# Slope Deflection Method

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a method to solve indeterminate statics

### Text of Slope Deflection Method

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

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