1. ERT 348 Controlled Environment Design 1
Moment Distribution
Ms Siti Kamariah Binti Md Sa’at
School of Bioprocess Engineering, UniMAP

2. Introduction
Such structures are indeterminate, i.e. there are more unknown variables than can be solved using only the three equations of equilibrium.
This section deals with continuous beams and propped cantilevers.
An American engineer, Professor Hardy Cross, developed a very simple, elegant and practical method of analysis for such structures called Moment Distribution.

3. Indeterminate structures

4. Bending (Rotational) Stiffness
A fundamental relationship which exists in the elastic behaviour of structures and structural elements is that between an applied force system and the displacements which are induced by that system,
Force = Stiffness x Displacement
P = k x δ
Stifness, k = P/δ
when δ = 1.0 (i.e. unit displacement) the stiffness is: ‘the force necessary to maintain a UNIT displacement, all other displacements being equal to zero.’

5. Bending (Rotational) Stiffness
The displacement can be a shear displacement, an axial displacement, a bending (rotational) displacement or a torsional displacement, each in turn producing the shear, axial, bending or torsional stiffness.
When considering beam elements in continuous structures using the moment distribution method of analysis, the bending stiffness is the principal characteristic which influences behaviour.

6. Example:
The force (MA) necessary to maintain this displacement can be shown (e.g. Using McCaulay’s Method) to be equal to (4EI)/L.
If the bending stiffness of the beam is equal to (Force/1.0), therefore k = (4EI)/L.
This is known as the absolute bending stiffness of the element.

7. General Principles & Definition
Member stiffness factor
Joint stiffness factor
The total stiffness factor of joint A is
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

8. General Principles & Definition
Member relative stiffness factor
Quite often a continuous beam or a frame will be made from the same material
E will therefore be constant
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

9. The moment MB is known as the carry-over moment.
When the beam element deforms due to the applied rotation at end A, an additional moment (MB) is also transferred by the element to the remote end if it has zero slope (i.e. is fixed).
The moment MB is known as the carry-over moment.

10. Carry-Over Moment
Carry-over (CO) factor
Solving for  and equating these eqn,
The moment M at the pin induces a moment of M’ = 0.5M at the wall
In the case of a beam with the far end fixed, the CO factor is +0.5

11. Carry-Over Moment
Therefore be stated that ‘if a moment is applied to one end of a beam then a moment of the same sense and equal to half of its value will be transferred to the remote end provided that it is fixed.’
If the remote end is ‘pinned’, then the beam is less stiff and there is no carry-over moment.

12. Pinned End ‘the stiffness of a pin-ended
beam is equal to ¾ × the stiffness of a fixed-end beam.’

13. Free Bending Moment
When a beam is free to rotate at both ends, no bending moment can develop at the supports, then the bending moment diagram resulting from the applied loads on the beam is known as the Free Bending Moment Diagram.

14. Fixed Bending Moment
When a beam is fixed at the ends (encastre) such that it cannot rotate, i.e. zero slope at the supports, then bending moments are induced at the supports and are called Fixed-End Moments.
The bending moment diagram associated only with the fixed-end moments is called the Fixed Bending Moment Diagram.

15. Fixed Bending Moment +

16. Example calculation 1:

17. 1 2 1 2

19. Propped Cantilever
The fixed-end moment for propped cantilevers (i.e. one end fixed and the other end simply supported) can be derived from the standard values given for encastre beams as follows.
Consider the propped cantilever, which supports a uniformly distributed load as indicated.

20. Propped Cantilever
The structure can be considered to be the superposition of an encastre beam with the addition of an equal and opposite moment to MB applied at B to ensure that the final moment at this support is equal to zero.

21. Example calculation 2:

22. Solutions:

23. Example: Solution

25. Example: Solution

26. Distribution Factor
Distribution Factor (DF)
That fraction of the total resisting moment supplied by the member is called the distribution factor (DF)

27. Consider 1 case

28. Stiffness of span BA = KBA = (I1/L1)
Stiffness of span BC = KBC = (I2/L2)
Total stiffness at the support = ∑K = KBA + KBC
The moment absorbed by beam BA
The moment absorbed by beam AB

29. Example Calculation 1:

30. Solution:
Note that the above results could also have been obtained if the relative stiffness factor is used

31. Solution:
As a result, portions of this moment are distributed in spans BC and BA in accordance with the DFs of these spans at the joint
Moment in BA is 0.4(8000) = 3200Nm
Moment in BC is 0.6(8000) = 4800Nm
These moment must be carried over since moments are developed at the far ends of the span

32. Solution

33. Example Calculation 2 (Example 12.2)
Determine the internal moment at each support of the beam. The moment of inertia of each span is indicated.
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

34. A moment does not get distributed in the overhanging span AB
Solution
A moment does not get distributed in the overhanging span AB
So the distribution factor (DF)BA =0
Span BC is based on 4EI/L since the pin rocker is not at the far end of the beam
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

35. Solution
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

36. -2000Nm is added to BC in order to satisfy this condition.
Solution
The overhanging span requires the internal moment to the left of B to be +4000Nm.
Balancing at joint B requires an internal moment of –4000Nm to the right of B.
-2000Nm is added to BC in order to satisfy this condition.
The distribution & CO operations proceed in the usual manner.
Since the internal moments are known, the moment diagram for the beam can be constructed.
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

37. Solution
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

38. Stiffness-Factor Modifications
Member pin supported at far end
As shown the applied moment M rotates end A by an amt 
To determine , the shear in the conjugate beam at A’ must be determined

39. Stiffness-Factor Modifications
Member pin supported at far end (cont’d)
The stiffness factor in the beam is
The CO factor is zero, since the pin at B does not support a moment
By comparison, if the far end was fixed supported, the stiffness factor would have to be modified by ¾ to model the case of having the far end pin supported

40. Stiffness-Factor Modifications
Symmetric beam & loading
The bending-moment diagram for the beam will also be symmetric
To develop the appropriate stiffness-factor modification consider the beam
Due to symmetry, the internal moment at B & C are equal
Assuming this value to be M, the conjugate beam for span BC is shown

41. Stiffness-Factor Modifications
Symmetric beam & loading (cont’d)
Moments for only half the beam can be distributed provided the stiffness factor for the center span is computed

42. Stiffness-Factor Modifications
Symmetric beam with asymmetric loading
Consider the beam as shown
The conjugate beam for its center span BC is shown
Due to its asymmetric loading, the internal moment at B is equal but opposite to that at C

43. Stiffness-Factor Modifications
Symmetric beam with asymmetric loading
Assuming this value to be M, the slope  at each end is determined as follows:

44. Example Calculation 3 (Example 12.4)
Determine the internal moments at the supports of the beam shown below. The moment of inertia of the two spans is shown in the figure.
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

45. The beam is roller supported at its far end C.
Solution
The beam is roller supported at its far end C.
The stiffness of span BC will be computed on the basis of K = 3EI/L
We have:
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

46. Solution
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd

47. The forgoing data are entered into table as shown.
Solution
The forgoing data are entered into table as shown.
The moment distribution is carried out.
By comparison, the method considerably simplifies the distribution.
The beam’s end shears & moment diagrams are shown.
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd