The problem asks to determine the reaction at the supports of a given structure using Castigliano's theorem. The structure is a continuous beam with supports at A, B, and C. There is a 10kN load at 4m from support A and a 2kN load at 4m from support B. The distances between supports are: A to B is 4m + 2m = 6m and B to C is 4m.

1. Problem Description

The problem asks to determine the reaction at the supports of a given structure using Castigliano's theorem. The structure is a continuous beam with supports at A, B, and C. There is a 10kN load at 4m from support A and a 2kN load at 4m from support B. The distances between supports are: A to B is 4m + 2m = 6m and B to C is 4m.

2. Solution Steps

Castigliano's theorem states that the partial derivative of the total strain energy $U$ with respect to a force $P$ is equal to the displacement $\delta$ in the direction of that force:

$\delta = \frac{\partial U}{\partial P}$

For statically indeterminate structures, we introduce a redundant reaction, say $R_B$ at support B, and treat it as an external force. Then, we apply Castigliano's theorem by setting the deflection at B to zero since the support doesn't move:

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

First, determine the reactions $R_A$ and $R_C$ in terms of $R_B$ by applying the equations of static equilibrium. Sum of moments about A equal to 0:

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

$10R_C = 40 + 6R_B + 20$

$R_C = 6 + 0.6R_B$

Sum of vertical forces equal to 0:

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

$R_A = 12 - R_B - R_C = 12 - R_B - (6 + 0.6R_B) = 6 - 1.6R_B$

Next, we determine the bending moment equations for each span.

Span AB (0 <= x <= 6):

$M_1(x) = R_A x - 10(x-4)$ (for $x > 4$)

$M_1(x) = (6 - 1.6R_B)x - 10(x-4)$ (for $x > 4$)

$M_1(x) = (6 - 1.6R_B)x$ (for $0 <= x <= 4$)

Span BC (0 <= x <= 4):

$M_2(x) = R_C x - 2(x)$

$M_2(x) = (6 + 0.6R_B)x - 2x = (4 + 0.6R_B)x$

The total strain energy $U$ is given by:

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

Applying Castigliano's theorem:

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

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

$\frac{\partial M_1}{\partial R_B} = -1.6x$ (for $0 <= x <= 4$)

$\frac{\partial M_1}{\partial R_B} = -1.6x$ (for $4 < x <= 6$)

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

$\int_0^4 \frac{(6 - 1.6R_B)x}{EI} (-1.6x) dx + \int_4^6 \frac{(6 - 1.6R_B)x - 10(x-4)}{EI} (-1.6x) dx + \int_0^4 \frac{(4 + 0.6R_B)x}{EI} (0.6x) dx = 0$

Assuming EI is constant,

$\int_0^4 (6 - 1.6R_B)x (-1.6x) dx + \int_4^6 ((6 - 1.6R_B)x - 10(x-4)) (-1.6x) dx + \int_0^4 (4 + 0.6R_B)x (0.6x) dx = 0$

$-1.6 \int_0^4 (6x^2 - 1.6R_B x^2) dx - 1.6 \int_4^6 (6x^2 - 1.6R_Bx^2 - 10x^2 + 40x) dx + 0.6 \int_0^4 (4x^2 + 0.6R_B x^2) dx = 0$

$-1.6 [2x^3 - \frac{1.6}{3} R_B x^3]_0^4 - 1.6 [2x^3 - \frac{1.6}{3} R_B x^3 - \frac{10}{3}x^3 + 20x^2]_4^6 + 0.6 [\frac{4}{3}x^3 + \frac{0.6}{3}R_Bx^3]_0^4 = 0$

$-1.6 [128 - \frac{102.4}{3}R_B] - 1.6 [(432 - \frac{345.6}{3}R_B - 720 + 720) - (128 - \frac{102.4}{3}R_B - \frac{640}{3} + 320)] + 0.6 [\frac{256}{3} + \frac{38.4}{3}R_B] = 0$

$-204.8 + 54.613 R_B - 1.6 [432 - 115.2 R_B -720+720 - 128 + 34.133 R_B +213.333 - 320] + 51.2 + 7.68 R_B = 0$

$-204.8 + 54.613 R_B - 1.6[304 - 81.067 R_B] + 51.2 + 7.68 R_B = 0$

$-153.6 + 62.293 R_B - 486.4 + 129.707 R_B = 0$

$192 R_B = 640$

$R_B = \frac{640}{192} = \frac{10}{3} = 3.33 kN$

3. Final Answer

The reaction at support B is $R_B = 3.33 kN$.