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 load applied at 4 m from A and a 2 kN load applied at 10 m from A. The distances between the supports are AB = 6 m and BC = 4 m. We need to find the reaction at support B.

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 load applied at 4 m from A and a 2 kN load applied at 10 m from A. The distances between the supports are AB = 6 m and BC = 4 m. We need to find the reaction at support B.

2. Solution Steps

Since we're using Castigliano's theorem, we need to introduce a dummy variable, let's call it $R_B$, which represents the reaction at support B. Castigliano's theorem states that the deflection at a point where a force is applied is equal to the partial derivative of the strain energy with respect to that force. Since support B is a roller support, the deflection at B is zero. Therefore, we have:

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

Where $U$ is the strain energy of the beam due to bending, given by:

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

Where $M$ is the bending moment, $E$ is the modulus of elasticity, and $I$ is the moment of inertia. Assuming EI is constant, we can write:

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

Which simplifies to:

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

First, we need to find the reactions at supports A and C in terms of $R_B$. We have the following equilibrium equations:

$\sum F_y = R_A + R_B + R_C - 10 - 2 = 0$

$R_A + R_B + R_C = 12$

$\sum M_A = R_B(6) + R_C(10) - 10(4) - 2(10) = 0$

$6R_B + 10R_C = 60$

We express $R_A$ and $R_C$ in terms of $R_B$. From the second equation:

$10R_C = 60 - 6R_B$

$R_C = 6 - 0.6R_B$

Substitute into the first equation:

$R_A + R_B + (6 - 0.6R_B) = 12$

$R_A + 0.4R_B = 6$

$R_A = 6 - 0.4R_B$

Now we define the bending moment in each section:

Section AB (0 < x < 6):

$M_1 = R_A x = (6 - 0.4R_B)x$

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

Section from A to the 10kN load (0 < x < 4):

$M_{1a} = R_A x = (6 - 0.4R_B)x$

$\frac{\partial M_{1a}}{\partial R_B} = -0.4x$

Section between 10kN load and B (4 < x < 6):

$M_{1b} = R_A x - 10(x-4) = (6 - 0.4R_B)x - 10x + 40 = (6 - 0.4R_B - 10)x + 40 = (-4 - 0.4R_B)x + 40$

$\frac{\partial M_{1b}}{\partial R_B} = -0.4x$

Section BC (0 < x < 4) where x is the distance from B:

$M_2 = R_C x = (6 - 0.6R_B)x$

$\frac{\partial M_2}{\partial R_B} = -0.6x$

Section from B to the 2kN load (0 < x < 4):

$M_{2a} = R_C x = (6 - 0.6R_B)x$

$\frac{\partial M_{2a}}{\partial R_B} = -0.6x$

Now we can calculate the integral. Divide the integral into parts. We will integrate from A to B, and then from B to C. Also we need to consider the points where the 10 kN and 2 kN forces are applied.

$\int_0^4 M_1 \frac{\partial M_1}{\partial R_B} dx + \int_4^6 M_{1b} \frac{\partial M_{1b}}{\partial R_B} dx + \int_0^4 M_2 \frac{\partial M_2}{\partial R_B} dx = 0$

$\int_0^4 (6 - 0.4R_B)x (-0.4x) dx + \int_4^6 ((-4 - 0.4R_B)x + 40) (-0.4x) dx + \int_0^4 (6 - 0.6R_B)x (-0.6x) dx = 0$

$-0.4\int_0^4 (6 - 0.4R_B)x^2 dx -0.4\int_4^6 ((-4 - 0.4R_B)x^2 + 40x) dx -0.6\int_0^4 (6 - 0.6R_B)x^2 dx = 0$

$-0.4 [(2x^3 - 0.4R_B\frac{x^3}{3})]_0^4 -0.4[ (-\frac{4}{3}x^3 - \frac{0.4R_B}{3}x^3 + 20x^2)]_4^6 -0.6 [(2x^3 - 0.6R_B \frac{x^3}{3})]_0^4 = 0$

$-0.4 [128 - \frac{25.6}{3}R_B] -0.4[(-288 - 28.8R_B + 720) - (-\frac{256}{3} - \frac{25.6}{3}R_B + 320)] -0.6 [128 - 12.8R_B] = 0$

$-51.2 + 3.4133R_B -0.4 [(432-28.8R_B) - (320 -8.533R_B)] -76.8 + 7.68R_B = 0$

$-51.2 + 3.4133R_B -0.4(112 - 20.26667R_B) -76.8 + 7.68R_B = 0$

$-51.2 + 3.4133R_B -44.8 + 8.106668R_B -76.8 + 7.68R_B = 0$

$19.199968 R_B = 172.8$

$R_B \approx 9.00$

3. Final Answer

The reaction at support B is approximately 9.00 kN.