Test: Structural Analysis & Steel Structures- 1 - GATE MCQ

Using Maxwell Reciprocal theorem
Virtual Work Done = Constant

(Load at Mid Point) (Deflection at Mid-Point due to load at B) = (Load at B) (Deflection at B due to load at Mid Point)

(0.5 W) (δ) = ( W) (Deflection at B)
Deflection at B = (δ)/2

Here we will apply the principle of superposition. For a simply supported beam, a concentrated load will form a triangle. The peak of triangle will be M as given in the question,

Area of free bending moment diagram = Area of triangle = 0.5 x M x L

Now, superposing the fixed end moments of supports say M1 and M2. The BMD by these fixed end moments will be a trapezoid with parallel sides having magnitude M1 and M2.

Area of fixed bending moment diagram = Area of trapezoid = - 0.5 x (M1 + M2) x L

As the slopes at the ends of a fixed beam is zero. So, the area of total B.M.D. )fixed end moment + free moment) will be zero. 
-0.5 x (M1 + M2) x L = 0.5 x M x L
(M1 + M2) = M

Point of contraflexure is the point where bending moment changes its sign. As it can be seen at point C it becomes negative to positive and at point D it becomes positive to negative, therefore, point C and D are points of contraflexure.

Fixed end moment due to concentrated load P is given by PL/8.
Since, the fixed end moment does not depend on the cross-sectional properties, hence, it will remain same.

The deflected shape of beam is shown in the figure below. It can be seen that the maximum deflection occurs between the load point and center of the beam.

Carry-over factor means the amount by which moment applied on one support is developed on another support. For a fixed adjacent support in a uniform beam carry over factor is ½. But here, since the cross-section is larger on side A, thus, moment of Inertia will be higher leading to higher flexural rigidity. Hence, support A can bear larger moment in comparison to B. 
Therefore, carry-over factor from A to B will be less than 0.5 while using moment distribution method.

Member BC will act as a zero force member unless the unit load is between A and D. This is because member AC and CD are collinear horizontal members and any vertical force, developed in member BC cannot exist due to violation of Fy =0 equilibrium equation Moreover when the unit load is at C entire unit load will be balanced by member BC. 
Therefore, the option D matches with the above description.