module1 c - rajesh sir

Structural Analysis - II

Indeterminate structures;Indeterminate structures;Force method of analysis

Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur

Dept. of CE, GCE Kannur Dr.RajeshKN

1

Module IModule IStatically and kinematically indeterminate structures

• Degree of static indeterminacy, Degree of kinematic indeterminacy, Force and displacement method of analysis

Force method of analysis

• Method of consistent deformation-Analysis of fixed and i b


continuous beams

• Clapeyron’s theorem of three moments-Analysis of fixed and continuous beams

• Principle of minimum strain energy-Castigliano’s second theorem-


2

p gy gAnalysis of beams, plane trusses and plane frames.

Types of Framed Structures• a. Beams: may support bending moment, shear force and axial force

yp

• b. Plane trusses: hinge joints; In addition to axial forces, a member CAN have bending moments and shear forces if it has


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loads directly acting on them, in addition to joint loads

• c. Space trusses: hinge joints; any couple acting on a member should have moment vector perpendicular to the axis of the member, since a truss member is incapable of supporting a twisting momentg

• d. Plane frames: Joints are rigid; all forces in the plane of the frame, all couples normal to the plane of the frame


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• e Grids: all forces normal to the plane of the grid all couples • e. Grids: all forces normal to the plane of the grid, all couples in the plane of the grid (includes bending and torsion)

• f. Space frames: most general framed structure; may support bending moment, shear force, axial force and torsion


5

g , ,

Deformations in Framed Structures

, ,x y zT M MThree couples: , ,x y zN V VThree forces:

•Significant deformations in framed structures:•Significant deformations in framed structures:

Structure Significant deformationsStructure Significant deformationsBeams flexuralPlane trusses axialSpace trusses axialPlane frames flexural and axialGrids flexural and torsionalSpace frames axial, flexural and torsional


6

Types of deformations in framed structures b) axial c) shearing d) flexural e) torsional


7

b) axial c) shearing d) flexural e) torsional

Equilibrium•Resultant of all actions (a force, a couple or both) must vanish for static equilibrium

q

0F∑ 0F∑ ∑

•Resultant force vector must be zero; resultant moment vector must be zero

0xF =∑ 0yF =∑ 0zF =∑0xM =∑ 0yM =∑ 0zM =∑∑ y∑ ∑

•For 2 dimensional problems (forces are in one plane and

0F =∑ 0F =∑ 0M =∑

•For 2-dimensional problems (forces are in one plane and couples have vectors normal to the plane),

0xF =∑ 0yF =∑ 0zM =∑


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Compatibility

•Compatibility conditions: Conditions of continuity of p y ydisplacements throughout the structure

•Eg: at a rigid connection between two members, the g g ,displacements (translations and rotations) of both members must be the same


9

Indeterminate Structures

Force method and Displacement method

• Force method (Flexibility method)

• Actions are the primary unknowns

St ti i d t i f k ti th • Static indeterminacy: excess of unknown actions than the available number of equations of static equilibrium

• Displacement method (Stiffness method)

• Displacements of the joints are the primary unknowns

• Kinematic indeterminacy: number of independent


Kinematic indeterminacy: number of independent translations and rotations

Static indeterminacy

• Beam:

• Static indeterminacy = Reaction components number of eqns • Static indeterminacy = Reaction components - number of eqns available 3E R= −

• Examples: • Examples:

• Single span beam with both ends hinged with inclined l d loads • Continuous beam• Propped cantilever• Fixed beam


Degree of Static Structure Degree of Static Indeterminacy

Structure

0

1

3


12

Degree of Static I d t iStructure

0

Indeterminacy

0

1

55

2


• Rigid frame (Plane): g ( )

• External indeterminacy = Reaction components - number of eqns available 3E R= −available

• Internal indeterminacy = 3 × closed frames

3E R

3I a=• Internal indeterminacy = 3 × closed frames 3I a=

• Total indeterminacy = External indeterminacy + Internal indeterminacy

( )3 3T E I R a= + = − +

• Note: An internal hinge will provide an additional eqn


14

Example 1 Example 2p p

( )3 3T E I R a= + = − +( )3 3T E I R a= + = − + ( )( )3 3 3 3 2 12= × − + × =( )2 2 3 3 0 1= × − + × =

Example 3 Example 4

( )( )

3 3T E I R a= + = − + ( )( )

3 3T E I R a= + = − +


( )3 2 3 3 3 12= × − + × = ( )4 3 3 3 4 21= × − + × =

Degree of Static I d t iStructure Indeterminacy

3

63


16

Degree of Static I d t iStructure Indeterminacy

2

1


• Rigid frame (Space): g ( p )

• External indeterminacy = Reaction components - number of eqns available 6E R= −available

• Internal indeterminacy = 6 × closed frames

6E R

Example 1

( )( )

6 6T E I R a= + = − +

( )4 6 6 6 1 24= × − + × =

If i l d f ti l t d t ti i d t i i t If axial deformations are neglected, static indeterminacy is not affected since the same number of actions still exist in the structure


Plane truss (general):

External indeterminacy = Reaction components - number of eqns available 3E R= −

Minimum 3 members and 3 joints. A dditi l j i t i 2 dditi l b

( )3 2 3 2 3m j j= + − = −Hence, number of members for stability,

Any additional joint requires 2 additional members.

( )2 3I m j= − −Hence, internal indeterminacy,

Total (Internal and external) indeterminacyTotal (Internal and external) indeterminacy

( )3 2 32

T E I R m jm R j

= + = − + − −

= + − 2m R j+

• m: number of members• R : number of reaction components


• j : number of joints• Note: Internal hinge will provide additional eqn

E l 1Example 1

2 9 3 2 6 0T m R j= + − = + − × =

3 3 3 0E R= − = − =

0I T E= − =

Example 22 15 4 2 8 3T R

p2 15 4 2 8 3T m R j= + − = + − × =

3 4 3 1E R= − = − =

2I T E= − =

Example 3 2 6 4 2 5 0T m R j= + − = + − × =

( ) 3 1 4 4 Hinge at 0 AE R= − + = − =


( )0I T E= − =

Example 4 2 7 3 2 5 0T m R j= + − = + − × =Example 4 2 7 3 2 5 0T m R j+ +

3 3 3 0E R= − = − =

0I T E 0I T E= − =

2 6 4 2 4 2T m R j= + − = + − × =

3 4 3 1E R= = =

Example 5

3 4 3 1E R= − = − =

1I T E= − =

2 11 3 2 6 2T m R j= + − = + − × =Example 6

3 3 3 0E R= − = − =

2I T E= − =


2I T E= =

• Wall or roof attached pin jointed plane truss (Exception to the • Wall or roof attached pin jointed plane truss (Exception to the above general case):

• Internal indeterminacy 2I m j= −

• External indeterminacy = 0 (Since, once the member forces are determined, reactions are determinable)

Example 1 Example 3Example 2Example 1 Example 3p

26 2 3 0

T I m j= = −= − × =

25 2 1 3

T I m j= = −= − × =2

7 2 3 1T I m j= = −= × =


22

7 2 3 1= − × =

• Space Truss:

External indeterminacy = Reaction components - number of eqns available

Minimum 6 members and 4 joints

6E R= −

( )6 3 4 3 6m j j= + − = −Hence, number of members for stability,

Minimum 6 members and 4 joints. Any additional joint requires 3 additional members.

( )6 3 4 3 6m j j+Hence, number of members for stability,

( )3 6I m j= − −Hence, internal indeterminacy,

Total (Internal and external) indeterminacy

( )6 3 6T E I R m j= + = − + − −

3m R j= + −


Example

3T m R j= +Total (Internal and external) indeterminacy 3T m R j= + −Total (Internal and external) indeterminacy

12 9 3 6 3T∴ = + − × =

6 9 6 3E R= − = − =


Kinematic indeterminacyKinematic indeterminacy

• joints: where members meet, supports, free ends• joints undergo translations or rotations

•in some cases joint displacements will be known, from the restraint conditions•the unknown joint displacements are the kinematically •the unknown joint displacements are the kinematically indeterminate quantities

odegree of kinematic indeterminacy: number of degrees of f dfreedom

Two types of DOF

• Nodal type DOFJ i DOF

Two types of DOF


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• Joint type DOF

• degree of kinematic indeterminacy (degrees of freedom) is d fi d defined as:

• the number of independent translations and rotations in a structure.

DOF = 1DOF = 2


DOF 2

• in a truss the joint rotation is not regarded as a degree of • in a truss, the joint rotation is not regarded as a degree of freedom. joint rotations do not have any physical significance as they have no effects in the members of the truss

• in a frame, degrees of freedom due to axial deformations can be neglected


Kinematic Degree f F dStructure of Freedom

2

11

22



4

If the effect of the cantilever portion is considered as a joint load at If the effect of the cantilever portion is considered as a joint load at the roller support on the far right, kinematic indeterminacy can be taken as 2.



2

3



6

2

5


Method of Consistent DeformationIllustration of the method

Problem

Released structure


32

Released structure

D fl ti f l d

45wLΔ

Deflection of released structure due to actual loads

384B EIΔ =

C Deflection of released structure due to redundant applied as a ppload


33

3BR LDeflection due o

48t B

BR L

IR

E=

3BR L

Δ48

BB EI

Δ =

4 35 BwL R L

Compatibility condition (or equation of superposition or equation of geometry) 5

384 48BwL R L

EI EI=

5B

wLR∴ =8B


34

•A general approach (applying consistent sign convention for loads and displacements):

RA l it l d di t BR•Apply unit load corresponding to

3

48BLEI

δ =Let the displacement due to unit load be

BR B BR δDisplacement due to is


35

445384B

wLEI

Δ = − (Negative, since deflection is downward)

0B B BR δΔ + = (Compatibility condition)

BB

B

RδΔ

= − 58BwLR∴ =

Bδ 8


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Example 1: Propped cantilever

AM

PP

A

2l 2l2l 2lA B

AV BVChoose VB as the redundant

P

AM

2l 2l

AV


37Released structure

Find deflection at B of the released structure

31 2 5PL l l l Pl− −⎛ ⎞Δ = + =⎜ ⎟Pl− . . .2 2 2 2 3 2 48B EI EI

Δ = + =⎜ ⎟⎝ ⎠2

PlEI

M diagramEIEI

Apply unit load on released structure corresponding to VApply unit load on released structure corresponding to VBand find deflection at B

3

3BlEI

δ =

1l

3EI


38

0R δΔ35 3 5B Pl EI PR −Δ

= = =0B B BR δΔ + = 3.48 16B

B

REI lδ

= = =

+ve sign indicates that VB is in the same direction of the unit load.i.e., in the upward direction.

P

2l 2l

5 32 16 16APl Pl PlM = − =Other reactions

2l 2l

516P11

16P

16

532Pl

Bending moment diagram

3Pl


3916

−

Example 2: Continuous beam10kN

A2m 2m

A B 4m C

10kN

A B C

V VVAV CVBV

Choose VB as the redundant


40

10kN2 6

ABΔ C

2m 6mB

Released structure2.5 kN7.5 kN

15EI

10EI

1 251 358 6+⎛ ⎞

A B CEI EI

4 m

Conjugate beam of the released structure (to find ∆ )

( )1 25

0.5 15 8 35EI EI=× × −

1 358 6 80.5 15 83EI EI+⎛ ⎞ ÷ =× × ×⎜ ⎟

⎝ ⎠

Conjugate beam of the released structure (to find ∆B)

1 73.333425 4 0 5 10 4⎛ ⎞∴Δ ⎜ ⎟ (numerically)

Dept. of CE, GCE Kannur Dr.RajeshKN41

25 4 0.5 10 43B EI EI

⎛ ⎞∴Δ = =× − × × ×⎜ ⎟⎝ ⎠

(numerically)

1kN1kN

4mA B 4m C4m 4m

Released structure with unit load corresponding to V (to find δ )Released structure with unit load corresponding to VB (to find δB)

3 38 10.667B

lδ = = = (numerically) (di i i ∆ i d d )48 48B EI EI EI

δ

0R δΔ + 73.333 6 875B EIR kN−Δ −

(direction is same as ∆B i.e., downwards)

0B B BR δΔ + = . 6.87510.667

BB

B

R kNmEIδ

= = = −

i i di t th t V i i th it di ti f th it l d-ve sign indicates that VB is in the opposite direction of the unit load.i.e., in the upward direction.


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10kNOther reactions

10kN

A B CB C

4.063AV kN= 0.938CV kN=6.875 kNA

Bending moment diagramg g

AB

C

8.122 kNm

A C3.752 kNm−


Example 3A:

2m3mCD

30kNBEI is constant

2m

DMD

A

2m6mC

DHD

30kNB

VD

2mB

HA Choose H as the redundant


44A

Choose HA as the redundant

D CD

To find ∆HA and δHA (using unit load method)

10kN2m3m

CD 2m6m

B

C

2mA

B 2mA

B1kNE=2x105N/mm2

I= 2x108mm4

Portion Origin Limits M m Mm m2

AB A 0-2 0 x 0 x2AB A 0 2 0 x 0 xBC A 2-4 -10(x-2) x -10x2+20x x2

CD A 0 3 20 4 80 16

( ) ( )4 3

21 110 20 80Mmdx x x dx dxΔ = = + +∫ ∫ ∫

CD A 0-3 -20 4 -80 16

( ) ( )2 0

10 20 80HA x x dx dxEI EI EI

Δ = = − + + −∫ ∫ ∫

[ ]43

321 1x⎡ ⎤ 306 67−


[ ]320

2

1 110 10 803HAx x x

EI EI⎡ ⎤

Δ = − + + −⎢ ⎥⎣ ⎦

306.67EI

−=

4 32 2 4 33 69 3316⎡ ⎤ ⎡ ⎤4 32 2

0 0

16HA

m dx x dx dxEI EI EI

δ = = +∫ ∫ ∫33

00

69.33163

xxEIEIEI

⎡ ⎤ ⎡ ⎤= + =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

0HA A HAH δΔ + =306.67 4.4.

69 33

32B

AEIH

EIkN

δ−Δ

= = =HA A HA 69.33B EIδ


46

Example 3B: 2ICD

2m6mCD

30kN

2m

IB

M

2m

A

26m

2IC

DHD

MD A

30kN

2m6m

IVD

Choose H as the redundant2m

B

H


Choose HA as the redundant

A

HA

2ID

2ICD

To find ∆HA and δHA (using unit load method)

30kN2m6m

I

CD 2m6m

I B

C

2mA

B 2mA

B1kNE=2x105N/mm2

I= 2x108mm4

Portion Origin Limits M m Mm m2

AB A 0-2 0 x 0 x2

BC A 2-4 -30(x-2) x -30x2+60x x2

CD A 0-6 -60 4 -240 16

( ) ( )4 6

21 130 60 240Mmdx x x dx dxΔ = = + +∫ ∫ ∫

CD A 0 6 60 4 240 16

( ) ( )2 0

30 60 2402HA x x dx dx

EI EI EIΔ = = − + + −∫ ∫ ∫

[ ]4 63 21 110 30 240⎡ ⎤Δ 920−


[ ]63 202

1 110 30 2402HA x x x

EI EI⎡ ⎤Δ = − + + −⎣ ⎦

920EI−

=

4 62 2 16d4 63 69 3338x⎡ ⎤ ⎡ ⎤2 2

0 0

162HA

m dx x dx dxEI EI EI

δ = = +∫ ∫ ∫3

00

69.33383

xxEIEIEI

⎡ ⎤ ⎡ ⎤= + =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

0H δΔ + =920. 13.269

69 333HA

AEIH kN

EIδ−Δ

= = =0HA A HAH δΔ + = 69.333AHA EIδ


Example 4: Two or more redundants

60kN

A B C

80kN 50kND

3m 3mA B C

3m 3m 3m 3m

60kN 80kN 50kN

3m 3mA B C D

3m 3m 3m 3m

Choose VB and Vc as the redundant


50

60kN 80kN 50kN

A B C D

BΔ CΔB C

1kN

A B Dδ

C

BBδ CBδ

1kN

A B D

δ

C

BCδCCδ


0B B BB C BCV Vδ δΔ + + = 0C B CB C CCV Vδ δΔ + + =

60kN

A B C

80kN 50kNDA B

3m 3m 3m 3m 3m 3m

A B C D150kNm

15

120kNm 60kNm

3m 15m525kN825kN

240kNm 240kNm

A B C D360kNm

9 9

240kNm 240kNm

9 m 9m1620kN1620kN

A B C D125kNm100kNm

50kNm


15m 3m687.5kN437.5kN

3420 8280 2325 14025BEIΔ = + + = 2790 8280 2850 13920CEIΔ = + + =

D2kNm4kNmA

B CD

12 m6m16kN20kN

2kNm

16kN20kN

96BBEIδ = 84CBEIδ =

A D2kNm 4kNmA

B CD

12m6m

20kN16kN

2kNm


84BCEIδ =96CCEIδ =

0V Vδ δΔ + + = 14025 96 84 0B CV V⇒ + + =0B B BB C BCV Vδ δΔ + + =

0C B CB C CCV Vδ δΔ + + =

14025 96 84 0B CV V⇒ + +

13920 84 96 0B CV V⇒ + + =C B CB C CC B C

82BV kN= − 73.25CV kN= −

( )82BV kN= ↑ ( )73.25CV kN= ↑

( )19.25AV kN= ↑ ( )15.5DV kN= ↑

( )B ( )C


Clapeyron’s theorem of three moments1. Uniform loading

w kN m w kN m1w kN m

1 1l IA B C

2w kN m

2 2l I

M

1 1,l I 2 2,l I

1

AMEI

B C

1

BMEI

2

BMEI

2

CMEI

A B CM/EI diagram for joint moments

+

B C

21 1

18w lEI

22 2

28w lEI


A B C

M/EI diagram for simple beam moments

To find slopes at B using Conjugate Beam Method:

21 1 2 2M l l M l l w l l⎡ ⎤⎡ ⎤

From span AB:

1 1 1 1 1 1 111 1

1 1 1

1 1 2 2. . . . . . .2 3 2 3 3 8 2

A BB

M l l M l l w l llV lEI EI EI

⎡ ⎤⎡ ⎤+= + ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

31 1 1 1

11 1 16 3 24

A BB BA

M l M l w lVEI EI EI

θ⇒ = + + =

From span BC:

22 2 2 2 2 2 2

22 22 2 2

1 1 2 2. . . . . . .2 3 2 3 3 8 2

B CB

M l l M l l w l llV lEI EI EI

⎡ ⎤⎡ ⎤+= + ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

32 2 2 2

2 3 6 24B C

B BCM l M l w lV

EI EI EIθ⇒ = + + =


2 2 23 6 24EI EI EI

BAθ

A B C

BA

BCθ

Deflected shape

0BA BC BA BCθ θ θ θ+ = ⇒ = −

332 2 2 21 1 1 1

3 6 246 3 24B CA B M l M l w lM l M l w l

EI EI EIEI EI EI⎛ ⎞

+ +⇒ + + = − ⎜ ⎟⎝ ⎠2 2 21 1 1 3 6 246 3 24 EI EI EIEI EI EI ⎝ ⎠

3 31 2l lM l M l w l w l⎛ ⎞1 21 2 1 1 2 2

1 21 2 1 2

24 4

A CB

l lM l M l w l w lMEI EIEI EI EI EI

⎛ ⎞+⇒ + + = − −⎜ ⎟⎝ ⎠


2. General loadingg1a

2a

AB

C

x x1x 2x

1 1 1 1

1 1 1 16 3A B

BAM l M l a x

EI EI EI lθ = + +

2 2 2 2

2 2 2 23 6B C

BCM l M l a x

EI EI EI lθ = + +

2 2 2 2

1 21 2 1 1 2 26 62A Cl lM l M l a x a xM ⎛ ⎞+⇒ + + = − −⎜ ⎟BA BCθ θ= −


1 21 2 1 1 2 2

2 BMEI EIEI EI EI l EI l

+⇒ + + = − −⎜ ⎟⎝ ⎠

BA BCθ θ

2. General loading with support settlement1a

2a ;A B

B C

δ δδ δ

><

AB

C

x x1x 2x

M l M l δ δ1 1 1 1

1 1 1 1 16 3A B A B

BAM l M l a x

EI EI EI l lδ δθ −

= + + +2 2 2 2

2 2 2 2 23 6B C C B

BCM l M l a x

EI EI EI l lδ δθ −

= + + +

BA BCθ θ= −

( ) ( )1 21 2 1 1 2 2

1 21 2 1 1 2 2 1 2

6 66 62 A B C BA CB

l lM l M l a x a xMEI EIEI EI EI l EI l l l

δ δ δ δ− −⎛ ⎞+⇒ + + = − − − −⎜ ⎟⎝ ⎠


1 21 2 1 1 2 2 1 2l l l l⎝ ⎠

Example 1:

20kN m 20kN m

4 mA B C

4 m

EI is constant

3 3l lM l M l l l⎛ ⎞

1 2l l=

3 31 21 2 1 1 2 2

1 21 2 1 2

24 4

A CB


⎛ ⎞++ + = − −⎜ ⎟⎝ ⎠

1 2w w=2244BwlM = −

0A CM M= = 0A CM M

2

8BwlM∴ = −

220 4 408

kNm×= − = −


8 8

Example 2:W

lA B Cl l D

EI is constant

EI EI

Span ABC0M =

EI is constant

3 3l lM l M l l l⎛ ⎞

1 2l l= 1 2EI EI= 0AM =

3 31 21 2 1 1 2 2

1 21 2 1 2

24 4

A CB


⎛ ⎞++ + = − −⎜ ⎟⎝ ⎠

4 0A B CM M M⇒ + + =

4 0M M⇒ + = ( )1


4 0B CM M⇒ + = ( )1

Wl1 Wla l=

lA B Cl l D

Span BCD4WlEI

2

2

. .2 4

2

a l

lx

=

=0M

( ) 1 1 2 26 64 a x a xl M M M+ +

2 20DM =

( )2 14 6 lWll M M l⎛ ⎞⎜ ⎟

( ) 1 1 2 24B C Dl M M Ml l

+ + = − −

( )2 4 6 . .22 4B Cl M M l⎛ ⎞+ = − ⎜ ⎟

⎝ ⎠

34 WlM M ( )248B CM M+ = − ( )2

Wl− WlM


10CWlM = 40BM =( ) ( )&1 2

40Wl

5Wl

A B C40

10Wl−

D5

10

Bending moment diagram


63

Example 3: 20kN m 20kN mp

6mA B C8mEI is constant

20kN m 20kN m

6mA B C8mImaginary

span A’Aspan A A

Span A’AB1

2

020

ww kN m

== ' 0AM = 1 2EI EI=

Span A AB

3 31 2' 1 2 1 1 2 22A B

A

l lM l M l w l w lMEI EI

⎛ ⎞++ + = − −⎜ ⎟⎝ ⎠


1 21 2 1 24 4A EI EIEI EI EI EI⎜ ⎟⎝ ⎠

320 6×20 62 6 64A BM M ×

× + × = −

2 180M M+ ( )12 180A BM M+ = −

Span ABC

( )1

1 2 20w w kN m= =3 31 21 2 1 1 2 2

1 21 2 1 2

24 4

A CB


⎛ ⎞++ + = − −⎜ ⎟⎝ ⎠

0CM =1 21 2 1 2⎝ ⎠

3 320 6 20 86 2 144 4A BM M × ×

+ × = − −4 4

3 14 1820A BM M+ = − ( )2

( ) ( )&1 2 28AM kNm= − 124BM kNm= −


Example 4 (Support settlement):

226320EI kN 87δ δEI is constant

20kN m 20kN m20kN m

226320EI kNm= 87B C mmδ δ= =EI is constant

6mA DC 3mB3m

EI EI EI= =0MSpan ABC

1 2EI EI EI= =0AM =

( ) ( )3 31 21 2 1 1 2 2 66 C BA BA Cl lM l M l w l w l δ δδ δ −−⎛ ⎞ ( ) ( )1 21 2 1 1 2 2

1 21 2 1 2 1 2

662

4 4C BA BA C

B

l lM l M l w l w lMEI EIEI EI EI EI l l

δ δδ δ⎛ ⎞++ + = − − − −⎜ ⎟⎝ ⎠

( ) ( )3 31 20 3 20 6 6 6 00.08718 626320 4 26320 4 26320 3 6B CM M × × ×−+ = − − − −

× ×


3 560.78B CM M+ = ( )1

Span BCD0DM =

( ) ( )3 31 21 2 1 1 2 2 6 6

24 4

B C D CB DC

l lM l M l w l w lMEI EIEI EI EI EI l l

δ δ δ δ− −⎛ ⎞++ + = − − − −⎜ ⎟⎝ ⎠

D

1 21 2 1 2 1 24 4EI EIEI EI EI EI l l⎜ ⎟⎝ ⎠

( ) ( )3 31 20 6 20 3 6 0 6 0.0876 18M M × × × −+( )6 1826320 4 26320 4 26320 6 3B CM M+ = − − − −

× ×

3 560.78B CM M+ = ( )2

140.195BM kNm= 140.195CM kNm=

Alternatively, from symmetry, B CM M=


673 560.78B CM M+ = 4 560.78 140.195B BM M kNm⇒ = ⇒ =

Examples 5 (Fixed Beam) xa p es 5 ( xed ea )

P

A BC

lA

P

A BC

A′ B′lAA′ B

Imaginary span A’A

Imaginary span BB’span A A span BB

1 21 2 1 1 2 26 62A Cl lM l M l a x a xM ⎛ ⎞++ +⎜ ⎟


68

1 21 2 1 1 2 2

1 21 2 1 1 2 2

2A CBM

EI EIEI EI EI l EI l++ + = − −⎜ ⎟

⎝ ⎠

Span A’AB1 l

' 0AM =4PlEI

21 . .2 4

Pla l

l

=

1 2EI EI=4EI

2 2lx =A B

6Pl6216A BPlM M⇒ + = −

S ABB’

( )1

' 0BM = 1 2EI EI=

Span ABB’2

1 1,8 2

Pl la x= =8 2

6216A BPlM M⇒ + = − ( )2

16

( ) ( )&1 2 A BPlM M= = −


69

( ) ( )&1 2 8A BM M

Examples 6 (Fixed Beam)

w

xa p es 6 ( xed ea )

lA B

CC


70

ENERGY PRINCIPLES BASED ON DISPLACEMENT FIELDENERGY PRINCIPLES BASED ON DISPLACEMENT FIELD

Principle of Minimum Total Potential

Castigliano’sTheorem (Part I)Total Potential

Energy (PMTPE)Theorem (Part I)

alternative forms ofalternative forms of

Principle of Stationary TotalPrinciple of Stationary TotalPotential Energy (PSTPE)


71

Principle of Stationary Total Potential Energy (PSTPE)p y gy ( )

When the displacement field in a loaded elastic structure is given all d bit t b ti i t i i tibilit dsmall and arbitrary perturbation, maintaining compatibility and

without disturbing the associated force field, then the first variationof the total potential energy is equal to zero, if the forces are in astate of static equilibrium.


72

Alternative form of Principle of Stationary Total Potential EnergyAlternative form of Principle of Stationary Total Potential Energy(PSTPE)

The total potential energy π in a loaded elastic structure expressed as afunction of n independent displacements D1, D2, … Dn in acompatible displacement field must be rendered stationary with thecompatible displacement field must be rendered stationary, with thepartial derivative of π with respect to every Dj being equal to zero, ifthe associated force field is to be in a state of static equilibrium.


Principle of Minimum Total Potential Energy (PMTPE)Principle of Minimum Total Potential Energy (PMTPE)

When the displacement field in a loaded linear elastic structure is givena small and arbitrary perturbation, maintaining compatibility andwithout disturbing the associated force field, then the first variationof the total potential energy is equal to zero, if the forces are in ap gy q ,state of static equilibrium.

Castigliano’s Theorem (Part I)

If the strain energy, U, in an elastic structure, subject to a system ofexternal forces in static equilibrium, can be expressed as a functionq , pof n independent displacements D1, D2, … Dn satisfyingcompatibility, then the partial derivative of U with respect to everyDj will be equal to the value of the conjugate force, Fj.


74

Dj will be equal to the value of the conjugate force, Fj.

ENERGY PRINCIPLES BASED ON FORCE FIELDENERGY PRINCIPLES BASED ON FORCE FIELD

Principle of Minimum Total Complementary Potential

Castigliano’sTheorem (Part II)

Complementary Potential Energy (PMTCPE)

Theorem of LeastWork

alternative forms of

Principle of StationaryPrinciple of StationaryComplementary Total PotentialEnergy (PSCTPE)


Principle of Stationary Total Complementary Potential EnergyPrinciple of Stationary Total Complementary Potential Energy(PSTCPE)

When the force field in a loaded elastic structure is given a small andarbitrary perturbation, maintaining equilibrium compatibility andwithout disturbing the associated displacement field, then the firstg p ,variation of the total complementary potential energy is equal tozero, if the displacements satisfy compatibility.


Alternative form of Principle of Stationary Total ComplementaryPotential Energy (PSTCPE)Potential Energy (PSTCPE)

The total complementary potential energy π* in a loaded elasticp y p gystructure expressed as a function of n independent forces F1, F2, …Fn in a statically admissible force field must be rendered stationary,with the partial derivative of π* with respect to every Fj being equalw t t e pa t a de vat ve o w t espect to eve y j be g equato zero, if the associated displacement field is to satisfycompatibility.


Principle of Minimum Total Complementary Potential EnergyPrinciple of Minimum Total Complementary Potential Energy(PMTCPE)

When the force field in a loaded linear elastic structure is given a smalland arbitrary perturbation, maintaining equilibrium compatibilityand without disturbing the associated displacement field, then theg p ,first variation of the total complementary potential energy is equalto zero, if the displacement satisfy compatibility.


78

Castigliano’s Theorem (Part II)Castigliano s Theorem (Part II)

If the complementary strain energy, U*, in an elastic structure, with akinematically admissible displacement field, is expressed as afunction of n independent external forces F1, F2, … Fn satisfyingequilibrium, then the partial derivative of U* with respect to every Fjq , p p y jwill be equal to the value of the conjugate displacement, Dj.

If the behaviour is linear elastic, U* can be replaced by U.

Thus, Castigliano’s Theorem (Part II) can otherwise be stated as:

“If U is the total strain energy in a linear elastic structure due toapplication of external forces F1 F2 F3 F at points A1 A2 A3 Aapplication of external forces F1, F2, F3, … Fn at points A1, A2, A3,…, Anrespectively in the directions AX1, AX2, AX3,…, AXn then thedisplacements at points A1, A2, A3,…, An respectively in the directionsAX AX AX AX ∂U/∂F ∂U/∂F ∂U/∂F ∂U/∂F


AX1, AX2, AX3,…, AXn are ∂U/∂F1, ∂U/∂F2, ∂U/∂F3,…, ∂U/∂Fnrespectively.”

Proof for Castigliano’s theorem (Part II)Proof for Castigliano s theorem (Part II)

Let x1, x2, x3,… xn be deflections at points A1, A2, A3,… An

due to F F F Fdue to F1, F2, F3,… Fn

Total strain energy, 1 1 2 2 3 31 1 1 12 2 2 2 n nU F x F x F x F x= + + + +… (1)

Let the load F1 alone be increased by 1Fδ

Let 1 2 3, , , nx x x xδ δ δ δ… be the additional deflections at points A1, A2, A3,… An

I i t iIncrease in strain energy,

1 1 1 2 2 3 312 n nU F F x F x F x F xδ δ δ δ δ δ⎛ ⎞= + + + + +⎜ ⎟

⎝ ⎠…

2⎝ ⎠

1 1 2 2 3 3 n nU F x F x F x F xδ δ δ δ δ= + + + +… , neglecting small quantities.


80

(2)

( )1 1 2 3, , , nF F F F Fδ+ …Let are acting on the original structure

( ) ( ) ( ) ( ) ( )1 1 1 1U U F F F F Fδ δ δ δ δ δ

Total strain energy,

( ) ( ) ( ) ( ) ( )1 1 1 1 2 2 2 3 3 31 1 1 12 2 2 2 n n nU+ U F F x x F x x F x x F x xδ δ δ δ δ δ= + + + + + + + + +…

(3)1 1 1 1U F F F F+ + + +(1) 1 1 2 2 3 32 2 2 2 n nU F x F x F x F x= + + + +…(1)

1 1 1 1U F F F Fδ δ δ δ δ⎛ ⎞⎜ ⎟(3) (1) 1 1 1 1 2 22 2 2 2 n nU x F F x F x F xδ δ δ δ δ⎛ ⎞= + + + +⎜ ⎟⎝ ⎠

…(3)-(1)

( )2 U x F F x F x F xδ δ δ δ δ= + + + + (4)

1 1 2 2 3 3 n nU F x F x F x F xδ δ δ δ δ= + + + +…(2)

( )1 1 1 1 2 22 n nU x F F x F x F xδ δ δ δ δ= + + + +… (4)


811 1U x Fδ δ=(4)-(2)

Uδ U∂1

1

U xF

δδ

=1 1

1

0, UWhen F xF

δ ∂→ =

∂

2 22 3

, , nn

U U USimilarly, x x xF F F∂ ∂ ∂

= = =∂ ∂ ∂

…

For example, in the case of bending,

2

2MU dxEI

⎛ ⎞= ⎜ ⎟

⎝ ⎠∫⎝ ⎠

U M M dxF EI F

δ ∂ ∂⎛ ⎞∴ = = ⎜ ⎟∂ ∂⎝ ⎠∫F EI F∂ ∂⎝ ⎠


82

Example 1: Using Castigliano’s Theorem, analyse the continuous

w kN/m

p g g , ybeam shown in figure.

AB C

/

LLL

w kN/m

AB C

/

LLRB

L

2BRwL −

2BRwL −

Let RB be the redundant.


83

B

2wxR⎛ ⎞22

Bx

wxRM xwL⎛ ⎞= −−⎜ ⎟⎝ ⎠

From A to B,

0Bδ = 0B B

U M M dxR EI R∂ ∂⎛ ⎞⇒ = =⎜ ⎟∂ ∂⎝ ⎠∫

2M xR∂ −

=∂ F th ti AC2BR∂

22M M∂ ⎛ ⎞⎛ ⎞

For the entire span AC

220 022 2

B

B

M M xR x wxdx dxwLxEI R EI

∂ −⎛ ⎞⎛ ⎞∴ = ⇒ =− −⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠∫ ∫

5B

wLR⇒ =


4BR⇒

Example 2: Using Castigliano’s Theorem analyse the frame shown Example 2: Using Castigliano s Theorem, analyse the frame shown in figure.

3 kN/3 kN/m

2mB C

3 kN/m

2mB C

2m1m

D2m

1m

D

A

D

A VD

HD

VD

V

HA

VA

Let HA be the redundant.


Let HA be the redundant.

3 kN/m

2m

1

B CFrom A to B,

x AM H x= −

2m1m

D

AH A

M xH∂

= −∂

A3 AHV = −

AH

From B to C,23⎛ ⎞HA

32DV

3 AHV +

232322

Ax A

xHM x H⎛ ⎞= − −+⎜ ⎟⎝ ⎠

32

AAV = +

22A

M xH∂

= −∂

F D t C M H x= M x∂


From D to C, x AM H x= −A

xH

= −∂

0M M dx∂⎛ ⎞∴ ⎜ ⎟∫ 0A

dxEI H

∴ =⎜ ⎟ ∂⎝ ⎠∫

2 2 1⎧ ⎫⎛ ⎞⎛ ⎞( )( ) ( ) ( )2 2 12

0 0 0

1 3 023 222 2

AA AA

xH x xdx dx dxH x H xx Hx xEI⎧ ⎫⎛ ⎞⎛ ⎞⇒ + + =− −−+ − −− −⎨ ⎬⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎩ ⎭∫ ∫ ∫

2 2 13 3 3 2 4 2 32 33 03 4A A A A A

AH x x H x H x x H x H xx H x x⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ + =+ − − − − + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦0 0 03 2 12 2 16 2 3A⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

8 8 8 4A AA AH HH H⎡ ⎤8 8 8 4 03 12 2 8 83 32 12 2

A AA AA A

H HH H H H⎡ ⎤+ + =+ − − − − + +⎢ ⎥⎣ ⎦

3 3.192

AA

HV kN= + =0.39AH kN=

H


3 2.82

AD

HV kN= − =

Example 4: Using Castigliano’s Theorem, analyse the truss shown p g g , yin figure. AE is constant.

CD CD10kN

CD

T10kNCD

3m 3m

T

4mA B

4mA B

T

m

. . 0P LPT AE∂

=∂∑ 6.25T kN=


Example 4: Do the above problem using the method of consistent Example 4: Do the above problem using the method of consistent deformation.

P kL′

10kN 2

P kLAETk L

′

= −∑

∑

CD

3m

k LAE∑

A B4m B

P P kT′= +


Note 1: If there are two internal redundants in a truss in method of Note 1: If there are two internal redundants in a truss, in method of consistent deformation,

2P k L k L k k L′ 21 1 1 2

1 2 0P k L k L k k LT TAE AE AE′

+ + =∑ ∑ ∑

22 1 2 2

1 2 0P k L k k L k LT TAE AE AE′

+ + =∑ ∑ ∑AE AE AE

Note 2: If there are both internal redundants external redundants in a truss, in method of consistent deformation,

21 1 1 0BP k L k L k k LT V

′+ + =∑ ∑ ∑1 0BT V

AE AE AE+ + =∑ ∑ ∑

2P k L k k L k L′


21

1 0B B BB

P k L k k L k LT VAE AE AE′

+ + =∑ ∑ ∑

Example 3: Using Castigliano’s Theorem, analyse the pin-jointed h i fi

3m 4m

truss shown in figure.

A B C3m 4m

400A =

A

3m 500A = 400A =

D

80kN

Internally indeterminate to degree 1.

Take force in BD as redundant.

y g

0P LP ∂∑Dept. of CE, GCE Kannur Dr.RajeshKN

91Assume force in BD is T . . 0P

T AE=

∂∑

cos45 sin 53.13AD CDF F=cos45 sin 53.13AD CDF F

cos45 cos53.13 80AD CDF F T+ + =T

57.14 0.714T−

64.64 0.808T−

sin 53.13 cos53.13 80CD CDF F T+ + =80kN

57.14 0.714CDF T= −

i 53 13( ) sin 53.1357.14 0.714 64.64 0.808cos45ADF T T= − = −

. . 0P LPT AE∂

=∂∑

( ) ( ) ( ) ( )4.243 57.14 0.714 5 64.64 0.80830.714 3 0.808 0500 400 400

T TTE E E− −⎛ ⎞ ⎛ ⎞⎛ ⎞× − + × + × − =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠


9249.09T kN∴ =

SummarySummaryStatically and kinematically indeterminate structures

• Degree of static indeterminacy, Degree of kinematic indeterminacy, Force and displacement method of analysis


• Method of consistent deformation-Analysis of fixed and continuous beams

• Clapeyron’s theorem of three moments-Analysis of fixed and continuous beams

• Principle of minimum strain energy-Castigliano’s second theorem-Analysis of beams, plane trusses and plane frames.


93

module1 c - rajesh sir

Engineering

gce kannur dr

static indeterminacy

t e i r

total indeterminacy

g external indeterminacy

axial forces

ti th static indeterminacy

g p external indeterminacy