Analysis:

To find out the reaction at the left end of the bar, we need to consider the equilibrium of the bar. The axial load of 5 kN applied at a distance of 2m from the left end will induce a moment at the left end of the bar. This moment will cause a reaction at the left end of the bar.

Solution:

Let's assume that the cross-sectional area of the bar is A, and the material of the bar is homogeneous and isotropic. We can use the formula for the bending moment induced by the axial load to find out the reaction at the left end of the bar.

M = P * e

where M is the bending moment induced by the axial load, P is the axial load, and e is the distance of the axial load from the neutral axis of the bar.

In this case, P = 5 kN and e = 2m. The neutral axis of the bar is at its centroid, which is at a distance of 2.5m from both ends of the bar. Therefore, e = 2 - 2.5 = -0.5m.

M = 5 kN * (-0.5m) = -2.5 kNm

The negative sign indicates that the bending moment is clockwise, which means it causes compression at the left end of the bar.

To find out the reaction at the left end of the bar, we need to consider the equilibrium of the bar. Since the bar is held rigidly between the walls, it cannot move horizontally. Therefore, the sum of horizontal forces acting on the bar must be zero.

Let's assume that the reaction at the left end of the bar is R1 and the reaction at the right end is R2. Then,

R1 + R2 = 0

Since the bar is in equilibrium, the sum of vertical forces acting on the bar must be zero as well. Therefore,

R1 + R2 = 5 kN

Solving these two equations simultaneously, we get

R1 = 2.5 kN

Therefore, the reaction at the left end of the bar is 2.5 kN.

Answer:

The correct option is (b) 2 kN.