The problem is to determine the reaction at support B of a continuous beam ABC using Castigliano's theorem. The beam is supported at A, B and C. There is a 10kN load acting downward between A and B, and a 2kN load acting downward between B and C. The distances are: A to the 10kN load is 4m, the 10kN load to B is 2m, B to the 2kN load is 4m, according to the image.

1. Problem Description

The problem is to determine the reaction at support B of a continuous beam ABC using Castigliano's theorem. The beam is supported at A, B and C. There is a 10kN load acting downward between A and B, and a 2kN load acting downward between B and C. The distances are: A to the 10kN load is 4m, the 10kN load to B is 2m, B to the 2kN load is 4m, according to the image.

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 in the direction of that force. In this case, since support B is a rigid support, the displacement at B is zero. Therefore,

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

where $R_B$ is the reaction force at support B.

First, we need to determine the bending moments in the beam.

Let $R_A$, $R_B$, and $R_C$ be the reactions at supports A, B, and C, respectively. Consider $R_B$ as the redundant reaction. We will express the moments in terms of $R_B$.

Taking sum of moments about A to zero:

$R_B (4+2) + R_C (4+2+4) = 10(4) + 2(4+2)$

$6 R_B + 10 R_C = 40 + 12 = 52$

$R_C = \frac{52 - 6 R_B}{10}$

Sum of vertical forces equal to zero:

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

$R_A + R_B + \frac{52 - 6 R_B}{10} = 12$

$R_A = 12 - R_B - \frac{52 - 6 R_B}{10} = \frac{120 - 10 R_B - 52 + 6 R_B}{10} = \frac{68 - 4 R_B}{10}$

Now we can find the bending moment equations for the different segments:

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

$M_1(x) = R_A x = \frac{68 - 4 R_B}{10} x$

Segment AB (4 <= x <= 6):

$M_2(x) = R_A x - 10 (x - 4) = \frac{68 - 4 R_B}{10} x - 10 (x - 4)$

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

$M_3(x) = R_C x = \frac{52 - 6 R_B}{10} x$

The total strain energy due to bending is given by:

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

where $EI$ is the flexural rigidity.

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

We need to calculate the partial derivatives of the bending moments with respect to $R_B$:

$\frac{\partial M_1}{\partial R_B} = \frac{-4}{10} x = -0.4 x$

$\frac{\partial M_2}{\partial R_B} = \frac{-4}{10} x = -0.4 x$

$\frac{\partial M_3}{\partial R_B} = \frac{-6}{10} x = -0.6 x$

Now, we can set up the integral:

$\int_0^4 \frac{M_1}{EI} \frac{\partial M_1}{\partial R_B} dx + \int_4^6 \frac{M_2}{EI} \frac{\partial M_2}{\partial R_B} dx + \int_0^4 \frac{M_3}{EI} \frac{\partial M_3}{\partial R_B} dx = 0$

Assuming EI is constant, it can be factored out.

$\int_0^4 (\frac{68 - 4 R_B}{10} x) (-0.4 x) dx + \int_4^6 (\frac{68 - 4 R_B}{10} x - 10 (x - 4)) (-0.4 x) dx + \int_0^4 (\frac{52 - 6 R_B}{10} x) (-0.6 x) dx = 0$

$-0.4 \int_0^4 (\frac{68 - 4 R_B}{10}) x^2 dx - 0.4 \int_4^6 (\frac{68 - 4 R_B}{10} x - 10 (x - 4)) x dx - 0.6 \int_0^4 (\frac{52 - 6 R_B}{10}) x^2 dx = 0$

$-0.4 \frac{68 - 4 R_B}{10} [\frac{x^3}{3}]_0^4 - 0.4 \int_4^6 (\frac{68 - 4 R_B}{10} x^2 - 10 (x^2 - 4x)) dx - 0.6 \frac{52 - 6 R_B}{10} [\frac{x^3}{3}]_0^4 = 0$

$-0.4 \frac{68 - 4 R_B}{10} (\frac{64}{3}) - 0.4 [\frac{68 - 4 R_B}{10} \frac{x^3}{3} - 10 (\frac{x^3}{3} - 2x^2)]_4^6 - 0.6 \frac{52 - 6 R_B}{10} (\frac{64}{3}) = 0$

$-0.4 \frac{68 - 4 R_B}{10} (\frac{64}{3}) - 0.4 [\frac{68 - 4 R_B}{10} (\frac{216}{3} - \frac{64}{3}) - 10 (\frac{216}{3} - 2(36) - (\frac{64}{3} - 2(16)))] - 0.6 \frac{52 - 6 R_B}{10} (\frac{64}{3}) = 0$

$-0.4 \frac{68 - 4 R_B}{10} (\frac{64}{3}) - 0.4 [\frac{68 - 4 R_B}{10} (\frac{152}{3}) - 10 (\frac{152}{3} - 72 + 32)] - 0.6 \frac{52 - 6 R_B}{10} (\frac{64}{3}) = 0$

$-0.4 \frac{68 - 4 R_B}{10} (\frac{64}{3}) - 0.4 [\frac{68 - 4 R_B}{10} (\frac{152}{3}) - 10 (\frac{152-120}{3} -40)] - 0.6 \frac{52 - 6 R_B}{10} (\frac{64}{3}) = 0$

$-0.4 \frac{68 - 4 R_B}{10} (\frac{64}{3}) - 0.4 \frac{68 - 4 R_B}{10} (\frac{152}{3}) + 4(\frac{32}{3}) + 4(\frac{32}{3}) - 0.6 \frac{52 - 6 R_B}{10} (\frac{64}{3}) = 0$

$-0.4 \frac{68 - 4 R_B}{10} (\frac{64}{3}) - 0.4 \frac{68 - 4 R_B}{10} (\frac{152}{3}) - 0.6 \frac{52 - 6 R_B}{10} (\frac{64}{3}) + \frac{256}{3} = 0$

Multiply by 30:

$-4 (68 - 4 R_B) (6.4) - 4 (68 - 4 R_B) (15.2) - 6 (52 - 6 R_B) (6.4) + 2560 = 0$

$-4 (435.2 - 25.6 R_B) - 4 (1033.6 - 60.8 R_B) - 6 (332.8 - 38.4 R_B) + 2560 = 0$

$-1740.8 + 102.4 R_B - 4134.4 + 243.2 R_B - 1996.8 + 230.4 R_B + 2560 = 0$

$576 R_B = 1740.8 + 4134.4 + 1996.8 - 2560 = 5312$

$R_B = \frac{5312}{576} = 9.222$

3. Final Answer

The reaction at support B is 9.222 kN.