heights, R would be calculated as follows:

B

A

P

D

C

4I

2 L3I

Rab

HA

Hd

RcdL2

Rab =

Rcd = L2

L1I

L1

To show the application to frames with sidesway, let us solve examples. EXAMPLE NO. 10:– Analyze the rigid frame shown by slope–deflection method.

A D

CB 2m 5m5KN

4I

I3m 3mI


SOLUTION:– Step 1: Relative Stiffness :–

Member I L IL Krel.

AB 1 3 13 × 21 7

BC 4 7 47 × 21 12

CD 1 3 13 × 21 7

Step 2: Relative Values of R :–

Rab = Rcd = ∆3 = Rrel or R (columns are of 3m length)

Mab = Mfab + Krelab (– 2 θa – θb + Rrel)

Mba = Mfba + Krelab (– 2 θb – θa + Rrel)

Other expressions can be written on similar lines.

NOTE :– In case of side sway, R values are obtained for columns only because the columns are supposed to prevent (resist) side sway not beams. Step 3: Fixed End Moments :–

Mfbc = 5 × 52 × 2

72 = 5.10 KN–m

Mfcb = −5 × 22 × 5

72 = – 2.04 KN–m

All other F.E.M. are zero because there are no loads on other Spans. i.e. Mfab = Mfba = 0 & Mfcd = Mfdc = 0 Step 4: Slope – deflection Equations :– Putting values of FEM’s while R will now appear as unknown.

Mab = 0 + 7 (– 2θa – θb + R)

Mba = 0 + 7 (– 2 θb – θa + R)

Mbc = 5.1 + 12 (– 2 θb – θc)

Mcb = – 2.04 + 12 ( – 2 θc – θb)

Mcd = 0 + 7 (– 2 θc – θd + R)

Mdc = 0 + 7 (– 2 θd – θc + R)


Step 5: Joint Conditions :– Joint A : θa = 0 (Fixed joint) Joint B : Mba + Mbc = 0 (Continuous joint) → (1) Joint C : Mcb + Mcd = 0 (Continuous joint) → (2) Joint D : θd = 0 (Fixed joint) Step 6: Shear Conditions :–

B

AMab

Mba Mcd

3m3m

D

C

Mdc3

Mab + Mba3

Mdc + McdHa = Hd =

Fx=0Ha + Hd = 0

NOTE: Shear forces are in agreement with direction of ∆. The couple constituted by shears is balanced by the direction of end moments. (Reactive horizontal forces constitute a couple in opposite direction to that of end momens). ∑ Fx = 0 Ha + Hd = 0 Write in terms of moments. Mab + Mba + Mdc + Mcd = 0 → (3) Apply equations (1), (2) & (3) and solve for θb, θc & R. Equation (3) is also called shear condition. Step 7: Simultaneous Equations :– Put θa and θd equal to zero in joint conditions for B and C in terms of end moments. Mba + Mbc = 0

so – 14 θb + 7 R + 5.1 – 24 θb – 12 θc = 0 → (1) Mcb + Mcd = 0

– 38 θb – 12 θc + 7 R + 5.1 = 0 – 2.04 – 24 θc – 12 θb – 14 θc + 7 R = 0

or – 12 θb – 38 θc + 7 R – 2.04 = 0 → (2)

Mab + Mba + Mdc + Mcd = 0

– 7 θb + 7 R – 14 θb + 7 R – 7 θc + 7 R – 14 θc + 7R=0

– 21 θb – 21 θc + 28 R = 0 or – 3 θb – 3 θc + 4 R = 0 → (3)

re-writing the equations again.

– 38 θb – 12 θc + 7 R + 5.1 = 0 → (1)

– 12 θb – 38 θc + 7 R – 2.04 = 0 → (2)

– 3 θb – 3 θc + 4 R = 0 → (3)


Subtract (2) from (1) – 38 θb – 12 θc + 7 R + 5.1 = 0

– 12 θb – 38 θc + 7 R – 2.04 = 0 – 26 θb + 26 θc + 7.14 = 0 → (4) Multiply (2) by 4 & (3) by 7 & subtract (3) from (2)

– 48 θb – 152 θc + 28 R – 8.16 = 0 → (2)

21 θb 21 θc ± 28 R = 0 → (3) – 27 θb – 131 θc – 8.16 = 0 → (5) From (4)

θb = 26 θc + 7.14

26 put in ( 5) and solve for θc

– 27 26 θc + 7.14

26 – 131 θc – 8.16 = 0 → (6)

– 27 θc – 7.415 – 131 θc – 8.16 = 0

– 158 θc – 15.575 = 0

θc = – 0.0986 rad.

From (6), θb = – 26 × 0.0986 + 7.14

26

θb = + 0.1760 rad.

From (1) – 38 (0.1760) – 12 (–0.0986)+7R+5.1 = 0

R = + 0.0580

So finally, we have. ∴ θa = 0 θb = + 0.1760 θc = – 0.0986 θd = 0 R = + 0.0580

END MOMENTS :–

Putting above values of rotations and R in slope deflection equations, we have. Mab = 7 (0– 0.176 + 0.058) = – 0.826 KN–m Mba = 7 (– 2 × 0.176 – 0 + 0.058) = – 2.059 KN–m Mbc = 5.1 + 12 (– 2 × 0.176 + 0.0986) = + 2.059 KN–m Mcb = – 2.04 + 12 (+ 2 × 0.0986 – 0.176) = – 1.786 KN–m Mcd = 7 (+ 2 × 0.0986 – 0 + 0.058) = + 1.786 KN–m Mdc = 7 ( 0 + 0.0986 + 0.058) = + 1.096 KN–m Draw SFD , BMD and sketch elastic curve.


SHEAR FORCE & B.M. DIAGRAMS :– By Parts

0.9622.059

2m 5m

5KN 1.7860.962

3.61 +3.571 +1.429 1.39+0.039 -0.039

3.61 1.392.059 1.786

C

DA

B

3m 3m

1.096

1.39

-0.9620.826

3.6

+0.962 +0.961

B C

5KN

2m 5m 1.786KN-m

3.61KN

2.059KN-m

3.61

+

1.391.39

S.F.D.

0.962

1.39KN

5.1610.862+

00

2.059 1.786B.M.D.

X=0.57m X=1.28m

Mx=-2.059 +3.61X=0X=0.57mMx=-1.786 +1.39X =0X =1.28m

/

/

B C

3.61

X=0

.86

m

2.05

9KN

-m

3m

00 0.

962

0.96

2 K

N0.

962

KN

B.M

.D.

+2.

059

X=0

.86m

Mx=

0.82

6-0.

962X

=0

AB

0.82

6 kN

-m

0.82

6


1.78

6-

+

+

00 0

0

1.09

6 B.M

.D.

0.96

1KN

0.96

1KN

1.09

6KN

-m1.

786K

N-m 3m

1.78

6

+

+

00 0

0

1.09

6

0.96

10.

961 B.

M.D

.

0.96

1KN

0.96

1KN

1.09

6KN

-m1.

786K

N-m 3m

CD

Super impositing member SFD’s and BMD’s.

B1

1

+

+

A D

C

3.61

0.962

0.962

S.F.D.

0.961

B.M.D.

2.059

2.059

1.786

1.786

1.096

5.161

0.862

1

A D

CB

1

+

++

SFD BMD

ELASTIC CURVE:-


EXAMPLE NO. 11:– Analyze the rigid frame shown by slope–deflection method.

A BC

D

E

F

I5m

7m

7m

2 2I I

3m I5m

20 KN

2m

I

SOLUTION:– Step 1: FIXED END MOMENTS :–

Mfbc = 20 × 22 × 5

72 = + 18.16 KN–m

Mfcb = 20 × 52 × 2

72 = – 20.41 KN–m

Mfad = Mfda = 0 | Mfbe = Mfeb = 0 | Mfab = Mfba = 0 | As there are no loads on these spans. Mfcf = Mffc = 0 |

Step 2: RELATIVE STIFFNESS:–

Member I L IL Krel.

AB 2 7 27 × 105 30

BC 2 7 27 × 105 30

AD 1 5 15 × 105 21

BE 1 3 13 × 105 35

CF 1 5 15 × 105 21


Step 3: RELATIVE VALUES OF R :–

Member ∆ L ∆L Rrel.

AB 0 7 0 0

BC 0 7 0 0

AD ∆ 5 ∆5 × 15 3 R

BE ∆ 3 ∆3 × 15 5 R

CF ∆ 5 ∆5 × 15 3 R

Step 4: SLOPE–DEFLECTION EQUATIONS :– Putting the values of fixed end moments.

Mab = 0 + 30 (– 2 θa – θb) = – 60 θa – 30 θb

Mba = 0 + 30 (– 2 θb – θa) = – 60 θb – 30 θa

Mbc = 8.16 + 30 (– 2 θb – θc) = 8.16 – 60 θb – 30 θc

Mcb = – 20.41 + 30(– 2 θc – θb) = – 20.41–60 θc–30 θb

Mad = 0 + 21 (– 2 θa – θd + 3 R) = – 42 θa + 63 R

Mda = 0 + 21 (– 2 θd – θa + 3 R) = – 21 θa + 63 R

Mbe = 0 + 35 (– 2 θb – θe + 5 R) = – 70 θb + 175 R

Meb = 0 + 35 (– 2 θe – θb + 5 R) = – 35 θb + 175 R

Mcf = 0 + 21 (– 2 θc – θf + 3 R) = – 42 θc + 63 R

Mfc = 0 + 21 (– 2 θf – θc + 3 R) = – 21 θc + 63 R


Joint A : Mad + Mab = 0 (Continuous joint) → (1)

Joint B : Mba + Mbc + Mbe = 0 (Continuous joint) → (2)

Joint C : Mcb + Mcf = 0 (Continuous joint) → (3)

Joint D : θd = 0 (Fixed end)

Joint E : θe = 0 (Fixed end)

Joint F : θf = 0 (Fixed end)


Step 6: SHEAR CONDITIONS :–

A B C

D F

E5m 5m

3m

Meb

Mfc

McfMbeMad

MdaHd=Mda+Mad

5

Hf= Mfc+Mcf5

He=Meb+Mbe3

∑ FX = 0 Hd + He + Hf = 0, Now put Hd, He and Hf in terms of end moments. We have

Mda + Mad

5 + Meb + Mbe

3 + Mfc + Mcf

5 = 0

or 3 Mda + 3 Mad + 5 Meb + 5 Mbe + 3 Mfc + 3 Mcf = 0 → (4) Step 7: SIMULTANEOUS EQUATIONS :– Putting end conditions in above four equations. We have

(Mad + Mab = 0)

so – 42 θa + 63 R – 60 θa – 30 θb = 0

– 102 θa – 30 θb + 63 R = 0 → (1)

Mba + Mbc + Mbe = 0

so – 60 θb – 30 θa + 8.16 – 60 θb – 30 θc – 70 θb + 175 R = 0

– 30 θa – 190 θb – 30 θc + 175 R + 8.16 = 0 → (2)

Mcb + Mcf = 0 so – 20.41 – 60 θc – 30 θb – 42 θc + 63 R = 0

– 30 θb – 102 θc + 63 R – 20.41 = 0 → (3)

3Mda + 3Mad + 5Meb + 5Mbe + 3Mfc + 3Mcf = 0 so, 3(–21 θa+63 R)+3(–42 θa+63 R) +5(–35 θb+175 R – 70 θb+175 R) +3(–21 θc+63 R – 42 θc+63 R)=0

–63 θa + 189 R – 126 θa + 189 R – 175 θb + 875 R – 350θb + 875 R – 63 θc + 189 R – 126 θc + 189 R = 0

– 189 θa – 525 θb – 189 θc + 2506 R = 0 → (4)

(not a necessary step). Writing in a matrix form to show that slope-deflection method is a stiffness method.

We get a symmetric matrix about leading diagonal.

– 102 θa – 30 θb + 0 + 63 R + 0 = 0

– 30 θa – 190 θb – 30 θc + 175 R + 8.16 = 0

0 – 30 θb – 102θc + 63 R – 20.41 = 0


–189 θa – 525 θb – 189θc + 2506 R = 0

–102 θa – 30θb + 63 R = 0 → (1) –30θa – 190θb – 30θc + 175 R + 8.116 = 0 → (2) –30θb – 102θc + 63 R – 20.41 = 0 → (3) –189θa – 525θb – 189θc + 2506 R = 0 (4) Solve the above equations, find end moments and hence draw, S.F, B.M, elastic curse diagrams. Solving aboving 4 equations, following values, are obtained.

θa = –0.024924, θb = 0.0806095, θc = –0.225801, R = –0.00196765.( use programmable calculator or Gausian elimination)

Putting these values in step 4, nodal moments may be calculated as follows: Mab = 0 + 30 (–2θa – θb) = –60θa – 30θb = –60(–0.024924) –30 (0.0806095) = 0.923 KN-m. Mba = –60θb – 30θa = –60(.0806095) –30(–0.024924) = –4.089 KN-m. Mbc = 8.16–60 (.0806095) –30 (–0.225801) = 10.097 KN-m. Mcb = –20.41 – 60 (–.225801) –30 (0.0806095) = 0.928 KN-m. Mad = –42 (–.024924) +63 (–.00196765) = 0.923 KN-m. Mda = –21(–.024924) +63 (–.00196765) = 0.3994 KN-m. Mbe = –70 (.0806095) +175 (–.00196765) = –5.987 KN-m. Meb = –35(0.0806095) + 175 (–.00196765) = 3.166 KN-m. Mef = –42(0.225801) +63 (–.001968) = –9.60 KN-m. Mfc = –21 (–0.2258) +63 (–.00197) = 4.12 KN-m. SFD, BMD and elastic curve can be sketched now as usual. 4.9. DOUBLE STOREYED FRAMES WITH SIDE SWAY( GENERALIZED TREATMENT) FOR R VALUES.

L1 L1

L2

P1 C 1 D 1

HE

Hb

B

P2

E

2

Ha

F

A

L3

21

2 2

1

HF

Rbc = Red = ∆1 – ∆2

L1

Rab = ∆2L2

Ref = ∆2L3

If L1 = L3

Then Rab = ∆2L2

, Ref = ∆2L1


4.9.1. SHEAR CONDITIONS FOR UPPER STOREY :–

C

B E

D

Mbc Med

L1 L1

Mcb Mde

Hb=Mbc+McbL1

He=Med+MdeL1

P1-Hb-He=0

P1

Hb He

ΣFX = 0 Hb and He can be written in terms of end moments as above. Applied load upto Section-1-1.

4.9.2. SHEAR CONDITIONS FOR LOWER STOREY :–

Ha=Mba+MabL2

HF=Mef+MfeL3A

B E

F

Mba

(P + P ) Ha Hf=01 2 - -

Mef

Mfe

L2 L3

Mab

P2

ΣFX = 0 Applied shear is to be considered upto Section 2-2. To demonstrate the application, let us solve the following question.

EXAMPLE NO. 12:- Analyze the following frame by slope – deflection method. Consider: I = 500 × 10–6m4 , E = 200 × 106 KN/m2 It is a double story frame carrying gravity and lateral loads.

8mA

F

8m2I

2I 6m

BE

DC10KN

24KN/m

5I24KN/m

2I2I6m

2

2

1

1

5I


SOLUTION :–

Step 1: Relative Stiffness:–

Member I L IL Krel

AB 2 8 28 × 24 6

BC 2 6 26 × 24 8

CD 5 8 58 × 24 15

DE 2 6 26 × 24 8

EF 2 6 26 × 24 8

BE 5 8 58 × 24 15

Step 2: Relative Values of R. For upper story columns

Rbc = Rde = ∆1 – ∆2

6 = R1 (Say)

Rab = ∆28 × 24 Ref =

∆26 × 24

Rab = 3 R 2 (say) Ref = 4 R2 (Say) Because lower story columns have different heights. Step 3: F.E.M :– F.E.M.s are induced in beams only as no loads act within column heights.

Mfbe = Mfcd = 24 × 82

12 = + 128 KN–m

Mfeb = Mfdc = – 128 KN–m Step 4: Slope – Deflection Equations :– Put values of FEM’s and R Values for columns. MAB = 0 + 6 ( –2θa – θb + 3 R2)

MBA = 0 + 6 ( – 2θb – θac + 3 R2)

MBC = 0 + 8 (–2θb– θc + R1)


MCB = 0 + 8 (–2θc – θb + R1)

MCD = 128 + 15 ( – 2θc – θd )

MDC = – 128 + 15 (– 2θd – θc)

MDE = 0 + 8 ( – 2θd – θe + R1)

MED = 0 + 8 ( – 2θe – θd + R1)

MEF = 0 + 8 ( – 2θe – θf + 4R2 )

MFE = 0 + 8 ( – 2θf – θe + 4R2 )

MBE = 128 + 15 (– 2θb – θe)

MEB = – 128 + 15 ( – 2θe – θb) Step 5: Joint Conditions :–

Joint A: θa = 0 (Fixed joint)

Joint B: MBA + MBC + MBE = 0 → (1)

Joint C: MCB + MCD = 0 → (2)

Joint D: MDC + MDE = 0 → (3)

Joint E: MED + MEB + MEF = 0 → (4)

Joint F: θf = 0 (Fixed joint) Step 6: Shear Conditions :– For Upper Storey :–

MCB M DE

C D

6m 6m

B EM BC MED

HB =MBC+MCB

6HE =

MED+MDE6

HdHc

Hb He

10

ΣFX = 0, 10 – Hb –He =0 putting values of Hb and He interms of end moments and simplifying, we get.

60 – MBC – MCB – MED – MDE = 0 → (5)


For Lower Storey.

8m 6m

FA

MBA MEF

EB

MFEMAB

HA =MAB+MBA

8HF =

MFE+MEF6

HeHb

Ha Hf

Σ FX = 0, 10 – Ha – Hf = 0

Putting the values of Ha and Hf in terms of end moments and simplifying, we get.

480 – 6 MAB – 6 MBA – 8 MFE – 8 MEF = 0 → (6) Now we have got six equations and Six unknowns. (θb, θc, θd, θe, R1, R2) Step 7: Simultaneous Equations :– Putting joint conditions in slope deflection equations we have. Mba + Mbc + Mbe = 0, – 12θB+18 R2 – 16θB – 8θC+8R1+128 – 30θB –15θE = 0 or – 58θB – 8 θC – 15θE + 8R1 + 18R2 + 128 = 0 → (1) Mcb + Mcd = 0 – 16θC – 8θB + 8R1 + 128 – 30θC – 15θD = 0 or – 8θB – 46θC – 15θD + 8R1 + 128 = 0 → (2) Mdc + Mde = 0 – 128 – 30θD – 15θC – 16θD – 8θE + 8 R1 = 0 or – 15θC – 46θD – 8θE + 8 R1 – 128 = 0 → (3) Med + Meb + Mef = 0 – 16θE – 8θD + 8 R1 – 128 – 30θE – 15θB – 16θE + 32R2 = 0 or – 15θB – 8θD – 62θE + 8 R1 + 32R2 – 128 = 0 → (4) Putting expressions of end moments in equations 5 and 6 , we have.

60 – (–16θB–8θC + 8 R1 – 16θC – 8θB + 8R1) – (– 16θE – 8θD + 8R1 – 16θD – 8θE + 8R1) = 0 or 60 + 16θB + 8θC – 8R1 + 16θC + 8θB – 8 R1 + 16θE + 8θD – 8R1 + 16θD + 8θE – 8R1 = 0 or 24θB + 24θC + 24θD + 24θE – 32R1 + 60 = 0 → (5)

480 – 6( – 6θB + 18R2 – 12θB + 18R2) – 8(–16θE + 32R2 – 8θE + 32R2) = 0 or 480 + 108θB – 216 R2 + 192 θE – 512 R2 = 0 or 108 θB + 192 θE – 728 R2 + 480 = 0 → (6) Solving above six equations, we have. θb=2.721 rad, θc=3.933 rad, θd= –3.225 rad, θe= –1.545 rad, R1=3.289 rad, R2=0.656 rad. Putting these in slope deflection equations, the values of end moments are. Mab= –4.518, Mba= –20.844, Mbc= –48.688. Mcb= –58.384, Mcd=58.384, Mdc=–90.245, Mde=90.272, Med= 76.816, Mef=45.696, Mfe=33.344 , Mbe= 69.53, Meb = –122.495 KN-m Now SFD, BMD and elastic curve can be sketched as usual.


EXAMPLE NO. 13:– Analyze the rigid frame shown by slope–deflection method. SOLUTION: It is a double storey frame carrying gravity loads only. Because of difference in column heights, it has become an unsymmetrical frame.

A

B

CD

F

E

5I

5I

2I 4m

2I2I

3 KN/m

3 KN/m

1 1

4m 2I

2 2

4m5m

5m Step 1: RELATIVE STIFFNESS. Member I L

IL Krel

AB 2 5 45 × 10 8

BC 2 4 24 × 10 5

CD 5 5 55 × 10 10

DE 2 4 24 × 10 5

Ef 2 4 24 × 10 5

BE 5 5 55 × 10 10

Step 2: F.E.M :– F.E.Ms. are induced in beams only as they carry u.d.l. No loads act within column heights.

Mfbe = Mfcd = 3 × 25

12 = + 6.25 KN–m

Mfeb = Mfdc = – 6.25 KN–m.


Step 3: RELATIVE VALUES OF R :–

Member ∆ L ∆L Krel

AB ∆2 5 ∆25 × 20 4 R2

BC (∆1–∆2) 4 ∆1 – ∆2

4 R1

CD 0 5 0 0

DE (∆1–∆2) 4 ∆1 – ∆2

4 R1

EF ∆2 4 ∆2 4 × 20 5 R2

BE 0 5 0 0

∆2 terms have been arbitrarily multiplied by 20 while ∆1 − ∆2

4 has been taken equal to R1.

Step 4: SLOPE – DEFLECTION EQUATIONS :–

By putting FEM’s and Krel Values.

Mab = 0 + 8 (– 2 θa – θb + 4 R2) = – 8 θb + 32 R2

Mba = 0 + 8 (– 2 θb – θa + 4 R2) = – 16 θb + 32 R2

Mbc = 0 + 5 (– 2 θb – θc + R1) = – 10 θb – 5 θc + 5 R1

Mcb = 0 + 5 (– 2 θc – θb + R1) = – 10 θc – 5 θb + 5 R1

Mcd = 6.25 + 10 (– 2 θc – θd) = 6.25 – 20 θc – 10 θd

Mdc = – 6.25 + 10 (– 2 θd – θc) = – 6.25 – 20 θd – 10 θc

Mde = 0 + 5 (– 2 θd – θe + R1) = – 10 θd – 5 θe + 5 R1

Med = 0 + 5 (– 2 θe – θd + R1) = – 10 θe – 5 θd + 5 R1

Mef = 0 + 5 (– 2 θe – θf + 5 R2) = – 10 θe + 25 R2

Mfe = 0 + 5 (– 2 θf – θe + 5 R2) = – 5 θe + 25 R2

Mbe = 6.25 + 10 (– 2 θb – θe) = 6.25 – 20 θb – 10 θe

Meb = – 6.25 + 10 (– 2 θe – θb) = – 6.25 – 20 θe – 10 θb

Step 5: JOINT CONDITIONS :– Joint A : θa = 0 (Fixed joint)

Joint B : Mba + Mbc + Mbe = 0 → (1)


Joint C : Mcb + Mcd = 0 → (2)

Joint D : Mdc + Mde = 0 → (3)

Joint E : Med + Meb + Mef = 0 → (4)

Joint F : θf = 0 (Fixed joint)

Step 6: SHEAR CONDITIONS :– Upper Storey

4m 4m

Mcb Mde

B E

C D

Med

HeHbMbc

He=Med+Mde

Hb=Mbc+Mcb4

4

HdHc

ΣFX = 0, Hb + He = 0 , Now putting their values → (5)

Mbc + Mcb

4 +

Med + Mde

4 = 0 Simplify

Mbc + Mcb + Med + Mde = 0 → (5)

A

B E

F HFHA

MBA MEF

MABMFE

5m 4m

Shear Condition: Lower Storey.

ΣFX = 0, Ha + Hf = 0

Ha =

Mab + Mba

5 , Hf =

Mfe+Mef

4 Simplify

4 (Mab + Mba) + 5 (Mfe + Mef) = 0 → (6)



Putting joint and shear conditions in above six equations and simplify.

Mba + Mbc + Mbe = 0 or – 16 θb + 32 R2 – 10 θb – 5 θc +5 R1 + 6.25 –20 θb–10θe = 0 – 46 θb – 5 θc – 10 θe + 5 R1 + 32 R2 + 6.25 = 0 → (1) Mcb + Mcd = 0 or – 10 θc – 5 θb + 5 R1 + 6.25 – 20 θc – 10 θd = 0 – 5 θb – 30 θc – 10 θd + 5 R1 + 6.25 = 0 → (2) Mdc + Mde = 0 or – 6.25 – 5 θd – 10 θc – 10 θd – 5 θe + 5 R1 = 0 – 10 θc – 30 θd – 5 θe + 5 R1 – 6.25 = 0 → (3) Med + Mcb + Mef = 0 or – 10 θe – 5 θd + 5 R1 – 6.25–20 θe–10 θb–10 θe + 25 R2 = 0 – 10 θb – 5 θd – 40 θe + 5 R1 + 25 R2 – 6.25 = 0 → (4) Mbc + Mcb + Med + Mde = 0 or – 10 θb – 5 θc + 5 R1 – 10 θc – 5 θb + 5 R1 – 10 θd – 5 θe + 5 R1 – 10 θe – 5 θd + 5 R1 = 0 – 15 θb – 15 θc – 15 θd – 15 θe + 20 R1 = 0 → (5) 4(Mab + Mba) + 5 (Mfe + Mef) = 0 or 4 (–8 θb + 32 R2 – 16 θb + 32 R2 ) +5(–10 θe + 25 R2 – 5 θe + 25 R2) = 0 – 96 θb – 75 θe + 506 R2 = 0 → (6) Solving above six equations (by programmable calculator) we have.

θb=0.141, θc=0.275, θd= –0.276, θe= –0.156, R1=0.01224, R2=0.003613. By Putting these in slope deflection equations, the values of end moments are.

Mab = –1.012, Mba = –2.14, Mbc = –2.846, Mcb = –3.5162, Mcd = 3.51, Mdc = –3.48, Mde = 3.52, Med = 2.8788, Mef = 1.65, Mfe = 0.87, Mbe = 4.99, Meb = –4.54

Now SFD, BMD and elastic curve can be sketched as usual.


CHAPTER FIVE

5. THE MOMENT − DISTRIBUTION METHOD

5.1. Introduction :− Professor Hardy Cross of University of Illinois of U.S.A. invented this method in 1930. However, the method was well-established by the end of 1934 as a result of several research publications which appeared in the Journals of American Society of Civil Engineers (ASCE). In some books, the moment-distribution method is also referred to as a Hardy Cross method or simply a Cross method. The moment-distribution method can be used to analyze all types of statically indeterminate beams or rigid frames. Essentially it consists in solving the linear simultaneous equations that were obtained in the slope-deflection method by successive approximations or moment distribution. Increased number of cycles would result in more accuracy. However, for all academic purposes, three cycles may be considered sufficient. In order to develop the method, it will be helpful to consider the following problem. A propped cantilever subjected to end moments.

A aMa

B

LMa

A B B.D.S. under redundant Ma,aa ba

0 0+

+

ab bb

0 0

BMb

B.D.S. under redundant Mb,

MbL

MaEI

MaL2EI

Mb E = Constt,I

2EIMbEI

MbEI

MaEI

Diagram Over Conjugate - beam

Diagram Over Conjugate - beam

aa = rotation at end A dueto moment at A.

ba = rotation at B dueto moment at A.

ab = rotation at A dueto moment at B.

bb = rotation at B dueto moment at B.( )

Note: Counterclockwise moment are considered (+ve) Geometry requirement at B :− θb = 0, or θba − θbb = 0 and (1) θb = θba − θbb =0 (Slope at B). Now calculate all rotations shown in diagram by using conjugate beam method.

θaa =

MaL

2EI × 23 L

L ( By conjugate beam theorem)

θaa = MaL3EI

THE MOMENT __ DISTRIBUTION METHOD 259

θab =

MbL

2EI × L3

L ( By conjugate beam theorem)

θab = MbL6EI

θba =

MaL

2EI ×

L

3L ( By conjugate beam theorem)

θba = MaL6EI

θbb =

MbL

2EI ×

2L

3L ( By conjugate beam theorem)

θbb = MbL3EI

Put θba & θbb in (1)

MaL6EI =

MbL3EI

or Mb = Ma2 (3)

If Ma is applied at A, then Ma/2 will be transmitted to the far end B. Also, θa = θaa − θab Geometry requirement at A. (2) Put values of θaa and θab, we have,

θaa = Ma.L3EI −

Mb.L6EI

= Ma.L3EI −

Ma.L12EI (by putting Mb =

Ma2 for above)

θaa = 3 Ma.L12EI

or θaa = Ma.L4EI . It can be written as

θaa = Ma

L

4EI

or Ma =

4EI

L θaa (4)


5.2. STIFFNESS FACTOR :− The term 4EI/L is called the stiffness factor “stiffness factor is defined as the moment required to be applied at A to produce unit rotation at point A of the propped cantilever beam shown.” 5.3. CARRY-OVER FACTOR:− The constant (1/2) in equation 3 is called the carry-over factor.

Mb = Ma2

MbMa =

12

“Carry-over factor is the ratio of the moment induced at the far end to the moment applied at near end for a propped cantilever beam.” Now consider a simply supported beam carrying end moment at A.

A Baa L

Ma

+

MaEI

MaL2EI

EI = Constt:

(M/EI Diagram)

θaa = MaL2EI ×

23 L

L = MaL3EI or Ma =

3θaa EIL

Compare this Ma with that for a propped cantilever beam. We find that Stiffness factor of a simple beam is 3/4th of the cantilever beam. So propped cantilever beam is more stiff. 5.4. DISTRIBUTION FACTOR :− Let us consider a moment applied at joint E as shown. Values shown are the stiflnesses of the members.

A

D

C

B

10,000ME4000 4000

10,000

Consider a simple structure shown in the diagram which is under the action of applied moment M.

For the equilibrium requirements at the joint, it is obvious that the summation of moments ( ∑ M ) should be zero at the joint. This means that the applied moment ‘M’ will be distributed in all the members meeting at that joint in proportion to their stiffness factor. (This called stiffness – concept)

Total stiffness factor = 28,000 = 10,000 + 10,000 + 4,000 + 4,000

So Mae = Mec = 40002800 × M =

17 M

Mbe = Med = 100002800 × M =

514 M. Therefore,

“ Distribution at any end of a member factor is the ratio of the stiffness factor of the member being considered to the sum of the stiffnesses of all the members meeting at that particular continuous joint.”


EXAMPLE NO. 1:- Now take the continuous beam as shown in the figure and analyze it by moment distribution method.

A

10m 10m

CB

20KN5m

5 KN/m

3I 4I

FIXED END MOMENTS :−

A B B41.67 25

16.67

4/73/7

41.67

A

Lockingmoment

41.67

7.14

16.67

16.67= net moment at B

9.53

B

41.67 25

25

C

25Locking moment = reactive moment

C41.67

A BB C

Mfab = 5 × 102

12 = + 41.67 KN−m

Mfba = − 41.67 KN−m

Mfbc = 20 × 52 × 5

102 = + 25 KN−m

Mfcb = − 25 KN−m

M = 16.67 is to be distributed. (Net moment at B support)

Total stiffness of members of joint B = 7

so Mab = 37 × M =

37 × 16.67 = 7.14 KN−m

and Mbc = 47 × M =

47 × 16.67 = 9.53 KN−m

The distribution factor at joint A is obviously equal to zero being a fixed joint. In the above diagram and the distribution factor at point C is infact 1 being an exterior pin support. (If we apply moment to the fixed support, same reactive moment will develop, so re−distribution moment is not created for all fixed supports and if a moment is applied at a pin support, we reactive moment develops.)

Fixed ended moments are sometimes referred to as the restraining moments or the locking moments. “The locking moments are the moments required to hold the tangents straight or to lock the joints against rotation”.


Consider the above diagram. Joint A is fixed joint. Therefore, the question of release of this joint does not arise. Now let us release joint to the net locking moments acting at joint B is 16.67 in the clockwise direction. After releasing the joint B, the same moment (16.67) will act at joint B in the counterclockwise direction. This net released moment will be distributed to various members framing into the joint B w.r.t. their distribution factors. In this case, 7.14 KN−m in the counterclockwise direction will act on member BA and 9.53 KN−m in the counterclockwise direction will act on member BC.

Now we hold the joint B in this position and give release to joint ‘C’. The rotation at joint ‘C’ should be such that the released moment at joint ‘C’ should be 25 KN−m. The same procedure is repeated for a desired number of cycles. The procedure explained above corresponds to the first cycle. 5.5. STEPS INVOLVED IN MOMENT DISTRIBUTION METHOD:−

The steps involved in the moment distribution method are as follows:−

(1) Calculate fixed end moments due to applied loads following the same sign convention and procedure, which was adopted in the slope-deflection method.

(2) Calculate relative stiffness.

(3) Determine the distribution factors for various members framing into a particular joint.

(4) Distribute the net fixed end moments at the joints to various members by multiplying the net moment by their respective distribution factors in the first cycle.

(5) In the second and subsequent cycles, carry-over moments from the far ends of the same member (carry-over moment will be half of the distributed moment).

(6) Consider this carry-over moment as a fixed end moment and determine the balancing moment. This procedure is repeated from second cycle onwards till

convergence For the previous given loaded beam, we attempt the problem in a tabular form..

K = IL =

310 × 10 = 3

and 410 × 10 = 4

Joints. A B C

Members. AB BA BC CB K 3 3 4 4 Cycle No. D. Factor 0 0.428 0.572 1

1

F.E.M. Balancing moment.

+ 41.67 0

− 41.67 + 7.14

+ 25 + 9.53

− 25 + 25

2 COM. Bal.

+ 3.57 0

0 − 5.35

+ 12.5 − 7.15

+ 4.77 − 4.77

3 COM. Bal.

− 2.67 0

0 + 1.02

− 2.385 + 1.36

− 3.575 + 3.575

∑ + 42.57 − 38.86 + 38.86 0


NOTE:- Balancing moments are, in fact, the distributed moments.

Now draw SFD , BMD and hence sketch elastic curve as usual by drawing free-body diagrams.

42.57 5

KN/m

38.86 38.8620KN

A 10m B B 10m C

+25 +25 +10 +10 __ due to applied loads+0.371 -0.371 +3.886 -3.886 __ due to end moments

25.371 +24.629 +13.886 6.114

38.515Ra

Rb

Rc __ net reaction at support considering both sides of a joint.

B5 KN/m 5m CA

10m10m

25.371 38.515 6.114

+ +25.371 13.886

24.629 6.114

+ +

SFD

B.M.D

1.973m 30.570

2.8

38.862.12m

42.57 POINTS OF CONTRAFLEXURES :− Near A: Span AB MX = 25.371 X − 42.57 − 2.5 X2 = 0 See free-body diagram

2.5 X2 − 25.371 X + 42.57 = 0

X = 25.371 ± (25.371)2 − 4 × 2.5 × 42.57

2 × 2.5

X = 2.12 m


Near B :− Mx′ = − 38.86 + 24.629 X′ − 2.5 X′2 = 0

2.5 X′ 2 − 24.629 X′ + 38.86 = 0

X′ = 24.629 ± (24.629)2 − 4 × 2.5 × 38.86

2 × 2.5

X′ = 1.973 m Span BC (near B) MX// = − 38.86 + 13.886X// = 0 X// = 2.8 m EXAMPLE NO. 2:− Analyze the following beam by moment-distribution method. Draw S.F. & B.M. diagrams. Sketch the elastic curve. SOLUTION :−

3KN/m 6KN/m 36KN

D2m 2m5m B 8m C

E = Constt:I

A

Step 1: FIXED END MOMENTS :−

Mfab = + 3 (5)2

12 = + 6.25 KN−m

Mfba = − 6.25 KN−m

Mfbc = + 6 × 82

12 = + 32 KN−m

Mfcb = − 32 KN−m

Mfcd = 36 × 22 × 2

42 + 18 KN−m Mfdc = − 18 KN−m Step 2: RELATIVE STIFFNESS :−


AB 1 5 15 × 40 8

BC 1 8 18 × 40 5

CD 1 4 14 × 40 10


STEP (3) DISTRIBUTION FACTOR :− Joint. D.F. Member. A 0 AB

B 813 = 0.615 BA

B 513 = 0.385 BC

C 515 = 0.333 CB

C 1015 = 0.667 CD

D

10

10+0 = 1 DC

Attempt and solve the problem now in a tabular form by entering distribution .factors and FEM’s. Joint A B C D Members. AB BA BC CB CD DC K 8 8 5 5 10 10 Cycle No. D.F. 0 0.615 0.385 0.333 0.667 1 1 F.E.M

Bal. + 6.25 0

−6.25 −15.836

+32 −9.914

− 32 +4.662

+ 18 +9.338

− 18 + 18

2 Com. Bal.

− 7.918 0

0 −1.433

+2.331 −0.897

−4.957 −1.346

+ 9 −2.697

+4.669 −4.669

3 Com. Bal.

− 0.7165 0

0 +0.414

−0.673 +0.259

−0.4485 + 0.927

−2.3345 +1.856

−1.3485 +1.3485

∑ − 2.385 −23.141 +23.11 −33.16 +33.16 0 Usually for academic purposes we may stop after 3 cycles. Applying above determined net end moments to the following segments of a continuous beam, we can find reactions easily.

2.38 23.11 23.11 33.16 33.16 36KN

5m 8m

3KN/m 6KN/m

A B B C C D

+7.5 +7.5 +24 +24 +18 +18 __ reaction due to applied load

-5.098 +5.098 -1.261 +1.261 +8.29 -8.29 __ reaction due to end moment

2.402 +12.598 + (22.739) +25.261 +26.29 9.71 __ net reaction of a support

Final Values considering bothsides of a support.35.337 51.551


3KN/m 6KN/m 36KN

2m 2m DCB

A2.38KN

2.402 35.337KN 51.557KN 9.71KN

22.739 26.29 26.29

a =0.8m

2.40+

0

+

0 S.F.D.b=3.79m

15.598

9.71 9.71

19.42

+

33.1623.11

0

2.380

3.34

X X X X

0Vb=22.739-6b=0b=3.79m

Va=2.402-3a=0a = 2.402 = 0.8m 3

BMD

POINTS OF CONTRAFLEXURES :−

Span AB (near A) MX = 2.38 + 2.402 X − 1.5 X2 = 0 1.5 X2 − 2.402 X − 2.38 = 0

X = 2.402 ± (2.402)2 + 4 × 1.5 × 2.38

2 x 1.5

X = 2.293 m Span BC (near B) MX′ = − 23.11 + 22.739 X′ − 3 X′2 = 0 3 X′2 − 22.739 X′ + 23.11 = 0

X′ = 22.739 ± (22.739)2 − 4 × 3 × 23.11

2 x 3

X′ = 1.21 m Span BC (near C) MX" = − 33.16 + 25.261 X" − 3 X"2 = 0 3 X" 2 − 25.261 X" + 33.16 = 0

X" = 25.261 ± (25.261)2 − 4 × 3 × 33.16

2 x 3

X" = 1.63 m Span CD (near C) MX"′= − 33.16 + 26.29 X"′ = 0 X"′ = 1.26m


5.6. CHECK ON MOMENT DISTRIBUTION :− The following checks may be supplied. (i) Equilibrium at joints.

(ii) Equal joint rotations or continuity of slope.

General form of slope-deflection equations is Mab = Mfab + Krel ( − 2 θa − θb ) → (1) Mba = Mfba + Krel (− 2 θb − θa) → (2) From (1)

θb = − ( Mab − Mfab)

Krel − 2 θa → (3)

Put (3) in (2) & solve for θa.

Mba = Mfba + Krel

2 (Mab − Mfab)

Krel + 4 θa − θa

Mba = Mfba + Krel

2 (Mab − Mfab) +3 θa Krel

Krel

(Mba − Mfba) = 2 (Mab − Mfab) + 3 θa Krel 3 θa Krel = (Mba − Mfba) − 2 (Mab − Mfab)

θa = (Mba − Mfba) − 2 (Mab − Mfab)

3 Krel → (4)

or θa = (Mba − Mfba) − 2 (Mab − Mfab)

Krel → (5)

θa = Change at far end − 2 (Change at near end)

Krel

or θa = 2 ( Change at near end) − (Change at far end)

−Krel

θa = (Change at near end)−1/2(change at far end)

− Krel

Put (4) in (3) & solve for θb.

θb = − (Mab − Mfab)

Krel − 2 (Mba − Mfba)

3 Krel + 4(Mab−Mfab)

3 Krel

= − 3 Mab + 3 Mfab − 2 Mba + 2 Mfba+4 Mab−4 Mfab

3 Krel


= (Mab − Mfab) − 2 (Mba − Mfba)

3 Krel

= 2 (Mba − Mfba) − (Mab − Mfab)

− 3 Krel

= (Mba − Mfba) − 1/2 (Mab − Mfab)

− 3/2 Krel


− 1.5 Krel


− Krel

θb = (Change at near end) − 1/2(Change at far end)

− Krel

These two equations serve as a check on moment – Distribution Method. EXAMPLE NO. 3:− Analyze the following beam by moment-distribution method. Draw shear force and

B.M. diagrams & sketch the elastic curve. SOLUTION :−

3KN

A B C D

2m1.2KN/m 8KN

1m 4m 5m 4m 2I 4I 3I

Step 1: FIXED END MOMENTS :− Mfab = Mfba = 0 ( There is no load on span AB)

Mfbc = + 1.2 × 52

12 = + 2.5 KN−m

Mfcb = − 2.5 KN−m

Mfcd = 8 × 22 × 2

42 = + 4 KN−m

Mfdc = − 4 KN−m


Step 2: RELATIVE STIFFNESS (K) :−

Span I L IL Krel

AB 2 4 24 × 20 10

BC 4 5 45 × 20 16

CD 3 4 34 × 20 15

Moment at A = 3 × 1 = 3 KN−m. (Known from the loaded given beam according to our sign convention.) The applied moment at A is counterclockwise but fixing moments are reactive moments. Step 3: D.F. Joint D.F. Members. A 1 AB

B 1026 = 0.385 BA

B 1626 = 0.615 BC

C 1631 = 0.516 CB

C 1531 = 0.484 CD

D

4

4 + 0 = 1 DC

Now attempt the promlem in a tabular form to determine end moments.

3KN 3 3 0.38 0.38

1mA B C DCB4m 5m 4m

8KN1.2KN/m4.94 4.94 2m

3.845 1.091 9.299

+3 +0.845 -0.845 +1.936 +4.064 +5.235 2.7650 +0.845 -0.845 -1.064 +1.064 +1.235 -1.235

+3 0 0 +3 +3 +4 +4

2.765

(due to applied loads)

(due to end moments)

(net reaction)


Insert Page No. 294-A


3KN1m

4m 5m

1.2KN/m 8KN2m

3.845 KN 1.091KN 9.299KN 2.765KN

+0 S.F.D.

1.936 5.235

++0.845 0.845

0

3 3 X=1.61m 2.7652.765

4.064

5.53

0 B.M.D.

X X1.940.38

3

0

1.936 - 1.2 x X = 0X=1.61 m for B in portion BC

++

X4.94

A B C D

LOCATION OF POINTS OF CONTRAFLEXURES :− MX = − 0.845 X +0.38 = 0 X = 0.45 m from B. in portion BA. MX′ = 4.064 X′ − 4.94 − 0.6 X′2 = 0 0.6X′2 − 4.064 X′ + 4.94 = 0

X′ = 4.064 ± (4.064)2 − 4 × 0.6 × 4.94

2 x 0.6

= 1.59 m from C in span BC MX" = − 4.94 + 5.235 X" = 0 X" = 0.94 m from C in span CD 5.7. MOMENT−DISTRIBUTION METHOD (APPLICATION TO SINKING OF SUPPORTS) :−

Consider a generalized differential sinking case as shown below: L

EI Constt:R BAMFab

MFba

B

0L/2

0 +

LMFba4 EI

LMFab4 EI

MFabEI

MFbaEI5/6L

B.M.D.

Bending moments areinduced due to differentialsinking of supports.


(1) Change of slope between points A and B (θab) = 0 ( First moment−area theorem )

(1) L

4EI Mfab − L

4EI Mfba = 0

or Mfab = Mfba

(2) ∆ = L

4EI Mfab

5

6 L − L

4EI Mfab

L

6 ( Second moment area theorem ), simplify.

6EI ∆ = 5L2 Mfab − L2 Mfab

4

= 4 L2 Mfab

4

6EI ∆ = L2Mfab

or Mfab = Mfba = 6EI ∆

L2 , where R = ∆L

Mfab = Mfba = 6EI R

L

Equal FEM’s are induced due to differential sinking in one span. The nature of the fixed end moments induced due to the differential settlement of the supports depends upon the sign of R. If R is (+ve) fizingmment is positive or vice versa. Care must be exercised in working with the absolute values of the quantity 6EIR/L which should finally have the units of B.M. (KN−m). Once the fixed end moments have been computed by using the above formula, these are distributed in a tabular form as usual. EXAMPLE NO.4:− Analyse the continuous beam shown due to settlement at support B by moment − distribution method. Apply usual checks & draw S.F., B.M. diagrams & hence sketch the elastic curve take E = 200 × 106 , I = 400 × 10−6 m4

A B C D2 4 3I I I

15mm

1m 4m B 5m 4m SOLUTION :− Step (1) F.E.M. In such cases, Absolute Values of FEM’s are to be calculated

Mfab = Mfba = 6EI∆

L2 = 6(200 × 106 )(2 × 400 × 10−6 )(+0.015)

42

= + 900 KN−m (positive because angle R = ∆L is clockwise).


Mfbc = Mfcb = 6 (200 × 106) (4 × 400 × 10−6)(−0.015)

52

= − 1152 KN−m (Because angle is counter clockwise) Mfcd = Mfdc = 0


Members. I L IL Krel.

AB 2 4 24 × 20 10

BC 4 5 45 × 20 16

CD 3 4 34 × 20 15

Step 3: D.F :− (Distribution Factors) Joint D.F. Members.

A 1 AB B 0.385 BA B 0.615 BC C 0.516 CB C 0.484 CD D 1 DC We attempt and solve the problem in a tabular form as given below:

Joint A B C D Members AB BA BC CB CD DC K 10 10 16 16 15 15 Cycle D.F. 1.0 0.385 0.615 0.516 0.484 0 1 FEM.

BAL. + 900 − 900

+ 900 + 97.02

− 1152 +154.98

− 1152 + 594.43

0 + 557.57

0 0

2 COM. BAL.

+ 48.51 − 48.51

− 450 + 58.82

+ 297.22 + 93.96

+ 77.49 − 39.98

0 − 37.51

+ 278.79 0

3 COM. BAL.

+ 29.41 − 29.41

− 24.255 + 17.03

− 19.99 + 27.21

+ 46.98 − 24.24

0 − 22.74

− 18.75 0

4 COM. BAL.

+ 8.515 − 8.515

− 14.705 + 10.328

− 12.12 + 16.497

+ 13.605 − 7.020

0 − 6.585

− 11.37 0

5 COM. BAL.

+ 5.164 − 5.164

− 4.258 + 2.991

− 3.51 + 4.777

+ 8.249 − 4.256

0 − 3.493

− 3.293 0

End Moment. 0 + 592.97 − 592.97 − 486.74 + 486.74 +245.38 (change) near end. − 900 − 307.03 + 559.03 + 665.26 + 486.74 + 245.38 −1/2(change) far end. + 153.515 + 450 − 332.63 − 279.515 − 122.69 − 243.37 ∑ − 746.485 + 142.97 + 226.4 + 385.745 +367.05 + 2.01

θ rel = ∑

−K + 74.65 − 14.30 − 14.15 − 24.11 − 24.47 − 0.134

θ checks have been satisfied. Now Draw SFD , BMD and sketch elastic curve as usual yourself.


5.8. APPLICTION TO FRAMES (WITHOUT SIDE SWAY) :− The reader will find not much of difference for the analysis of such frames. EXAMPLE NO. 5:− Analyze the frame shown below by Moment Distribution Method.

A

B C3I

2m16KN

2m

1.5m 2

1.5m

I8 KN

SOLUTION :− Step 1: F.E.M :−

Mfab = + 8 × 1.52 × 1.5

32 = + 3 KN−m

Mfba = − 8 × 1.52 × 1.5

32 = − 3 KN−m

Mfbc = + 16 × 22 × 2

42 = + 8 KN−m

Mfcb = − 8 KN−m Step 2: RELATIVE STIFFNESS (K) :−

Members. I L IL Krel

AB 2 3 23 × 12 8

BC 3 4 34 × 12 9

Step 3: D.F :− (Distribution Factors) Joint. D.F., Member. A 0 AB

B 0.47 BA

B 0.53 BC

C 0 CB


Example is now solved in a tabular form as given below:

Joint A B C Members AB BA BC CB K 8 8 9 9 Cycle D.F. 0 0.47 0.53 0 1 Fem.

Bal. +3 0

− 3 −2.35

+8 −2.65

− 8 0

2 Com. Bal.

−1.175 0

0 0

0 0

−1.325 0

3 Com. Bal.

0 0

0 0

0 0

0 0

∑ +1.175 −5.35 +5.35 −9.325 (Change) near end −1.825 −2.35 −2.65 −1.325 −1/2(change)far end +1.175 +0.5875 +0.6625 +1.325 Sum 0 −1.7625 −1.9875 0 θrel=Sum/(−K) 0 +0.22 +0.22 0

θ Checks have been satisfied. DETERMINATION OF SUPPORT REACTIONS, SFD AND BMD.

CB

A

B

8KN

5.35

1.5m

1.5m1.825

+8 +8

16KN2m 2m

+1.175 5.175

+4

- 1.175 2.825

-0.994 7.006

-0.994 8.994

5.35 9.325

+4

7.006

7.006 B,M. & S.F. DIAGRAMS :−

B

0 0

C2m 2m

16KN 9.325 KN-m5.35KN-m

7.006KN

+

+

8.994 8.994

S.F.D.

8.994KN

00

5.35

Mx=7.006X-5.35=0 x=0.764mMx=8.994 X-9.325=0 x=1.057 m

8.662

7.006

9.325

B.M.D.X X


BA(rotated member)

Note: It is a column rotated through 90.

+ CBS.F.D.

3.825

7.006

5.175 8.994

+

+ + CBBMD

1.825

5.35 9.325

8.662

ELASTIC CURVE


EXAMPLE NO.6:− Analyze the frame shown in the fig. by Moment Distribution Method.

A 2m 4m B 4m 2m C20KN 20KN

4m 2I 6m

2I

2I 4m

5I 5I

D

EF

6m 6m SOLUTION :− Step 1: F.E.M :−

Mfab = + 20 × 42 × 2

62 = + 17.778 KN−m

Mfba = − 20 × 22 × 4

62 = − 8.889 KN−m

Mfbc = + 20 × 22 × 4

62 = + 8.889 KN−m

Mfcb = − 20 × 42 × 2

62 = − 17.778 KN−m

Mfad = MFda = 0

Mfbe = Mfeb = 0 There are no loads on these spans.

Mfcf = Mffc = 0


Members. I L IL Krel

AB 5 6 56 × 12 10

BC 5 6 56 × 12 10

AD 2 4 24 × 12 6

BE 2 6 26 × 12 4

CF 2 4 24 × 12 6


Step 3: Distribution Factor (D.F):− Joint Member D.F. A AD 0.375

A AB 0.625

B BA 0.417

B BE 0.166

B BC 0.417

C CB 0.625

C CF 0.375

F FC 0

E EB 0

D DA 0

Now we attempt the problem in a tabular form. Calculation table is attached Draw SFD, BMD and sketch elastic curve now.

BA

13.33 +6.67

20KN2m 4m

- 1.296 12.034

+1.296 7.966

6.667 14.4472.5 CB

13.33 +6.67

20KN2m4m

- 1.29612.034

+ 1.2967.966

14.447 6.6672.5

A 6.667+2.5

+2.53.334

D

4m

B

6m

E

6.6672.5

2.5

3.334

4m

C

F

2.5 2.5

12.034 15 12.034

12.034 15 12.034


Insert Page No. 304-A


B.M. & SHEAR FORCE DIAGRAMS :−

0 0

2m 4m20KN 14.4446.667

12.034KN

+

+

17.401

7.966

S.F.D. (KN)

7.966KN

0

6.667

Mx=12.034 x-6.667= 0 x=0.554m

Mx=7.966 x -14.444= 0 x=1.813 m

12.034

14.444

X X

A B

+

0+

B.M.D. (KN-m)

0

2m 4m

20KN14.444 6.667

12.034KN

+

+

7.966S.F.D. (KN)

7.966KN

0

12.034

14.444

1.813m 0.554m

B C

6.667

0+

0+

B.M.D. (KN-m)

B

E

6m

Mx=3.334 - 2.5 x=0 x=1.334m

002.

52.

5

2.53.

334

3.33

4

6.66 6.

667

+X

AD

0 0

0 02.5

2.5

2.53.

334

3.33

4

3.33

4

6.66

7

6.66

7+

+

CF


D

EF

A B C

Elastic Curve

EXAMPLE NO. 7:- Analyze the following frame by Moment Distribution Method. SOLUTION:− This is a double story frame carrying gravity and lateral loads and hence would be able to sway both at upper and lower stories.

C D

I

5m

3m

E

2KN/m

2KN/m3m

2 2

2 I 2I

A

3KN/m B

II

Step 1: F.E.Ms Due to applied loads :−

Mfab = 3 × 32

12 = + 2.25 KN−m

Mfba = − 2.25 KN−m

Mfbc = 3 × 32

12 = + 2.25

Mfcb = − 2.25 KN−m.

Mfbe = Mfcd = 2.52

12 = + 4.167 KN−m

Mfeb = Mfdc = − 4.167 KN−m Mfde = Mfcd = 0 Mfef = Mffe = 0


Step 2: Relative Stiffness :−

Member I L IL Krel

AB 2 3 23 × 15 10

BC 2 3 23 × 15 10

DE 2 3 23 × 15 10

EF 2 3 23 × 15 10

CD 1 5 15 × 15 3

BE 1 5 15 × 15 3

Step 3: F.E.Ms. Due to side Sway of upper storey:−

C

B

A F

2I 3m

E

R

5m

R

1 D

I

1

Mfbc = Mfcb = + 6EI ∆

L2 = + 6E(2I ) ∆

32 × 900 = + 1200 (Note: 900 value is an arbitrary multiplier)

Mfde = Mfed = + 6 EI ∆

L2 = + 6 E(2 I) ∆

32 × 900 = + 1200 (Because R is clockwise)

Step 4: F.E.Ms. Due To Side Sway Of Lower Storey :−

3m 2I

3m 2I

C I D5m

B

A F

ER

2-R-R

2

R


Mfbc = Mfcb =Mfde = Mfed = − 6E(2I) ∆

9 × 900 = − 1200

(R is counter clockwise so negative)

Mfab = Mfba = + 6EI(2I) ∆

9 × 900 = + 1200 (R is clockwise, So positive)

Mfef = Mffe = + 6EI(2I) ∆

9 × 900 = + 1200 (R is clockwise, So positive)

Determination Of Shear Co-efficients (K1, K2) for upper and lower stories :−

MCB MDE

C D

3m 3m

B EMBC M ED

HB = 4.5+

HBMBC+MCB

3HE =

HEMED+MDE

3

3KN/m

Upper Storey:

Shear Conditions : 1. Upper story Hb + He =0 (1) where Hb and He values in terms of end

moments are shown in the relavant diagram. 2. Lower storey Ha + Hf = 0 (2)

FA

MBA MEFEB

MFEMAB

HF =

HF

MFE+MEF3HA

= 4.5+

HA

MAB+MBA3

3m 3m3KN/m

Lower Storey

Where Ha and Hf values in terms of end moments are shown in the relavant diagram Now we attempt the problem in a tabular form. There would be three tables , one due to loads(Table−A), other due to FEMs of upper story (Table−B) and lower story (Table−C). Insert these three tables here. Now end moment of a typical member would be the sum of moment due


to applied loads ± K1 × same end moment due to sway of upper story ± K2 × same end moment due to sway of lower story. Picking up the values from tables and inserting as follows we have. Mab = 1.446 − K1(143.66) + K2 (1099 .625).

Mba = − 3.833 − K1 (369.4) + K2 (1035.46)

Mbc = − 0.046 + K1 (522.71) − K2 (956.21)

Mcb = − 4.497 + K1 (314.84) − K2 (394.38).

Mcd = + 4.497 − K1 (314.84) + K2 (394.38)

Mdc = − 3.511 − K1 (314.84) + K2 (394.38)

Mde = + 3.511 + K1 (314.84) − K2 (394.38)

Med = + 2.674 + K1 (522.71) − K2 (956.29).

Mef = + 1.335 − K1 (369.4) + K2 (1035.46)

Mfe = + 0.616 − K1 (193.66) + K2 (1099.625).

Mbe = + 3.878 − K1 (153.32) − K2 (79.18)

Meb = 4.009 − K1 (153.32) − K2 (79.18) Put these expressions of moments in equations (1) & (2) & solve for K1 & K2. − 0.046 + 522.71 K1 − 956.21 K2 − 4.497 + 314.84 K1 − 394.38 K2 +2.674+522.71 K1 −956.29 K2+3.511+314.84 K1 −394.38 K2 = 13.5 1675.1 K1 − 2701.26 K2 − 11.858 = 0 → (3) 1.446 − 143.66 K1 + 1099.625 K2 − 3.833 − 369.4 K1 + 1035.46 K2 +0.646−193.66 K1+1099.625 K2+1.335−369.4K1+1035.46K2 = 40.5 − 1076.12 K1 + 4270.17 K2 − 40.936 = 0 → (4) From (3)

K2 =

1675.10 K1 − 11.858

2701.26 → (5)

Put K2 in (4) & solve for K1

− 1076.12 K1 + 4270.17

1675.10 K1 − 11.858

2701.26 − 40.936 = 0

− 1076.12 K1 + 2648 K1 − 18.745 − 40.936 = 0 1571.88 K1 − 59.68 = 0 K1 = 0.03797

From (5) ⇒ K2 = 1675.1 (0.03797) − 11.858

2701.26

K2 = 0.01915


Putting the values of K1 and K2 in above equations , the following end moments are obtained. FINAL END MOMENTS :− Mab = 1.446 − 0.03797 x 143.66 + 0.01915 x 1099.625 = + 17.05KN−m

Mba = + 1.97 KN−m

Mbc = + 1.49 KN−m.

Mcb = − 0.095 KN−m.

Mcd = + 0.095 KN−m

Mdc = − 7.91 KN−m

Mde = + 7.91 KN−m

Med = + 4.21 KN−m

Mef = + 7.14 KN−m

Mfe = + 14.32 KN−m

Mbe = − 3.46 KN−m

Meb = − 11.35 KN−m

These values also satisfy equilibrium of end moments at joints. For simplicity see end moments at joints C and D. Space for notes:


Insert Page No. 309−A−B


Insert Page No. 309−C


CHAPTER SIX

6. KANIS METHOD OR ROTATION CONTRIBUTION

METHOD OF FRAME ANALYSIS

This method may be considered as a further simplification of moment distribution method wherein the problems involving sway were attempted in a tabular form thrice (for double story frames) and two shear co-efficients had to be determined which when inserted in end moments gave us the final end moments. All this effort can be cut short very considerably by using this method. → Frame analysis is carried out by solving the slope − deflection equations by successive approximations. Useful in case of side sway as well. → Operation is simple, as it is carried out in a specific direction. If some error is committed, it will be eliminated in subsequent cycles if the restraining moments and distribution factors have been determined correctly. Please note that the method does not give realistic results in cases of columns of unequal heights within a storey and for pin ended columns both of these cases are in fact extremely rare even in actual practice. Even codes suggest that RC columns framing into footings or members above may be considered more or less as fixed for analysis and design purposes.

Case 1. No side sway and therefore no translation of joints derivation. Consider a typical member AB loaded as shown below:

A B

L

P1 P2 MbaMab

ba

Tangent at B

Tangent at A Elastic Curve

A GENERAL BEAM ELEMENT UNDER END MOMENTS AND LOADS General Slope deflection equations are.

Mab = MFab + 2EIL ( − 2θa − θb ) → (1)

Mba = MFba + 2EIL ( − θa −2θb ) → (2)

equation (1) can be re-written as Mab = MFab + 2 M′ab + M′ba → (3) where MFab = fixed end moment at A due to applied loads.

and M′ab = rotation contribution of near end A of member AB = − EIL (2θa)

= − 2EI θa

L = − 2E k1 θa → (4) where

k1=

I1L1

M/ba = rotation contribution of for end B of member AB.

So M/ba = − 2 EI θb

L = − 2Ek1 θb → (5)

KANIS METHOD OF FRAME ANALYSIS 289

Now consider a generalized joint A in a frame where members AB, AC, AD.........meet. It carries a moment M.

EA

B

D

Ck3

k1

k2

k3

M

For equilibrium of joint A, ∑Ma = 0 or Mab + Mac + Mad + Mae..................= 0 Putting these end moments in form of eqn. (3) or ∑MF (ab, ac, ad) + 2 ∑M′ (ab, ac, ad ) + ∑M′ (ba, ca, da) = 0 Let ∑MF (ab, ac, ad) = MFa (net FEM at A) So MFa + 2 ∑M′ (ab, ac, ad) + ∑M′ (ba, ca, da) = 0 → (6)

From (6), ∑M′ (ab, ac, ad) = − 12 [(MFa + ∑M′ (ba, ca, da)] → (7)

From (4), ∑M′ (ab, ac, ad) = − 2Ek1 θa − 2 Ek2 θa − 2 Ek3 θa + ............... = − 2 Eθa ( k1 + k2 + k3) = − 2 Eθa (∑k), ( sum of the member stiffnesses framing in at joint A)

or θa = − ∑M′ (ab, ac, ad)

2E (∑k) → (8)

From (4), M′ab = − 2 Ek1 θa. Put θa from (8), we have

M′ab = − 2E k1

−

∑M′ (ab, ac, ad)2E (∑k) =

k1

∑k [ ∑M′ (ab, ac, ad)]

From (7), Put ∑M′ (ab, ac, ad)

So M′ab = k1

∑k

−

12 (MFa + ∑M′ (ba, ca, da))


or M′ab = − 12

k1

∑k [ MFa + ∑M′ (ba, ca, da)]

on similar lines M′ac = − 12

k2

∑k [ MFa + ∑M′ (ba, ca, da)]

and M/ad = − 12

k3

∑k [ MFa + ∑M′ (ba, ca, da)]

rotation contribution of near sum of the rotations contributions of far end of member ad. ends of members meeting at A. Sum of rotation factors at near end of members ab, ac, ad is

− 12

k1

∑k − 12

k2

∑k − 12

k3

∑k = − 12

k1 + k2 + k3 + .........

∑k

= − 12 , [sum of rotation factors of different members meeting at a

joint is equal to – 12 ]

Therefore, if net fixed end moment at any joint along with sum of the far end contribution of members meeting at that joint are known then near end moment contribution can be determined. If far end contributions are approximate, near end contributions will also be approximate. When Far end contributions are not known (as in the first cycle), they can be assumed to be zero. 6.1. RULES FOR CALCULATING ROTATION CONTRIBUTIONS :__ Case-1: Without sides way. Definition: “Restrained moment at a joint is the algebraic sum of FE.M’s of different members meeting at that joint.” 1. Sum of the restrained moment of a joint and all rotation contributions of the far ends of members meeting at that joint is multiplied by respective rotation factors to get the required near end rotation contribution. For the first cycle when far end contributions are not known, they may be taken as zero (Ist approximation). 2. By repeated application of this calculation procedure and proceeding from joint to joint

in an arbitrary sequence but in a specific direction, all rotation contributions are known. The process is usually stopped when end moment values converge. This normally happens after three or four cycles. But values after 2nd cycle may also be acceptable for academic.

6.2. Case 2:__ With side sway (joint translations) In this case in addition to rotation contribution, linear displacement contributions ( Sway contributions ) of columns of a particular storey are calculated after every cycle as follows:


6.2.1. For the first cycle. (A) → Linear Displacement Contribution ( LDC) of a column = Linear displacement factor (LDF) of a

particular column of a story multiplied by [storey moment + contributions at the ends of columns of that story]

Linear displacement factor (LDF) for columns of a storey = − 32

Linear displacement factor of a column = − 32

k ∑k Where k=stiffness of the column being

considered and Σk is the sum of stiffness of all columns of that storey.

6.2.2. (B) → Storey moment = Storey shear x 13 of storey height.

6.2.3. (C) → Storey shear : It may be considered as reaction of column at horizontal beam / slab

levels due to lateral loads by considering the columns of each sotrey as simply supported beams in vertical direction. “If applied load gives + R value (according to sign conversion of slope deflection method), storey shear is +ve or vice versa.”

Consider a general sway case.

h RR

P

6.3. SIGN CONVENSION ON MOMENTS:− Counter-clockwise moments are positive and clockwise rotations are positive. For first cycle with side sway.

(D) Near end contribution of various = respective rotation contribution factor × [Restrained moment + members meeting at that joint. far end contributions]

Linear displacement contributions will be calculated after the end of each cycle for the columns only.

FOR 2ND AND SUBSEQUENT CYCLES.

(E) → Near end contributions of various = Respective rotation contribution factor × [Restrained members meeting at a joint. moment + far end contributions + linear displacement contribution of columns of different storeys meeting at that joint].


6.4. Rules for the Calculation of final end moments (sidesway cases) (F) For beams, End moment = FEM + 2 near end contribution + Far end contributions. (G) For columns, End moment. = FEM + 2 near end contribution + Far end contribution + linear displacement contribution of that column for the latest cycle. 6.5. APPLICATION OF ROTATION CONTRIBUTION METHOD (KANI’S METHOD)

FOR THE ANALYSIS OF CONTINUOUS BEAMS Example No.1: Analyze the following beam by rotation contribution method. EI is constant.

A

B C D

7k/ft 6k/ft

16 24 12

36K

EI = constt.

Note. Analysis assumes continuous ends with some fixity. Therefore, in case of extreme hinged supports in exterior spans, modify (reduce) the stiffness by 3/4 = (0.75).for a hinged end. Step No. 1. Relative Stiffness.

Span I L IL Krel K modified.

AB 1 16 1

16 × 48 3 3

BC 1 24 1

24 2 2

CD 1 12 1

12 4 x (3/4) 3

(exterior or discontinuous hinged end) Step No.2. Fixed end moments.

Mfab = + wL2

12 = + 3 × 162

12 = + 64 K-ft.

Mfba = − 64

Mfbc = + 6 × 242

12 = + 288

Mfcb = − 288

Mfcd = + Pa2bL2 =

+ 36 × 62 × 6122 = + 54

Mfdc = − 54


Step No.3. Draw Boxes, enter the values of FEMs near respective ends of exterior boxes and rotation contribution factors appropriately (on the interior side).

FEMs+64 -64

-67.2-83.92-84.48

A-0.5( 3

3+2) -0.5( 3

3)

-0.3+224

-0.2 -0.2

restraining moment =algebraic sum of FEMmeeting at that joint is extendin inner box..

+288 -288-44.8 +55.76-55.95 +60.95-57 +61.94

+54 -54+83.64 -14.82+91.43 -18.71+92.9 -19.45

-0.5-54-234

C DB

-0.3000

* * * * * *

* = Distribution factors. A C( Far end contribution) B D( Far end contributions) FIRST CYCLE ↓ ↓ ↓ ↓ Joint B: − 0.3 (+224 + 0 + 0) = − 67.2 (Span BA) Joint C: − 0.2(− 234 − 44.8 + 0) = +55.76 (Span CB) and − 0.2 ( 224 + 0 + 0) = − 44.8 (Span BC) and − 0.3(− 234 − 44.8 + 0) = +83.64 (Span CD) Joint D: − 0.5(− 54 +83.64) = − 14.82 (Span DC) 2nd cycle: A C ( Far end contributions) B D (far end contributions) ↓ ↓ ↓ ↓ Joint B. − 0.3 (+ 224+0 +55.76) = − 83.92 Joint C: − 0.2 (− 234 − 55.95 − 14.82) = 60.95 − 0.2 (+224+0 +55.76) = − 55.85 − 0.3 (− 234 − 55.95 − 14.82) = 91.43 Joint D. − 0.5 ( − 54 + 91.43) = − 18.715 3rd cycle: Singular to second cycle procedure. We stop usually after 3 cycles and the answers can be further refined by having another couple of cycles. (Preferably go up to six cycles till difference in moment value is 0.1 or less). The last line gives near and far end contribution.

Step No. 4. FINAL END MOMENTS

For beams. End moment = FEM + 2near end cont. + Far end contribution.

Mab = + 64 + 2 x 0 − 84.48 = − 20.48 k − ft.

Mba = − 64 − 2 x 84.48 + 0 = − 232.96 k − ft.

Mbc = + 288− 2 x 57 + 61.94 = +235.9 k − ft.

Mcb = − 288 + 2 x 61.94 − 57 = − 221.12

Mcd = + 54 + 2 x 92.9 − 19.45 = + 220.35

Mdc = − 54 − 2 x 19.45 + 92.9 = zero

The beam has been analyzed and we can draw shear force and bending moment diagrams as usual.


6.6. Rotation Contribution Method: Application to frames without side sway. Example No 2:

Analyze the following frame by Kanis method ( rotation Contribution Method )

A B

D

C2I3I

6 10

9K

2I 10

12

1 k/ft

Step No. 1 Relative Stiffness.

Span I L IL Krel K modified.

AB 3 16 3

16 × 240 45 45

BC 2 12 2

12 × 240 40

3

4 30 (Exterior hinged end)

BD 2 10 2

10 × 240 48 48 .

∑103

Step No.2. FEM’s

Mfab = 9 × 6 × 102

162 = + 21.1 K-ft

Mfba = 9 × 10 × 62

162 = − 12.65

Mfbc = 1 × 122

12 = + 12

Mfcb = − 12

Mfbd = Mfdb = 0 ( No load within span BD)


Step No. 3. Draw Boxes, enter values of FEM’s, rotation contribution factors etc.

A

CB

+21.1 -12.650 +0.119 -0.97 -1.03

rot. cont.factor.

-0.195

-0.65-0.183 -0.122

+12 -12+0.079 +5.96-0.647 +6.32-0.69 +6.345

-12-0.5

+0.126-1.03-1.10

0

FEM's

00

D(rotation contribution factor)

*

* * * *

*

*

Apply all relevant rules in three cycles. Final end moments may now be calculated.

For beams. End moment = FEM + 2 x near end contribution. + Far end contribution For Columns : End moment = FEM + 2 x near end contribution + Far end contribution + Linear displacement contribution of that column. To be taken in sway cases only. Mab = 21.1 + 2x0 –1.03 = + 20.07 K−ft Mba = −12.65 –2 x 1.03 + 0 = −14.71 Mbc = +12 –2 x 0.69 + 6.345 = 16.965 Mbd = 0 – 2x1.1 +0 = −2.2 Mcb = −12 + 2x 6.345 –0.69 = 0 Mdb = 0 + 2x0−1.10 = −1.10 Equilibrium checks are satisfied. End moment values are OK. Now SFD and BMD can be drawn as usual. Example No. 3: Analyse the following frame by rotation Contribution Method. SOLUTION:- It can be seen that sway case is there.

10

A

B C

D

20

I

4I155

16k

I


Step No. 1. Relative Stiffness.

Member. I L IL Krel

AB 1 10 1

10 × 10 1

BC 4 20 4

20 × 10 2

CD 1 10 1

10 × 10 1

Step No. 2. FEM’s

MfBC = + 16 × 5 × 152

202 = + 45

MfCB = − 16 × 52 × 15

202 = − 15

All other fixing moments are zero. Step No.3 Draw Boxes, enter FEM’s and rotation Contribution factors etc. Apply three cycles.

+45 -0.333 -0.333+45 -15-14.98 +9.98-18.93 +10.67-19.57 +10.47

-15

-0.167-0.167

0 0

000 0

FEMs

-7.51-9.49-9.80

-3/2(1/2)= -0.75

Linear disp.factors

AD

+1.8825+3.105+3.41

B C

LDCLDC

Rotation factorRotation factor

+5.0+5.35+5.25

-0.75

*

*

*

*

**

LDF

* = rotation factors.

See explanation of calculations on next page. Note: After applying the first cycle as usual, calculate linear displacement contribution for columns of all storeys. Repeat this calculation after every cycle. Linear displacement contribution (LDC) of a column=Linear displacement factor [ story moment + contribution of column ends of that storey) Storey moment is zero because no horizontal load acts in column and there is no storey shear. ↓ After 1st cycle: Linear Disp. Cont = − 0.75 [ 0 + 5.0 − 7.5 + 0 + 0] = + 1.8825 → For 2nd cycle onwards to calculate rotation contribution, apply following Rule:− Rotation contribution = rotation contribution factor [restrained moment + far end contributions + linear displacement contribution of columns. of different. storeys meeting at that joint.]


2nd cycle. A C( Far ends) ↓ ↓ Joint B. − 0.167 [ +45 + 0 + 9.98 + 1.8825 ] = − 9.49 (Span BA)

and − 0.333 [ −−−−−−− do −−−−−−−− ] = − 18.93 (Span BC)

Joint C. − 0.333 [ − 15 − 18.93 + 0 + 1.8825 ] = + 10.67 (Span CB) and − 0.167 [ −−−−−−−− do −−−−−−−− ] = + 5.35 (Span CD)

After 2nd cycle. Linear displacement contribution is equall to

storey moment.

↓ = − 0.75 [ 0 − 9.49 + 0 + 5.35 + 0 ] = + 3.105 After 3rd cycle. After 3rd cycle , linear displacement. contribution of columns is equall to storey moment.

↓

= − 0.75 [ 0 − 9.80 + 5.25 + 0 + 0 ] = 3.41 Calculate end moments after 3rd cycle. For beams: End moment = FEM + 2 near end contribution. + Far end contribution. For columns. End moment = FEM + 2 near end contribution + Far end contribution. + linear displacement. contribution of that column. Applying these rules

Mab = 0 + 0 − 9.80 + 3.41 = − 6.3875 k.ft.

Mba = + 0 − 2 × 9.80 + 0 + 3.41 = + 16.19

Mbc = + 45 − 2 × 19.57 + 10.47 = + 16.33

Mcb = − 15 + 2 × 10.47 − 19.57 = 13.63

Mcd = 0 + 2 × 5.25 + 0 + 3.41 = 13.91

Mdc = 0 + 2 × 0 + 5.25 + 3.41 = 8.66 By increasing number of cycles the accuracy is increased.


Example No 4 : Solve the following double story frame carrying gravity and lateral loads by rotation contribution method.

3 KN/m B

C D

E

FA

2 KN/m

( )I

5mI

3m 2I

3m 2I

2 KN/m

( )I

2I2I

2I 2I

SOLUTION :− If this is analyzed by slope-deflection or Moment distribution method, it becomes very lengthy and laborious. This becomes easier if solved by rotation contribution method. Step 1: F.E.Ms.

Mfab = + 3 × 32

12 = + 2.25 KN−m

Mfba = − 2.25 KN−m

Mfbc = + 2.25 KN−m

Mfcb = − 2.25 KN−m

Mfcd = 2 × 52

12 = + 4.17 KN−m

Mfdc = − 4.17 KN−m

Mfbe = + 4.17 KN−m

Mfeb = − 4.17 KN−m.

Mfde = Mfed = 0

Mfef = Mffe = 0

Step 2: RELATIVE STIFFNESS :−

Span I L

IL K

AB 2 3 23 × 15 10

BC 2 3 23 × 15 10


BE 1 5 15 × 15 3

CD 1 5 15 × 15 3

DF 2 3 23 × 15 10

EF 2 3 23 × 15 10

LINEAR DISPLACEMENT FACTOR = L.D.F. of a column of a particular storey.

L.D.F. = − 32

K∑K

Where K is the stiffness of that column & ∑K is the stiffness of columns of that storey. Assuming columns of equal sizes in a story. ( EI same)

L.D.F1 = − 32 ×

10(10 + 10) = − 0.75 (For story No. 1)

L.D.F2 = − 32 ×

10(10 + 10) = − 0.75 (For story No. 2)

Storey Shear :− This is, in fact, reaction at the slab or beam level due to horizontal forces. If storey shear causes a (−ve) value of R, it will be (−ve) & vice versa. For determining storey shear the columns can be treated as simply supported vertical beams. (1) Storey shear = − 9 KN ( For lower or ground story. At the slab level of ground story) (2) Storey shear = − 4.5 ( For upper story ). At the slab level of upper story root) Storey Moment ( S.M) :− S.M. = Storey shear + h/3 where h is the height of that storey.

SM1 = − 9 × 33 = − 9 ( lower story )

S.M2 = − 4.5 × 33 = − 4.5 ( Upper story )

Rotation Factors The sum of rotation factors at a joint is − ½. The rotation factors are obtained by dividing the value − ½ between different members meeting at a joint in proportion to their K values.


µab = − 12

k1

∑k

µac = − 12

k2

∑k etc.

Rotation Contributions:− The rule for calculating rotation contribution is as follows.

Sum the restrained moments of a point and all rotation contribution of the far ends of the members

meeting at a joint. Multiply this sum by respective rotation factors to get the required rotation

contribution. For the first cycle far end contribution can be taken as zero.

Span K Rotation factor. AB 10 0 (Being fixed end)

BC 10 − 12

10

23 = − 0.217

BE 3 − 0.5

3

23 = − 0.065

BA 10 − 0.5

10

23 = − 0.217

CB 10 − 0.385 CD 3 − 0.115 DC 3 − 0.115 DE 10 − 0.385 ED 10 − 0.217 EB 3 − 0.065 EF 10 − 0.217 FE 10 0 (Being fixed end) Now draw boxes, enter FEMs values, rotation factors etc. As it is a two storeyed frame, calculations on a single A4 size paper may not be possible. A reduced page showing calculation is annexed.


CRestrainingMoment1.92

cb = -0.385

cd=-0.115

F.E.M.= +4.17R.C.-0.12-0.25-0.52-0.76-0.95-1.09-1.19-1.26

R.C.=Rotation Con-tribution.

-0.39-0.89-1.74-2.55-3.18-3.65-3.99-4.23

-6.50-6.30-6.00-5.61-5.05-4.24-2.46-0.9

R.C.

Linear Dis-placementfactor(L.D.F)-0.75

Linear Displacement ContributionL.D.C.2.76.719.8712.2514.0015.316.2116.21

F.E.M. = +2.25

-4.17=F.E.M.

R.C.0.490.13-0.11-0.3-0.45-0.56-0.64-0.7

-4.17 Ddc=-0.115

de=-0.385

R.C. F.E.M. = 01.650.45-0.35-1.00-1.50-1.87-2.14-2.34

-4.93-4.69-4.37-3.88-3.23-2.33-1.10.55

R.C. F.E.M. = 0

L.D.C.2.76.719.8712.2514.0015.3016.2116.21

L.D.F.= -0.75

bc=-0.217

B 4.17

ba=-0.217

=-0.065be

F.E.M. = +4.17

R.C.-0.27-0.89-1.27-1.51-1.68-1.89-1.95-1.95

F.E.M. = -2.25

L.D.C.79.811.6812.9613.8714.5314.9915.00

R.C.-0.9-2.96-4.24-5.05-5.61-6.00-6.30-6.50

L.D.F.=-0.75

F.E.M. = +2.25

A

-4.17 E

F.E.M. = 0

L.D.F.= -0.75

F.E.M. = -4.17

ed=-0.217

ef=-0.217

eb=-0.065

R.C. F.E.M. = 0

R.C.0.16-0.33-0.70-0.97-1.16-1.31-1.41-1.48

R.C.0.55-1.1-2.333.23-3.88-4.37-4.69-4.93

F.E.M. = 0

L.D.C.79.811.6812.9613.8714.5314.9915.00

F

Double – storey frame carrying gravity and lateral loads – Analysed by Rotation Contribution Method.


First Cycle :− Near end contribution = Rotation factor of respective member (Restrained moment + far end contributions). Joint B = R.F. ( 4.17 ) C = R.F. ( 1.92 − 0.9 ) D = R.F. (− 4.17 − 0.12) E = R.F. (− 4.17 + 1.65) After First Cycle :− Linear Displacement Contribution :−= L.D.F.[Storey moment + Rotation contribution at the end of columns of that storey]. L.D.C1 = − 0.75 (− 9 − 0.9 + 0.55) = 7 L.D.C2 = − 0.75 ( 4.5 − 0.9 − 0.39 + 0.55 + 1.65) = 2.7 For 2nd Cycle And Onwards :− Near end contribution = R.F.[Restrained moment + Far end contribution + Linear displacement

contributions of columns of different storeys meeting at that joint]

Joint B= R.F. (4.17 + 0.16 − 0.39 + 7 + 2.7 )

C= ″ (1.92 + 0.49 − 2.96 + 2.7)

D= ″ (− 4.17 − 0.25 + 0.55 + 2.7)

E= ″ (− 4.17 + 0.45 − 0.89 + 2.7 + 7 ).

After 2nd Cycle :− L.D.C1 = − 0.75 (− 9 − 2.96 − 1.1) = 9.8 L.D.C2 = − 0.75 (− 4.5 − 2.96 − 0.83 − 1.1 + 0.45) = 6.71 3rd Cycle :− Joint B= R.F. ( 4.17 − 0.33 − 0.83 + 9.8 + 6.71)

C= ″ ( 1.92 + 0.13 − 4.24 + 6.71 )

D= ″ (− 4.17 − 1.1 − 0.52 + 6.71)

E= ″ (− 4.17 − 1.27 − 0.35 + 9.8 + 6.71)


After 3rd Cycle :− L.D.C1 = − 0.75 (− 9 − 4.24 − 2.33) = 11.68 L.D.C2 = − 0.75 (− 4.5 − 1.74 − 4.24 − 0.35 − 2.33) = 9.87 4th Cycle :− Joint B= R.F. ( 4.17 − 0.70 − 1.74 + 11.68 + 9.87)

C= ″ ( 1.92 − 0.11 − 5.05 + 9.87)

D= ″ (− 4.17 − 0.76 − 2.33 + 9.87 )

E= ″ (− 4.17 − 1 − 1.51 + 9.87 + 11.68).

After 4th Cycle :− L.D.C1 = − 0.75 (− 9 − 5.05 − 3.23) = 12.96 L.D.C2 = − 0.75 (− 4.5 − 5.05 − 2.55 − 1.00 − 3.23) = 12.25 5th Cycle :− Joint B= R.F. (4.17 − 0.97 − 2.55 + 12.25 + 12.96)

C= ″ ( 1.92 − 0.3 − 5.61 + 12.25)

D= ″ (− 4.17 − 0.95 − 3.23 + 12.25 )

E= ″ (− 4.17 − 1.5 − 1.68 + 12.25 + 12.96)

After 5th Cycle :− L.D.C1 = − 0.75 (− 9 − 5.61 − 3.88) = 13.87 (ground storey) L.D.C2 = − 0.75 (− 4.5 − 5.61 − 3.18 − 1.5 − 3.88 ) = 14 (First Floor) 6th Cycle :− Joint B = R.F. (4.17 − 1.16 − 3.18 + 14 + 13.87 )

C = ″ (1.92 − 0.05 − 6 + 14)

D = ″ (− 4.17 − 3.88 − 1.09 + 14)

E = ″ (− 4.17 − 1.87 − 1.68 + 14 + 13.87)


After 6th Cycle :− L.D.C1 = − 0.75 ( − 9 − 6 − 4.37) = 14.53 L.D.C2 = − 0.75 (− 4.5 − 6 − 3.65 − 1.87 − 4.37) = 15.3 7th Cycle :− Joint B = R.F. (4.17 − 1.31 − 3.65 + 15.3 + 14.53)

C = ″ (1.92 − 0.56 − 6.30 + 15.3)

D = ″ (− 4.17 − 1.19 − 4.37 + 15.3)

E = ″ (− 4.17 − 1.89 − 2.14 + 15.3 + 14.53)

After 7th Cycle :− L.D.C1 = − 0.75 (− 9 − 6.30 − 4.69 ) = 14.99

L.D.C2 = − 0.75 (− 4.5 − 6.3 − 3.99 − 2.14 − 4.69 ) = 16.21

8th Cycle :− Joint B = R.F. (4.17 − 1.41 − 3.99 + 16.21 + 14.99)

C = ″ (1.92 − 6.5 − 0.64 + 16.21)

D = ″ (− 4.17 − 4.69 − 1.26 + 16.21)

E = ″ (− 4.17 − 2.34 − 1.95 + 16.21 + 14.99)

After 8th Cycle :− L.D.C1 = − 0.75 (− 9 − 6.5 − 4.93) ≅ 15

L.D.C2 = − 0.75 (− 4.5 − 6.5 − 4.23 − 4.93 − 2.34).≅ 16.21

FINAL END MOMENTS :−

(1) Beams or Slabs :− = F.E.M + 2 (near end contribution) + far end contribution of that particular beam or slab. (2) For Columns :−

= F.E.M + 2 (near end contribution) + far end contribution of that particular column + L.D.C. of that column. Applying these rules we get the following end moments.


END MOMENTS :− Mab = 2.25 + 2 × 0 − 6.5 + 15 = + 10.75 KN−m

Mba = − 2.25 − 2 (6.5) − 1 + 15 = − 0.25 ″

Mbc = 2.25 − 2 × 6.5 − 4.23 + 16.21 = + 1.23 ″

Mbe = 4.17 − 2 (1.95) − 1.48 = − 1.21 ″

Mcb = − 2.25 − 2 × 4.23 − 6.5 + 16.21 = − 1 ″

Mcd = 4.17 − 2 × 1.26 − 0.7 = + 0.95≅+1 ″

Mdc = − 4.17 − 2 × 0.7 − 1.26 = − 6.83 ″

Mde = 0 − 2 × 2.34 − 4.93 + 16.21 = + 6.60 ″

Med = 0 − 2 × 4.93 − 2.34 + 16.21 = + 4.01 ″

Meb = − 4.17 − 2 × 1.48 − 1.95 = − 9.08 KN−m

Mef = 0 − 2 × 4.93 + 15 = + 5.14 ″

Mfe = 0 − 2 × 0 − 4.93 + 15 = + 10.07 ″ Now frame is statically determinate and contains all end moments. It can be designed now. Space for notes:


CHAPTER SEVEN

7. INTRODUCTION TO COLUMN ANALOGY METHOD

The column analogy method was also proposed by Prof. Hardy Cross and is a powerful technique to analyze the beams with fixed supports, fixed ended gable frames, closed frames & fixed arches etc., These members may be of uniform or variable moment of inertia throughout their lengths but the method is ideally suited to the calculation of the stiffness factor and the carryover factor for the members having variable moment of inertia. The method is strictly applicable to a maximum of 3rd degree of indeterminacy. This method is essentially an indirect application of the consistent deformation method.

The method is based on a mathematical similarity (i.e. analogy) between the stresses developed on a column section subjected to eccentric load and the moments imposed on a member due to fixity of its supports. *(We have already used an analogy in the form of method of moment and shear in which it was assumed that parallel chord trusses behave as a deep beam). In the analysis of actual engineering structures of modern times, so many analogies are used like slab analogy, and shell analogy etc. In all these methods, calculations are not made directly on the actual structure but, in fact it is always assumed that the actual structure has been replaced by its mathematical model and the calculations are made on the model. The final results are related to the actual structure through same logical engineering interpretation.

In the method of column analogy, the actual structure is considered under the action of applied loads and the redundants acting simultaneously on a BDS. The load on the top of the analogous column is usually the B.M.D. due to applied loads on simple spans and therefore the reaction to this applied load is the B.M.D. due to redundants on simple spans considers the following fixed ended loaded beam.

(d) Loading on top of analogous column, Ms diagram, same as(b).

L

1 (Unity)

(e) X-section of analogous column.

(f) Pressure on bottom of analogous column, Mi diagram.

E =Constt.I(a) Given beam under loads

Ma MBA

P1 P2

BL

0 0

00

MBMA (c) B.M.D. due to

redundants,

plottedon

the

compression

side on simple span

(b) B.M.D. due toapplied loads,plotted on thecompressin side.

WKN/m

on simple span

MaMb

COLUMN ANALOGY METHOD 307

The resultant of B.M.D’s due to applied loads does not fall on the mid point of analogous column section which is eccentrically loaded. Msdiagram = BDS moment diagram due to applied loads. Mi diagram = Indeterminate moment diagram due to redundants. If we plot (+ve) B.M.D. above the zero line and (−ve) B.M.D below the zero line (both on compression sides due to two sets of loads) then we can say that these diagrams have been plotted on the compression side. (The conditions from which MA & MB can be determined, when the method of consistent deformation is used, are as follows). From the Geometry requirements, we know that (1) The change of slope between points A & B = 0; or sum of area of moment diagrams between

A & B = 0 (note that EI = Constt:), or area of moment diagrams of fig.b = area of moment diagram of fig..c.

(2) The deviation of point B from tangent at A = 0; or sum of moment of moment diagrams between A

& B about B = 0, or Moment of moment diagram of fig.(b) about B = moment of moment diagram of fig.(c) about B. Above two requirements can be stated as follows.

(1) Total load on the top is equal to the total pressure at the bottom and; (2) Moment of load about B is equal to the moment of pressure about B), indicates that the analogous column is on equilibrium under the action of applied loads and the redundants. 7.1. SIGN CONVENTIONS:−

It is necessary to establish a sign convention regarding the nature of the applied load (Ms − diagram) and the pressures acting at the base of the analogous column (Mi−diagram.) 1. Load ( P) on top of the analogous column is downward if Ms/EI diagram is (+ve) which means that

it causes compression on the outside or (sagging) in BDS vice-versa. If EI is constant, it can be taken equal to units.

Inside

Outside

CT

2. Upward pressure on bottom of the analogous column ( Mi − diagram) is considered as (+ve). 3. Moment (M) at any point of the given indeterminate structure ( maximum to 3rd degree) is given by

the formula.

M = Ms − Mi, which is (+ve) if it causes compression on the outside of members.


EXAMPLE NO. 1:− Determine the fixed−ended moments for the beam shown below by the method of column analogy. SOLUTION: Choosing BDS as a simple beam. Draw Ms diagram. Please it on analogous column.

A

W/Unit length.

B

L

EI=Constt.

2WL8

WL12 WL

12

+0 0

Ms-diagram(B.M.D. due to appliedloads on B.D.S.)Loading on top ofanalogous column.3

3

L

1X-section ofanalogous column

0

0

WL12

2

WL12

2

WL12

2

WL12

2WL2/24

+

Mi-diagamPressure on bottom ofanalogous column.(uniform asresultant falls on the mid point of analogous column section

(Final BMD) M = Ms - Mi

Pressure at the base of the column = PA

A = L × I (area of analogous column section).

= WL3

12(Lx1)

Mi = WL2

12 . In this case, it will be uniform as resultant of Ms

diagram falls on centroid of analogous column)

(MS)a = 0 , (Ms at point A to be picked up for M-s diagram) Ma = (Ms − Mi)a , (net moment at point A)

= 0 − WL2

12

Ma = − WL2

12

Mb = (Ms−Mi)b =

0 −

WL2

12 = −WL2

12

Mc = (Ms − Mi)c = WL2

8 − WL2

12

Mc = 3 WL2 − 2 WL2

24 = WL2

24 . Plot these values to get M = Ms − Mi diagram.

The beam has been analyzed.


EXAMPLE NO. 2:- SOLVING THE PREVIOUS EXAMPLE, IF B.D.S. IS A CANTILEVER SUPPORTED AT ‘A’.

A BEI=Constt.

L

00 Ms-diagram(It creates hagging so load acts upwards)The resultant of Ms diagram does not fall onthe centroid of analogous column.

L/4 L/2

L/4 3/4L

Lyo

M

yo

W L6

WL6

3

3

WL2

2

X-section ofanalogous column. Carrying eccentric load of WL /6

3

Eccentric load wL3/6 acts on centre of

analogous column x-section with an associated moment as well(Eccentric load = Concentric load plus accomprying moment)

W/unit-length

1

Centroidal axis

Area of Ms diagram A = bh

(n+1) = L × WL2

2(2+1) = WL3

6

X′ = b

(n+2) = L

(2+2) = L4 (from nearest and)

Alternatively centroid can be located by using the following formula)

X = ∫ MXdX∫ MdX

∫ MdX = L

∫o

−

WX2

2 dX = − W2

X3

3 L|o = −

WL3

6 ( Same as above)

∫ MXdX = L

∫o

−

WX2

2 XdX = L

∫o −

WX3

2 dx

= − W2

X4

4 L|o = −

WL4

8

X−

= ∫ MXdX∫ MdX


X−

= − WL4

8 × 6

(−WL3) = 34 L. (from the origin of moment

expression or from farthest end) NOTE : Moment expression is always independent of the variation of inertia. Properties of Analogous Column X−section :− 1. Area of analogous column section, A = L × 1 = L

2. Moment of inertia, I yo yo = L 3

12

3. Location of centroidal column axis, C = L2

A e’=M =

WL3

6

L

4 = WL4

24 , ( L4 is distance between axis yo− yo and the centroid of Ms diagram

where the load equal to area of Ms diagram acts.)

(Mi)a = PA ±

McI (P is the area of Ms diagram and is acting upwards so negative

C = L2 and I =

L3

12 )

= −WL3

6 . L − WL4 . L . 12

24 . 2 . L3 (Load P on analogous column is negative)

= − WL2

6 − WL2

4 ( Reaction due to MC/I would be having the same

direction at A as that due to P while at B these

= −2WL2 − 3 WL2

12 two would be opposite)

= − 512 WL2

(Ms)a = − WL2

2

Ma = (Ms − Mi)a

= − WL2

2 + 5

12 WL2

= − 6 WL2 + 5 WL2

12

Ma = − WL2

12


Mb = (Ms − Mi)b

(Mi)b = PA ±

McI

= − WL3

6 × L + WL4 × L × 12

24 × 2 × L3

= −WL2

6 + WL2

4

= − 2WL2 + 3 WL2

12

= WL2

12

(Ms)b = 0

Mb = (Ms − Mi)b = 0 − WL2

12 = − WL2

12

Same results have been obtained but effort / time involved is more for this BDS). EXAMPLE NO. 3:− Determine the F.E.Ms. by the method of column analogy for the following loaded beam. 3.1 SOLUTION:− CASE 1 ( WHEN BDS IS A SIMPLE BEAM )

P

ba

LPab

L

+

L+a3

L+b3

Pab2

12

(Pab)L

Pab2xL=

e

M

L

1

Ms-diagram

x-section of analogous column

e = L2 −

L + a

3 = 3 L − 2 L − 2a

6 =

L − 2 a

6 ( The eccentricity of load w.r.t

mid point of analogous column)

M =

Pab

2

L − 2 a

6 = Pab12 (L − 2a)


Properties of Analogous Column X − section . 1. A = L × 1 = L

2. I = L3

12

3. C = L2

(Mi)a = PA ±

McI

= Pab2 L +

Pab 12 (L − 2a) ×

L × 12 2 × L3

= Pab2 L +

Pab2 L2 (L − 2a)

= PabL + PabL − 2 Pa2b

2 L2

= 2 PabL − 2 Pa2b

2 L2

(Mi)a = PabL − Pa2b

L2

= Pab (L − a)

L2 ∴ a + b = L

b = L − a

= Pab . b

L2

(Mi)a = Pab2

L2

(Ms) a = 0 Net moment at A = Ma = (Ms − Mi) a

= 0 − Pab2

L2

Ma = − Pab2

L2


The (−ve) sign means that it gives us tension at the top when applied at A.

(Mi)b = PA ±

MCI

= Pab2L −

Pab12L2 (L − 2a) ×

L × 122 × L3

= Pab2L −

Pab2L2 (L − 2a)

= PabL − PabL + 2Pa2b

2L2

= 2Pa2b2L2

(Mi)b = Pa2bL2

(Ms)b = 0

Mb = (Ms − Mi)a = 0 − Pa2bL2

Mb = − Pa2b

L2

The minus sign means that it gives us tension at the top. EXERCISE 3.2:- If B.D.S. is a cantilever supported at A:− We solve the same exercise 3.1 but with a different BDS.

A a b

P

BL

EI=Constt

0 0

Pa

e Pa2

2

M

LL/2

1

22Pa

12 Pa(a) =

Ms-diagram (load equal to area ofMs diagram acts upwards)

The upper eccentric load has been nowplaced on centroid axis of analogous columnsection plus accompaying moment.

x-section of analogous column underload and accompaying moment at columncentroidal aixis.

L2

a3


e = L2 −

a3 =

3L − 2a

6

Pe = M = Pa2

2

3L − 2a

6 = Pa2 (3L − 2a)

12

Properties of Analogous Column section :− A = L , I = L3

12 , C = L2

(Mi)a = PA ±

MCI

= − Pa2

2L − Pa2 (3L − 2a) . L . 12

12 . 2 . L3 (Due to upward P= Pa2/2, reaction at A

and B is downwards while due to moment,

= − Pa2

2L − Pa2 (3L − 2a)

2L2 reaction at B is upwards while at A it is

downwards. Similar directions will have

= −Pa2L − 3Pa2L + 2Pa3

2L2 the same sign to be additive or vice−versa)

= −4 Pa2L + 2Pa3

2L2

= −2Pa2L + Pa3

L2

= Pa2 (a − 2L)

L2

= −Pa2 (2L − a)

L2 , We can write 2L − a = L + L − a = L + b

(Mi)a = −Pa2 (L + b)

L2

(Ms)a = − Pa Ma = (Ms − Mi)a

= −Pa + Pa2(L + b)

L2

= − PaL2 + Pa2 L + Pa2b

L2


= − PaL (L − a) + Pa2 b

L2

= − PabL + Pa2 b

L2

= − Pab (L − a)

L2

= − Pab . b

L2

Ma = − Pab2

L2 ( Same result as was obtained with a different BDS)

(Mi)b = PA ±

MCI

= − Pa2

2L + Pa2 (3L − 2a)

2L2

= − Pa2 L + 3Pa2L − 2Pa3

2L2

= 2 Pa2 L − 2Pa3

2L2

= Pa2 L − Pa3

L2

= +Pa2 (L − a )

L2

(Mi)b = Pa2 b

L2

(Ms)b = 0 Mb = (Ms − Mi)b

= 0 − Pa2 b

L2

Mb = − Pa2 b

L2 ( Same result as obtained with a different BDS)


EXAMPLE NO.4:− Determine the F.F.Ms. by the method of column analogy for the following loaded beam. SOLUTION:− Choosing cantilever supported at B as BDS.

L8

A B

L/2 L/2

w/unit length

E = ConsttI

0

If B.D.S. is a cantileversupported at b.

0

WL48

WL48

3

3

= WL2

x L4

WL8

2

M3/8 L

1

LAnalogous columnsection.

e=

Ms-diagram

Eccentricity = e = L2 −

L8 =

4L − L8 =

3L8

Moment = Pe = M = WL3

48 × 3L8 =

WL4

128 Where P = Area of Ms diagram=WL3

48 =

bh

n+1

Properties of Analogous column section. A = L, I =

L3

12 and C = L2

Step 1: Apply P= Area Of BMD(Ms diagram ) due to applied loads in a BDS at the center of analogous column section i.e. at L/2 from either side.

Step 2: The accompanying moment Pe, where e is the eccentricity between mid point of analogous column section and the point of application of area of Ms diagram, is also applied at the same point along with P.

Step 3: Imagine reactions due to P and M=Pe. At points A and B, use appropriate signs.

(Mi)a = PA ±

MCI ( Subtractive reaction at A due to P)

= − WL3

48.L + WL4 × L × 12128 × 2 × L3 ( P is upwards, so negative. Reactions due to this P

at A and B will be downwards and those due to moment term will be upward at A and downward

= − WL2

48 + 3WL2

64 at B. Use opposite signs now for A)

= − 4WL2 + 9WL2

192

= + 5 WL2

192


(Ms)a = 0 ( Inspect BMD drawn on simple determinate span) Ma = (Ms − Mi)a

= 0 − 5WL2

192

Ma = − 5WL2

192

(Mi)b = PA ±

MCI ( Additive reactions at B as use negative sign with

McI term)

= − 4WL2 − 9WL2

192

= − 13 WL2

192

(Ms)b = − WL2

8

Mb = (Ms − Mi)b

= − WL2

8 + 13 WL2

192 =

−24 WL2 + 13 WL2

192

Mb = −11192 WL2

The beam is now statically determinate etc. EXAMPLE NO. 5:− Determine the F.E. M’s by the method of column analogy for the following loaded beam. SOLUTION:−

A= bh

n+1

= WL4 192

X= b

n+2

=

L

2(3+2)

X= L

10

AL/2 L/2

B

EI=Constt:WL192

WL192

L10 00

eM

4

L

1xWxL 2 2

x L 3

(L) 2

= WL 24

3

1

W/Unit length

Analogous column section

Ms-diagram ( )

4

WL3

24

L 2 x

A = 4

e = L2 −

L10 =

5L − L10 =

4L10 =

25 L

M =

WL4

192 ×

2 L

5 = WL5

480

Comment [A1]:


Properties of Analogous column section.

A = L , I = L3

12, C = L2

(Mi)a = PA ±

MCI

(Mi)a = − WL4

192L + WL5 × L × 12480 × 2 × L3 (Downward reaction at A due to P and upward reaction at A due to M)

= − WL3

192 + WL3

80

= − 80WL3 + 192 WL3

15360

= 112 WL3

15360 ( Divide by 16)

(Mi)a = 7 WL3

960

(Ms)a = 0 Ma = (Ms − Mi)a

Ma = 0 − 7

960 WL3 = − 7960 WL3

(Mi)b = PA ±

MCI

= − WL3

192 − WL3

80

= − 80 WL3 − 192 WL3

15360

= − 272 WL3

15360

= − 17 WL3

960

(Ms)b = − WL3

24

Mb = (Ms − Mi) b


= − WL3

24 + 17

960 WL3

= − 40 WL3 + 17 WL3

960

Mb = − 23 WL3

960

Note : After these redundant end moments have been determined, the beam is statically

determinate and reactions , S.F, B.M, rotations and deflections anywhere can be found.

7.2. STRAIGHT MEMBERS WITH VARIABLE CROSS − SECTION. EXAMPLE NO. 6:− Determine the fixed−end moments for the beam shown by the method of column analogy SOLUTION:__ BDS is a simple beam.

3kn/m90kn

4mBA

6m 10m

3x168b90P2

3.83mMsEI

dia. dueto U.D.L.only. 0 a

= 96

0

3kn/m

6m 10m 24kn24knM=24x6-3 x (6)2=90kn-m2

12m90kn4m

67.5kn90x416

=22.5kn

1M=22.5x6 =135kn-m

Analogouscolumnx-section.

6.85m9.15m

1/2

yo P3x

4m

P48m 8m

P1 90x12x416= 270

67.5

16+43 =6.67m

dia dueto pointload only.

MsEI

2I=2 I=1

M

45

135

C0 (reactions due to UDL)

(reactions due toconcentrated load)

The above two MsEI diagrams will be taken full first and then load corresponding to areas of these

diagrams on left 6m distance will be subtracted. (P2 and P4 will be subtracted from P1 and P3 respectively). In this solution, two basic determinate structures are possible.

(1) a simply supported beam.

(2) a cantilever beam.


This problem is different from the previous one in the following respects. (a) Ms − diagram has to be divided by a given value of I for various portions of span. (b) The thickness of the analogous column X − section will also vary with the variation of

inertia. Normally, the width 1/EI can be set equal to unity as was the case in previous problem, when EI was set equal to unity.

(c) As the dimension of the analogous column X − section also varies in this case, we will have

to locate the centroidal axis of the column and determine its moment of inertia about it. (1) SOLUTION:- By choosing a simple beam as a B.D.S.

P1 = 23 × 16 × 96 = 1024 KN ( Load corresponding to area of entire BMD due to UDL)

∫ MdX = 6

∫o (24X − 1.5 X2) dX (Simply supported beam moment due to UDL of left 6/ portion)

= 12X2 − 0.5X3 6|o = 12 × 36 − 0.5 × 216 = 432 − 108 = 324

area of abc = 324

∫ MXdX = 6

∫o (24X − 1.5X2) XdX

= 6

∫o (24X2 − 1.5X3) dX

= 243 X3 −

1.54 X4

6|o = 8 × 63 −

1.54 × 64

= 1242

X = ∫ MxdX∫ MdX =

1242342 = 3.83 m from A. (of left 6/ portion of BMD)

P2 = 12 ( area abc) =

3242 = 162 KN( To be subtracted from Ms diagram )

P3 =

12 × 16 × 270 = 2160 KN ( Area of BMD due to concentrated Load)

P4 =

12 × 6 × 67.5 = 202.5 KN ( To be subtracted from Ms diagram )


Properties of Analogous column x − section.

Area = A = 1 × 10 + 12 × 6 = 13 m2

X = ∫XdA

A = (1 × 10) 5 + (1/2 × 6 × 13)

13 from R.H.S.

= 6.85 m ( From point B) . It is the location of centroidal axis Yo−Yo.

Iy0 y0 = 1 × 103

12 + 10(1.85)2 + 0.5 × 63

12 + (0.5 × 6) × (6.15)2 = 240 m4

by neglecting the contribution of left portion about its own centroidal axis. Total load to be applied at the centroid of analogous column x − section. = P1 + P3 − P2 − P4

= 1024 + 2160 − 162 − 202.5 = 2819.5 KN Applied Moment about centroidal axis = M = + 1024 (1.15) − 2160 (0.18) − 162 (5.32) − 202.5 (5.15) = − 1116 KN−m , clockwise (Note: distance 5.32 = 9.15 − 3.83 (and 5.15 = 9.15 − 4) The (−ve) sign indicates that the net applied moment is clockwise.

(Mi)a = PA ±

MCI ( subtractive reactions at A)

= 2819.5

13 − 1116 × 9.15

240 , (Preserve at A due to McI is downwards so negative).

= + 174.34 KN−m (Ms)a = 0 Ma = (Ms − Mi)a = 0 − 174.34 = − 174.34 KN−m

(Mi)b = 2819.5

13 + 1116 × 6.85

240 , ( Note the difference in the values of C for points A and B.)

= + 248.74 KN−m

(Ms)b = 0

Mb = (Ms − Mi)b

= 0 − 248.74

= − 248.74 KN−m

The −ve sign with Ma & Mb indicates that these cause compression on the inside when applied of these points.


EXAMPLE NO.7:− Determine the F.E.Ms. by the method of column analogy. SOLUTION:−

1. Choosing a simple beam as a B.D.S.

A IC B

90kn

4m3kn/m

2 CI3m 6m

1.95

m

P2 P3

2.58

6.5 mf

P1

2.25

a

3x132

8 =63.45445

m

e

P5 2m

83

41.5124.62

P62.67

mdiagram due to point load.

diagram due to U.D.L.

MsEI

MsEI

3KN/m

19.5

13m

19.5

yo

yo

1/2 Analogous columnx-section

x877.6kn-m

175.9knP4

6.66m6.34m

1

90x9x413 =249.23

(13+4)3 =5.67m

b d

½

2 CI

c

27

249.23

(BDS under UDL)

(M3)L = 19.5 × 3 − 1.5(3)2 = 45 KN−m ( 3m from A )

(M4)R = 19.5 × 4 − 32 (4)2 = 54 KN−m ( 4m from B)

90

4m9m27.69 62.307

(BDS under point load)

(M3)L = 27.69 × 3 = 83 KN−m ( 3m from A) ( M4)R = 62.307 × 4 = 249.22 (4m from B)

∫ MdX = area abc = 3

∫o (19.5 X − 1.5 X2) dX


= 19.5

2 X2 − 1.53 X3

3|o = 74.25

∫ MXdX = 3

∫o (19.5 X2 − 1.5 X3) dX =

19.53 X3 −

1.54 X4

3|o

= 145.12

X = 145.1274.25 = 1.95.m ( From point A as shown )

Area def = ∫ MdX = 4

∫o (19.5X − 1.5 x2) dX = 124

∫ MXdX = 4

∫o (19.5 X2 − 1.5 x3) dX

= 19.5X3

3 − 1.54 X4

4|o

= 320

X = 320124 = 2.58 m ( From point B )

P1 = 23 × 63.4 × 13 = 549.5 KN( Due to entire BMD due to UDL )

P2 = 12 (area abc) =

12 (74.25) = 37.125 KN ( To be subtracted )

P3 = 12 (area def) =

12 (124) = 62 KN ( To be subtracted )

P4 = 12 × 249.23 × 13 = 1620 KN ( Entire area of BMD due to point load)

P5 = 12 × 41.5 × 3 = 62.25 KN ( To be subtracted )

P6 = 12 × 4 × 124.62 = 249.23 KN ( To be subtracted )


Properties of Analogous column x − section.

A = 12 × 4 + 1 × 6 +

12 × 3 = 9.5m2

X = (0.5 × 4) × 2 + (1 × 6) × 7 + (0.5 × 3) × (11.5)

9.5

X = 6.66 ( From point B) meters

Iyoyo = 0.5 × 43

12 + (0.5 × 4)(4.68)2 + 1 × 63

12 + (1 × 6)(0.34)2

+ 0.5 × 32

12 + (1.5)(4.84)2

= 101.05 Total concentric load on analogous column x – section to be applied at centroidal column axis ) P = P1 − P2 − P3 + P4 − P5 − P6 = 549.5 − 37.125 − 62 + 1620 − 62.25 − 249.23 = 1759 KN Total applied moment at centroid of analogous column due to above six loads is = 549.5 (0.16) + 37.125 (4.39) − 62(4.08) + 1620 (0.99) + 62.25 (4.34) − 249.2 (3.99) = + 877.6 clockwise.

(Mi)a = PA ±

MCI ( Reactions due to P and M are subtractive at A)

= 17599.5 −

877.6 × 6.34101.05

= + 130 KN−m (Ms)a = 0 Ma = (Ms − Mi)a = 0 − 130 = − 130 KN−m

(Mi)b = PA ±

MCI

= 17599.5 +

877.6 × 6 × 6.66101.05 ( Reactions due to P and M are additive at B)

= + 243 KN−m (Ms)b = 0


Mb = (Ms − Mi)b = 0 − 243

Mb = − 243 KN−m Now the beam has become determinate. EXAMPLE NO. 7:- (2) Choosing cantilever supported at B as a B.D.S. Let us solve the loaded beam shown below again.

P3=

A

3KN/m 90KN

B

3m 6m 4m

P1=1098.53.25m

fa2.25m

b

d

60.79

P2=6.75

13.5

e121.5126.75

g

3x 13x13/2

=253.5367

1.33m

180360

1.33mPs=360KN

1/2

1/2

6.66myo

yo

6.34m

1089.75Kn

3894KN-m

A =

bhn+113x253.5

3=1098.5

=

A= bhn+1 = 4x360

2 =720

X' = bn+2

= 134 =3.25

X' = bn+2

= 43 =1.33

3KN/m 4m

B

253.5

10m

39

3m

A

1

C

Analogous column section

Ms/EI diagram due to point load

Ms/EI diagram due to u.d.l(2nd degree curve)

P4 = 720

2I I 2I

BDS under UDL

P1

P4=

Area abc = ∫ MdX = 3

∫o

−

32 X2 dX


= − 1.5 X3

3 3|o = 0.5 × 33 = − 13.5 ( Upwards to be subtracted)

∫ MXdX = 3

∫o (1.5X3)dX = −

1.5X4

4 3|o

= − 30.375 Location of centroidal axis from B: ( 1/2 × 3 + 1 × 6+1/2 × 4)X′ =(1/2 × 4 × 2+1 × 6 × 7+1/2 × 3 × 11.5) 9.5X’= 63.25 0r X’ = 6.66m from B or 6.34 m from A. (already done also)

location of centroid of area abc = X = − 30.375

− 13.5 = 2.25 m ( From A)

Area defg = ∫ MdX = 4

∫o (39X − 253.5 − 1.5X2)dX

Moment expression taken from B considering BDS under UDL.

= 39 X2

2 − 253.5 X − 1.53 X3

4|o

= − 734 (Area is always positive).

∫ MXdX = 4

∫o (39X2 − 253.5X − 1.5X3)dX

= 39X3

3 − 253.5X2

2 − 1.5X4

4 4|o

= − 1292

X = − 1292− 734

X = + 1.76 m From B (Centroid of area defg) P1 = 1098.5 KN ( Area of entire BMD due to UDL )

P2 = 12 (area abc) =

12 (13.5) = 6.75 K( To be subtracted)

P3 = 12 ( area defg) =

12 (734) = 367 KN( To be subtracted )

P4 = 720 KN( Area of entire BMD due to point Load )

P5 = 12 × 180 × 4 = 360 KN


Total concentric load on analogous column X − section is P = P1 + P2 + P3 − P4 + P5

= − 1098.5 + 6.75 + 367 − 720 + 360 = − 1084.75 KN( It is upward so reactions due to this will be downward) Total applied moment at centroid of column = − 6.75 (6.34 − 2.25) + 1098.5 (6.66 − 3.25) − 367 (6.66 − 1.76) + 720 (6.66 −1.33) − 360 (6.66 − 1.33) = 3894 KN−m (anticlockwise) Properties of Analogous column X − section. A =

12 × 4 + 1 × 6 +

12 × 3 = 9.5

X = 6.66 meters From B as in previous problem. Iyoyo = 101.05 m4 as in previous problem.

(Mi)a = PA ±

MCI ( Reactions are subtractive at A)

= − 1084.75

9.5 + 3894 × 6.34

101.05

(Mi)a = + 130 KN−m ( Same answer as in previous problem ) (Ms)a = 0 Ma = (Ms − Mi)a Ma = ( 0 − 130) = − 130 KN−m

(Mi)b = PA ±

MCI ( Reactions are additive at B )

= − 1084.75

9.5 − 3894 × 6.66

101.05

= − 370.83 KN−m (Ms)b = − 253.5 − 360 = − 613.5 KN−m Mb = (Ms − Mi)b = − 613.5 + 370.83 Mb = − 243 KN−m Now beam is determinate. Please note that the final values of redundant moments at supports remain the same for two BDS. However, amount of effort is different.


7.3. STIFFNESS AND CARRYOVER FACTORS FOR STRAIGHT MEMBERS WITH CONSTANT SECTION:__ For the given beam, choose a simple beam as BDS under Ma and Mb

A

L

Ma=K a Mb=(COF)Ma

BE =Constt:I

00 L/3 2/3L

0 0

a

a

L

L/32/3L

Ma

MaEI

= MaL 2EI

MbL

Mb

1

x L x

a

12

+

ba

M/E Loading on theIconjugate beam for asingle BDS.

Reaction on theconjugatebeam.

Analogouscolumnsection.

__

L/2

2EI

EI

EI

EI

Ma

A

BDS under Ma

B

Mb

A

BDS under Mb

B

By choosing a B.D.S. as simple beam under the action of Ma and Mb, we can verify by the use of conjugate beam method that θb = 0. In this case, we are required to find that how much rotation at end A is required to produce the required moment Ma. In other words, θa (which is in terms of Ma and Mb can be considered as an applied load on the analogous column section). The moments computed by using the

formula PA ±

MCI will give us the end moments directly because in this case Ms diagram will be zero.

So, M = Ms − Mi = 0 − Mi = − Mi. Properties of analogous column section:− A =

LEI , I =

1EI

L3

12 = L3

12EI

factor Downward load on analogous column = θa at A.

Accompanying moment = θa × L2 ( About centroidal column axis )

and C = L2 for use in above formula.


Ma = PA +

MCI

= θa EI

L + θa × L × L × 12EI

2 × 2 × L3 ( Reactions are additive at A and are upwards)

= θa EI

L + 3θa EI

L

Ma = 4 EI

L θa

Where 4 EI

L = Ka

Where Ka = stiffness factor at A.

Mb = PA ±

MCI ( Reactions are subtractive at B)

= θa EI

L − 3θa EI

L

= − 2θa EI

L

= − 2EI

L . θa

The (−ve) sign with Mb indicates that it is a (−ve) moment which gives us tension at the top or

compression at the bottom.

(COF) a → b Carry−over factor from A to B = MbMa =

24 = +

12

“BY PUTING θA EQUAL TO UNITY , MA & MB WILL BE THE STIFFNESS FACTORS AT

THE CORRESPONDING JOINTS”. STIFFNESS FACTOR IS THE MOMENT REQUIRED TO

PRODUCE UNIT ROTATION.

In the onward problems of members having variable X-section, we will consider θa = θb = 1

radians and will apply them on points A & B on the top of the analogous column section. The resulting

moments by using the above set of formulas will give us stiffness factor and COF directly.


EXAMPLE NO. 8:− Determine the stiffness factors at A & at B and the carry-over factors from A to B and from B to A for the straight members with variable X-sections shown in the figure below.. SOLUTION:− Draw analogous column section and determine its properties.

2 2I I I4m 6m 6m

1 rad1 rad 7.73

7.73m 8.27m

1 1

1

3

11

6 2

1

12EI

1A = x 6 + x 6 + x4

= + +

=

A

A

B

BAnalogous column section

2EI

2EI

EIEIEI

EI

EI

EI 2EI

Centroidal axis

Taking moments of areas about point B.

X = (0.5 × 6) × 3 + (6 × 1) × 9 + (4 × 0.5) × 14

11

X = 8.27 meters from B.

I = 0.5 × 63

12 + (0.5 × 6) × (5.27)2 + 1 × 63

12 + (1 × 6) ×

(0.73)2 + 0.5 × 43

12 + (0.5 × 4) × (5.73)2

I = 181.85

EI

Consider loads acting at centroid of analogous column and determine indeterminate moments at A and B.

Ma = PA ±

MCI

= PA +

MCI =

1 × EI11 +

7.73 × 7.73 × EI181.85

Ma = 0.419 EI = 0.419 × 16 EIL , (by multiplying and dividing RHS by L)

Ma = 6.71 EIL

Ka = 6.71


Mb = EI11 −

7.73 × 8.27 × EI181.85 ×

16L (by multiplying and dividing by L)

= − 4.17 EIL

(COF)A→B = MbMa =

4.176.71 = 0.62

(COF)A→B = 0.62 Now applying unit radian load at B. This eccentric load can be replaced by a concentric load Plus accompanying moment.

8.271 rad

1 rad

8.277.73 Considering eccentric 1 rad load to be acting at centroid of section alongwith moment.

Ma =

EI

11 − (8.27 × 7.73 × EI)

181.85 16L , (multiplying and dividing by L)

Ma = − 4.17 EIL

Mb =

EI

11 + (8.27 × 8.27 × EI)

181.85 16L (multiplying and dividing by L)

Mb = 7.47 EIL

Kb = 7.47

(COF)b→a Carry−over factor from B to A = MaMb =

4.177.47

(COF)b→a = 0.56


7.4. APPLICATION TO FRAMES WITH ONE AXIS OF SYMMETRY:− EXAMPLE NO. 9:- Analyze the quadrangular frame shown below by the method of column analogy. Check the solution by using a different B.D.S. SOLUTION:−

12KNB C

DA

6m 2I 2I 6m

5I

10m

Axis of Symmetry w.r.t. geometry

The term “axis of symmetry” implies that the shown frame is geometrically symmetrical (M.O.I. and support conditions etc., are symmetrical) w.r.t. one axis as shown in the diagram. The term does not include the loading symmetry (the loading can be and is unsymmetrical). Choosing the B.D.S. as a cantilever supported. at A.

12KNB C

DA

6m 6m

10m

72 kN-m

Ms-diagram

5I

2I 2I

AD

CB 5I

6m 2I6m 2I

2Force=108

EI36

- DiagramMsEI

12 kN-m

EI


According to our sign convention for column analogy, the loading arising out of negative MsEI giving tension

on outside will act upwards on the analogous column section. Sketch analogous column section and place load.

x x

B C

AD

y

Mxx5m 5m

Myy3.73m

2m

1/212

108EI

15

y=2.27m

(1) Properties of Analogous Column Section:−

A =

1

2 × 6 × 2 + 15 × 10 =

8EI

y =

1

5 × 10 × 1

10 + 2

1

2 × 6 × 3 1EI

8EI

= 2.27 m about line BC. (see diagram)

Ixx = 2

0.5 × 63

12 +

1

2 × 6 x (0.73)2 + 10 × (1/5)3

12 + (0.2 × 10) × (2.27)2

= 31.51

EI m4

Iyy = 0.2 × 103

12 + 2

6 × 0.53

12 + (6 × 0.5) × (5)2

= 167EI m4

Mxx = 108 × 1.73 = 187EI clockwise.

Myy = 108 × 5 = 540EI clockwise.

Applying the formulae in a tabular form for all points. Imagine the direction of reactions at exterior frame points due to loads and moments. Ma = ( Ms− Mi)a

( Mi)a = PA ±

Mx yIx ±

My XIy


POINT Ms P/A Mx y

Ix My X

Iy Mi M =

Ms−Mi A − 72 − 13.5 − 22.14 − 16.17 − 51.81 − 20.19 B 0 − 13.5 + 13.47 − 16.17 − 16.20 + 16.20 C 0 − 13.5 + 13.47 + 16.17 + 16.14 − 16.14 D 0 − 13.5 − 22.14 + 16.17 − 19.47 + 19.47

Note: Imagine the direction of reaction due to P, Mx and My at all points A, B, C and P. Use appropriate signs. Repeat the analysis by choosing a different BDS yourself. EXAMPLE NO. 10:− Analyze the quadrangular frame shown by the method of column analogy.

A

B C

D

6m6m

5I

3KN/m

10m

2I 2I

Choosing B.D.S. as a cantilever supported at A.

B C

DA

150K n-m

3KN/m

30BDS under loads


Draw Ms−diagram by parts and then superimpose for convenience and clarity.

30

150

B

A

150150

150

150

30 B C

C

C

D

B

Free Body Diagrams 3 KN/m

30

150

150

150B

3KN/m

C

D150A

Ms-Diagram

6m

3m

450

CB

2.575

75

10m

MsEI

- Diagram

30

100

A D

For Portion BC

Area = bb

n+1 = 10 × 302 + 1 =

3003 = 100

X' = b

n+2 = 10

2 + 2 = 104 = 2.5 from B.

Note: As BMD on portions BC and AB are negative the loads equal to their areas will act upwards.

Now sketch analogous column section carrying loads arising from MEI contributions.


2.275m450

B2.25m 100 y

C1/5

XX

3.725

6m 0.725m

3m

1/2 1/210m

D

My

Mx

yAnalogus colmun section

Properties of analogous column section:−

A = 2

1

2 × 6 + 15 × 10 =

8EI (as before)

y =

1

5 × 10 × 1

10 + 2

6 ×

12 × 3

8 = 2.275 about line BC (as before)

Ix = 2

1

2 × 63 +

1

2 × 6 × (0.725)2 +

10 ×

1

5

3

+

10 ×

15 × (2.275)2

= 31.51

EI m4 (as before)

Iy = 2

6 × 0.53

12 + (6 × 0.5) × 52 + 0.2 × 103

12

= 166.79

EI m4 (as before)

Mx = 450 × 0.725 − 100 × 2.275 = 95.75 KN−m Clockwise My = 450 × 5 + 100 × 2.75 = 2525 KN−m clockwise. P = 100 + 450 = 550 KN

Now this eccentric load P and MX and My are placed on column centroid. Applying the formulae in a tabular form. Ma = ( Ms− Mi)a


and ( Mi)a = PA ±

Mx yIx ±

My xIy

POINT Ms P/A Mx . y

Ix My . x

Iy Mi M =

Ms−Mi A − 150 − 68.75 − 11.32 − 75.69 − 155.76 5.76 B − 150 −68.75 + 6.91 − 75.69 − 137.53 −12.47 C 0 −68.75 + 6.91 + 75.69 13.85 −13.85 D 0 −68.75 −11.32 + 75.69 −4.38 4.38

EXAMPLE NO. 4:− Determine stiffness factors corresponding to each end and carry-over factors in both directions of the following beam. SOLUTION:−

2m 1.5m 2m 1m 2m5I 2I 4I I 3I

A B

Sketch analogous column section.

1/5 ½ ¼ 1/EI 1/3EI

4.74m 3.76m

yo

o

Properties of Analogous Column Section :− A =

15 × 2 +

12 × 1.5 +

14 × 2 + 1 × 1 +

13 × 2

A = 3.32EI

Taking moment about B of various segments of column section.

X = 13 × 2 × 1 + 1 × 1 × 2.5 +

14 × 2 × 4 +

12 × 1.5 × 5.75 +

15 × 2 × 7.5

3.32

X = 12.4725

3.32

X = 3.76 m from B.


Iyoyo = 13 ×

23

12 +

1

3 × 2 × (2.76)2 + 1 × 13

12 + (1 × 1)(2.26)2

+

1

4 × (2)3

12 +

1

4 × 2 (0.24)2 +

1

2 × (1.5)3

12

+

1

2 × 1.5 (1.99)2 +

1

5 × (2)3

12 +

1

5 × 2 (3.74)2

= 19.53

EI

1. Determination of stiffness factor at A (ka) and carry-over factor from A to B. Apply unit load at

A and then shift it along with moment to centroidal axis of column as shown below:

1 rad

A B8.5m

1

4.74 BA

4.74 3.76

=

Ma =

PA ±

MCI

= 1 × EI

3.32 + 4.74 × 4.74 EI

19.53

= 1.45 EI , multiply and divide by L

Ma = 1.45 × 8.5 × EIL = 12.33

EIL

Ka = 12.33

Mb = EI

3.32 − 4.74 × 3.26 × EI

19.53

= − 0.61 EI = − 0.61 × 8.5 × EIL = − 5.19

EIL (multiply and divide by L)

Mb = − 5.19 EIL

(COF)a → b = MbMa =

5.1912.33 = 0.42

(COF)a → b = 0.42


2. Determination of stiffness factor at B (Kb) and carry-over from B to A. Apply a unit load at B and them shift it along with moment to centroidal axis of column as shown below:

Ma = PA ±

McI

1 rad

A B8.5m

1

3.76 BA

4.74 3.76

=

Ma = EI

3.32 − 3.76 × 4.74 × EI

19.53

= −0.61EI , multiply and divide by L.

= − 0.61 × 8.5 × EIL = −5.19

EIL

Mb = PA ±

McI

= EI

3.22 + 3.76 × 3.76 × EI

19.53

=1..03 EI = 1.03 × EIL × 8.5 , multiply and dividing by L.

Mb = 8.76 EIL

Kb = 8.76

(COF)b → a = MaMb =

5.198.76 = 0.6

(COF) b → a = 0.6


EXAMPLE NO.12:− Analyze the following gable frame by column analogy method. SOLUTION :−

3 kN/m

3 m

7 m

A E

B DC

14 m

I

3I 3I

I

Choosing a simple frame as BDS

A

B D

E

3kN/m

7.62

7

21 21B.D.S under loads

C

73.573.5

A E

B DC

7.62

Ms-diagram

A

B

C

D

EMs diagramEI

24.5 24.5

4.375

4.76 2.86x

2.86


Taking the B.D.S. as a simply supported beam.

MX = 21X – 1.5X2 , taking X horizontally. MX = Mc at X = 7m Mc = 21 × 7 – 1.5 X 72 = 73.5 KN−m

Sin θ = 3

7.62 = 0.394

Cos θ = 7

7.62 = 0.919

P1 = P2 = 23 × 24.5 × 7.62 = 124.46

P = P1 + P2 = 248.92

∫ MX dX = 7

∫o (21 X − 1.5X2) dX =

21

2 X2 − 1.53 X3

7 o= 343

∫ (MX)X dX = 7

∫o (21 X2 − 1.5X3)dX =

21

3 X3 − 1.54 X4

7 o

= 7 × 73 − 1.54 × 74 = 1500.625

X = ∫ (MX) X dX

∫ MX dX = 1500.625

343

X = 4.375 Horizontally from D or B. Shift it on the inclined surface.

Cos θ = 4.375

a

a = 4.375Cos θ =

4.3750.919

a = 4.76


Now draw analogous column section and place loads on top of it.

1

A

E

1

4.83 mMx

XX

2.17 m

3mB

C

1/3D

2.86

4.76

124.46124.46

PROPERTIES OF ANALOGOUS COLUMN SECTION

A = 2 (1 × 7) + 2

1

3 × 7.62 = 19.08 m2

Y = 2[(1 × 7) × 3.5] + 2

1

3 × 7.62 × 8.5

19.08 = 49 + 43 − 18

19.08

Y = 4.83 m from A or E

Ix = 2

1 × 73

12 + (1 × 7) (4.83 − 3.5)2

+ 2

1

3 × (7.62)3

12 × ( 0.394 )2 + 13 (7.62) ( 1.5 + 2.17)2 ,

the first term in second square bracket is bL3

12 Sin2θ

= 154.17 So Ix ≅ 154 m4

Now Iy = 2

7 × 13

12 + (7 × 1) × 72

+ 2

1

3 × (7.62)3

12 × (0.919 )2 +

1

3 × 7.62 × (3.5)2 ,


the first term in second square bracket is bL3

12 Cos2θ

=770.16 So Iy ≅ 770 m4 Total load on centroid of analogous column P = P1 + P2 = 124.46 + 124.46 = 248.92 KN Mx = 2 × [124.46 × 4.05 ] , 4.05 = 2.17 + 4.76 Sinθ = 2.17 + 4.76 × 0.394.

Mx = 1007 (clockwise).

My = 0 (because moments due to two loads cancel out)

Applying the general formulae in a tabular form for all points of frame. Ma = ( Ms− Mi)a

( Mi)a = PA ±

Mx yIx ±

My XIy

Point Ms (A)

P/A (1)

Mx .YIx

(2)

My .XIy

(3)

(B)=Mi (1)+(2) +(3)

M = Col (A)−(B)

A 0 + 13.05 − 31.58 0 − 18.53 + 18.53 B 0 + 13.05 + 14.19 0 + 27.24 − 27.24 C + 73.5 + 13.05 + 33.81 0 + 46.86 +26.64 D 0 + 13.05 + 14.19 0 + 27.24 − 27.24 E 0 + 13.05 − 31.58 0 − 18.53 + 18.53

EXAMPLE NO. 13:- Analyze the frame shown in fig below by Column Analogy Method.

B C

A D

2kN/m

10kN

4m

3m

2I

3I

2I

Choosing the B.D.S. as a cantilever supported at A.


MA = 10 x 1.5 + 2 x 4 x 42

MA = 31 KN−m

318

10

2kN/m

10 kN

B.D.S

B C

A D

Draw Free Body Diagrams and sketch composite BMD:−

10

10

831

2kN/m

B

15

A

31

15

15

15

10

10

CB1.5 1.5

C

D

no B.M.D4m

31

15

15

10

Ms-diagram15.5

7.5

5

10

MsEI diagram

,


Properties Of Analogous Column Section :− Sketch analogous column section and show loads on it. BMD along column AB is split into a rectangle and other second degree curve.

A =

1

2 × 4 × 2 +

1

3 × 3 = 5 m2

Y =

3 ×

13 ×

1

6 + 2

1

2 × 4 × 2

5

Y = 1.63 m From line BC

Ix = 3 ×

1

3

3

12 +

1

3 × 3 × (1.63)2 + 2

0.5 + 43

12 + (0.5 × 4) × (0.37)2

12

= 8.55 m4

Iy =

1

3 × (3)3 + 2

4 × 0.53

12 + (4 × 0.5) × (1.5)2

= 9.83 m4

2.37 m

1,63 m

1/3

D

½½

I

y3m

yP1

1.00.5

B C

P2

P3

0.37X4m X


Total load on top of analogous column section acting at the centroid. P = 3.75 + 30 + 10.67 = 44.42 KN upward.

P1 = 12 × 1.5 × 5 = 3.75, P2 = 7.5 × 4 = 30, P3 =

4 × 7.52 + 1 = 10

X' = 44 = 1 meters for A.

MX = − 3.75 x 1.63 + 30 x 0.37 + 10.67 x 1.37

= 19.61 KN-m clockwise. My = 10.67 × 1.5 + 30 × 1.5 + 3.75 × 1 = 64.76 clockwise. Applying the general formulae in a tabular form for all points of frame. Ma = ( Ms− Mi)a

( Mi)a = PA ±

Mx yIx ±

My XIy

Point Ms P/A

(1) MxIx . y

(2)

MyIy . X

(3)

Mi (1)+(2) + (3)

M Ms − Mi

A − 31 −8.88 − 5.44 − 9.88 − 24.2 − 6.8 B − 15 − 8.88 + 3.74 − 9.88 −15.02 + 0.02 C 0 − 8.88 + 3.74 + 9.88 + 4.74 − 4.74 D 0 − 8.88 − 5.44 + 9.88 − 4.44 + 4.44

EXAMPLE NO. 14:- Analyze the following beam by column analogy method. SOLUTION :−

Choosing B.D.S as cantilever supported at B

3kN/m10kN

I1.5I3I

Ms-diagram

due to u.d.l. only

7224

6( c)

(b)(a)

962m2m4m

321m 96 1.33


40 Ms diagram due to concentrated load only Slectch analogous column section and determine its proteins

a b d

ce 24

2472

161.5m

26

P2=1.33

3.21mP3=18.67

2.14

Ms-diagramEI due to u.d.l

P1

40

P4=801.33

MSEI diagram due to point load.

Slectch analogous column section and determine its properties.

1/1.5

o

yo3.224.78

1/3 1Analogouscolumnsection

2.14P6

P4

1.33

P3P21.5

3.21

P1 = 24 × 4

3 + 48 × 4

2 + 24 × 4 = 224 KN. Corresponding to full Ms diagram, due to u.d.l.

Location of P1 from B 224 × X = 96 × 1.33 + 96 × 2 + 32 × 5

X = 2.14 meters

P4 = 12 × 4 × 40 = 80 KN, Corresponding to full Ms diagram due to point load.

Note: Area of 32 and its location of Ms diagram due to u.d.l. has been calculate d by formula e used in moment – area Theorems.

area (abc) = ∫ MXdX = 2

∫o −1.5X2 dX =

− 1.5 X3

3

2 o = −4

∫(MX) X dX = 2

∫o −1.5X3dX =

− 1.5 X4

4

2 o = − 6

X = − 6− 4 = 1.5m from A

area (bcde) = ∫ (MX) dX = 4

∫o − 1.5X2dX −

2

∫o − 1.5 X2 dX


=

− 1.5

X3

3

4 o −

− 1.5

X3

3

2 o = − 28

∫ (MX)X dX = 4

∫o− 1.5 X3dX −

2

∫o − 1.5 X3dX = − 90

X = −90−28

= 3.21 meters from A (centroid of area bcde)

P3 = 1

1.5 (area bcde) = 1

1.5 (28) = 18.67 KN , P2 = 13 area abc =

13 × 4 = 1.33

P4 = 80 KN Total concentric load on analogous column section. P = − P1 + P2 + P3 − P4

= − 224 + 1.33 + 18.67 − 80 = − 284 KN (upward)

Total applied moment = M = − 224 × 1.68− 80 × 1.89 − 18.67 × 1.57 − 1.33 × 33 × 3.28

= − 426.79 KN-m (It means counter clockwise)

This total load P and M will now act at centroid of analogous column section. Properties of Analogous Column Section.

A = 13 × 2 +

11.5 × 2 + 1 × 4 = 6

X = (1 × 4) × 2 +

2 ×

11.5 × 5 +

1

3 × 2 × 7

6

= 3.22 from B.

Iyoyo = 1 × 43

12 + (1 × 4)(1.22)2 +

1

1.5 × 23

12 +

1

1.5 × 2 (1.78)2


+

1

3 × 23

12 +

1

3 × 2 (3.78)2

12 = 25.70 m4

(Mi)a = PA ±

McI

= − 284

6 + 426.79 × 4.78

25.7

= + 32.05 KN-m (Ms)a = 0 Ma = (Ms − Mi)a = 0 − 32.05 Ma = − 32.05 KN−m

(Mi)b = PA −

McI

= − 284

6 − 426.79 × 3.22

25.7

= − 100.81 (Ms)b = − 72 − 40 = − 112 Mb = (Ms − Mi)b = − 112 + 100.81 Mb = − 11.19 KN−m The beam has been analyzed. It is now statically determinate.


CHAPTER EIGHT

8. PLASTIC ANALYSIS OF STEEL STRUCTURES

8.1. Introduction: Although the terms Plastic analysis and design normally apply to such procedures for steel structures within the yield flow region, at almost constant stress, however the Idea may also be applied to reinforced concrete structures which are designed to behave elastically in a ductile fashion at ultimate loads near yielding of reinforcement. The true stress-strain curve for a low grade structural steel is shown in fig. 1 while an idealized one is shown in fig. 2 which forms the basis of Plastic Analysis and Design.

A

B

C D

E

F

Stress AB-ElasticBC-Yeild pointsCD-Plastic Strain flowDE-StrainhardeningEF-Failure

Stress

A

(B,C) Plastic D

Elastic

Strain Strain Fig 1: Fig 2:

ff

8.2. Advantages of Plastic Analysis 1. Relatively simpler procedures are involved. 2. Ultimate loads for structures and their components may be determined. 3. Sequence and final mode of failure may be known and the capacity at relevant stages may be

determined. 8.3. Assumptions in Plastic bending 1. The material is homogeneous and isotropic. 2. Member Cross-section is symmetrical about the axis at right angles to the axis of bending. 3. Cross-section which were plane before bending remain plane after bending. 4. The value of modulus of Elasticity of the material remains the same in tension as well as in

compression. 5. Effects of temperature, fatigue, shear and axial force are neglected. 6. Idealized bi-linear stress-strain curve applies. 8.4. Number of Plastic Hinges “ The number of Plastic Hinges required to convert a structure or a member into a mechanism is one more than the degree of indeterminacy in terms of redundant moments usually. Thus a determinate structure requires only one more plastic hinge to become a mechanism, a stage where it deflects and rotates continuously at constant load and acquires final collapse. So Mathematically N = n+ 1 where N = Total number of Plastic hinges required to convert a structure into a mechanism. and n = degree of indeterminacy of structure in terms of unknown redundant moments.

PLASTIC ANALYSIS METHOD 351

8.5. Plastic Hinge. It is that cross-section of a member where bending stresses are equal to yield stresses σ= σy= fy. It has finite dimensions.

From bending equation σ = MyI or σy =

MpCI or σy =

MpZp so Hp = Zp σy

From elastic bending σy =

MI or

σIy = M where

Iy = Z

So M= σZ and Z is elastic section modules and is equal to the first moment of area about N.A Z = ∫A ydA. 8.6. Plastic moment of a rectangular section. Consider a simple rectangular beam subject to increasing bending moment at the centre. Various stress-strain stages are encountered as shown below.

case A: M<My case B: M=My case C case D

∈< y∈ ∈ y=∈ ∈ y>∈ ∈ y>>∈

∈ y>>∈

D

By= y= y=< yC

T

2 D 3

C

T

D 2

y=y= y>

Var ious Stress-strain distr ibutions Case A - Stresses and strains are within elastic range. Case B - Stresses and strains at yield levels only at extreme fibers Case C - Ingress of yielding within depth of section. Case D - Full plastification of section. On the onset of yielding σ = σy and M = My = σy.Z. On full plastification σ = σy and M = Mp = σy.Zp. or Zp = ∫A yda (First moment of area about equal area axis). All compact sections as defined in AISC manual will develop full plastification under increasing loads realizing Mp. However local buckling of the compression flange before yielding has to be avoided by providing adequate lateral support and by applying width / thickness checks as was done during the coverage of subject of steel structures design. Case B. Stresses and Strains at yield at extreme fibres only. Consult corresponding stress and strain blocks. M = Total compression × la = Area × σ × la where Area = Area in compression (from stress block). σ = Average compression stress. la = Lever arm i.e. distance b/w total compressive and tensile forces.

So M =

BD

2

σy + o

2 . 23 D


In general M = Cjd or Tjd , where C and T are total compressive and tensile forces respectively which have to be equal for internal force equilibrium.

or My = σy BD2

6 , but BD2

6 = Z

Z= Elastic Section modules =

IC =

BD3

12 ÷ D2

So My = σy.Z. = BD2

6

Case D: Full plastification, σ = σy upto equal area axis.

M = Cla =

B.

D2 (σy)

D2 where la is lever arm

= σy . BD2

4 or Zp = BD2

4 , where ZP = Plastic section Modules.

or Mp = σy . Zp or Zp = A2 [y1 + y2] (first moment of areas about equal area axis)

and y1 + y2 = D/2 (distance from equal area axis to the centroids of two portions of area.) Case C: Moment Capacity in Elasto - Plastic range. Extreme fibres have yielded and the yielding ingresses in the section as shown by the stress – distribution.

1

1

2

2

Z

Z

D2

D2

y

ycase C : Stress-Distribution

C1

C2

T2

T1

la2 la1

wherela1 = lever axis b/w C1 and T1la2 = lever axis b/w C2 and T2C1 = Av.stress X area of element No.1C2 = Av-stress acting on element No.2 x area of element 2.

M = [C1la1 + C2 . la2 ] (A) , la1=

Z+

D2 − Z

2 2 =D2+ Z

2

C1 = (σy) B

D

2 − Z la2 =

2

3 × Z × 2 = 43 × Z

C2 =

σy + o

2 Z . B = σy ZB2 and so, putting values of C1 , C2 la1 and la2 in equation A above.


M = σy . B

D

2 − Z

D

2 + Z + σy

Z.B

2 x 43 Z , Simplifying

M = σy . B

D2

4 − Z2 + 23 σy BZ2

= σy . B

D2

4 − Z2 + 23 Z2

Mr = σy . B

D2

4 − Z2

3 , where Mr is moment of resistance.

Mp = Mr = σy . B

3D2 − 4Z2

12 __ For rectangular section.

Calculating on similar lines, Plastic moment for various shapes can be calculated. 8.7. Shape Factor (γ) It is the ratio of full plastic moment Mp to the yield moment My. It depends on the shape of Cross-section for a given material.

Shape Factor = γ = MpMy =

σy . Zp σy . Z or γ =

ZpZ (Ratio of Plastic section modulus to

Elastic Section Modulus). 8.8. Calculation of Shape Factor for different Sections.

D

B

(1)

(2)

y1

y2

D/2

B

dyy

8.8.1 For rectangular section.

I = BD3

12 , IC = Z , C =

D2

So Z = BD3 × 212 × D =

BD2

6

Zp = A2 [y1 + y2] =

BD2

D

4 + D4 or alternatively, Zp = ∫A ydA.

= BD2

4 = 2 D/2

∫o

y . Bdy

γ = ZpZ =

BD2 × 64 × BD2 =

64 = 1.5 = 2B

D/2

∫o

ydy.

γ = 1.5 so [Mp is 1.5 times My] or Zp = BD2

4


8.8.2 For Circular Cross-section

I = πD4

64 , A = π4 D2

Z = IC =

πD4

64 × 2D =

πD3

32 ,

Zp = A2 [y1 + y2]

D

b

y

dy

(a) Cross-Section (b) StrainDistribution

(c) Stress at fullplastification Distribution

= πD2

8

2D

3π + 2D3π , r =

D2 , y1 =

4r3π =

4 × D3π × 2 =

2D3π

Zp = D3

6 γ = ZpZ =

D3 × 326 × πD3 =

326π ≅ 1.7

γ = 1.7 , [Mp is 1.7 times My] 8.8.3 Hollow Circular Section

dD

2D3 2d

3

d

D

I = π64 (D4 − d4)

IC = Zmin =

π64 (D4 − d4) .

2D

Zmin = π

32D (D4 − d4)

Zp = A2 [ y1 + y2] , putting values. putting values Ay = A1y1 + A2y2

= π8 (D2 − d2)

2 ×

23π

(D3 − d3)(D2 − d2)

π8 (D2 − d2) y =

πD2

8 . 2D3π −

πd2

8 × 2d3π

Zp =

D3 − d3

6 π8 (D2 − d2) y =

D3

I2 − d3

12


γ =

D3 − d3

6 × 32D

(D4 − d4) π Putting Z and Zp y = 8

12π (D3 − d3)(D2 − d2)

γ = 326π

D(D3 − d3)(D4 − d4) y =

23π

(D3 − d3)D2 − d2

for N-A or equal area axis. For D = 10” d = 8” γ = 1.403 For I - Section:

D d

B

b/2

T2

T1

C2

C1

stressdistribution

straindistribution

la2la1

y

As = Z = IC and C =

D2

I = (BD3 − bd3)

12 , My = σy . Z = σy (BD3 − bd3)

6D , Putting value of Z from (1)

Z = IC =

(BD3 − bd3)12

2D Mp = C1 la1 + C2 la2

Z =

BD3 − bd3

6D (1) la1=

d

2 +

D-d

2½

2 = (D+ d)/2,

la2 =

D

2 −

D−d

2½ × 2=

d2

Mp= σy . B (D − d)

2 (D + d)

2 + σy . d2 (B − b)

d2

Mp = σy

B

4 (D2 − d2) + d24 (B − b)

γ = MpMy =

σy(BD2 − bd2)4 ×

BDσy(BD3 − bd3) Mp = σy

BD2 − bd2

4

γ = 3D2

(BD2 − bd2)(BD3 − bd3) if B = 4”

b = 3.75” D = 8” , shape factor γ = 1.160 d = 7.5”


Similarly for T-section, Equilateral Triangle and hollow rectangular section the values of shape-factor are 1.794, 2.343 and 1.29 respectively. For diamond shape its value is 2.0. 8.9. Significance of Shape Factor Zp is First moment of area about equal area axis. 1. It gives an indication of reserve capacity of a section from on set of yielding at extreme fibres

to full plastification. 2. If My is known,, Mp may be calculated. 3. A section with higher shape factor gives a longer warning before collapse. 4. A section with higher shape factor is more ductile and gives greater deflection at collapse. 5. Greater is the γ value, greater is collapse load factor λc 8.10. Collapse load of a structure. Collapse load is found for a structure by investigating various possible collapse mechanisms of a structure under conceivable load systems. For any given mechanism, possible plastic hinge locations are determined by noting the types of loads and support conditions remembering that under increasing loads, the plastic hinges would form in a sequence defined by corresponding elastic moments at the possible plastic hinge locations. “ Collapse loads are usually the applied loads multiplied by collapse load factor λc . λc is defined as the ratio of the collapse load to the working load acting on any structure / element” . The value of λc may indicate a margin of safety for various collapse mechanisms and steps can be taken in advance to strengthen the weaker structural elements before erection. Benefit of ` strength reserve’ is obtained due to increased moments of resistance due to plastification. The reserve of strength is large if the section widens out near the vicinity of neutral surface. 8.11. Assumptions made in Plastic Theory.

The plastic analysis is primarily based on following assumptions.

1. For prismatic members,, the value of Mp is independent of magnitude of bending moment.

2. The length of plastic hinge is limited to a point.

3. Material is very ductile and is capable of undergoing large rotations / curvatures at the constant moment without breaking.

4. The presence of axial force and shear force does not change the value of Mp.

5. The structure remains stable until the formation of last plastic hinge and serviceability would not be impaired till such time.

6. Loads acting on structure are assumed to increase in proportion to each other.

7. Continuity of each joint is assumed. 8.12. Fundamental Theorems of Plastic Collapse. When degree of redundancy increases beyond 2 or 3 in situations where collapse mechanism is not very clear, we try to pick up collapse load with the help of three fundamental theorems.

a. Lower bound theorem or static theorem.

b. Upper bound theorem or kinematic theorem.

c. Uniqueness theorem.


8.12.1 Lower Bound theorem “ A Load computed on the basis of bending moment distribution in which moment nowhere exceeds Mp is either equal to or less than the true collapse load” . 8.12.2 Upper bound theorem “ A load computed on the basis of an assumed mechanism is either equal to or greater than true collapse load” . When several mechanisms are tried, the true collapse load will the smallest of them. 8.12.3. Uniqueness theorem “ A load computed on the basis of bending moment distribution which satisfies both plastic moment and mechanism conditions is true plastic collapse load” .

Mp

StaticTheorems.

Curvature.

KinematicTheorems.

Moment

True

8.13. Methods of analysis

Basically there are two methods of analysis.

a. Equilibrium Method.

b. Mechanism Method.

8.13.1. Equilibr ium Method

Normally a free bending moment diagram on simple span due to applied loads is drawn and B.M.D due to reactants is superimposed on this with due regard to their signs leaving the net moment distributed. Then by making the moment values equal to Mp values at the known potential plastic hinge locations, a revised diagram can be drawn. Then by splitting the simple span moment due to applied loads in terms of relevant Mp, the values of collapse load can be determined.

8.13.2. Mechanism Method

In this approach, a mechanism is assumed and plastic hinges are inserted at potential plastic hinge locations. At plastic hinges the corresponding rotations and deflections are computed to write work equations which may be written as follows.

Work done by external loads = Actual loads x Average displacements = Work absorbed at Plastic hinges (internal work done) = Mp. θ


Typically Σ W. δ = Σ Mp . θ.

In both methods, the last step is usually to check that M < Mp at all sections.

8.14. Values of Collapse loads for different loaded structures.

Beam Under loads Collapse load Pc or Wc

PL/2

4

MpL

PL/2

8

MpL

W

16

MpL2

P2/3L

9

MpL

PL/2

6

MpL

L P

1

MpL

W

11.65

MpL2

8

MpL2

L3 P P/

6

MpL

PL/3

6

MpL


8.15. In order to explain the above procedure, Let us solve examples. Analysis of a Continuous beam by Mechanism Method. EXAMPLE NO. 1:- Consider the beam loaded as shown. Three independent possible collapse mechanisms along with potential plastic hinge locations are shown. SOLUTION: degree of indeterminacy in terms of moments = n = 2 ( moments at A and B) No of Plastic hinges required = 2 + 1 = 3

A4

20K 20

B

20

4

C

8

12

K K

4

First possible beam mechanism for span AB.

4

/2

1.5 Second possible beam mechanism for span AB.

3

2

8

Possible beam mechanism for span BC

2

Real Hinge

Write work equations for all mechanisms and find corresponding Mp values. Mechanism (1)

20 × 4 θ + 20 × 2θ = Mp . θ + Mp.1.5θ + Mp θ2

120 θ = 3 Mp θ Mp = 40 K-ft. Mechanism (2) 20 × 4 θ + 20 × 8 θ = Mp . θ + Mp.3θ + Mp . 2θ


240 θ = 6 Mp θ Mp = 40 K-ft Mechanism (3) 20 × 4 θ = Mp. θ + Mp . 2 θ + 0 × θ 80 θ = 3 Mp . θ Mp = 26.67 K-ft. Minimum Collapse load or Max. Mp will be the collapse mechanism So Mp = 40 K-ft.( Corresponding to mechanisms 1 and 2) 8.16. EXAMPLE NO.2:-Find the collapse load for the following continuous beam loaded as shown. SOLUTION: Do elastic analysis by three moment equation to find Mb and Mc. Apply the equation twice to spans AB and BC and then BC and CD. (In this case, noting symmetry and concluding that Mb = Mc, only one application would yield results).

85.33 36 85.33

16 9 16

8m 6m 8m

AB C D

2T/m

(Simple span B.M.D. due to loads)4m 3m

By using three-moment equation

8

I Ma + 2Mb

8

I + 6I + Mc

6

I = − 6 × 85.33 × 4

8 − 6 × 36 × 3

6

Ma = 0 , 34 Mb = 364 So Mb = Mc = 10.70 T − m ( By symmetry)

8.17. Maximum bending moment in a member car rying UDL

A B

W=wL

C

ML

ML

MR

MR

L/2L/2

R1 R2

Mc Mmax

yo

xo zo

B.M.D

Consider a general frame element subjected to Udl over its span alongwith end moments plot BMD.


After derivation we find the location of maximum moments Xo, Yo and MC.

In some books, plastic huge is stated to form in the centre of span. However, the formulae given below are very precise and give correct location of plastic huges due to u.d.l.

Where, ML = Moment at left of element

MR = Moment at right of element

MC = Moment at centre of element

Xo , Zo , yo = Location of max. moment from left, right and centre respectively as shown on BMD.

yo = MR − ML

WL =

10.70 − o2 x 8 = 0.6687 m (1)

MC = WL2

8 + (MR − ML)

2 = 2(8)2

8 +

10.70

2

MC = 21.35 T− m (2)

Mmax = Mc + WL . yo2

2L = 21.35 + 2 × 8 (0.6687)2

2 x 8

Mmax = 21.79 T − m

Xo = 4MC − 3MR − ML

WL =

4 (21.35) − 3 (10.7) − 02 × 8 = at 3.313 m from A and D.

Plastic hinges would form first at a distance Xo = 3.313 m from points A and D and then at points B and C.

Now determine collapse load by mechanism method. SOLUTION: No internal work is absorbed at real hinges.


AB 2T/m C D

8m 6 8m

3.13m

3.13 0.707

1.707

3

2

First possible collapse mechanism of span AB.

Second possible collapse mechanism of span AB.

Real Hinge

For first Mechanism

(2 × 8) 3.313 θ

2 = Mp × 1.707 θ + 0.707 θ Mp + 0

So Mp = 10.98 T − m For second Mechanism

Mp . θ + Mp . θ + Mp . 2 θ = (2 × 6)

30

2

Mp = 4.5 T − m

So Mp = 10.98 T − m or Load factor λ = Mp

10.98

8.18. Types of Collapse Three types of collapses are possible as described below.

1. Complete collapse

2. Partial collapse

3. Over complete collapse. 8.18.1. Complete Collapse If in a structure, there are R redundancies and collapse mechanism contains (R + 1) plastic hinges, it is called a complete collapse provided the structure is statically determinate at collapse.


8.18.2. Par tial Collapse: If in a structure, the number of plastic hinges formed at collapse do not render the structure as statically determinate it is called a partial collapse. 8.18.3. Over Complete Collapse If in a structure there are two or more mechanisms which give the same value of collapse load (or collapse load factor λc) then this type of collapse is known as overcomplete collapse. 8.19. Analysis of Frames

In portal frames, three types of mechanisms are possible.

1. Beam Mechanisms (due to gravity loads)

2. Sway Mechanisms (due to lateral loads).

3. Combined Mechanisms (both loads). Step 1: Draw frame in thickness in two lines i.e., solid lines and broken lines. Solid lines are “ outside” of frame and broken lines are “ inside” of frame. Step 2: Nodal moments creating compression on out sides are positive or vice-versa.

+

+

Outside OutsideInside

Step 3: Hinge cancellation at joints occur when rotations of different signs are considered and mechanisms are combined. EXAMPLE NO. 3:- Analyse the frame shown below SOLUTION: 1, 2, 3, 4 and 5 are possible plastic Hinge locations. Three independent mechanisms are possible

Beam mechanisms, Sway mechanisms and Combined mechanisms are possible.


1

5

2 5 520

15

4

5

3

1. Beam Mechanism Write work equation ( Fig A ) 20 λ.5 θ = M2 (− θ) + M3 (2 θ) + M4 (− θ) 100 λ = − M2 + 2M3 − M4 by taking θ as common above. (1) Remember that work is always positive. putting M2 = Mp M3 = Mp M4 = Mp in equation (1), we have 100 λ = 4 Mp or [λ = 0.04 Mp]

20

5

2

2

3

4

1 5(a) Beam mechanism of element 2-4

+

+

5 520

152

3

1 5

(b) Sway Mechanism of Columns

44


+

5 520

15-2

25

2

3

1 5

(c) = a + b combined mechanism 2. Sway Mechanism: 15 λ.5 θ = M1(− θ) + M2 θ + M4 ( − θ) + M5 (θ) 75 λ = − M1 + M2 − M4 + M5 (2) M1, M2, M4 and M5 are all equal to Mp 75 λ = 4 Mp or [ λ = 0.053 Mp] 3. Combined Mechanism: 20 λ . 5 θ + 15 λ . 5 θ = M1 (−θ) + M2 (0) + M3 (2 θ) + M4 ( − 2 θ) + M5 (θ) 175 λ = − M1 + 2M3 − 2 M4 + M5 (3) all these moments are equal to Mp 175 λ = 6 Mp , [ λ = 0.034 Mp ] or Mp = 29.15 λ. Keeping in mind the definition of a true mechanism [one giving highest value of Mp in terms of Pc or lowest value of Pc in terms of Mp or λ ] Combined mechanism is the true collapse mechanism. So λc = 0.0343 Mp It will be a complete collapse if the structure is statically determinate and moment anywhere does not exceed Mp value since there are n + 1 plastic hinges in the true collapse mechanism Note: “ Moment checks are normally applied at those plastic hinge positions which are not included in the true collapse mechanism” . In the true collapse mechanism which is combined mechanism in this case, moments at points 1, 3, 4 and 5 are equal to Mp, we need to find and check moment value at point 2 only in this case. The generalized work equations 1 and 2 in terms of moments may be used for the purpose alongwith their signs. 100 λ = − M2 + 2M3 − M4 (1) 75 λ = − M1 + M2 − M4 + M5 (2) Noting that λ = 0.0343 Mp eqn (1) becomes 100 × 0.0343 Mp = − M2 + 2Mp + Mp so M2 = − 0.431 Mp < Mp − O.K. eqn (2) becomes 75 (0.0343 Mp) = + Mp + M2 + Mp + Mp so M2 = − 0.42755 Mp < Mp − O.K. Net value of M2 = algebraic sum of equations 1 and 2 as combined mechanism is combination of case A and case B.


M2 = ( − 0.431 − 0.427 ) Mp = − 0.858 Mp < Mp − O.K. If at this stage a higher load factor is specified by the designer, there is no need to revise the frame analysis and following formula can be applied to get increased Mp value.

(Mp) new = specified new collapse load factor

present calculated collapse load factor x (Mp Present)

8.20. EXAMPLE NO. 4:- Par tial or incomplete collapse: Find collapse load factor for the following loaded frame. Mp is 80 KN-M for all members.

Mp=80KN-m

7.5m 7.5m37.5

3 412.5

5m

2

1 5 SOLUTION: Draw three possible independent collapse mechanisms. Write work equation and find 1, 2, 3, 4 and 5 possible plastic hinge locations. 1. Beam Mechanism: (35.5 λ) 7.5 θ = − M2 θ + M3 2 θ + M4 (−θ) 281.25 λ = − M2 + 2M3 − M4 (1) moment at 2, 3 and 4 is equal to Mp. so 281.25 λ = 4 Mp (work is always + ve) or λ = 1.1377 2. Sway Mechanism: (12.5 λ) 5 θ = + M1 (− θ) + M2 (θ) + M4 (−θ) + M5 (θ) 62.5 λ = − M1 + M2 − M4 + M5 (2), Moment at 1,2,4 and 5 is Mp.

62.5 λ = 4 Mp or λ = 4

62.5 × 80 = 5.12

λ = 5.12

37.57.5

22

3

4

(a) Beam mechanism

+

5 537.5

12.5

-2

2

7.52

3

1 5

(c) Combined mechanism

4

(b) Sway Mechanism 3. Combined Mechanism: (37.5 λ) (7.5θ)+ (12.5 λ) (5θ)= M1 (−θ) + M2 × 0 + M3 (2θ) + M4 (−2θ) + M5 (θ) 343.75 λ = − M1 + 2M3 − 2M4 + M5 (3) Moment at 1,3,4 and 5 is Mp


343.75 λ = 6 Mp or λ = 6 x 80343.75 = 1.396

λ = 1.396. Therefore, according to kinematic theorem, beam mechanism containing 3 Plastic hinges (one less than required) is the collapse mechanism for this frame with 3 redundancies. (N= n+ 1)= 3+ 1= 4 are required.; Note: In partial or incomplete collapse, only a part of the structure becomes statically determinate. Check moments at locations (1) and (5) with λ = 1.1377 , M2 , M3 , M4 = Mp substituting is eqn (2). 62.5 λ = −M1 + M2 − M4 + M5 or 62.5 (1.1377) = − M1 + Mp + Mp + M5 − 88.937 = M5 − M1 (4) or M1 − M5 = 88.937 (4) Putting same values in eqn (3) 343.75 (1.137) = − M1 + 2Mp + 2Mp + M5 = − M1 + M5 + 4 × 80 70.84 = M5 − M1 (5) Values of M1 and M5 cannot be found from either of equations (4) and (5) as this is incomplete or partial collapse. Instead of a unique answer on values of M1 and M5 which do not violate yield criteria, different pairs of possible values of M1 and M5 can be obtained satisfying equations 4 and 5. Therefore, according to Uniqueness theorem beam mechanism is the true collapse mechanism. It is a partial collapse case. 8.21. EXAMPLE NO. 5:- Overcomplete collapse Determine λc for the following loaded frame.

3m 3m36

3 424

6m

2

1 5

Mp42 42

63

SOLUTION: Sketch possible independent collapse mechanisms. Notice that locations where beam and

column meets, plastic huge is formed in weaker member near the joint.


363

2

2

3

4

(a) Beam mechanism

+

6 536

24-22

32

3

1 5

(c) Combined mechanism(a + b)

(d) Another Combined mechanism (b+c)

4+

+

+

+ +

6(

+(

)

)

) )

3 2

2

- (

(b) Sway mechanism

1. Beam Mechanism: Fig A (36λ) 3φ = − M2 φ + M3 (2φ) − M4 φ 108 λ = − M2 + 2M3 − M4 (1) All are equal to respective Mp. Putting values. 108 λ = 42 + 2 x 63 + 42 λ = 1.944 2. Sway Mechanism Fig B. (24λ) 6θ = M1 (−θ) + M2 (θ) + M4 (−θ) + M5(θ) 144 λ = − M1 + M2 − M4 + M5 (2) 144 λ = 42 + 42 + 42 + 42 or λ = 1.166 3. First Combined Mechanism Fig C (24 λ) (6φ) + (36λ) (3φ) = M1 (−φ) + M2 (0) + M3 (2φ) + M4 (−2φ) + M5 (φ) 252 λ = − M1 + 2M3 − 2M4 + M5 (3)

λ = 294252 λ = 1.166

4. Second Combined Mechanism Fig D (36 λ)3φ+ 24λ (θ+ φ)6= M1 (−θ −φ)+ M2 (θ)+ M3 (2φ) + M4 (θ + 2φ) + M5 (θ + φ) φ ≅ θ 396 λ = − M1 + M2 + 2M3 − 2M4 + 2M5 396 λ = 2(42) + 42 + 2(63) + 3 x 42 + 2 x 42


λ = 462396 = 1.166

λ = 1.166. Note: In overcomplete collapse, more than one mechanism give the same value of collapse load factor. Any or both of the collapse mechanisms can contain extra number of plastic hinges than those required for complete collapse. So in this case fig c and d mechanisms give the same value. This was the case of over complete collapse. Space for notes:

THE THREE MOMENT EQUATION 369

CHAPTER NINE

9. THE THREE MOMENT EQUATION

Most of the time we are concerned with the classical analysis of statically determinate structures. In this chapter we shall consider the analysis of statically indeterminate (externally) beams due to applied loads and due to settlement of supports. It must be remembered that supports for beams may be walls or columns. As we know that for the analysis of statically indeterminate systems, compatibility of deformations is also essential requirements in addition to considerations of equilibrium and statics. By compatibility it is understood that deformations produced by applied loads should be equal to those produced by redundants. It has been already mentioned that reactions occur at supports in various directions if

(i) There is some action (applied load) in that direction.

(ii) There is restraint offered by support in that directions

Action and reactions are equal in magnitude but opposite in direction. In the structural analysis it is sometimes customery to think that rotations are generally associated with moments and deflections or translations are associated with loads. It must also be kept in mind that we never analyze actual structural systems or sub-systems, it is only the idealized ones which are analyzed. Representing beams and columns by just a straight line located on their centroidal axis is also a sort of idealization on the structural geometry. Reactions and loads are, therefore, also idealized and are shown by a sort of line loads acting on a point.

The three-moment equation is a good classical analysis tool in which support moments produced by the loads as well as by the differential settlements can be easily calculated by using second-moment area theorem which states that

“ The deviation of a point A on the elastic curve w.r.t any other point B on the elastic curve is

equal to 1EI multiplied by the moment of area of B.M.D’ s between those two points.” The moments of

B.M.D’ s are taken about a line passing through the point of loaded beam where deviation is being measured.

The method is essentially based on continuity (equality) of slopes on the either side of a support by reducing an indeterminate system to its determinate equivalents as follows by using supperposition.

= +

An indeterminate beam under applied loads and redundant moments is equated to corresponding detemrinate system carrying these two effects separately. Let-us derive the three-moment equation.

Consider a generalized two-span beam element under the action of applied loads and redundant support moments acting on BDS.


W

A/

A 1

A

C/

C

C1

ha

B

Tangent at B

Fig (a)

I1 I2

L1L1

hc

θb θb

A1 A2

a1 a2

Mb

Ma Mc

A3

A4 A5

A6

L /31

2/3 L1 2/3 L2

L /32

Fig (b)

Fig ( c)

BMD due toapplied loadson simple spans

Generalizedredundant momentdiagram

Fig(a) is an indeterminate beam subjected to applied load (udl in this case) which has shown settlement such that support B is at a lower elevation than support at A and C and difference of elevation w.r.t intermediate support B is ha and hc. The angle θB on either side of support B must be equal. Fig(b) is B.M.D. due to applied load on simple spans where A1 is Area of B.M.D. on span L1 and A2 is area of B.M.D. on span L2. a1 and a2 are the locations of centroids of B.M.D’ s on L1 and L2 from left and right supports respectively. So invoking continuity of slopes and knowing that for small angels θ = tanθ.

AA1

L1 =

CC1

L2


Evaluate AA1 by second Moment Area Method. We know that AA1 = AA/ − A1A/ = ha − deviation of point A/ on the elastic curve from the tangent drawn at point B on the elastic curve.

= ha − 1

EI1

A1a1 + A3 ×

L1

3 + A4 × 23 L1

expressing A3 and A4 in terms of moments

AA1 = ha − 1

EI1

A1a1+

L1

3 × 12 MaL1 +

23 L1 ×

12 MbL1

= ha − 1

EI1

A1a1 +

MaL12

6 + MbL12

3 divide by L1

AA1

L1 =

ha

L1 −

1EI1

A1a1

L1 +

MaL1

6 + MbL1

3 (1)

Now evaluate CC1

L2 on similar lines. We have from geometry

CC1 = C1C/ − CC/

= (deviation of point C/ from tangent at B) − hc

= 1

EI2

A2a2 + A5 ×

23 L2 + A6 ×

L2

3 − hc

expressing A5 and A6 in terms of Moments

CC1 = 1

EI2

A2a2 +

23 L2 ×

12 MbL2 +

L2

3 × 12 MCL2 − hc

= 1

EI2

A2a2 + Mb

L22

3 + MC L22

6 − hc divide by L2

CC1

L2 =

1EI2

A2a2

L2 +

Mb L2

3 + MC L2

6 − hc

L2 (2)

Equating (1) and (2), we have

ha

L1 −

1EI1

A1a1

L1 +

Ma L1

6 + Mb L1

3 = 1

EI2

A2a2

L2 +

Mb L2

3 + Mc L2

6 − hc

L2


Multiply by 6E and simplify, we have after re-arrangement

Ma

L1

I1 + 2Mb

L1

I1 +

L2

I2 + Mc

L2

I2 = −

6 A1a1

I1L1 −

6 A2a2

I2L2 +

6 Eha

L1 +

6 Ehc

L2

The above equation is called three-moment equation. 9.1. Analysis of Continuous Beams by three-Moment Equation. We apply three moment equation to two spans at a time which gives us one equation. With the successive applications, the required member of equations are obtained and are solved simultaneously. EXAMPLE: Analyze the continuous beam shown below by three-Moment equation. Take E = 20 × 106 KN/m2 and Ic = 40 × 10-6 m4.

12 KNA B C D9.6 KN/m

32 KN

3m

LoDo

Io =

Lo6m8m6m2m

2Ic 4Ic 2Ic

Fig (a)

Fig (b)A3A2

3m4m409.6 144

A = 01

9.6 x 82

8= 76.8

32 x 64

= 48

BMD

SOLUTION: When a fixed support at either end is encountered, an imaginary hinged span of length Lo and Interia Io = ∞ is added to conform to acted support conditons and to make the method applicable in similar situations. The same has already been done in Fig(a). Fig (b) is the BMD’ s on simple spans, their Areas and its locations. Apply three-moment equation to spans AB and BC at a time. We have

Ma

6

2Ic + 2Mb

6

2Ic + 8

4Ic + Mc

8

4Ic = − 6 × 0 − 6 × 409.6 × 4

4Ic × 8

Simplify and multiplying by Ic both sides of equation, we get. 3Ma + 10 Mb + 2 Mc = − 307.2 put Ma = − 24 KN-m 10 Mb + 2 Mc = − 235.2 divide by 10 Mb + 0.2 Mc = − 23.52 (1)


Now apply three-moment equation to spans BC and CD

Mb

8

4Ic + 2 Mc

8

4Ic + 6

3Ic + MD

6

3Ic = − 6 × 409.6 × 4

4Ic × 8 − 6 × 144 × 3

3Ic × 6

Simplify and multiply by Ic, we have, 2 Mb + 8 Mc + 2 MD = − 307.2−144 = − 451.3 divide by 2 Mb + 4 Mc + MD = − 225.625 (2) Now apply three-moment equation to spans CD and DDO

Mc

6

3Ic + 2 MD

6

3Ic + Lo∞ + Mdo

Lo

∞ = − 6 × 144 × 3

3Ic × 6

Simplify and multiply by Ic both sides of equation. 2 Mc + 4 MD = − 144 divide by 2 Mc + 2 MD = − 72 (3) We have obtained three equations from which three-Unknowns Mb, Mc and MD can be calculated. Subtract equation (2) from (1) Mb + 0.2 Mc = − 23.52 Mb + 4 Mc + MD = − 225.625 − 3.8 Mc − MD = 202.105 (4) Multiply equation (4) by (2) and add in equation (3) − 7.6 Mc − 2MD = 404.21 Mc + 2 MD = − 72 − 6.6 MC = 332.21 So Mc = − 50.3 KN-m put Mc in equation (1), we get Mb = − 13.46 KN-m put Mc in (3), we get MD = − 10.85 KN-m.

Finally Mb = − 13.46 KN-m Mc = − 50.3 KN-m MD = − 10.85 KN-m Checks: The above calculated values of moments are correct if they satisfy the continuity of slope requirements. Slopes at any intermediate support point can be calculated from the two adjacent spans by using conjugate beam method. While applying checks, it is assumed that reader is well conversant with the conjugate beam method. Before we could apply checks, it is necessary to plot reactant moment diagram (support-moments) to get their contribution in slope calculation. Here is the statement of conjugate beam theorem number one again.

“ The shear force at any point on the conjugate beam loaded with MEI diagram is the slope at the

corresponding point in the actual beam carrying applied loads.” In applying the conjugate -beam method, we must use the original sign convention for shear force as applied in strength of Materials subject. (i.e., “ left up, right-down, positive)


6m8m6m2m

A B C D

A5

A4 A6 A8

A7 A9

13.45

50.3

10.85OO

24

Fig ( c)BMD divided into convenient shapes.

Fig(c) is the reactant moment diagram The areas of positive BMD’ s act as loads in downward direction to which reactions are upwards. The areas of negative BMD’ s act as loads in upward direction to which support reactions are downwards. The direction of reaction is accounted for in the signs appropriately.

A4 = 13.45 × 6 = 80.7 A7 = 8(50.3 − 13.45)

2 = 146.2

A5 = 6(24 − 13.45)

2 = 31.65 A8 = 10.85 × 6 = 65.1

A6 = 13.45 × 8 = 107.6 A9 = 6(50.3 − 10.85)

2 = 118.35

Checks. SPAN AB

S.F at A = θa = 1EI

−

A42 −

23 A5 =

12EIc

− 80.7

2 − 23 × 31.65

θa = − 30.725

EIc (There is no check on this value as, it is not a continuous support)

θb = 1

2EIc

A4

2 + 13 A5 =

12EIc

80.7

4 + 31.65

3

= 25.45EIc Clockwise.

SPAN BC

θb = 1

4EIc

A2

2 − A62 −

13 A7 =

14EIc

409.6

2 − 107.3

2 − 13 × 147.5

θb = 25.46EIc Clockwise


θc = 1

4EIc

− A2

2 + A62 +

23 A7 =

14EIc

− 409.6

2 + 107.3

2 + 23 × 147.5

θc = − 13.18

EIc

SPAN CD

θc = 1

3EIc

A3

2 − A82 −

23 A9 =

13EIc

144

2 − 65.1

2 − 23 × 118.33

θc = − 13.16EIc

θD = 1

3EIc

−

A32 +

A82 +

13 A9 =

13EIc

−

1442 +

65.12 +

118.333

θD = 0 (Fixed end) All slope values have been satisfied. This means calculated support moment values are correct. Now beam is statically determinate we can construct SFD and BMD very easily. We have seen that numerical values of E and I are required in this case only if one is interested in absolute values of θ. However, these values are required while attempting a support settlement case. Determine reactions and plot SFD and BMD.


12 KNA B C D9.6 KN/m

32 KN

3m

6m8m6m2m

13.76 32.031 KN 69.203 5.806

O

1.76

33.79

12 3.52m

43.009

5.806

o

26.194 KN

SFD

BMD

-2413.44

50.308

10.856

EXAMPLE-2: Analyze the continuous beam shown below by three moment equation if support at B

sinks by 12 mm. Take E = 20 × 106 KN/m2; Ic = 40 × 10-6 m4.

A B C D

Do

Io =

Lo6m8m6m2Ic 4Ic 3Ic

Fig (b)Reactant moment diagramA to A are areas of adjusted BMD.1 5

12mm

B/

O O

0.82.0

A A1 A2 C

A4

-1.6A5

A3


SOLUTION: As the extreme right support is fixed, an imaginary Hinged span of length Lo and Ic = ∞ has already been added to make the method applicable and to conform to the support characteristic at D. Now it is a sort of continuous support. Only analysis due to differential settlement at B is required. Had there been some applied loads also, those could have been considered at the same time also. Now EI = 20 × 106 × 40 × 10−6 = 800 KN-m2. we also know that Ma = 0 and MDO = 0 being extreme hinge supports. Spans AB and BC When we consider these spans and compare them with the derivation, we find that situation is

similar so both ha and hc terms are positive and equal to 12 mm using three-moment equation.

Ma

6

2Ic + 2Mb

6

2Ic + 8

4Ic + Mc

8

4Ic = 6E × 12 × 10-3

6 + 6E × 12 × 10−3

8

put Ma = 0, simplify and multiply by Ic 2Mb (3+ 2) + Mc (2) = EIC × 12 × 10-3 + 0.75 EIC × 12 × 10-3 put EI = 800 10 Mb + 2 Mc = 9.6 + 7.2 = 16.8 divide by 10 Mb + 0.2 Mc = 1.68 (1) Spans BC and CD Comparing these two spans with the derivation, we notice that ha term is equal to − 12mm and hc term is zero.

Ma

8

4Ic + 2Mc

8

4Ic + 6

3Ic + Md

6

3Ic = 6E(− 12 × 10-3)

8 + 0

Simplify and multiply by Ic 2 Mb + 8 Mc + 2 Md = − 7.2 divide by 2

Mb + 4 Mc + Md = − 3.6 (2) Spans CD and DDO There is no load and settlement on these two spans so right handside of equation is zero

Mc

6

3Ic + 2Md

6

3Ic + Lo∞ + Mdo

Lo

∞ = 0


We know that Mdo = 0; Lo∞ = 0

Simplify and multiply by Ic 2 Mc + 4 Md = 0 divide by 2

Mc + 2 Md = 0 (3) Above three linear simultaneous equations which are solved. Subtract (2) from (1) Mb + 0.2 Mc = 1.68 Mb + 4 Mc + Md = − 3.6 − 3.8 Mc − Md = 5.26 (4) Now multiply equation (4) by 2 and add to equation (3) − 7.6 Mc − 2 Md = 10.56 Mc + 2 Md = 0 − 6.6 Mc = 10.56 Mc = − 1.6 KN-m

Md = − Mc2 = + 0.8

Mb = 2 KN-m Plot end moment diagram. Add and subtract equal areas on spans BC and CD and apply conjugate beam method.

A1 = 12 × 6 × 2 = 6

A2 = 12 × 8 × 2 = 8

A3 = 12 × 6 × 0.8 = 2.4

A4 = 12 × 8 × 1.6 = 6.4

A5 = 12 × 6 × 1.6 = 4.8

Compute slopes at supports. θa = Slope due to settlement (configuration) + due to end moments

= 12 × 10−3

6 + 1

2EIc

A1

3 = 12 × 10−3

6 + 1

1600 6

3 = 3.25 × 10−3 rad.

Span AB


θb = 12 × 10−3

6 + 1

2EIc

−

23 A1 =

12 × 10−6

6 + 1

1600

−

23 × 6

= -5 × 10−4 rad. Span BC

θb = 12 × 10−3

8 + 1

4EIc

2

3 A2 − 13 A4 =

12 × 10−3

8 + 1

4 × 800

2

3 × 8 − 13 × 6.4

θb = − 5 × 10−4 rad.

θc = 12 × 10−3

8 + 1

4EIc

−

13 A2 +

23 A4

θc = − 1 × 10−3 rad. Span CD

θc = 0 + 1

3EIc

1

3 A3 − 23 A5 =

13 × 800

1

3 × 2.4 − 23 × 4.8

θc = −1 × 10−3 rad.

θd = 0 + 1

3EIc

−

23 A3 +

13 A5 = 0 +

13 × 800

−

23 × 2.4 +

13 × 4.8

θd = 0 (Fixed end) Checks on slopes have been satisfied so computed moment values are correct. Now beam is determinate. SFD and BMD can be plotted. Resolve same problem, for a differential sinking of 12 mm at support C. we get the following equations. Mb + 0.2 Mc = − 0.72 (1) Mb + 4 Mc + Md = 8.4 (2) Mc + 2 Md = − 4.8 (3) Solution gives Mc = + 3.49 Md = − 4.145 Mb = − 1.418 apply continuity checks and plot SFD and BMD. Unsolved Examples: Solve the following loaded beams by three-moment equations.


70 KN3m

A B C

8m 12m

EI = Constt. Final equations: Ma + 0.5 Mb = − 90.312 (1) Ma + 5 Mb + 1.5 Mc = − 213.12 (2) Mb + 2 Mc = 0 (3) End Moment Values: Mc = 16.41 Mb = − 32.82 Ma = − 73.91

24 KN/m

A

B C D

6m 72 KN

4m

Lo

1.5 m

6m12m6m3Ic 10Ic 2Ic

E

60 KN

16 KN/m24 KN

Final Equations: 2 Ma + Mb = − 216 (1) 2 Ma + 6.4 Mb + 1.2 Mc = − 1555.2 (2) 1.2 Mb + 8.4 Mc = − 1495.2 (3) End moment values: Ma = − 0.361 KN-m Mb = − 215.28 Kn-m Mc = − 147.25 Kn-m

A B C D

6m12m6m

3Ic 10Ic 2Ic

15 mm E = 200 x 10 KN/m6 2

Ic = 400 x 10 m-6 4

Final Equations: 2 Ma + Mb = − 600 (1) 2 Ma + 6.4 Mb + 1.2 Mc = 1800 (2)


1.2 Mb + 8.4 Mc = − 600 (3) End moment values: Ma = − 537.69 KN-m Mb = 475.38 Mc = − 139.34 KN-m

A

3 KN/m 20 KN

3m8m5m2I I 2I

15 KN

8mI

B C

End moment values: Ma = − 75 KN-m Mb = 21.75 Mc = − 60 KN-m

A

9.6 KN/m 32 KN

6m8m2I 4Ic 3Ic

12 KN

6m

B C 3m D2m

Final equations: 10 Mb + 2 Mc = − 235.2 (1) 2 Mb + 8 Mc = − 451.2 (2) End moment values: Ma = − 24 KN-m Mb = − 12.88 Mc = − 53.18 Md = 0

A

9.6 KN/m 32 KN

6m8m2I 4Ic 3Ic

12 KN

6m

B C 3m D2m


Final equations: 10 Mb + 2 Mc = − 235.2 (1) 2 Mb + 8 Mc + 2 MD = − 451.2 (2) 2 Mc + 4 MD = − 144 (3) End moment values: Ma = − 24 KN-m Mb = − 13.455 Mc = − 50.33 Md = − 10.835

A

6m8m2Ic 4Ic 3Ic

6m

B C 3m D

2m

4.5 mm

Final equations: 10 Mb + 2 Mc = 6.3 (1) 2 Mb + 8 Mc + 2 Md = − 2.7 (2) 2 Mc + 2 MD = 0 (3) End moment values: Ma = 0 Mb = 0.7714 Mc = − 0.707 Md = 0.707

A B C3m3

64 KN

9m

EI = Constt. Final equations: 2 Ma + Mb = − 144 (1) 2 Ma + 10 Mb + 3 Mc = − 288 (2) Mb + 2 Mc = 0 (3) End moment values:


Mb = -19.2 Mc = 9.6 Ma = − 62.4

A

6m8m2Ic 4Ic 3Ic

6m

B C 3m D

4.5 mm3 mm

Ic = 400 x 10 m-6 4

E = 200 x 10 KN/m6 2

Final equations: Mb + 0.2 Mc = 5.4 (1) Mb + 4 Mc + MD = − 1.5 (2) Mc + 2 MD = − 12 (3) End moment values: Ma = 0 Mb = 5.45 Mc = − 0.27 MD = − 5.86

INFLUENCE LINES 383

CHAPTER TEN

10. INFLUENCE LINES

This is also another very useful technique in classical structural analysis. Influence lines are plotted for various structural effects like axial forces, reactions, shear forces, moments and thrust etc. As structural members are designed for maximum effects, ILD’ s help engineer decide the regions to be loaded with live load to produce a maxima at a given section.

“ An influence line is a graphical representation of variation of a particular structural effect at a given section for all load positions on its span.”

Two methods, viz, static method and virtual displacement method are used for the construction of ILD’ s. Mostly it is the later method which is prefered. All structures in general and Railway and Highway bridges in particular are frequently subjected to various types of moving loads. As influence lines describe variation at a particular section for all load positions on span, the effects of moving loads can be calculated very easily. It must be remembered that a system of moving loads moves as a unit. For Railway bridges standard cooper’ s E-60 and E-72 loadings are used whereas for highway bridges AASHTO lane loadings and truck loadings or sometimes tank loadings are used. When dealing with calculations regarding moving loads the problem is how to place the system so as to produce maximum effects at a given section. Sometimes mathematical criteria are used for the live load purpose and sometimes simple inspection is made. In each case influence lines help us simplify the things.

10.1. Statical Method of Constructing Influence L ines

In this method, a load may be placed at several positions within span/(s) and a mathematical expression for a particular structural effects at a section is set-up. By placing limits of X (the distance), the shape and ordinates of influence lines (called influence co-efficients also) can be determined.

For example consider the cantilever loaded below and let moment at fixed end A be represented by its influence line.

For a generalized load position as defined by distance X in the diagram, moment at A is.

BA

PX

l

LI.L.D. for Ma

Ma = − P (L − X) 0 < X < L

Minus sign with P shows a negative moment at A for all load positions (consider sign convention for moments)


For X = 0 (load at point B) moment at A is − PL. Influence co-efficient is L at B. If X = L load is at A so moment at A is zero. Influence co-efficent is zero. In between A and B, moment at A varies linearly, joining the points, ILD for Ma is obtained. Now even if several loads are placed on the cantilever, Ma is simply the sum of all loads when multiplied by corresponding ordinates.

If a cantilever supports a ud.l, the above I.L.D for Ma is applicable. Consider a strip of width dX located at a distance X from free end,

wA

wdXb

L

XdX

y

L - X

Ma = b

∫o WydX = w

b

∫o ydX

Where b

∫o ydX is area of I.L.D between limits zero to b.

10.2. Influence L ines for beam Reactions:

ILD’ s for reactions in case of simple beams and compound beams (determinate beams resting over several supports) can be drawn by using the already described procedure. Consider a simple beam with a single load sitting at any moment of time as shown

From statics it can be shown that

BA

PX

L

Ra = PXL

Rb = P(L - X)L

(L - X)L

yi lI.L.D. for Rb

I.L.D. for Ral

yi

X/L

Ra = PXL and Rb = P

L − X

L 0 < X < L

INFLUENCE LINES 385

When X = 0 (load at B); Ra = 0 and Rb = P (by putting limits in above expressions)

When X = L (load at A); Ra = P and Rb = 0 (by putting limits in above expressions)

Instead of maximum co-efficients equal to P it is costomary to have them equal to 1 so that these could be evaluated by the product of loads and respective ordinates and these diagrams become valid for several loads. So algebraically

Ra = ∑ Pi yi

Rb = ∑ Pi yi

10.3. Pr incipal of Vir tual Displacements:

Consider a simple beam under the action of load P as shown. Ra can be found by virtual displacements by imagining that support at A has been removed and beam is under the action of load P and Ra. Under the action of Ra, beam is displaced as A/B. The virtual work equation is

BA

PX

L

Ra

y

A/

Ra × AA/ − Py = 0 (Force × displacement)

So Ra = PyAA/ where y is the displacement due to Ra under P.

If AA/ = 1, Ra = Py A result already obtained.

This procedure of drawing ILDs’ is more useful for the complicated cases.

10.4. Reactions for Compound Beams:

A beam resting over several supports which has been made determinate by the availability of inserted hinges at suitable points is called a compound beam. The following Rules must be kept in mind while constructing ILD’ s for such cases.

1. Points of I.L.D corresponding to supports should show zero displacement except where virtual displacement is given (in case of reactions).

2. Portion of the beam between hinges which are straight before virtual displacements should remain straight after virtual displacement.

3. If a beam is continuous over two consecutive support and there is a hinge after these two supports, that portion of beam behaves a unit in case the virtual displacement is given elsewhere.

4. Portions of beam between pins which is straight before virtual displacement, shall remain


straight after virtual displacement.

Considering these guidelines given, draw influence lines for reactions for the following beam.

I. L. D for Ra

I. L. D for Rb

I. L. D for Rc

I. L. D for Rd

I. L. D for Re

I +

+ I

+ I

++ I

++ I

A F B G C H D E

If positive areas of above diagrams are loaded, upward reactions at corresponding support will occur or vice-versa.

Construct Influence lines for reactions for the following compound beam by virtual displacements.

INFLUENCE LINES 387

I. L. D for Ra

I. L. D for Rb

I. L. D for Rc

I. L. D for Rd

I. L. D for Re

I +

+ I

+ I

++I

+I

A G B H C I D E J F

+

I. L. D for Rf

I

Evaluation of maximum upward and down reaction due to concentrated loads and udl can be done by using the basic principles described already.

If several moving loads, from right to left direction, approach left hand support of a simple beam, the left reaction continues to increase and becomes maximum till leading wheel is at the left support. This corresponding first maxima will decrease immediately if the load falls off and leaves the span from left upon further advance, reaction at left support will start increasing and will become maximum again when second wheel is at the left support. So there will be as many maxima as is the number of loads.

Evaluation of reactions due to live load udl is rather simple as the span portion required to be loaded for maximum upward and downward support reactions are obvious by the simple inspection. Of course positive areas if loaded will give maximum upward reactions and vice-versa.


10.5. Influence L ines for Shear Force:

In structural analysis, normally we develop the methods by considering simple cases and some generalized conclusions are drawn which can then be applied to more complicated cases. So consider the following simple beam wherein a moving load (right to left) occupies the position shown at any instant of time.

Using left-up and write-down as sign convention for positive shear force.

A

PX

BC

a b

Ra Rb

For all load positions to right of point C, the shear force for at C (Vc) is equal to + Ra.

Vc = Ra

It means that for load position between point B and C, the Shape of ILD for SF at C will be the same as the shape of ILD for + Ra.

For all load positions to left of point C, the shear force at C (Vc) is equal to − Rb.

Vc = − Rb

It means that for load position between point A and C, the shape of ILD for SF at C will the same as shape of ILD for −Rb. Knowing that positive ILD is drawn above the reference line and negative ILD is drawn below the reference line, we obtain the ILD for Vc as shown below with the help of ILD’ s for reactions (+ Ra1 − Rb)

A

PX

BC

a b

Ra RbL

l b/L

a/L

+

l

I. L. D. for + Ra

I. L. D. for - Rb

I. L. D. for Vc

INFLUENCE LINES 389

Mathematically

Ra = PXL 0 < X < L

Rb = P (L − X)

L 0 < X < L

At X= 0, load is at B and Vc is zero. At x= b, load is at C and Vc = + Ra = PbL or

bL if P= 1.

The ordinates aL and

bL can be obtained by using similar triangles. Now inspect the ILD for Vc.

For a right to left advance of load system, Vc keeps on increasing till the “ leading load is at the section” , when leading load just crosses the section, Vc drops by the magnitude of load and this process continues. So we can write that for maximum SF at a section, “ the load should be at that section” . This is the first criterion of calculation of Vmax. Now the question comes to mind that which load among the moving load system should be placed at the section? To address this question, we have noted, that change in SF at a section, ∆V, is equal to change in Ra (∆Ra) minus the load leaving the Section. (Pn)

So, ∆V = ∆Ra − Pn

If W is sum of all the loads on the span L before advance of a, it can be shown that

∆Ra = WaL

So, ∆V = WaL − Pn

Any load which reverses this expression, should be brought back and placed at that section to realize the maximum SF at that section. So a change in the sign of above expression can be regarded as the second criterion for maximum shear force at a section.

It can also be shown that loads entering or leaving the span as a result of any particular advance do not affect the above expression very significantly.

The above method is called the statical method. The same shape of ILD for Vc can be obtained by virtual displacement method also.

A BC

a b

Ra RbL

b/L

a/L

+ I. L. D. for Vc

V V


Now imagine that resistance to vertical displacement at C has been destroyed (imagine a sort of cut at the section) and the vertical shear force as shown (opposite to sign convention for positive shear force). The area enclosed between the original position before virtual displacement and the deformed position after virtual displacement is the ILD for Vc.

10.6. Influence L ine Diagrams for Bending Moment:

Again we consider the simple beam under the action of a simple moving load as shown. Let it be required to construct ILD for Mc.

A

PX

BC

a b

Ra RbL

I. L. D. for Mcab/L

ILD for Ra x a

ILD for Rb x b

+a

b

If the load is between points B and C.

Mc = Ra × a = PXL × a 0 < X < b

at X = 0; load at B, Mc = 0. If X = b;

Mc = PabL

=

abL if P = 1

It means that for portion BC, the shape of ILD for Mc is the same as the shape of ILD for Ra multiplied by distance a.

If the load is between points A and C

Mc = Rb × b = P(L − X)

L × b b < X < L

At X = b, load is at C; , Mc = Rb × b

So Mc = PabL

=

abL if P = 1

INFLUENCE LINES 391

It means that for portion AC, the shape of ILD for Mc is the same as the shape of ILD for Rb multiplied by b.

At X = L; Load at A; Mc = 0 The same shape of ILD for Mc can be obtained by virtual displacements also.

A

PX

BC

a b

I. L. D. for Mcy

M M

a

b

Ra Rb

dx

M M

Idealized section at Cbefore virtual displacements

Section at C after virtualdisplacement

The virtual work equation is

work done by loads = work done by the moments.

P × y = M × δθ.

Or M = Pyδθ

So, if δθ = 1; the moment at Section C for a single load system will be load multiplied by corresponding influence ordinate (influence co-efficient) while constructing ILD’ s by virtual displacements, loads are not considered. Now construct ILD for Mc by virtual displacements.

At Section C, we imagine that the beam resistance to moments which produce rotations has been destroyed while resistance to shear and axial loads is intact. This situation is obtained by considering that at Section C; there is a sort of hinge (one degree of freedom system). On this hinge the moments are


applied on two sides of hinge as shown alone. The segments of beam rotate and the displaced beam position is ILD for Mc.

The one-degree of freedom system such as a hinge is further explained in diagrams shown which

illustrate the movement. This procedure can now be applied to more complicated cases where statical

approach may be laborious.

The method of virtual displacements can be applied to more complicated cases like compound

beams etc., by considering the basic ideas established in this chapter.

A

+

++

+

E B C F D1 2 3

1 2 3

ILD for M1 - 1

ILD for M2 - 2

ILD for M3 - 3

10.7. Evaluation of Mmax at a Section

In case of a simple beam supporting a moving load system, the maximum moment at a section is obtained when

1. One of the loads is at the section.

2. In case of several moving loads, that load shall be placed at the Section, for producing maximum moment at that Section, which reverses the average loading on two portions of span adjacent to Section.

INFLUENCE LINES 393

Average loading on any portion = sum of all loads on that portion

length of portion

10.8. Absolute Maximum bending Moment

In case of a series of moving loads traverse on a beam, the absolute maximum bending moment occurs near the mid span under the adjusted position of that load which gave us maximum bending moment at mid span. Procedure is as follows:

1. Apply the criteria of maximum bending moment at mind span to find the load which is to be placed at mid span.

2. For this position of loads find the position of resultant of all loads on span.

3. Move the system slightly so that mid-span is bisected by the resultant of all loads on span and the load which gives us maximum bending moment at mid-span.

4. Find absolute maximum bending moment. It will occur under displaced position of that load which gave us maximum bending moment at mid-span.

Considering that invariably loads would be magnified for design purpose and appreciating that the numerical difference between the values of maximum mid-span bending moment and absolute maximum bending moment is insignificant, evaluation of absolute maximum bending moment for a given moving load system appears to be of theoretical interest only. How interested students can evaluate it for only moving load system by considering the above four points and guidelines contained in this chapter.

10.9. Girders with Floor beams (Panelled girders)

Normally in bridge construction, moving loads are hardly applied to the main girders directly but instead following arrangement is used for the load transfer.

The moving load system comes on the stringers which transfer it to the main girder through floor beams in form of concentrated loads (Reactions of floor beams). So main girder is subjected to concentrated loads only. For large spans the main girder may be of steel, poured in-situ reinforced concrete or pre-stressed concrete. Points a, b, c, …. F are called panel points and the distance between any two panel points is called a panel.

With the above mentioned load-transfer mechanisms, it can be easily seen that ILD’ s for main reactions remain same as that for a simple beam as discussed already.


As there will be no load on the main girder except floor beam reactions, it is stated that for a given load position, the shear force within a panel remains constant so we can talk of shear force in panels rather that shear force at a section (panel and becomes a section). Let us now construct ILD’ s for shear force for various panels of girder already shown.

INFLUENCE LINES 395

4/5 ILD for + Ra

ILD for Vab

ILD for Vef

( + )

( - )

ILD for + RaILD for - Rb

( - )

( + )

0.4

0.4

ILD for Vcd

ILD for - Rb

ILD for Rb x b

ILD for Ra x a

3a/5

b/5

yb yc+

a

b

ILD for Mmn

d x 4d5d

6 d52d x 3d

5d=

+

ILD for Mc

I

I

A five panels main girder is shown for which various ILD’ s have been sketched.


10.10. ILD For Vab (ILD for shear in end panel)

If a load P is placed at a distance X from panel point b, then reactions at panel points a and b

will be PXd and

P(d − X)d respectively.

Pa = Panel point load at a or reaction of floor beam at a = Pxd , 0 < X < d

Pb = Panel poiint load at b or reaction of floor beam at b = P(d − X)

d 0 < X < d.

if X = 0, load P will be at b, then Pa = 0 and Pb = P

if X = d, load P will be at a, then Pa = P and Pb = 0 So, Vab ≡ 0

In between a and b, shear force varies linearly.

Now inspect the shape of ILD for Vab, it resembles with the shape of ILD for moment at point b considering the panelled girder as a simple beam. So to evaluate (Vab)max, criteria of max bending moment at a section b (reversal of average loading expression) will be applied.

10.11. ILD for Vef (ILD for shear in other end panel)

The construction of ILD for Vef is same as that for Vab and same arguments apply. Inspecting this diagram, it is clear that the shape resembles with ILD for bending moment at e if panelled girder was treated as a simple beam. So to evaluate (Vef)max, the criteria for maximum bending at point e shall be applied.

10.12. ILD for Ved (ILD for shear in intermediate panel)

Considering the load P on panel cd acting at a distance X from panel point d.

Pd = Panel point load at d or floor-beam reaction at d = P(d − X)

d , 0 < X < d.

Pc = Panel point load at c or floor-beam reaction at c = P(X)

d , 0 < X < d.

If load is to right of d; Vcd = + Ra So, ILD for Vcd for this region will be the same as that for Ra. If load is to left of C, Vcd = − Rb. So for this region shape of ILD for Vcd will be the same as the shape of ILD for − Rb. Now third possibility is load actinig on span CD itself as shown.

Inspecting the expressions for panel point loads at d and c stated above, we observe that the shear Vcd within the panel varies linearly. So joining the ordinates under points C and D by a straight line will complete ILD for Vcd.

10.13. Evaluation of (Vcd)max (Maximum shear force in intermediate panel)

If a moving load is advanced at point d in a direction from right to left, considering W/ is resultant of all loads on span CD, the following criteria can be easily developed as a consequence of variation of shear force is panel CD due to an advance.

WL >

W/

d

Any load which reverses the above criteria shall give (Vcd)max.

INFLUENCE LINES 397

10.14. ILD for Mmn

Section mn is located within panel bc. Same technique can be applied for constructing ILD for Mmn. If load P is to right of panel point C.

Mmn = Ra × a.

It means that if load is between points c and f, the shape of ILD for Mmn will be the same as shape of ILD for Ra multiplied by a. If load P is to left of panel point b, then.

Mmn = Rb × b.

It means that if load is between points a and b, then shape of ILD for Mmn will be the same as shape of ILD for Rb multiplied by b. Now consider load within panel bc with P acting at a distance X from c.

Pb = PXd and Pc =

P(d − X)d 0 < X < d.

then Mmn = Pb yb + Pc yc = PXd yb +

P(d − X)d yc 0 < X < d.

So between the panel, the moment varies linearly. Therefore joing the ordinates of ILD for Mmn at b and c by a straight line, we complete the ILD for Mmn.

Now it is understood that SF is generally maximum near the support while moment is generally maximum near the mid-span. So ILD for Mmn can also be used to evaluate corresponding maxima. If criteria of maximum bending moment is applied at a section corresponding to bigger ordinate, then (Mmn)max can be calculated for a moving load system.

10.15. ILD for Mc

At the panel points, the load is directly transmitted to the main girder and the panel girder behaves as a simple beam at the panel points. So ILD for Mc will be drawn considering the girder as a simple beam.

10.16. Influence L ines for axial forces in Truss Members:

As before, let us consider a simple case of parallel chord truss carrying loads at its lower chord. The conclusions obtained are general and can be extended to non-parallel chord trusses.


1

2

3

4

AG C F E

D

B

h

( - )

( + )

( + )

( - )

( + )

I

θ θ

ILD for S1

When a moving load system traverses the bottom chord of this trussed bridge, it is known that forces in top chord members will be compressive in nature while that in bottom chord will be tensile in nature. The forces in chord members are a function of moment divided by truss height. For a chord member take “ moment at the point where other two members completing the same triangle meet divided by height of truss.” This has already been established in this book when discussing method of moments and shears. So applying this S1 is a compressive force, so assigned a negative sign, equal to moment at C divided by the height of truss. So considering the truss as a simple beam, draw an ILD for Mc and divide it by the height of Truss. (S1)max can be evalutated by applying the criteria of maximum bending moment (Average loadings) at point C considering the truss as a simple beam.

INFLUENCE LINES 399

ILD for S3

It is a tensile force equal to moment at D divided by height of Truss. (S3)max can be evalauted by applying the criteria of maximum bending moment at point D.

ILD for S2

It is known that axial force in an inclined member is ± V

± Cosθ . Minus before Cosθ shall be taken

if the angle “ between inclined member and vertical” is counterclockwise. Now if the load is right of D, SF applicable to member 2 is + Ra. So corresponding portion of ILD for + Ra is taken. This is divided by − Cosθ. If the load is to left of C, SF applicable to member 2 is − Rb. So corresponding portion of ILD for − Rb is taken. This is again divided by − Cosθ. In between the panel SF varies linearly so we can join the corresponding points.

The shape of ILD for S2 resembles with the shape of ILD for intermediate panel shear in a panelled girder. So (S2)max can be evaluated by applying the criteria of maximum intermediate panel shear.

ILD for S4

If the load is at E or right of E, Force in member 4 is zero and if load is at or to left of point C, again the force in member 4 is zero. If the load is at F, the same will be the tensile force in member. Using these boundary conditions, ILD for S4 is constructed. Now inspect its shape. It resembles with the shape of ILD for moment at F (or D) in an equivalent simple beam of span CE. So (S4)max can be evaluated by applying the criteria of maximum bending moment (average loading criteria) at F (or D).

10.17. Influence lines for moment and hor izontal thrust in a three hinged arch.

We know that H = µcyc and

Mx = µx − Hy.

Where y will be the rise of arch at a distance X from origin (usually a support).


y yc

x

C

A

H

B

H

L

Va Vb

x(L - x)L

Ly4yc

L4yc

( + )

( + )( - )

ILD for H

ILD for Mx

( + )

Shaded area

Influence line for any structural effect can be drawn by following the formula for that structural effect.

10.17.1. ILD for hor izontal thrust H

Horizontal thrust H is developed at the springings (supports) of an arch. Examine the formula

for H

H =

µcyc . So ILD for H will be obtained if ILD for moment at centre is drawn, considering the

arch to be a simple bam, and is then divided by yc. The peak ordinate of ILD for H will be Lµyc. (H)max

due to a moving load system can be obtained by applynig the criteria of maximum bending moment at the centre.

10.17.2. ILD for Moment in the arch

From the Eddy’ s theorem we know that bending moment in the arch at a distance x from support is

Mx = µx − Hy

where µx = simple span bending moment at a distance X.

INFLUENCE LINES 401

So as a first step, we construct ILD for simple span bending moment at a distance X. Then we subtract the ILD for Hy. The net area between these two diagrams is the ILD for moment in the arch as shown.

10.18. Standard Leadings

For the design of Railway bridges standard Cooper’ s E-60 and E-72 loadings consisting of two locomotives each weighing 213 tons on 18 axles each followed by infinite udl representing compartments is considered. Structural affects obtained for a E loading can be used to get the same for another E loading by simply multiplying them with the ratio of E loadings.

Original E-60 or E-72 loadings are in kip-ft. system as follows:

15↓ 8/

30↓ 5/

30↓ 5/

30↓ 5/

30↓ 9/

4 of 19.5 ↓ 5 ↓ 6 ↓ 5 ↓ 8/

15↓ 8/

4 of 30 ↓ 5/ ↓ 5/ ↓ 5/ ↓ 9/

4 of 19.5 ↓ 5/ ↓ 6/ ↓ 5/ ↓ 5/

3/ft

Above wheel loads are in kips per rail or tonnes per track. (1 Ton = 2 Kips ; small ton)

Converting E-72 loading in SI Units we have IK = 5 KN approximately.

80 KN↓ 2.44

4 of 160 KN↓ 1.52 ↓ 1.52 ↓ 1.52 ↓ 1.52 ↓ 2.74

4 of 104 KN↓ 1.52 ↓ 1.83 ↓ 1.52 ↓ 2.44

80↓ 2.44

4 of 104 KN

↓ 1.52 ↓ 1.52 ↓ 1.52 ↓ 2.42 4 of 104 KN

↓ 1.52 ↓ 1.83 ↓ 1.52 ↓ 1.52 53 KN/m

Cooper’ s E-72 loading in SI-units is shown above and E-60 below:

66.75↓ 2.44

4 of 133.5 KN↓ 1.52 ↓ 1.52 ↓ 1.52 ↓ 2.74

4 of 86.77 KN↓ 1.52 ↓ 1.83 ↓ 1.52 ↓ 2.44

66.75↓ 2.44

4 of 133.5 KN↓ 1.52 ↓ 1.52 ↓ 1.52 ↓ 2.74

4 of 86.77 KN↓ 1.52 ↓ 1.83 ↓ 1.52 ↓ 1.52

43.8 KN/m

Distance between loads is in meters. For highway bridges AASHTO HS24-44 loading is internationally considered and it consists of a Tractor and Semi-trailer with three axles carrying 0.2W, 0.4W and 0.4W respectively. These loads when converted into kips are 16k, 32k and 32k. Standard AASHTO lane loading is probably 100 lbs/ft2.

However, in our country, due to circumstances 70 ton tank loading or Truck-train loading described in Pakistan Highway code can be used.

We shall use railway loadings only. Let us solve some typical problems now.

Example No.1: In a girder with Floor beams having five equal panels of length 9 meters each. Determine (a) Maximum positive and negative end panel shears. (b) Maximum Shear in the first intermediate panel from left hand end. The live load is Coopers E-72 loading.


( + )

( - )

( - )

( + )

ILD for Vab

ILD for Vef

ILD for Vbc

0.80

0.6

0.2

0.80

SOLUTION: 1. Maximum positive End Panel Shears (Vab)max

Advance loads at section B and use criteria W/

d < WL

Portion ab Portion bf

809 <

2498.8745 after 1st advance.

2409 <

2338.8745 after 2nd advance

4009 <

2178.8745 after 3rd advance

5609 >

2018.8745 after 4th advance.

It means that once 3rd load of 160 KN crosses point b, the criterion is reversed so for maximum end panel shear, 3rd load of 160 KN should be placed at point b. Now place the system of loads accordingly and compute corresponding ordinates.

INFLUENCE LINES 403

3.52 2.44 1.52 1.52 1.52 2.74 1.52 1.83 1.52 2.44 2.44 1.52 1.52 1.52 2.44 1.52 1.83 1.52 1.52

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 8.6 m

4 of 160 4 of 104 4 of 160 4 of 1048080

y1

y4

y19

0.8

ILD for Vab

Ordinates Under Loads:

y1 = 0.3128 y2 = 0.5297 y3 = 0.6648

y4 = 0.80 y5 = 0.766 y6 = 0.7053

y7 = 0.6715 y8 = 0.6308 y9 = 0.597

y10 = 0.5428 y11 = 0.488 y12 = 0.4548

y13 = 0.421 y14 = 0.387 y15 = 0.333

y16 = 0.299 y17 = 0.2586 y18 = 0.2248

y19 = 0.191

(Vab)max = 80 × 0.3128 + 160 (0.5297 + 0.6648 + 0.8 + 0.766)

+ 104 (0.7053 + 0.6715 + 0.6308 + 0.597)

+ 80 × 0.5428 + 160 (0.488 + 0.4548 + 0.421 + 0.387)

+ 104 (0.333 + 0.299 + 0.2586 + 0.2248) + 12 × 8.6 × 0.191 × 53

= 25.024 + 441.68 + 271.62 + 43.42 + 280.128 + 116 + 43.52

= 1221.4 KN.

Similarly (Vef)max = -1145 KN (Do the Process yourself)

We have to observe a similar Process for evaluation of (Vef)max as was used for (Vab)max. The loads will be advanced at point e and average loadings on portions ae and ef will be compared. The


load which produces reversal after advance should be brought back and placed at section e for (Vef)max.

Evaluation of (Vbc)max

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

4 of 160 4 of 104 4 of 160 4 of 1048080KN

y1

y4

y18

0.6

0.82m

f

y14

ed

y5+c

ba

ILD for Vbc( )-0.2

2.25 6.75m

I.L.D for Vbc

Once loads are advanced from right to left at C, the following criteria shall be used to evaluate maximum intermediate panel shear (Vbc)max

WL >

W/

d

Portion bc portion cf

809 <

206445 after 1st advance.

2409 <

216845 after 2nd advance

4009 <

227245 after 3rd advance

5609 >

2315.4645 after 4th advance.

So maximum positive SF in panel bc will be obtained when 3rd wheel of 160 KN is placed at point c. Now place loads as shown above and determine corresponding ordinates of ILD. Multiply loads and ordinates by giving due care to signs of ILD, we obtain (Vbc)max.

Now from similar triangles, influence co-efficients y1,...... y18 are:

y1 = 0.113 y2 = 0.33 y3 = 0.465

y4 = 0.6 y5 = 0.566 y6 = 0.505

y7 = 0.472 y8 = 0.431 y9 = 0.397

y10 = 0.343 y11 = 0.289 y12 = 0.255

y13 = 0.221 y14 = 0.187 y15 = 0.126

INFLUENCE LINES 405

y16 = 0.093 y17 = 0.052 y18 = 0.018

So, (Vbc)max = 80 × 0.113 + 160 (0.33 + 465 + 0.6 + 0.566)

+ 104 (0.505 + 0.472 + 0.431 + 0.397) + 80 × 0.343

+ 160 (0.289 + 0.255 + 0.221 + 0.187)

+ 104 (0.126 + 0.093 + 0.052 + 0.018)

(Vbc)max = 720.34 KN

EXAMPLE NO.2: Determine the maximum bending moment at a cross-section 9.1m from left hand for

a beam of span 27.3m. The moving live load is 117 KN/m having a length of 6m.

SOLUTION:

Sketch ILD for moment at the indicated section.

Now let us assume that the given position of Udl gives us(Mc)max at a distance X from C as

shown. Determine Ra for this position

∑Mb = 0

Ra × 27.3 = 702 (3 + 12.2 + X)


Ra = 390.84 + 25.71 X

Moment at C = Mc = Ra × 9.1 − 117 X2

2

Mc = (390.84 + 25.71) 9.1 − 117 X2

2

Simplify

Mc = 3556.64 + 233.96 X − 58.5 X2

If BM at C is maximum, then

d McdX = Vc = 0

233.96 X − 2 × 58.5 X = 0

X = 2m

Now compute y1 and y2 from similar triangles of ILD

18.227.3 =

y1

7.1 → y1 = 4.733 m

9.127.3 =

y2

14.2 → y2 = 4.733 m

So (Mc)max = ud.l × area of ILD Under UDL

= 117 (6 × 4.733 + 12 × 6 × 1.327)

= 3788.3 KN-m

INFLUENCE LINES 407

EXAMPLE NO. 3: Calculate maximum bending moment at Section mn and pq of a five panel bridge. Each panel is of 9m.

Five loads of 160 KN each spaced at 1.52m travel from right to left.

1.52 1.52 1,521.52

160 160 160 160 160

8.1

6.3

y2y1 y4

y3

5 of 160

ILD for Mmn

ILD for Mpq

99

Evaluation of (Mmn)max

It is recommended that criteria of maximum bending moment be applied at maximum ordinate of 8.1 corresponding to Panel point C. Now comparing average loadings on portion ac and cf, we find that 3rd load reverses the criterion as it crosses. So it must be placed at point C. Determine ordinates


8.127 =

y3

25.48 → y3 = 7.644, y4 = 7.188, y1 = 6.3 + 1.496 = 7.796

y2 = 6.3 + 1.192 = 7.492

(Mmn)max = 160 (7.492 + 7.796 + 8.1 + 7.644 + 7.188) = 38.22 × 160

= 6115.2 KN-m

The reader is also suggested to calculate (Mmn)max. by coinciding the resultant of moving load system with the maximum ordinate. Place the loads accordingly. Compute influence co-efficients and multiply loads with respective ordinates to compute (Mmn)max. Compare this value with the previous one.

(Mpq)max

As ILD for Mpq is symmetrical about centre-line (mid span), Arrange the loads such that the resultant falls on mid-span. All five loads shall be accommodated and will have an ordinate of 9.

(Mpq)max = 160 (9 + 9 + 9 + 9 + 9) = 7200 KN-m

Impor tant:

The instructor is advised to work with lesser number of loads, usually five to seven, in the class and Establish the procedure. The students can then be given assignments involvinig standard trains etc., for clarification of their concepts.

EXAMPLE NO. 4:

A simple beam has a clear span of 27.5 m. Construct ILD for SF at a section 6.1m from left support. How should Coopers-E-60 loading be placed to calculate maximum shear force at this section?

SOLUTION: Draw ILD for Vc. Advance the loads at section C. We shall show the load position required for (Vc)max only.

( - )

( + )

66.754 of 133.5 4 of 86.77 3 of 133.5

66.751.28 m

CA B

6.1 m 21.4 m0.778

y1

y2

y9y12

ILD for Vc

0.222 Computing influence co-efficients y1......y12 from similar triangles.

INFLUENCE LINES 409

y1 = − 0.133, y2 = 0.722, y3 = 0.667,

y4 = 0.612, y5 = 0.512 y6 = 0.4566,

y7 = 0.3901, y8 = 0.335, y9 = 0.246,

y10 = 0.157 y11 = 0.10, y12 = 0.0466

In order to have (Vc)max, at least one load should be at C. To decide which load should be placed at C, reversal in the sign of following equation is sought.

∆V = WaL − Pn

W = Sum of all the loads on span before advance.

a = any particular Advance

L = Span

Pn = magnitude of Load crossing the section due to an advance.

____ For the first advance

∆V = 1281.58 × 2.44

27.5 − 66.75 = + 46.96 KN.

It shows that SF at C has increased due to 1st advance.

____ For second advance.

∆V = 1415 × 1.524

27.5 − 133.5 = − 55.08 KN.

It shows that if second advance at C is made, Vc decreases. So for (Vc)max, position before 2nd advance (after 1st advance) is required. For this position above influence co-efficients have been computed.

(Vc)max = 66.75 (− 0.133) + 133.5 (0.778 + 0.722 + 0.667 + 0.612)

+ 86.77 (0.512 + 0.4566 + 0.3901 + 0.335)

+ 66.75 (0.246) + 133.5 (0.157 + 0.1 + 0.046)

= 567.37 KN

EXAMPLE NO. 5:- Calculate the maximum bending moment at the points C and D if five loads of


160 KN each spaced at 1.52 m cross-the bean from right to left.

CA B

7 m 7 m 14 m

5.25

y1 y2 y3 y4 y5

(+)

1.521.521.521.521.52

5 of 160 KN

y1 y2 y3 y4 y5(+)

ILD for Mc

ILD for Md

D

MCMax

Line-up all loads upto point C (theoretically slightly to right of C). Give advances at point C and compare average loading in portion AC and BC due to various advances.

Portion Ac Portion Bc

1607 <

4 × 16021 after 1st advance.

2 × 160

7 > 3 × 160

21 after 2nd advance.

So, as the second load of 160 KN crosses ponit C, reversal is obtained. So for (Mc)max, this load

INFLUENCE LINES 411

should be brought back and placed at C (position before 2nd advance or after 1st advance). Compute influence co-efficients.

y1 = 4.11, y2 = 5.25, y3 = 4.87

y4 = 4.49, y5 = 4.11

(Mc)max = 160 (4.11 + 5.25 + 4.87 + 4.49 + 4.11) = 3652.8 KN-m

(Md)max

This section is mid span of beam. Clearly applying the criteria of maximum bending moment at D (comparing Average loadings on AB and BD), we get

Span AD Span BD

16014 <

4 × 16014 after 1st advance

2 × 160

14 < 3 × 160

14 after 2nd advance

3 × 160

14 > 2 × 160

14 after 3rd advance.

So position before 3rd advance (or after 2nd advance) will give us (Md)max. Place the loads accordingly and compute influence co-efficients.

y1 = y5 = 5.48 y2 = y4 = 6.24 y3 = 7

So, (Md)max = 160 (5.48 + 6.24 + 7 + 6.24 + 5.48)

= 4870.4 KN-m

EXAMPLE NO.6:

Calculate maximum axial forces induced in members 1, 2, 3 and 4 of truss already shown if five loads of 150 KN each spaced at 1.52m corsses at the bottom chord from right to left. Take h = 2m and span = 5d = 10 meters.


SOLUTION:

The corresponding ILD’ s for S1.....S4 have already been plotted. Now we will use those diagrams to calculate maxima. See the Truss of article 9.16.

160

( + )

( + )

( - )

1.521.521.521.52

5 of 160 KN

1.521.521.521.52

5 of 160 KN

1 m

1 m

+0.565

( + )

1.52 1.53

y1 y3

INFLUENCE LINES 413

S1max.

The shape of ILD for S1 resembles with the shape of ILD for Mc in an equivalent simple beam. So giving advances at C (now forget the truss and play with ILD’ s only). Apply the criterion for maximum moment at C. Portion Ac Portion Bc

1604 <


2 × 160

4 = 3 × 160

6 after 2nd advance.

Considering equality as a reversal, S1max will be obtained for position before second advance (or after 1st advance). Place loads accordingly and compute influence co-efficients. y1 = .744, y2 = 1.2 y3 = 0.896 y4 = 0.592 y5 = 0.288 So, S1max = 160 (0.744 + 1.2 + 0.896 + 0.592 + 0.288) = − 595.2 KN (It is a compressive force) S3max Inspect the shape of ILD for S3. It resembles with the shape of ILD for moment at D considering the truss to be a simple beam. So apply the criterion of maximum moment at D. Portion AD Portion BD

1606 <

3 × 1604 (last load not on span) after 1st advance.

2 × 160

6 < 3 × 160

4 After 2nd advance.

3 × 160

6 = 2 × 160

4 After 3rd Advance.

So for S3max, position before 3rd advance is valid (After second advance). Place the loads accordingly and compute influence co-efficients. y1 = 0.592, y2 = 0.893, y3 = 1.2, y4 = 0.744, y5 = 0.288 (S3)max = 160 (0.592 + 0.893 + 1.2 + 0.744 + 0.288) = 594.72 KN (It is a tensile force).

S2max Inspect the shape of ILD for S2. It resembles with the shape of ILD for as shear force in a intermediate panel of a panelled girder. So for evaluating S2max, we apply the criterion of maximum intermediate panel shear. Advance is made at D or F.

W/

d < WL

1602 =


So for S2max, the leading load should be placed at maximum ordinate, only three loads will be


acting on portion BD. y1 = − 0.565 y2 = − 0.3503 y3 = − 0.1356 (S2)max = 160 (− 0.565 − 0.3503 − 0.1356) = − 168.144 KN (It is a compressive force) S1max y1 = y3 = 0.24 y1 = 1 S1max = 160 (0.24 + 1 + 0.24) = 236.8 KN (It is a tensile force) 10.19. Influence L ines for Statically Indeterminate Structures: The same procedure can be adopted for constructing ILDs’ for indeterminate structures. However, compatibility and redundants have to be considered as demonstrated earlier. INFLUENCE LINE DIAGRAM FOR INDETERMINATE BEAMS (By method of virtual displacement) Influence line diagram for Shear . In virtual work for shear the B.M. does not do any work only shear force does the work. Case 1: Let us investigate ILD at a section of a simple beam. The section is at a distance a from A and at b from B support. This has already been done.

INFLUENCE LINES 415

P = 1.0a

v

c cA B

b

C//

C/

Ra Rb

c

b2*

2*

1*

1*

* = 1/L

a

RBRA

* = 1/L C/

b/L

a/L

y

C//

c

This is ordinate of ILD under load A

By Vir tual Work: Both the lines are parallel therefore, its work done by Moment is equal to zero.

θ*1 = θ*2 = θ Va θ* + Vb θ*

Vir tual Work: (Vir tual displacement) (i) total displacement equal to 1 unit. aθ* + bθ* = 1 (ii) total B.M. equal to zero. V(aθ* + bθ*) − Mθ* + Mθ* − Py* = 0 put aθ + bθ = 1 V(1*) − Py* = 0 If we take P = 1 V = y*

Or θ = 1L

Case 2: I.L.D for bending moment at the same section. Write work equation and equate to zero. Mθ*1 + Mθ*2 − Va θ*1 + Vb θ*2 − Py* = 0


or M (θ*1 + θ*2) − 0 − Py* = 0

M (θ*) = Py* or M = Pyθ . If P = 1 and θ = 1 radian.

than M = y* So aθ*1 = bθ*2 Or θ*1 + θ*2 = 1

⇒ θ*1 + ab θ*1 = 1 ⇒ θ*1 =

bL

θ*2 = aL

P = 1V

MRA RBb

a

2* = a/L

* = 1 radC//

C/

C C

ywyw

1* = b/LA B

1* = b 2*a

L

b

a

c

a/Lb/L

RA RB

We have obtained ILD for B.M at X in a simple beam Let us now consider the shown conjugate beam.

INFLUENCE LINES 417

2m 4m

1.0 = P

A C D B

Ra RbRc 6m6m

4m 6m

0.60/10 0/10

0/10

0.4

2/40 B

D2/10

C

Applying same concepts we get following ILD


X (L - X)

P = 1

A B

MB

RBRA

P

aL

Consider a propped cantilever

If support at A is removed,this will be deflected shape.

L-x B.M.D due to load on BDS as cantilever supported at B.

1/3 (L - x)

Applying moment area thereon, deflection at part A due to loads is

∆XL = 1EI

P

2 (l − X)2

l −

13 (l − X)

(+)L

fXX

Ra = 1(deflected shape) of BDS

Now consider load underredundant Ra = 1

B.M.D. for Ra = 1

L/3

L /22

Applying moment area thereon, deflection at A due to Ra = 1

fXX = 1EI

1

2 (l)2 2l

3 = l 3

3EI

Equation for compatibility ∆al − fXX Ra = 0 because A is a support. Net deflection should be zero.

INFLUENCE LINES 419

Ra = Salfxx , Ra =

P(l − X)2 (2l + X)2l 3 after putting values of Sal and fxx

Rb = 1 − Ra (equilibrium requirement)

So we get Rb = X(3l 2 − X2)

2l 3

We know

Mb = Ra × L − P (l − x) . Put value of Ra and simplify

Mb = PX(l 2 − X2)

2l 2 This expression will help in plotted ILD for Mb

ILD for Ra

X (L - X)P = 1

A B

Mb

RbRa 10m

Ra = P(l − X)2 (2l + X)

2 l 3

When X = 0 ⇒ Ra = 1.0 (put in above equation for Ra)

When X = 5 ⇒ Ra = 516 (put in above equation for Ra)

5/16

3rd-degree curve1.0ILD for Ra

O Simplify ILD for Rb can be plotted as below:

Rb

1.0ILD for Rb

Putting boundary conditions in the Mb expression ILD for Mb is obtained.


Mb = PX ( - X )2ll

2 2

2

3/16 l

ILD for Mb Ral − P(l − X) + Mb = 0 Mb = 1(l − X) − Ral

10.20. ILD for shear at Section mn:

XP

A B

Mb

RbRa 10m b = 6ma = 4m

m

n

m

A B

Mb

RbRa

m

n

c

1.0

( + )

1.0

1.0

Vmn

I.L.D. for Ra x a

I.L.D. for Ra x b

Load to right ofmn, Vmn = Ra x ait mean ILD for Vmnwill be same as ILDfor Ra multiplied bya for this portion

Load on left of mn

Vmn = Rb x bfor this portion, ILD forVmn is same is ILDfor Rb x b

10.21. ILD for Mmn Consider a hinge where ILD for moment desired.

P = 1

A C

10m

B

6m

INFLUENCE LINES 421

XP = 1

A C

RcRa L1

X P = 1

State-I

X

1.0

L2

B

Rb

Rb

B

Rb = 1.0State-II

y

P

ba

y

X

BC

C//

B//

Ra l

Rb l

Pab l

( ) PbX l

Primary structure or BDSunder load P = 1and redundant Rb at B.

Rb δbb − Py = 0 P = 1

Rb = yδbb

Compatibility equation at point B.

δbb

We know this is ILD formoment at B in asimple beam.


y = PbX6EIl (l

2 − b2 − X2) (X = 0 − a)

y = PaX6EIl (l

2 − a2 − X2) (X = 0 − b)

y = l2X(l2 − l22 − X2)

6EIl

δbb = l2X (l2 − l22 − l12)

6EIl

δbb = l12 l22

3EIl

and

Rb =

X (l2 − l12 − X2)2l12 l2

X = 0 − l1 with Origin at A

Rb = X (l2 − l12 − X2)

(2l12 l22) X = 0 to l2

Origin at C

INFLUENCE LINES 423

Now assume same values of spans and re-calculate.

P = 1

A C

L1 = 10m L2 = 6m

B

P = 1

A CB

I.L.D for Rb

1.0

y

A CB

( + )1.0

1.0

1.0

1.0 B

l2

l

Ra δaa − Py = 0

Ra =

y

δaa

I.L.D. for Ra

We knowl + l = L1 2

Compatibility at A

+

Rb = X (l2 − l22 − X2)

(2 × l12 × l2) = X (162 − 62 − X2)

2 × 102 × 6 by putting values of L1l1 and l2


X Rb Ra Rc

0 0

1 0.1825

2 0.36

3 0.5275

4 0.68

5 0.8125

6 0.92

7 0.997 Calculate Calculate

8 1.04 yourself yourself

9

10

0 Calculate

1 yourself

2

3

4

ILD for Ra can be obtained from ILD for Rb. Taking moments about C is equality to zero.

Ral + Rb × l2 − P(l − X) = 0

So Ra = P

l − X

l − Rb l2

l

and Rb = (l1 − X)

2l12 l (2l1 l − l1 X − X2)

INFLUENCE LINES 425

P = 1.0

A B

Mb

RbRa X

Ma

(L - X)

P

A B

MbMa

RbRa

State-I

yB State-IIA

1.0

θaa

1.0 2EIl

4EIl

θa = 0

4EIl

2EIl

(-)

(+)

1 rad

1 rad

At fixed support,

BDS underredundantmoment.


CHAPTER ELEVEN

11. THREE HINGED ARCHES

These are Curved Structures which are in use since ancient times. These were mostly used in buildings and the abatements used to be very thick. As our analysis capacity increased due to faster computers, it is now possible to understand behaviour of arches for various support, load and material conditions. These days arch bridges either in Reinforced concrete or the pre-stressed concrete are becoming a common sight due to asthetics of curved surfaces.

Arches when loaded by gravity loads, exhibit appreciable compressive stresses. At supports, horizontal reaction (thrust) is also developed which reduces the bending moment in the arch.

Aches can be built in stone, masonry, reinforced concrete and steel. They can have a variety of end conditions like three hinged arches, two hinged arches and find arches. Considering the geometry these can be segmental, parabolic and circular. An arch under gravity loads generally exhibits three structural actions at any cross-section within span including shear force, bending moment and axial compressive force. The slope of centerline of arch keeps on varying along span so above mentioned three structural actions also vary along span.

11.1. Eddy’ s theorem: The bending moment at any point on the arch is the difference between simple span bending moment and product Hy” .

Where H is the horizontal thrust at supports (springings), y is the rise of arch at a distance X from the origin.

Shape of simple span bending moment diagram due to applied loads is also called linear arch. Hy may also be termed as equation of centerline of actual arch multiplied by a constant (H).

Consider the following arch carrying the loads P1, P2 and P3. The shaded area is the BMD.

X

y

A BH H

VbVa

Hy

given arch y

linear archa P1

P2

P3

Bending moment at X is

MX = VaX − Hy − P1(X − a)

MX = µX − Hy. (Eddy’ s theorem)

THREE HINGED ARCHES 425

Where µz = Va × x − P1(X − a) = Simple span bending moment considering the arch to be a simple beam.

The inclined axial force (normal thrust) also contributes towards vertical shear force in addition to applied loads and reactions.

11.2. Three-hinged arch:

If an arch contains three hinges such that two hinges are at the supports and the third one anywhere within span, it is called a three hinged arch. This type of arch is statically determinate wherein reactions, horizontal thrust and all internal structural actions can be easily determined by using the laws of equilibrium and statics. If the third hinge is provided at the highest point, it is called crown of the arch.

Consider a three hinged arch with third hinge at the crown, then

MX = µX − Hy (1) becomes at center

Mc = µc − Hyc = 0

SO H = µc

Yc (2)

Cutting the arch as shown, and projecting forces along axis 1−1 and 2−2 and putting V = Va − P1 we have.

P = H Cosθ + VSinθ (3) along 1−1

Q = H Sin θ − Vcos θ (4) along 2 − 2

AH

C

B

P 2 V

Va

H

H

I

I

2

θ

11.3. Parabolic Arch If a three-hinged parabolic arch carries udl over its span, the arch will carry pure compression and no SF or BM. This is because the shape of linear arch (BMD due to loads) will be the same as shape of actual arch.

For a parabolic arch having origin at either of springings, the equation of centre line of arch at a distance X from origin where rise is y will be.

y = C.X (L − X) (5) constant C will be evaluated from boundary conditions.

at X = L2 , y = yc. we get

Yc = C. L2 .

L2 or C =

4 yc

L2

So y = 4 yc

L2 . X (L − X) (6)


The slope θ can be calculated from

dydX = tan θ =

4 yc

L2 (L − 2X) (7)

11.4. Circular Arch:

If arch is a part of Circle, it is convenient to have origin at the centre.

Consider triangle OEF OE2 = EF2 + OF2 Or R2 = X2 + (R −yc + y)2 (8)

and we also have from triangle ADO

L2

4 + (R − yc)2 = R2

yc (2R − yc) = L2

4 (9)

As span and central rise are usually known, Radius of arch R can be calculated from (9)

A

E

C

BDy

R

X

F

O

Rθ

L

Equation (8) can bge written as y = R2 − X2 − (R − yc)

Now once the basic equations for parabolic and circular arches have been established, let us solve some numericals.

EXAMPLE NO. 1

Analyze a three-hinged arch of span 20m and a central rise of 4m. It is loaded by udl of 50 KN/m over its left half. Calculate maximum positive and negative moments if

(i) The arch is parabolic

(ii) The arch is circular

SOLUTION: 1. Arch is Parabolic

∑ Ma = 0

Vb × 20 = 50 × 10 × 5

Vb = 250020 = 125 KN

Va + Vb = 50 × 10 = 500 KN

So Va = 500 − Vb = 500 − 125

= 375 KN

50 KN/m

A B

C

y4m

X

20RaVa=375 Vb=125

H=312.5 H=312.5


H = µc

yc =

125 × 104 = 312.5 KN

H = 312.5 KN

Ra = Va2 + H2

= 3752 + 312.52 = 140625 + 9765.25

and Rb = Vb2 + H2= 1252 + 312.52

Rb = 15625 + 97656.25

Rb = 113281.25 = 336.57 KN

= 238281.25 = 488.14 KN

Tanθa = VaH =

375312.5 = 1.2

Tanθb = VbH =

125312.5 = 0.4

θa = 50.19o θb = 21.800

Maximum positive Moment It is expected in portion AC. Write generalize MX expression.

MX = 375X − 50X2

2 − 312.5y

Now y = 4yc

L2 (L − X) = 4 × 4202 X(20 − X) = 0.04 (20X − X2)

y = 0.8 − 0.04X2 So MX = 375X − 25X2 − 312.5 [0.8X − 0.04X2]

= 375X − 25X2 − 250X + 12.5X2 Simplifying MX = 125X − 12.5X2 dMX

dX = VX = 0 = 125 − 25X

X = 5m from A. Putting Value of X in MX expression above. So Mmax = 125 × 5 − 12.5 × 52

= 625 − 312.5

Mmax = 312.5 KN-m

Maximum negative moment:

It would occur in portion BC at a distance x from B.

MX = 125X − 312.5y , Putting equation of y.

= 125X − 312.5 (0.8X − 0.04X2)

MX = 125X − 250X + 12.5X2

MX = − 125X + 12.5X2

dMxdX = VX = 0 = − 125 + 25X


X = 5m from B.

So putting value of X in MX expression above.

Mmax = − 125 × 5 + 12.5(5)2

= − 625 + 312.5

Mmax = − 312.5 KN−m

SOLUTION: Considering Circular Arch

EXAMPLE NO.2: Now or Solve the following loaded three hinged Circular Arch

A B

C

y4m

X

20m

Va=375 Vb=125

H=312.5 H=312.5

50 KN/m

Step 1. Reactions:

As before reactions are same.

Step 2. Equation of Circular Arch

The general equation is (X − h)2 + (y − k)2 = r2

h and k are co-ordinates at the centre and r is radius of Circle. There are three unknown in above equation, Viz, h, k and r and these can be determined from the following boundary conditions. Origin is at point A.

Boundary conditions

1. At X = 0, y = 0 It gives (−h)2 + (−k)2 = r2

h2 + k2 = r2 (1)

2. At X= 20, y = 0 It gives (20 − h)2 + (−k2) = r2

400 + h2 − 40h + k2 = r2 (2)

3. At X= 10, Y = 4 It gives (10 − h)2 + (4 − k)2 = r2

100 + h2 − 20h + 16 + k2 − 8k = r2

116 + h2 − 20h + k2 − 8k = r2 (3)

Or

Subtract (1) from (2) we get

400 − 40h = 0

h = 10


Put value of h in (1) and 3

100 + k2 = r2 (1)

116 + 100 − 200 + k2 − 8k = r2 (3)

or 16 + k2 − 8k = r2 (3) 16 + k2 − 8k = 100 + k2 (by putting Value of r2 from 1) 8k = 16 − 100 = − 84

k = −848 = − 10.5

Putting k = − 10.5 in (3) we get r2 = 16 + (− 10.5)2 + 8 x 10.5 = 16 + 110.25 + 84 = 210.25 So r = 14.5 meters.

Putting Values of h, k and r in general equation, we get

(X − 10)2 + (y + 10.5)2 = 14.52 Simplify it, we get.

y = − 10.5 + 14.52 − (X − 10)2

(y + 10.5)2 = 14.52 − (X − 10)2

= − 10.5 + 210.25 − X2 − 100 + 20X

y = − 10.5 + 110.25 − X2 + 20X (4)

We know, yc (2r − yc) = L2

4 (5)

and

y = r2 −

L

2 − X2

− (r − yc) (6) These equations are same as were

used in derivation earlier.

Alternatively to avoid evaluation of constants each time, equations (5) and (6) can be used.

Equation (6) is the equation of Centre-line of Circular arch.

Step 3: Calculation of Maximum moment.

Maximum positive moment occurs in span AC. Write MX expression

MX = 375X − 50X2

2 − 312.5 y put y from (4) above.

= 375X − 25X2 − 312.5 [− 10.5 + 110.25 − X2 + 20X ]

MX = 375X − 25X2 + 3281.25 − 312.5 110.25 − X2 + 20X


Now maximum moment occurs where shear force is zero. So

dMX

dX = VX = 375 − 50X −

312.5 (− 2X + 20)2 110.25 − X2 + 20X

= 0

375 − 50X = 312.5 (−X + 10)110.25 − X2 + 20X

divide by 50

7.5 − X = 6.25 (10 − X)

110.25 − X2 + 20X multiply by − 1 , We get

X − 7.5 = 6.25 (X − 10)

110.25 − X2 + 20X

(X − 7.5) 110.25 − X2 + 20X = 6.25 (X − 10) square both sides

(X − 7.5)2 (110.25 − X2 + 20X) = 6.252 (X − 10)2 Simplify

(X2 − 15X + 56.25) (110.25 − X2 + 20X) = 39.0625 (X2 − 20X + 100)

or 110.25X2 − X4 + 20X3 − 1653.75X + 15X3 − 300X2 = 39.0625X2 − 781.25X + 3906.25

+ 6201.56 − 56.25X2 + 1125X

Simplifying

− X4 + 35X3 − 285.0625X2 + 252.5X + 2295.3125 = 0

or X4 − 35X3 + 285.0625X2 − 252.5X − 2295.3125 = 0

Now it is considered appropriate to solve this equation by Modified Newton – Raphson iteration solutions which in general is

Xn+ 1 = Xn + f (xn)f / (Xn) (A)

So f (X) = X4 − 35X3 + 285.0625X2 − 252.5X − 2295.3125

And differentiate, f/ (X) = 4X3 − 105X2 + 570.125X − 252.5

In general, it is recommended that first root Xn should be always taken at 1 because it converges very fast. However, knowing that B. M will be maximum near the middle of portion AC, we take Xn = 2 (to reduce number of iterations possibly) and solve in the following tabular form. Evaluate f(X) and f/(Xn) expressions.

Iteration Number Xn f(Xn) f/ (Xn) Xn+ 1 from A above

1 2 −1924.06 499.75 5.85

2 5.85 147.251 290.1629 5.3425

3 5.3425 − 30.3142 406.3845 5.417

4 5.417 − 0.58794 390.546 5.418


So we get Xn and Xn + 1 as same after 4th iteration.

So X = 5.418 m put this in MX expressions

Mmax = 375 (5.418) − 25 (5.418)2 + 3281.25 − 312.5 110.25 − 5.4182 + 20 × 5.418

= 280.066 KN-m

Maximum negative moment in the arch

Let us assume that it occurs in portion BC at a distance X from A (10 < X < 20)

MX = 125 (20 − X) − 312.5 (− 10.5 + 110.25 − X2 + 20X ) Simplify

= 2500 − 125X + 3281.25 − 312.5 110.25 − X2 + 20X

or MX = 5781.25 − 125X − 312.5 110.25 − X2 + 20X

Maximum moment occurs where SF is zero, So differentiate MX expression w.r.t. X.

dMX

dX = 0 = − 125 − 312.5 (− 2X + 20)

2 110.25 − X2 + 20X

or 0 = − 125 + 312.5(X − 10)110.25 − X2 + 20X

125 110.25 − X2 + 20X = 312.5 (X − 10) squaring both sides. We have,

15625 (110.25 − X2 + 20X) = 97656.25 (X2 − 20X + 100) Simplify

110.25 − X2 + 20X = 6.25 (X2 − 20X + 100)

0 = 7.25X2 − 145X + 514.75 dividing by 7.25

X2 − 20X + 71 = 0 Solve this quadretic equation.

X = 20 ± 400 − 284

2

X = 20 ± 10.77

2 = 15.385m from A Put this value of X in MX expression above.

So Mmax = 5781.25 − 125 × 15.385 − 312.5 110.25 − (15.385)2 + 20 (15.385)

= 5781.25 − 1923.125 − 312.5 181.257 = − 349.115KN.m


11.5. Der ivation for center -line of a parabolic arch with suppor ts at different levels.

A

C

yc

hy

BX

L/2 L/2

The general form of a parabola is

y = aX2 + bX + c

Evaluate constants a, b and c by putting boundary conditions in above equation

At X = o; Y = o, (Point B) So C = o (1)

At X = L; Y = h, (Point A) So h = aL2 + bL (2)

At X = L2 ;

Y = yc + h, (Point C) So yc + h =

aL2

4 + bL2 (3) multiply by 4

h = aL2 + bL (2)

Equation (3) can also be written as

4(yc + h) = aL2 + 2bL (3) Subtract (3) from (2), we have

h − 4(yc + h) = − bL or b =

4L (yc + h) −

hL

Put this value of b in (2) and solve for a

h = aL2 + 4 (yc + h) − h or a = 2 h − 4(yc + h)

L2 )

a = −2 h − 4yc

L2


Now all constant have been evaluated in general terms. Put Values of a, b and c in general equation; we have

y = − 2X2 (h + 2yc)

L2 + X (4yc + 3h)

L . This is the generalized equation for a parabolic arch

with supports at different levels. Test this derived equation and see whether boundary conditions are satisfied.

− At X = o; y = o, put this in above equation. It is satisfied − At X = L, y = h, put this in above equation. It is satisfied

− At X = L2, y = h + yc, put this in above equation. It is also satisfied.

If supports are at the same level, h = o Put this in above equation, we get

y = − 4ycX2

L2 + 4 ycX

L or y = 4ycXL2 (L − X), after simplification.

and dy

dX = 4yc

L2 (L − 2X)

These two equations have already been used. Now we solve some Example. EXAMPLE NO.3:- Solve the following 3 hinged parabolic loaded arch with supports at different levels as shown.

A

C

yc = 9m

h = 3m

B

Vb45m

40 KN/m

H

H

Va

45m

y is the distance between hinges at A and C.c

∑Fy = 0

VA + Vb = 40 (45) = 1800 KN (1)

∑Mc = O , VA (45) − 9 H − 40 (45)

45

2 = 0 (2) Moments at C of forces on its left.

45VA − 9H − 40500 = 0

45Vb − 12H = 0 (3) Moment at C of forces on its right

Divide Equation (2) by 9

5VA − H − 4500 = 0 (2)


Multiply this Equation by 12 and subtract equation (3) from it.

60 VA − 12H − 54000 = 0 (2)

45VB − 12H = 0 (3)

_______________________________

60 VA − 54000 − 45VB = 0 (4)

Multiply Equation (1) by 45 and add in equation (4)

45VA + 45VB = 81000 (1)

60 VA − 45VB = 54000

Adding we get.

105 VA = 135000

or VA = 135000

105

VA = 1285.7 KN , put this value in equation (1)

so VB = 514.3 KN

We know, 45VB − 12H = 0 (3) , from this

H = 45 × VB

12

= 1928.63 KN (after putting value of Vb)

H = 1928.63 KN

New calculate (Mac)max and (Mbc)max

Keeping “ B” as origin , at a distance X from B in portion AC moment expression is

(Mac) = Va (90 − X) − H(y) − 402 (90 − X)2

y = X (4yc + 3h)

L − 2X2 (3 + 18)

902 . This equation was derived in previous article.

If h = 3m , y = X2 −

7X2

1350 (A) If h = 0 , y = 0.4X − X2

225 (B)

(Mac) = 1285.7 (90 − X) − 1928.63

0.4X −

X2

225 − 20 (90 − X)2 − (i)

after putting values of h and yc in above equation for y.

= 115713 − 1285.7X − 771.45X + 8.57X2 − 20 (8100 + X2 − 180X)

dMacdX = 0 = −1285.7 − 771.45 + 17.14X − 40X + 3600 Simplify

0 = + 1542.85 − 22.86X

X = 1542.85/22.86 = 67.5m (This value should be more than 45)


Putting this Value in Equation (i)

(Mac)max = 1285.7 (22.5) − 1928.63 (27 − 20.25) − 20 (22.5)2

= 5785 KN-m

(Mbc)max = 514.3X − 1928.63 × y . Moment at a distance X from B.

= 514.3X − 1928.63 ×

X

2 − 7X2

1350 (After putting equation for y) and values

of yc, h and L and using equation A.

= 514.3X − 964.315X + 10X2

dMbc

dX = 0 = 514.3 − 964.315 + 20X

= 20X = 450

X = 22.5m , Mbc = −5062.68 KN−m (after putting value of X above)

11.6. Development of Generalized equation of three hinged circular arch with suppor t at different levels.

A

C

yc=9m

y = 3m y

BX

L/2 = 45 L/2 = 45

General Equation of Circle is

(X − h)2 + (y − k)2 = R2

at X = 0, y = 0 ~ h2 + k2 = R2 − (1)

at X = 45 , y = yc + δ = 12

Putting (45 − h)2 + (12 − k)2 = R2 Simplifying it.

2025 − 90h + h2 + 144 − 24k + k2 = R2


2025 + 144 − 90h − 24k + h2 + k2 = R2 (2) Simplifying

2169 − 90h − 24k + h2 + k2 = R2 (2)

at X = 90 , y = 3 [ point A ]

(90 – h)2 + (3 – k)2 = R2 Simplifying

Put these values, 8100 − 180h + h2 + 9 + k2 − 6k = R2

8109 − 180h − 6k + h2 + k2 = R3 (3)

Equating (1) with (2) and multiply resulting equation by 2 and then equation (1) and (3)

2 [2169 − 90h − 24k = 0] ~ (4) or 4338 − 180h − 48k = 0 (4)

8109 − 180h − 6k = 0 ~ (5)

Subtract (4) from (5), we have

3771 + 42k = 0

k = − 3771

42

k = − 89.79

Put in Eq (4)

2169 − 90h – 24 (− 89.79) = 0

h = 4323.86

90

h = 48.04

Now from (1)

(48.04)2 + (−89.79)2 = R2

R = 101.83m

Now write equation of center-line of arch.

y = R2 − (X − h)2 + k

= (101.83)2 − (X − 48.04)2 + (− 89.79)

y = (10369.35 − X2 − 2307.84 + 96.08X) – 89.79 (A)

Point B: At X = 0 , y = 0 (see diagram now)

Point C: At X = 45 , y = 12

Point A: At X = 90 , y = 3

So Eq ,(A) has been correctly derived.


EXAMPLE NO. 4: Calculate maximum moments in portion AC & BC for the following 3-hinged loaded Circular each.

A

C

9m

3m

B

514.3

40 KN/m

1928.63

1285.7

1928.63X

y

SOLUTION: Reactions will be same as Previous.

1. Calculation of (Mac)max.

Write moment expression for use previously developed equation. Consider forces on left of section.

MX = 1285.7 (90 − X) − 402 (90 − X)2 − 1928.63 ( 10369.35 − X2 − 2307.84 + 96.08X − 89.79)

MX = 115713−1285.7X−20(8100−180X+ X2)+ 173171.7−1928.63 (10369.35−X2−2307.84+ 96.08X)

MX = 126884.69 − 20X2 + 2314.3X − 1928.63 (8061.51 − X2 + 96.08X)½ - (B) differentiate w.r.t.

dMX

dX = 0 = − 40X + 2314.3 − 964.315 (−2X + 96.08)

(8061.51 − X2 + 96.08X)½

(40X − 2314.3) (8061.51 − X2 + 96.08X)½ = 964.315 (2X − 96.08)

(40X − 2314.3)2 (806151 − X2 + 96.08X) = [1928.63 (X − 48.04)]2

Squaring and simplifying, we get.

(1600X2+ (2314.3)2−185144X) (8061.51−X2+ 96.08X)= 1928.6322 (X2+ 2307.84−96.08X) Simplifying

1298416X2 − 1600X4 + 153728X3 + 4.32 × 1010 − 5355984.5X2 + 514602989.8X

−1.49254021×109X+ 185144X3−17.78863×106X2= 3719613.68X2+ 8.5843×109−357380482.1X Simplifying

− 13965812.2X2 − 1600X4 + 338872X3 − 620556738X + 3.46157 × 1010= 0

8728.63X2 + X4 − 211.8X3 + 387848X − 2163412.5 = 0

f(X) = X4 − 211.8X3 + 8728.63X2 + 387848X − 21634812.5 differentiate it.

f/ (X) = 4X3 − 635.4X2 + 17457.26X + 387848

To cut-short, Let X = 55 (Because it is portion AC and X has to be more than 45)


So f(X) = (55)4 − 211.8 × (55)3 + 8728.63 × (55)2 + 387848 (55) − 21634812.5

= 1333.25

f/(X) = 4(55)3 − 635.4(55)2 + 17457.26(55) + 387848

= 91412.3

Xn − fx

f/(x) = Xn+ 1

= 55 − 13333.2591412.3

= 55 − 0.146

= 54.85

Now X = 54.85 (use this update value now)

f(X) = −560.16

f/(X) = 93833.35

Xn+ 1 = Xn − f(n)f/(n)

= 54.85 − (−560.16)93833.35

Xn+ 1 or X = 54.855969. The value of X has converged now.

Putting this value of X in Equation B to find (Mac)max

(Mac)max = 126884.69 − 20 (54.855969)2 + 2314.3 (54.855969)

− 1928.63 (8061.51 − (54.855969)2 + 96.08 × 54.855969)½

(Mac)max = 193552.7 KN−m

(Mbc)max. Working on similar lines (Mbc)max can be calculated now.

MX = 514.3X − 1928.63 y putting equation of center line of arch y.

= 514.3X − 1928.63 [ 10369.35 − X2 − 2307.84 + 96.08X − 89.79] (C)

dMX

dX = 0 = 514.3 − 1928.63

2 × (−2X + 96.08)

(10369.35 − X2 − 2307.84 + 96.08X)½ = 0

or 514.3 + 1928.63 (X − 48.04)

(8061.51− X2 + 96.08X)½ = 0 ,

− 514.3 (8061.51 − X2 + 96.08X)1/2 = 1928.63 (X − 48.04)

Squaring both sides of equation.

264504.5 [ (8061.51 − X2 + 96.08X)1/2]2 = [1928.63 (X − 48.04)]2

we get.


264504.5 (8061.51 − X2 + 96.08X) = (1928.63)2(X2 + 2307.84 − 96.08X) Simplifying

2132305672 − 264504.5X2 + 25413592.36X= 3719613.68X2 + 8584273228 − 357380482.1X

− 6451967556 − 3984118.18X2 + 382794074.1 on further simplification, we get

− X2 + 96.08X − 1619.422 = 0 (after dividing by 3984118.18)

or X2 − 96.08X + 1619.422 = 0 solving this quadretic equation where.

a = 1, b = −96.08, C = 1619.422

X = −b ± b2 − 4ac

2a

X = 96.08 ± (96.08)2 − 4(1) (1619.422)

2

= 96.08 ± 52.47

2

X = 96.08 + 52.47

2 Or 96.08 − 52.47

2

= 74.24 or 1.80

Therefore, (X = 74.24 is not applicable so not accepted as a root.

X = 21.80m

Put this value of X in equation (C), we have (Mbc)max.

Putting in (c)

(Mbc)max = 514.3X − 1928.63 10369.35 − X2 − 2307.84 + 96.08X + 173171.69

= 11211.74 − 189698.0911 + 173171.69

= − 5314.67 KN-m

We have solved some representative problems. Using the guidance given in this chapter a student should be able to solve any problem on three hinged parabolic or circular arches, whether supports are at the same level or not.

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