Lecture notes on Matrix Structural Analysis from the University of Ain Shams
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JIHVFull description
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Structural Analysis BookFull description
structures
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by vaidyanathan and perumal
A brief guide to computer aided analysis for civil/structural engineers. Discusses basic concepts, modelling, finite element analysis, shell structures, arches, pre-stressed concrete, 2D/3D …Full description
Structural Analysis BookFull description
Moment distribution method B Y D R . M A HD H D I D A M G HA HA N I
Introduction
y
y
y
y
It
was developed by Prof. Hardy Cross (one of America's most brilliant engineers) in 1932 It
is also known as Hardy cross method
This method was originally though of to analyse reinf reinfor orced ced concre concrete te stru struct ctur ures es A very powerful method of analysing indeterminate cont contin inuo uous us be beam amss and and fram frames es
Introduction
y
y
y
y
It
was developed by Prof. Hardy Cross (one of America's most brilliant engineers) in 1932 It
is also known as Hardy cross method
This method was originally though of to analyse reinf reinfor orced ced concre concrete te stru struct ctur ures es A very powerful method of analysing indeterminate cont contin inuo uous us be beam amss and and fram frames es
Indeterminate
y
y
structure
How many unknowns does the beam have? How many equations do we have (knowns) to solve the unknowns?
150 kN 15 kN/m
10 kN/m 3m
A I
8m
B
I
6m
C
D I
8m
Example
Back to our problem y
y
y
y
y
Fix all the joints (assume all the joints in the structure have no ability to rotate) Calculate the Fixed End Moments (FEM) Allow the joints that were fixed artificially to rotate freely The unbalanced moments created must be balanced out based on the relative stiffness of members and carry over factor Sum up the moments
See it in practice 150 kN 15 kN/m
10 kN/m 3m
A
B
I
8m
D
C
I
I
6m
8m
We are taking anti-clockwise moments as negative
15 kN/m
-80 kN.m
+80 kN.m -112.5kN.m 3m
A 8m
B
150 kN
B 6m
112.5 kN.m -53.33 kN.m
C
10 kN/m 53.33 kN.m
C 8m
D
How fixed end moment are calculated In beam AB Fixed end moment at A = -wl 2/12 = - (15)(8)(8)/12 = - 80 kN.m Fixed end moment at B = +wl2/12 = +(15)(8)(8)/12 = + 80 kN.m In beam BC Fixed end moment at B = - (Pab2)/l2 = - (150)(3)(3)2/62 = -112.5 kN.m Fixed end moment at C = + (Pab 2)/l2 = + (150)(3)(3)2/62 = + 112.5 kN.m In beam AB Fixed end moment at C = -wl 2/12 = - (10)(8)(8)/12 = - 53.33 kN.m Fixed end moment at D = +wl 2/12 = +(10)(8)(8)/12 = + 53.33kN.m You can get FEMs from next slide for various types of loadings
Fixed End Moment Table -
-
-
-
-
-
-
Fixed End Moment Table
R elease
y
y
the fixed joints
Now we allow those joints that were artificially fixed to rotate freely Due to the joint release, the fixed end moments on either side of joints B, C and D act in the opposite direction now, and cause a net unbalanced moment to occur at the joint 150 kN
15 kN/m
10 kN/m
3m A
B
I
8m R eleased moments
un balanced moment
C
I
I
6m -80.0
+112.5
+32.5
D
8m -112.5
+53.33
-59.17
-53.33
-53.33
Unbalanced
y
y
y
moment
The joint moments are distributed to either side of the joint B, C or D, according to their relative stiffnesses These distributed moments also modify the moments at the opposite side of the beam span, i.e. at joint A in span AB, at joints B and C in span BC and at joints C and D in span CD. Modification is dependent on the carry -over factor (which is equal to 0.5 in this case) 150 kN
15 kN/m
10 kN/m
3m A
B
I
8m R eleased moments
un balanced moment
C
I
I
6m -80.0
+112.5
+32.5
D
8m -112.5
+53.33
-59.17
-53.33
-53.33
Essential
y
concepts
Before continuing we need to know about
Stiffness Distribution factor Carry-over factor
Stiffness y
Stiffness = R esistance offered by member to a unit displacement or rotation at a point, for given support constraint conditions K !
4 EI L
Flexural stiffness of a member
y
Note: The above stiffness is obtained assuming that the opposite support is fixed, if it is not the case you may use 3 4 EI 3 EI K !
4
v
L
!
L
Distribution factor y
For each member at each node is DF member !
K member
§ K Summation of flexural stiffness of all connected members at a particular joint
y
Note: unbalanced moment is distributed between members based on DF of each member, i.e. M member ! M unbalanced v DF member
Carry-over factor y
If M A
is applied to the beam below it causes; U A !
y
M A L
4 EI Then, half of this moment goes to end B; 1 ! ( ) M A M B 2 MB
