Analysis:
Given:
- Span of the beam (L) = 6 m
- Flexural rigidity (EI) = 4200 kN m^2
- Displacement at support B (δ) = 25 mm = 0.025 m

We need to find the moment induced at the end A due to the downward displacement at support B.

Assumptions:
- The beam is uniform and has a constant flexural rigidity along its length.
- The beam is loaded in such a way that it remains horizontal throughout the loading process.
- The downward displacement at support B is small enough that it does not cause significant changes in the beam's shape or behavior.

Derivation:
- The equation governing the behavior of a beam under bending is the Euler-Bernoulli beam equation:

d²M/dx² = -w(x)

where M is the bending moment, x is the distance along the beam, and w(x) is the distributed load per unit length.

- The solution to this equation for a simply supported beam subjected to a uniformly distributed load is given by:

M(x) = -w₀x²/2 + C₁x + C₂

where w₀ is the uniformly distributed load, C₁ and C₂ are constants of integration.

- For a simply supported beam with a span of L and a uniformly distributed load of w₀, the bending moment at the supports is zero. Therefore, we can determine the values of C₁ and C₂ using the following conditions:

M(0) = 0 (at support A)
M(L) = 0 (at support B)

Using these conditions, we can solve for C₁ and C₂.

- Substituting the values of C₁ and C₂ into the equation for M(x), we get:

M(x) = -w₀x²/2 + 2w₀Lx - w₀L²/2

This equation represents the bending moment along the length of the beam.

- To find the moment induced at the end A (M_A), we substitute x = 0 into the equation:

M_A = -w₀(0)²/2 + 2w₀L(0) - w₀L²/2

= -w₀L²/2

The bending moment at the end A is given by -w₀L²/2.

Solution:
In this problem, the downward displacement at support B causes a clockwise rotation at that point. According to the sign convention for bending moments, a clockwise rotation induces a negative bending moment. Therefore, the moment induced at the end A is in the opposite direction, i.e., anticlockwise.

Given:
- Displacement at support B (δ) = 0.025 m

Since the beam is simply supported, the displacement at the end A is zero. Therefore, the moment induced at the end A can be calculated using the equation:

M_A = -w₀L²/2

To find the value of w₀, we can use the formula:

δ = 5w₀L^4 / (384EI)

Substituting the given values of δ, L, and EI into the equation, we can solve for w₀. Once we have