The problem asks us to determine the reaction at support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a $10 kN$ point load applied downwards at a distance of $4m$ from A and a $2 kN/m$ uniformly distributed load applied over the span BC which is $4m$. The distance between A and B is $4m+2m = 6m$, and the distance between B and C is $4m$.

1. Problem Description

The problem asks us to determine the reaction at support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a $10 kN$ point load applied downwards at a distance of $4m$ from A and a $2 kN/m$ uniformly distributed load applied over the span BC which is $4m$. The distance between A and B is $4m+2m = 6m$, and the distance between B and C is $4m$.

2. Solution Steps

We will use Castigliano's second theorem to determine the reaction at the support B. Let $R_B$ be the reaction force at support B. The theorem states that the partial derivative of the strain energy $U$ with respect to the reaction force $R_B$ is equal to the displacement at the support B, which is zero in this case.

$\frac{\partial U}{\partial R_B} = \Delta_B = 0$

To apply Castigliano's theorem, we need to express the bending moment in terms of the external loads and the reaction at B. First, consider the entire beam. Summing moments about support A, we get:

$R_C * 10 - R_B * 6 - 10 * 4 - 2 * 4 * (6+4/2) = 0$

$10R_C - 6R_B - 40 - 56 = 0$

$10R_C = 6R_B + 96$

$R_C = 0.6R_B + 9.6$

Summing forces in the vertical direction, we have:

$R_A + R_B + R_C - 10 - 2*4 = 0$

$R_A + R_B + R_C = 18$

$R_A = 18 - R_B - R_C = 18 - R_B - (0.6R_B + 9.6) = 8.4 - 1.6R_B$

Now, we need to find the bending moment equations for different sections of the beam.

Section 1: $0 \le x \le 4$ (from A to the point load)

$M_1(x) = R_A x = (8.4 - 1.6R_B)x$

Section 2: $4 \le x \le 6$ (from A to B)

$M_2(x) = R_A x - 10(x-4) = (8.4 - 1.6R_B)x - 10(x-4) = (8.4 - 1.6R_B)x - 10x + 40 = (-1.6R_B - 1.6)x + 40$

Section 3: $0 \le x \le 4$ (from C to B)

$M_3(x) = R_C x - 2x^2/2 = R_C x - x^2 = (0.6R_B + 9.6)x - x^2$

Now we apply Castigliano's theorem:

$\frac{\partial U}{\partial R_B} = \int_0^L \frac{M}{EI} \frac{\partial M}{\partial R_B} dx = 0$

Assuming EI is constant, we can write:

$\int_0^L M \frac{\partial M}{\partial R_B} dx = 0$

The integral can be split into three parts according to the three sections:

$\int_0^4 M_1 \frac{\partial M_1}{\partial R_B} dx + \int_4^6 M_2 \frac{\partial M_2}{\partial R_B} dx + \int_0^4 M_3 \frac{\partial M_3}{\partial R_B} dx = 0$

$\frac{\partial M_1}{\partial R_B} = -1.6x$

$\frac{\partial M_2}{\partial R_B} = -1.6x$

$\frac{\partial M_3}{\partial R_B} = 0.6x$

Now we can calculate each integral:

$\int_0^4 (8.4 - 1.6R_B)x (-1.6x) dx = \int_0^4 (-13.44x^2 + 2.56R_B x^2) dx = [-13.44x^3/3 + 2.56R_B x^3/3]_0^4 = -13.44(64/3) + 2.56R_B (64/3) = -286.72 + 54.61R_B$

$\int_4^6 ((-1.6R_B - 1.6)x + 40) (-1.6x) dx = \int_4^6 (2.56R_B x^2 + 2.56x^2 - 64x) dx = [2.56R_B x^3/3 + 2.56 x^3/3 - 32x^2]_4^6 = (2.56R_B (216/3 - 64/3) + 2.56(216/3 - 64/3) - 32(36-16)) = 2.56R_B (50.67) + 2.56(50.67) - 32(20) = 129.71R_B + 129.71 - 640 = 129.71R_B - 510.29$

$\int_0^4 ((0.6R_B + 9.6)x - x^2) (0.6x) dx = \int_0^4 (0.36R_B x^2 + 5.76x^2 - 0.6x^3) dx = [0.36R_B x^3/3 + 5.76 x^3/3 - 0.6x^4/4]_0^4 = 0.36R_B (64/3) + 5.76(64/3) - 0.6(256/4) = 7.68R_B + 122.88 - 38.4 = 7.68R_B + 84.48$

Adding these three integrals and setting equal to 0:

$-286.72 + 54.61R_B + 129.71R_B - 510.29 + 7.68R_B + 84.48 = 0$

$192R_B - 712.53 = 0$

$R_B = 712.53/192 = 3.71 kN$

3. Final Answer

The reaction at support B is $3.71 kN$.