The problem is to determine the reactions at the supports of a given beam structure using Castigliano's theorem. The beam has three supports (A, B, and C). There is a point load of 10 kN acting downward and a uniformly distributed load of 2 kN/m. The distances between the supports and the loads are also given. The distance from support A to the 10 kN point load is 4 m, from the point load to support B is 2 m, and from support B to support C is 4 m.

1. Problem Description

The problem is to determine the reactions at the supports of a given beam structure using Castigliano's theorem. The beam has three supports (A, B, and C). There is a point load of 10 kN acting downward and a uniformly distributed load of 2 kN/m. The distances between the supports and the loads are also given. The distance from support A to the 10 kN point load is 4 m, from the point load to support B is 2 m, and from support B to support C is 4 m.

2. Solution Steps

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

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

For a support, the displacement is zero. Therefore, we introduce a redundant reaction at, say, support B, and call it $R_B$. Then, the condition that $\frac{\partial U}{\partial R_B} = 0$ can be used to find $R_B$. The total strain energy due to bending is given by:

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

Therefore, the partial derivative of $U$ with respect to $R_B$ is:

$\frac{\partial U}{\partial R_B} = \int \frac{M}{EI} \frac{\partial M}{\partial R_B} dx = 0$

Since EI is constant for the entire beam, it can be taken out of the integral. Therefore, the equation reduces to:

$\int M \frac{\partial M}{\partial R_B} dx = 0$

Let $R_A$ and $R_C$ be the reactions at supports A and C, respectively. Taking the entire beam as a free body, we have:

$R_A + R_B + R_C = 10 + 2 * 6 = 22$

Taking moments about point A:

$R_B * 6 + R_C * 10 = 10*4 + 2*6*3 = 40 + 36 = 76$

Now, we consider the bending moment equations in different sections of the beam. Let's consider three sections:

1. From A to the point load (0 <= x <= 4):

$M_1 = R_A * x$

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

2. From the point load to B (0 <= x <= 2):

$M_2 = R_A *(4+x) - 10x$

$\frac{\partial M_2}{\partial R_B} = \frac{\partial R_A}{\partial R_B} (4+x)$

3. From B to C (0 <= x <= 4):

$M_3 = R_C * x - 2*x*\frac{x}{2} = R_C * x - x^2$

$\frac{\partial M_3}{\partial R_B} = \frac{\partial R_C}{\partial R_B} x$

From the equilibrium equations, we have:

$R_A = \frac{76 - 10R_C}{6}$

$R_A + R_B + R_C = 22$

$R_A + R_C = 22 - R_B$

Therefore $\frac{76 - 10R_C}{6} + R_C = 22 - R_B$

$76 - 10R_C + 6R_C = 132 - 6R_B$

$4R_C = 6R_B - 56$

$R_C = \frac{3}{2}R_B - 14$

$R_A = 22 - R_B - (\frac{3}{2}R_B - 14) = 36 - \frac{5}{2}R_B$

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

$\frac{\partial R_C}{\partial R_B} = \frac{3}{2}$

We now substitute for the bending moment equation derivatives.

$\int_0^4 (36 - \frac{5}{2}R_B)x * (-\frac{5}{2}) dx + \int_0^2 ((36 - \frac{5}{2}R_B)(4+x) - 10x) (-\frac{5}{2})dx + \int_0^4 ((\frac{3}{2}R_B - 14)x - x^2) (\frac{3}{2}) dx = 0$

Solving the integrals and the equation:

$-\frac{5}{2}\int_0^4 (36x - \frac{5}{2}R_B x)dx -\frac{5}{2} \int_0^2 ((36 - \frac{5}{2}R_B)(4+x) - 10x) dx + \frac{3}{2}\int_0^4 (\frac{3}{2}R_B x - 14x - x^2) dx = 0$

$-\frac{5}{2} [\frac{36x^2}{2} - \frac{5}{2}R_B \frac{x^2}{2}]_0^4 -\frac{5}{2} \int_0^2 (144 + 36x - 10R_B - \frac{5}{2}R_B x -10x) dx + \frac{3}{2} [\frac{3}{2} R_B \frac{x^2}{2} - \frac{14x^2}{2} - \frac{x^3}{3}]_0^4 = 0$

$-\frac{5}{2}[288 - 10 R_B] - \frac{5}{2} \int_0^2 (144 + 26x - 10R_B -\frac{5}{2} R_B x) dx + \frac{3}{2} [12 R_B - 112 - \frac{64}{3}] = 0$

$-\frac{5}{2} [288 - 10 R_B] -\frac{5}{2}[144x + 13x^2 - 10R_B x - \frac{5}{4}R_B x^2 ]_0^2 + \frac{3}{2}[12 R_B - 112 - \frac{64}{3}] = 0$

$-\frac{5}{2}[288 - 10 R_B] - \frac{5}{2}[288 + 52 - 20R_B - 5R_B] + \frac{3}{2}[12R_B - 112 - \frac{64}{3}] = 0$

$-1440 + 25 R_B - 1440 - 130 + 50R_B + \frac{25}{2} R_B + 18 R_B - 168 - 32 = 0$

$25 R_B + 50 R_B + 12.5 R_B + 18 R_B = 1440 + 1440 + 130 + 168 + 32$

$105.5 R_B = 3210$

$R_B = 30.4265 \approx 30.43 kN$

Then:

$R_C = \frac{3}{2} * 30.43 - 14 = 45.645 - 14 = 31.645 \approx 31.65 kN$

$R_A = 36 - \frac{5}{2}*30.43 = 36 - 76.075 = -40.075 \approx -40.08 kN$

Note: The negative sign for $R_A$ signifies that the reaction acts downward.

3. Final Answer

$R_A = -40.08$ kN

$R_B = 30.43$ kN

$R_C = 31.65$ kN