The problem asks 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 at 4m from A. There is a uniformly distributed load of 2 kN/m between the 10kN load and C. The length from A to the 10kN load is 4m. The length from the 10kN load to B is 2m. The length from B to C is 4m.

1. Problem Description

The problem asks 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 at 4m from A. There is a uniformly distributed load of 2 kN/m between the 10kN load and C. The length from A to the 10kN load is 4m. The length from the 10kN load to B is 2m. The length from B to C is 4m.

2. Solution Steps

First, recognize that the beam is statically indeterminate to the first degree. We will consider the reaction at B, $R_B$, as the redundant reaction.

According to Castigliano's theorem, the deflection at the support B is given by:

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

Where $U$ is the strain energy of the beam.

The strain energy due to bending is given by:

$U = \int \frac{M^2}{2EI} dx$

Therefore,

$\frac{\partial U}{\partial R_B} = \int \frac{M}{EI} \frac{\partial M}{\partial R_B} dx = 0$

We need to find the bending moment $M$ as a function of $x$ and $R_B$ for different sections of the beam.

Let $R_A$ and $R_C$ be the vertical reactions at A and C, respectively.

From the equilibrium equations:

$\Sigma F_y = 0$: $R_A + R_B + R_C - 10 - 2*4 = 0$

$R_A + R_B + R_C = 18$

$\Sigma M_A = 0$: $10(4) + 2(4)(4+2+4/2) - R_B(4+2) - R_C(4+2+4) = 0$

$40 + 48 - 6R_B - 10R_C = 0$

$6R_B + 10R_C = 88$

$3R_B + 5R_C = 44$

Now, we will express $R_A$ and $R_C$ in terms of $R_B$:

$5R_A + 5R_B + 5R_C = 90$

$5R_A = 90 - 5R_B - 5R_C$

$5R_A = 90 - 5R_B - (44 - 3R_B)$

$5R_A = 46 - 2R_B$

$R_A = 9.2 - 0.4R_B$

$5R_C = 44 - 3R_B$

$R_C = 8.8 - 0.6R_B$

Now we analyze different sections:

Section 1: $0 \le x \le 4$ (from A to 10kN force)

$M_1(x) = R_A * x = (9.2 - 0.4R_B)x$

$\frac{\partial M_1}{\partial R_B} = -0.4x$

Section 2: $4 \le x \le 6$ (from 10kN force to B)

$M_2(x) = R_A * x - 10(x-4) = (9.2 - 0.4R_B)x - 10(x-4) = (9.2-0.4R_B-10)x + 40 = (-0.8-0.4R_B)x + 40$

$\frac{\partial M_2}{\partial R_B} = -0.4x$

Section 3: $6 \le x \le 10$ (from B to C)

$M_3(x) = R_C (10-x) + \frac{2*(10-x)^2}{2} = (8.8 - 0.6R_B)(10-x) + (10-x)^2$

$M_3(x) = (8.8 - 0.6R_B)(10-x) + (100-20x+x^2)$

$\frac{\partial M_3}{\partial R_B} = -0.6(10-x)$

$\int_0^4 \frac{M_1}{EI} \frac{\partial M_1}{\partial R_B} dx + \int_4^6 \frac{M_2}{EI} \frac{\partial M_2}{\partial R_B} dx + \int_6^{10} \frac{M_3}{EI} \frac{\partial M_3}{\partial R_B} dx = 0$

Since $EI$ is constant,

$\int_0^4 M_1 \frac{\partial M_1}{\partial R_B} dx + \int_4^6 M_2 \frac{\partial M_2}{\partial R_B} dx + \int_6^{10} M_3 \frac{\partial M_3}{\partial R_B} dx = 0$

$\int_0^4 (9.2 - 0.4R_B)x (-0.4x) dx + \int_4^6 ((-0.8-0.4R_B)x + 40)(-0.4x) dx + \int_6^{10} ((8.8 - 0.6R_B)(10-x) + (10-x)^2)(-0.6(10-x)) dx = 0$

$\int_0^4 (-3.68x^2 + 0.16R_B x^2) dx + \int_4^6 (0.32x^2 + 0.16R_B x^2 - 16x) dx + \int_6^{10} (-5.28(10-x)^2 + 0.36R_B(10-x)^2 - 0.6(10-x)^3) dx = 0$

$[-3.68x^3/3 + 0.16R_B x^3/3]_0^4 + [0.32x^3/3 + 0.16R_B x^3/3 - 8x^2]_4^6 + [-5.28(10-x)^3/3 + 0.36R_B(10-x)^3/3 + 0.6(10-x)^4/4]_6^{10} = 0$

$[-3.68(64)/3 + 0.16R_B (64)/3] + [0.32(216)/3 + 0.16R_B(216)/3 - 8(36) - (0.32(64)/3 + 0.16R_B(64)/3 - 8(16))] + [0 - (-5.28(64)/3 + 0.36R_B(64)/3 + 0.6(256)/4)] = 0$

$-78.61 + 3.41R_B + [23.04 + 11.52R_B - 288 - (6.83 + 3.41R_B - 128)] + [5.28(64)/3 - 0.36R_B(64)/3 - 0.6(256)/4] = 0$

$-78.61 + 3.41R_B + 23.04 + 11.52R_B - 288 - 6.83 - 3.41R_B + 128 + 112.64 - 7.68R_B - 38.4 = 0$

$-78.61 + 23.04 - 288 - 6.83 + 128 + 112.64 - 38.4 + (3.41 + 11.52 - 3.41 - 7.68)R_B = 0$

$-148.16 + 3.84 R_B = 0$

$R_B = 148.16/3.84 = 38.58$

3. Final Answer

The reaction at support B is 38.58 kN.