THEORY OF STRUCTURES CHAPTER 3 : SLOPE DEFLECTION …

by Saffuan Wan Ahmad

THEORY OF STRUCTURESCHAPTER 3 : SLOPE DEFLECTION

(FOR FRAME)PART 2

bySaffuan Wan Ahmad

Faculty of Civil Engineering & Earth [email protected]

For updated version, please click on http://ocw.ump.edu.my


Chapter 3 : Part 2 – Slope Deflection

• Aims– Determine the end moment for frame using Slope Deflection Method.

• Expected Outcomes :– Able to identify the frame – with or without side sway.– Able to determine end moment at critical points.

• References– Mechanics of Materials, R.C. Hibbeler, 7th Edition, Prentice Hall – Structural Analysis, Hibbeler, 7th Edition, Prentice Hall– Structural Analysis, SI Edition by Aslam Kassimali,Cengage Learning– Structural Analysis, Coates, Coatie and Kong – Structural Analysis - A Classical and Matrix Approach, Jack C. McCormac

and James K. Nelson, Jr., 4th Edition, John Wiley


FRAME

WITH SIDE SWAY

Displace to the side when the body or the loading acting on it is

nonsymmetric

WITHOUT SIDE SWAY

Properly restrainedSymmetric with

respect both loading and geometry


DEPEND ON CONFIGURATION

SUPPORT SYSTEMLOADING SYSTEM

MEMBER SIZES


FRAME WITH SIDESWAY

FRAME WITHOUT SIDESWAY


ANALYSIS - FRAME WITHOUT SIDESWAY.

By using SDM, determine the bending moment at critical points. Assume EI is constant.

EXAMPLE 1


SOLUTION

FIXED END MOMENT

0 FSR

FRS

FQP

FPQ MMMM

kNm .75338

)6(458

kNm 75.338

)6(458

PLM

PLM

FRQ

FQR


SLOPE DEFLECTION EQUATION 0 SP

FPQ

QPPQ M

LEI

LEI

LEIM

2

624

3Q

PQ

EIM

32 Q

QP

EIM

FQP

PQQP M

LEI

LEI

LEI

M

2

624


75.3333

2 RQ

QREIEI

M

75.3333

2 QR

RQ

EIEIM

FRQ

QRRQ M

LEI

LEI

LEIM

2

624

FQR

RQQR M

LEI

LEI

LEI

M

2

624


32 R

RSEIM

3R

SREIM

FRS

SRRS M

LEI

LEI

LEIM

2

624

FSR

RSSR M

LEI

LEI

LEIM

2

624


EQUILIBRIUM AT JOINT

0QM

0RM

1

2

75.3333

40

RQ

QRQP

EIEIMM

75.3333

4

0

QR

RSRQ

EIEI

MM


75.3375.33

34

31

31

34

R

Q

EIEI

BY USING CALCULATOR

75.3375.33

R

Q

EIEI


SUBSTITUTING INTO SDE

kNmMkNmM

kNmMkNmM

kNmMkNmM

SR

RS

RQ

QR

QP

PQ

25.115.22

5.225.22

5.2225.11


Determine end moment at critical point.

10kN/m

A

B C

D

3m 3m

6m

EI

2EI 2EI

FOOD OF MIND


SOLUTION

FIXED END MOMENT

0 FDC

FCD

FBA

FAB MMMM

kNmMM FCB

FBC 30

SLOPE DEFLECTION EQUATION 0 DA

FABBAAB M

LLEIM

32)2(2

32 B

ABEIM


FBAABBA M

LLEIM

32)2(2

34 B

ABEIM

FBCCBBC M

LLEIM

322

3033

2 CB

BCEIEIM


FCBBCCB M

LLEIM

322

FCDDCCD M

LLEIM

32)2(2

34 C

CDEIM

3033

2 BC

CBEIEIM


FDCCDDC M

LLEIM

32)2(2

32 C

DCEIM

EQUILIBRIUM AT JOINT

00

00

CDCB

C

BCBA

B

MMM

MMM


3030

231

312

C

B

EIEI

Solving by using Calculator

1818

C

B

EIEI


kNmMkNmMkNmMkNmM

kNmMkNmM

DC

CD

CB

BC

BA

AB

12242424

2412

SUBSTITUTING INTO SDE


ANALYSIS OF FRAME WITH SIDE SWAY.

DEPEND ON CONFIGURATION

SUPPORT SYSTEMLOADING SYSTEM

MEMBER SIZES


FUNDAMENTAL ASSUMPTIONS1. That axial deformation is ignored

2. That transverse end displacement do not affect the member lengthB B’ C C’

EI

EI

3EI

B B’ C C’

EI

EI

3EI

B’BA A’ C C’


DETERMINE END MOMENT AT CRITICAL POINT. ASSUME EI IS CONSTANT.

200 kN

6m

4m

5m

C

D

A

B

EXAMPLE 2


200 kN

6m

4m

5m

C

D

A

B


SOLUTION

0 FDC

FCD

FCB

FBC

FBA

FAB MMMMMM

Fixed End Moment

83

2

EIEIM B

AB

SLOPE DEFLECTION EQUATION

83

EIEIM BBA


52

54 CB

BCEIEIM

52

55 BC

CBEIEIM

63

EIEIM C

DC

632

EIEIM C

CD


0BM

Equilibrium at joint

0 BCBA MM

05

25

48

3

CB

BEIEIEIEI

08

35

25

9

EIEIEI CB 1


0CM

0 CDCB MM

063

25

25

4

EIEIEIEI CBC

0615

225

2

EIEIEI CB 2


0H

0200 DA HH

4

0)4(0

BAABA

ABAAB

B

MMH

HMMM

Consider member AB

A

B

AH

BH BAM

ABM


6

0)6(0

CDDCD

DDCCD

C

MMH

HMMM

Consider member CD

DH

CH CDM

DCM


48004466

20064

CDDCABBA

CDDCABBA

MMMM

MMMM

Insert the value……

48006

353

129

EIEIEI C

B 3


MATRIX FORM

480000

635

3129

61

1522

52

83

52

59

EIEIEI

C

B

Solving by using calculator

86.125066.7578.243

EIEIEI

C

B


Substituting into SDE

kNmM

kNmM

kNmM

kNmM

kNmM

kNmM

DC

CD

CB

BC

BA

AB

26.1836

88.1250366.75

04.1586

88.12503

)66.75(2

04.1585

)78.243(25

)66.75(4

29.2255

)66.75(25

)78.243(4

29.2258

)88.1250(378.243

16.3478

)88.1250(32

78.243


THANKS


Author Information

Mohd Arif Bin SulaimanMohd Faizal Bin Md. Jaafar

Mohammad Amirulkhairi Bin ZubirRokiah Binti Othman

Norhaiza Binti GhazaliShariza Binti Mat Aris

THEORY OF STRUCTURES CHAPTER 3 : SLOPE DEFLECTION …

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