Solution:

Given: Span of the cantilever = L, Concentrated load at midspan = W

To find: Reaction at end B

We can solve this problem by using the principle of superposition. We will first consider the cantilever without any load and calculate the reactions at both ends. Then, we will apply the load and calculate the additional reactions due to the load. Finally, we will add the two sets of reactions to get the final reactions.

Reactions without load:

As per the principle of statics, the sum of vertical forces and the sum of moments about any point must be zero.

∑Fy = 0 (sum of vertical forces)

RA + RB = 0 ...(i)

∑MA = 0 (sum of moments about point A)

RB.L = 0

RB = 0

RA = 0

So, the reactions without load are RA = 0 and RB = 0.

Reactions due to load:

Now, we will apply the load W at midspan. Due to this load, the cantilever will deflect downwards at midspan. Let the deflection at midspan be δ.

The deflection at midspan can be calculated using the following formula:

δ = (W.L^3)/(48.E.I)

where E is the modulus of elasticity of the cantilever material and I is the moment of inertia of the cantilever about its neutral axis.

As the cantilever is propped at end A, the slope and deflection at A must be zero. So, we can assume that the cantilever is simply supported at A and calculate the slope and deflection at B due to the concentrated load.

The slope at B can be calculated using the following formula:

θB = (W.L)/(24.E.I)

The deflection at B can be calculated using the following formula:

δB = (5.W.L^4)/(384.E.I.L)

The reactions due to the load can be calculated by considering the deflection at B as the displacement of the cantilever at end B.

RB = (W/2) + (2.δB/L)

RA = W - RB

RB = (W/2) + (2.5.W.L^3)/(384.E.I.L^2)

RA = W - (W/2) - (2.5.W.L^3)/(384.E.I.L^2)

RA = (5.W)/(16)

RB = (11.W)/(16)

Therefore, the correct option is (c) 5W/16.