subject-analysis of structure (cet-401)

SUBJECT-ANALYSIS OF STRUCTURE

(CET-401)

BRANCH-CIVIL ENGG.

SEMESTER-4TH

ER.DEEPALI BARIK

4TH SEM CIVIL SUB – ANALYSIS OF STRUCTURE CET -401

2

CHAPTER – 1

Short questions (2 marks)

Q1. Define statically determinate and statically indeterminate structure

2015, 2(a)

Ans: Statically determinate Structure : - Conditions of equilibrium are sufficient to fully

analyse the structure, i.e. called as statically determinate structure.

Statically Indeterminate structure:- Conditions of equilibrium are insufficient analyse

the structure fully, i.e. called as statically indeterminate structure.

Q.2. What is a deficient frame ? Write down the equation 2013(s), 1(a),

2016(s)

Ans: A deficient frame is an imperfect frame. In which the number of members are less

than (2J-3)

Equation n < (2J – 3)

Where, n = number of members

J = number of joints.

Q.3. What do you mean by a portal frame 2016, 7(a)

Ans: The portal frame is an example for a statically indeterminate structure. This frame can

be analysed by moment distribution method, slope deflection method etc.

Q.4. How a truss differs from a beam? 2015, 1(a) , 2013, 1(b)

Ans: The only difference between truss and beam is that a

Truss transmits the force only in the axial direction.

Beam transmits force both in axial & vertical direction.

Q.5. Explain how method of Joint differs over the method of section in plane

truss ? 2014, 7(a)

Ans: Only two unknown forces can be found by method of joints where as three unknown

forces can be found by method of sections.

Q.6. Write down the equation for the maximum deflection of a simple

supported beam of Span ‘l’ which a central point load (w) . 2017 (s)

2(a)

Ans:- Maxm Deflection , Yc= 𝑤𝑙3

48𝐸𝐼

Where W=point load


3

L= span of beam

E= Modulus of elasticity

I = moment of inertia

Q.7.Write down the statement of second the orem of moment area

method. 2017 3(a)

Ans: Mohr’s theorem-II

It states ‘the intercept taken on a vertical reference line of tangents at any

two points on an elastic curve, is equal to the moment of the B.M. diagram

between these points about the reference line divided by EI.’

5 MARKS

Q.1. By using method of section, find out the axial forces along member1, 2 and

3 as shown in the figure. 2014, 7(c)

Ans: Considering RHS of the section X – X

ME = 0 FCB × EC + 6 × 4 = 0


4

Q.2. Find the forces in various member of the truss as shown in figure and

tabulate the results 2015 1(b)

Ans:

Let Q be the inclination of the member BD with vertical.

= 30o, DA = 3 m

Joint A, resolving vertically

PAB sin 60o = 10000 N = 10 kN

PAB = 11.55 kN (Tensile)

Resolving horizontally,

PAD + PAB cos 60o = 0.

AAD = –5.78 kN (compressive)

Joint B,

CB

B EA

EA

EA EB

EB

EB

EB

( )6 4or F ( )6 / tan 30 3.46 kN

4 / tan30

(compress)

4M 0 F cos 30 6 4 6 8

cos 30

72or F 18 kN (Tension)

4

V 0 F sin 30 F sin30 12 0

or 18 0.5 F 0.5 12 0

or 9 12 E 0.5

3or F ( )6kN(compress)

0.5


5

Resolving vertically,

PBA cos 30o + PBP cos 30o = 0.

PBA cos 30o = – PBD cos 30o.

PBA = – PBD {We know PBA = PAB)

PBD = –11.55 kN (compressive)

Resolving horizontally,

PBC = PBD sin 30o + PBA sin 3oo

PBC = –11.55 × ½ + 11.55 × ½ = 0.

Q.3. Find out forces in all the members with their nature as tensile or

compressive as shown in figure below. 2015 (s) 3(c)

Ans:

Member Tensile compressive

AB

AD

BC

BD

11.55

0

5.78

0

11.55

D

D

D

A

Support reactions,

R 6 4 4.5 2 1.5

R 6 21

21R 3.5kN

6

Member forces R (2 4) 3.5 2.5kN

Jo int D.


6

DC D

DDC

DE DC

DE

V 0

F sin60 R 0

RF 4.04kN(compression)

sin60

H 0

F F cos 60 0

F 2kN(Tension)

Jo int A

A AB

AAB

o

AE AB

AE

V 0

R F sin60 0

RF 2.89kN(compression)

sin60

H 0

F F cos 60 0

F 1.45kN(Tension)

Jo int E

EB EC

EB EC

o o

EA EB CD EC

o

EA ED EC

BC

V 0

F sin60 F sin60 0

F F

H 0

F F cos 60 F F cos 60 0

F F F cos 60 0.

F 0.55(comprssion)

Jo intB


7

7 MARKS

Q.1. The girder is loaded at ‘B’ & ‘C’ as shown in the fig. find the forces in all

members of the girder & indicate the nature of forces.2014, 1(c)

BC AB BE

BC

H 0

H F cos 60 F cos 60 0

F 1.17(Tensile)

D

D

A

o

CD D

CD

CD

o

CD ED

ED

ED

Support reactions

R 8 6 2 4 6 36

36or R 4.5 kN

8

R (6 4) 4.5 5.5 kN.

Member forces

Jo int D

V 0 F sin60 R 0

or F 0.866 4.5

4.5or F ( )5.2 kN (compression)

0.866

H F cos 60 F 0

or 5.2 0.5 F 0

or F 2.6 kN(

o

AB A

AB

AB

o

AB AE

AB

AE

Tension)

Jo int A

V 0 F sin60 R 0

or F 0.866 5.5 0

5.5or F ( )6.35 kN (comprssion)

0.866

H F cos 60 F 0

or 6.35 0.5 F 0

or F 3.18 kN (Tension)

B C

DAo60

o60 o60 o60

aR bRE8m


8

Q.2. Find the force in the member BC shown in following fig. Find out the force on

members using method of section. 2016(s) 1(c)

Ans: Section 1 -1 as shown in fig.

To find the forces in BC. Consider the left part of section 1-1 & take moments about A

Q.3. Find out the force in members using method of section 2013, (3)

Ans:

o 0

EB EC

EB EC

EB EC

o 0

AB ED FB EC

0

EC

EC

o o

BC BE AB

BC

B

V 0 F sin60 F sin60 0

or F F 0

or F F 0

H 0 F F F cos 60 F cos 60 0

or 3.18 2.6 2F cos 60 0

0.58or F 0.58 kN (Tension)

2 0.5

Jo int B

H 0

F F cos 60 F cos 60 0

F ( 0.58) 0.5 6.35 0.5 0

F

C 2.89kN (compression)

B C

A

5m

o60o30

10kN

A B

C

5m

o60o30

10kN

A B

C

o60

10kN

PCB

o

CB

o

CB

CB

CB

V 0

10 AC cos 60 P AC 0

10AC cos 60P

AC

1P 10 5kN

2

P 5 kN (compression)


9

Taking moments of the forces acting in the left part of the truss only about the

joint C and equating the same, PAB × l sin 60o = R1 × l

Now taking moments of the forces acting in the left part of the truss only about

the joint A and equating the same.

Taking moments of the force acting in the right part of the truss only about

the joint B and equating the same.

Now taking moments of the forces acting in the left part of the truss only

about the joint A and equating the same

1 1AB 1o o

R l R lP 1.16R KN(compression)

l sin60 lsin60

o

BC 1

11

BC 1oo

l lP tan60 R

4 4

lR

R4P 0.57 R KN(Tension)l tan60

tan604

o

AC 2

1 2AC 2o o

P lsin 30 R l

R l RP 2 R KN(compression)

lsin30 sin30

o

BC 2

22

BC 2oo

3l 3lP tan30 R

4 4

3lR

R4P 1.57 R KN(Tension)3l tan30

tan304


10

CHAPTER:2

2 MARKS

Q.2. State the slope and deflection equation for a member by (i) moment area method (ii)

Double integration method. 2014, (2a)

Ans:i) Moment area method

ii) Double integration method

Q.3. Write down the relationship between the slope deflection and radius of

curvature. 2016, 3(a) , 2013 , 1(e)

Ans: The relation between the slope, deflection and radius of curvature is

Q.4. Write down the equation of deflection at free end of a cantilever beam of

span ‘l’ subjected to a point load ‘w’ at free end.2016, 4(a) 2013 1(c)

Ans: Equation of deflection,

Where w = point load

L = span of beam

E = Modulus of elasticity

I = M.I. of the beam

Q.5. Write down the statement of second theorem of moment area method.

2016, 5(a), 2013 1(g)

Ans: Moment area method

x2

x

x1

x

x x

x1

MdSlope equation

EI

M dDeflectionequationz

EI

x

dySlopeequation EI. Md f '(x)

dx

DeflectionequationEI.Y f '(x)dx f(x)

2

2

d yM EI

dx

3

B

WlY

3EI w

B

Byl

A


11

Mohr’s theorem-II It states the intercept taken on a vertical reference line of

tangents at any two points on an elastic curve, is equal to the moment of the B.M.

diagram between these points about the reference line divided by EI.

Q Write down the equation for the maximum deflection of a simple

supported beam of Span ‘l’ which a central point load (w) . 2017 (s)

2(a)

Ans:- Max3 Deflection , Yc= 𝑤𝑙3

48𝐸𝐼

Where W=point load

L= span of beam

E= Modulus of elasticity

I = moment of inertia

Q. Write down the statement of second the orem of moment area

method. 2017 3(a)

Ans: Mohr’s theorem-II

It states ‘the intercept taken on a vertical reference line of tangents at any two

points on an elastic curve, is equal to the moment of the B.M. diagram

between these points about the reference line divided by EI.

Q) Calculate the maxn slope & deflection in case of a S/s beam of span

6m subjected to a point load 10kn at the middle of the span EI

constant?2017 5 (b)

ANS.


12

In this case as the load is symmetrically placed the deflection will be maxm at

the mid span & slope shall be maxm at the ends.

Qmax = 𝐴

𝐸𝐼

Area of shaded.

∆=1

2 ×

𝑤𝑙

4×

𝑙

2

=1

2×

10×6

4×

6

2

=360

16 =22.s

Qmax = 22.5

EI

Ymax = 𝐴𝑥

𝐸𝐼

2

.16 3

Wl l

EL

=

3

3

48

10 (6) 45

48

Wl

EI

x

xEI EI


13

5 MARKS

Q.1. A cantilever beam of 3 m long carried a point load of 40 kN at its free end.

Find the slope and deflection of the cantilever under the load. Take flexural

rigidity for the cantilever beam as 25 × 1012 N-mm2 2014, 1(b)

Ans: Given data

Length, L = 3m = 3000 mm

Point load, W = 40 kN = 40,000 N.

Flexural rigidity EI = 25 × 1012 N – mm2

Q.2. A simply supported beam of span 8 meters is subjected to a point load of 60

kN at the centre. Determine the maximum deflection at the centre by using

moment area method. Take EI of the beam section as 10 × 102 N-mm2.

2014, 2(b)

Ans: Given data

Span (l) 8 m = 8 × 103 mm

Point load (w) = 60 kN = 60 × 103 N

Flexural rigidity (EI) = 10 × 1012 N-mm2

Maximum deflection of the beam at its centre

2

B

2 11

12 13

3 3

B 12

WL(i) Slopeat the free end is i

2EI

40,000 3000 3.6 10

2 25 10 5 10

0.0072 rad.

(ii) DEflection at the free end is

WL 40,000 3000y 14.4mm

3EI 3 25 10

3

3 33

c 12

60 10 8 10WLy 64mm

48EI 48 10 10


14

Q. A cantilever beam 120 mm wide and 160 mm deep is 2 m long. Determine the slope and deflection at the free end of the beam. When it carries a point

load of 30 kN at its free end. Take E for the cantilever beam as 200 GPa. 2014, 6(b)

Ans: Width, b = 120 mm

Depth, d = 160 mm Span, 1 = 2 m = 2000 mm Point load , w = 30 kN = 30 × 103 N

E = 200 GPa = 200 × 103 N/m3 Slope at the free end We know that moment of inertia of the beam section

Q.4. Derive the slope and deflection of simply supported beam with a central

point load ‘W’ with span length ‘L’ by double integration method 2014,7(b)

Ans: Simply supported beam with a central load

3 36 4

2 3 2

B 3 6

3 3 3

B 3 6

bd 120 160I 40.96 10 mm

12 12

and slope at the free end

wl 30 10 2000i 0.0073rad

2EI 2 200 10 40.96 10

Deflection of the free end

wl 30 10 2000y 9.76mm

3EI 3 200 10 40.96 10

x

2

2

2

1

WM x

2

d y WEI x

dx 2

dy WxEI C

dx 4

2

1

2

1

2 2

3 2

2

2

3 2

2

A

2

1 dyAt x , 0

2 dx

(As beam isloaded symmetrically)

wlC

16

wlC

16

dy wx wlEI

dx 4 16

wx wlEly x C

12 16

At x 0, y 0

C 0

wx wlEly x

12 16

x 0

wlslope

16EI

At x l / 2

wlDefletion yc

48EI


15

Q.5. Find the reaction at the propped end of the cantilever as shown in figure.

2015 , 3(b)

Ans:

Downward deflection of the cantilever due to fore ‘P’

Since both the deflection are equal.

Shear force dia

Shear force at B.

Let M be the pt at a distance x from B, where shear force charges sign.

This the shear force is zero at a distance 31/8 from B

Bending moment dia.

Bending moment will be maximum at M., Where shear force changes sign

B

PL

A

B

PL

A

W /unit run

3

a

Ply

3EI

3 4Pl Wl

3EI 8EI

3Wl 3WP

8 8

B

A

3wlF

8

5wlF

8

x 35x 31 3x

1 x 5

31x

8

B

2 2

A

M 0

3wl wl wlM .1

8 2 8


16

Bending moment at any section x at a distance x from the propped end B.

Find out the pt. of contraflexure, let up equate this bending moment to zero

Q. 6. Derive expression for slope and deflection of a cantilever carrying point

load at its free end using moment area method 2015 7(b)

Ans:

Let l be the length of cantilever. Let the point load W applied at B.

Let the slope at B be b.

Area of bending moment diagram

Let the deflection of B with respect to A be yb.

2 2

m

3wl 31 w 31 9wlM

8 8 2 8 128

2

m

3wl waM .x

8 2

23wl wx.x 0

8 2

31x

4

b

Area of bending moment diagram between A & B

EI

2

2

b

1 w,wl

2 2

w

2EI

b

3 3

b

Axy

EI

2x

3

w 2 wy .

2EI 3 3EI


17

7 MARKS

Q.7. Calculate maximum slope and deflection in case of a cantilever of span 6m

subjected to a point load of 10 kN at free end and UDL of 10 kN/m over half

span from fixed end. Take EI constant. 2013 2(b), 2016 2(c)

Ans:

Given load at free end (W) = 10KN = 10 × 103 N. length = AB(l) = 6m = 6 × 103 mm,

Udl AC(w) = 10 kN/m = 10 N/mm lngth AC(l1) = 3m = 3 × 103 mm. E = 200 GPa =

200 × 103 N/mm2, I = 100 × 106 mm4.

Slope at the free end :

2 2 3

B

23 3

3 6

33 3 33

3 6 3 6

Wl Wl w(l l)i

2EI 6EI 6EI

10 10 6 10

2 200 10 100 10

10 6 10 3 1010 6 10

6 200 10 100 10 6 200 10 100 10

0.009 (0.018 1.75) 1.7

3 4

B

4 3

1 1

3 43 3 3

3 6 3 6

43 3

3 6

77

Deflectionat the free end :

wl wly

3EI 8EI

W l l W l l l

8EI 6EI

10 10 6 10 10 6 10

3 200 10 100 10 8 200 10 100 10

10 6 10 3 10

8 200 10 100 10

33 3 3

3 3

10 6 10 3 10 6 10

6 200 10 100 10

36 81 (75.93 94500)

117 (94575.93) 94458.93


18

Q.8. Find out the prop. Reactional of a cantilever beam of span 6m subjected to

an UDL 10 kN/m2 over whole span. The beam is propped at free end.

2013,2(c)

Q.9. Write down the assumption in slope deflection method. (2013 )

Ans: Consider a bam AV subjected to a bending moment. As a result of loading let the

beam deflect from ACB to ADB into a circular arc, as shown in figure below.

Let L = Length of the beam AB.

M = Bending moment.

R = Radius of curvature of the bent up beam

I = Moment of inertia of the beam section.

E =Modulus of elasticity of beam material.

Y = Deflection of the beam.

P = Slope of the beam.

From the geometry of a circle, we know that

Given : length 6m, load10 KN/m

We know that proportion reaction.

3wl 3 10 6 180P 22.5KN

8 8 8

B

6m

A

10kN /m

22

2

AC CB EC CD

l l(2R y) y

2 2

l2 Ry y 2Ry

4

ly

8R

M Ewe know that

I R

EIR

M


19

Now substituting this value of R in equation (i)

From the geometry of the figure, we find that the slope of the beam I at A or B is also

equal to angle AOC.

Since the angle i is very small, therefore sin i may be taken equal to

Again substituting the value of R in equation (ii)

Q.1. Derive an expression for the slope and deflection of a simply supported

beam subjected to a UDL of w/unit length. 2014 2(c)

Ans: Simply supported beam with uniformly distributed load

AC Lsin i

OA 2R

2 2l mly

EI 8EI8

M

li radians

2R

l l mli radians

EI2R 2EI2

M

2

x

2 2 2

2

wl wxM x

2 2

d y wl wx wlx wxEI x

dx 2 2 2 2

2 3

1

3 3

1

3

1

2 3 3

dy wlx wxEI C

dx 4 6

1 dyx , 0 As beam is loaded symmetrically

2 dx

wl wl0 C

16 48

wlC

24

dy wlx wx wlEI

dx 4 6 24


20

Q.2. Derive the slope and deflection of a cantilever beam with a point load at its

free end by double integration method. 2014 6(c)

Ans: Deflection curve

Cantilever with a load at the free end

Substituting in a, C1 = 0

Substituting in b, C2 = 0

x

2

2

2

1

2 2

1 2

M W(1 x)

d yEI [ W(1 x)] Wl Wx

dx

dy WxEI Wlx C ............(a)

dx 2

Wlx WxEly Wlx C x C ................(b)

2 6

at x 0

dy0 and y 0

dx

2

2 3

2 22

B

3 3

B

3

dy WxEI Wlx

dx 2

Wlx Wxand Ely

2 6

at x 1

1 Wl WlSlope Wl

EI 2 2EI

1 Wl WlDeflection, y

EI 2 6

Wl

3EI


21

Q.3. Derive the expression for maximum slope and maximum deflection in case

of a simply supported beam of span 6m subjected to an electrically placed

point load 30 kN at distance 2m and 4m respectively from supports. Take EI

constant. 2013 (4)

Ans:

Given Span l = 6m = 6 × 103 mm

Point load (wt) = 30 kN = 30 × 103 N

A = 2m = 2 × 103 mm, b = 4mm = 4 × 103 mm

EI = 26 × 1012 N-mm2

Slope at A

2 2

A

3 32

3 2 3

12 3

6 6

8

6

8

2 2

B

3 3

wbi (l b )

6EIL

(30 10 ) (4 10 )(6 10 ) 4 10

6 (26 10 )(6 10 )

12(36 10 ) (16 10 )

6 (26 10 ) 6

12(20 10 )

36 (26 10 )

12 20.0025 rad

36 (260)

Slope at B

wai (l a )

6EIL

(30 10 ) (2 10 )

6 (2

12 3

3 2 3 2

6 6

8

6

8

6 10 )(6 10 )

(6 10 ) (2 10 )

6(36 10 ) (4 10 )

6 (26 10 ) 6

6(32 10 )

36 (26 10 )


22

Q.4. Derive an expression for slope and deflection of a simply supported beam

subjected to a central point load. 2015 1(c)

Ans: Let us consider a simply supported beam subjected to a central point load.

In order to attain a single expression for bending moment which will apply across the

complete beam in this case it is convenient to take the origin for x at the centre, then :

2

2 2 2

C

3 3 3

12 3

3 2 3 2 3

5

6 6 6

5

6 2 1920.00205

36 26 10 93600

wabDeflection (l a b )

6EIL

(30 10 ) (2 10 )(4 10

6 (26 10 )(6 10 )

(6 10 ) (2 10 ) (4 10 )

3 2 4

6 (26 10 ) 6

(36 10 ) (4 10 ) (16 10 )

24(32 10

36 (26 10 )

6 6

6

5

) (20 10 )

24(16 10 )

36 (26 10 )

24 160 38404.102mm

36 26 936

2

xx 2

2

2 2

d y w L WL WxM EI x

dx 2 2 4 2

dy WL WxEI x A

dx 4 4

{Integrating both side)

WLx WxEIy Ax B

8 12

{Again by int egrating}

3 3

3 3 3

2 3 3

3

max

2

max

dyAt x 0, 0, A 0

dx

L WL WLx my0, 0

2 32 48

WL WL WLB

96 32 48

1 WLx Wx WLy

EI 8 12 48

WLSlope, y at the centre

48EI

dy WLDeflection, at the ends.

dx 16EI


23

Q.Find slope and deflection for a s/s beam with a UDL over the span by

double integration method 7(c) 2017

Consider a secn x-x at a distance x from the end A

2 2

2

2 3

2 2

2 2

int

12 2 2 3

wl xMx X x w X x X

d Y wlx wxEI

dx

egrating we get

dy wl x w xEI c

dx


24

1. ., , 0

2,

3 30 1

16 48

3 3 3 3 31

16 48 48

31

24

2 3 3

4 6 24

dyi e x

dx

wl wlC

wl wl wl wlC

wlC

dy x wx wlEI wl

dx

Slope at “A”

Putting X=0, we 𝑊𝑙3

24

𝜃A = −𝑤𝑙3

24𝐸𝐼

Integrating the slope eq1 we get

3 4 3

3

4

24 3 6 4 24

0, 0, 2 0

3 4 3

12 24 24

2

4 3max

12 8 24 16 24 2

5 4max

384

5

384

n n

n

WL x w x wlEUY x c

when x y C

wlx wx wl xEIy Defl en

lat mid span x

Wl x l wl wl x lwe getEI y

x x x

wlY

EI

wlDownward defl Yman

EI


25

CHAPTER:3

2 MARKS

Q.1. State any two advantages to fixed beam. 2015 , 3(a)

Ans: i. The fixed beam is subjected to a lesser maximum bending moment than the

simply supported beam carrying the same, load.

ii. For the same loading maximum deflection a fixed beam is less than that if the

simply supported beam.

Q.2. Write down the expression for fixed and moment for a fixed beam of span ‘l’

subjected to a point load ‘w’ at distance ‘a’ and ‘b’ from both ends

2013 (s) (New) 1(f)

Ans: The expressions are

2

2

2

2

wabMa

l

wa bMb

l


26

5 MARKS

Q.1. A fixed beam AB of span 6 m is subjected to two point load of 20 kN and 30

kN at a distance of 2m and 4 m from A. Calculate fixing moment at A and B.

2015 2(b)

Ans:

Q.2. A fixed beam AB of span 6m is subjected to a UDL of 2 kN/m. Determine the

S.F. & B.M.D., its diagram 2014, 3(b)

Ans:

Consider any section ‘X’ at a distance x from the left end ‘A’. The shear force

at the section is :

2

2

2 2

2 2

2

2

2 2

2 2

WabFixing moment at A

20 2 4 30 2 231.12kN m.

6 6

Wa bFixing moment at B

20 2 4 30 4 235.56kN m.

6 6

A B

WL 2 6R R 6kN

2 2

x A

A A

B A

f R w.x 6 2.x.

wxAt A, x 0, hence F R

2

2 66 6kN( )

2

At B,x 6, hence F R w.x

6 2 6 6kN.


27

Simply supported bending moment at the section ‘X’ at a distance ‘x’ from left end ‘A’ is given by

7 MARKS

c A

2

2

C

2

2 2

L 6At C, x 3

2 2

Hence f R w.x

6 2 3 0.

26 6 6 0

2

w L w LAt, C, x ,hence M . .

2 2 2 2 2

6 2 66 18 9 9 kN m( )

2 2 2

Fixed end BM at A and B

Wl 2 6( ) 6kNm

12 12

x A

22

A

2

B

xM R w.x.

2

2x6x 6x x

2

The value of B.M. at different po int s are :

w w.0At A, x 0, hence M 0 0

2 2

At B, x L 6

w wHence M .L L

2 2


28

Q.1. A fixed beam is subjected to an UDL over whole span. Derive the expression

for fixed end moments. 2013, 2(d) 2016, 4(c)

Ans:

Let mA = Fixing moment at A and

MB = Fixing moment at B

Since the beam is symmetrical therefore mA and mB will also be equal.

Now equating the areas of the two diagrams.

We know that maximum positive bending moment at the centre of the beam = wl2/8.

Net positive bending moment at the centre of the beam

Shear force diagram, Let RA = Reaction at A, and RB = Reaction at B.

Equating the clockwise moments and anticlockwise moments about A.

Deflection of the beam,

We know that bending moment at any section X, at a distance x from A,

2 3

A

2

A

2

B

2 wl wlm l l.

3 8 12

wlm

12

wlSimilarly, m

12

2 2 2wl wl wl

8 12 24

B 0 B

2

B

A

lR l m m w l

2

wlR

8

wlSimilarly, R

2


29

Integrating the above equation

Where C1 is the first constant of integration. We know that when x = 0, the dy/dx = 0

Therefore C1 = 0

Integrating the equation (ii) once again

Where C2 is the second constant of integration, We know that when x = 0, then y = 0

Therefore C2 = 0

We know that the maximum deflection occurs at the centre of the beam. Therefore

substituting x = l/2 in the above equation.

Point of contraflexures.

The points of contrflexures may be found out by equation (i) to zero.

2 2'

x x x

2 2 2

2

wl wx wlM x

2 2 12

d y wlx wx wlEI ...................(i)

dx 2 2 12

2 3 2

1

dy wlx wx wl xEI C

dx 4 6 12

2 3 2dy wlx wx wl xEI .................(ii)

dx 4 6 12

2 3

2

dy wlx wxEI _ C

dx 4 6

3 4 2 2wlx wx wl xEI.y ..............(iii)

12 24 24

3 4 22

c

4 4 4 4

4

c

wl l w l wl lEI.y

12 2 24 2 24 2

wl wl wl wl

96 384 96 384

wly

384EI


30

Solving this quadratic equation for x,

CHAPTER:4

2 MARKS

Q.1. Explain theorem of three moment, 2014, 5(a), 2015, 5(a)

Ans: Theorem of three moments: If a beam has a support, the end ones being fixed, then

the same number of equations required to determine the support moments may be

obtained from the consecutive pairs of spans.

Theorem of three moments states that

Where MA, MB, MC = Support moments at A, B and C.

2 2

22

22

wlx wx wl0

2 2 12

llx x 0

6

lx lx 0

6

22 4l

l l6x

2

l l

2 2 3

0.5l 0.289l 0.789 l and 0.211 l.

1 1 2 2A B c

1 1 1 2

1 1 2 2

1 1 2 2

l l l lM 2M M

I I I I

A x A x6

I l I l


31

A1, A2 = Area of B.M. diagram for the given loading (s/s)

X1,x2 = Distance of centroid of areas A1 and A2 from A and C respectively.

l1,l2 = Span AB and BC respectively

I1, I2 = MI. of span AB and BC respectively.

Q.2. Write down the expression for three moment equation with usual meaning

2013, 2(i)

Ans: Three moment equation:-

6 MARKS

Q.1. A continuous beam is simply supported over two spans, such that AB = 6m

and BC = 4m. It carries uniformly distributed load of 2 kN/m over span BC

and a point load of 5 kN at the centre of span AB. Determine the support

moment over B by applying theorem of three moments. 2014,2016

Ans:

1 1 2 2A 1 B 1 2 0 2

1 2

6a x 6a xM l 2M (l l ) M l

l l

AB

C

5kN 2kN

1 1 2 2A 1 B 1 2 C 2

2 2

D


l l

wl 5 6M 7.5 kNm

4 4


32

Bending moment at the mid of the span BC

Now using three moment equation.

Q.2. A continuous beam ABC is simply supported over support ‘A’, ‘B’ and ‘C’. the

span AB is 5m and BC is 6m. It is subjected to a point load 30 kN at mid

span of AB and an UDL 10 kN/m over whole span of BC. Find out the

moment at support ‘B’. The beam is of uniform cross-section.2013 2(c)

Ans:

Let mA = Fixing moment at A.

MB = Fixing moment at B.

MC = Fixing moment at C.

2 2

1 1

2 2

wl 2 44kNm

8 8

1 2 1 3a x 3 7.5 3 3 7.5 3

2 3 2 3

22.5 45 67.5

2a x 4 4 2 21.33

3

1 1 2 2A 1 B 1 2 C 2

1 2

B

B

B


l l

6 67.5 6 21.330 2M (6 4) 0

6 4

20M (67.5 31.99) 99.495

M 4.97 Nm

D

2 2

2

1 1

wl 30 5m 37.5 KN.m

4 4

bending moment at the mid of the span BC

wl 10 6 10 36 36045.00 KN.m

8 8 8 8

We find that

1 2 2.5 1 2.5a x 2.5 37.5 2.5 37.5 2.5

2 3 2 3

78.12 (46.87 3.33)

78.12 156.07 23

2 2

4.19

2a x 45 6 3 540

3


33

Now using three moments equation

Shear force diagram

Let RA = Reaction at A

RB = Reaction at B

RC = Reaction at C

Taking moments about B.

7 MARKS

Q.1. A continuous beam ABC 10m long rests on three support A, B and C at the

same level and is loaded as shown in figure.

Determine the moments over the beam and draw the bending moment

diagram. Also calculate the reactions at the support and draw shear force

diagram using theorem of three moment.

Ans: Applying the theorem of three moments of the span AB and AC

1 1 2 2A 1 b 1 2 c 2

1 2

b

B

B

6a x 6a xm l 2m l l m l

l l

6 234.19 6 5400 2m (5 6) 0

5 6

22m (281.02 540) 821.02

821.02m 37.31

22

B A A

A

A

C

C

C

B

R R 5 30 2.5 37 /31 R 5 75

R 5 37.31 75

37.31 75 37.69R 7.53kN

5 5

SimilarlyR 6 (60 3) 37.31

R 6 37.31 (60 3)

37.31 (60 3)R 23.78

6

R (30 10 6) (7.53 23.78) 90 31.31 58.69


34

Length of AB = l1 = 6 m

Length of BC = l2 = 4m

MA = Fixing moment of A

MB = Fixing moment of B

MC = Fixing moment of C.

Let us consider the beam AB as simply supported so bending moment at D.

Bending moment at the mid of the span BC.


RB = Reaction at B

RC = Reaction at C

Taking moment about B.

MB = RA × 6 – 3 × 4

4.002 = RA × 6 – 12

RA = {4.002 + 12} × 1/6 = 2.667 kN

Again MB = RC × 4 – 4 × 2

4.002 × RC × 4 – 8

RC = (4.002 + 8) × ¼ = 3.0005 kN.

D

1

wab 3 2 4M 4 kN m

l 6

2 22

1 1 2 2A 1 B 1 2 C 2

1 2

Wl 3 46 kN m.

8 8

Using three moment equation.


l l

2

1

1

2

B 1 2

A C

B

1a 6 4 12m

2

6 2 8x 2.67m.

3 3

x 2m.

6 12 2.67 6 16 22M (l l )

6 4

80.04 M M 0 kN m

80.04M 4.002 kN m.

2 (6 4)


35

fy=0

RA + RB + RC = 3 + 4

2.667 + RB + 3.0005 = 7

RB = 7 – (2.667 + 3.0005)

RB = 1.3325 kN.

Q.2. A continuous beam ABCD is simply supported over three spans, such that AB

= 6m. BC = 8 m and CD = 5m. It carries UDL of 4 kN/m in span AB, 3 kN/m

in span BC and 2 kN/m in span CD. Find the support moments B and C and

draw the SF and BM diagrams. 2014 4(c)

Ans:

Applying the theorem of three moments for the spans AB and BC.

Since A is the simply supported end of the girder.

Ma = 0

28MB + 8MC = 216 + 384 = 600

14MB + 4MC = 300 ……………………(1)

Consider the span BC and CD.

Applying the theorem of three moments for these spans.

Since D is the simply supported end.

Md = 0

8Mb + 26MC = 384 + 52.5 = 446.5

4Mb + 13MC = 223.25 ……………………..(2)

14Mb + 4Mc = 300

4Mb + 13Mc = 223.25

a B C

3 3

M 6 2M (6 8) M 8m

4 6 3 8

4 4

3 3

B C d

3 8 2 5M 8 2M (8 5) M 5

4 4


36

10Mb – 9Mc = 76.75

10 Mb –76.75 + 9Mc

4Mb + 150 .79 = 223.25

4Mb = 72.45

Mb = 18.11 kN –m

Maximum free bending moment for span AB.

Maximum free bending moment for span CD

B.M. at c = 15.01 × 14 + Vb × 8 – 4 × 6 × 11 – 3 × 8 × 4 = 11.599

Vb = 11.599 – 210.14 + 264 + 96

= 11.599 – 210.14 + 360

Vb = 161.459 kN

Vd = 2.68kN

Vc = total load – (Va + Vb + Vd)

= (4 × 6 + 3 × 8 + 2 × 5) –(15.01+161.459+2.68)

= (24 +24 + 10) – 179.149 = 58 – 179.149 = –121.149

CB

CC

76.75 9MM

10

76.75 9M4 13M 223.25

10

24 618kN m

8

2

2

a

a

a

2 56.25kN m

8

4 6B.M atB V 6 18.11

2

18.11 72V

6

V 15.01kN

2

d

d

2 5B.M. at c V 5 11.599

2

V 5 11.599 25


37

Q.3. A continuous beam ABC with fixed end at ‘A’ simply supported over support

‘B’ and ‘C’. the span AB = 6m and BC = 5m. Span AB is subjected to a point load

20 kN at 2m from support ‘A’ and UDL 5 kn/m over whole span BC. Find out the

reactions and moment at supports. Draw the shear force and bending moment

diagram for the same. 2016 5(c) 2013 (s)

Ans:

Support moments at A, B and C.

Let mA = Support moment at A

MB = Support moment at B

MC = Support moment at C

Bending moment under the 20 KN load at AB

Consider the beam BC as a simply supported beam. Therefore bending moment at the

mid of span BC.

Geometry of the above bending moment diagram. We find that for the span OA and

AB.

= [142.19+(26.66) (4.67)

=(142.19 + 124.50)=266.69

Similarly for the spans AB and BC.

1

wab 20 2 426.66 KN.m

l 6

2

2wl 5 515.63 KN.m

8 8

0 0

1 1

a x 0

1 2 4 1 2a x 26.66 4 26.66 2 4

2 3 2 3

1 1

2 2

A B

A B

A B

1 2 4 1 2a x 26.66 4 26.66 2 4 266.69

2 3 2 3

2a x 15.63 5 2.5 130.25

3

12m 6m 266.69

6(2m m ) 266.69

266.692m m 44.45..........(i)

6


38

Now using three moments equation for the span AB and BC.

6mA + 22mB = –[266.69+156.30]

6mA + 22mB = 422.99 ……………….. (ii)

Solving equation (i) and (ii)

Equation (i) × 3 = 6mA + 3mB = 133.35

Equation (ii) × 1 = 6mA + 22mB = – 422.99

Put the value of mB in equation (i)

2 mA + (–29.28) = – 44.45

2 mA – 29.28 = – 44.453

2 mA = – 44.45 + 29.28

= – 15.17

mA = – 7.58 Kn-m

mB = – 29.28 KN-m

mC = 0

Shear force diagram


RB = Reaction at B

RC = Reaction at C

Taking moment about B

–29.28 = RC × 5 – (25 × 2.5) – 29.28 = 5RC – 62.5

5 RC = –29.28 + 62.5 = 33.22

1 1 2 2A 1 B 1 2 C 2

1 2

A B

6a x 6a xm l 2m (l l ) m l

l l

6 266.69 6 130.25m 6 2m (6 5) 0

6 5

B

B

19m 556.34

556.34M 29.28

19

A

15.17m 7.58

2

C

33.22R 6.65 KN

5


39

Now taking moment about A.

– 7.58 = RB × 6 – (25 × 8.5) – (20 × 2) + 6.65 × 11

– 7.58 = 6RB – 212.5 – 40 + 73.15

6RB = –7.58 + 212.5 + 40 – 73.15 = – 80.73 + 252.50 = 171.77

CHAPTER:5

Q.1. Define carry over factor , Stiffness factor 2013 1(j) 2015 4(a)

Ans: Carry over factor: The ratio of moment produced at a joint to the moment applied

at the other joint, without displacing , it is called Carry over Factor.

Stiffness Factor: Moment required to rotate end by unit angle, when rotation is

permitted at that end, is called stiffness factor.

Q.2. state the Stiffness Factor for a beam fixed at one end & freely supported at

the other. 2014 4(a)

Ans: The stiffness factor at fixed end,

K1=4EI/L

The Stiffness Factor at J/S end

K2=3EI/L

Q.3. Find the other State the stiffness factor for a beam fixed at our end

& freely supported at. 2017 4(a)

Ans : K=4𝐸𝐼

𝑙

(one end fixed s Free support at the other)

Q.4. State the stiffness factor for a beam fixed at one end & freely

supported at the other 2017 4 (a)

Ans:- K = 4𝐸𝐼

𝐿 (𝑂𝑛𝑒 𝑒𝑛𝑑 𝐸𝑖𝑥𝑒𝑑 & 𝐹𝑟𝑒𝑒𝑙𝑦 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑜𝑡ℎ𝑒𝑟)

B

A

171.77R 28.62KN

6

R 20 2 28.62

40 28.62 11.38 KN


40

7 MARKS

Q.1. A continuous beam ABC with fixed end at ‘A’ S/S over support ‘B’ and ‘C’.

the span AB = 6m & BC = 5m, Span AB is subjected to a point load 20 kN at 2m

from support ‘A’ & UDL 5 kN/m over whole Span BC. Find outthe reactions

moment of supports by suing moment distribution method.

2016 5(c) 2014 (s)

Ans:

Let us assume the continuous beam ABC to be split up into filxed beams AB, BC.

In span AB, fixing moment at A

In span BC, fixing moment at B

Now let us find out he distribution factor at B. From the geometry of the figure

we find that the stiffness factor for BA.

2 2

2 2

2 2

2

wab 20 2 (4)17.77kNm

l (6)

Fixing moment at B,

wab 20 2 48.88 kNm

l 6

2 2

2

wl 5 510.41kNm

12 12

wlFixing moment of c 10.41kNm

12

BA

BC

4EI 4E I 2k EI

l 6 3

4EI 4E I 4k EI

l 5 5

Distribution factor for BA and BC

2 4EI EI

5EI 6EI3 5and and2 4 2 4 11 11

EI EI EI EI3 5 3 5

–17.77 8.88

5/11

–10.41 10.41

6/11

Fixed end moment

–10.41

-5.41

Release c.

Carry over

-17.77 8.88

3.06

1.53

0 0

-15.62 0

3.67

0 0

Initial moments

Distribute

Carryover istribute

Final moment –11.95 0 –16.24 11.94


41

8 MARKS

Q.1. Analyse the portal frame shown in figure using moment distribution method

and draw the bending moment diagram 2015, 7(c)

Ans:

FAB FBA CD FDC

2

FBC

FCB

Fixed End Moment

M M M M 0

W 2 6 6M 6 kNm

12 12

M 6 kNm.


42

Free moment diagram of BC parabola with maximum ordinate

Draw in compression side of member. Difference of these two diagram is BM

diagram. Positive and negative are marked upon whether cousing tension/compression on

doted side.

Q.2. A rectangular portal frame of uniform flexural rigidity EI, carries a UDL of 20

kN/m as shown in fig. Draw the bending moment diagram and sketch the

deflected curve. If L = 4 m and EIAB = EIBC = EICD.

2014(s)

Ans: Given data

Length of AB =4m

Length of Bc = 2 L = 2 × 4 = 8 m2

Length of CD = 4m

Load on BC (A) = 20 kN/m

EIAB = EIBC = EICD

Support Reactions:

Let MAB = Moment at Aa

MBA = Moment at B in span BA.

MBC = Moment at B in span BC.

MCB = Moment at C in span CB.

2 6118kNm

8


43

MCB = Moment at C in span CD.

MDC = Moment at D.

(i) Fixed end moments: Let us assume the frame to be made up of fixed beams AB, BC

and CD from the geometry of the figure, we find that fixed end moment at A.

(ii) Slope Deflection Equation : Since frame is fixed at A and D. therefore the slopes iA

and iD will be equal to zero.

Moment at ‘A’ in the span AB,

' '

AB BA

2 2'

BC

2 2'

CB

' '

CB DC

M M 0

wL 20 8M 106.67kN m

12 12

wL 20 8M 106.67kN m

12 12

M M 0

'ABAB A B AB

B

B

'BABA B A BA

'

B A BA

B

B

2EIM 2i i M

2EI(0 i ) 0

4

EI i

2

2EIM 2i i M

2EI2i i M

4

2EI(2i 0) 0

4

EI i

Moment at 'B ' in span BC,

'BCBC B C BC

B C

B C

'CBCB C B CB

C B

C B

2E IM (2i i ) M

2I I(2i i ) 106.67

63

EI(2i i )106.67

3

2E IM (2i i ) M

2EI(2i i ) 106.67

82

EI(2i i )106.67

2


44

Moment at ‘C’ in span CD.

'CDCD C D CD

C

C

'DCDC D C DC

C

C

2E IM (2i i ) M

2EI(2i 0) 0

4

EI i

2EIM (2i i ) M

2EI(0 i ) 0

4

EI i

2


45


46


47

Q.3. Analyze a symmetrical rectangular portal frame using deflection method of

horizontal span 5m subjected to an UDL 10 kN/m over whole span and

height of 3.5 m. 2013, (7)

Ans:

The ends A and D of the frame ABCD are fixed and therefore

A = D = 0

As the portal frame is symmetrical and loaded symmetrically rotation 3 = c

and there will be no way i.e. = 0.

The fixed moments are :

The slope deflection equations in terms of unknown are :

For equilibrium the sum of the moments at joint B is zero.

BC

CB

10 5 5M 20.83kNm

12

M 20.83kNm

BAB B

BBA B

BC 3 B

B

2EI2EIM ( 0)

3.5 3.5

4EI2EIM ( 2 )

3.5 3.5

2EIM (2 ) 20.83

5

2EI 20.83

5


48

iii) Equilibrium equations: Since the joints B and C are in equilibrium, so first equation

MBA + MBC equal to zero.

(EI × iB) + [EI (2 iB + iC) – 106.67] = 0

(EI × iB) + (2EI × iB) + (EI × iC) – 106.67 = 0

3 (EI × iB) + (EI × iC) = 106.67.

Now equation MCB + MCD = 0

[EI(2iC + iB) + 106.67] + (EI × iC) = 0

2(EI × iC) + (EI × iB) + 106.67 + (EI × iC) = 0

3(EI × iC) + (EI × iB) + 106.67 = 0

By symmetry, iB = iC, substituting these values, we get 2EI × iB = 106.67

AB BC

BB

B

B

B

M M 0

2EI 2EI 20.83 0

3.5 5

1 12EI 20.83 0

3.5 5

2EI 0.48 20.83

21.69

EI

BAB

BBA

BC B

CB B

2EI 2EI 21.69M 12.39kNm

3.5 3.5 EI

4EI 2EI 21.69M 24.78kNm

3.5 3.5 EI

2M EI 20.83

5

2 21.69EI 20.83 12.15kNm

5 EI

2 2 21.69M EI 20.83 EI 20.83

5 5 EI

12.15kNm

B

C

C

106.67EI i 53.335

2

2EI i 106.67

106.67EI i 53.335

2


49

Final moments :

The bending moment at the mid of the span BC, buy considering it as a simply

supported beam.

CHAPTER:6

2 MARKS

Q.1. In case of a column whose both ends are hinged , what will be its equivalent

length. 2015, 4(a)

Ans: In Case of a column whose both ends are hinged. Then its equivalent length will be

same as its actual length le = l

BAB

BA B

B CBC

B C

C BCB

C B

EI iM

2

53.33526.667 kN m.

2

M EI i 53.335 kN m.

EI(i i )M 106.67

3

(2EIi ) (EIi )106.67

3

106.67 ( 53.335)106.67

3

88.89 kN m.

EI(2i i )M 106.67

2

2EIi EIi106.67

3

106.67 53.335106.67

2

80.0

0 kN m.CD C

CDC

M EI i

53.335 kN m.

EI iM

2

53.33526.667 kN m.

2

2 2w 20 8160 kN m.

8 8


50

5 MARKS

Q.1. A steel rod 5m long and 40 mm dia. is used as a column with one end fixed

& other free. Determine the cripping load by Euler’s formula. Take l as 200

GPa. 2016, 7(b)

Ans: Given data l = 5 m = 5 × 103 mm

D = 40 mm

L = 200 GPa = 200 × 103 N/mm2

One end fixed other free le = 2l

Bu Euler’s formula

Q.2. State different end conditions of column and write down the relation

between equivalent length and actual length in each case. 2015, 6(b)

Ans: In actual practice there are a number of end conditions, for columns. But we shall

study the Euler’s column theory on the following four types of end conditions, which

are important from the subject point of view.

2

Eulers 2

4 44

2 3

Eulers 3 2

EIP

le

d (40)I 125663.7mm

64 64

200 10 125663.7P

(2 5 10 )

24805N

24.85 N

Types of End Conditions Relation between equivalent

Length (Le) & actual length

(ℓ)

1. Both ends hinged

2. One end fixed and the

other free

3. Both ends fixed

4. One end fixed and the

other hinged

Le = ℓ

Le = 2ℓ

Le = ℓ/2

Le = ℓ/√2


51

Q3.- State different end conditions of calumn and write down the

relation between equivalent length , actual length in each case.

20176(b)

Ans

End conditions

1) Both ends hinged

2) One end fixed and the other free

3) Both ends fixed

4) One end fixed and the other hinged

Relation between equivalent length (Le) and actual length (L)

Le=l

Le=2l

Le=𝑙

2

L2=𝑙

√2

CHAPTER:7

2 MARKS

Q.1. What do you mean by three hinged arch ? 2013, 1(d), 2016, 6(a)

Ans: Vector diagram L Diagram showing the magnitude of forces along with direction is

called vector dig. Polar diagram: Diagram showing magnitude of forces is called polar

diagram.

5 MARKS


52

Q.1. A three hinged symmetrical arch of span 7m is subjected on UDL of w over

whole span. Draw the S.F. & B.M.D. for the arch 2013, 2(f)

Ans:

Let the rise of the arch = hm , Due to symmetry

VA = VB = ½ × total load.

CHAPTER-7

5 MARKS

Q.1. A three-hinged parabolic arch of span 40m, and rise 10 m is carrying a UDL

as shown in the fig. Find the horizontal thrust at the springing. 2014,5(b)

A

2

2

2

A

2

2 2

2

2

1 w 7w.L 3.5w

2 2

Taking moment about c, we get

1 L LO V . Hh w .

2 2 4

WL L wLHh

2 2 8

wL 49wH

8 8

At any sec tion dis tance x from A

wLM V x Hy

2

4hx(L x)But in parabolic arxh y

L

wL wL 4hx(L x) wLM x .

2 8 L 2

wLx w wLx(L x)

2 2

02


53

Ans: Given data,

Span (l) = 40m and central rise (yC) = 10m.

H = Horizontal thrust at the springing.

VA = Vertical reaction at A.

VB = Vertical reaction at B.

Vertical reaction VB at B can be calculated by taking moments about A and equating

anti-clockwise moments with clockwise moments.

The beam moment at C due to external loading

c = VB × 20 = 150 × 20 = 3000 kN-m.

Horizontal thrust,

Q.2.

Find out the reaction at A and B and draw the bending moment diagram for

the parabolic arch as shown in figure. 2015, 6(c)

B

B

V 40 (30 20) 10 6000

6000V 150kN.

40

c

c

3000H 300kN

y 10


54

Ans:

Bending moment diagram is a triangle with maximum ordinate at load point.

At mid-span the net bending moment is zero. Ordinate of the beam moment diagram

is zero. Ordinate of bending moment diagram at mid span is :

Parabola drawn with central ordinate equal to 60 kNm.

B

A

A

A

A B

B

M 0

V 20 200(20 6) 0

V 20 2800 0

2800V 140kN

20

V V 200kN

V 200 140 60kN

200 6(20 6) 200 6 1484kNm

20 20

c

84 1060kN

14

M 0(In Arch),Hh 60kN

60 60H 12kN.

h 5

BM Bending Moment 2

4hx(L x)Y Y Hy H


55

subject-analysis of structure (cet-401)

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