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 three supports (A, B, and C). There is a point load of 10 kN acting between support A and B and a uniformly distributed load of 2 kN/m acting between support B and C. The distances between the supports and loads are specified: 4 m between A and the point load, 2 m between the point load and B, and 4 m between B and C.

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 three supports (A, B, and C). There is a point load of 10 kN acting between support A and B and a uniformly distributed load of 2 kN/m acting between support B and C. The distances between the supports and loads are specified: 4 m between A and the point load, 2 m between the point load and B, and 4 m between B and C.

2. Solution Steps

To determine the support reactions using Castigliano's theorem, we need to find the strain energy of the beam and then differentiate it with respect to the reactions. Because the beam has multiple supports, it's statically indeterminate. We can assume that the reactions at A and C are the redundants.

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

Let's consider the beam as two sections AB and BC.

Section AB:

The length of the section is $4 + 2 = 6$ m. The point load is at 4 m from A.

Let's express the bending moment $M_1(x)$ at any point $x$ from A in section AB (where $0 \le x \le 6$):

$M_1(x) = R_A x - 10 <x-4>$

where $<x-a>$ is the Macaulay bracket, which is zero if $x < a$ and equals $(x-a)$ if $x > a$.

Section BC:

The length of the section is 4 m.

Let's express the bending moment $M_2(x)$ at any point $x$ from C in section BC (where $0 \le x \le 4$):

$M_2(x) = R_C x - \frac{2x^2}{2} = R_C x - x^2$

The reaction at B can be determined using the equilibrium equation for the entire beam:

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

$R_B = 18 - R_A - R_C$

Using Castigliano's theorem:

The deflection at the supports A and C is zero.

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

The strain energy $U$ in the beam due to bending is given by:

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

where M is the bending moment, E is Young's modulus, and I is the moment of inertia.

Since $\frac{\partial U}{\partial R_A} = 0$, we have:

$\int_0^6 \frac{M_1}{EI} \frac{\partial M_1}{\partial R_A} dx + \int_0^4 \frac{M_2}{EI} \frac{\partial M_2}{\partial R_A} dx = 0$

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

$\frac{\partial M_2}{\partial R_A} = 0$

So, we have:

$\int_0^6 M_1 x dx = 0$

$\int_0^6 (R_A x - 10 <x-4>) x dx = 0$

$\int_0^6 (R_A x^2 - 10x <x-4>) dx = 0$

$R_A \int_0^6 x^2 dx - 10 \int_4^6 x(x-4) dx = 0$

$R_A [\frac{x^3}{3}]_0^6 - 10 \int_4^6 (x^2 - 4x) dx = 0$

$R_A (\frac{6^3}{3}) - 10 [\frac{x^3}{3} - 2x^2]_4^6 = 0$

$72 R_A - 10 [(\frac{216}{3} - 2*36) - (\frac{64}{3} - 2*16)] = 0$

$72 R_A - 10 [(72-72) - (\frac{64}{3} - 32)] = 0$

$72 R_A - 10 [0 - (\frac{64-96}{3})] = 0$

$72 R_A - 10 (\frac{32}{3}) = 0$

$72 R_A = \frac{320}{3}$

$R_A = \frac{320}{3*72} = \frac{320}{216} = \frac{40}{27}$ kN

Similarly, since $\frac{\partial U}{\partial R_C} = 0$, we have:

$\int_0^6 \frac{M_1}{EI} \frac{\partial M_1}{\partial R_C} dx + \int_0^4 \frac{M_2}{EI} \frac{\partial M_2}{\partial R_C} dx = 0$

$\frac{\partial M_1}{\partial R_C} = 0$

$\frac{\partial M_2}{\partial R_C} = x$

So, we have:

$\int_0^4 M_2 x dx = 0$

$\int_0^4 (R_C x - x^2) x dx = 0$

$\int_0^4 (R_C x^2 - x^3) dx = 0$

$R_C \int_0^4 x^2 dx - \int_0^4 x^3 dx = 0$

$R_C [\frac{x^3}{3}]_0^4 - [\frac{x^4}{4}]_0^4 = 0$

$R_C (\frac{4^3}{3}) - \frac{4^4}{4} = 0$

$R_C \frac{64}{3} = \frac{256}{4} = 64$

$R_C = \frac{64 * 3}{64} = 3$ kN

$R_B = 18 - R_A - R_C = 18 - \frac{40}{27} - 3 = 15 - \frac{40}{27} = \frac{405 - 40}{27} = \frac{365}{27}$ kN

3. Final Answer

$R_A = \frac{40}{27}$ kN

$R_B = \frac{365}{27}$ kN

$R_C = 3$ kN