The problem asks us to determine the reactions at the supports of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a point load of 10 kN applied at a distance of 4m from support A. There is a uniformly distributed load of 2 kN/m applied between the second support and the end. The distance between A and B is 4+2=6m and between B and C is 4m. The total length of the beam is 6+4=10m.

1. Problem Description

The problem asks us to determine the reactions at the supports of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a point load of 10 kN applied at a distance of 4m from support A. There is a uniformly distributed load of 2 kN/m applied between the second support and the end. The distance between A and B is 4+2=6m and between B and C is 4m. The total length of the beam is 6+4=10m.

2. Solution Steps

Castigliano's second theorem states that the partial derivative of the total strain energy with respect to an applied force is equal to the displacement at the point of application of that force in the direction of that force. Since the supports A, B, and C are fixed, the vertical displacement at each support is zero. Thus, we can apply Castigliano's theorem to find the reactions at the supports.

Let $R_A$, $R_B$, and $R_C$ be the reactions at supports A, B, and C, respectively. Since we have three unknowns and only two equations of equilibrium, we need to use Castigliano's theorem to obtain a third equation.

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

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

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

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

First, consider the free body diagram of the beam. Summing vertical forces:

$R_A + R_B + R_C = 10 + 2*4 = 18$ kN

Summing moments about A:

$R_B*6 + R_C*10 = 10*4 + 2*4*(6+4/2) = 40 + 8*8 = 40 + 64 = 104$ kNm

We now use Castigliano's theorem. Consider the moment at a distance $x$ from A.

For $0 < x < 6$:

$M(x) = R_A x - 10\langle x-4 \rangle^1$

For $6 < x < 10$:

$M(x) = R_A x - 10\langle x-4 \rangle^1 + R_B(x-6)$

Where $\langle x-a \rangle^n = (x-a)^n$ if $x > a$ and 0 if $x < a$.

$U = \int_0^{10} \frac{M^2}{2EI} dx$

We differentiate with respect to $R_A$:

$\frac{\partial U}{\partial R_A} = \int_0^{10} \frac{2M}{2EI} \frac{\partial M}{\partial R_A} dx = \int_0^{10} \frac{M}{EI} \frac{\partial M}{\partial R_A} dx = 0$

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

$\int_0^{10} M x dx = 0$

$\int_0^6 (R_A x - 10\langle x-4 \rangle)x dx + \int_6^{10} (R_A x - 10\langle x-4 \rangle + R_B(x-6))x dx = 0$

$\int_0^6 (R_A x^2 - 10x(x-4))dx + \int_6^{10} (R_A x^2 - 10x(x-4) + R_B x(x-6)) dx= 0$

$\int_0^4 R_A x^2 dx + \int_4^6 (R_A x^2 - 10x(x-4))dx + \int_6^{10} (R_A x^2 - 10x(x-4) + R_B x(x-6)) dx= 0$

$\int_0^4 R_A x^2 dx + \int_4^6 (R_A x^2 - 10x^2 + 40x) dx + \int_6^{10} (R_A x^2 - 10x^2 + 40x + R_B x^2 - 6R_B x) dx= 0$

$[R_A \frac{x^3}{3}]_0^4 + [R_A \frac{x^3}{3} - \frac{10x^3}{3} + 20x^2]_4^6 + [R_A \frac{x^3}{3} - \frac{10x^3}{3} + 20x^2 + R_B \frac{x^3}{3} - 3R_B x^2]_6^{10} = 0$

$R_A\frac{64}{3} + R_A(\frac{216}{3} - \frac{64}{3}) - \frac{10}{3}(216-64) + 20(36-16) + R_A (\frac{1000}{3} - \frac{216}{3}) - \frac{10}{3}(1000-216) + 20(100-36) + R_B(\frac{1000}{3} - \frac{216}{3}) - 3R_B(100-36)=0$

$R_A \frac{64}{3} + R_A \frac{152}{3} - \frac{10*152}{3} + 400 + R_A \frac{784}{3} - \frac{10*784}{3} + 20*64 + R_B\frac{784}{3} - 3R_B(64) = 0$

$R_A(\frac{64+152+784}{3}) + R_B(\frac{784}{3}-192) = \frac{10*152}{3} - 400 + \frac{10*784}{3} - 1280$

$R_A(\frac{1000}{3}) + R_B (\frac{784-576}{3}) = \frac{1520+7840}{3} - 1680$

$\frac{1000}{3} R_A + \frac{208}{3} R_B = \frac{9360-5040}{3}$

$1000 R_A + 208 R_B = 4320$

$125 R_A + 26 R_B = 540$

We have $R_A + R_B + R_C = 18$ and $6 R_B + 10 R_C = 104$. Then $R_C = \frac{104-6R_B}{10}$.

$R_A + R_B + \frac{104-6R_B}{10} = 18$

$10R_A + 10R_B + 104 - 6R_B = 180$

$10R_A + 4R_B = 76$

$5R_A + 2R_B = 38$

We also have $125 R_A + 26 R_B = 540$. Multiply the first equation by

1

3. $65R_A + 26 R_B = 494$

$125 R_A + 26 R_B = 540$

$60 R_A = 46$

$R_A = \frac{46}{60} = \frac{23}{30} = 0.766667$ kN

$5*(\frac{23}{30}) + 2 R_B = 38$

$\frac{23}{6} + 2 R_B = 38$

$2 R_B = 38 - \frac{23}{6} = \frac{228-23}{6} = \frac{205}{6}$

$R_B = \frac{205}{12} = 17.083333$ kN

$R_C = \frac{104 - 6(\frac{205}{12})}{10} = \frac{104 - \frac{205}{2}}{10} = \frac{208 - 205}{20} = \frac{3}{20} = 0.15$ kN

Check: $0.766667 + 17.083333 + 0.15 = 18$.

3. Final Answer

$R_A = 0.767$ kN

$R_B = 17.083$ kN

$R_C = 0.15$ kN