Slope Deflection by Hibbeler

85
Chapter 11 Displacement Method of Analysis Slope Deflection Method

description

Slope Deflection for beams and frames by Hibbeler.

Transcript of Slope Deflection by Hibbeler

Page 1: Slope Deflection by Hibbeler

Chapter 11

Displacement Method of Analysisp ySlope Deflection Method

Page 2: Slope Deflection by Hibbeler

Displacement Method of Analysis• Two main methods of analyzing indeterminate structure

– Force method • The method of consistent deformations & the equation of three moment 

Th i k f• The primary unknowns are forces or moments

– Displacement methodTh l d fl ti th d & th t di t ib ti th d• The slope‐deflection method & the moment distribution method

• The primary unknown is displacement (slope & deflection)

• It is particularly useful for the analysis of highly statically indeterminate• It is particularly useful for the analysis of highly statically indeterminate structure 

• Easily programmed on a computer & used to analyze a wide range of indeterminate structures   

Page 3: Slope Deflection by Hibbeler

Displacement Method of AnalysisDisplacement Method of Analysis

• Degree of Freedom– The number of possible joint rotations & independent joint 

translations in a structure is called the degree of freedom of the structure

– In three dimensions each node on a frame can have at most three linear displacements & three rotational displacements.

– In two dimension each node can have at most two linear displacements & one rotational displacement.

DOF = n J– R

‐ n number of possible joint’s movements

‐ J number of jointsj

‐ R number of restrained  movements

Page 4: Slope Deflection by Hibbeler

Displacement Method of AnalysisDisplacement Method of Analysis

• Degree of Freedom

DOF = n J– R‐ n number of possible joint’s movements

‐ n = 2 for two dimensional truss structures

‐ n = 2 for three dimensional truss structures

‐ n= 3 for two dimensional frame structures

‐ n= 6 for three dimensional frame structures

‐ J number of joints

‐ R number of restrained  movements

• Neglecting Axial deformationeg ect g a de o at o

DOF = n J– R – m

‐ m number of members

Page 5: Slope Deflection by Hibbeler

Displacement Method of AnalysisDisplacement Method of Analysis

Page 6: Slope Deflection by Hibbeler

Slope‐Deflection EquationsSlope Deflection Equations

•DerivationDerivation 

Page 7: Slope Deflection by Hibbeler

Slope‐Deflection Equations

A l Di l A θ• Angular Displacement at A, θA

1 1 2 0.2 3 2 3

AB BAM L M LL LEI EI

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦' 0.AM =∑

1 1 2 0.2 3 2 3

BA ABA

M L M LL L LEI EI

θ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞− + =⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦' 0.BM =∑

4AB A

EIML

θ=

2EI2BA A

EIML

θ=

Page 8: Slope Deflection by Hibbeler

Slope‐Deflection Equations

A l Di l B θ• Angular Displacement at B, θB

4BA B

EIML

θ=

2AB B

EIML

θ=

L

Page 9: Slope Deflection by Hibbeler

Slope‐Deflection EquationsR l i Li Di l ∆• Relative Linear Displacement ∆

1 2 1 0.2 3 2 3

M L M LL LEI EI

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞− + ∆ =⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦' 0.BM =∑ 2 3 2 3EI EI⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

6AB BA

EIM M M −= = = ∆2AB BAM M M

L∆

Page 10: Slope Deflection by Hibbeler

Slope‐Deflection EquationsFi d E d M• Fixed‐End Moment 

1 12 0PL ML L⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥0.yF =∑ 2 0.

2 2L L

EI EI⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

PLFixed-End

8PLM =moment

Page 11: Slope Deflection by Hibbeler

Slope‐Deflection Equationsp q

( )M FEM( )AB ABM FEM=

( )BA BAM FEM=

Page 12: Slope Deflection by Hibbeler

Slope‐Deflection Equations

• Fixed‐End Moment 

Page 13: Slope Deflection by Hibbeler

Slope‐Deflection Equations

Sl d fl i i• Slope‐deflection equations– The resultant moment (adding all equations together)

I ⎡ ⎤∆⎛ ⎞ ⎛ ⎞2 2 3 ( )AB A B ABIM E FEML L

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

2 2 3 ( )BA B A BAIM E FEML L

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

• Lets represent the member stiffness as k = I/L &

• The span ration due to displacement as ψ = ∆/L

L L⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

p p ψ /• Referring to one end of the span as near end (N) & the other end as the far end (F).

Rewrite the equations– Rewrite the equations 

[ ]2 2 3 ( )N N F NM EK FEMθ θ ψ= + − +

[ ]2 2 3 ( )F F N FM EK FEMθ θ ψ= + − +

Page 14: Slope Deflection by Hibbeler

Slope‐Deflection Equations

Sl d fl i i f i d E d S• Slope‐deflection equations for pin supported End Span– If the far end is a pin or a roller support

[ ]2 2 3 ( )N N F NM EK FEMθ θ ψ= + − +

[ ]0 2 2 3 0F NEK θ θ ψ= + − +

– Multiply the first equation by 2 and subtracting the second equation f itfrom it

[ ]3 ( )N N NM EK FEMθ ψ= − +

Page 15: Slope Deflection by Hibbeler
Page 16: Slope Deflection by Hibbeler

Slope‐Deflection EquationsSlope Deflection Equations• MN, MF = the internal moment in the near & far end of the span. 

Considered positive when acting in a clockwise direction

E k d l f l ti it f t i l & tiff k I/L• E, k = modulus of elasticity of material & span stiffness k = I/L

• θN, θF = near & far end slope of the span at the supports in radians. Considered positive when acting in a clockwise direction

• ψ = span ration due to a linear displacement ∆ / L. if the right end of   a member sinks with respect the the left end the sign is positive

• Steps to analyzing beams using this method– Fined the fixed end moments of each span (both ends left & right)

– Apply the slope deflection equation on each span & identify the unknowns 

– Write down the joint equilibrium equationsWrite down the joint equilibrium equations

– Solve the equilibrium equations to get the unknown rotation & deflections

– Determine the end moments and then treat each span as simply supported beam subjected to given load & end moments so you can workout the reactions & drawsubjected to given load & end moments so you can workout the reactions & draw the bending moment & shear force diagram

Page 17: Slope Deflection by Hibbeler

E l 1

Example 1

• Example 1– Draw the bending moment & shear force diagram. 

4Fi d E d M t

2 .BAFEM t m=

22 6

4t2t/m

Fixed End Moment4 4 2 .

8ABFEM t m×= − = −

A B C2II22 6 6 .12BCFEM t m×

= − = −4m 6m

6 .CBFEM t m=

Slope Deflection Equations

A B C2II

θ 0[ ]2 2 0 2

4AB A BIM E θ θ⎛ ⎞= + − −⎜ ⎟

⎝ ⎠0.5 2AB BM EIθ= −

[ ]2 2 0 2BA A BIM E θ θ⎛ ⎞= + − +⎜ ⎟ 2BA BM EIθ= +

θA= 0.

θA = 0.[ ]2 2 0 2

4BA A BM E θ θ+ +⎜ ⎟⎝ ⎠

BA Bθ

[ ]22 2 0 66BC B CIM E θ θ⎛ ⎞= + − −⎜ ⎟

⎝ ⎠

4 63BC BM EIθ= −

θC = 0.

[ ]22 2 0 66CB B CIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠

2 63CB BM EIθ= +

θC = 0.

Page 18: Slope Deflection by Hibbeler

Joint Equilibrium Equations

0BA BCM M+ =

Joint Equilibrium Equations

1 74

Joint B

1.7B EI

θ =

1 7

42 6 0.3B BEI EIθ θ+ + − =

Substituting in slope deflection equations 1.70.5 2 1.15 .ABM EI t mEI

= − = −

1.7 2 3.7 .BAM EI t mEI

= + =EI

3.7 .BCM t m= −

2 1.7 6 7.1 .CBM EI t m= + =4t

6 7.1 .3CBM EI t m

EI+

Computing The Reactions

R

1.15

R

3.7

1.15 4 2 3.7 1 36R t+ × −= = RA RB1

1.364AR t= =

1 4 1.36 2.64BR t= − =

Page 19: Slope Deflection by Hibbeler

2 6 3 3 7 7 1 2t/2

2 6 3 3.7 7.1 5.436BR t× × + −

= =

2 6 5.43 6.57CR t= × − =3.7

2t/m7.1

RB2 RC5.43

1.36 ++

6 572.64 --

+

6.57

7.13.7

1.15

++---

3.791.57 +

Page 20: Slope Deflection by Hibbeler

• Example 1: Determine the internal moments in the beam at the supports

Example 2Example 1: Determine the internal moments in the beam at the supports. 

Fixed End Moment2

2

60 4 2 26.67 .6ABFEM kN m× ×

= − = −2

2

60 2 4 53.33 .6BAFEM kN m× ×

= =230 6 135 .BCFEM kN m×

= − = −

660kN

30kN/m

A C

4m

8BC

θ 0Slope Deflection Equations

6m 6m

A B CII

θA= 0.

θA= 0.[ ]2 2 0 26.67

6AB A BIM E θ θ⎛ ⎞= + − −⎜ ⎟

⎝ ⎠

1 26.673AB BM EIθ= −

[ ]2 2 0 53.33BA A BIM E θ θ⎛ ⎞= + − +⎜ ⎟

2 53.33BA BM EIθ= +[ ]2 2 0 53.336BA A BM E θ θ+ +⎜ ⎟

⎝ ⎠53.33

3BA BM EIθ +

[ ]3 0 1356BC BIM E θ⎛ ⎞= − −⎜ ⎟

⎝ ⎠

1 1352BC BM EIθ= −

Page 21: Slope Deflection by Hibbeler

Joint Equilibrium Equations

0BA BCM M+ =

Joint Equilibrium Equations

7

Joint B

2 1 7 81.676 BEIθ =2 153.33 135 0

3 2B BEI EIθ θ+ + − =

70BEIθ =

1 (70) 26.67 3.33 .ABM kN m= − = −

Substituting in slope deflection equations

(70) 26.67 3.33 .3ABM kN m

2 (70) 53.33 100 .3BAM kN m= + =

1 (70) 135 100 .2BCM kN m= − = −

Page 22: Slope Deflection by Hibbeler

Example 3E l 2• Example 2: Determine the internal moments in the beam at the supports

– Support A; downward movement of 0.3cm & clockwise rotation of 0.001 rad. Support B; downward movement of 1.2cm. Support C downward movement of 0.6cm. EI = 5000 t.m2

A B C

Fixed End Moment0.AB BA BC CBFEM FEM FEM FEM= = = =

1 2 0 3− 6m 6m

A B C

II

AB BA BC CB

Displacements 0.001Aθ = +1.2 0.3 0.0015

600ABψ = =

0.6 1.2 0.001600BCψ −

= = −

Slope Deflection Equations

[ ]2 5000 2 0.001 3 0.00156AB BM θ×⎛ ⎞= × + − ×⎜ ⎟

⎝ ⎠⎝ ⎠

[ ]2 5000 2 0.001 3 0.00156BA BM θ×⎛ ⎞= + − ×⎜ ⎟

⎝ ⎠

Page 23: Slope Deflection by Hibbeler

⎛ ⎞[ ]3 5000 0.0016BC BM θ×⎛ ⎞= +⎜ ⎟

⎝ ⎠

Joint Equilibrium EquationsJoint B

0BA BCM M+ =

7 0.004 0Bθ − =

0.00057B radθ =

Page 24: Slope Deflection by Hibbeler

Substituting in slope deflection equations

[ ]2 5000 2 0.001 0.00057 3 0.0015 3.22 .6ABM t m×⎛ ⎞= × + − × = −⎜ ⎟

⎝ ⎠

[ ]2 5000 2 0.00057 0.001 3 0.0015 3.93 .6BAM t m×⎛ ⎞= × + − × = −⎜ ⎟

⎝ ⎠

[ ]3 5000 0 00057 0 001 3 93M t m×⎛ ⎞= + =⎜ ⎟[ ]0.00057 0.001 3.93 .6BCM t m= + =⎜ ⎟

⎝ ⎠

Page 25: Slope Deflection by Hibbeler

Example 1

0.0AB BAFEM FEM= =

0 0θ θ= = 0.0A Cθ θ= =

Page 26: Slope Deflection by Hibbeler

Slope Deflection Equationsp q

Page 27: Slope Deflection by Hibbeler

E ilib i E tiEquilibrium Equations

Page 28: Slope Deflection by Hibbeler

Example 2

Page 29: Slope Deflection by Hibbeler
Page 30: Slope Deflection by Hibbeler
Page 31: Slope Deflection by Hibbeler

Example 3

Page 32: Slope Deflection by Hibbeler
Page 33: Slope Deflection by Hibbeler

Example 4

Page 34: Slope Deflection by Hibbeler
Page 35: Slope Deflection by Hibbeler
Page 36: Slope Deflection by Hibbeler

• Analysis of Frames Without Sway– The side movement of the end of a column in a frame is called SWAY.

Page 37: Slope Deflection by Hibbeler

D t i th t t h j i t f th f EI i t tExample 5

– Determine the moment at each joint of the frame. EI is constant

Fixed End Moment0.AB BAFEM FEM= =

25 24 8 80FEM kN m× ×= = 80 .

96BCFEM kN m= − = −

25 24 8 80 .96CBFEM kN m× ×

= =

0.CD DCFEM FEM= =

θ = 0I⎛ ⎞Slope Deflection Equations

θA= 0.[ ]2 0. 0. 012AB BIM E θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠0.1667AB BM EIθ=

θA= 0[ ]I⎛ ⎞ θA 0.[ ]2 2 0 0. 012BA BIM E θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠0.333BA BM EIθ=

Page 38: Slope Deflection by Hibbeler

[ ]I⎛ ⎞[ ]2 2 0 808BC B CIM E θ θ⎛ ⎞= + − −⎜ ⎟

⎝ ⎠0.5 0.25 80BC B CM EI EIθ θ= + −

[ ]2 2 0 808CB C BIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠0.5 0.25 80CB C BM EI EIθ θ= + +

8⎝ ⎠θD= 0.[ ]2 2 0 0. 0

12CD CIM E θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠0.333CD CM EIθ=

θ = 0I⎛ ⎞ θD= 0.[ ]2 0 0. 012DC CIM E θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠0.1667DC CM EIθ=

Joint Equilibrium Equations

0BA BCM M+ =

Joint B

0 333 0 5 0 25 80 0EI EI EIθ θ θ+ + 0 833 0 25 80EI EIθ θ+ =

0 333 0 5 0 25 80 0EI EI EIθ θ θ

0.333 0.5 0.25 80 0.B B CEI EI EIθ θ θ+ + − = 0.833 0.25 80B CEI EIθ θ+ =

0CB CDM M+ =

Joint C

0 833 0 25 80EI EIθ θ0.333 0.5 0.25 80 0.C C BEI EI EIθ θ θ+ + + = 0.833 0.25 80C BEI EIθ θ+ = −

Page 39: Slope Deflection by Hibbeler

Substituting in slope deflection equations

Two equation & two unknown 137.1

B C EIθ θ= − =

22.9 .45.7 .

AB

BA

M kN mM kN m

==

Substituting in slope deflection equations 45.7− 45.7−

45.7 .45.7 .

45.7 .45 7

BA

BC

CB

M kN mM kN mM kN mM kN m

= −

==

82.345.7 .22.9 .

CD

DC

M kN mM kN m

= −

= −

22.9 22.9

Page 40: Slope Deflection by Hibbeler

Example 6

Page 41: Slope Deflection by Hibbeler
Page 42: Slope Deflection by Hibbeler
Page 43: Slope Deflection by Hibbeler
Page 44: Slope Deflection by Hibbeler

Example 6b

– Draw the bending moment diagram

30kN20kNk /Fixed End Moment

2I IIA

B C D

48kN/m

220 1 2× ×

Fixed End Moment2

2

20 2 1 4.44 .3ABFEM kN m× ×

= − = −

2II

EF

2

20 1 2 8.89 .3BAFEM kN m× ×

= =

248 4 64 .12BCFEM kN m×

= − = −

4m3m

1m

F1264 .CBFEM kN m=

0BE EB CF FCFEM FEM FEM FEM= = = =4m 2m3m

30 2 60 .CDM kN m= − × = −

θ 0Slope Deflection Equations

⎛ ⎞ θA= 0.[ ]2 2 0 4.44

3AB A BIM E θ θ⎛ ⎞= + − −⎜ ⎟

⎝ ⎠

2 4.443AB BM EIθ= −

Page 45: Slope Deflection by Hibbeler

θ 0⎛ ⎞ 4θA= 0.[ ]2 2 0 8.89

3BA A BIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠

4 8.893BA BM EIθ= +

2I⎛ ⎞[ ]22 2 0 644BC B CIM E θ θ⎛ ⎞= + − −⎜ ⎟

⎝ ⎠2 64BC B CM EI EIθ θ= + −

[ ]22 2 0 64CB B CIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠2 64CB B CM EI EIθ θ= + +[ ]

4CB B C⎜ ⎟⎝ ⎠

CB B C

[ ]2 2 0 0BE B EIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠

43BE BM EIθ=

θE= 0.[ ]

3BE B E⎜ ⎟⎝ ⎠ 3BE B

[ ]2 2 0 03EB E BIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠

23EB BM EIθ=

θE= 0.

[ ]22 2 0 04CF C FIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠2CF CM EIθ=

θF= 0.

⎛ ⎞ θ 0[ ]22 2 0 0

4FC F CIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠FC CM EIθ=

θF= 0.

Page 46: Slope Deflection by Hibbeler

Equilibrium EquationsEquilibrium Equations

0BA BC BEM M M+ + =

Joint B

4.67 55.11B CEI EIθ θ+ =4 48.89 2 64 03 3B B C BEI EI EI EIθ θ θ θ+ + + − + =

Joint C0CB CF CDM M M+ + =

4 4B CEI EIθ θ+ = −2 64 2 60 0B C CEI EI EIθ θ θ+ + + − =

Two equation & two unknown 4.18CEIθ = −12.70BEIθ =

S b tit ti i l d fl ti tiSubstituting in slope deflection equations 2 12.7 4.44 4.03 .3ABM kN m= × − = 2 12.7 4.18 64 42.77 .BCM kN m= × − − = −

4 12.70 8.89 25.833BAM kNm= × + = 12.7 2 ( 4.18) 64 68.34 .CBM kN m= + × − + =

Page 47: Slope Deflection by Hibbeler

4 12 70 16 94M kN 2 ( 4 18) 8 35M EI kN

68 3

12.70 16.94 .3BEM kN m= × =

2 12.70 8.47 .3EBM kN m= × =

2 ( 4.18) 8.35 .CFM EI kN m= − = −

4.18 .FCM kN m= −

6068.35

42.77

25.83

8.3516.944.03

8.47

7.1

4.18

Page 48: Slope Deflection by Hibbeler

Slope Deflection (Frame with Sway)Slope Deflection (Frame with Sway) • Analysis of Frames with Sway

Page 49: Slope Deflection by Hibbeler

D th b di tExample 7 

Fixed End Moment

– Draw the bending moment diagram. EI constant

Fixed End MomentAs there is no span loading in any of the member FEM for all the members is zero

2 0 3 0IM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎢ ⎥

Slope Deflection Equations

members is zero

2 0. 3 012 12AB BM E θ⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

1 16 24BEI EIθ= − ∆

2 2 0. 3 012 12BA BIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

1 11 13 24BEI EIθ= − ∆

Page 50: Slope Deflection by Hibbeler

⎡ ⎤0.2 2 3 015 15BC B CIM E θ θ⎡ ⎤⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

4 215 15B CEI EIθ θ= +

[ ]2 2 0 0CB C BIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠

2 4B CEI EIθ θ= +[ ]

15CB C B⎜ ⎟⎝ ⎠ 15 15B C

2 2 0 3 0.18 18CD CIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

2 19 54CEI EIθ= − ∆

18 18⎝ ⎠ ⎝ ⎠⎣ ⎦

2 0 3 018 18DC CIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

9 541 19 54CEI EIθ= − ∆

Equilibrium Equations

0BA BCM M+ =Joint B ΣMB = 0

1 1 4 2 0.3 24 15 15B B CEI EI EI EIθ θ θ− ∆ + + =

14.4 3.2 0B CEI EI EIθ θ− ∆ + + = 1

0CB CDM M+ =Joint C ΣMC = 0

Page 51: Slope Deflection by Hibbeler

2 4 2 12 4 2 1 0.15 15 9 54B C CEI EI EI EIθ θ θ+ + − ∆ =

7.2 26.4 0B CEI EI EIθ θ− ∆ + + = 2Three unknown & just two equations so we need another equilibrium equation. Let take ΣFx = 0.40 0.A DH H− − =

12BA AB

AM MH +

= −

CD DCM MH +18

CD DCDH = −

( ) ( )40 18 1.5 0.BA AB CD DCM M M M× + + + + =

1 1 1 1⎛ ⎞1 1 1 1720 1.53 24 6 24B BEI EI EI EIθ θ⎛ ⎞+ − ∆ + − ∆⎜ ⎟

⎝ ⎠2 1 1 1 0.9 54 9 54C CEI EI EI EIθ θ⎛ ⎞+ − ∆ + − ∆ =⎜ ⎟

⎝ ⎠9 54 9 54⎜ ⎟⎝ ⎠

0.162 0.75 0.333 720B CEI EI EIθ θ− ∆ + + = − 3

Page 52: Slope Deflection by Hibbeler

Now solve the three equationNow solve the three equation 1 14.4 3.2 01 7.2 26.4 0

0 162 0 75 0 333 720B

EIEIEIθθ

− ∆⎧ ⎫⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥− =⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎩ ⎭⎩ ⎭

136.2CEIθ =

438.9BEIθ =0.162 0.75 0.333 720CEIθ⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎣ ⎦ ⎩ ⎭⎩ ⎭ 6756.6EI ∆ =

Substituting in slope deflection equations

208 .ABM kN m= −

135 .BAM kN m= −

135 .BCM kN m=

95M kN m= 95 .CBM kN m=

95 .CDM kN m= −

110 .DCM kN m= −

Page 53: Slope Deflection by Hibbeler

-+ +

135 95

95

-+ +135

-

+208

+110

Page 54: Slope Deflection by Hibbeler

h b dExample 7 

Fixed End Moment

– Draw the bending moment diagram. EI constant

220 15 375FEM kN m×= =

Fixed End Moment0.AB BAFEM FEM= =

375 .12BCFEM kN m= − = −

220 15 375 .12BCFEM kN m×

= =

0.CD DCFEM FEM= =

I ⎡ ⎤∆⎛ ⎞ ⎛ ⎞Slope Deflection Equations

2 0. 3 012 12AB BIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

1 16 24BEI EIθ= − ∆6 24

Page 55: Slope Deflection by Hibbeler

I ⎡ ⎤∆⎛ ⎞ ⎛ ⎞ 1 12 2 0. 3 012 12BA BIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

1 13 24BEI EIθ= − ∆

0.2 2 3 375IM E θ θ⎡ ⎤⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎢ ⎥4 2 375EI EIθ θ+2 2 3 375

15 15BC B CM E θ θ= + − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦375

15 15B CEI EIθ θ= + −

[ ]2 2 0 37515CB C BIM E θ θ⎛ ⎞= + − +⎜ ⎟

⎝ ⎠

2 4 37515 15B CEI EIθ θ= + +

2 2 0 3 0.18 18CD CIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞

2 19 54CEI EIθ= − ∆

2 0 3 018 18DC CIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Equilibrium Equations

1 19 54CEI EIθ= − ∆

Equilibrium Equations

0BA BCM M+ =Joint B ΣMB = 0

1 1 4 2 375 0EI EI EI EIθ θ θ∆ + + =375 0.3 24 15 15B B CEI EI EI EIθ θ θ− ∆ + + − =

14.4 3.2 9000B CEI EI EIθ θ− ∆ + + = 1

Page 56: Slope Deflection by Hibbeler

Joint C ΣMC = 00CB CDM M+ =

Joint C ΣMC 0

2 4 2 1375 0.15 15 9 54B C CEI EI EI EIθ θ θ+ + + − ∆ =

Three unknown & just two equations so we need another equilibrium equation. Let take ΣFx = 0.

7.2 26.4 20250B CEI EI EIθ θ− ∆ + + = − 2

40 0.A DH H− − =

12BA AB

AM MH +

= −

18CD DC

DM MH +

= −

( ) ( )40 18 1.5 0.BA AB CD DCM M M M× + + + + =

1 1 1 1720 1.53 24 6 24B BEI EI EI EIθ θ⎛ ⎞+ − ∆ + − ∆⎜ ⎟

⎝ ⎠2 1 1 1 0.C CEI EI EI EIθ θ⎛ ⎞+ − ∆ + − ∆ =⎜ ⎟ 0.9 54 9 54C CEI EI EI EIθ θ+ ∆ + ∆⎜ ⎟

⎝ ⎠0.162 0.75 0.333 720B CEI EI EIθ θ− ∆ + + = − 3

Page 57: Slope Deflection by Hibbeler

Now solve the three equationNow solve the three equation 1 14.4 3.2 90001 7.2 26.4 20250

0 162 0 75 0 333 720B

EIEIEIθθ

− ∆⎧ ⎫⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥− = −⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎩ ⎭⎩ ⎭

R1-R2 0.162R1-R3

1 14.4 3.2 90000 7.2 23.2 29250B

EIEIθ

− ∆⎧ ⎫⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥− =⎨ ⎬ ⎨ ⎬⎢ ⎥

0.162 0.75 0.333 720CEIθ⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎣ ⎦ ⎩ ⎭⎩ ⎭

0.22R2-R3

0 1.58 0.185 2178B

CEIθ⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎣ ⎦ ⎩ ⎭⎩ ⎭

1 14.4 3.2 9000EI− ∆⎧ ⎫⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬

4257 804 7EIθ0. 7.2 23.2 292500. 0. 5.29 4257

B

C

EIEIθθ

⎪ ⎪ ⎪ ⎪⎢ ⎥− =⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎣ ⎦ ⎩ ⎭⎩ ⎭

4257 804.75.29CEIθ = = −−1469.6BEIθ =

9587.2EI ∆ =S b i i i l d fl i i 9587.2EI ∆Substituting in slope deflection equations 1 11467.2 9587.2 155 .6 24ABM kN m= − = −

1 11 11469.6 9587.2 90 .3 24BAM kN m= − =

Page 58: Slope Deflection by Hibbeler

4 24 21469.6 ( 804.7) 375 90 .15 15BCM kN m= + − − = −

2 41469.6 ( 804.7) 375 356 .15 15CBM kN m= + − + =

2 1( 804.7) 9587.2 356 .9 54CDM kN m= − − = − 356

1 1( 804.7) 9587.2 267 .9 54DCM kN m= − − = −

90356

-90

356-

155

-

267

Page 59: Slope Deflection by Hibbeler

Example 8

Page 60: Slope Deflection by Hibbeler
Page 61: Slope Deflection by Hibbeler
Page 62: Slope Deflection by Hibbeler
Page 63: Slope Deflection by Hibbeler

Example 9 – Draw the bending moment diagram. EI constant

Fixed End MomentFixed End MomentAs there is no span loading in any of the member FEM for all the members is zero

12 0. 3 05 5AB BIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Slope Deflection Equations

5 5⎝ ⎠ ⎝ ⎠⎣ ⎦

12 65 25BEI EIθ= − ∆

⎡ ⎤⎛ ⎞ ⎛ ⎞12 2 0 3 05 5BA BIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

14 65 25BEI EIθ= − ∆5 25

Page 64: Slope Deflection by Hibbeler

0I ⎡ ⎤⎛ ⎞ ⎛ ⎞ 4 202 2 3 07 7BE B EIM E θ θ⎡ ⎤⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

4 27 7B EEI EIθ θ= +

02 2 3 07 7EB E BIM E θ θ⎡ ⎤⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

2 47 7B EEI EIθ θ= +

7 7EB E B⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ 7 7

12 0. 3 05 5FE EIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

12 65 25EEI EIθ= − ∆

12 2 0 3 05 5EF EIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

14 65 25EEI EIθ= − ∆

I ⎡ ⎤∆⎛ ⎞ ⎛ ⎞ 4 2 622 2 3 05 5BC B CIM E θ θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

24 2 65 5 25B CEI EI EIθ θ= + − ∆

22 2 3 05 5CB C BIM E θ θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

22 4 65 5 25B CEI EI EIθ θ= + − ∆

5 5⎝ ⎠ ⎝ ⎠⎣ ⎦ 5 5 25

02 2 3 07 7CD C DIM E θ θ⎡ ⎤⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

4 27 7C DEI EIθ θ= +

⎣ ⎦02 2 3 0

7 7DC D CIM E θ θ⎡ ⎤⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

2 47 7C DEI EIθ θ= +

Page 65: Slope Deflection by Hibbeler

I ⎡ ⎤∆⎛ ⎞ ⎛ ⎞ 4 2 622 2 3 05 5DE D EIM E θ θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

24 2 65 5 25D EEI EI EIθ θ= + − ∆

22 2 3 05 5ED E DIM E θ θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

22 4 65 5 25D EEI EI EIθ θ= + − ∆

5 5⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ 5 5 25

Equilibrium EquationsJoint B ΣMB = 0

0BA BC BEM M M+ + =B

1 24 6 4 2 6 4 2 05 25 5 5 25 7 7B B C B EEI EI EI EI EI EI EIθ θ θ θ θ− ∆ + + − ∆ + + =5 25 5 5 25 7 7

1 2380 70 50 42 42 0B C EEI EI EI EI EIθ θ θ+ + − ∆ − ∆ = 1

Joint E ΣME = 00EB ED EFM M M+ + =

Joint E ΣME 0

2 12 4 2 4 6 4 6 07 7 5 5 25 5 25B E D E EEI EI EI EI EI EI EIθ θ θ θ θ+ + + − ∆ + − ∆ =7 7 5 5 25 5 25

1 250 70 380 42 42 0B D EEI EI EI EI EIθ θ θ+ + − ∆ − ∆ = 2

Page 66: Slope Deflection by Hibbeler

Joint C ΣM 00CB CDM M+ =

Joint C ΣMC= 0

22 4 6 4 2 05 5 25 7 7B C C DEI EI EI EI EIθ θ θ θ+ − ∆ + + =5 5 25 7 7

270 240 50 42 0B C DEI EI EI EIθ θ θ+ + − ∆ = 3

Joint D ΣMD = 00DC DEM M+ =

Joint D ΣMD 0

22 4 4 2 6 07 7 5 5 25C D D EEI EI EI EI EIθ θ θ θ+ + + − ∆ =

250 240 70 42 0C D EEI EI EI EIθ θ θ+ + − ∆ = 4

40 0H H

Top story ΣFX = 0H H40 0B EH H− − =

5CB BC

BM MH +

= −

M M+

HB HE

5DE ED

EM MH +

= −

Page 67: Slope Deflection by Hibbeler

2 4 6 4 2 62 2

2 4 6 4 2 62005 5 25 5 5 25B C B CEI EI EI EI EI EIθ θ θ θ+ + − ∆ + + − ∆

2 24 2 6 2 4 6 05 5 25 5 5 25D E D EEI EI EI EI EI EIθ θ θ θ+ + − ∆ + + − ∆ =

B tt t ΣF 0

26 6 6 6 4.8 1000B C D EEI EI EI EI EIθ θ θ θ+ + + − ∆ = − 5

40 80 0A FH H+ − − =

Bottom story ΣFX = 0

BA ABA

M MH += −

5AH

5EF FE

FM MH +

= −

4 6 2 6600 EI EI EI EIθ θ∆ ∆1 16005 25 5 25B BEI EI EI EIθ θ+ − ∆ + − ∆

1 14 6 2 6 05 25 5 25E EEI EI EI EIθ θ+ − ∆ + − ∆ =

HA HF130 30 24 15000B EEI EI EIθ θ+ − ∆ = − 6

Page 68: Slope Deflection by Hibbeler

6 k d 6 ti6 unknown and 6 equation

380 70 0 50 42 42 050 0 70 380 42 42 0

B

C

EIEIθθ

− − ⎧ ⎫⎡ ⎤ ⎧ ⎫⎪ ⎪⎢ ⎥ ⎪ ⎪− − ⎪ ⎪⎢ ⎥ ⎪ ⎪

171.79BEIθ =

79.80CEIθ =

70 240 50 0 0 42 00 50 240 70 0 42 06 6 6 6 0 4 8 1000

C

D

E

EIEIEI

θθ

⎪ ⎪⎢ ⎥ ⎪ ⎪⎪ ⎪⎢ ⎥ ⎪ ⎪−

=⎨ ⎬ ⎨ ⎬⎢ ⎥− ⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ∆− −

79.80DEIθ =

171.79EEIθ =

1054 46EI ∆ =1

2

6 6 6 6 0 4.8 100030 0 0 30 24 0 15000

EIEI⎪ ⎪ ⎪ ⎪⎢ ⎥ ∆⎪ ⎪ ⎪ ⎪⎢ ⎥ ∆− −⎣ ⎦ ⎩ ⎭⎩ ⎭

1 1054.46EI ∆ =

2 837.29EI ∆ =

Substituting in slope deflection equations

184.4 .ABM kN m= −

115.6 .BAM kN m= −

147.2 .BEM kN m=

147.2 .EBM kN m=

184.4 .FEM kN m= −

115.6 .EFM kN m= −

31.6 .BCM kN m= −

68 4M kN m= −

68.4 .CDM kN m=

68 4M kN m=

68.4 .DEM kN m= −

31 6M kN m= −68.4 .CBM kN m= − 68.4 .DCM kN m= 31.6 .EDM kN m= −

Page 69: Slope Deflection by Hibbeler

--+

+

68.4

68.4

68.468.4

+

147.2-

-+

+

+

-

147 2

31.631.6

115.6115.6

- +

147.2

+

184.4 184.4

Page 70: Slope Deflection by Hibbeler

Example 10– Draw the bending moment diagram. EI constant

Degree of freedomDOF = 3x4-6-3 = 3

That means we got three unknown & we need three equations

Before we start let us discus the relative displacement (∆) of each span

Page 71: Slope Deflection by Hibbeler

∆CDThe relative displacement (∆) for span AB is equal ∆AB – 0. = ∆AB (clockwise)

The relative displacement (∆) for span BC is

B C

∆AB

The relative displacement (∆) for span BC is equal 0. – ∆BC = –∆BC (counterclockwise)

The relative displacement (∆) for span CD

∆BC

A 60o

is equal 0. – (–∆CD)= ∆CD (clockwise)

take ∆ ∆Let us build a relationship between ∆AB, ∆BC & ∆CD

D

take ∆AB = ∆∆BC = ∆AB × sin30 = 0.5∆∆CD = ∆AB × cos30 = 0.866∆

So in the slope deflection equations we will use;∆ as the relative displacement of span AB.

∆∆BC

∆CD60o

Bθ =30o

–0.5∆ as the relative displacement of span BC. 0.866∆ as the relative displacement of span CD.

∆AB60o

Page 72: Slope Deflection by Hibbeler

Fixed End Moment

22 4 2 667FEM t m×= − = −

Fixed End Moment0.AB BAFEM FEM= =

2.667 .12BCFEM t m= =

2.667.CBFEM m=

0.CD DCFEM FEM= =CD DC

2 0. 3 03 3AB BIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Slope Deflection Equations2 23 3BEI EIθ= − ∆

3 3AB B⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ 3 3B

2 2 0 3 03 3BA BIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

4 23 3BEI EIθ= − ∆

0.52 2 3 2.6674 4BC B CIM E θ θ⎡ ⎤− ∆⎛ ⎞ ⎛ ⎞= + − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

1 3 2.6672 16B CEI EI EIθ θ= + + ∆ −

0.52 2 3 2 667IM E θ θ⎡ ⎤− ∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥1 3 2 667EI EI EIθ θ= + + ∆ +2 2 3 2.667

4 4CB B CM E θ θ= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦2.667

2 16B CEI EI EIθ θ= + + ∆ +

Page 73: Slope Deflection by Hibbeler

⎡ ⎤⎛ ⎞ ⎛ ⎞0.8662 2 0 3 06 6CD CIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

2 0.8663 6CEI EIθ= − ∆

0.8662 0 3 06 6DC CIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

1 0.8663 6CEI EIθ= − ∆

6 6DC C⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ 3 6

Equilibrium EquationsJoint B ΣMB = 0

0BA BCM M+ =B

4 2 1 3 2.667 03 3 2 16B B CEI EI EI EI EIθ θ θ− ∆ + + + ∆ − =3 3 2 16112 24 23 128B CEI EI EIθ θ+ − ∆ = 1

Joint C ΣMC = 00CB CDM M+ =

1 3 2 0.8662.667 02 16 3 6B C CEI EI EI EI EIθ θ θ+ + ∆ + + − ∆ =

24 80 2.072 128B CEI EI EIθ θ+ + ∆ = − 2

Page 74: Slope Deflection by Hibbeler

Third Eq ilibri m Eq ations (Method 1)Third Equilibrium Equations (Method 1)

6.92m6.92m8.08.0

( ) ( )

( )

11 12.923 6

8 2 0.0

DC CDAB BAAB DC

M MM MM M ++ ⎛ ⎞⎛ ⎞+ − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− =

21 22 12.92 11.92 96 0AB BA CD DCM M M M− − − − − =3 03 0 2 02 0 2 02 0

( )

34.5 3.1 38.375 144B CEI EI EIθ θ− − ∆ = − 3

C C

6.0m6.0m

3.03.0 2.02.0 2.02.0

3.03.0

6.06.0

Page 75: Slope Deflection by Hibbeler

Third equation ΣF = 0Third Equilibrium Equations (Method 2)

0A DH H− − =Third equation ΣFX = 0

From the free body diagram for column CDM M+

CCDM

BBAM

6CD DC

DM MH +

= −

Free body diagram for column AB AHA 1.5m

2.6m

D HD

ABM

VA

( )2.6 1.5 0A BA AB AH M M V× + + + × =

Free body diagram for Beam BC

DCM

CBCM 2t/m

( )4 2 4 2 0B BC CBV M M× + + − × × =

( ) 44

BC CBA B

M MV V

+= = − +

VBCBM

B( ) ( )1.5 42.6 2.6 4

BC CBBA ABA

M MM MH

⎡ ⎤+− +∴ = − − +⎢ ⎥

⎣ ⎦

⎡ ⎤( ) ( )1.5 4 0.2.6 2.6 4 6

BC CBBA AB CD DCM MM M M M⎡ ⎤++ ++ − + + =⎢ ⎥

⎣ ⎦

Page 76: Slope Deflection by Hibbeler

( ) ( ) ( )24 9 10 4 144M M M M M M( ) ( ) ( )24 9 10.4 144BA AB BC CB CD DCM M M M M M+ − + + + = −

4 2 2 2 124 93 3 3 3 2B B B CEI EI EI EI EI EIθ θ θ θ⎛ ⎞ ⎛− ∆ + − ∆ − +⎜ ⎟ ⎜

⎝ ⎠ ⎝⎞

2 0.866 1 0.86610.4 144C CEI EI EI EIθ θ⎛ ⎞+ − ∆ + − ∆ = −⎜ ⎟⎝ ⎠

3 1 32.667 2.66716 2 16B CEI EI EI EIθ θ ⎞+ ∆ − + + + ∆ + ⎟

3 6 3 6C C⎜ ⎟⎝ ⎠

34.5 3.1 38.375 144B CEI EI EIθ θ− − ∆ = − 3

Solving the three equation 112 24 23 12824 80 2 072 128

BEIEIθθ

−⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬⎢ ⎥

3.11BEIθ =

2 71EIθ24 80 2.072 12834.5 3.1 38.375 144

CEIEIθ⎢ ⎥ = −⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥− − ∆ −⎣ ⎦ ⎩ ⎭ ⎩ ⎭

2.71CEIθ = −

6.77EI ∆ =

Page 77: Slope Deflection by Hibbeler

Solving the three equation

112 24 23 12824 80 2 072 128

BEIEIθθ

−⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬24 80 2.072 128

34.5 3.1 38.375 144CEI

EIθ⎪ ⎪ ⎪ ⎪⎢ ⎥ = −⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥− − ∆ −⎣ ⎦ ⎩ ⎭ ⎩ ⎭

3 11θ 3.11BEIθ =

2.71CEIθ = −

6 77EI ∆ = 6.77EI ∆

Page 78: Slope Deflection by Hibbeler

Substituting in slope deflection equations 2.44 .ABM t m= −

0 36M t m= −2.78

0.36 .BAM t m= −

0.36 .BCM t m= B C_

2.780 362.78 .CBM t m=

2 78M t m= −

_

+0.36

0.36

2.442.78 .CDM t m=

1.88 .DCM t m= − A≈2.8

D+

1.88

Page 79: Slope Deflection by Hibbeler

Example– Draw the bending moment diagram. EI constant

Before we start let us discus the relative displacement (∆) of each span

Page 80: Slope Deflection by Hibbeler

The relative displacement (∆) for span AB is equal ∆AB – 0. = ∆AB (clockwise)

The relative displacement (∆) for span BC is equal

∆BC = δ1+ δ2

B C

∆CD

δThe relative displacement (∆) for span BC is equal(– δ2) – δ1 = – (δ1+ δ2) = – ∆BC (counterclockwise)

The relative displacement (∆) for span CD A D

δ1

δ2

is equal 0. – (–∆CD)= ∆CD (clockwise)

take ∆ ∆Let us build a relationship between ∆AB, ∆BC & ∆CD

A D∆AB

take ∆AB = ∆∆BC = 2(∆AB × cosθ) = 2∆ × 5/8.6 = 1.163∆∆CD = ∆AB = ∆ ∆BC

∆CD

θB

δ2θ

So in the slope deflection equations we will use;∆ as the relative displacement of span AB.

∆ABθ

δ1θ

–1.163∆ as the relative displacement of span BC. ∆ as the relative displacement of span CD.

δ1 = δ2 & ∆CD = ∆AB because of the symmetry in the geometry

Page 81: Slope Deflection by Hibbeler

Fixed End Moment

24 7 16 33FEM kN m×= − = −

Fixed End Moment0.AB BAFEM FEM= =

16.33 .12BCFEM kN m= =

16.33 .CBFEM kN m=

0.CD DCFEM FEM= =CD DC

2 0. 3 08 6 8 6AB BIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Slope Deflection Equations2 6

8 6 73 96BEI EIθ= − ∆8.6 8.6AB B⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ 8.6 73.96B

2 2 0 3 08.6 8.6BA BIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

4 68.6 73.96BEI EIθ= − ∆

1.1632 2 3 16.337 7BC B CIM E θ θ⎡ ⎤− ∆⎛ ⎞ ⎛ ⎞= + − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

4 2 6.978 16.337 7 49B CEI EI EIθ θ= + + ∆ −

1.1632 2 3 16 33IM E θ θ⎡ ⎤− ∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥2 4 6.978 16 33EI EI EIθ θ= + + ∆ +2 2 3 16.33

7 7CB C BM E θ θ= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦16.33

7 7 49B CEI EI EIθ θ= + + ∆ +

Page 82: Slope Deflection by Hibbeler

⎡ ⎤⎛ ⎞ ⎛ ⎞2 2 0 3 08.6 8.6CD CIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

4 68.6 73.96CEI EIθ= − ∆

2 0 3 08 6 8 6DC CIM E θ⎡ ⎤∆⎛ ⎞ ⎛ ⎞= + − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

2 68 6 73 96CEI EIθ= − ∆

8.6 8.6DC C⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ 8.6 73.96

Equilibrium EquationsJoint B ΣMB = 0

0BA BCM M+ =B

4 6 4 2 6.978 16.33 08.6 73.96 7 7 49B B CEI EI EI EI EIθ θ θ− ∆ + + + ∆ − =8.6 73.96 7 7 49103.65 28.57 6.13 1633B CEI EI EIθ θ+ + ∆ = 1

Joint C ΣMC = 00CB CDM M+ =

2 4 6.978 4 616.33 07 7 49 8.6 73.96B C CEI EI EI EI EIθ θ θ+ + ∆ + + − ∆ =

28.57 103.65 6.13 1633B CEI EI EIθ θ+ + ∆ = − 2

Page 83: Slope Deflection by Hibbeler

Third equation ΣF = 0 BAM

6 0A DH H− − =Third equation ΣFX = 0

Free body diagram for column AB B

BAM

AHA 5m

7m( )7 5 0A BA AB AH M M V× + + + × =

Free body diagram for Beam BC

( )7 4 7 3 5 0V M M× + + × × =ABM

VA

( )7 4 7 3.5 0B BC CBV M M× + + − × × =

( ) ( )M MM M ⎡ ⎤

( ) 147

BC CBA B

M MV V

+= = − + C

B

BCM 4kN/m

From the free body diagram for column CD

( ) ( )5 147 7 7

BC CBBA ABA

M MM MH

⎡ ⎤++∴ = − − − +⎢ ⎥

⎣ ⎦VB

CBMVC

CCDM

C

DCM

5

7m

D

( )7 5 0D CD DC DH M M V× + + − × =

( )7 4 7 3.5 0C BC CBV M M× − + − × × =

( )M M+HD

5m D

VD

( ) 147

BC CBD C

M MV V

+= = +

Page 84: Slope Deflection by Hibbeler

( ) ( )5M M M M⎡ ⎤+ +

( ) ( ) ( ) ( )6 49 7 5 7 5 0M M M M M M M M× + + − + + + − + =

( ) ( )5 147 7 7

CD DC BC CBD

M M M MH

⎡ ⎤+ +∴ = − + +⎢ ⎥

⎣ ⎦6 0A DH H− − =Q

( ) ( ) ( ) ( )6 49 7 5 7 5 0BA AB BC CB CD DC BC CBM M M M M M M M× + + − + + + − + =

( ) ( ) ( )7 7 10 294BA AB CD DC BC CBM M M M M M+ + + − + = −

2 6 4 6 2 67 7B B CEI EI EI EI EI EIθ θ θ⎛ ⎞ ⎛− ∆ + − ∆ + − ∆⎜ ⎟ ⎜7 78.6 73.96 8.6 73.96 8.6 73.96B B CEI EI EI EI EI EIθ θ θ∆ + ∆ + ∆⎜ ⎟ ⎜

⎝ ⎠ ⎝4 6 4 2 6.97810 16.33

8.6 73.96 7 7 49C B CEI EI EI EI EIθ θ θ⎞ ⎛+ − ∆ − + + ∆ −⎟ ⎜⎠ ⎝

2 4 6 978 ⎞

3.688 3.688 5.12 294B CEI EI EIθ θ− − − ∆ = − 3

2 4 6.978 16.33 2947 7 49B CEI EI EIθ θ ⎞+ + + ∆ + = −⎟

Solving the three equation 103.65 28.57 6.13 163328 57 103 65 6 13 1633

BEIEIθθ

⎡ ⎤ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬

18.897BEIθ =

24 603EIθ =28.57 103.65 6.13 16333.688 3.688 5.12 294

CEIEIθ⎢ ⎥ = −⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥− − − ∆ −⎣ ⎦ ⎩ ⎭ ⎩ ⎭

24.603CEIθ = −

61.532EI ∆ =

Page 85: Slope Deflection by Hibbeler

Substituting in slope deflection equations 0.6 .ABM kN m= −

3 8M kN m= 3.8 .BAM kN m=

3.8 .BCM kN m= − _3.8

16.4316.43

3 816.43 .CBM kN m=

16 43M kN m= −

_

+3.8

≈14.4

B C

16.43 .CDM kN m=

10.71 .DCM kN m= −0.6

A D10.71

A D