In such a case the stiffness of beam BC gets modified. This will be discussed in the next section. In the above problem the convergence may be improved if we leave the hinged end C unlocked after the first cycle. The iteration procedure is terminated when the change in beam end moments is less than say 1%. The complete procedure is shown in Fig.18.2e. The whole procedure of locking and unlocking the joints C and B successively has to be continued till both joints B and C are balanced simultaneously. However joint C is not balanced due to the carry over moment -1.287 kN.m that is developed when the joint B is allowed to rotate. Now, it is seen that joint B is balanced. This is shown in Fig.18.2d along with the carry over moments. These distributed moments restore the equilibrium of joint B. Now the distributed moments M BC and M BA are obtained by multiplying the unbalanced moment with the corresponding distribution factors and reversing the sign. When joint B is unlocked, it will rotate under an unbalanced moment equal to algebraic sum of the fixed end moments(+5.0 and -1.5 kN.m) and a carry over moment of +2.5 kN.m till distributed moments are developed to restore equilibrium. When joint C rotates, a carry over moment of +2.5 kN.m is developed at the B end of member BC.These are shown in Fig.18.2c. Now joint C is relocked and a line is drawn below +5 kN.m to indicate equilibrium. This in turn develops a beam end moment of +5 kN.m (M CB). This is the distributed moment and thus restores equilibrium. Note that joint C starts rotating under the unbalanced moment of 5 kN.m (counterclockwise) till a moment of -5 kN.m is developed (clockwise) at the joint. In this diagram B and C are assumed to be locked. In Fig.18.2b the fixed end moments and distribution factors are shown on a working diagram. In the case of fixed joint, it does not rotate and hence no distribution moments are developed and consequently distribution factor is equal to zero. The distribution moments are developed only when the joints rotate under the action of unbalanced moment. The sum of distribution factor at a joint, except when it is fixed is always equal to one. Note that distribution factor is dimensionless. Note that counterclockwise moments are taken as positive.īefore we start analyzing the beam by moment-distribution method, it is required to calculate stiffness and distribution factors. Assume that supports are unyielding.Īssuming that supports B and C are locked, calculate fixed end moments developed in the beam due to externally applied load. A continuous prismatic beam ABC (see Fig.18.2a) of constant moment of inertia is carrying a uniformly distributed load of 2 kN/m in addition to a concentrated load of 10 kN.
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