Module2 rajesh sir

Structural Analysis - II

Slope deflection methodSlope deflection methodMoment distribution method

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

Dept. of CE, GCE Kannur Dr.RajeshKNDept. of CE, GCE Kannur Dr.RajeshKN

1

Module IIModule II

Displacement method of analysis

• Slope deflection method-Analysis of continuous beams and


Slope deflection method Analysis of continuous beams and frames (with and without sway)

• Moment distribution method- Analysis of continuous beams and frames (with and without sway)and frames (with and without sway).

Dept. of CE, GCE Kannur Dr.RajeshKN

2

Displacement method

Example 1: Propped cantilever (Kinematically indeterminate to first degree)

p

degree)

• Required to get Bθ

•degrees of freedom: one

• Kinematically determinate structure is obtained by restraining all Kinematically determinate structure is obtained by restraining all displacements (all displacement components made zero - restrained structure)


3

Restraint at B causes a reaction of MB as shown.

2

12BwLM =12

The actual rotation at B is Bθ

To induce a rotation of at B, it is required to apply a moment of MB anticlockwise.

BθBθ

4B B

EIM θ=

pp y

B BML

θ

2 4 J i ilib i i

3wL

2 412 BwL EI

Lθ=

Joint equilibrium equation (or equation of action superposition)


448BwL

EIθ∴ =

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

• Restrained structure: Restraint at B causes a reaction of MB

2L (N t th i ti

Restrained structure: Restraint at B causes a reaction of MB.

2

12BwLM = (Note the sign convention:

clockwise positive)

Bθ•Apply unit rotation corresponding to

4EIBmLet the moment required for this unit rotation be

4B

EImL

= −

anticlockwise


Bθ B Bm θ• Moment required to induce a rotation of is

0B B BM m θ+ = (Joint equilibrium equation)

2 4 0wL EI θ3

BB

wLMθ∴ = − =

B B B (Joint equilibrium equation)

012 BL

θ− = 48BB EIm

θ∴

m (Moment required for unit rotation) is the stiffness coefficient here Bm (Moment required for unit rotation) is the stiffness coefficient here.


66

Sign convention for moments Sign convention for moments (for Slope Deflection and Moment Distribution methods)

• A support moment acting in the anticlockwise direction will be taken as positive (reactive moment is clockwise)

• A support moment acting in the clockwise direction will be taken as negative (reactive moment is anticlockwise)

+ve

A B

ve+ve−

A B

support moments

reactive moment

reactive moment


support moments

anticlockwise support moment (reactive moment is clockwise)

anticlockwise support moment (reactive moment is clockwise)moment is clockwise) moment is clockwise)

ve+ ve+

A B

Moment distribution distribution/slope deflection sign deflection sign convention

veve+

A B

ve−ve+Usual sign convention for drawing BMD

A B


clockwise support moment (reactive moment is

anticlockwise support moment (reactive moment is clockwise)moment is

anticlockwise)moment is clockwise)

ve+ve−

A B

Moment distribution/slope deflection sign deflection sign convention

ve− ve−

A B

ve veUsual sign convention for drawing BMD

A B


clockwise support moment (reactive moment is

clockwise support moment (reactive moment is moment is

anticlockwise)moment is anticlockwise)

ve− ve−

A B

Moment distribution distribution/slope deflection sign deflection sign convention

ve− ve+

A B

ve ve+Usual sign convention for drawing BMD

A B


Sign convention for slopes and deflections

A l k i t ti i t k iti d ti l k i

g p(for Slope Deflection and Moment Distribution methods)

• A clockwise rotation is taken as positive and anticlockwise rotation as negative

ve−Aθ

ve+Bθ

•If one end of a beam settles, the rotation at both ends are taken as positive if the beam as a whole rotates clockwise, and negative if the beam as a whole rotates anticlockwise

Bθ

beam as a whole rotates anticlockwise

Aθve+ δ ve+B

ve−δ ve−


Bθve+ Aθ ve−

Sl d fl ti th dSlope deflection method


Introduction

• This method is based on the relationships of end moments with slopes and deflections (called slope-deflection equations) for each p ( p q )member.

Approach to solve problems

• The slope-deflection equations are written for each member.

pp p

• Joint equilibrium conditions are written.

• Solving the joint equilibrium conditions, unknown displacements are found out.

• Substituting these unknown displacements back in the slope-deflection equations, we get the unknown end moments.


13

Derivation of fundamental equations

BAMM =BAABM

Aθ Bθ

AFEM( )1

+ABFEM BAFEM( )1

( )2

ABM ′

Aθ′ Bθ′

+( )2

BAM ′A

M ′′ M ′′

+( )3


ABM BAM

( ) Ends assumed as fixed (zero rotation) This requires restraining ( )1 Ends assumed as fixed (zero rotation). This requires restraining moments (fixed end moments) FEMAB and FEMBA. External loads are acting.

( )2 Rotations are forced at ends. This requires moments M’AB and M’BA

( )3 If there is a support settlement, moments M’’AB and M’’BA will be induced.


( )2

M′ BAM′Aθ′ Bθ′

ABM BA

=

BAM′2Bθ′2Aθ′M′

1Aθ′ 1Bθ′ +BA

ABM′

BAM′+ABM′

+BAM lθ′−′

EI+AB

EI

1 3AB

AM l

EIθ

′′ =

2 6BA

A EIθ′ =

l′ M l′

Conjugate beams


1 6AB

BM lEI

θ′−′ = 2 3

BAB

M lEI

θ′ =

AB BAM l M lθ θ θ′ ′

′ ′ ′ AB BAM l M lθ θ θ′ ′

′ ′ ′1 2 3 6AB BA

A A A EI EIθ θ θ′ ′ ′= + = − 1 2 6 3

AB BAB B B EI EIθ θ θ′ ′ ′= + = − +

( )3 Rotation at the end A due tosupport settlement

δABM ′′BAM ′′

support settlement= Rotation at the end B due tosupport settlement

BA

=lδl

Total rotations at the ends are:

3 6AB BA

A AM l M l

l EI EI lδ δθ θ

′ ′′= + = − +

3 6BA AB

B BM l M l

l EI EI lδ δθ θ

′ ′′= + = − +


17

3 6l EI EI l 3 6l EI EI l

2 32EIM δθ θ⎛ ⎞′ ⎜ ⎟

Solving the above two equations, we get:

2 32EIM δθ θ⎛ ⎞′ = +⎜ ⎟2AB A BMl l

θ θ⎛ ⎞′ = + −⎜ ⎟⎝ ⎠

2BA B AMl l

θ θ= + −⎜ ⎟⎝ ⎠

Hence the final moments at the supports are:

2 32EIM FEMδθ θ⎛ ⎞= + +⎜ ⎟

Hence the final moments at the supports are:

2AB A B ABM FEMl l

θ θ= + − +⎜ ⎟⎝ ⎠

2 3EI δ⎛ ⎞2 32BA B A BAEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

These are the slope deflection equations


18

Fixed end moments

8PL−

8PL

+2L 2L2L 2L


19

Illustration of the method

B3

5kN 8kN2 5

Example 1

AB

C3m 2.5m

5m 5m

Problem structure

A B CB

Problem structure

A B

2 4kN 3 6kN

C

5kNm 5kNm

B

2.4kNm 3.6kNm 5kNm 5kNm


Fixed end moments (reactive)

Fixed end moments2 2

2 2

5 3 2 2.45AB

PabFEM kNml

− − × ×= = = −

2 2

2 2

5 3 2 3.65BA

Pa bFEM kNml

× ×= = =

8 5 58 8BCPlFEM kNm− − ×

= = = −8 5 5

8 8CBPlFEM kNm×

= = =8 8

Known displacements

0A Cθ θ= = 0A B Cδ δ δ= = =


22

Slope deflection equations

2 32AB A B ABEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 2.45AB BEIM θ⇒ = −

2 32EIM FEMδθ θ⎛ ⎞= + +⎜ ⎟ ( )2 2 3 6EIM θ⇒ +2BA B A BAM FEMl l

θ θ= + − +⎜ ⎟⎝ ⎠

( )2 3.65BA BM θ⇒ = +

( )2 2 55BC BEIM θ⇒ = −2 32BC B C BC

EIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠ 5l l⎝ ⎠

2 32CB C B CBEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 55CB BEIM θ⇒ = +


23

Joint equilibrium condition

0BA BCM M+ =

( ) ( )2 22 3.6 2 5 0B BEI EIθ θ⎛ ⎞ ⎛ ⎞⇒ + + − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠( ) ( )2 3.6 2 5 0

5 5B Bθ θ⇒ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1.6 1.4 0BEIθ − = 0.875B EI

θ⇒ =

( )2 2 0.8752.4 2.4 2.055 5AB BEI EIM kNm

EIθ ⎛ ⎞= − = − = −⎜ ⎟

⎝ ⎠( )

5 5AB B EI⎜ ⎟⎝ ⎠


24

⎛ ⎞2 0.8752 3.6 4.35BAEIM kNm

EI⎛ ⎞= × + =⎜ ⎟⎝ ⎠

2 0.8752 5 4.3BCEIM kNm⎛ ⎞= × − = −⎜ ⎟

⎝ ⎠2 5 4.3

5BCM kNmEI

×⎜ ⎟⎝ ⎠

2 0.875 5 5.355CBEIM kNm

EI⎛ ⎞= + =⎜ ⎟⎝ ⎠

A B C4.3kNm2.05kNm 4.3kNm 5.35kNm


25

A B C4 3kNm2 05kNm 4 3kNm 5.35kNmA B C4.3kNm2.05kNm 4.3kNm

AB C

2 05

2.6 5.175

4.3

2.055.35


26

Procedure to solve problems

• The slope-deflection equations are written for each member.

l b d• Joint equilibrium conditions are written.

• Solving the joint equilibrium conditions, unknown displacements g jare found out.

• Substituting these unknown displacements back in the slope-g p pdeflection equations, we get the unknown end moments.


27

Example 240kN 80kN60kN

20kN m

A B C D

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


Fixed end moments

2 220 6 40 6 9012 8 12 8AB BAwl PlFEM FEM kNm× ×

− = = + = + =

220 6 60 6 10512 8BC CBFEM FEM kNm× ×

− = = + =12 8

80 6 60FEM FEM kNm×= = =

K di l t

608CD DCFEM FEM kNm− = = =

Known displacements

0A B C Dδ δ δ δ= = = =A B C D




δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 2 906AB A BEIM θ θ⇒ = + −

2 32BA B A BAEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 2 906BA B AEIM θ θ⇒ = + +

( )2 2 1056BC B CEIM θ θ⇒ = + −2 32BC B C BC

EIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )6

2 32EIM FEMδθ θ⎛ ⎞⎜ ⎟

BC B C BCl l⎜ ⎟⎝ ⎠

( )2 2 105EIM θ θ32CB C B CBM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 1056CB C BM θ θ⇒ = + +

2EI2 3EI δ⎛ ⎞ ( )2 2 606CD C DEIM θ θ⇒ = + −

2 3EI δ⎛ ⎞

2 32CD C D CDEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2EI


30

2 32DC D C DCEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 2 606DC D CEIM θ θ⇒ = + +

Joint equilibrium conditions

0ABM = ( )2 2 90 06 A BEI θ θ⇒ + − =6

0.667 0.333 90A BEI EIθ θ+ =

( )190 0.333 135 0.50.667

BA B

EIEI EIθθ θ−= = −

( )2 2 60 0EI θ θ0DCM = ( )2 60 06 D Cθ θ⇒ + + =

0 667 0 333 60EI EIθ θ

( )2

0.667 0.333 60D CEI EIθ θ+ = −

60 0.333 90 0 5CEIEI EIθθ θ− −= =

Dept. of CE, GCE Kannur Dr.RajeshKN31

( )290 0.50.667D CEI EIθ θ= = − −

0BA BCM M+ = 0BA BCM M+

( ) ( )22 2 105 02 90EIEI θ θθ θ ⎛ ⎞⎛ ⎞⇒ + + =+ +⎜ ⎟ ⎜ ⎟( ) ( )2 105 02 9066 B CB A θ θθ θ⇒ + + − =+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0 333 1 333 0 333 15EI EI EIθ θ θ+ +0.333 1.333 0.333 15A B CEI EI EIθ θ θ+ + =

( )0 333 1 333 0 333 15135 0 5 EI EIEI θ θθ + + =( )1

( )

( )0.333 1.333 0.333 15135 0.5 B CB EI EIEI θ θθ + + =−( )1

( )31.166 0.333 30B CEI EIθ θ+ = −


32

0CB CDM M+ = 0CB CDM M+

( ) ( )22 2 60 02 105EIEI θ θθ θ ⎛ ⎞⎛ ⎞

⎜ ⎟ ⎜ ⎟( ) ( )2 60 02 10566 C DC B θ θθ θ ⎛ ⎞⎛ ⎞⇒ + + − =+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0.333 1.333 0.333 45B C DEI EI EIθ θ θ+ + = −

( )0 333 1 333 0 333 90 0 5 45EI EI EIθ θ θ+ + − − = −( )2

( )40.333 1.167 15B CEI EIθ θ+ = −

( )0.333 1.333 0.333 90 0.5 45B C CEI EI EIθ θ θ+ + − − = −( )2

( )4B C

( ) 0 333 0 388 0 111 103 EI EIθ θ ( )5( ) 0.333 0.388 0.111 103 B CEI EIθ θ× ⇒ + = − ( )5

( ) 1.167 0.388 1.362 17.54 B CEI EIθ θ× ⇒ + = − ( )6


C

7 5( ) ( ) 1.251 7.56 5 CEIθ− ⇒ = − 7.5 6.01.251CEIθ −

⇒ = = −

0.333 1.167 6.0 15 24B BEI EIθ θ+ ×− = − ⇒ = −( )4

( )135 0.5 14724AEIθ = − =−

90 0.5 6.0 87DEIθ = − − ×− = −D

4 2 90EI EIθ θ⎛ ⎞⎜ ⎟( )2 2 90EIM θ θ 90

6 6B AEI EIθ θ⎛ ⎞= + +⎜ ⎟⎝ ⎠

( )2 906BA B AM θ θ⇒ = + +

⎛ ⎞4 224 147 906 6

⎛ ⎞= ×− + × +⎜ ⎟⎝ ⎠


123=

( )2EI 4 2⎛ ⎞( )2 2 1056BC B CEIM θ θ⇒ = + − 4 2 105 123

6 6B CEI EIθ θ⎛ ⎞= + − = −⎜ ⎟⎝ ⎠

( )2 2 1056CB C BEIM θ θ⇒ = + + 4 2 105 93

6 6C BEI EIθ θ⎛ ⎞= + + =⎜ ⎟⎝ ⎠

( )2 2 606CD C DEIM θ θ⇒ = + −

4 2 60 936 6C DEI EIθ θ⎛ ⎞= + − = −⎜ ⎟

⎝ ⎠


Example 3


Fixed end moments

10 8 108 8AB BAPlFEM FEM kNm×

− = = = =

2 25 8 26.66712 12BC CB CD DCwlFEM FEM FEM FEM kNm×

− = = − = = = =12 12

Known displacements

0A B C Dδ δ δ δ= = = =


37



δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )4 2 108AB A BEIM θ θ⇒ = + −


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )4 2 108BA B AEIM θ θ⇒ = + +

( )6 2 26 667EIM θ θ⇒ = + −2 32BC B C BCEIM FEMδθ θ⎛ ⎞= + − +⎜ ⎟ ( )2 26.667

8BC B CM θ θ⇒ = + −

2 32EIM FEMδθ θ⎛ ⎞= + +⎜ ⎟

2BC B C BCM FEMl l

θ θ+ +⎜ ⎟⎝ ⎠

( )6 2 26 667EIM θ θ2CB C B CBM FEMl l

θ θ= + − +⎜ ⎟⎝ ⎠

( )2 26.6678CB C BM θ θ⇒ = + +

6EI2 3EI δ⎛ ⎞ ( )6 2 26.6678CD C DEIM θ θ⇒ = + −

2 3EI δ⎛ ⎞


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠



δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )6 2 26.6678DC D CEIM θ θ⇒ = + +


0ABM = ( )4 2 10 08 A BEI θ θ⇒ + − = 10 0.5A BEI EIθ θ⇒ = −

0DCM = 17.778 0.5D CEI EIθ θ⇒ = − −( )6 2 26.667 08 D CEI θ θ⇒ + + =

50 0M M+ − = ( ) ( )4 62 10 2 26.667 50B A B CEI EIθ θ θ θ⎛ ⎞ ⎛ ⎞⇒ + + + + − =⎜ ⎟ ⎜ ⎟50 0BA BCM M+ = ( ) ( )2 10 2 26.667 508 8B A B Cθ θ θ θ⇒ + + + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 5 0 5 0 75 66 667EI EI EIθ θ θ⇒ + +2.5 0.5 0.75 66.667B A CEI EI EIθ θ θ⇒ + + =

( )2.5 0.5 0.75 66.66710 0.5B CBEI EIEIθ θθ⇒ + + =−

( )1

( )2.5 0.5 0.75 66.66710 0.5B CBEI EIEIθ θθ⇒ + +

2.25 0.75 61.667B CEI EIθ θ⇒ + =


2.25 0.75 61.667B CEI EIθ θ⇒ +

0CB CDM M+ = ( ) ( )6 62 26 667 2 26 667 0EI EIθ θ θ θ⎛ ⎞ ⎛ ⎞⇒ + + + + − =⎜ ⎟ ⎜ ⎟0CB CDM M+ ( ) ( )2 26.667 2 26.667 08 8C B C Dθ θ θ θ⇒ + + + + − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3 0 75 0 75 0EI EI EIθ θ θ⇒ + + =3 0.75 0.75 0C B DEI EI EIθ θ θ⇒ + + =

( )3 0.75 0.75 17.778 0.5 0C B CEI EIθ θ θ⇒ + + − − =

2.625 0.75 13.333C BEIθ θ⇒ + = ( )2

( )12.25 0.75 61.667B CEI EIθ θ⇒ + =

28.421BEIθ =

3 041EIθ = −( )10 0.5 10 0.5 28.421A BEI EIθ θ= − = −

3.041CEIθ = − 4.211AEIθ⇒ = −

17.778 0.5 17.778 0.5 3.041D CEI EIθ θ= − − = − − ×−


C

16.258DEIθ⇒ = −

( )4 2 10 0 5 10EIM EI EIθ θ θ θ+ + + +( )2 10 0.5 108BA B A B AM EI EIθ θ θ θ= + + = + +

28.421 0.5 4.211 10 36.32kNm= + ×− + =

( )6 2 26.667 1.5 0.75 26.6678BC B C B CEIM EI EIθ θ θ θ= + − = + −8

1.5 28.421 0.75 3.041 26.667 13.68kNm= × + ×− − =

( )6 2 26.667 1.5 0.75 26.6678CB C B C BEIM EI EIθ θ θ θ= + + = + +

( )6 2 26 667 1 5 0 75 26 667EIM EI EIθ θ θ θ

1.5 3.041 0.75 28.421 26.667 43.42kNm= ×− + × + =

( )2 26.667 1.5 0.75 26.6678CD C D C DM EI EIθ θ θ θ= + − = + −

1.5 3.041 0.75 16.258 26.667 43.42kNm= ×− + ×− − = −


Example 4

90kN 20kN30kN m90kN

A B C D

2.5m 2.5m

E2I I 2.4I

7.5m 5m 5m 3m

20 3 60DEM kNm= − × = −




δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )4 1507.5AB BEIM θ⇒ = −


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )4 2 1507 5BA BEIM θ⇒ = +l l⎝ ⎠ 7.5

( )2 2 62.5BC B CEIM θ θ⇒ = + −2 32BC B C BC

EIM FEMδθ θ⎛ ⎞= + − +⎜ ⎟ ( )5BC B C


2BC B C BCM FEMl l

θ θ+ +⎜ ⎟⎝ ⎠

( )2 2 62 5EIM θ θ32CB C B CBM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 62.55CB B CM θ θ⇒ = + +

( )4.8 2EIM θ θ2 32EIM FEMδθ θ⎛ ⎞⎜ ⎟

2 3EI δθ θ⎛ ⎞ ( )4.8 2EIM θ θ

( )25CD C DM θ θ⇒ = +2 32CD C D CD

EIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠


44


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )25DC D CM θ θ⇒ = +

Fixed end moments

2 2 2 2

2 2 2 2

90 2.5 5 90 2.5 5 1507.5 7.5AB BA

Pab Pa bFEM FEM kNml l

× × × ×− = = + = + =

2 230 5 62.5BC CBwlFEM FEM kNm×

− = = = =12 12BC CB

0FEM FEM= =

Known displacements

0CD DCFEM FEM= =

0A B C Dδ δ δ δ= = = =

Known displacements

0Aθ =


45


0BA BCM M+ =

( ) ( )4 22 150 2 62.5 07.5 5B B CEI EIθ θ θ⎛ ⎞ ⎛ ⎞⇒ + + + − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

1.867 0.4 87.5B CEI EIθ θ⇒ + = − ( )1

0CB CDM M+ =

2 4 8EI EI⎛ ⎞ ⎛ ⎞( ) ( )2 4.82 62.5 2 05 5B C C DEI EIθ θ θ θ⎛ ⎞ ⎛ ⎞⇒ + + + + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

0.4 2.72 0.96 62.5B C DEI EI EIθ θ θ⇒ + + = − ( )2


( )4.8 2 60 0EI θ θ⎛ ⎞⇒ + − =⎜ ⎟0M M+ = ( )2 60 05 D Cθ θ⇒ + − =⎜ ⎟

⎝ ⎠0DC DEM M+ =

1 92 0 96 60EI EIθ θ⇒ + =1.92 0.96 60D CEI EIθ θ⇒ + =

31.25 0.5D CEI EIθ θ⇒ = −

( ) ( )0.4 2.72 0.96 31.25 0.5 62.52 B C CEI EI EIθ θ θ⇒ + + − = −

( )30.4 2.24 92.5B CEI EIθ θ⇒ + = −

39.532BEIθ = − 34.235CEIθ = −

( )31.25 0.5 31.25 0.5 48.36834.235D CEI EIθ θ= − = − =−


4 4EI ( ) ( )4 4150 39.532 150 171.0847.5 7.5AB BEIM kNmθ∴ = − = − − = −

( ) ( )4 42 150 2 39.532 150 107.8337.5 7.5BA BEIM kNmθ= + = ×− + =

( ) ( )2 22 62.5 2 39.532 34.235 62.5 107.825 5BC B CEIM kNmθ θ= + − = ×− − − =5 5

( ) ( )2 22 62 5 39 532 2 34 235 62 5 19 3EIM kNmθ θ= + + = − + ×− + =( ) ( )2 62.5 39.532 2 34.235 62.5 19.35 5CB B CM kNmθ θ= + + = − + ×− + =

( ) ( )4.8 4.82 2 34 235 48 368 19 3EIM kNθ θ

4 8 4 8EI

( ) ( )2 2 34.235 48.368 19.35 5CD C DM kNmθ θ= + = ×− + = −


( ) ( )4.8 4.82 2 48.368 34.235 605 5DC D CEIM kNmθ θ= + = × − =

Example 5

Support B settles by 10 mm. 6 4200 , 50 10E GPa I mm= = ×

kN6 6 4 4 22200 10 50 10 10kNEI m kNm

m−= × × × =


Fixed end moments

208AB BAPlFEM FEM kNm− = = =

2

1612BC CBwlFEM FEM kNm− = = =12C C

K di l t

0A Cδ δ= =

Known displacements

10B mmδ =

0Cθ =




δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

24 3 102 208 8AB A BEIM θ θ

−⎛ ⎞×⇒ = + − −⎜ ⎟

⎝ ⎠


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

24 3 102 208 8BA B AEIM θ θ

−⎛ ⎞×⇒ = + − +⎜ ⎟

⎝ ⎠l l⎝ ⎠ 8 8⎝ ⎠

26 3 102 16EIM θ−⎛ ⎞×

⇒ = + −⎜ ⎟2 32BC B C BCEIM FEMδθ θ⎛ ⎞= + − +⎜ ⎟ 2 16

8 8BC BM θ⇒ = +⎜ ⎟⎝ ⎠

2 3EI δ⎛ ⎞

2BC B C BCM FEMl l

θ θ+ +⎜ ⎟⎝ ⎠

26 3 10EI −⎛ ⎞×2 32CB C B CBEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

6 3 10 168 8CB BEIM θ

⎛ ⎞×⇒ = + +⎜ ⎟

⎝ ⎠


51


0ABM =24 3 102 20 0

8 8A BEI θ θ

−⎛ ⎞×⇒ + − − =⎜ ⎟

⎝ ⎠⎝ ⎠26 100.5 20 0

32A BEIEI EIθ θ

−×⇒ + − − =

322

34

20 6 100.5 3.875 1010 32A Bθ θ

−−×

⇒ + = + = × ( )1

0BA BCM M+ =

2 24 3 10 6 3 102 20 2 16 08 8 8 8B A BEI EIθ θ θ

− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞× ×⇒ + − + + + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

420 16 9.375 100.5 1.5B A BEI EIθ θ θ −⎛ ⎞ ⎛ ⎞⇒ + = − ×+ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


B A BEI EI⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

20 16 44 4

20 162.5 0.5 9.375 1010 10B Aθ θ −⇒ + + − = − ×

4 44 42.5 0.5 9.375 10 4 10B Aθ θ − −⇒ + = − × − ×

( )242.5 0.5 13.375 10B Aθ θ −+ = − ×

30.5 3.875 10A Bθ θ −+ = × ( )1

31.456 10Bθ−= − ×34.603 10Aθ

−= ×


24 3 10EI −⎛ ⎞×4 3 102 208 8BA B AEIM θ θ

⎛ ⎞×= + − +⎜ ⎟

⎝ ⎠

4 24 10 3 10⎛ ⎞( )4 2

3 34 10 3 102 1.456 10 4.603 10 208 8

−− −⎛ ⎞× ×

= − × + × − +⎜ ⎟⎝ ⎠

9.705 kNm=

26 3 102 168 8BC BEIM θ

−⎛ ⎞×−= − −⎜ ⎟

⎝ ⎠

4 236 10 3 102 1.456 10 16

8 8

−−⎛ ⎞× ×−

= ×− × − −⎜ ⎟⎝ ⎠

9.705 kNm= −

26 3 10 168 8CB BEIM θ

−⎛ ⎞×−= − +⎜ ⎟

⎝ ⎠

⎝ ⎠

8 8⎜ ⎟⎝ ⎠

4 236 10 3 101 456 10 16

−−⎛ ⎞× ×−

= − × − +⎜ ⎟ 33 21 kNm=


54

1.456 10 168 8

= − × − +⎜ ⎟⎝ ⎠

33.21 kNm=

Slope Deflection for frames:No sideswayy


Example 6Example 6

B10kN 2kN m

AB C

22I I

332m

3m I

3m3m

D

I

D


Fixed end moments

2 2

2 2

10 2 3 7.25AB

PabFEM kNml

− − × ×= = = −

2 2

2 2

10 2 3 4.85BA

Pa bFEM kNml

× ×= = =

2 22 3 1 5wlFEM FEM kN×

2 25l

3 1.512 12BC CBwlFEM FEM kNm− = = = = 0BD DBFEM FEM= =

Known displacements

0A B C Dδ δ δ δ= = = =

0A Dθ θ= =




δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )4 7.25AB BEIM θ⇒ = −


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )4 2 4.85BA BEIM θ⇒ = +l l⎝ ⎠ 5

( )2 2 1 5EIM θ θ⇒ = + −2 32BC B C BCEIM FEMδθ θ⎛ ⎞= + − +⎜ ⎟ ( )2 1.5

3BC B CM θ θ⇒ = + −

2 3EI δ⎛ ⎞

2BC B C BCM FEMl l

θ θ+ +⎜ ⎟⎝ ⎠

2EI2 32CB C B CBEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 2 1.53CB C BEIM θ θ⇒ = + +

⎛ ⎞2 32BD B D BDEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 23BD BEIM θ⇒ =


582 32DB D B DBEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )23DB BEIM θ⇒ =


0CBM = ( )2 2 1.5 03 C BEI θ θ⇒ + + =

1.333 0.667 1.5C BEI EIθ θ⇒ + = − ( )1

0BA BC BDM M M+ + =

( ) ( ) ( )4 2 2 02 4.8 2 1.5 25 3 3B B C BEI EI EIθ θ θ θ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⇒ + + =+ + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠5 3 3⎝ ⎠ ⎝ ⎠ ⎝ ⎠

4.267 0.667 3.3B CEI EIθ θ⇒ + = − ( )2

0.6481BEIθ = − 0.801CEIθ = −


( )4EI ( )4( )4 7.25AB BEIM θ= − ( )4 0.6481 7.2 7.718

5kNm= − − = −

( )4 2 4.85BA BEIM θ= + ( )4 2 0.6481 4.8 3.763

5kNm= ×− + =

( )2 2 1.53BC B CEIM θ θ= + − ( )2 2 0.6481 0.801 1.5 2.898

3kNm= ×− − − = −

( )2 23BD BEIM θ= ( )2 2 0.6481 0.864

3kNm= ×− = −( )

3BD B

2EI

( )3

2


60( )2

3DB BEIM θ= ( )2 0.6481 0.432

3kNm= − = −

Slope Deflection for frames:S dSidesway


P BAM CDMPB

CBA CD

2l

DH D1l

DCM

AH A

ABMAB

AB BAA

M MH += CD DCM MH +

= 0H H P


1AH

l2

DHl

= 0A DH H P+ + =

M CDM

BCBAM CDM

a

Pl

2l

DH D1l

DCM

AH A

ABMAB

CD DCM MH += 0H H P

AB BAA

M M PaH + −=


632

DHl

= 0A DH H P+ + =1

AHl

CCDM

BC

BAMCDM

l2l

w

DH

D

1l

D

DCM

AH A

ABM

22 2CD DC

DM M wlH + +

= 2 0A DH H wl+ − =AB BAA

M MHl+

=

ABM


642

D l1A l

Example 7


Fixed end moments

2 216 1 4 10.24BCPabFEM kNm− − × ×

= = = −

Fixed end moments

2 2 10.245BCFEM kNm

l

2 216 1 4 2 56Pa bFEM kN× ×2 2

16 1 4 2.565CB

Pa bFEM kNml

= = =

Known displacements

0A Dθ θ= =

δ=Let horizontal movement (sidesway)


66



δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2 35 5BEI δθ⎛ ⎞= −⎜ ⎟

⎝ ⎠


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2 325 5BEI δθ⎛ ⎞= −⎜ ⎟

⎝ ⎠

2 32BC B C BCEIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )2 2 10.24B CEI θ θ= + −



( )2EI

( )5 B C

32CB C B CBM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2 3EI δ⎛ ⎞

( )2 2 2.565 C BEI θ θ= + +

2 3EI δ⎛ ⎞

2 3EI δ⎛ ⎞


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2 3EI δ⎛ ⎞

2 325 5CEI δθ⎛ ⎞= −⎜ ⎟

⎝ ⎠


67


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2 35 5CEI δθ⎛ ⎞= −⎜ ⎟

⎝ ⎠


0BA BCM M+ = ( )2 3 22 02 10.24B B CEI EIδθ θ θ

⎡ ⎤⎛ ⎞ ⎡ ⎤⇒ − + =+ −⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦⎝ ⎠⎣ ⎦BA BC ( )

5 5 5B B C⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦⎝ ⎠⎣ ⎦

( )1( )1.6 0.4 0.24 10.24B CEI θ θ δ+ − = ( )

0M M+ ( ) 2 32 2 02 2 56EIEI δθθ θ

⎡ ⎤⎛ ⎞⎡ ⎤⇒ ++ + ⎜ ⎟⎢ ⎥⎢ ⎥

( )1.6 0.4 0.24 10.24B CEI θ θ δ+

0CB CDM M+ = ( ) 2 02 2.565 55 CC B θθ θ⎡ ⎤⇒ + − =+ + ⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦

( )2( )1.6 0.4 0.24 2.56C BEI θ θ δ+ − = −


0H H+ = 0AB BA CD DCM M M M+ +⇒ + 5l l0A DH H+ =

1 2

0AB BA CD DC

l l⇒ + = 1 2 5l l= =

2 3 2 3 2 3 2 32 2 05 5 5 5 5 5 5 5B B C CEI EI EI EIδ δ δ δθ θ θ θ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⇒ − + − + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

3 3 3 32 2 05 5 5 5B B C Cδ δ δ δθ θ θ θ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⇒ − + − + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

0.8 0B Cθ θ δ⇒ + − = ( )3

845BEIθ =

272CEIθ −=

487

EIδ =105BEIθ

105CEIθ 7


69

2 3EIM δθ⎛ ⎞= ⎜ ⎟2 845 3 48 1 5733 kNm⎛ ⎞= − =⎜ ⎟5 5AB BM θ= −⎜ ⎟

⎝ ⎠

2 3EI δ⎛ ⎞

1.57335 105 5 7

kNm= − =⎜ ⎟⎝ ⎠

2 845 3 48⎛ ⎞2 325 5BA BEIM δθ⎛ ⎞= −⎜ ⎟

⎝ ⎠2 845 3 482 4.8385 105 5 7

kNm⎛ ⎞= × − =⎜ ⎟⎝ ⎠

( )2 2 10.245BC B CEIM θ θ= + −

2 845 2722 10.24 4.8385 105 105

kNm⎛ ⎞= × − − = −⎜ ⎟⎝ ⎠

( )2 2 2.565CB C BEIM θ θ= + +

2 272 8452 2.56 3.7075 105 105

kNm−⎛ ⎞= × + + =⎜ ⎟⎝ ⎠

( )5

2 325 5CD CEIM δθ⎛ ⎞= −⎜ ⎟

⎝ ⎠

5 105 105⎝ ⎠

2 272 3 482 3.7075 105 5 7

kNm−⎛ ⎞= × − × = −⎜ ⎟⎝ ⎠

2 3EIM δθ⎛ ⎞= −⎜ ⎟

5 5CD C⎜ ⎟⎝ ⎠ 5 105 5 7⎜ ⎟

⎝ ⎠

2 272 3 48 2.682 kNm−⎛ ⎞= − × = −⎜ ⎟


5 5DC CM θ= ⎜ ⎟⎝ ⎠

2.6825 105 5 7

kNm×⎜ ⎟⎝ ⎠

Example 8

10 kN B C

p

4m

4m

4mEI

A D

4m 4mEI EI

• No fixed end moments

• Known displacements: 0θ =• Known displacements: 0Aθ =

δ=• Let horizontal movement (sidesway)


0DCM =



δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2 34 4AB BEIM δθ⎛ ⎞⇒ = −⎜ ⎟

⎝ ⎠


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2 324 4BA BEIM δθ⎛ ⎞⇒ = −⎜ ⎟

⎝ ⎠

( )2 24BC B CEIM θ θ⇒ = +2 32BC B C BC

EIM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )4



( )2 2EIM θ θ32CB C B CBM FEMl l

δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

( )24CB C BM θ θ⇒ = +

2 3EI δ⎛ ⎞2 3EI δ⎛ ⎞ 2 324 4CD C DEIM δθ θ⎛ ⎞⇒ = + −⎜ ⎟

⎝ ⎠

2 3EI δ⎛ ⎞


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2 3EI δ⎛ ⎞


72


δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠

2 324 4DC D CEIM δθ θ⎛ ⎞⇒ = + −⎜ ⎟

⎝ ⎠


0DCM = 2 32 04 4D CEI δθ θ⎛ ⎞⇒ + − =⎜ ⎟

⎝ ⎠324D Cδθ θ⇒ + =

4 4⎝ ⎠

( )1

438 2

CD

δ θθ⇒ = −

0BA BCM M+ = ( )2 3 22 024 4 4B B CEI EIδθ θ θ

⎡ ⎤⎛ ⎞ ⎡ ⎤⇒ − + =+⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦⎝ ⎠⎣ ⎦4 4 4⎣ ⎦⎝ ⎠⎣ ⎦

( )2344B Cδθ θ⇒ + =

316 4

CB

δ θθ⇒ = −

0CB CDM M+ = ( ) 2 32 2 024 4C DC BEIEI δθ θθ θ

⎡ ⎤⎛ ⎞⎡ ⎤⇒ + + − =+ ⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦

4 16 4

CB CD ( )4 44 C DC B ⎜ ⎟⎢ ⎥⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦

3 3 34 C Cδ θ δ θ δθ ⎡ ⎤⇒ + + =⎢ ⎥ 3 25 0 1875 0θ δ =


( )34

16 4 8 2 4Cθ⇒ + − + − =⎢ ⎥⎣ ⎦3.25 0.1875 0Cθ δ− =

0H H P 0AB BA CD DCM M M M P+ +⇒ + +0A DH H P+ + =

1 2

0AB BA CD DC Pl l

⇒ + + =

M M M 10 04 4

AB BA CDM M M+⇒ + + = 40AB BA CDM M M⇒ + + = −

2 3 2 3 2 32 2 404 4 4 4 4 4B B C DEI EI EIδ δ δθ θ θ θ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⇒ − + − + + − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣ ⎦

80 93 24B C D EIδθ θ θ −

⇒ + + = +4EI

3 3 80 93 216 4 8 2 4

C CC EI

δ θ δ θ δθ −⎡ ⎤ ⎡ ⎤⇒ − + + − = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( )4

16 4 8 2 4C EI⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

800.75 1.3125Cθ δ −− =


74

( )40.75 1.3125C EIθ δ

( )3.25 0.1875 0CEI θ δ− =( )3 ( )C

( )0.75 1.3125 80CEI θ δ− = −3.636

63.03CEI

EIθδ

==

( )

( )4

3 CDEI EI δ θθ ⎛ ⎞= −⎜ ⎟

⎝ ⎠3 63.03 3.636 21 818EIθ ×

= − =8 2DEI EIθ ⎜ ⎟⎝ ⎠

3 CEI EI δ θθ ⎛ ⎞= ⎜ ⎟

21.8188 2DEIθ = − =

3 63.03 3.636 10 909EIθ ×16 4BEI EIθ = −⎜ ⎟⎝ ⎠

10.90916 4BEIθ = − =

( )2 3 0.5 10.909 0.75 63.03 18.1824 4AB BEIM kNmδθ⎛ ⎞= − = − × = −⎜ ⎟

⎝ ⎠

( )2 32 0.5 2 10.909 0.75 63.03 12.727BA BEIM kNmδθ⎛ ⎞= − = × − × = −⎜ ⎟


75

( )2 0.5 2 10.909 0.75 63.03 12.7274 4BA BM kNmθ × ×⎜ ⎟

⎝ ⎠

( ) ( )2EI ( ) ( )2 2 0.5 2 10.909 3.636 12.7274BC B CEIM kNmθ θ= + = × + =

( ) ( )2 2 0.5 2 3.636 10.909 9.0914CB C BEIM kNmθ θ= + = × + =

( )2 32 0.5 2 3.636 21.818 0.75 63.03 9.091CD C DEIM kNmδθ θ⎛ ⎞= + − = × + − × = −⎜ ⎟ ( )2 0.5 2 3.636 21.818 0.75 63.03 9.0914 4CD C DM kNmθ θ+ × + ×⎜ ⎟

⎝ ⎠


76

M t di t ib ti th dMoment distribution method


Stiffness, Carry-over factor and Distribution factor

Beam hinged at both ends

AB

M

ABθBAθ

(Applied moment)

M

( )2 2EIM θ θ( )2 2 0BA BA AB

EIML

θ θ= + =

)

( )2AB AB BAML

θ θ= +( )

L

1θ θ⇒ = −2BA ABθ θ⇒ = −

2 1 32EI EIM θ θ θ⎛ ⎞⎜ ⎟


78

22AB AB AB ABM

L Lθ θ θ⎛ ⎞∴ = − =⎜ ⎟

⎝ ⎠

3M EI3AB

AB

M EILθ

=

3EIi.e., the moment required at A to induce a unit rotation at A is

(when the far end B is free to rotate)

3EIL

This moment, i.e., moment required to induce a unit rotation, is called stiffness (denoted by k).is called stiffness (denoted by k).


Beam hinged at near end and fixed at far end

A BABθ

0BAθ =

A BM

(Applied

2EI

moment)

4M EI( )2 2 0AB AB

EIML

θ= +4AB

AB

M EILθ

⇒ =

i.e., the moment required at A to induce a unit rotation at A is

(when the far end B is fixed against rotation)

4EIL

(when the far end B is fixed against rotation)


80

( )2EI 2 AB ABEI M MM L⎛ ⎞⎜ ⎟( )2 0BA AB

EIML

θ= +2

4 2AB AB

BAEI M MM LL EI

⎛ ⎞= =⎜ ⎟⎝ ⎠

A moment applied at the near end induces at a fixed far end a moment equal to half its magnitude, in the same direction.

Half of moment applied at the near end is carried over to the fixed far end.

Carry over factor is 1/2.


Several members meeting at a joint

1 13E IM kθ θ= =1 11

M kL

θ θ= =

2 24E I2 22 2

2

4E IM kL

θ θ= =

3 33 3

3

3E IM kL

θ θ= =

4 44 4

4E IM kL

θ θ= =4L

1 2 3 4 1 2 3 4: : : :: : : :M M M M k k k k


1 2 3 4 1 2 3 4: : : :: : : :k k k k

11

1 2 3 4

kM Mk k k k

=+ + +

1k Mk

=∑

ii

kM Mk

=∑

A moment applied at a joint, where several members meet, will be distributed amongst the members in proportion to their stiffnessdistributed amongst the members in proportion to their stiffness.

ii

kM Mk

=∑

distribution factordistribution factor


Illustration of the method

B3

5kN 8kN2 5

Example 1

AB

C3m 2.5m

5m 5m

Problem structure

A B CB

Problem structure

A B C

5kNm 5kNm2.4kNm 3.6kNm 5kNm 5kNm


Fixed end moments (reactive)

AB C1.4kNm

Unbalanced moment

AB C0.7kNm0.7kNm Unbalanced moment

distributed amongst members

B 0 35kNmA

B C0.35kNm0.35kNm

Distributed moments carried over to far


ends of members

A B C0 5 0 5 di t ib ti f t0.5 0.5

-2.4 +3.6 -5.0 +5.0 Fixed End Moments0 0 distribution factors

+0.7 +0.7

+0 35 +0 35

Distribution

Carry over+0.35 +0.35 Carry over

Distribution

-2.05 +4.3 -4.3 +5.35 Final Moments


5 35kNA B C4.3kNm2.05kNm 4.3kNm 5.35kNm

AB CA C

2.055 35

4.35.35


Example 240kN 80kN60kN

20kN m

A B C D

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

Fi d d t

2wl Pl

Fixed end moments

220 6 40 6× ×12 8AB BAwl PlFEM FEM− = = +

2wl Pl

20 6 40 6 60 30 9012 8

kNm× ×= + = + =

220 6 60 6× ×

80 6Pl ×12 8BC CBwl PlFEM FEM− = = +

20 6 60 6 60 45 10512 8

kNm× ×= + = + =


80 6 608 8CD DCPlFEM FEM kNm×

− = = = =

A B C D

0 5 0 5 0 5 0 51 1 Di t ib ti f t0.5 0.5-90 +90 -105 +105 -60 +60 Fixed End Moments

0.5 0.51 1 Distribution factors

+90 +7.5 +7.5 -22.5 -22.5 -60

+3.5 +45 -11.25 +3.5 -30 -11.25Distribution

Carry over

-3.5 -16.875 -16.875 +13.25 +13.25 +11.25

8 434 1 75 +6 625 8 434 +5 625 +6 625

Distribution

Carry over-8.434 -1.75 +6.625 -8.434 +5.625 +6.625

+8.434 -2.438 -2.438 +1.405 +1.405 -6.625

Carry over

Distribution

0 +121.48 -121.44 +92.221 - 92.22 0 Final Moments


A B C D

-90 +90 -105 +105 -60 +60 Fixed End Moments0.571 0.429 Distribution factors0.429 0.571

90 +90 105 +105 60 +60

+90 +45 -30 -60Release A& D, and carry over

0 +135 -105 +105 -90 0

-12.87 -17.13 -8.565 -6.435

Initial moments

Distribution

-4.283 -8.565

+1 837 +2 445 4 89 3 674

Carry over

Di t ib ti+1.837 +2.445 4.89 3.674

+2.445 1.223

Distribution

Carry over-1.049 -1.396 -0.698 -0.524

0 +122.92 -122.92 +93.29 - 93.29 0 Final Moments

Distribution


Example 3


91

A B C DA B C DDistrib. factors 1 0.2727 0.7273 0.6667 0.3333 0

Fixed-end moments -14.700 +6.300 -8.333 +8.333 -12.500 +12.500

Release A andRelease A and Carry over +14.700 +7.350

Initial moments 0.00 13.65 -8.333 +8.333 -12.500 +12.500moments

Dist 1 -1.450 -3.867 +2.779 +1.388CO 1 1.39 -1.934 +0.694Dist 2 -0.379 -1.011 1.29 0.644CO 2 0.645 -0.506 0.322Dist 3 -0.176 -0.469 0.338 0.168Final

moments 0 +11.645 -11.645 +10.3 -10.3 +13.516


92


93

Example 3

2 25 8l2 25 8 26.66712 12wl kNm×

= =10 8 108 8Pl kNm×

= =


Distribution factors

( )( ) ( )

1 3 2 83 42 8 3 8BA

EIKDFK K EI EI

= =+ +

0.333=( ) ( )1 2 3 42 8 3 8K K EI EI+ +

( )2 4 3 80 667

EIKDF = = =

( )

( ) ( )1 2

0.6673 42 8 3 8BCDF

K K EI EI= = =

+ +

( )( ) ( )

4 3 83 43 8 3 8CB

EIDF

EI EI=

+0.571=

0.429=( )( ) ( )

3 3 83 43 8 3 8CD

EIDF

EI EI=

+( ) ( )


A B C D

0.333 0.667 0.571 0.4291 1 Distribution factorsJoint couple dist.16.65 33.35

Fixed End Moments

Carry over16.675-10 +10 -26.667 +26.667 -26.667 +26.667

Release A& D, and carry overInitial moments

+10 +5.0 -13.333 -26.667

0.0 +15.0 -26.667 +43.342 -40.0 0.0

Carry over

Distribution3.885 7.782 -1.905 -1.437-0.953 +3.891

Carry over

Distribution0.317 0.636 -2.218 -1.673-1.109 0.318

Distribution

Final Moments

0.369 0.74 -0.181 -0.137

0 +36.22 13.78 +43.28 -43.25 0


90kN 20kN30kN m90kN

Example 4

A B C D

2.5m 2.5m

E2I I 2.4I

7.5m 5m 5m 3m

2 230 5 62.512 12wl kNm×

= =2 2

2 2

90 2.5 5 1007.5

Pab kNml

× ×= =

2 2

2 2

90 2.5 5 507.5

Pa b kNml

× ×= = 20 3 60kNm× =

( )( ) ( )

4 2 7.54 42 7 5 5BA

EIDF

EI EI=

+

( )( ) ( )

4 54 42 7.5 5BC

EIDF

EI EI=

+0.571=0.429=

7.5l

( ) ( )4 42 7.5 5EI EI+ ( ) ( )

( )4 5EIDF 0 357 ( )3 2.4 5EI


( )( ) ( )4 35 2.4 5CBDFEI EI

=+ 0.643=0.357= ( )

( ) ( )3 2.4 5

4 35 2.4 5CDEI

DFEI EI

=+

A B C D

0.571 0.429-150.0 150.0 -62.5 62.5 0.0 0.0 -60.0 FEM

0.357 0.6430 DFs1 0

30.0 60.0

-150.0 150.0 -62.5 62.5 30.0 60.0 -60.0

Balance D, and carry overInitial moments

-49.96 -37.54 -33.02 -59.48

-24.98 -16.51 -18.77 CO

Dist

9.43 7.08 6.7 12.07

4.72 3.35 3.54 CO

Dist

-1.91 -1.44 -1.26 -2.28

-0.96 -0.63 -0.72

Dist

CO0.36 0.27 0.26 0.46

-171.2 107.92 -107.92 +19.23 -19.23 60.0 -60.0 Final Moments

Dist


Example 5Support B settles by 10 mm. 6 4200 , 50 10E GPa I mm= = ×

( )3 2 8EIDF ( )4 3 8EI

DF0 333 0 667( )( ) ( )3 42 8 3 8BADF

EI EI=

+( )

( ) ( )3 42 8 3 8BCDFEI EI

=+

0.333= 0.667=


2 23 8l20 8Pl × 2 23 8 1612 12wl kNm×

= =20 8 20

8 8Pl kNm×

= =

2

68ABPl EIFEM

Lδ−

= −8 L

6 6 12 3

2

6 2 200 10 50 10 10 10 1020− −× × × × × × × ×

= − − 2208

20 18.75 38.75 kNm= − − = −

2

68BAPl EIFEM

Lδ

= −8 L

6 6 12 3

2

6 2 200 10 50 10 10 10 10208

− −× × × × × × × ×= −


2820 18.75 1.25 kNm= − =

22

2

612BCwl EIFEM

Lδ−

= +

6 6 12 3

2

6 3 200 10 50 10 10 10 10168

− −× × × × × × × ×= − +

8

16 28.125 12.125 kNm= − + =

2

2

612CBwl EIFEM

Lδ

= + 212CB L6 6 12 36 3 200 10 50 10 10 10 1016

− −× × × × × × × ×= + 216

8= +

16 28 125 44 125 kNm= + =


16 28.125 44.125 kNm= + =

A B C0 333 0 6670.333 0.667

-38.75 +1.25 12.125 44.125 Fixed End Moments

1 0

Release A and 38.75 19.375

0 0 20 625 12 125 44 125

Release A, and carry over

Initial Moments0.0 20.625 12.125 44.125

-10.906 -21.844 0 Distribution

-10.922 Carry over

Distribution

0.0 + 9.719 -9.719 +33.203 Final Moments


Example 6

Support B settles by 10 mm. 6 4200 , 50 10E GPa I mm= = ×


( )3 2 8EIDF =

( )4 3 8EIDF =0 333= 0 667=

( ) ( )3 42 8 3 8BADFEI EI

=+ ( ) ( )3 42 8 3 8BCDF

EI EI=

+0.333 0.667=

2 23 8 16wl kN×20 8 20Pl kN×16

12 12kNm= =20

8 8kNm= =

6Pl EIδ2

68ABPl EIFEM

Lδ

= − −

6 6 12 36 2 200 10 50 10 10 10 106 6 12 3

2

6 2 200 10 50 10 10 10 10208

− −× × × × × × × ×= − −

20 18 75 38 75 kN20 18.75 38.75 kNm= − − = −

6Pl EIFEM δ= − 28BAFEM

L6 6 12 36 2 200 10 50 10 10 10 1020

− −× × × × × × × ×=


2208

= −

20 18.75 1.25 kNm= − =

2

2

612BCwl EIFEM

Lδ−

= + 212 L6 6 12 3

2

6 3 200 10 50 10 10 10 1016− −× × × × × × × ×

= − + 28

16 28.125 12.125 kNm= − + =

2

2

612CBwl EIFEM

Lδ

= + 212CB L6 6 12 36 3 200 10 50 10 10 10 1016

− −× × × × × × × ×= + 216

8= +

16 28 125 44 125 kNm= + =


16 28.125 44.125 kNm= + =

A B C0.333 0.667

12 -5.0 -101 0

Joint couple dist.

6 -5

-38 75 +1 25 12 125 44 125 Fixed End Moments

Carry over

-38.75 +1.25 12.125 44.125

38.75 19.375

Fixed End Moments

Release A, and carry over

-0.0 26.625 12.125 39.125

-12.904 -25.834 0

Initial Moments

Distribution

-12.917 Carry over

Di t ib ti

12.0 + 8.721 -23.709 +26.208

Distribution

Final Moments


Moment Distribution for frames:No sideswayy


Example 7Example 7

B10kN 2kN m

AB C

22I I

332m

3m I

3m3m

D

I

D


Fixed end moments

2 2

2 2

10 2 3 7.2ABPabFEM kNm− − × ×

= = = −2 2 7.5AB kNm

l

2 210 2 3 4 8Pa bFEM kN× ×2 2 4.8

5BAFEM kNml

= = =

2 22 3 1.512 12BC CBwlFEM FEM kNm×

− = = = = 0BD DBFEM FEM= =



( )( ) ( ) ( )

4 2 54 3 42 5 3 3BA

EIDF

EI EI EI=

+ +

( )( ) ( ) ( )

3 34 3 42 5 3 3BC

EIDF

EI EI EI=

+ +


( ) ( ) ( )4 3 42 5 3 3EI EI EI+ + ( ) ( ) ( )

0.407= 0.254=

( )( ) ( ) ( )

4 34 3 42 5 3 3BD

EIDF

EI EI EI=

+ +( ) ( ) ( )4 3 42 5 3 3EI EI EI+ +

0.339=


AB BA BD BC CB

0.407 0.339 0.254-7.2 4.8 0.0 -1.5 1.5 Fixed End Moments0 1 Distribution factors

-0.75 -1.5-7.2 4.8 0.0 -2.25 0.0

Release C, and carry overInitial moments

-1.04 -0.864 -0.648

-0.52 Carry over

DistributionInitial moments

-7.72 3.72 -0.864 -2.9 0.0

y

Final MomentsDistribution

0.864 0.4322DBM kNm−

= = −


Moment Distribution for frames: Moment Distribution for frames: sidesway


• Assume sway is prevented by giving a support at CAssume sway is prevented by giving a support at C.

• This causes a reaction R at C, along with end-moments M.

• This reaction R has to be cancelled out, since actually sway is not prevented.

M

RR


• It is required to find out what member-end-moments cause this reaction R, in the absence of actual external loads (Let this unknown moment be MBA)

MBA M’BA

R’R’

• Let us assume that a moment of M’ ( = 100kNm say) is causing a


• Let us assume that a moment of M BA ( = 100kNm, say) is causing a reaction of R’.

• If M’BA= MBA R+R’ must be zero If M BA MBA , R+R must be zero.

And the total moment will be M + M’BA .

• But since M’BA≠ MBA , R+ R’ ×(MBA /M’BA) =0

• Therefore, MBA /M’BA= – R / R’ = C1 i.e., MBA = C1 × M’BA

• Hence the total moment is

M M M C M’ M M’ R / R’ M + MBA = M+ C1 ×M’BA = M – M’BA × R / R’


RR’

1 0R C R′+ =


• When a moment of M’BA ( = 100kNm, say) is applied, what will be the BA ( , y) pp ,value of M’CD?

Ratio of sway moments at column heads

21 1

22 2

BA

CD

M I LM I L′=

′Both column bases hinged:

21 1

22 2

BA

CD

M I LM I L′=

′Both column bases fixed:

21 1

2

2BAM I LM I L′=

′One column base hinged and the other fixed:2 2CDM I Lg


Both column bases hinged

3 31 1 2 2

3 3P PEI EI

δ = =δδPC

g

1 23 3EI EIB C

3 3EI EIδ δ1 2 1 2

1 23 31 2

3 3,EI EIP Pδ δ= =

1 1 2 2,BA CDM P M P′ ′= =AD

2P

2M P I′

1P

1 1 1 12

2 2 2 2

BA

CD

M P IM P I

= =′


, 0, 0AB DCAlso M M′ ′= =

Both column bases fixed

PC

δδ

1 22 21 2

6 6,BA CDEI EIM Mδ δ′ ′= =

B C

1 21 2

21 1

22 2

BA

CD

M IM I′=

′AD

2P

1P

, ,BA AB CD DCAlso M M M M′ ′ ′ ′= =


One column base fixed and the other hinged

32 2

3PEI

δ =

g

δδPC 23EI

23EI δ

B C

22 3

2

3EIP δ=1 2

1 22 22 2

1 2

6 3,BA CDEI EIM M Pδ δ′ ′= = =A

D

2P1 2

21 1

2

2BAM I′=

1P

22 2CDM I′

0Al M M M′ ′ ′


, , 0BA AB DCAlso M M M′ ′ ′= =

Example 8

1 0R C R′+ =

R R’


( )4 5EI( )( ) ( )

4 54 45 5BA BC CB CD

EIDF DF DF DF

EI EI= = = =

+0.5=

2 2

2 2

16 1 4 10.245BC

PabFEM kNml

− − × ×= = = −

5l

2 216 1 4 2 56Pa bFEM kN× ×2 2 2.56

5CBFEM kNml

= = =


To find M values

A B C D

0 5 0 5 0 5 0 50 DF00.5 0.5-10.24 2.56

5.12 5.12 -1.28 -1.28FEM

Di t

0.5 0.50 DFs0

2.56 -0.64 2.56 -0.64

0.32 0.32 -1.28 -1.28CO

Dist

Dist0.16 -0.64 0.16 -0.64

0.32 0.32 -0.08 -0.08 DistCO

Dist

0.16 -0.04 0.16 -0.04

0.02 0.02 -0.08 -0.08

2.88 5.78 -5.78 +2.72 -2.72 -1.32 Final Moments


2.88 5 5.78xA− × = −

( )1.73xA⇒ = →

R

( )x →

1.32 5 2.72xD− + × =

( )0.81xD⇒ = ←

1.73 0.81 0R− − =

( )0 92R ( )0.92R = ←

Assume M’BA= -100 kNm

21 1

2BAM I L

M I L′=

′

Ratio of sway moments at column heads for both column bases fixed:


2 2CDM I LHence M’CD= -100 kNm Also, M’AB= M’DC= -100 kNm

To find M’BA values

A B C DA B C D

0.5 0.5-100.0 -100.0 -100.0 -100.0 FEM

0.5 0.50 DFs0

50.0 50.0 50.0 50.0

25.0 25.0 25.0 25.0

FEM

CO

Dist

-12.5 -12.5 -12.5 -12.5

-6.25 -6.25 -6.25 -6.25

CO

CO

Dist6.25 6.25 6.25 6.25

3.125 3.125 3.125 3.125

1.563 1.563 1.563 1.563DistCO

CO-0.781 -0.781 -0.781 -0.781

-79.687 -60.156 60.157 60.156 -60.157 -79.687 Final Moments

Dist


80 5 60xA− + × =

( )28xA⇒ = ←R′

80 5 60xD− + × =

( )28D⇒ = ←( )28xD⇒ = ←

28 28 0R− − + =

( )56R′ = →

0 921 0R C R′+ = 10.92 56 0C⇒− + = 1

0.92 0.016456

C∴ = =


All the end-moments are to be found as:

1FINAL BAM M C M′= +

All the end moments are to be found as:

R=0.92 R’=56R =56


13.01 2.99


Example 9 Δ Δ

20 kN B C

4mB C

4m

4m

4mEI

A DEI EI

DA

( )4 4EIDF DF= =

( )4 4EIDF =

( )( ) ( )4 44 4BA BCDF DFEI EI

= =+

0.5=

( )( ) ( )4 34 4CBDFEI EI

=+

0.571= 0.57

( )( ) ( )

3 44 34 4CD

EIDF

EI EI=

+


( ) ( )0.429=

To find M values

No fixed end moment since there is no member force (only a joint force of 20 kN).

To find M values

(o y a jo t o ce o 0 N).

R= -20 kN

Assume fixed end moment due to sway M’BA as -10 kNm.

Ratio of sway moments at column heads for One column base hinged and the other fixed: 2

1 12BAM I L′ 1 12

2 2

BA

CDM I L=

′ 2=

5CDM kNm′∴ = −


Also, M’AB= -10 kNm, M’DC= 0

To find M’BA values

A B C D

0.5 0.5 0.571 0.4290 DFs0.5 0.5-10 -10 -5

5 5 2.86 2.14FEM

Dist

0.571 0.429 DFs

2.5 1.43 2.5

-0.72 -0.71 -1.43 -1.07CO

Dist

Dist-0.36 -0.72 -0.36

0.36 0.36 0.21 0.15 DistCO

0.18 0.1 0.18

-0.05 -0.05 -0.1 -0.08 DistCO

-7.68 -5.41 5.41 3.86 -3.86 Final Moments


( )7.68 4 5.41xA− + × = ( )3.27xA⇒ = ←

4 3 86D × = ( )0 965D⇒ = ←4 3.86xD × = ( )0.965xD⇒ = ←

3 27 0 965 0R′+ ( )4 235R′⇒ = →3.27 0.965 0R− − + = ( )4.235R⇒ = →

201 0R C R′+ = 120 4.235 0C⇒− + = 1

20 4.7234.235

C∴ = =

All the end-moments are to be found as:

1 1FINAL BA BAM M C M C M′ ′= + =


4 723 7 68M × 36 273 kNm=4.723 7.68ABM = ×−

4 723 5 41M = ×−

36.273 kNm= −

25 551 kNm= −4.723 5.41BAM = ×

4 723 5 41M = ×

25.551 kNm

25 551 kNm=4.723 5.41BCM = ×

4.723 3.86CBM = ×

25.551 kNm=

18.231 kNm=.7 3 3.86CB

4.723 3.86CDM = ×− 18.231 kNm= −4.723 3.86CDM

0DCM = 0DCM


SummarySummary


• Slope deflection method-Analysis of continuous beams and


Slope deflection method Analysis of continuous beams and frames (with and without sway)

• Moment distribution method- Analysis of continuous beams and frames (with and without sway)and frames (with and without sway).


136

Module2 rajesh sir

Engineering

Transcript of Module2 rajesh sir