Department of Civil Engineering

University of Engineering and Technology, Taxila, Pakistan

Analysis of Statically

Indeterminate Structures

by the Displacement Method:

Moment-Distribution Method

Theory of Structure - I

2

Introduction

In the previous lesson we discussed the slope-deflection method. In the slope-deflection analysis, the unknown displacements (rotations and translations) are related to the applied loading on the structure. The slope-deflection method results in a set of simultaneous equations of unknown displacements. The number of simultaneous equations will be equal to the number of unknowns to be evaluated. Thus one needs to solve these simultaneous equations to obtain displacements and beam end moments. Today, simultaneous equations could be solved very easily using a computer. Before the advent of electronic computing, this really posed a problem as the umber of equations in the case of multistory building is quite large.

3

Introduction

The moment-distribution method proposed by Hardy Cross in 1932, actually solves these equations by the method of successive approximations. In this method, the results may be obtained to any desired degree of accuracy. Until recently, the moment-distribution method was very popular among engineers. It is very simple and is being used even today for preliminary analysis of small structures. It is still being taught in the classroom for the simplicity and physical insight it gives to the analyst even though stiffness method is being used more and more. Had the computers not emerged on the scene, the moment-distribution could have turned out to be a very popular method.

4

Moment-Distribution Method

SIGN CONVENTION

We will establish the same sign convention as that established for the slope-deflection equations:

Clockwise moments that act on the member are considered positive (+), whereas

counterclockwise moments are negative (-).

5

Moment-Distribution Method

DEFINITION OF TERMS

30 kN 15 kN/m 2.5 kN

K

DF

FEM

1 DM

COF

2 DM

COF

3 DM

COF

4 DM

FM

SBR

EMR

V

FR

A B C D

Moment-Distribution Method

DEFINITION OF TERMS

1. Stiffness, K. The amount of force/moment required to produce a unit deformation/rotation.

Absolute Stiffness, K = 4EI/L

Relative Stiffness, K = EI/L

2. Distribution Factor, DF. A value which determines the appropriate amount of moment to be distributed.

Distribution Factor, DF = K/(ƩKj)

where, K = stiffness of the member

ƩKj = summation of stiffness meeting at a joint

Distribution of Moment, DM = - DF(ƩMj)

Caryy-Over Factor, COF = ½(DMadj)

Moment-Distribution Method

DEFINITION OF TERMS

3. Fixed End Moment, FEM. Moments developed at the end of the member due to applied loads (and/or settlement of supports) considering or idealizing the beam has fixed supports. (See list of formulas for FEM)

4. Final Moment, FM. Final moment at the end the idealized beam after the distribution of moments.

5. Simple Beam Reaction, SBR. Reaction at supports of an idealized simple beam.

6. End Moment Reaction, EMR. Reaction due to net moment of an idealized simple span.

7. Shear, V. Shear force computed as Ʃ(SBR + EMR).

8. Final Reaction, FR. Final support reaction which is equal to Ʃ(Vj).

Moment-Distribution Method

12

Moment-Distribution Method (Beams)

Example 1: Determine the reactions at supports of the beam shown.

30 kN 15 kN/m 2.5 kN

A B C D 4 m 8 m 5 m

K 10 x 0.75 = 7.5 5 x 1.0 = 5 8 x 0.75 = 6

DF

FEM

1 DM

COF

2 DM

COF

3 DM

COF

4 DM

FM

SBR

EMR

V

FR

1.0

-15

15

0

0

0

0

0

0

0

15

-17.415

-2.415

-2.415 kN

0.6

15

39

7.5

6.207

0

1.668

0

0.283

69.658

15

17.415

32.415

0.4

-80

26

-17.845

4.138

-2.78

1.112

-0.471

0.188

-69.658

60

2.059

62.059

0.455

80

-35.689

13

-5.559

2.069

-0.941

0.556

-0.253

53.183

60

-2.059

57.941

1.0

1.563

-1.563

0

0

0

0

0

0

0

1.25

-10.636

-9.386

-9.386 kN

0.545

-1.563

-42.748

-0.782

-6.659

0

-1.128

0

-0.303

-53.183

1.25

10.637

11.887

94.475 kN 69.826 kN

13

Moment-Distribution Method

Example 2: Determine the reactions at supports of the beam shown.

3 kN/m 3 kN/m

5 kN

A B C D 2 m 3 m 2 m

K 3 2 2

DF

FEM

1 DM

COF

2 DM

COF

3 DM

COF

4 DM

FM

SBR

EMR

V

FR

0

-1.833

0

-0.314

0

-0.064

0

-0.013

0

-2.224

2

0.294

2.294

2.294 kN

0.6

1.833

-0.627

0

-0.128

0

-0.025

0

-0.005

1.048

2

0.294

1.706

0.4

-0.788

-0.418

0.213

-0.085

0.042

-0.017

0.009

-0.004

-1.048

0.75

-0.302

0.448

0.4

2.588

0.425

-0.209

0.084

-0.043

0.017

-0.009

0.004

2.857

3.75

0.302

4.052

0

2.85

0

0.319

0

0.063

0

0.013

0

3.245

3

0.097

3.097

3.097 kN

0.6

-3.65

0.637

0

0.125

0

0.026

0

0.005

-2.857

5

0.097

4.903

2.154 kN 8.955 kN

1 m 1 m 3 m 2 m

14

Moment-Distribution Method

Example 3: Determine the reactions at supports of the beam shown.

1.5 kN/m 15 kN

A B C

4 m 4 m

K

DF

FEM

1 DM

COF

2 DM

COF

3 DM

COF

4 DM

FM

SBR

EMR

V

FR

12 m

Moment-Distribution Method (Frames)

1. Frames without Sidesway

2. Frame with Sidesway

Factors that may induce sidesway:

1. Horizontal load

2. Settlement of support

3. Unequal stiffness

4. Unequal legs

5. Unsymmetrical loading

A frame will not sidesway, or be displaced to the left or right, provided it is properly restrained. Also, no sidesway will occur in an unrestrained frame provided it is symmetric with respect to both loading and geometry.