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 supports at A, B, and C. The beam is subjected to a uniformly distributed load of 2 kN/m and a point load of 10 kN. The distances between the supports are: AB = 4+2 = 6 m and BC = 4 m. The distances are as follows: distance from A to the 10 kN load is 4 m.

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 supports at A, B, and C. The beam is subjected to a uniformly distributed load of 2 kN/m and a point load of 10 kN. The distances between the supports are: AB = 4+2 = 6 m and BC = 4 m. The distances are as follows: distance from A to the 10 kN load is 4 m.

2. Solution Steps

Castigliano's theorem states that the partial derivative of the total strain energy with respect to a force gives the displacement in the direction of that force. Since the supports are fixed, the displacements at the supports are zero. We can use this information to find the reactions at the supports.

Let $R_A$, $R_B$, and $R_C$ be the vertical reactions at supports A, B, and C, respectively.

Let's consider $R_A$ and $R_C$ as redundant reactions.

We will have two spans: AB and BC. The total length of the beam is 10 m.

The uniformly distributed load is 2 kN/m. The point load of 10 kN is acting at 4m from support A.

Since the displacements at supports A and C are zero, we can write the following equations based on Castigliano's theorem:

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

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

Where $U$ is the total strain energy of the beam.

Let's divide the beam into two spans: AB and BC. Let's denote the moment at any point in the spans AB and BC as $M_1$ and $M_2$ respectively.

$M_1 = R_A x - 10 <x-4>^1 - \frac{2x^2}{2}$

$M_2 = R_C (10-x) - \frac{2(10-x)^2}{2}$ , where x starts from A

Where $<x-4>^1$ = (x-4) if x > 4 or 0 if x <

4.

Castigliano's second theorem says:

$\delta_i = \frac{\partial U}{\partial P_i}$ = $\int_0^L \frac{M}{EI} \frac{\partial M}{\partial P_i} dx$

$\frac{\partial U}{\partial R_A} = \int_0^{6} \frac{M_1}{EI} \frac{\partial M_1}{\partial R_A} dx + \int_{6}^{10} \frac{M_2}{EI} \frac{\partial M_2}{\partial R_A} dx = 0$

Since $\frac{\partial M_2}{\partial R_A} = 0$,

$\int_0^{6} \frac{M_1}{EI} \frac{\partial M_1}{\partial R_A} dx = 0$

$\frac{\partial M_1}{\partial R_A} = x$

$\frac{\partial U}{\partial R_C} = \int_0^{6} \frac{M_1}{EI} \frac{\partial M_1}{\partial R_C} dx + \int_{6}^{10} \frac{M_2}{EI} \frac{\partial M_2}{\partial R_C} dx = 0$

Since $\frac{\partial M_1}{\partial R_C} = 0$,

$\int_{6}^{10} \frac{M_2}{EI} \frac{\partial M_2}{\partial R_C} dx = 0$

$\frac{\partial M_2}{\partial R_C} = (10-x)$

Therefore,

$\int_0^{6} (R_A x - 10 <x-4>^1 - x^2) x dx = 0$

$\int_{6}^{10} (R_C (10-x) - (10-x)^2) (10-x) dx = 0$

$R_A \int_0^6 x^2 dx - 10 \int_4^6 (x-4) x dx - \int_0^6 x^3 dx = 0$

$R_A (\frac{6^3}{3}) - 10 \int_4^6 (x^2 - 4x) dx - \frac{6^4}{4} = 0$

$72 R_A - 10 [\frac{x^3}{3} - 2x^2]_4^6 - 324 = 0$

$72 R_A - 10 [(\frac{216}{3} - 72) - (\frac{64}{3} - 32)] - 324 = 0$

$72 R_A - 10 [(72 - 72) - (\frac{64}{3} - \frac{96}{3})] - 324 = 0$

$72 R_A - 10 [\frac{32}{3}] - 324 = 0$

$72 R_A = 324 + \frac{320}{3} = \frac{972 + 320}{3} = \frac{1292}{3}$

$R_A = \frac{1292}{3 * 72} = \frac{1292}{216} = 5.98 \approx 6 kN$

$R_C \int_6^{10} (10-x)^2 dx - \int_6^{10} (10-x)^3 dx = 0$

$R_C [\frac{-(10-x)^3}{3}]_6^{10} - [\frac{-(10-x)^4}{4}]_6^{10} = 0$

$R_C [0 + \frac{4^3}{3}] - [0 + \frac{4^4}{4}] = 0$

$R_C \frac{64}{3} - \frac{256}{4} = 0$

$R_C \frac{64}{3} = 64$

$R_C = 3 kN$

$R_A + R_B + R_C = 2 * 10 + 10 = 30$

$6 + R_B + 3 = 30$

$R_B = 30 - 9 = 21 kN$

3. Final Answer

$R_A = 6$ kN

$R_B = 21$ kN

$R_C = 3$ kN