We are asked to determine the reaction at support B of a continuous beam ABC using Castigliano's theorem. The beam is subjected to a 10 kN vertical load at a distance of 4 m from support A and a 2 kN vertical load at a distance of 6 m from support A (4 m from support B). The lengths of the beam segments are AB = 4 m and BC = 4 m.

1. Problem Description

We are asked to determine the reaction at support B of a continuous beam ABC using Castigliano's theorem. The beam is subjected to a 10 kN vertical load at a distance of 4 m from support A and a 2 kN vertical load at a distance of 6 m from support A (4 m from support B). The lengths of the beam segments are AB = 4 m and BC = 4 m.

2. Solution Steps

Let $R_A$, $R_B$, and $R_C$ be the vertical reactions at supports A, B, and C, respectively.

Let $x$ be the distance from support A.

First, we need to find an expression for the bending moment $M(x)$ in terms of an assumed redundant reaction. We'll consider $R_B$ as the redundant reaction.

From static equilibrium:

$\sum F_y = 0$:

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

$R_A + R_B + R_C = 12$

$\sum M_A = 0$:

$R_B * 4 + R_C * 8 - 10 * 4 - 2 * 6 = 0$

$4R_B + 8R_C = 40 + 12 = 52$

$R_B + 2R_C = 13$

From these equations, we have

$R_A + R_C = 12 - R_B$

$2R_C = 13 - R_B$

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

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

Now, we need to find the bending moment $M(x)$ as a function of $x$.

For $0 \le x \le 4$:

$M(x) = R_A * x = \frac{11 - R_B}{2} * x$

For $4 \le x \le 8$:

$M(x) = R_A * x + R_B * (x - 4) - 10 * (x-4) = \frac{11 - R_B}{2} * x + R_B * (x - 4) - 10 * (x-4)$

$M(x) = \frac{11x - R_B x + 2R_B x - 8R_B - 20x + 80}{2} = \frac{-9x + R_B x - 8R_B + 80}{2}$

$M(x) = \frac{(-9 + R_B)x - 8R_B + 80}{2}$

Castigliano's Theorem:

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

Since $EI$ is constant, we can write:

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

For $0 \le x \le 4$:

$M(x) = \frac{11 - R_B}{2} * x$

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

For $4 \le x \le 8$:

$M(x) = \frac{(-9 + R_B)x - 8R_B + 80}{2}$

$\frac{\partial M}{\partial R_B} = \frac{x - 8}{2}$

$\int_0^4 \frac{11 - R_B}{2} * x * (-\frac{x}{2}) dx + \int_4^8 \frac{(-9 + R_B)x - 8R_B + 80}{2} * (\frac{x - 8}{2}) dx = 0$

$-\frac{1}{4} \int_0^4 (11x^2 - R_B x^2) dx + \frac{1}{4} \int_4^8 ((-9 + R_B)x - 8R_B + 80)(x-8) dx = 0$

$-\int_0^4 (11x^2 - R_B x^2) dx + \int_4^8 ((-9 + R_B)x - 8R_B + 80)(x-8) dx = 0$

$-\int_0^4 (11x^2 - R_B x^2) dx = -[\frac{11x^3}{3} - \frac{R_B x^3}{3}]_0^4 = -(\frac{11*64}{3} - \frac{64 R_B}{3}) = -\frac{704}{3} + \frac{64}{3} R_B$

$\int_4^8 ((-9 + R_B)x - 8R_B + 80)(x-8) dx = \int_4^8 ((-9 + R_B)x^2 - 8(-9+R_B)x - 8R_B x + 64R_B + 80x - 640) dx$

$= \int_4^8 ((-9+R_B)x^2 + (72-8R_B - 8R_B + 80)x + 64R_B - 640) dx$

$= \int_4^8 ((-9+R_B)x^2 + (152 - 16R_B)x + 64R_B - 640) dx$

$= [\frac{(-9 + R_B)x^3}{3} + \frac{(152 - 16R_B)x^2}{2} + (64R_B - 640)x]_4^8$

$= [\frac{(-9 + R_B)x^3}{3} + (76 - 8R_B)x^2 + (64R_B - 640)x]_4^8$

$= [\frac{(-9+R_B)512}{3} + (76 - 8R_B)64 + (64R_B - 640)8] - [\frac{(-9+R_B)64}{3} + (76-8R_B)16 + (64R_B - 640)4]$

$= \frac{(-9+R_B)512}{3} - \frac{(-9+R_B)64}{3} + (76 - 8R_B)64 - (76 - 8R_B)16 + (64R_B - 640)8 - (64R_B - 640)4$

$= \frac{(-9+R_B)448}{3} + (76-8R_B)48 + (64R_B - 640)4$

$= \frac{-4032 + 448R_B}{3} + 3648 - 384R_B + 256R_B - 2560$

$= \frac{-4032 + 448R_B + 32544 - 1152R_B + 768R_B - 7680}{3} = \frac{20832 + 64 R_B}{3}$

$-\frac{704}{3} + \frac{64}{3} R_B + \frac{20832 + 64 R_B}{3} = 0$

$20128 + 128 R_B = 0$

$128R_B = -20128$

$R_B = -157.25$ kN. This result does not seem physically possible, since all loads point downwards. We must have made an error.

Let's reconsider the moments.

For $4 \le x \le 8$:

$M(x) = R_A * x - 10(x-4)$.

$M(x) = (\frac{11-R_B}{2})x - 10(x-4) = (\frac{11-R_B}{2})x - 10x + 40 = (\frac{11-R_B-20}{2})x + 40 = (\frac{-9-R_B}{2})x + 40$.

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

$\int_0^4 (\frac{11-R_B}{2})x (-\frac{x}{2}) dx + \int_4^8 (\frac{-9-R_B}{2}x + 40)(-\frac{x}{2}) dx = 0$

$-\frac{1}{4}\int_0^4 (11x^2 - R_Bx^2) dx - \frac{1}{4} \int_4^8 (-9x^2 - R_Bx^2 + 80x) dx = 0$

$\int_0^4 (11x^2 - R_Bx^2) dx + \int_4^8 (-9x^2 - R_Bx^2 + 80x) dx = 0$

$\frac{11x^3}{3} - \frac{R_B x^3}{3}\Big|_0^4 = \frac{704}{3} - \frac{64 R_B}{3}$

$-3x^3 - \frac{R_B x^3}{3} + 40x^2\Big|_4^8 = (-3(512) - \frac{R_B 512}{3} + 40(64)) - (-3(64) - \frac{R_B 64}{3} + 40(16))$

$= (-1536 - \frac{512 R_B}{3} + 2560) - (-192 - \frac{64R_B}{3} + 640) = 1024 - \frac{512 R_B}{3} - 448 + \frac{64 R_B}{3} = 576 - \frac{448R_B}{3}$

$\frac{704}{3} - \frac{64R_B}{3} + 576 - \frac{448R_B}{3} = 0$

$704 - 64R_B + 1728 - 448R_B = 0$

$2432 - 512 R_B = 0$

$512 R_B = 2432$

$R_B = \frac{2432}{512} = 4.75 kN$.

3. Final Answer

The reaction at support B is 4.75 kN.