slope deflection method

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a method to solve indeterminate statics

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DISPLACEMENT METHOD OF ANALYSIS: SLOPE DEFLECTION EQUATIONS! ! ! ! ! ! ! ! !

General Case Stiffness Coefficients Stiffness Coefficients Derivation Fixed-End Moments Pin-Supported End Span Typical Problems Analysis of Beams Analysis of Frames: No Sidesway Analysis of Frames: Sidesway

1

Slope Deflection Equationsi

P

w

j

k

Cj

settlement = j

Mij

i

P

w

j

Mji

i j

2

Degrees of FreedomM

AL

B

1 DOF:

P B

A

C

2 DOF: ,

3

StiffnesskBA

kAA

1

ALk AA = k BA = 4 EI L 2 EI L

B

4

kAB 1 Lk BB = k AB = 4 EI L 2 EI L

kBB

A

B

5

Fixed-End ForcesFixed-End Moments: Loads PPL 8 P 2

L/2 L

L/2

PL 8 P 2

wwL2 12wL 2

L

wL2 12wL 2

6

General Case

i

P

w

j

k

Cj

settlement = j

Mij

i

P

w

j

Mji

i j

7

Mij

i

P

w

j

iL

Mji

4 EI 2 EI i + j = M ij L L

jMji =

settlement = j

2 EI 4 EI i + j L L

i(MFij)

j

+(MFji) settlement = j P w (MFji)Load

+(MFij)Load

M ij = (

4 EI 2 EI 2 EI 4 EI ) i + ( ) j + ( M F ij ) + ( M F ij ) Load , M ji = ( ) i + ( ) j + ( M F ji ) + ( M F ji ) Load 8 L L L L

Equilibrium Equationsi

P

w

j

k

Cj

Mji Mjij

Cj M

jk

Mjk

+ M j = 0 : M ji M jk + C j = 0

9

Stiffness CoefficientsMij Mji

i

j

L

i

j

kii =

4 EI L

k ji =

12 EI L

2 EI L

i

+k jj = 4 EI L j

kij =

1

10

Matrix Formulation

M ij = (

4 EI 2 EI ) i + ( ) j + ( M F ij ) L L 2 EI 4 EI ) i + ( ) j + ( M F ji ) L L

M ji = (

M ij (4 EI / L) ( 2 EI / L) iI M ij F M = + (2 EI / L) ( 4 EI / L) j M ji F ji

[k ] =

kii k ji

kij k jj

Stiffness Matrix11

Mij

i

P

w

j

L

i

Mji

[ M ] = [ K ][ ] + [ FEM ]

Mij Mji

j

j

([ M ] [ FEM ]) = [ K ][ ]

[ ] = [ K ]1[ M ] [ FEM ]

i

j

+(MFij) (MFji)

Fixed-end moment Stiffness matrix matrix

[D] = [K]-1([Q] - [FEM]) +(MFij)Load P w (MFji)Load Displacement matrix Force matrix

12

Stiffness Coefficients Derivation: Fixed-End SupportMii

ij

Mj Real beam

LL

Mi + M j

Mi + M j

L/3

M jL 2 EI

L

Mj EI

Conjugate beamMi EI

+ M 'i = 0 : (

MiL 2 EI

M j L 2L MiL L )( ) + ( )( ) = 0 2 EI 3 2 EI 3 M i = 2 M j (1)M L MiL ) + ( j ) = 0 (2) 2 EI 2 EI

From (1) and (2); 4 EI ) i L 2 EI ) i Mj =( L Mi = (

+ Fy = 0 : i (

13

Stiffness Coefficients Derivation: Pinned-End SupportMii

iLMi Lj

jMi L

Real beam

2L 3

Conjugate beam

Mi EI

i+ M ' j = 0 : (

MiL 2 EI

j+ Fy = 0 : (MiL ML ) ( i ) + j = 0 3EI 2 EI

M i L 2L )( ) i L = 0 2 EI 3 ML i = ( i ) 3EI

j =(MiL 3EI ) Mi = L 3EI

MiL ) 6 EI14

i = 1 = (

Fixed end moment : Point LoadP Real beam A A MM EI

Conjugate beam BM EI

L

BM EI

M

ML 2 EI

PPL2 16 EIPL 4 EI

ML 2 EI

M EI

PL2 16 EI

PL ML ML 2 PL2 + Fy = 0 : + = 0, M = 2 EI 2 EI 16 EI 8 15

PPL 8P 2

L P/2P 2

PL 8

PL/8 -PL/8-

P/2

-PL/8

-PL/8

-PL/16-

-PL/16 PL/4+

-PL/8

PL PL PL PL + = + 16 16 4 8

16

Uniform load w A MM EI

Real beam A

Conjugate beam BM EI M EI

L

B

M

ML 2 EI

ML 2 EI

M EI

w

wL3 24 EI

wL2 8 EI

wL3 24 EI

wL2 ML ML 2 wL3 + Fy = 0 : + = 0, M = 2 EI 2 EI 24 EI 12 17

Settlements Mi = Mj LMi + M j

Real beam

Mj A Mi + M j LM EI M EI

Conjugate beam

M EI

B

M

L

ML 2 EI

M

ML 2 EI

M EI

ML L ML 2 L )( ) + ( )( ) = 0, 2 EI 3 2 EI 3 M= 6 EI L218

+ M B = 0 : (

Pin-Supported End Span: Simple CaseP A4 EI 2 EI A + B L L

w B

L

2 EI 4 EI A + B L L

AA

+P

Bw

B

(FEM)AB A B = 0 = (4 EI / L) A + (2 EI / L) B + ( FEM ) AB

(FEM)BA

M AB

(1) ( 2)

M BA = 0 = (2 EI / L) A + (4 EI / L) B + ( FEM ) BA 2(2) (1) : 2 M BA = (6 EI / L) B + 2( FEM ) BA ( FEM ) BA

M BA = (3EI / L) B + ( FEM ) BA

( FEM ) BA 2

19

Pin-Supported End Span: With End Couple and SettlementP MA A4 EI 2 EI A + B L L

w B

L

2 EI 4 EI A + B L L

AA (MF AB)load (MFAB)

P

B

w

B (MF BA)load

A A

B B

(MF BA)

M AB = M A =

E lim inate A by

4 EI 2 EI F F A + B + ( M AB )load + ( M AB ) (1) L L 2 EI 4 EI F F A + B + ( M BA )load + ( M BA ) (2) M BA = L L 2(2) (1) 3EI 1 1 M F F F : M BA = B + [( M BA )load ( M AB )load ] + ( M BA ) + A 2 2 2 2 L

20

Fixed-End MomentsFixed-End Moments: Loads PPL 8

L/2

L/2

PL 8

P L/2 L/2PL 1 PL 3PL + ( )[( )] = 8 2 8 16

wL2 12

wL2 12

wL2 1 wL2 wL2 )] = + ( )[( 12 2 12 8

21

Typical ProblemA

P1

CB w

P2 C

L1

B

L2wL2 12

PL 8

P L

PL 8

w L

wL2 12

0 PL 4 EI 2 EI M AB = A + B + 0 + 1 1 8 L1 L1 0 EI PL 2 EI 4 M BA = A + B + 0 1 1 8 L1 L1 0 2 P2 L2 wL2 4 EI 2 EI M BC = B + C + 0 + + L2 L2 8 12 0 2 P2 L2 wL2 2 EI 4 EI B + C + 0 + M CB = 8 12 L2 L2

22

P1 A L1 MBA B

CB w

P2 C L2

CB M BC

B

M BA = M BC

PL 2 EI 4 EI A + B + 0 1 1 8 L1 L12

P L wL 4 EI 2 EI = B + C + 0 + 2 2 + 2 L2 L2 8 12

+ M B = 0 : C B M BA M BC = 0 Solve for B

23

P1 MAB A L1

MBA B

CB w

P2 C M CB

MBC L2

Substitute B in MAB, MBA, MBC, MCB 0 PL 4 EI 2 EI A + B + 0 + 1 1 M AB = L1 L1 8 0 EI PL 2 EI 4 M BA = A + B + 0 1 1 8 L1 L1 0 2 4 EI 2 EI P2 L2 wL2 B + C + 0 + + M BC = 8 12 L2 L2 0 2 P2 L2 wL2 2 EI 4 EI B + C + 0 + M CB = L2 L2 8 12

24

P1 MAB A Ay L1

MBA B

CB w

P2 MCB Cy C

MBC L2

By = ByL + ByR C MCB

P1 MAB A Ay B

B MBA MBC ByR

P2

L1

ByL

L2

Cy

25

Example of Beams

26

Example 1 Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant.

10 kN A 4m 4m B 6m

6 kN/m C

27

10 kN A 4m PL 8 P 4m PL 8 FEM MBA [M] = [K][Q] + [FEM] 0 B 6m wL2 30

6 kN/m C

w

wL2 20 MBC

B

4 EI 2 EI (10)(8) A + B + 8 8 8 0 2 EI 4 EI (10)(8) M BA = A + B 8 8 8 0 4 EI 2 EI (6)(6 2 ) B + C + M BC = 6 6 30 0 2 EI 4 EI (6)(6) 2 B + C M CB = 6 6 20 M AB =

+ M B = 0 : M BA M BC = 04 EI 4 EI (6)(6 2 ) + =0 ( ) B 10 + 8 6 30 2.4 B = EI Substitute B in the moment equations:

MAB = 10.6 kNm, MBA = - 8.8 kNm,

MBC = 8.8 kNm MCB = -10 kNm 28

10 kN 8.8 kNm 10.6 kNm A 4m 4m B 8.8 kNm 6m

6 kN/m C 10 kNm

MAB = 10.6 kNm, MBA = - 8.8 kNm,

MBC = 8.8 kNm MCB= -10 kNm 10 kN 2m 18 kN 6 kN/m B 8.8 kNm B 10 kNm 8.8 kNm

10.6 kNm

A

Ay = 5.23 kN

ByL = 4.78 kN

ByR = 5.8 kN

Cy = 12.2 kN

29

10 kN 10.6 kNm A 4m 4m B 6m

6 kN/m C 10 kNm

5.23 kN

12.2 kN

4.78 + 5.8 = 10.58 kN 5.23 + - 4.78 10.3 M (kNm) -10.6 Deflected shape + -8.8 2.4 B = EI

5.8 + -12.2 x (m) -10 x (m)30

V (kN)

x (m)

Example 2 Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant. 10 kN A 4m 4m B 6m 6 kN/m C

31

10 kN A 4m PL 8 P 4m PL 8 FEM 10 [M] = [K][Q] + [FEM] 0 4 EI 2 EI (10)(8) A + B + M AB = (1) 8 8 8 10 2 EI 4 EI (10)(8) A + B (2) M BA = 8 8 8 4 EI 2 EI 0 (6)(6 2 ) B + C + (3) M BC = 6 6 30 2 EI 4 EI 0 (6)(6) 2 B + C (4) M CB = 6 6 20 6 EI 2(2) (1) : 2 M BA = B 30 8 3EI B 15 (5) M BA = 8 B 6m wL2 30

6 kN/m C

w

wL2 20

32

MBA BM BC

MBC

4 EI (6)(6 2 ) = B + 6 30

(3)

Substitute A and B in (5), (3) and (4): MBA = - 12.19 kNm

M CB =M BA =

2 EI (6)(6) B 6 20

2

(4)

MBC = 12.19 kNm MCB = - 8.30 kNm

3EI B 15 (5) 8 + M B = 0 : M BA M BC = 0

3EI 4 EI (6)(6 2 ) + = 0 (6 ) ( ) B 15 + 8 6 30 7.488 B = EI

Substitute B in (1) : 0 =

4 EI 2 EI A + B 10 8 8 23.74 A = EI

33

10 kN 12.19 kNm A 4m 4m B 12.19 kNm 6m

6 kN/m C 8.30 kNm

MBA = - 12.19 kNm, MBC = 12.19 kNm, MCB = - 8.30 kNm 2m 10 kN A B B 12.19 kNm 12.19 kNm Ay = 3.48 kN ByL = 6.52 kN ByR = 6.65 kN Cy = 11.35 kN 18 kN 6 kN/m C 8.30 kNm

34

10 kN A 3.48 kN 4m 4m B 6m

6 kN/m C 11.35 kN

6.52 + 6.65 = 13.17 kN 3.48 6.65 x (m) - 6.52 14 M (kNm) -12.2 Deflected shape 23.74 A = EI

V (kN)

-11.35 x (m)

B =

7.49 EI

-8.3 x (m)35

Example 3 Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant.

10 kN A 2EI 4m 4m B 3EI 6m

4 kN/m C

36

10 kN A 2EI (10)(8)/8 (10)(8)/8 4m 4m 3EI B (4)(62)/12 6m

4 kN/m C (4)(62)/12

0M AB

10 4(2 EI ) 2(2 EI ) (10)(8) = A + B + (1) 8 8 8 102(2 EI ) 4(2 EI ) (10)(8) A + B 8 8 15 8 (2)

M BA =

2(2) (1) 3(2 EI ) (3 / 2)(10)(8) : M BA = B ( 2a ) 2 8 8 12 4(3EI ) (4)(6 2 ) B + (3) M BC = 6 12

37

10 kN A 3EI 2EI (3/2)(10)(8)/8 B (4)(62)/12 4m 6m 4m

4 kN/m C (4)(62)/12

15 3(2 EI ) (3 / 2)(10)(8) M BA = B ( 2a ) 8 8 12 2 4(3EI ) (4)(6 ) B + (3) M BC = 6 12 M BA M BC = 0 : 2.75EI B = 12 + 15 = 3

B = 1.091 / EIM BA = M BC M CB 3(2 EI ) 1.091 ( ) 15 = 14.18 kN m 8 EI 4(3EI ) 1.091 ( ) 12 = 14.18 kN m = 6 EI 2(3EI ) = B 12 = 10.91 kN m 6

38

10 kN A

4 kN/m C 10.91 B 14.18 6m 3EI

2EI 14.18 4m 4m

MBA = - 14.18 kNm, MBC = 14.18 kNm, MCB = -10.91 kNm

10 kN A B 14.18 kNm 140.18 kNm

24 kN 4 kN/m 10.91 kNm C ByR = 12.55 kN Cy = 11.46 kN

Ay = 3.23 kN

ByL = 6.73 kN

39

10 kN A 3.23 kN 4m 2EI 4m B 3EI 6m

4 kN/m C 10.91 kNm 11.46 kN

V (kN)

3.23

+

6.77 + 12.55 = 19.32 kN 12.55 2.86 + -6.73 12.91 5.53 + -14.18 + -

x (m) -11.46

M (kNm)

x (m) -10.91 x (m)

Deflected shape

B = 1.091/EI

40

Example 4 Draw the quantitative shear , bending moment diagrams and qualitative deflected curve for the beam shown. EI is constant.

10 kN 12 kNm A 2EI 4m 4m B 3EI 6m

4 kN/m C

41

10 kN 12 kNm A

4 kN/m C

2EI 3EI 2/12 = 12 1.5PL/8 = 15 B wL wL2/12 = 12 4m 6m 4m -3.273/EI 4(2 EI ) 2(3EI ) (10)(8) A + B + M AB = 8 8 8 7.21 A = EI 3.273 M BA = 0.75 EI ( ) 15 = 17.45 kN m EI 3.273 M BC = 2 EI ( ) + 12 = 5.45 kN m EI 3.273 M CB = EI ( ) 12 = 15.27 kN m EI 0

M BA =

3(2 EI ) B 15 (1) 8

M BC = M CB =

4(3EI ) B + 12 (2) 6

2(3EI ) B 12 (3) 6 12 kNm MBA MBA

B MBC MBCJo int B : M BA M BC 12 = 0 (0.75 EI 15) (2 EI B + 12) 12 = 0

B =

3.273 EI

42

3.273 ) 15 = 17.45 kN m EI 3.273 M BC = 2 EI ( ) + 12 = 5.45 kN m EI 3.273 M CB = EI ( ) 12 = 15.27 kN m EI M BA = 0.75EI (

10 kN A 2.82 kN B 5.45 kNm 17.45 kNm 7.18 kN

24 kN 4 kN/m 15.27 kNm 13.64 kN

10.36 kN

C

10 kN 12 kNm A 2.82 kN B 17.54 kN

4 kN/m C 15.27 kNm43

13.64 kN

10 kN 12 kNm A 2.82 kN 4m 2EI 4m B 17.54 kN 6m 10.36 V (kN) 2.82 + -7.18 M (kNm) 11.28 + -5.45 -17.45 Deflected shape 7.21 A = EI

4 kN/m C 15.27 kNm 3EI 13.64 kN A = 7.21 EI 3.273 B = EI

+

3.41 m

x (m) -13.64

7.98 + -15.27 x (m) x (m)

B =

3.273 EI

44

Example 5 Draw the quantitative shear, bending moment diagrams, and qualitative deflected curve for the beam shown. Support B settles 10 mm, and EI is constant. Take E = 200 GPa, I = 200x106 mm4.

12 kNm A

10 kN B

6 kN/m C

2EI 4m 4m 10 mm

3EI 6m

45

12 kNm A

10 kN B

6 kN/m C

2EI6 EI L2

3EI 4m6 EI L2

4m A

10 mm

6m6 EI L2

6 EI L2

B w

PL 8

P

B

[FEM]PL 8

wL 30

2

C wL2 30

[FEM]load -12 4(2 EI ) 2(2 EI ) 6(2 EI )(0.01) (10)(8) M AB = A + B + + (1) 8 8 82 8 2(2 EI ) 4(2 EI ) 6(2 EI )(0.01) (10)(8) M BA = A + B + ( 2) 8 8 82 8 0 4(3EI ) 2(3EI ) 6(3EI )(0.01) (6)(6 2 ) M BC = B + C + (3) 6 6 62 30 0 2(3EI ) 4(3EI ) 6(3EI )(0.01) (6)(6) 2 M CB = B + C (4) 2 6 6 6 30

46

12 kNm A

10 kN B

6 kN/m C

2EI 4mM AB = M BA

3EI 4m 10 mm 6m

4(2 EI ) 2(2 EI ) 6(2 EI )(0.01) (10)(8) A + B + + (1) 8 8 82 8 2(2 EI ) 4(2 EI ) 6(2 EI )(0.01) (10)(8) = A + B + ( 2) 8 8 82 8

Substitute EI = (200x106 kPa)(200x10-6 m4) = 200x200 kN m2 :M AB = M BA 4(2 EI ) 2(2 EI ) A + B + 75 + 10 (1) 8 8 2(2 EI ) 4(2 EI ) = A + B + 75 10 (2) 8 8 16.5

2(2) (1) 3(2 EI ) : M BA = B + 75 (75 / 2) 10 (10 / 2) 12 / 2 (2a ) 8 247

12 kNm A

10 kN B

6 kN/m C

2EI 4m 4m 10 mm

3EI 6m

MBA = (3/4)(2EI)B + 16.5 MBC = (4/6)(3EI)B - 192.8 + MB = 0: - MBA - MBC = 0 MBAB

(3/4 + 2)EIB + 16.5 - 192.8 = 0

MBC

B = 64.109/ EISubstitute B in (1):

A = -129.06/EI

Substitute A and B in (5), (3), (4): MBA = 64.58 kNm, MBC = -64.58 kNm MCB = -146.69 kNm48

12 kNm A

10 kN B

6 kN/m C 146.69 kNm

64.58 kNm 4m MBA = 64.58 kNm, MBC = -64.58 kNm MCB = -146.69 kNm 12 kNm A 10 kN 64.58 kNm B 4m

64.58 kNm 6m

2m 18 kN 64.58 kNm 6 kN/m 146.69 kNm B C Cy = 47.21 kN

Ay = 11.57 kN

ByL = -1.57 kN

ByR = -29.21 kN

49

12 kNm A 11.57 kN 4m

10 kN

6 kN/m C 146.69 kNm

2EI 4m

B

3EI 6m 47.21 kN

1.57 + 29.21 = 30.78 kN V (kN) 11.57 + 1.57 -29.21 12 58.29 + -146.69 Deflected shape A = -129.06/EI 10 mm x (m) 64.58 -47.21 x (m) x (m)

A = -129.06/EI B = 64.109/ EI

M (kNm)

B = 64.109/ EI

50

Example 6 For the beam shown, support A settles 10 mm downward, use the slope-deflection method to (a)Determine all the slopes at supports (b)Determine all the reactions at supports (c)Draw its quantitative shear, bending moment diagrams, and qualitative deflected shape. (3 points) Take E= 200 GPa, I = 50(106) mm4.

6 kN/m B 2EI 3m C 1.5EI 3m

12 kNm A

10 mm

51

6 kN/m B 2EI 3m6(32 ) = 4.5 12 6(1.5 200 50)(0.01) 32 = 100 kN m

12 kNm A

C

1.5EI 3m 6 kN/m

10 mm

4.5 C C MFw A0.01 m

MF (1)

100 kN m

M CB = M CA =M AC =

4(2 EI ) C 3

A

12

4(1.5EI ) 2(1.5 EI ) C + A + 4.5 + 100 (2) 3 32(1.5EI ) 4(1.5 EI ) C + A 4.5 + 100 (3) 3 3 ( 2a )52

2(2) (2) 3(1.5EI ) 3(4.5) 100 12 : M CA = C + + + 3 2 2 2 2

6 kN/m B 2EI 3mM CB =M CA

12 kNm A

C

1.5EI 3m

10 mm

4(2 EI ) C (1) 3 3(1.5 EI ) 3(4.5) 100 12 = C + + + 3 2 2 2

( 2a )

MCB

MCA

Equilibrium equation:M CB + M CA = 0

(8 + 4.5) EI 3(4.5) 100 12 C + + + =0 C 3 2 2 2 15.06 C = = 0.0015 rad EI 2(1.5 EI ) 15.06 4(1.5 EI ) Substitute C in eq.(3) 12 = ( )+ A 4.5 + 100 (3) EI 3 3 34.22 A = = 0.0034 rad EI

53

6 kN/m B 2EI 3m C 1.5EI 3m

12 kNm A

10 mm

C =M BC M CB

15.06 34.22 = 0.0015 rad A = = 0.0034 rad EI EI 2(2 EI ) 2( 2 EI ) 15.06 = C = ( ) = 20.08 kN m 3 3 EI 4(2 EI ) 4(2 EI ) 15.06 = C = ( ) = 40.16 kN m 3 3 EI

20.08 kNm B20.08 kN

C

40.16 kNm

40.16 + 20.08 = 20.08 kN 3 18 kN

6 kN/m 40.16 kNm C 26.39 kN

12 kNm

A

8.39 kN

54

6 kN/m

12 kNm A

C = 0.0015 rad A = 0.0034 rad20.08 kNm

B 2EI 3m B 20.08 kN C

C

1.5EI 3m 40.16 kNm 6 kN/m

10 mm

20.08 kN 40.16 kNm

12 kNm

V (kN) M (kNm) -20.08 20.08 -40.16

C 26.39 kN 26.39 +

A

8.39 kN 8.39 x (m)

12

x (m)

Deflected shape

C = 0.0015 rad

x (m) A = 0.0034 rad

55

Example 7

For the beam shown, support A settles 10 mm downward, use the slope-deflection method to (a)Determine all the slopes at supports(b)Determine all the reactions at supports (c)Draw its quantitative shear, bending moment diagrams, and qualitative deflected shape. Take E= 200 GPa, I = 50(106) mm4.

6 kN/m B 2EI 3m C 1.5EI 3m

12 kNm A

10 mm

56

6 kN/m B 2EI 3m C 1.5EI 3mEI C 4 EI C 3

12 kNm A

10 mm

C B6( 2 EI ) C 4 EI C = 2 3 3

C C6 kN/m

C

AA

6(1.5 EI ) C = EI C 2 3

4.5 100 (1) (2)

C

6(32 ) = 4.5 12 6(1.5 200 50)(0.01) 32 = 100 kN m

M BC M CB M CA =

2(2 EI ) 4 EI = C C 3 3 4(2 EI ) 4 EI = C C 3 3

C0.01 m

A

12 2(1.5 EI ) 4(1.5 EI ) M AC = C + A + EI C 4.5 + 100 (4) 3 3 2(3) (4) 3(1.5 EI ) EI 3(4.5) 100 12 : M CA = C + C + + + (3a) 2 3 2 2 2 2

4(1.5 EI ) 2(1.5EI ) C + A + EI C + 4.5 + 100 (3) 3 3

57

6 kN/m B 2EI 3m Equilibrium equation: MBC MCA C 1.5EI 3m

12 kNm A

10 mm

18 kN 6 kN/m

12 kNm

B By

C(C y ) CB = (

MCB

M BC + M CB (C y ) CA ) 3 MCB MCA

Ay M + 12 + 18(1.5) M CA + 39 = CA = 3 3M C = 0 : M CB + M CA = 0 (1*)

C

A

(Cy)CB

C

C y = 0 : (C y ) CB + (C y ) CA = 0 (2*)

(Cy)CA

Substitute in (1*) 4.167 EI C 0.8333EI C = 62.15 (5) Substitute in ( 2*) 2.5 EI C + 3.167 EI C = 101.75 (6) From (5) and (6) C = 25.51 / EI = 0.00255 rad C = 52.27 / EI = 5.227 mm58

6 kN/m B 2EI 3m Solve equation 25.51 = 0.00255 rad EI

12 kNm A

C

1.5EI 3m

10 mm

C =

Substitute C and C in (1), (2) and (3a)M BC = 35.68 kN m M CB = 1.67 kN m M CA = 1.67 kN m

52.27 C = = 5.227 mm EI

Substitute C and C in (4)

A =

2.86 = 0.000286 rad EI

6 kN/m 35.68 kNm B 2EI 3m C 1.5EI 3m

12 kNm A

B y = 18 5.55 = 12.45 kN

Ay =

18(4.5) 12 35.68 6 = 5.55 kN59

6 kN/m 35.68 kNm B 2EI 3m C 1.5EI 3m

12 kNm A

12.45 kN

10 mm 5.55 kN

V (kN) 12.45 C = 0.00255 rad + C = 5.227 mm

0.925 m x (m) -5.55 1.67 14.57 + 12 x (m)

A = 0.000286 rad M (kNm)-35.68 Deflected shape

C = 5.227 mm

C = 0.00255 rad

x (m)

A = 0.000286 rad

60

Example of Frame: No Sidesway

61

Example 6 For the frame shown, use the slope-deflection method to (a) Determine the end moments of each member and reactions at supports (b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame. 10 kN B 3EI 2EI 12 kN/m C 3m 40 kN 3m A 1m 6m

62

10 kN B

36/2 = 18

12 kN/m C

Equilibrium equations 10 MBC MBA10 M BA M BC = 0 (1*)

3m 40 kN 3m

2EI 2 36 (wL /12 ) =36 PL/8 = 30 3EI PL/8 = 30 6m

A 1m

Substitute (2) and (3) in (1*)10 3EI B + 30 54 = 0

Slope-Deflection EquationsM AB = 2(3EI ) B + 30 (1) 6

B =

14 4.667 = (3EI ) EI 4.667 in (1) to (3) EI

M BA =M BC

4(3EI ) B 30 (2) 6

Substitute B =

3( 2 EI ) B + 36 + 18 (3) = 6

M AB = 25.33 kN m M BA = 39.33 kN m M BC = 49.33 kN m63

10 kN B 49.33 39.33 3EI 25.33 6m 2EI

12 kN/m C10

20.58 -39.3 -49.33 27.7 -25.33 Bending moment diagram

3m 40 kN 3m

MAB = 25.33 kNm MBA = -39.33 kNm MBC = 49.33 kNm

A 1m

12 kN/m B B 40 kN A 17.67 kN 25.33 C 49.33 39.33 27.78 kN

B B B = -4.667/EI

Deflected curve

64

Example 7 Draw the quantitative shear, bending moment diagrams and qualitative deflected curve for the frame shown. E = 200 GPa. 25 kN B 240(106) mm4 5m 120(106) mm4 A 3m 3m C 180(106) 60(106) mm4 D 4m 5 kN/m E

65

25 kN PL/8 = 18.75 18.75 B 240(106) mm4 C 5m 120(106) mm4 A 0 4(2 EI ) 2(2 EI ) M AB = A + B 5 5 0 2(2 EI ) 4(2 EI ) M BA = A + B 5 5 4( 4 EI ) 2(4 EI ) M BC = B + C + 18.75 6 6 2(4 EI ) 4(4 EI ) M CB = B + C 18.75 6 6 3( EI ) C M CD = 5 3(3EI ) M CE = C + 10 4 3m 3m

5 kN/m 180(106) 6.667+ 3.333 60(106) mm4 D 4mM BA + M BC = 0 8 16 8 ( + ) EI B + ( ) EI C = 18.75 (1) 5 6 6 M CB + M CD + M CE = 0 16 3 9 8 ( ) EI B + ( + + ) EI C = 8.75 (2) 6 5 4 6 From (1) and (2) : B = 5.29 EI

E (wL2/12 ) = 6.667

C =

2.86 EI66

Substitute B = -1.11/EI, c = -20.59/EI below4(2 EI ) 0 2(2 EI ) M AB = A + B 5 5 0 2(2 EI ) 4(2 EI ) M BA = A + B 5 5 4( 4 EI ) 2(4 EI ) M BC = B + C + 18.75 6 6 2(4 EI ) 4(4 EI ) B + C 18.75 M CB = 6 6 3( EI ) C M CD = 5 3(3EI ) C + 10 M CE = 4

MAB = 4.23 kNm MBA = 8.46 kNm MBC = 8.46 kNm MCB = 18.18 kNm MCD = 1.72 kNm MCE = 16.44 kNm

67

MAB = -4.23 kNm, MBA = -8.46 kNm, MBC = 8.46 kNm, MCB = -18.18 kNm, MCD = 1.72 kNm, MCE = 16.44 kNm 25 kN 20 kN E 2.2 kN 5.89 kN 16.44 kNm 3m 3m C C 2.54 kN B 2.54 kN 2.54-0.34 =2.2 kN 8.46 kNm 18.18 kNm (20(2)+16.44)/4 (25(3)+8.46-18.18)/6 14.12 kN = 14.11 kN = 10.88 kN 10.88 kN B 8.46 kNm (8.46 + 4.23)/5 = 2.54 kN 14.12+14.11=28.23 kN 1.72 kNm (1.72)/5 = 0.34 kN B

5m 2.54 kN 4.23 kNm 10.88 kN

5m 0.34 kN

A

A

28.23 kN

68

10.88+

14.11 1.18 m+ -

24.18 1.29 m 2.33 m 0.78 m+

-2.54-

2.82 m+

-

1.72 -8.46-

3.46+

-14.12

-5.89 -8.46 -

1.18 m

-18.18+

-16.44

-2.54 Shear diagram

0.34

1.67m Moment diagram

4.23 0.78 m

1.29 m 2.33 m

B = 5.29/EI

1.18 m

C = 2.86/EI1.67m Deflected curve69

Example of Frames: Sidesway

70

Example 8 Determine the moments at each joint of the frame and draw the quantitative bending moment diagrams and qualitative deflected curve . The joints at A and D are fixed and joint C is assumed pin-connected. EI is constant for each member 3m 1m B 10 kN C

3m A D

71

Overview B 1m 10 kN C Unknowns

B and Boundary Conditions

3m A Ay MAB Ax 3m MDC D D Dy

A = D = 0 Equilibrium Conditions - Joint B B MBA MBC

x

M B = 0 : M BA + M BC = 0 (1*)

- Entire Frame Fx = 0 : 10 Ax Dx = 0 (2*)+

72

(0.375EI)

C(0.375EI)

MAB B C

1m

B 10 kN(5.625)load

10 kN

3m(1.875)load

4m(1/2)(0.375EI) (0.375EI)

4m D

A

(0.375EI)

D

A

Ax MBA

Dx MDC

3m

Slope-Deflection Equations 5.625 0.375EI 2( EI ) 10(3)(12 ) 6 EI M AB = B + + 2 (1) 4 42 4 5.625 0.375EI 2 4( EI ) 10(3 )(1) 6 EI M BA = B + 2 (2) 4 42 4M BC = 3( EI ) B 3 (3)

+ M B = 0 : ( M AB + M BA ) 4 Ax = 0.375EI B + 0.1875EI + 1.563 (5) Ax =+ M C = 0 : Dx = M DC = 0.0468EI (6) 473

1 M DC = 0.375EI 0.375EI = 0.1875EI (4) 2

Equilibrium Conditions:M BA + M BC = 0 (1*)

Solve equation Substitute (2) and (3) in (1*) 2EI B + 0.375EI = 5.625 ----(7)

10 Ax Dx = 0 (2*)

Slope-Deflection Equations: Substitute (5) and (6) in (2*) 2( EI ) M AB = B + 5.625 + 0.375EI (1) 0.375EI B 0.235EI = 8.437 (8) 4 4( EI ) M BA = B 5.625 + 0.375EI (2) From (7) and (8) can solve; 4 5.6 44.8 3( EI ) B = = M BC = B (3) EI EI 3 5.6 44.8 M DC = 0.1875EI (4) Substitute B = and = in (1)to (6) EI EI Horizontal reaction at supports:Ax = 0.375EI B + 0.1875EI + 1.563 (5)Dx = 0.0468EI (6)

MAB = 15.88 kNm MBA = 5.6 kNm MBC = -5.6 kNm MDC = 8.42 kNm Ax = 7.9 kN Dx = 2.1 kN

74

B 1m 10 kN

C 5.6 MAB = 15.88 kNm, MBA = 5.6 kNm, MBC = -5.6 kNm, MDC = 8.42 kNm, Ax = 7.9 kN, Dx = 2.1 kN, 8.42 D 2.1 kN

3m A 15.88 7.9 kN 3m 5.6 5.6 7.8 B C

= 44.8/EI B

= 44.8/EI C

B = -5.6/EI

A

15.88

D

8.42

A

D Deflected curve75

Bending moment diagram

Example 9 From the frame shown use the slope-deflection method to: (a) Determine the end moments of each member and reactions at supports (b) Draw the quantitative bending moment diagram, and also draw the qualitative deflected shape of the entire frame. B 10 kN 2m A 4m 3m EI 2 EI 2.5 EI D 4m C pin

76

Overview B 2EI 10 kN 2m A B EI MAB Ax Ay 4m

CD

C

C BC MDC D Dy 4m Dx

Unknowns

B and Boundary Conditions A = D = 0 Equilibrium Conditions - Joint B B MBA MBC

2.5EI

3m

M B = 0 : M BA + M BC = 0 (1*)

- Entire Frame Fx = 0 : 10 Ax Dx = 0 (2*)+

77

Slope-Deflection Equation C B CD BC 5 B 2EI C 10 kN 36.87 4m 2.5EI EI 2m PL/8 = 5 D A 4m 3m 0.375EI B B

C

= / cos 36.87 = 1.25 C CD BC = tan 36.87 = 0.75 36.87 M AB =

2( EI ) B + 0.375EI + 5 (1) 4 (6)(2.5EI)(1.25)/(5)2 = 0.75EI 4( EI ) M BA = B + 0.375EI 5 (2) C 4 CD= 1.25 3(2 EI ) C M BC = B 0.2813EI (3) 4

A 6EI/(4) B2=

M DC = 0.375EI (4)

D 0.75EI 0.375EI C C 0.5625EI

(1/2) 0.75EI

(6)(2EI)(0.75)/(4) 2 = 0.5625EI BC= 0.75 78

B (1/2) 0.5625EI

Horizontal reactions B 10 kN 2m A 4m MBC MBA B 10 kN 3m B C MBC 4 C A Ax = (MBA+ MAB-20)/4 -----(5) MAB D Dx= (MDC-(3/4)MBC)/4 ---(6) MDC79

2 EI C EI

pin

2.5 EI D

4m

MBC 4

MBC/4

Equilibrium Conditions:M BA + M BC = 0 (1*)

Solve equations Substitute (2) and (3) in (1*)2.5EI B + 0.0938 EI 5 = 0 (7)

10 Ax Dx = 0 (2*) Slope-Deflection Equation:M AB =

Substitute (5) and (6) in (2*)

2( EI ) 6 EI 0.0938EI B + 0.334EI 5 = 0 (8) B + 5 + 2 (1) 4 4 From (7) and (8) can solve; 6 EI 4( EI ) M BA = B 5 + 2 (2) 4 4 1.45 14.56 B = = 3(2 EI ) 3(2 EI )(0.75) EI EI M BC = B (3) 2 4 4 1.45 14.56 3(2.5EI )(1.25) Substitute B = and = in (1)to (6) M DC = (4) EI EI 52 Horizontal reactions at supports: MAB = 15.88 kNm MBA = 5.6 kNm ( M BA + M AB 20) (5) Ax = MBC = -5.6 kNm 4 MDC = 8.42 kNm 3 Ax = 7.9 kN M DC M BC 4 Dx = (6) Dx = 2.1 kN 4 80

1.91 10 kN 2m A

1.91 B 2 EI C

pin

MAB = 11.19 kNm MBA = 1.91 kNm MBC = -1.91 kNm 5.46 4m 1.73 0.478 kN MDC = 5.46 kNm Ax = 8.28 kNm Dx = 1.72 kNm

2.5 EI EI 11.19 kNm 8.27 kN 0.478 kN 4m 1.91 3m

D

B 1.91 5.35 A

C

B

B=1.45/EIC

B=1.45/EI11.19 5.46 D A D Deflected shape

Bending-moment diagram

81

Example 10 From the frame shown use the moment distribution method to: (a) Determine all the reactions at supports, and also (b) Draw its quantitative shear and bending moment diagrams, and qualitative deflected curve. 20 kN/m B 3m 3EI 3m 2EI 4EI A D 4m C

pin

82

Overview B 20 kN/m 3EI 2EI 4EI C Unknowns

B and Boundary Conditions 4m

3m

A = D = 0 Equilibrium Conditions

A [FEM]load 3m D

- Joint B B MBA MBC

M B = 0 : M BA + M BC = 0 (1*)

- Entire Frame Fx = 0 : 60 Ax Dx = 0 (2*)+

83

Slope-Deflection Equation 3m B 20 kN/m wL2/12 2EI 3EI = 15 4EI C 6(2EI)/(3) = 1.333EI2

B 1.5EI

C

3m

wL2/12 = 15 A [FEM]load 0M AB M BA

4m A D

6(2EI)/(3) 2 = 1.333EI (1/2)(1.5EI) [FEM]

6(4EI)/(4) 2 = 1.5EI

D 4(2 EI ) 2(2 EI ) = A + B + 15 + 1.333EI = 1.333EI B + 15 + 1.333EI ----------(1) 3 3 0 2( 2 EI ) 4(2 EI ) = A + B 15 + 1.333EI = 2.667 EI B 15 + 1.333EI ----------(2) 3 33(3EI ) B = 3EI B ----------(3) 3 0 3( 4 EI ) = D + 0.75 EI = 0.75 EI ----------(4) 4

M BC =M DC

84

Horizontal reactions MAB C B 1.5 m 60 kN 1.5 m A Ax MBA D Dx MDC + MC = 0:Dx = M DC = 0.188EI (6) 4+ M B = 0 :

4m

M BA + M AB + 60(1.5) 3 Ax = 1.333EI B + 0.889 EI + 30 (5) Ax =

85

Equilibrium ConditionsM BA + M BC = 0 (1*)

Solve equation Substitute (2) and (3) in (1*)5.667 EI B + 1.333EI = 15 (7)

60 Ax Dx = 0 (2*) Equation of momentM AB = 1.333EI B + 15 + 1.333EI (1) M BA = 2.667 EI B 15 + 1.333EI (2)M BC = 3EI B (3)

Substitute (5) and (6) in (2*) 1.333EI B 1.077 EI = 30 (8)

From (7) and (8), solve equations;

B =

5.51 EI

=

34.67 EI in (1)to (6)

M DC = 0.75 EI ( 4)

Substitute B =

Horizontal reaction at supportAx = 1.333EI B + 0.889 EI + 30 (5)

34.67 5.51 and = EI EI M AB = 53.87 kN m M BA = 16.52 kN m M BC = 16.52 kN m

Dx = 0.188EI (6)

M DC = 26.0 kN m

Ax = 53.48 kN Dx = 6.52 kN86

B 20 kN/m

3m C 16.52 kNm 3m 53.87 kNm 4m 53.48 kN 26 kNm 6.52 kN C B CM AB = 53.87 kN m M BA = 16.52 kN m M BC = 16.52 kN m

M DC = 26.0 kN m

Ax = 53.48 kN D Dx = 6.52 kN

A

5.55 kN

5.55 kN 16.52 16.52 B

53.87 A Moment diagram 26. D A Deflected shape D

87