The problem is 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. It has a 10 kN point load applied at the center of the span between supports A and B, and a 2 kN/m uniformly distributed load on the span between supports B and C. The span lengths are: AB = 4m + 2m + 4m = 10m and BC = 4m. The locations of supports A, B, and C are at x=0, x=10, x=14 respectively.

1. Problem Description

The problem is 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. It has a 10 kN point load applied at the center of the span between supports A and B, and a 2 kN/m uniformly distributed load on the span between supports B and C. The span lengths are: AB = 4m + 2m + 4m = 10m and BC = 4m. The locations of supports A, B, and C are at x=0, x=10, x=14 respectively.

2. Solution Steps

Since the prompt says to determine the reactions and to use Castigliano's Theorem, it is likely referring to the method of least work. The beam is statically indeterminate. We will consider the reaction at B as the redundant reaction ($R_B$).

Castigliano's second theorem states:

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

Where $\delta_i$ is the displacement in the direction of the force $P_i$ and $U$ is the strain energy.

Since support B is a rigid support, the vertical displacement at B is zero. Therefore, $\delta_B = 0$.

Thus, according to Castigliano's theorem:

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

The strain energy due to bending is given by:

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

Therefore,

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

Or simply,

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

First, determine the support reactions at A and C. Let $R_A$ and $R_C$ be the vertical reactions at A and C, respectively. Sum of moments about C = 0:

$R_A(14) - 10(9) + R_B(4) - 2(4)(2) = 0$

$14 R_A + 4 R_B = 90 + 16 = 106$

$R_A = \frac{106 - 4R_B}{14} = \frac{53 - 2R_B}{7}$

Sum of vertical forces = 0:

$R_A + R_B + R_C - 10 - 2(4) = 0$

$R_A + R_B + R_C = 18$

$R_C = 18 - R_B - R_A = 18 - R_B - (\frac{53 - 2R_B}{7}) = \frac{126 - 7R_B - 53 + 2R_B}{7} = \frac{73 - 5R_B}{7}$

Now we divide the beam into two sections, AB and BC.

For section AB (0 <= x <= 10), measuring from A:

$M(x) = R_A x - 10(x-4) H(x-4) = (\frac{53 - 2R_B}{7})x - 10(x-4) H(x-4)$

Where H(x-4) is the Heaviside step function (0 for x<4 and 1 for x>=4).

$\frac{\partial M}{\partial R_B} = -\frac{2}{7}x$

For section BC (0 <= x <= 4), measuring from C:

$M(x) = R_C x - 2x(\frac{x}{2}) = (\frac{73 - 5R_B}{7})x - x^2$

$\frac{\partial M}{\partial R_B} = -\frac{5}{7}x$

The integration is performed as follows:

$\int_0^{10} M \frac{\partial M}{\partial R_B} dx + \int_0^{4} M \frac{\partial M}{\partial R_B} dx = 0$

$\int_0^{10} [(\frac{53 - 2R_B}{7})x - 10(x-4) H(x-4)] (-\frac{2}{7}x) dx + \int_0^{4} [(\frac{73 - 5R_B}{7})x - x^2] (-\frac{5}{7}x) dx = 0$

$-\frac{2}{49} \int_0^{10} (53 - 2R_B)x^2 dx + \frac{20}{7} \int_4^{10} (x^2 - 4x) dx -\frac{5}{49} \int_0^{4} (73 - 5R_B)x^2 dx + \frac{5}{7} \int_0^{4} x^3 dx = 0$

$-\frac{2}{49} (53 - 2R_B) [\frac{x^3}{3}]_0^{10} + \frac{20}{7} [\frac{x^3}{3} - 2x^2]_4^{10} -\frac{5}{49} (73 - 5R_B) [\frac{x^3}{3}]_0^{4} + \frac{5}{7} [\frac{x^4}{4}]_0^{4} = 0$

$-\frac{2}{49} (53 - 2R_B) \frac{1000}{3} + \frac{20}{7} [(\frac{1000}{3} - 200) - (\frac{64}{3} - 32)] -\frac{5}{49} (73 - 5R_B) \frac{64}{3} + \frac{5}{7} \frac{256}{4} = 0$

$-\frac{2000}{147} (53 - 2R_B) + \frac{20}{7} [\frac{400}{3} + \frac{32}{3}] -\frac{320}{147} (73 - 5R_B) + \frac{5}{7} (64) = 0$

$-\frac{2000}{147} (53 - 2R_B) + \frac{20}{7} [\frac{432}{3}] -\frac{320}{147} (73 - 5R_B) + \frac{320}{7} = 0$

$-\frac{106000}{147} + \frac{4000}{147}R_B + \frac{20}{7} (144) -\frac{23360}{147} + \frac{1600}{147}R_B + \frac{320}{7} = 0$

$\frac{5600}{147}R_B = \frac{106000}{147} + \frac{23360}{147} - \frac{2880}{7} - \frac{320}{7}$

$\frac{5600}{147}R_B = \frac{129360}{147} - \frac{3200}{7} = \frac{129360 - 67200}{147} = \frac{62160}{147}$

$R_B = \frac{62160}{147} \cdot \frac{147}{5600} = \frac{62160}{5600} = 11.1 \mathrm{kN}$

$R_A = \frac{53 - 2(11.1)}{7} = \frac{53 - 22.2}{7} = \frac{30.8}{7} = 4.4 \mathrm{kN}$

$R_C = \frac{73 - 5(11.1)}{7} = \frac{73 - 55.5}{7} = \frac{17.5}{7} = 2.5 \mathrm{kN}$

3. Final Answer

$R_A = 4.4 \mathrm{kN}$

$R_B = 11.1 \mathrm{kN}$

$R_C = 2.5 \mathrm{kN}$