4.1 Moment distribution methodStiffness factor Carry over factor and Distribution factor | unit 4 moment distribution method

Unit - 4

Moment Distribution Method

4.1 Moment distribution method-Stiffness factor, Carry over factor and Distribution factor

Moment distribution method:

Moment Distribution Method (MDM) is an iteration procedure developed by Cross Hardy in 1930. Therefore, this method is sometimes called as the 'Cross Hardy Method'.

This method starts with the assumption that all points in an indeterminate structure are fixed temporarily against rotation and displacement.

The fixed end moment (FEM) induced by the loading on the spans are written down at each joint.

Then, the joint is released one at a time in succession. Because of the release of joints, unbalanced moments arise at the joints.

These unbalanced moments distributed at the joints in proportion to the stiffness of the members meeting at the joints.

Then these distributed moments are carried over to the far end of the members. Because of the carry-over moments at joints, again unbalanced moments are created which are further distributed and carried over.

In this manner the iteration is performed in successive cycles of operation till the value of joints becomes significant. At this stage, iteration is stopped and the moments at each joint is algebraically summed up.

The sum of the moments at the end of each span at a joint is the required final end moment.

MDM method is quite different than slope deflection method though it is a displacement or stiffness method. Without calculating values of displacement end moments are obtained by successive iteration.

2. Stiffness factor

It is ratio of moment of inertia to the length of the beam is called stiffness factor of beam.

Moment required to rotate an end by a unit angel when rotation is permitted at that end is called stiffness of the beam

Stiffness of the beam AB=

Types of far end support	K
If far end is fixed or continuous support	I/L
2. If far end is simply supported Roller support, Internal hinged and overhang	3I/4L

3. Carry over factor:

It is defined as the ratio of moment developed at far or adjacent end of the moment applied at the near end.

The ratio of carryover moment to applied moment is called carry-over factor.

Carryover factor=

Types of toes and support

Fixed support	½
2. Intermediate support	½
3. Rigid joint	½
4. Roller hinged, internal	0
5. Overhanging	0
6. If for the end is fixed	1/l
7. If for the end is simply roller internal hinged, I overhung	3I/4L

4. Distribution factor

It is ratio of the stiffness of member to the total stiffness of the joint where all the members are meeting.

A moment which tends to rotate without translation a joint to which several members are connected will be divided amongst connected member in proportion to their stiffness.

Distribution factor (DF)

Thus, distribution factor is the ratio of the stiffness of member to the total stiffness of the joint where all members are meet.

5. Application of MD method to continuous beam with fixed ends: -

Assuming all ends are fixed, find fixed end moments

Calculate the distribution factor for all member meetings at the joint.

Balance a joint by distributing a balancing moment.

Carryover half the distributed movement to the far end of the members.

Repeat step 3 and step 4 till distributed moments are negligible.

Sum up all the moments at a particular end of the members to get the final moment.

Examples: To find stiffness

By conjugate method to find stiffness.

Then consider at the B end

Carryover factor =

Put

Distribution factor =>

Key takeaway:

It is ratio of moment of inertia to the length of the beam is called stiffness factor of beam.

It is defined as the ratio of moment developed at far or adjacent end of the moment applied at the near end.

It is ratio of the stiffness of member to the total stiffness of the joint where all the members are meeting

4.2 Application of Moment distribution method of analysis to beams with and without joint translation and yielding of support

Steps for analysis of moment distribution method

Find fixed end moment

Find the distribution factor

Draw moment distribution table

You get final moments

Draw BMD&SFD

Distribution factor is=

K=stiffness -it depends on the support condition.

1. Far end fixed &intermediate support =

2. The far end simply supports =

COF - support

O - Simply support

½ - fixed

½ - intermediate

Problems:

Analyze the given beam by M.D. method (10 marks item)

AB=>

Step 1) Fixed end moment

Step 2) Distribution factor

Draw distribution table

Joint	Member	K
B	BA BC			0.4 0.6

Step 3) Moment Distribution

1) Moment distribution Table

		0.4	0.6
Member	AB BA		BC CD
Fixed end Balancing	120 - 120 42		15 -15 63
COF Balancing	210 0		31.5
Final moments	141 -78		78 16.5

Step 4) Draw SFD and BMD

SFD

BMD

2. Analyze the given beam ABCD by Moment Distribution method

Step 1) Find fixed moment

KNM

Step 2) Distribution factor

JOINT	MEMBER	K
B	BA BC		5EI	0.2 0.8
C	CB CD		4.5EI	0.89 0.11

Step 3) Moment Distribution

Moment Distribution Table

		0.2		0.8	0.89	0.11
Members	AB BA 15 -15		BC CB 13.33 -13.33			CD DC 8.89 -4.44 2.22 + 4.44
Fixed End
Balancing
Initial Moments Balancing	15 -15 0.331		13.33 -13.33 1.34 1.98			11.1 0 0.25
COF Balancing	0.165 -0.19		0.99 0.67 -0.79 -0.59			-0.07
COF Balancing	-0.09 0.058		-0.29 -0.39 0.22 0.34			0.04
Final Moments	15.075 -14.801		14.8 -11.32			11.32 0

KNM

Step 4) Draw BMD

BMD

3. Analyze the given beam by M.D. method

Step 1) Find fixed end moment

Step 2) Distribution Factor

Joint	Members	K
B	BA BC		2.5 EI	0.2 0.8
C	CB CD			0.66 0.33

Step 3) Moment Distribution

Moment distribution table

		0.2		0.8	0.66	0.33
Members	AB BA		BC CB			CD DC
Fixed End	60 -60 -60 -30		30 -30			13.33 -6.66 3.33 +6.66
Balancing
Initial Moments Balancing	0 – 90 12		30 -30 48 8.8			16.66 0 4.40
COF Balancing	-0.88		4.4 24 -3.52 -15.84			-1.92
COF Balancing	1.58		-7.92 -1.76 6.33 1.16			0.58
COF Balancing	- 0.11		0.58 3.16 -0.46 -2.08			-1.04
COF Balancing	0.20		-1.04 -0.23 0.83 0.15			0.07
Final Moments	0 -77.21		77.21 -12.64			12.75 0

Step 4) Draw BMD

BMD

4. Analyze the given frame by MD method

Step 1) Find Fixed End Moments

Step 2) Distribution Factor

Distribution table

joint	Member	K
B	BA		1.41	0.47
	BC		1.41	0.53

Step 3) Moment Distribution

Moment distribution table

		0.47	0.53
Member	AB BA		BC CB
External moment Fixed end Balancing	60 - 60		135 2.5 -2.5 -1.25 2.5
Initial Moment Balancing	60 -60 27.61		1.25 135 31.13
COF Balancing	13.8 0 0		0 0
Final moments	73.8 -32.39		32.39135

BMD

Problem-based on sink

1) Analyze the given beam ABCD having pt B is sink by 10mm EI

For AB Span

KNM

BC= +VE

CD=0 No loading

Step 2) Distribution factors

Joint	Members	K
B	BA BC		1.66	0.60 0.40
C	CB CD			0.40 0.60

Step 3) Moment Distribution Table

			0.6	0.4			0.4		0.6
Members	AB	BA			BC	CB		CD		DC
Fixed end Balance	34.25	-5.75 -28.74			53.66 -19.16	-66.33 26.53		0 39.79		0
COF Balancing	+14.37	-7.95			13.26 -5.30	-9.58 +3.83		+5.74		19.89
COF Balancing	-3.97	-1.14			1.91 -0.76	-2.65 1.06		1.59
COF Balancing	-0.57	-0.31			0.53 -0.21	0.38 0.15		0.22		0.79
Final Moments	15.34	-43.89			43.89	-46.61		-46.61		20.68

2. Draw BMD by Moment Distribution Method and C support is sink by 25mm

For AB

For BC mm

For CD mm

Step 2) Distribution factor

Joint	Members	K
B	BA BC		0.33 0.66
C	CB CD		0.66 0.33

Step 3) Moment Distribution Table

			0.33	0.66			0.66		0.33
Members	AB	BA			BC	CB		CD		DC
Fixed end	40	-40			149.16	135.83		-40.83 42.91		-85.83 +85.83
Final Balance	40	-40 -36.02			149.16 -72.04	135.83 -91.02		2.08 -45.51		0
COF Balancing	-18.01	15.02			-45.51 30.04	-36.02 23.77		11.89
COF Balancing	7.505	-3.92			11.89 -7.85	15.02 -9.91		-4.957
COF Balancing	-1.96	1.64			-4.96 3.28	-3.93 2.59		1.296
Final Moments	29.28	-63.28			64.01	36.33		-35.20		0

Key takeaway:

Find fixed end moment

Find the distribution factor

Draw moment distribution table

You get final moments

Draw BMD&SFD

4.3 Application to non-sway rigid jointed rectangular portal frames, shear force and bending moment diagram.

Non- sway

Problems:

Analyze the frame shown in fig. by MDM and Draw BMD

Step 1) Fixed End Moments

Step 2) Distribution Factor

Joint	Members	K
B	BA BD BC		2.33	0.28 0.43 0.28

Step 3) Moment Distribution Factor

Moment Distribution Table

		0.28	0.43	0.28
Members	AB	BA	BD		BC	CB	DB
Fixed end Balance	90	-90 20.3	10 31.17		7.5 20.3	-1.5	-10
Cot Balancing	10.15	0 0	0 0		0 0	10.15	15.58
Final	100.15	-69.7	42.17		27.8	2.65	5.58

Step 4) Draw BMD and SFD

BMD

2. Determine Moments of the given figure by moment Distribution Method

Step 1) Non-sway

Step 2) Fixed End moments

knm

Step 3) Distribution factor

Joint	Members	K
B	BA BC		0.43 0.57
C	CB CD		0.64 0.36

Step 4) Moment Distribution Factor

Moment Distribution table

		0.43	0.57			0.64	0.36
Members	AB	BA		BC	CB			CD	DC
Fixed end	0	0		22.5	-22.5			0	0
Initial Balance	0	0 -0.96		22.5 -1.28	-22.5 +1.14			0 +0.81	0
COF Balancing	-0.48	0.24		+0.57 0.32	-0.64 0.40			0.23
COF Balancing	0.12	-0.086		0.20 -0.114	0.16 -0.10			-0.057
Final Moments	-0.36	-0.806		0.806	-3.57			-0.637	0

3. Analyses the given frame as shown. By MDM

Step 1) Fixed end moment

Step 2) Distribution Factor

Joint	Members	K
B	BA BD BC		3.455 EI	0.40 0.28 0.32

Step 3) Moment Distribution Factor

Moment Distribution Table

			0.40	0.28	0.32
Members	AB	BA		BD		BC	CB	DB
Fixed end	106.66	-53.33		10 5		40 20	-40 40	-10 +10
Initial Balance	106.66	-53.33 -8.66		15 -6.06		60 -6.43	0	0
COF Balancing	-4.33	0 0		0 0		0 0
Final Moments	102.35	-61.99		8.94		53.07	0	0

Step 5) Draw BMD

Problem-based on non-sway frame

4. Analyze the non-sway frame as shown in fig. by MDM

Step 1) Fixed end moment

Step 2) Distribution factor

Joint	Members	K
B	BA BC	=	0.70 0.30
C	CB CD		0.30 0.70

Step 4) Moment distribution factor

Moment Distribution table

		0,70	0.30			0.30	0.70
Members	AB	BA		BC	CB			CD	DC
Fixed end	30 -30	-30 -15		90	-90			30 15	-30 30
Initial Balance	0	-45 -31.5		90 -13.5	-90 13.5			45 31.5	0
COF Balancing		-4.72		6.75 -2.02	-6.75 2.02			4.72
COF Balancing		-0.7		1.01 -0.3	-1.01 0.3			0.7
COF Balancing		-0.1		0.15 -0.04	-0.15 0.04			0.1
Final Moments	0	-82.2		82.2	82.2			-82.2	0

Step 4) Draw BMD

Key takeaway:

Steps:

Fixed end moment

Distribution factor

Moment distribution factor

Draw BMD and SFD

4.4 Sway analysis of rigid jointed rectangular single bay single storey portal frames using Moment distribution method (Involving not more than three unknowns)

Concept: The carry over for fixed support

For Hinge support end moment

is not known.

Problems:

Analyze the given frame by moment distribution method take EI Constant

Step 1) Find Distribution Factor

Joint	Member	K	ƩK		D.F.
B	BA	=3EI/3=1 EI		0.33
	BC	= 4EI/2 = 2 EI		0.67
C	CB	= 4EI/2=2EI= 2EI		0.67
	CD	= 4EI/4= EI		0.33

Step 2) Non sway moment distribution moment Table

Members	AB BA	BC CB	CD DC
Fixed End Balancing	0 0 0	0 0 0 0	0 0 0
Final Moments	0 0	0 0	0 0

Step 3) Assumption For sway Frame

Step 4: Draw Sway Moment Distribution Table

			0.33	0.67		0.67	0.33
Member	AB	BA		BC CB			CD DC
FEM Balancing	0	8		0	0		9	9
		-2.67		-5.33	-6.00		-3.00
COF Balancing		1		-3	-2.67			1.5
				2	1.78		0.85
COF Balancing		-0.30		0.69	1
				-0.39	-0.67		-0.33
COF Balancing		0.11		-0.33	-0.30			0.17
				0.22	0.20		0.10
COF Balancing		-0.03		0.10	0.11			0.05
				-0.07	-0.07		0.05
	0	6.11		-6.11	-6.62		6.62	7.83

Horizontal thrust at A

Actual Sway force of 12 kN

Step 6) Final Moments table:

Member	AB BA			BC CB		CD DC
Non-Sway Moment	0 0			0 0		0 0
Sway Moment	0	6.11	-6.11		-6.62	6.62	7.83
Actual Sway	0	+12.98	-12.98		-14.06	+14.06	+16.6
Final Moment (Addition of Non-Sway Moment +corrected moment by sway)	0	+12.98	-12.98		-14.06	+14.06	+16.6

Correct Horizontal A reaction

2. A two hinged portal frame ABCD consist of a vertical columns AB, DC at 4m height if the beam BC of 3m the frame carried a vertical point load of 120kN on the beam at a distance of 1m from B find the reaction at support.

Solution:

Non-Sway analysis:

Step 1) Fixed end moment

Step 2) Draw Distribution Factor

Joint	Member	K	ƩK	D.F.
B	BA	3EI/L= 3EI/4 = 0.75 EI	2.08 EI	0.36
B	BC	4EI/L= 4EI/3=1.33 EI	2.08 EI	0.64
C	CB	4EI/L = 4EI/3=1.33EI	2.08 EI	0.64
C	CD	3EI/L= 3EI/4=0.75 EI	2.08 EI	0.36

Step 3) Draw Non sways moment distribution Table

		0.36	0.64	0.64	0.36
Member	AB BA		BC CB			CD DC
FEM Balancing	0 -19.2		53.33 -26.66 -34.13 17.06			0 0 9.6
COF Balancing	-3.07		8.53 -17.07 -5.46 10.92			6.15
COF Balancing	-1.96		5.46 -2.73 -3.49 1.75			0.98
Final Moments	0 -24.23		24.24 -16.73			16.73 0

Horizontal reaction

Sway force = 1.87 kN

Sway analysis

Step 4) Initial equivalent moments are negative.

		0.36	0.64	0.64	0.36
Member	AB BA		BC CB			CD DC
FEM Balancing	0 -10 3.6		0 0 6.4 6.4			-10 0 3.6
COF Balancing	- 1.15		3.2 3.2 -2.05 -2.05.			-2.05
COF Balancing	0.35		- 1.03 -1.03 0.66 0.66			0.35
Final Moments	0 -7.2		7.2 7.2			-7.2 0

Horizontal reaction

Resolving force horizontally = 1.81 +1.81 =3.63kN

Actual Sway = 3

Sway 0 7.27	7.27 7.27	-7.27 0
0 – 3.64	+3.64 3.64	-3.64 0
Non-Sway 0 21.17	-21.17 21.17	-21.17 0

Analysis of frame with sway

3. Analyses the rigid frame shown in fig by M.D. method.

Step1) Fixed end moment

Fixed end moment in all the members = 0

Step 2) Distribution factor

D.F

B	BA	2.33EI	0.57
B	BC	2.33EI	0.43
C	CB	2.5EI	0.4
C	CD	2.5EI	0.6

Conclude S = 80 kN

Sway Force

Consider FBD for column

SWAY ANALYSIS

Consider Sway on right hand side

Consider

Moment distribution maybe carried out.

A B C D

10	0.5 10	0.5	0.5	0.5 20	20
	-5	-5	-10	-10
-2.5	2.5	-5	-2.5		-5
-2.5	2.5	2.5	1.25	1.25
1.25	-0.32	0.63	1.25		0.63
1.25	-0.32	-0.32	-0.62	-0.62
-0.16	+0.16	-0.31	-0.16		-0.31
-0.16	+0.16	0.16	0.08	0.08
0.08	-0.02	0.04	0.08		0.04
0.08	-0.02	-0.02	-0.04	0.04
+8.67	-7.32	-7.32	-10.66	10.66	15.35

Sway Correction Factor

Arbitrary Sway	+8.67	+7.32	-7.332	-10.66	+10.66	+15.37
Actual Sway	8.67×6.37= 53.88	7.3×6.2= 45.44	-45.44
Non-Sway	+20.42	-12.44	+12.44	+3.35	-3.35	-1.77
Final	+74.30	+33.05	-33.05	-62.75	+62.7	+93.62

Key takeaway:

Step 1) Fixed end moment

Step 2) Distribution factor

Step 3) Moment Distribution factor

Step 4) Draw BMD and SFD

References:

1. Structural Analysis: Deodas Menon---Narosa Publishing House.

2. Structural Analysis: Thandavamoorthy---Oxford University Press.

3. Structural Analysis: A Matrix Approach by Pundit and Gupta, McGraw Hills.

4. Structural Analysis by Hibbler, Pearson Education.

5. Structural Analysis: M. M. Das, B. M. Das---PHI Learning Pvt Ltd. Delhi.

6. Fundamentals of Structural Analysis: 2nd ed---West---Wiley.

7. Theory of Structures: Vol. I & II by B. C. Punmia, Laxmi Publication.

8. Theory of Structures: Vol. I & II by Perumull & Vaidyanathan, Laxmi Publication.

9. Fundamentals of Structural Analysis: K. M. Leet, Vang, Gilbert—McGraw Hills

10. Matrix Methods for structural engineering. by Gere, Weaver.

11. Introduction to the Finite element method, Dr. P.N. God bole, New Age Publication, Delhi.

12. Finite element Analysis, S.S. Bhavikatti, New Age Publication, Delhi.

13. Basic Structural Analysis: Wilbur and Norris.

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