The problem is to determine the reaction at support B of a continuous beam ABC using Castigliano's theorem. The beam is subjected to a $10 kN$ point load and a uniformly distributed load of $2 kN/m$. The spans AB and BC are $6 m$ and $4 m$ respectively. The distances are such that the point load is $4 m$ from A and $2 m$ from B, and the uniformly distributed load starts at B and continues to C.

1. Problem Description

The problem is to determine the reaction at support B of a continuous beam ABC using Castigliano's theorem. The beam is subjected to a $10 kN$ point load and a uniformly distributed load of $2 kN/m$. The spans AB and BC are $6 m$ and $4 m$ respectively. The distances are such that the point load is $4 m$ from A and $2 m$ from B, and the uniformly distributed load starts at B and continues to C.

2. Solution Steps

First, we need to introduce a redundant reaction at support B, which we will call $R_B$. Then we need to find an expression for the bending moment along the beam as a function of the applied loads (including $R_B$) and the distance $x$. After that, we'll apply Castigliano's theorem.

Castigliano's Theorem states that the partial derivative of the total strain energy $U$ with respect to a force (or reaction) is equal to the displacement at the point of application of the force in the direction of the force. In this case, we want to find the reaction $R_B$ such that the vertical displacement at B is zero.

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

Where the strain energy $U$ is given by:

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

Therefore,

$$\frac{\partial U}{\partial R_B} = \int \frac{2M}{2EI} \frac{\partial M}{\partial R_B} dx = \frac{1}{EI} \int M \frac{\partial M}{\partial R_B} dx = 0$$

Since $EI$ is constant, we just need to solve

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

We'll break the problem into two sections: AB and BC.

* Section AB ($0 \le x \le 6$):

Let's consider the left end A. The reaction at A ($R_A$) and the reaction at C ($R_C$) can be obtained from static equilibrium. Summing moments about C gives us:

$R_A (10) + 10(4) = R_B(4) + (2)(4)(2)$

$10R_A + 40 = 4R_B + 16$

$10R_A = 4R_B - 24$

$R_A = 0.4R_B - 2.4$

Summing forces vertically:

$R_A + R_B + R_C = 10 + 2(4)$

$0.4 R_B - 2.4 + R_B + R_C = 18$

$R_C = 20.4 - 1.4 R_B$

For $0 \le x \le 4$:

$M = (0.4R_B - 2.4)x$

$\frac{\partial M}{\partial R_B} = 0.4x$

For $4 \le x \le 6$:

$M = (0.4R_B - 2.4)x - 10(x-4)$

$\frac{\partial M}{\partial R_B} = 0.4x$

* Section BC ($0 \le x \le 4$):

Consider the right end C.

$M = (20.4 - 1.4R_B)x - \frac{2x^2}{2}$

$M = (20.4 - 1.4R_B)x - x^2$

$\frac{\partial M}{\partial R_B} = -1.4x$

Therefore, the integral becomes:

$$\int_0^4 (0.4R_B - 2.4)x (0.4x) dx + \int_4^6 [(0.4R_B - 2.4)x - 10(x-4)] (0.4x) dx + \int_0^4 [(20.4 - 1.4R_B)x - x^2] (-1.4x) dx = 0$$

$$\int_0^4 (0.16R_B x^2 - 0.96 x^2) dx + \int_4^6 [0.16R_B x^2 - 0.96 x^2 - 4x^2 + 16x] dx + \int_0^4 [-28.56 x^2 + 1.96R_B x^2 + 1.4x^3] dx = 0$$

$$[0.16R_B \frac{x^3}{3} - 0.96 \frac{x^3}{3}]_0^4 + [0.16R_B \frac{x^3}{3} - 4.96 \frac{x^3}{3} + 8x^2]_4^6 + [-28.56 \frac{x^3}{3} + 1.96R_B \frac{x^3}{3} + 1.4 \frac{x^4}{4}]_0^4 = 0$$

$$\frac{0.16R_B (64)}{3} - \frac{0.96(64)}{3} + [\frac{0.16R_B (216 - 64)}{3} - \frac{4.96 (216-64)}{3} + 8(36-16)] + [-28.56 \frac{64}{3} + 1.96R_B \frac{64}{3} + 1.4 \frac{256}{4}] = 0$$

$$\frac{10.24}{3} R_B - 20.48 + \frac{24.32}{3}R_B - \frac{757.76}{3} + 160 - \frac{1827.84}{3} + \frac{125.44}{3} R_B + 89.6 = 0$$

$$(\frac{10.24}{3} + \frac{24.32}{3} + \frac{125.44}{3}) R_B = 20.48 + \frac{757.76}{3} - 160 + \frac{1827.84}{3} - 89.6$$

$$\frac{160}{3} R_B = 20.48 + 252.59 - 160 + 609.28 - 89.6 = 632.75$$

$$R_B = \frac{632.75 * 3}{160} = 11.86 kN$$

3. Final Answer

The reaction at support B is approximately 11.86 kN.