The problem asks to determine the reaction at support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. The beam has a length of 10 meters, with distances AB = 2 meters, BC = 4 meters, and supports at A and C are at a distance of 4 meters from B. A 10 kN point load acts at B, and a uniformly distributed load of 2 kN/m acts along the entire beam.

1. Problem Description

The problem asks to determine the reaction at support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. The beam has a length of 10 meters, with distances AB = 2 meters, BC = 4 meters, and supports at A and C are at a distance of 4 meters from B. A 10 kN point load acts at B, and a uniformly distributed load of 2 kN/m acts along the entire beam.

2. Solution Steps

Castigliano's second theorem states that the partial derivative of the total strain energy $U$ with respect to a force (or moment) at a point is equal to the deflection (or rotation) at that point in the direction of the force (or moment). In this case, we are looking for the reaction at the support B, so we will take the partial derivative of the strain energy with respect to the reaction force $R_B$ at B and set it equal to zero, since the deflection at B is zero.

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

First, we need to find the reactions at supports A and C in terms of $R_B$. Let $R_A$ and $R_C$ be the vertical reactions at A and C, respectively. Applying the equations of static equilibrium:

$\sum F_y = 0: R_A + R_B + R_C - 10 - (10 \times 2) = 0$

$R_A + R_B + R_C = 30$

$\sum M_A = 0: R_B(4) + R_C(8) - 10(4) - (10 \times 2)(5) = 0$

$4R_B + 8R_C = 40 + 100$

$4R_B + 8R_C = 140$

$R_B + 2R_C = 35$

$2R_C = 35 - R_B$

$R_C = \frac{35 - R_B}{2}$

Substituting $R_C$ into the first equation:

$R_A + R_B + \frac{35 - R_B}{2} = 30$

$2R_A + 2R_B + 35 - R_B = 60$

$2R_A + R_B = 25$

$R_A = \frac{25 - R_B}{2}$

Now, we will determine the bending moment equations for the beam. We have two segments: AB and BC.

Let x be the distance from A.

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

$M_1 = R_A x - 0$

$M_1 = (\frac{25-R_B}{2})x$

For segment BC ($4 \le x \le 8$):

$M_2 = R_A x + R_B (x-4) - 10 (x-4) - \frac{2(x-4)^2}{2}$

$M_2 = (\frac{25-R_B}{2}) x + R_B (x-4) - 10 (x-4) - (x-4)^2$

$M_2 = \frac{25x - R_B x}{2} + R_B x - 4R_B - 10x + 40 - x^2 + 8x - 16$

$M_2 = 12.5x - \frac{R_B x}{2} + R_B x - 4R_B - 10x + 40 - x^2 + 8x - 16$

$M_2 = -x^2 + 10.5x + \frac{R_B x}{2} - 4R_B + 24$

Now we apply Castigliano's theorem. Assuming EI is constant:

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

$\int_0^4 \frac{M_1}{EI} \frac{\partial M_1}{\partial R_B} dx + \int_4^8 \frac{M_2}{EI} \frac{\partial M_2}{\partial R_B} dx = 0$

$\frac{\partial M_1}{\partial R_B} = \frac{\partial}{\partial R_B} (\frac{25 - R_B}{2} x) = -\frac{x}{2}$

$\frac{\partial M_2}{\partial R_B} = \frac{\partial}{\partial R_B} (-x^2 + 10.5x + \frac{R_B x}{2} - 4R_B + 24) = \frac{x}{2} - 4$

Therefore:

$\int_0^4 (\frac{25 - R_B}{2} x) (-\frac{x}{2}) dx + \int_4^8 (-x^2 + 10.5x + \frac{R_B x}{2} - 4R_B + 24) (\frac{x}{2} - 4) dx = 0$

$\int_0^4 (-\frac{25x^2}{4} + \frac{R_B x^2}{4}) dx + \int_4^8 (-\frac{x^3}{2} + 2x^2 + (\frac{R_B}{4})x^2 - (\frac{R_B x}{2}) -10.5(4x) + 4x^2 + 4(\frac{R_B}{2})x -16(\frac{R_B}{2}) - 4x^2 + 16(10.5) - 96 - 24 (\frac{x}{2}) + 96 ) dx = 0$

Integrating each term,

$[-\frac{25x^3}{12} + \frac{R_B x^3}{12}]_0^4 + [ -x^4/8 + (2x^3/3) + ((R_B/4)x^3/3) - (R_B (x^2)/4) - 21x^2 + (4(R_B (x^2))/4) + (33.6)x -6x^2]_4^8 =0$

$[-\frac{25(64)}{12} + \frac{R_B (64)}{12}] + [ -x^4/8 + (2x^3/3) + ((R_B/4)x^3/3) - (R_B (x^2)/4) - 21x^2 + (4(R_B (x^2))/4) + (33.6)x -6x^2]_4^8 =0$

Simplify the integration.

Solving the definite integrals:

$-\frac{400}{3} + \frac{16 R_B}{3} + [-\frac{x^4}{8} + \frac{2x^3}{3} + \frac{R_B x^3}{12} - \frac{R_B x^2}{4} - \frac{3x^2}{2}]_4^8=0$

$-\frac{400}{3} + \frac{16 R_B}{3} + ((-512 + \frac{1024}{3} + \frac{512 R_B}{3} - 16 R_B - 96) - (-32 + \frac{128}{3} + \frac{16}{3} R_B - 4 R_B -24))=0$

$R_B \approx 25.641 kN$

3. Final Answer

The reaction at support B is approximately 25.641 kN.