The problem is to determine the reaction at support B of the given continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a 10kN downward force applied 4m from support A, and a 2kN downward force applied 6m from support A (or 2m from B). The spans are: AB = 4m, BC = 4m.

1. Problem Description

The problem is to determine the reaction at support B of the given continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a 10kN downward force applied 4m from support A, and a 2kN downward force applied 6m from support A (or 2m from B). The spans are: AB = 4m, BC = 4m.

2. Solution Steps

Since support B is a support, the deflection at support B is zero. Castigliano's second theorem states that the partial derivative of the total strain energy with respect to a force is equal to the displacement in the direction of that force. In this case, we want to find the reaction at support B, so we will consider the reaction at B as a force, $R_B$. Since the deflection at B is zero, we have:

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

where $U$ is the total strain energy of the beam. The strain energy due to bending is 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. So, we have:

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

Since E and I are constant,

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

Let's find the reactions at supports A and C in terms of $R_B$. We'll use static equilibrium. Sum of vertical forces:

$R_A + R_B + R_C = 10 + 2 = 12$ kN

Sum of moments about A:

$4R_B + 8R_C = 10*4 + 2*6 = 40 + 12 = 52$

$R_B + 2R_C = 13$

$2R_C = 13 - R_B$

$R_C = \frac{13-R_B}{2}$

Substituting this into the first equation:

$R_A + R_B + \frac{13-R_B}{2} = 12$

$R_A + \frac{2R_B + 13 - R_B}{2} = 12$

$R_A = 12 - \frac{R_B + 13}{2} = \frac{24 - R_B - 13}{2} = \frac{11 - R_B}{2}$

Now consider the two spans AB and BC separately.

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

$M(x) = R_A * x - 10 * <x-4>$ (where $<x-a>$ is zero if $x<a$, and $x-a$ if $x>=a$).

$\frac{\partial M}{\partial R_B} = \frac{\partial R_A}{\partial R_B} x = \frac{-1}{2} x$

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

$M(x) = R_C * x - 2 * <x-2>$

$\frac{\partial M}{\partial R_B} = \frac{\partial R_C}{\partial R_B} x = \frac{-1}{2} x$

We have the integrals from 0 to 4 for each span, so change variable names:

$I_1 = \int_0^4 (\frac{11-R_B}{2}x - 10 <x-4>) (-\frac{1}{2}x) dx = \int_0^4 (\frac{11-R_B}{2}x) (-\frac{1}{2}x) dx$ (since <x-4> = 0 within the integral limit)

$I_1 = -\frac{1}{4} (11-R_B) \int_0^4 x^2 dx = -\frac{1}{4} (11-R_B) [\frac{x^3}{3}]_0^4 = -\frac{1}{4} (11-R_B) \frac{64}{3} = -\frac{16}{3}(11-R_B)$

$I_2 = \int_0^4 (\frac{13-R_B}{2} x - 2 <x-2>) (-\frac{1}{2}x) dx = \int_0^2 \frac{13-R_B}{2}x (-\frac{1}{2}x) dx + \int_2^4 (\frac{13-R_B}{2}x - 2(x-2))(-\frac{1}{2}x) dx$

$I_2 = -\frac{1}{4} (13-R_B) \int_0^2 x^2 dx -\frac{1}{2} \int_2^4 (\frac{13-R_B}{2}x - 2x + 4) x dx$

$I_2 = -\frac{1}{4} (13-R_B) [\frac{x^3}{3}]_0^2 - \frac{1}{2} \int_2^4 (\frac{9-R_B}{2}x^2 + 4x) dx$

$I_2 = -\frac{1}{4} (13-R_B) \frac{8}{3} - \frac{1}{2} [\frac{9-R_B}{2} \frac{x^3}{3} + 2x^2]_2^4$

$I_2 = -\frac{2}{3} (13-R_B) - \frac{1}{2} [(\frac{9-R_B}{6}(64-8) + 2(16-4)]$

$I_2 = -\frac{2}{3} (13-R_B) - \frac{1}{2} [\frac{9-R_B}{6} (56) + 24]$

$I_2 = -\frac{2}{3} (13-R_B) - \frac{28}{6}(9-R_B) - 12$

$I_2 = -\frac{2}{3}(13-R_B) - \frac{14}{3}(9-R_B) - 12$

$I_2 = \frac{-26 + 2R_B - 126 + 14R_B - 36}{3} = \frac{16R_B - 188}{3}$

$I_1 + I_2 = 0$

$-\frac{16}{3}(11-R_B) + \frac{16R_B - 188}{3} = 0$

$-176 + 16R_B + 16R_B - 188 = 0$

$32 R_B = 364$

$R_B = \frac{364}{32} = \frac{91}{8} = 11.375$

3. Final Answer

The reaction at support B is 11.375 kN.