The problem asks us to determine the reaction at support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. It is subjected to a 10 kN load located 4m from support A, and a 2 kN load located 6m from support A (or 4m from support B, 1m from support C). The distances between supports are: AB = 6m (4+2) and BC = 5m (4+1).

1. Problem Description

The problem asks us to determine the reaction at support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. It is subjected to a 10 kN load located 4m from support A, and a 2 kN load located 6m from support A (or 4m from support B, 1m from support C). The distances between supports are: AB = 6m (4+2) and BC = 5m (4+1).

2. Solution Steps

Castigliano's second theorem states that the partial derivative of the total strain energy $U$ with respect to a force $P$ is equal to the displacement at the point of application of the force in the direction of the force. Mathematically, $\frac{\partial U}{\partial P} = \Delta$. For supports, we seek the reaction force, and the displacement is zero. Thus $\frac{\partial U}{\partial R_B} = 0$. We'll assume constant EI. The strain energy $U$ is given by:

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

Therefore, $\frac{\partial U}{\partial R_B} = \int \frac{2M}{2EI} \frac{\partial M}{\partial R_B} dx = \frac{1}{EI} \int M \frac{\partial M}{\partial R_B} dx = 0$

Since EI is constant and not zero, we need to find the bending moment M as a function of x, then take its partial derivative with respect to $R_B$, and finally integrate $M \frac{\partial M}{\partial R_B}$ over the length of the beam. We will consider two sections: Section 1: AB and Section 2: BC.

First, we need to determine the support reactions. Let's denote the vertical reactions at A, B, and C as $R_A$, $R_B$, and $R_C$, respectively. Taking the sum of vertical forces to be zero gives:

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

Taking moments about point A:

$6R_B + 11R_C = 10*4 + 2*6 = 40 + 12 = 52$

Now, we introduce a dummy variable $R_B$ at support B.

For section AB (0 <= x <= 6), taking moments about the cut at a distance x from A:

$M_1(x) = R_A x - 10 \langle x - 4 \rangle^1$, where $\langle x - a \rangle^n$ is the Macaulay bracket, equal to $(x-a)^n$ when $x > a$ and 0 when $x < a$.

For section BC (0 <= x' <= 5), taking moments about the cut at a distance x' from C:

$M_2(x') = R_C x' - 2x'$, where x' is the distance from C. Note that $x' = 11 - x$.

$R_A + R_B + R_C = 12$

$6R_B + 11R_C = 52$

$R_A = 12 - R_B - R_C$

$R_C = (52 - 6R_B) / 11$

$R_A = 12 - R_B - (52 - 6R_B) / 11 = (132 - 11R_B - 52 + 6R_B) / 11 = (80 - 5R_B) / 11$

$M_1(x) = \frac{(80 - 5R_B)}{11}x - 10 \langle x - 4 \rangle^1$

$M_2(x') = \frac{(52 - 6R_B)}{11}x' - 2x'$

$\frac{\partial M_1}{\partial R_B} = \frac{-5x}{11}$

$\frac{\partial M_2}{\partial R_B} = \frac{-6x'}{11}$

$\int_0^6 (\frac{(80 - 5R_B)}{11}x - 10 \langle x - 4 \rangle^1) (\frac{-5x}{11}) dx + \int_0^5 (\frac{(52 - 6R_B)}{11}x' - 2x') (\frac{-6x'}{11}) dx' = 0$

$\int_0^6 (\frac{-400x^2 + 25R_B x^2}{121} + \frac{50x}{11}\langle x - 4 \rangle^1) dx + \int_0^5 (\frac{-312x'^2 + 36R_B x'^2 + 132x'^2}{121}) dx' = 0$

$\int_0^6 (\frac{-400x^2 + 25R_B x^2}{121} + \frac{50x}{11}(x - 4)) dx$ from 4 to 6 $+ \int_0^5 (\frac{-180x'^2 + 36R_B x'^2}{121}) dx' = 0$

$(\frac{-400x^3/3 + 25R_B x^3/3}{121})|_0^6 + (\frac{50(x^3/3 - 2x^2)}{11})|_4^6 + (\frac{-180x'^3/3 + 36R_B x'^3/3}{121})|_0^5 = 0$

$(\frac{-400*216/3 + 25R_B *216/3}{121}) + (\frac{50(216/3 - 2*36) - 50(64/3 - 2*16)}{11}) + (\frac{-180*125/3 + 36R_B *125/3}{121}) = 0$

$(\frac{-28800 + 1800R_B}{121}) + (\frac{50(72 - 72) - 50(64/3 - 32)}{11}) + (\frac{-7500 + 1500R_B}{121}) = 0$

$(\frac{-28800 + 1800R_B}{121}) + (\frac{0 - 50(64/3 - 96/3)}{11}) + (\frac{-7500 + 1500R_B}{121}) = 0$

$(\frac{-28800 + 1800R_B}{121}) + (\frac{-50(-32/3)}{11}) + (\frac{-7500 + 1500R_B}{121}) = 0$

$(\frac{-28800 + 1800R_B}{121}) + (\frac{1600/3}{11}) + (\frac{-7500 + 1500R_B}{121}) = 0$

$(\frac{-28800 + 1800R_B}{121}) + (\frac{1600}{33}) + (\frac{-7500 + 1500R_B}{121}) = 0$

Multiplying by 121:

$-28800 + 1800R_B + 1600*121/33 - 7500 + 1500R_B = 0$

$-36300 + 3300R_B + 1600 * 11 / 3 = 0$

$3300R_B = 36300 - 17600/3 * 3/3 = (108900-17600)/3 = 91300/3$

$R_B = (91300/3) / 3300 = 91300/(3*3300) = 91300 / 9900 = 913 / 99 \approx 9.22$

3. Final Answer

$R_B \approx 9.22 kN$