The problem asks to determine the reaction at the support B of a continuous beam ABC using Castigliano's theorem. The beam is subjected to a 10 kN load at a distance of 4 m from support A and a 2 kN load at a distance of 4 m from support C. The distance between A and B is 4 m, and the distance between B and C is 4 m.

1. Problem Description

The problem asks to determine the reaction at the support B of a continuous beam ABC using Castigliano's theorem. The beam is subjected to a 10 kN load at a distance of 4 m from support A and a 2 kN load at a distance of 4 m from support C. 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 is equal to the displacement at the point of application of that force in the direction of the force. In this case, since support B does not deflect, the vertical displacement at B is zero. Therefore, we can express the reaction at B as a redundant force, say $R_B$, and then calculate the strain energy $U$ in terms of $R_B$. Taking the partial derivative of $U$ with respect to $R_B$ and setting it equal to zero allows us to solve for $R_B$.

First, let's find the reactions at A and C in terms of $R_B$. Let $R_A$ and $R_C$ be the vertical reactions at supports A and C, respectively. Taking the sum of vertical forces equal to zero:

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

$R_A + R_B + R_C = 12$ ---- (1)

Taking moments about point A equal to zero:

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

$4R_B + 8R_C = 40 + 16$

$4R_B + 8R_C = 56$

$R_B + 2R_C = 14$ ---- (2)

From (1) and (2), we can express $R_A$ and $R_C$ in terms of $R_B$:

From (2): $2R_C = 14 - R_B$

$R_C = 7 - \frac{R_B}{2}$

Substituting $R_C$ into (1):

$R_A + R_B + 7 - \frac{R_B}{2} = 12$

$R_A + \frac{R_B}{2} = 5$

$R_A = 5 - \frac{R_B}{2}$

Now we have $R_A = 5 - \frac{R_B}{2}$ and $R_C = 7 - \frac{R_B}{2}$.

Next, we need to find the bending moment equations for segments AB and BC.

For segment AB ($0 \le x \le 4$):

$M_{AB} = R_A * x = (5 - \frac{R_B}{2})x = 5x - \frac{R_B x}{2}$

For segment BC ($0 \le x \le 4$): Let x be the distance from C.

$M_{BC} = R_C * x - 2x = (7 - \frac{R_B}{2})x - 2x = 7x - \frac{R_B x}{2} - 2x = 5x - \frac{R_B x}{2}$

The strain energy $U$ due to bending is given by:

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

So, $U = \int_0^4 \frac{(5x - \frac{R_B x}{2})^2}{2EI} dx + \int_0^4 \frac{(5x - \frac{R_B x}{2})^2}{2EI} dx = 2\int_0^4 \frac{(5x - \frac{R_B x}{2})^2}{2EI} dx$

$U = \frac{1}{EI} \int_0^4 (25x^2 - 5R_B x^2 + \frac{R_B^2 x^2}{4}) dx$

Applying Castigliano's theorem:

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

$\frac{\partial U}{\partial R_B} = \frac{1}{EI} \int_0^4 (-5x^2 + \frac{R_B x^2}{2}) dx = 0$

$\int_0^4 (-5x^2 + \frac{R_B x^2}{2}) dx = 0$

$[-\frac{5x^3}{3} + \frac{R_B x^3}{6}]_0^4 = 0$

$-\frac{5(4^3)}{3} + \frac{R_B (4^3)}{6} = 0$

$-\frac{5(64)}{3} + \frac{64R_B}{6} = 0$

$\frac{64R_B}{6} = \frac{320}{3}$

$64R_B = \frac{320*6}{3} = 320 * 2 = 640$

$R_B = \frac{640}{64} = 10$ kN

3. Final Answer

The reaction at support B is 10 kN.