Moment Distribution Method
The method consists in solving indirectly the equations of equilibrium as formulated in slope deflection method without finding the displacements. This is an iterative procedure. This is also known as relaxation method.
Displacement Method of Analysis (Moment Distribution Method)
The moment distribution method is the most suitable manual method for the analysis of continuous beams and plane frames. The method was presented by Prof. Hardy Cross of the USA in 1929.
The method consists in solving indirectly the equations of equilibrium as formulated in the slope deflection method without finding the displacements. This is an iterative procedure. This is also known as a relaxation method.
In this method the analysis begins by assuming each joint in the structure to be fixed. Then, by unlocking and locking each joint in succession, the internal moments at the joints are distributed and balanced until the joints have rotated to the final and nearly positions. The following examples has been given to illustrate the basic concept.
Sign Convention
The clockwise moment is taken as +ve and the anti-clockwise moment is taken as -ve i.e., the sign convention is the same as that in slope deflection method for member end moment calculation.
Read Also- Slope Deflection Method
Member Stiffness Factor
Stiffness M/θ = K = 4EI/l and θ = Ml/4EI
Stiffness M/θ = K = 3EI/l and θ = Ml/3EI
Joint Stiffness Factor
As joint A is rigidly connected, all member at joint A will rotate by the same amount θA. If a moment M is applied at joint A the moment gets distributed in all the connected members. As all connected members rotate by same amount θA at A. We have-
M = (4EI3θA/l3 + 4EI2θA/l2 + 4EI1θA/l1)
M/ θA = K1 + K2 + K3 = Joint Stiffness Factor (K)
Distribution Factor (D.F.)
If a moment M is applied at joint A due to which the joint rotates by θA then,
Moment Distributed in AD= 4EI3θA/l3 = KAD θA = MDAD
Moment Distributed in AC= 4EI2θA/l2 = KAC θA = MDAC
Moment Distributed in AB= 3EI1θA/l1 = KAB θA = MDAB
MDAD : MDAC : MDAB = I3/l3 : I2/l2 : 3I1/4l1
Where,
I3/l3 = Relative stiffness of AD
I2/l2 = Relative stiffness of AC
3I1/4l1 = Relative stiffness of AB
Relative Stiffness when far end is fixed = I/l
Relative Stiffness when far end is hinged = 3I/4l
Distribution Factor of a Member = Stiffness of member/Total Stiffness of all members at the joint
Carry Over Factor
When M moment is applied at pin, Moment carried over to fixed far end= M/2
Carry Over Factor= Carried over moment/Applied moment
However, In the following case, carry over factor = 0
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