The problem asks us to determine the reaction at support B of a beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a 10 kN load acting downwards at a distance of 4 meters from A, and a 2 kN load acting downwards at a distance of 6 meters from A. The distances between supports are: AB = 2 meters, BC = 4 meters, and AC = 6 meters.

1. Problem Description

The problem asks us to determine the reaction at support B of a beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a 10 kN load acting downwards at a distance of 4 meters from A, and a 2 kN load acting downwards at a distance of 6 meters from A. The distances between supports are: AB = 2 meters, BC = 4 meters, and AC = 6 meters.

2. Solution Steps

We will use Castigliano's second theorem to determine the reaction at support B. Castigliano's second theorem states that the partial derivative of the total strain energy $U$ with respect to the force at a point where the displacement is zero is equal to zero.

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

First, let's determine the reactions at A and C in terms of $R_B$. We have three equilibrium equations:

1. Sum of vertical forces equals zero:

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

$R_A + R_B + R_C = 12$ (Equation 1)

2. Sum of moments about A equals zero:

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

$2R_B + 6R_C = 40 + 12$

$2R_B + 6R_C = 52$

$R_B + 3R_C = 26$ (Equation 2)

Solving equations (1) and (2) for $R_A$ and $R_C$ in terms of $R_B$:

From Equation 2:

$3R_C = 26 - R_B$

$R_C = \frac{26 - R_B}{3}$

Substitute this into Equation 1:

$R_A + R_B + \frac{26 - R_B}{3} = 12$

$3R_A + 3R_B + 26 - R_B = 36$

$3R_A + 2R_B = 10$

$R_A = \frac{10 - 2R_B}{3}$

Now we need to find the bending moment equations for the two sections of the beam (AB and BC).

Section AB ($0 \le x \le 2$):

$M_1 = R_A * x = \frac{10 - 2R_B}{3} x$

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

We measure $x$ from point B to point C.

$M_2 = R_C * x = \frac{26 - R_B}{3} x$

However, we need to include the 2kN force from the original support, which is located 2m from support C, resulting in a range of $0 < x < 2$ and $2 < x < 4$. Therefore, we have:

When $0 \le x \le 2$:

$M_2 = R_C * x = \frac{26 - R_B}{3} x$

When $2 < x \le 4$:

$M_3 = R_C * x - 2(x - 2) = \frac{26 - R_B}{3} x - 2x + 4$

Now apply Castigliano's Theorem:

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

We assume EI is constant, thus:

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

We have three integrals to evaluate:

$\int_0^2 M_1 \frac{\partial M_1}{\partial R_B} dx + \int_0^2 M_2 \frac{\partial M_2}{\partial R_B} dx + \int_2^4 M_3 \frac{\partial M_3}{\partial R_B} dx = 0$

Calculate the partial derivatives with respect to $R_B$:

$\frac{\partial M_1}{\partial R_B} = \frac{\partial}{\partial R_B}(\frac{10 - 2R_B}{3} x) = -\frac{2x}{3}$

$\frac{\partial M_2}{\partial R_B} = \frac{\partial}{\partial R_B}(\frac{26 - R_B}{3} x) = -\frac{x}{3}$

$\frac{\partial M_3}{\partial R_B} = \frac{\partial}{\partial R_B}(\frac{26 - R_B}{3} x - 2x + 4) = -\frac{x}{3}$

Now, we solve the integrals.

$\int_0^2 (\frac{10 - 2R_B}{3} x)(-\frac{2x}{3}) dx + \int_0^2 (\frac{26 - R_B}{3} x)(-\frac{x}{3}) dx + \int_2^4 (\frac{26 - R_B}{3} x - 2x + 4)(-\frac{x}{3}) dx = 0$

$-\frac{2}{9}\int_0^2 (10 - 2R_B)x^2 dx - \frac{1}{9}\int_0^2 (26 - R_B)x^2 dx - \frac{1}{9}\int_2^4 (\frac{26 - R_B}{3} x^2 - 2x^2 + 4x) dx = 0$

$-\frac{2}{9} (10 - 2R_B)[\frac{x^3}{3}]_0^2 - \frac{1}{9} (26 - R_B)[\frac{x^3}{3}]_0^2 - \frac{1}{9}[\frac{26 - R_B}{9}x^3 - \frac{2x^3}{3} + 2x^2]_2^4 = 0$

$-\frac{2}{9} (10 - 2R_B)(\frac{8}{3}) - \frac{1}{9} (26 - R_B)(\frac{8}{3}) - \frac{1}{9}[(\frac{26 - R_B}{9})(64 - 8) - \frac{2}{3}(64 - 8) + 2(16 - 4)] = 0$

Multiply by -9 to remove negative values and simplify:

$\frac{16}{3}(10 - 2R_B) + \frac{8}{3}(26 - R_B) + (\frac{26 - R_B}{9}(56) - \frac{2}{3}(56) + 2(12)) = 0$

$\frac{160}{3} - \frac{32}{3}R_B + \frac{208}{3} - \frac{8}{3}R_B + \frac{1456}{9} - \frac{56}{9}R_B - \frac{112}{3} + 24 = 0$

Multiply by 9 to get rid of fractions:

$480 - 96R_B + 624 - 24R_B + 1456 - 56R_B - 336 + 216 = 0$

$2480 - 176R_B = 0$

$176R_B = 2480$

$R_B = \frac{2480}{176} = \frac{310}{22} = \frac{155}{11} \approx 14.09$

3. Final Answer

The reaction at support B is approximately 14.09 kN.

$R_B = 14.09 kN$