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 a uniformly distributed load of $2 kN/m$ and a point load of $10 kN$. The beam has three supports labeled A, B, and C. The distances between the supports are 4m, 2m, and 4m respectively.

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 a uniformly distributed load of $2 kN/m$ and a point load of $10 kN$. The beam has three supports labeled A, B, and C. The distances between the supports are 4m, 2m, and 4m respectively.

2. Solution Steps

Castigliano's second theorem states that the partial derivative of the total strain energy $U$ with respect to a force $P_i$ applied at a point is equal to the displacement at that point in the direction of the force. Mathematically, this is expressed as:

$\delta_i = \frac{\partial U}{\partial P_i}$

Similarly, the partial derivative of the total strain energy $U$ with respect to a moment $M_i$ applied at a point is equal to the rotation at that point in the direction of the moment. Mathematically, this is expressed as:

$\theta_i = \frac{\partial U}{\partial M_i}$

For a beam, the strain energy due to bending is given by:

$U = \int_0^L \frac{M^2}{2EI} dx$

where $M$ is the bending moment, $E$ is the modulus of elasticity, $I$ is the moment of inertia, and $L$ is the length of the beam.

Since the supports are fixed or hinged, the displacements at the supports are zero. Let $R_A$, $R_B$, and $R_C$ be the vertical reactions at supports A, B, and C respectively. We can then write the following equations based on Castigliano's theorem:

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

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

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

Since the problem does not provide the flexural rigidity $EI$ value of the beam, we cannot derive exact numbers for the reactions at supports. We will need to make simplifying assumptions. Assume the beam is continuous over all three spans. We will denote $x$ to be the distance from point A.

The beam can be broken into 3 sections to calculate the bending moment $M$.

- Section 1 (0 <= x <= 4): $M_1 = R_A * x - (2 * x^2)/2 = R_A * x - x^2$

- Section 2 (4 <= x <= 6): $M_2 = R_A * x - x^2 + R_B(x-4) - 10*(x-4)$

- Section 3 (6 <= x <= 10): $M_3 = R_A * x - x^2 + R_B(x-4) - 10*(x-4) + R_C*(x-6)$

To solve this problem we would have to calculate:

$\frac{\partial U}{\partial R_A} = \int_0^4 \frac{M_1}{EI} \frac{\partial M_1}{\partial R_A} dx + \int_4^6 \frac{M_2}{EI} \frac{\partial M_2}{\partial R_A} dx + \int_6^{10} \frac{M_3}{EI} \frac{\partial M_3}{\partial R_A} dx = 0$

$\frac{\partial U}{\partial R_B} = \int_0^4 \frac{M_1}{EI} \frac{\partial M_1}{\partial R_B} dx + \int_4^6 \frac{M_2}{EI} \frac{\partial M_2}{\partial R_B} dx + \int_6^{10} \frac{M_3}{EI} \frac{\partial M_3}{\partial R_B} dx = 0$

$\frac{\partial U}{\partial R_C} = \int_0^4 \frac{M_1}{EI} \frac{\partial M_1}{\partial R_C} dx + \int_4^6 \frac{M_2}{EI} \frac{\partial M_2}{\partial R_C} dx + \int_6^{10} \frac{M_3}{EI} \frac{\partial M_3}{\partial R_C} dx = 0$

Also, we know that from equilibrium, $R_A + R_B + R_C = (2 kN/m * 10m) + 10kN$ or $R_A + R_B + R_C = 30$

Solving those 4 equations with 4 unknowns gives us our result.

3. Final Answer

The reactions at the supports can be found by solving the above equations obtained from Castigliano's theorem and static equilibrium. Because the $EI$ value is not given, we can't provide numerical results for $R_A, R_B,$ and $R_C$. However, in principle, we can set up the equations for the problem.