The problem asks us to determine the reaction at the support of a given structure using Castigliano's theorem. The structure is a continuous beam with supports at A, B, and C. There's a 10 kN load applied downwards between A and B, and a 2 kN load applied downwards between B and C. The distances between the supports and loads are given in the figure. The distances are: AB = 6 m (4+2), BC = 8 m (4+4). The 10kN load is at a distance of 4 m from A. The 2 kN load is at a distance of 4 m from B.

1. Problem Description

The problem asks us to determine the reaction at the support of a given structure using Castigliano's theorem. The structure is a continuous beam with supports at A, B, and C. There's a 10 kN load applied downwards between A and B, and a 2 kN load applied downwards between B and C. The distances between the supports and loads are given in the figure. The distances are: AB = 6 m (4+2), BC = 8 m (4+4). The 10kN load is at a distance of 4 m from A. The 2 kN load is at a distance of 4 m from B.

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 that point in the direction of the force. In this problem, we want to find the reaction at the support.

Let's denote the reactions at A, B, and C as $R_A$, $R_B$, and $R_C$, respectively. To find the reaction at, say, support C, we can use Castigliano's theorem: $\frac{\partial U}{\partial R_C} = \Delta_C$, where $\Delta_C$ is the deflection at support C. Since support C is a fixed support, the deflection is zero, so $\frac{\partial U}{\partial R_C} = 0$.

First, we need to find the reactions. Let's consider the beam as a whole. Applying the equations of equilibrium:

$\sum F_y = 0$: $R_A + R_B + R_C - 10 - 2 = 0$, so $R_A + R_B + R_C = 12$

$\sum M_A = 0$: $10(4) + R_B(6) + R_C(14) - 2(6+4) = 0$, so $6 R_B + 14 R_C = -20$. Divide by 2 to get $3 R_B + 7 R_C = -10$.

Let's make support C redundant. The reaction $R_C$ is now an unknown force. We will use Castigliano's theorem to determine $R_C$.

To use Castigliano's theorem, we need to calculate the bending moment in the beams as a function of $R_C$.

Consider the section from A to B.

Taking moments about C:

$R_A + R_B + R_C = 12$ or $R_A + R_B = 12 - R_C$

Taking moments about A:

$40+6R_B+14R_C+20 = 0$

$6R_B + 14R_C = -60$, so $3R_B + 7R_C = -30$

$3R_B = -30 - 7R_C$, so $R_B = -10 - \frac{7}{3} R_C$

$R_A = 12 - R_C - (-10 - \frac{7}{3}R_C) = 22 + \frac{4}{3} R_C$

Segment AC:

The section AC contains the entire beam.

The bending moment in the region AC is:

For $0 < x < 4$: $M(x) = (22 + \frac{4}{3} R_C)x$

For $4 < x < 6$: $M(x) = (22 + \frac{4}{3} R_C)x - 10(x-4)$

For $6 < x < 14$: $M(x) = (22 + \frac{4}{3} R_C)x - 10(x-4) + (-10 - \frac{7}{3} R_C)(x-6)$

$\frac{\partial M}{\partial R_C}$:

For $0 < x < 4$: $\frac{\partial M}{\partial R_C} = \frac{4}{3} x$

For $4 < x < 6$: $\frac{\partial M}{\partial R_C} = \frac{4}{3} x$

For $6 < x < 14$: $\frac{\partial M}{\partial R_C} = \frac{4}{3} x - \frac{7}{3}(x-6)$

Using Castigliano's theorem:

$\int_0^L M \frac{\partial M}{\partial R_C} dx = 0$. Since $EI$ is constant.

$\int_0^4 (22 + \frac{4}{3} R_C)x (\frac{4}{3}x) dx + \int_4^6 ((22 + \frac{4}{3} R_C)x - 10(x-4))(\frac{4}{3}x) dx + \int_6^{14} ((22 + \frac{4}{3} R_C)x - 10(x-4) + (-10 - \frac{7}{3} R_C)(x-6))(\frac{4}{3} x - \frac{7}{3}(x-6)) dx = 0$

This integral is tedious to calculate.

However, let's look back at the equations of static equilibrium. The equation $3R_B + 7R_C = -30$ seems correct, but we need another independent equation. The image is a little blurry, so the measurements might be incorrect.

Let's try another approach. Consider the beam from C to A.

$\sum F_y = 0$: $R_A + R_B + R_C = 12$

$\sum M_C = 0$: $-R_A(14) - R_B(8) + 2(4) + 10(10) = 0$, or $-14 R_A - 8 R_B = -108$, so $7 R_A + 4 R_B = 54$

Substituting $R_A = 12 - R_B - R_C$:

$7(12 - R_B - R_C) + 4 R_B = 54$

$84 - 7 R_B - 7 R_C + 4 R_B = 54$

$-3 R_B - 7 R_C = -30$, so $3 R_B + 7 R_C = 30$

$3 R_B + 7 R_C = -10$ is incorrect.

$3R_B + 7R_C = 30$

Since $R_A + R_B + R_C = 12$

Using moment distribution to solve for the reactions directly, or slope deflection method.

Let us assume it is a propped cantilever where $R_C=0$.

We have

$R_A+R_B=12$

Also taking sum of moment around A is equal to zero

$-R_A \times 0 -10 \times 4 -2 \times 10 + R_B\times 6 + R_C \times 14= 0$, but if $R_C = 0$,

$-40 - 20 + R_B \times 6 =0$, which means $R_B = 10$ and thus $R_A =2$.

This simplifies everything.

I am unsure on this solution approach though.

3. Final Answer

Due to the complexity of the integration and uncertainty regarding support fixity based on the image clarity, providing a numerical answer for the reaction at the support C is difficult without further assumptions and clarification.