The problem asks to determine the reactions at the supports of a given structure using Castigliano's theorem. The structure is a beam with three supports A, B, and C. The spans are 4m, 2m, and 4m respectively. There is a 10kN point load acting at a distance of 4m from support A, and a 2kN/m uniformly distributed load acting along the entire beam.

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 beam with three supports A, B, and C. The spans are 4m, 2m, and 4m respectively. There is a 10kN point load acting at a distance of 4m from support A, and a 2kN/m uniformly distributed load acting along the entire beam.

2. Solution Steps

Castigliano's theorem states that the partial derivative of the total strain energy with respect to a force is equal to the displacement in the direction of that force. For a fixed support, the displacement is zero. We'll use this to determine the reactions.

Let $R_A$, $R_B$, and $R_C$ be the reactions at supports A, B, and C respectively. We assume $R_B$ is redundant.

Therefore, $\frac{\partial U}{\partial R_B} = 0$, where U is the total strain energy.

First, calculate the reactions $R_A$ and $R_C$ in terms of $R_B$.

Taking summation of moments about point A:

$M_A = 0 = 10*4 + 2*10*5 - R_B*4 - R_C*10$

$40 + 100 = 4R_B + 10R_C$

$140 = 4R_B + 10R_C$

$10R_C = 140 - 4R_B$

$R_C = 14 - 0.4R_B$

Taking summation of vertical forces:

$\sum F_y = 0 = R_A + R_B + R_C - 10 - 2*10$

$R_A = 30 - R_B - R_C = 30 - R_B - (14 - 0.4R_B) = 16 - 0.6R_B$

Let's consider the bending moment at any section x.

We will use the method of sections to calculate the bending moment M(x). The expression for the strain energy for the whole beam is $\int \frac{M(x)^2}{2EI}dx$. So, $\frac{\partial U}{\partial R_B} = \int \frac{M(x)}{EI} \frac{\partial M(x)}{\partial R_B} dx=0$

Span AB (0 < x < 4):

$M(x) = R_A x - 0$ if $x<4$

$M(x) = R_A x - 10*(x-4) - 2x^2/2$ if $x > 4$

Then $M(x) = (16 - 0.6R_B)x - x^2$ since the point load isn't applicable.

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

$\int_0^4 M(x) \frac{\partial M}{\partial R_B} dx = \int_0^4 ((16 - 0.6R_B)x - x^2)(-0.6x) dx = \int_0^4 (-9.6x^2 + 0.36R_B x^2 + 0.6x^3)dx = [-3.2x^3 + 0.12R_Bx^3 + 0.15x^4]_0^4 = -3.2(64) + 0.12R_B(64) + 0.15(256) = -204.8 + 7.68R_B + 38.4 = -166.4 + 7.68R_B$

Span BC (0 < x < 2): Distance from B

$M(x) = R_C (4+x) - 10(2+x) - 2(6+x)^2/2 + 2(4+x)^2/2 + R_B(x) = (14 - 0.4R_B)(4+x) - 10(2+x) - (36 + 12x + x^2) + (16 + 8x + x^2) + R_Bx$

$56 + 14x - 1.6R_B - 0.4R_Bx - 20 - 10x -36 - 12x - x^2 + 16 + 8x + x^2 + R_Bx = R_Bx - 12x +16 - 0.4R_Bx - 1.6 R_B = 16 - 12x + 0.6R_Bx - 1.6 R_B$

$\frac{\partial M}{\partial R_B} = 0.6x - 1.6$

$\int_0^2 M(x) \frac{\partial M}{\partial R_B} dx = \int_0^2 (16 - 12x + 0.6R_Bx - 1.6 R_B)(0.6x - 1.6) dx = \int_0^2(9.6x -25.6 - 7.2x^2 + 19.2x + 0.36R_Bx^2 - 0.96R_Bx - 0.96R_Bx + 2.56R_B)dx = \int_0^2 (-7.2x^2 + 28.8x -25.6 + 0.36R_Bx^2 - 1.92R_Bx + 2.56R_B) dx = [-2.4x^3 + 14.4x^2 - 25.6x + 0.12R_Bx^3 - 0.96R_Bx^2 + 2.56R_Bx]_0^2 = -2.4(8) + 14.4(4) - 25.6(2) + 0.12R_B(8) - 0.96R_B(4) + 2.56R_B(2) = -19.2 + 57.6 - 51.2 + 0.96R_B - 3.84R_B + 5.12R_B = -12.8 + 2.24R_B$

Span CA (0 < x < 4): Distance from C

$M(x) = R_C x - 2x^2/2 = (14 - 0.4R_B)x - x^2$

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

$\int_0^4 M(x) \frac{\partial M}{\partial R_B} dx = \int_0^4((14 - 0.4R_B)x - x^2)(-0.4x) dx = \int_0^4 (-5.6x^2 + 0.16R_B x^2 + 0.4x^3) dx = [-\frac{5.6}{3}x^3 + \frac{0.16}{3}R_Bx^3 + \frac{0.4}{4} x^4 ]_0^4 = -\frac{5.6}{3}(64) + \frac{0.16}{3}R_B(64) + 0.1(256) = -119.467 + 3.413R_B + 25.6 = -93.867 + 3.413R_B$

Summing all the partial derivatives:

$\frac{\partial U}{\partial R_B} = -166.4 + 7.68R_B -12.8 + 2.24R_B -93.867 + 3.413R_B=0$

$-273.067 + 13.333R_B = 0$

$R_B = \frac{273.067}{13.333} = 20.48 kN$

$R_C = 14 - 0.4R_B = 14 - 0.4(20.48) = 14 - 8.192 = 5.808kN$

$R_A = 16 - 0.6R_B = 16 - 0.6(20.48) = 16 - 12.288 = 3.712kN$

3. Final Answer

$R_A = 3.712 kN$

$R_B = 20.48 kN$

$R_C = 5.808 kN$