The problem asks us to determine the reactions at the supports of a given structure using Castigliano's theorem. The structure is a continuous beam with three supports (A, B, and C). There is a point load of 110 kN acting at a distance of 4m from support A, and a uniformly distributed load of 2 kN/m acting over the entire span between supports B and C, which is 4m. The span between A and B also is 4m+2m=6m.

1. Problem Description

The problem asks us to determine the reactions at the supports of a given structure using Castigliano's theorem. The structure is a continuous beam with three supports (A, B, and C). There is a point load of 110 kN acting at a distance of 4m from support A, and a uniformly distributed load of 2 kN/m acting over the entire span between supports B and C, which is 4m. The span between A and B also is 4m+2m=6m.

2. Solution Steps

To solve this problem using Castigliano's theorem, we need to determine the bending moment equation for the entire beam and then apply the theorem to find the support reactions. Since the supports A and C are simple supports, the reactions are vertical reactions. The support at B is also a simple support, so it provides a vertical reaction as well. Let $R_A$, $R_B$, and $R_C$ be the vertical reactions at supports A, B, and C, respectively.

First, we'll consider the beam as two segments: AB and BC. Let's denote $x_1$ as the distance from A towards B, and $x_2$ as the distance from C towards B.

Segment AB ($0 \le x_1 \le 6$):

The bending moment $M_1(x_1) = R_A x_1 - 110 \langle x_1 - 4 \rangle$, where $\langle x \rangle = x$ if $x > 0$ and 0 if $x \le 0$.

Segment BC ($0 \le x_2 \le 4$):

The bending moment $M_2(x_2) = R_C x_2 + 2 x_2^2 / 2 = R_C x_2 + x_2^2$.

Now, we need to apply the equations of static equilibrium to relate the reactions.

Sum of vertical forces = 0: $R_A + R_B + R_C = 110 + 2*4 = 118$.

Consider moment about A=0: $R_B * 6 + R_C * 10 = 110 * 4 + (2 * 4)*8 = 440+64=504$

$6 R_B + 10 R_C = 504$

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. Since the supports are fixed, the deflections are zero at A, B, and C. We only need to consider Castigliano's theorem for one of the supports, B for instance:

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

For a beam like this one, it will be hard to apply Castigliano's theorem directly to find the reactions.

Alternatively we can view this as a superposition problem with 3 reaction forces R_A, R_B, R_C. We have already used two equilibrium equations.

The final equation should arise from the fact that there should be zero deflection at point B.

$R_A + R_B + R_C = 118$

$6 R_B + 10 R_C = 504$

From the second equation:

$R_B = (504 - 10 R_C)/6 = 84 - (5/3) R_C$

Substitute $R_B$ into the first equation:

$R_A + 84 - (5/3)R_C + R_C = 118$

$R_A = 34 + (2/3) R_C$

3. Final Answer

The reactions at the supports can be expressed in terms of $R_C$ as follows:

$R_A = 34 + (2/3) R_C$

$R_B = 84 - (5/3) R_C$

$R_C = R_C$

Without further information or the flexural rigidity (EI) of the beam, it is impossible to find the exact values of the reactions $R_A$, $R_B$, and $R_C$.

We need a third equation (usually arising from the deflection condition) to uniquely determine the reaction forces.

Final Answer: Cannot be uniquely determined. Need further information about deflection.