The problem asks 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. There is a 10 kN point load and a 2 kN/m uniformly distributed load. The lengths of the segments are 4m, 2m, and 4m, respectively.

1. Problem Description

The problem asks 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. There is a 10 kN point load and a 2 kN/m uniformly distributed load. The lengths of the segments are 4m, 2m, and 4m, respectively.

2. Solution Steps

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

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

For finding reactions at supports, the displacement is zero. Thus,

$R_i = \frac{\partial U}{\partial \Delta_i} = 0$

where $R_i$ is the reaction force at support $i$ and $\Delta_i$ is the displacement at support $i$ (which is zero). This means that the derivative of the strain energy with respect to the redundant reaction is zero.

The total strain energy $U$ for a beam is given by:

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

Where $M$ is the bending moment as a function of $x$, $E$ is the modulus of elasticity, and $I$ is the moment of inertia.

To solve this indeterminate problem, we can consider $R_B$ as the redundant reaction at support B. We can express the bending moment $M$ as a function of $x$ and $R_B$. Then, we can apply Castigliano's theorem:

$\frac{\partial U}{\partial R_B} = \int \frac{2M}{2EI} \frac{\partial M}{\partial R_B} dx = \int \frac{M}{EI} \frac{\partial M}{\partial R_B} dx = 0$

Since $EI$ is constant, it can be taken out of the integral:

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

Let's define the sections of the beam:

Section 1: A to point load (0 <= x <= 4)

Section 2: Point load to B (0 <= x <= 2)

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

$R_A + R_B + R_C = 10 + 2(10) = 30$

$R_A(10) + R_B(6) = 10(2) + 2(10)(5) = 20 + 100 = 120$

Solving for $R_A$:

$R_A = \frac{120 - 6 R_B}{10} = 12 - 0.6 R_B$

Solving for $R_C$:

$R_C = 30 - R_A - R_B = 30 - (12 - 0.6 R_B) - R_B = 18 - 0.4 R_B$

Section 1 (0 <= x <= 4):

$M_1 = R_A x = (12 - 0.6 R_B) x$

$\frac{\partial M_1}{\partial R_B} = -0.6x$

Section 2 (0 <= x <= 2): x is measured from the 10kN point load

$M_2 = R_A(4+x) - 10x = (12 - 0.6R_B)(4+x) - 10x = 48 + 12x - 2.4R_B - 0.6R_Bx - 10x = 48 + 2x - 2.4 R_B - 0.6 R_B x$

$\frac{\partial M_2}{\partial R_B} = -2.4 - 0.6x$

Section 3 (0 <= x <= 4):

$M_3 = R_C x + 2*x^2 /2 = (18-0.4R_B)x + x^2= 18x - 0.4 R_B x + x^2$

$\frac{\partial M_3}{\partial R_B} = -0.4x$

Now, we can set up the integral:

$\int_0^4 (12-0.6R_B)x(-0.6x) dx + \int_0^2 (48 + 2x - 2.4R_B - 0.6 R_B x)(-2.4 - 0.6x) dx + \int_0^4 (18x - 0.4R_Bx + x^2)(-0.4x) dx = 0$

Integrating each part:

$(-0.6) \int_0^4 (12x^2 - 0.6 R_B x^2) dx + \int_0^2 [(48+2x)(-2.4-0.6x)+(-2.4R_B - 0.6 R_B x)(-2.4 - 0.6x)] dx + (-0.4) \int_0^4 (18x^2 - 0.4 R_B x^2 + x^3) dx = 0$

$(-0.6) [\frac{12x^3}{3} - \frac{0.6R_B x^3}{3}]_0^4 + \int_0^2 [-115.2 - 28.8x - 4.8x - 1.2x^2 + 5.76R_B + 1.44 R_B x + 1.44 R_B x + 0.36 R_B x^2] dx + (-0.4)[\frac{18x^3}{3} - \frac{0.4 R_B x^3}{3} + \frac{x^4}{4}]_0^4 = 0$

$(-0.6) [4(4)^3 - 0.2 R_B (4)^3] + \int_0^2 [-115.2 - 33.6x - 1.2x^2 + 5.76 R_B + 2.88 R_B x + 0.36 R_B x^2] dx + (-0.4)[6(4)^3 - \frac{0.4 R_B (4)^3}{3} + \frac{(4)^4}{4}] = 0$

$(-0.6) [256 - 12.8 R_B] + [-115.2x - \frac{33.6x^2}{2} - \frac{1.2x^3}{3} + 5.76 R_B x + \frac{2.88 R_B x^2}{2} + \frac{0.36 R_B x^3}{3}]_0^2 + (-0.4) [384 - 8.533 R_B + 64] = 0$

$-153.6 + 7.68 R_B + [-115.2(2) - 16.8(2^2) - 0.4(2^3) + 5.76 R_B (2) + 1.44 R_B (2^2) + 0.12 R_B (2^3)] - 0.4[448 - 8.533 R_B] = 0$

$-153.6 + 7.68 R_B + [-230.4 - 67.2 - 3.2 + 11.52 R_B + 5.76 R_B + 0.96 R_B] - 179.2 + 3.413 R_B = 0$

$-153.6 + 7.68 R_B -300.8 + 18.24 R_B - 179.2 + 3.413 R_B = 0$

$-633.6 + 29.333 R_B = 0$

$R_B = \frac{633.6}{29.333} = 21.60 \text{ kN}$

$R_A = 12 - 0.6(21.60) = 12 - 12.96 = -0.96 \text{ kN}$

$R_C = 18 - 0.4(21.60) = 18 - 8.64 = 9.36 \text{ kN}$

3. Final Answer

$R_A = -0.96 \text{ kN}$

$R_B = 21.60 \text{ kN}$

$R_C = 9.36 \text{ kN}$