THEORY OF STRUCTURES CHAPTER 3 : MOMENT DISTRIBUTION …

by Saffuan Wan Ahmad

THEORY OF STRUCTURESCHAPTER 3 : MOMENT DISTRIBUTION

(FOR FRAME)PART 4

bySaffuan Wan Ahmad

Faculty of Civil Engineering & Earth [email protected]

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


Chapter 3 : Part 4 – Moment Distribution

• Aims– Determine the end moment for frame using Moment Distribution Method

• Expected Outcomes :– Able to do moment distribution for frame.

• 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


MDM for Frame without SIDE-SWAY• MDM- solving indeterminate structures is a process in which the

moment in the members are determined by successive approximation.• Does not result in moment diagram but it provides the magnitude and

sense of the internal moments at joint – to obtain the shear and bending moment.

• TERM USED– Fixed end moment (FEM)– Carry over factor– Stiffness or resistance to rotation of a member

Clockwise moments are considered positiveWhereas, counterclockwise is negative


MDM for Frame without SIDE-SWAYSummary• Stiffness for member: end is pinned

equal to:

• Stiffness for member: end is fixed equal to:

LEIK 3

LEIK 4


Frame WITHOUT Sidesway

Determine the final moment for frame ABCD shown below. Hence,Draw the SFD and BMD.EI is indicate in the figure.

100kN

Cable / Tie Rod

3m 3m

3m

A

B C

D

4EI

2EI

EI

EXAMPLE 1


Fixed End Moment (MF)

kNmMkNmL

PabM

MMMM

FCB

FBC

FDC

FCD

FBA

FAB

7575

0

2

2


Distribution Factor (DF)

JOINT MEMBER K K DF

A AB 0

BBA 0.8

BC 0.2

CCB 0.5

CD 0.5

D DC 0

3)4(4 EI

3)4(4 EI

6)2(4 EI

6)2(4 EI

34EI

34EI

3

16EI

3

4EI

320EI

38EI


Member AB BA BC CB CD DC

DF 0 0.8 0.2 0.5 0.5 0

FEM 0 0 -75 75 0 0

BalCO

030

600

15-18.75

-37.57.5

-37.50

0-18.75

BalCO

07.5

150

3.75-1.88

-3.751.88

-3.750

0-1.88

BalCO

00.75

1.50

0.37-0.47

-0.940.19

-0.940

0-0.47

Bal 0 0.38 0.1 -0.1 -0.1 0

End Moment 38.25 76.88 -76.88 42.28 -42.29 -21.10

Table Moment Distribution


Shear Force and Bending Moment Diagram

A

BC

D

_

_

+_

+

38.25

55.58

44.20

21.10 38.25

A

B C

D

76.85

90.50

21.00

42.23

SFD BMD


Frame WITHOUT Sidesway

Determine the final moment for frame ABCDF with overhanging member BEshown below. Draw the BMD.EI is indicate in the figure.

5 kN 15 kN 10 kN

5 kN/m

A D

EB C F

2m 2m 2m 2m3m

4m4EI

4EI 4EI

4EI

3EI

EXAMPLE 2


Fixed End Moment (MF)

kNmM

kNmM

kNmM

MM

kNmM

kNmM

kNmM

kNmM

FBE

FFC

FCF

FDC

FCD

FCB

FBC

FBA

FAB

10)2(5

2.7

8.4

0

75

5.7

67.6

67.6



JOINT MEMBER K K DF

B

BA 0.5

BC 0.5

BE 0

C

CB

CD

CF

4)4(4 EI

4)4(4 EI

0

4)4(4 EI

4)4(4 EI

5)3(3 EI

EI8

549EI


JOINT A B C D F E

Member AB BA BC BE CB CD CF DC FC EB

DF 0 0.5 0.5 0 0 1.0 0

FEM -6.67 6.67 -7.5 10.0 7.5 0 -4.8 0 7.2 0

BalCO

-4.59 -4.49 0 -7.2

BalCOBalCOBal

End Moment

0 0

Table Moment Distribution **Note:Any no. divide by infinity =0


5 kN 15 kN 10 kN

5 kN/m

A D

EB C F

2m 2m 2m 2m3m

4m4EI

4EI 4EI

4EI

3EI

9.13

1.78

11.7410 6.34

0.65

1.29

7.63


9.13

1.78

A

B C

D

FE

11.74

10

15

6.34

7.63

12

1.29

10

BMD


MDM for Frame with SIDE-SWAY• The frames that are non-symmetrical or subjected to non-symmetrical

loadings have a tendency to SIDE-SWAY• Application of this technique is illustrated as below.

R

P

Virtual Prop Force(No side-sway)

R’

Virtual Prop Force is REMOVED(side-sway)

∆ ∆

P

Horizontal displacement are EQUAL


Summary• Stiffness for member: end is pinned

equal to:

• Stiffness for member: end is fixed equal to:

LEIK 3

LEIK 4


Summary• Moment produced at each member

when one end of member is displaced relative to other and both ends are FIXED

• Moment produced at the near end of a member if the remote end is displaced relative to the near end with remote end held in position but allowed to rotate

26

LEIM

23

LEIM

L

L


A portal frame ABCD as shown in Figure below is subjected to point load of 100 kN atmember BC. EI is constant. Analyze using Moment Distribution Method.

3m

A

B

3m 3m 4m

C

D

100 kN

EXAMPLE 3


SOLUTION EXAMPLE 3:

DA

B C

100 kN

DA

BC

100 kN

=

+∆

DA

BC

∆BC

∆CD

∆


Case 1:Fixed End Moment (MF): Non-sway Analysis

kNmMM

MMMMF

CBF

BC

FDC

FCD

FBA

FAB

75

0



JOINT MEMBER K K DF

A AB 0

BBA 0.67

BC 0.33

CCB 0.53

CD 0.47

D DC 1

34EI

34EI

64EI

64EI

53EI

53EI

3

4EI

53EI

EI2

1519EI



DF 0 0.67 0.33 0.53 0.47 1

MF 0 0 -75 75 0 0

BalCO 25.2

50.3 24.7-19.9

-39.7512.4

-35.25-17.63

BalCO

06.7

13.3 6.6-3.3

-6.63.3

-5.8 17.63

BalCO

01.1

2.2 1.1-0.85

-1.70.55

-1.6

Bal 0 0.57 0.28 -0.29 -0.26

End Moments 33.0 66.37 -66.37 42.91 -42.91 0

Table Moment Distribution (Non-sway Analysis)


Horizontal Reactions

kNHH

M

B

B

A

12.33037.660.33)3(

,0

A

33.0

66.37

B

HA

HB


66.37

HB HC

VB VC

42.91100

kNVVVF

kNVV

M

B

CB

y

C

C

B

9.530100

,01.46

091.42)3100(37.66)6(,0


kNHH

M

C

C

D

77.75091.42)41.46()3(

,0

- 42.91

D

C

HD

HC

46.1

VD


33.12 kN75.77 kNR

kNRR

HHRF

R

CB

H

65.42077.7512.33

00

find To


Case 2:Fixed End Moment (MF): Sway Analysis

DA

BC

∆ ∆

∆BC

∆CD3

434tan

BCBC

35

54sin

CDCD

BC


5533

92

666

96

366

,

2

)35

(

2

2

)34

(

2

22

EIEILEIM

EIEILEIMM

EIEILEIMM

therefore

CDS

CBS

BCS

BAS

ABS


9:10:305

:9

2:9

6::

:,45

,,,

EIEIEIMMM

thereforeEIassume

DCCDS

CBBCS

BAABS



DF 0 0.67 0.33 0.53 0.47 1

MF 30 30 -10 -10 9 0

BalCO -6.7

-13.4 -6.60.27

0.53-3.3

0.470.24

BalCO

0-0.09

-0.18 -0.090.88

1.75-0.5

1.55

BalCO

0-0.3

-0.59 -0.290.14

0.27-0.15

0.23

Bal 0 -0.09 -0.05 0.08 0.07

Assume Sway Moment

22.91 15.74 -15.74 -11.32 11.32 0

Table Moment Distribution (Sway Analysis)


Horizontal Reactions

kNHH

M

B

B

A

88.12074.1591.22)3(

,0

A

22.91

15.74

B

HA

HB


-15.74

HB HC

VB VC

-11.32100

kNVV

M

C

C

B

51.4032.1174.15)6(

,0


kNHH

M

C

C

D

79.90)451.4(32.11)3(

,0

11.32

D

C

HD

HC

4.51

VD


12.88 kN9.79 kNR

kNRR

HHRF

R

CB

H

67.22079.988.12

00

find To


Correction Factor and Final Moment

88.167.2265.42

'

RRASM

A B C DAssume sway moment

22.91 15.74 -15.74 -11.32 11.32 0

Actual sway moment (ASM)

43.07 29.59 -29.59 -21.28 21.28

(Non-sway moment)

33.0 66.37 -66.37 42.91 -42.91 0

Final Moments

76.07 95.96 -95.96 21.63 -21.63 0

**ASM x Assume sway moment


THANKS


Author Information

Mohd Arif Bin SulaimanMohd Faizal Bin Md. Jaafar

Mohammad Amirulkhairi Bin ZubirRokiah Binti Othman

Norhaiza Binti GhazaliShariza Binti Mat Aris

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THEORY OF STRUCTURES CHAPTER 3 : MOMENT DISTRIBUTION …