CHAPTER 3Moment Distribution

Objectives1. Statically Indeterminate Moment Distribution (MOM-DIS)2. joint equilibrium 3. sidesway, no sidesway

Method of Analyzing Indeterminate Structure

Moment Distribution Developed by Prof Hardy Cross 1930 joint moment

Principles and Definition joint

fixed unlocking locking joint moment joint

General Principles

Sign Convention slope-deflection

Fixed-End Moments (FEMS)

Member Stiffness Factor

Joint stiffness factorLEIK 4

Derivation of Slope-Deflection Eqs

Moment due to angular displacement @A, A

MAB disp conjugate bm method

4

2

AB A

BA A

EIMLEIML

General Principles Distribution Factor (DF)

moment

KKDF

KK

MMDF

i

iii

General Principles Member relative stiffness factor

modulus

Carry-over factor

LIKR

AA 2 ; 4

LEIM

LEIM BAAB

ABBA MM 5.0

Example 1

DF

kNmwLFEM

kNmwLFEM

CB

BC

800012

)(

800012

)(

2

2

Example 1 contd

assume Joint B fixed locked FEM@B span BC apply 8000 N.m joint joint

Example 2

internal moment @ support

Example 2 contd

Stiffness-Factor Modification

MOM DIS fixed support carry over moment

factor MOM DIS

pin support

3AB A

EIML

Stiffness-Factor Modification

(contd) member pin support Stiffness factor

Carry over factor = 0 pin support moment

fixed pin support stiffness factor 3/4

3EIKL

Stiffness-Factor Modification

Symmetric BM & Loading

BMD internal MOM @ B, C M span BC

Stiffness-Factor Modification

Symmetric BM & Loading (contd)

MOM stiffness factor

LEIK

LEIM

EIMLV

LLEIMLVM

B

BC

2

22

'

02

)(' - 0'

Stiffness-Factor Modification

Symmetric BM & Asymmetric Loading

BM internal MOM @ B MOM @C

Stiffness-Factor Modification

Symmetric BM & Asymmetric Loading (contd)

MOM = MLEIK

LEIM

EIMLV

LLEIMLL

EIMLV

M

B

B

C

6

66

'

0622

16

522

1)(' -

0'

Moment Distribution for Frames:No Sidesway

BM

Example 3 internal moment @joint, @E pinned, @D fixed

Example 3 contd

Moment Distribution for Frames: Sidesway internal moment @joint moment

distribution principle of superposition frame joint support

R sidesway

Moment Distribution for Frames: SideswayMulti-story Frames Multistory frameworks may have several independent joints disp Consequently, the moment distribution analysis using the above

techniques will involve more computation The structure shown can have 2 independent joint disp since the

sidesway of the first story is independent of any disp of the second story

Moment Distribution for Frames: SideswayMulti-story Frames These disp are not known initially The analysis must proceed on the basis of superposition 2 restraining forces R1 and R2 are applied The fixed end moments are determined & distributed Using the eqn of eqm, the numerical values of R1 and R2 are then determined The restraint at the floor of the first story is removed & the floor is given a disp This disp causes fixed end moment (FEMs) in the frame which can be assigned

specific numerical values By distributing these moments & using the eqn of eqm, the associated numerical

values of R1 and R2 can be determined In a similar manner, the floor of the second story is then given a disp With reference to the restraining forces we require equal but opposite application of

R1 and R2 to the frame such that:

111

222

"'''

"'''

RCRCR

RCRCR

Moment Distribution for Frames: SideswayMulti-story Frames

Simultaneous solution of these eqn yields the values of C and C These correction factors are then multiplied by the internal joint

moments found from moment distribution The resultant moments are found by adding these corrected

moments to those obtained for the frame

Example 4

moment @joint, EI

Example 4 contd

moment @joint, EI

Example 4 contd

moment @joint, EI

Example 4 contd

moment @joint, EI

Example 4 contd