The problem is 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's a downward point load of 10 kN located 2 m from A and a downward point load of 2 kN located 4 m from B. The distance between A and B is 4 m and the distance between B and C is 4 m.

1. Problem Description

The problem is 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's a downward point load of 10 kN located 2 m from A and a downward point load of 2 kN located 4 m from B. The distance between A and B is 4 m and the distance between B and C is 4 m.

2. Solution Steps

Castigliano's second theorem states that the partial derivative of the total strain energy $U$ with respect to a force (or moment) is equal to the displacement (or rotation) at the point of application of that force (or moment) in the direction of the force (or moment).

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

Since support B does not deflect, $\Delta_B = 0$. Thus, we have:

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

The strain energy $U$ for a beam subjected to bending is given by:

$U = \int_0^L \frac{M(x)^2}{2EI} dx$

Therefore:

$\frac{\partial U}{\partial R_B} = \int_0^L \frac{2M(x)}{2EI} \frac{\partial M(x)}{\partial R_B} dx = \frac{1}{EI} \int_0^L M(x) \frac{\partial M(x)}{\partial R_B} dx = 0$

We can split the beam into two segments: AB and BC. Let's denote the distance from A as x1 for segment AB, and the distance from B as x2 for segment BC.

Reactions at A and C can be written in terms of Rb as $R_A$ and $R_C$.

Sum of forces: $R_A + R_B + R_C = 10 + 2 = 12 kN$

Sum of moments about A: $R_B * 4 + R_C * 8 = 10 * 2 + 2 * 6 = 20 + 12 = 32$

$4R_B + 8R_C = 32$, so $R_B + 2R_C = 8$

From sum of forces $R_A = 12 - R_B - R_C$. Substitute $R_C = (8-R_B)/2 = 4 - R_B/2$ to get:

$R_A = 12 - R_B - (4 - R_B/2) = 8 - R_B/2$

Segment AB (0 <= x1 <= 4):

$M_1(x_1) = R_A x_1$ for $0 <= x_1 <= 2$

$M_1(x_1) = R_A x_1 - 10(x_1-2)$ for $2 <= x_1 <= 4$

Substituting $R_A = 8 - R_B/2$ :

$M_1(x_1) = (8 - R_B/2)x_1$ for $0 <= x_1 <= 2$

$M_1(x_1) = (8 - R_B/2)x_1 - 10(x_1-2) = (8 - R_B/2)x_1 - 10x_1 + 20 = (-2 - R_B/2)x_1 + 20$ for $2 <= x_1 <= 4$

Segment BC (0 <= x2 <= 4):

$M_2(x_2) = R_C x_2$ for $0 <= x_2 <= 4$

Substituting $R_C = 4 - R_B/2$:

$M_2(x_2) = (4 - R_B/2)x_2$ for $0 <= x_2 <= 4$

Derivatives with respect to $R_B$:

$\frac{\partial M_1(x_1)}{\partial R_B} = -x_1/2$ for $0 <= x_1 <= 2$

$\frac{\partial M_1(x_1)}{\partial R_B} = -x_1/2$ for $2 <= x_1 <= 4$

$\frac{\partial M_2(x_2)}{\partial R_B} = -x_2/2$ for $0 <= x_2 <= 4$

Now, integrate:

$\int_0^4 M \frac{\partial M}{\partial R_B} dx = \int_0^2 ((8 - R_B/2)x_1)(-x_1/2) dx_1 + \int_2^4 ((-2 - R_B/2)x_1 + 20)(-x_1/2) dx_1 + \int_0^4 ((4 - R_B/2)x_2)(-x_2/2) dx_2 = 0$

$\int_0^2 (-\frac{8x_1^2}{2} + \frac{R_B x_1^2}{4}) dx_1 + \int_2^4 (\frac{2x_1^2}{2} + \frac{R_B x_1^2}{4} - \frac{20x_1}{2}) dx_1 + \int_0^4 (-\frac{4x_2^2}{2} + \frac{R_B x_2^2}{4}) dx_2 = 0$

$\int_0^2 (-4x_1^2 + \frac{R_B x_1^2}{4}) dx_1 + \int_2^4 (x_1^2 + \frac{R_B x_1^2}{4} - 10x_1) dx_1 + \int_0^4 (-2x_2^2 + \frac{R_B x_2^2}{4}) dx_2 = 0$

$[-4x_1^3/3 + R_B x_1^3/12]_0^2 + [x_1^3/3 + R_B x_1^3/12 - 5x_1^2]_2^4 + [-2x_2^3/3 + R_B x_2^3/12]_0^4 = 0$

$[-32/3 + 8R_B/12] + [(64/3 + 64R_B/12 - 80) - (8/3 + 8R_B/12 - 20)] + [-128/3 + 64R_B/12] = 0$

$-32/3 + 2R_B/3 + 64/3 + 16R_B/3 - 80 - 8/3 - 2R_B/3 + 20 -128/3 + 16R_B/3 = 0$

$(-32 + 64 - 240 - 8 + 60 - 128)/3 + (2 + 16 - 2 + 16)R_B/3 = 0$

$-284/3 + 32R_B/3 = 0$

$32 R_B = 284$

$R_B = 284/32 = 71/8 = 8.875 kN$

3. Final Answer

$R_B = 8.875 kN$