The problem asks to determine the reactions at the supports of the given structure using Castigliano's theorem. The structure is a continuous beam with supports at A, B, and C. There's a point load of 10 kN acting downward and a uniformly distributed load of 2 kN/m. The distances between the supports are: AB = 4m + 2m = 6m and BC = 4m.

1. Problem Description

The problem asks to determine the reactions at the supports of the given structure using Castigliano's theorem. The structure is a continuous beam with supports at A, B, and C. There's a point load of 10 kN acting downward and a uniformly distributed load of 2 kN/m. The distances between the supports are: AB = 4m + 2m = 6m and BC = 4m.

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 is equal to the displacement or rotation in the direction of the force or moment at the point of application.

Since the supports do not settle (i.e., the vertical displacement is zero at each support), we can apply Castigliano's theorem to find the reactions at the supports. We will introduce a redundant reaction at support B, denoted as $R_B$. Then the vertical displacement at B is zero, so $\frac{\partial U}{\partial R_B} = 0$.

First, let's define the coordinates.

x = 0 at A.

We have two sections, AB and BC.

Section AB (0 <= x <= 6)

Section BC (0 <= x' <= 4), where x' = x - 6

We need to express the bending moment $M$ as a function of $x$, $R_A$, $R_B$, and $R_C$ and other loads. Let us find $R_A$ and $R_C$ in terms of $R_B$. Taking the moment about A yields:

$R_B \cdot 6 + R_C \cdot 10 - 10 \cdot 2 - 2 \cdot 10 \cdot 5 = 0$

$6 R_B + 10 R_C = 20 + 100 = 120$

$3 R_B + 5 R_C = 60$

$R_C = 12 - \frac{3}{5}R_B$

Sum of vertical forces: $R_A + R_B + R_C = 10 + 2 \cdot 4 = 10 + 8 = 18$

$R_A + R_B + 12 - \frac{3}{5} R_B = 18$

$R_A = 6 - \frac{2}{5} R_B$

Now, we calculate the bending moments:

For section AB (0 <= x <= 2):

$M(x) = R_A \cdot x = (6 - \frac{2}{5} R_B)x$

For section AB (2 <= x <= 6):

$M(x) = R_A \cdot x - 10(x-2) = (6 - \frac{2}{5} R_B) x - 10(x-2) = (6 - \frac{2}{5} R_B - 10)x + 20 = (-4 - \frac{2}{5}R_B) x + 20$

For section BC (0 <= x' <= 4, x=x'+6):

$M(x') = R_C x' - 2 \frac{x'^2}{2} = (12 - \frac{3}{5} R_B) x' - x'^2$

Castigliano's theorem states: $\frac{\partial U}{\partial R_B} = \int \frac{M}{EI} \frac{\partial M}{\partial R_B} dx = 0$

$\int_0^6 M \frac{\partial M}{\partial R_B} dx + \int_6^{10} M \frac{\partial M}{\partial R_B} dx=0$.

The expressions are getting very complex. There appears to be an error in the diagram. The uniformly distributed load extends past support B. So, let's consider the alternative interpretation that the uniformly distributed load extends only over the segment BC.

For section AB (0 <= x <= 2): $M(x) = R_A x = (6 - \frac{2}{5} R_B) x$, $\frac{\partial M}{\partial R_B} = - \frac{2}{5} x$

For section AB (2 <= x <= 6): $M(x) = R_A x - 10(x-2) = (6 - \frac{2}{5} R_B) x - 10(x-2) = (-4 - \frac{2}{5}R_B) x + 20$, $\frac{\partial M}{\partial R_B} = - \frac{2}{5}x$

For section BC (0 <= x' <= 4, x=x'+6): $M(x') = R_C x' - 2 \frac{x'^2}{2} = (12 - \frac{3}{5} R_B) x' - x'^2$, $\frac{\partial M}{\partial R_B} = - \frac{3}{5}x'$

$\int_0^2 (6 - \frac{2}{5} R_B)x (-\frac{2}{5}x) dx + \int_2^6 ((-4 - \frac{2}{5}R_B)x + 20) (-\frac{2}{5}x) dx + \int_0^4 ((12 - \frac{3}{5} R_B) x' - x'^2) (-\frac{3}{5}x') dx = 0$

$-\frac{2}{5} \int_0^2 (6 - \frac{2}{5} R_B) x^2 dx - \frac{2}{5} \int_2^6 ((-4 - \frac{2}{5}R_B)x^2 + 20x) dx - \frac{3}{5} \int_0^4 ((12 - \frac{3}{5} R_B) x'^2 - x'^3) dx = 0$

$-\frac{2}{5} (2x^3 - \frac{2}{15}R_B x^3)|_0^2 - \frac{2}{5} (-\frac{4}{3}x^3 - \frac{2}{15}R_B x^3 + 10x^2)|_2^6 - \frac{3}{5} (4x'^3 - \frac{3}{20} R_B x'^3 - \frac{1}{4} x'^4)|_0^4 = 0$

$-\frac{2}{5} (16-\frac{32}{15}R_B) - \frac{2}{5} [ (-\frac{4}{3} (216-8) - \frac{2}{15}R_B(216-8) + 10(36-4)] - \frac{3}{5} [4(64) - \frac{3}{20}R_B(64) - \frac{1}{4} (256)] = 0$

$-\frac{32}{5} + \frac{64}{75} R_B -\frac{2}{5} [ -\frac{4}{3} (208) - \frac{2}{15}R_B (208) + 320 ] -\frac{3}{5} [256 - \frac{192}{20} R_B - 64 ] = 0$

$-\frac{32}{5} + \frac{64}{75} R_B + \frac{832}{15} + \frac{416}{75} R_B - \frac{640}{5} - \frac{576}{5} + \frac{144}{25} R_B + \frac{192}{20}= 0$

$\frac{64}{75} R_B + \frac{416}{75} R_B + \frac{432}{75} R_B = \frac{32}{5} - \frac{832}{15} + \frac{640}{5} + \frac{576}{5} - \frac{192}{20}$

$\frac{912}{75} R_B = \frac{96}{15} - \frac{832}{15} + \frac{1920}{15} + \frac{1728}{15} - \frac{144}{15}$

$\frac{912}{75} R_B = \frac{2768}{15}$

$R_B = \frac{2768}{15} \cdot \frac{75}{912} = \frac{2768 \cdot 5}{912} = \frac{13840}{912} = \frac{865}{57} \approx 15.175$

$R_C = 12 - \frac{3}{5} R_B = 12 - \frac{3}{5} \frac{865}{57} = 12 - \frac{519}{57} = \frac{684 - 519}{57} = \frac{165}{57} = \frac{55}{19} \approx 2.895$

$R_A = 6 - \frac{2}{5} R_B = 6 - \frac{2}{5} \frac{865}{57} = 6 - \frac{346}{57} = \frac{342-346}{57} = - \frac{4}{57} \approx -0.07$

These results do not make sense. Something has to be reconsidered.

3. Final Answer

Due to the complexity of calculations and difficulty of interpreting the image, I am unable to provide the final answer. The approach would involve writing down bending moment equations for sections AB and BC, taking partial derivatives with respect to the redundant reaction Rb, setting the integration to 0, and then solving for Rb. Knowing Rb, Ra and Rc can be found.