module2 rajesh sir
DESCRIPTION
GCE KannurTRANSCRIPT
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
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Module IIModule II
Displacement method of analysis
• Slope deflection method-Analysis of continuous beams and
Displacement method of analysis
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
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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)
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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)
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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
Dept. of CE, GCE Kannur Dr.RajeshKN
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.
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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
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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
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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
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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
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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−
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Bθve+ Aθ ve−
Sl d fl ti th dSlope deflection method
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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.
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Derivation of fundamental equations
BAMM =BAABM
Aθ Bθ
AFEM( )1
+ABFEM BAFEM( )1
( )2
ABM ′
Aθ′ Bθ′
+( )2
BAM ′A
M ′′ M ′′
+( )3
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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.
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( )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
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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
′ ′′= + = − +
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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
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Fixed end moments
8PL−
8PL
+2L 2L2L 2L
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Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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= = =
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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⇒ = +
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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⎜ ⎟⎝ ⎠
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⎛ ⎞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
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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
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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.
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Example 240kN 80kN60kN
20kN m
A B C D
3m 3m 3m 3m 3m 3m
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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
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Slope deflection equations
2 32AB A B ABEIM FEMl l
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
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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+ = −
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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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
⎛ ⎞= ×− + × +⎜ ⎟⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎛ ⎞= + − = −⎜ ⎟
⎝ ⎠
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Example 3
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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= = = =
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37
Slope deflection equations
2 32AB A B ABEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
( )4 2 108AB A BEIM θ θ⇒ = + −
2 32BA B A BAEIM FEMl l
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⎛ ⎞
2 32CD C D CDEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
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2 32DC D C DCEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
( )6 2 26.6678DC D CEIM θ θ⇒ = + +
Joint equilibrium conditions
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⇒ + =
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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= − − = − − ×−
Dept. of CE, GCE Kannur Dr.RajeshKN
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= ×− + ×− − = −
Dept. of CE, GCE Kannur Dr.RajeshKN41
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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= − × = −
Dept. of CE, GCE Kannur Dr.RajeshKN
Slope deflection equations
2 32AB A B ABEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
( )4 1507.5AB BEIM θ⇒ = −
2 32BA B A BAEIM FEMl l
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
2 32EIM FEMδθ θ⎛ ⎞⎜ ⎟
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⎛ ⎞= + − +⎜ ⎟⎝ ⎠
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44
2 32DC D C DCEIM FEMl l
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 =
Dept. of CE, GCE Kannur Dr.RajeshKN
45
Joint equilibrium conditions
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
Dept. of CE, GCE Kannur Dr.RajeshKN
( )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= − = − =−
Dept. of CE, GCE Kannur Dr.RajeshKN
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= + = ×− + = −
Dept. of CE, GCE Kannur Dr.RajeshKN
( ) ( )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−= × × × =
Dept. of CE, GCE Kannur Dr.RajeshKN
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 =
Dept. of CE, GCE Kannur Dr.RajeshKN
Slope deflection equations
2 32AB A B ABEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
24 3 102 208 8AB A BEIM θ θ
−⎛ ⎞×⇒ = + − −⎜ ⎟
⎝ ⎠
2 32BA B A BAEIM FEMl l
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
⎛ ⎞×⇒ = + +⎜ ⎟
⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
51
Joint equilibrium conditions
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 −⎛ ⎞ ⎛ ⎞⇒ + = − ×+ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
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
−= ×
Dept. of CE, GCE Kannur Dr.RajeshKN
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=
Dept. of CE, GCE Kannur Dr.RajeshKN
54
1.456 10 168 8
= − × − +⎜ ⎟⎝ ⎠
33.21 kNm=
Slope Deflection for frames:No sideswayy
Dept. of CE, GCE Kannur Dr.RajeshKN
Example 6Example 6
B10kN 2kN m
AB C
22I I
332m
3m I
3m3m
D
I
D
Dept. of CE, GCE Kannur Dr.RajeshKN
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= =
Dept. of CE, GCE Kannur Dr.RajeshKN
Slope deflection equations
2 32AB A B ABEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
( )4 7.25AB BEIM θ⇒ = −
2 32BA B A BAEIM FEMl l
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⇒ =
Dept. of CE, GCE Kannur Dr.RajeshKN
582 32DB D B DBEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
( )23DB BEIM θ⇒ =
Joint equilibrium conditions
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 = −
Dept. of CE, GCE Kannur Dr.RajeshKN
( )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
Dept. of CE, GCE Kannur Dr.RajeshKN
60( )2
3DB BEIM θ= ( )2 0.6481 0.432
3kNm= − = −
Slope Deflection for frames:S dSidesway
Dept. of CE, GCE Kannur Dr.RajeshKN
P BAM CDMPB
CBA CD
2l
DH D1l
DCM
AH A
ABMAB
AB BAA
M MH += CD DCM MH +
= 0H H P
Dept. of CE, GCE Kannur Dr.RajeshKN
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 + −=
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
642
D l1A l
Example 7
Dept. of CE, GCE Kannur Dr.RajeshKN
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)
Dept. of CE, GCE Kannur Dr.RajeshKN
66
Slope deflection equations
2 32AB A B ABEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
2 35 5BEI δθ⎛ ⎞= −⎜ ⎟
⎝ ⎠
2 32BA B A BAEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
2 325 5BEI δθ⎛ ⎞= −⎜ ⎟
⎝ ⎠
2 32BC B C BCEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
( )2 2 10.24B CEI θ θ= + −
2 32EIM FEMδθ θ⎛ ⎞⎜ ⎟
BC B C BCl l⎜ ⎟⎝ ⎠
( )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⎛ ⎞
2 32CD C D CDEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
2 3EI δ⎛ ⎞
2 325 5CEI δθ⎛ ⎞= −⎜ ⎟
⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
67
2 32DC D C DCEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
2 35 5CEI δθ⎛ ⎞= −⎜ ⎟
⎝ ⎠
Joint equilibrium conditions
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+ − = −
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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−⎛ ⎞= − × = −⎜ ⎟
Dept. of CE, GCE Kannur Dr.RajeshKN
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)
Dept. of CE, GCE Kannur Dr.RajeshKN
0DCM =
Slope deflection equations
2 32AB A B ABEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
2 34 4AB BEIM δθ⎛ ⎞⇒ = −⎜ ⎟
⎝ ⎠
2 32BA B A BAEIM FEMl l
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 32EIM FEMδθ θ⎛ ⎞⎜ ⎟
BC B C BCl l⎜ ⎟⎝ ⎠
( )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⎛ ⎞
2 32CD C D CDEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
2 3EI δ⎛ ⎞
Dept. of CE, GCE Kannur Dr.RajeshKN
72
2 32DC D C DCEIM FEMl l
δθ θ⎛ ⎞= + − +⎜ ⎟⎝ ⎠
2 324 4DC D CEIM δθ θ⎛ ⎞⇒ = + −⎜ ⎟
⎝ ⎠
Joint equilibrium conditions
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 =
Dept. of CE, GCE Kannur Dr.RajeshKN
( )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 −− =
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎛ ⎞= − = × − × = −⎜ ⎟
Dept. of CE, GCE Kannur Dr.RajeshKN
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+ × + ×⎜ ⎟
⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
76
M t di t ib ti th dMoment distribution method
Dept. of CE, GCE Kannur Dr.RajeshKN
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⎛ ⎞⎜ ⎟
Dept. of CE, GCE Kannur Dr.RajeshKN
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).
Dept. of CE, GCE Kannur Dr.RajeshKN
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)
Dept. of CE, GCE Kannur Dr.RajeshKN
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.
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
5 35kNA B C4.3kNm2.05kNm 4.3kNm 5.35kNm
AB CA C
2.055 35
4.35.35
Dept. of CE, GCE Kannur Dr.RajeshKN
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× ×= + = + =
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
Example 3
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
92
Dept. of CE, GCE Kannur Dr.RajeshKN
93
Example 3
2 25 8l2 25 8 26.66712 12wl kNm×
= =10 8 108 8Pl kNm×
= =
Dept. of CE, GCE Kannur Dr.RajeshKN
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=
+( ) ( )
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
( )( ) ( )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
Dept. of CE, GCE Kannur Dr.RajeshKN
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=
Dept. of CE, GCE Kannur Dr.RajeshKN
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
− −× × × × × × × ×= −
Dept. of CE, GCE Kannur Dr.RajeshKN
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= + =
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
Example 6
Support B settles by 10 mm. 6 4200 , 50 10E GPa I mm= = ×
Dept. of CE, GCE Kannur Dr.RajeshKN
( )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
− −× × × × × × × ×=
Dept. of CE, GCE Kannur Dr.RajeshKN
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= + =
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
Dept. of CE, GCE Kannur Dr.RajeshKN
Moment Distribution for frames:No sideswayy
Dept. of CE, GCE Kannur Dr.RajeshKN
Example 7Example 7
B10kN 2kN m
AB C
22I I
332m
3m I
3m3m
D
I
D
Dept. of CE, GCE Kannur Dr.RajeshKN
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= =
Dept. of CE, GCE Kannur Dr.RajeshKN
Distribution factors
( )( ) ( ) ( )
4 2 54 3 42 5 3 3BA
EIDF
EI EI EI=
+ +
( )( ) ( ) ( )
3 34 3 42 5 3 3BC
EIDF
EI EI EI=
+ +
Distribution factors
( ) ( ) ( )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=
Dept. of CE, GCE Kannur Dr.RajeshKN
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−
= = −
Dept. of CE, GCE Kannur Dr.RajeshKN
Moment Distribution for frames: Moment Distribution for frames: sidesway
Dept. of CE, GCE Kannur Dr.RajeshKN
• 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
Dept. of CE, GCE Kannur Dr.RajeshKN
• 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
Dept. of CE, GCE Kannur Dr.RajeshKN
• 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’
Dept. of CE, GCE Kannur Dr.RajeshKN
RR’
1 0R C R′+ =
Dept. of CE, GCE Kannur Dr.RajeshKN
• 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
Dept. of CE, GCE Kannur Dr.RajeshKN
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
= =′
Dept. of CE, GCE Kannur Dr.RajeshKN
, 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′ ′ ′ ′= =
Dept. of CE, GCE Kannur Dr.RajeshKN
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′ ′ ′
Dept. of CE, GCE Kannur Dr.RajeshKN
, , 0BA AB DCAlso M M M′ ′ ′= =
Example 8
1 0R C R′+ =
R R’
Dept. of CE, GCE Kannur Dr.RajeshKN
( )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
= = =
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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:
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
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∴ = =
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
13.01 2.99
Dept. of CE, GCE Kannur Dr.RajeshKN
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=
+
Dept. of CE, GCE Kannur Dr.RajeshKN
( ) ( )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′∴ = −
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
( )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′ ′= + =
Dept. of CE, GCE Kannur Dr.RajeshKN
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
Dept. of CE, GCE Kannur Dr.RajeshKN
SummarySummary
Displacement method of analysis
• Slope deflection method-Analysis of continuous beams and
Displacement method of analysis
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
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