The problem asks to determine the reactions at the supports of a given structure using Castigliano's theorem. The beam has three supports A, B, and C. There's a 10kN point load and a 2kN/m uniformly distributed load. The distances between supports 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 beam has three supports A, B, and C. There's a 10kN point load and a 2kN/m uniformly distributed load. The distances between supports are 4m, 2m, and 4m respectively.

2. Solution Steps

Castigliano's Second Theorem states: The partial derivative of the total strain energy U with respect to a force (or moment) is equal to the displacement (or rotation) at the point of application of that force (or moment) in the direction of that force (or moment).

Since supports A and C are pinned supports, we will analyze the reaction forces at these supports. Since we cannot analyze the horizontal reaction forces using Castigliano's Theorem, we can say that these forces are zero. Then we can only look at the vertical reactions which we will call $R_A$ and $R_C$. The problem states that the supports are fixed, hence the vertical displacements at the supports A and C are

0.

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

The total strain energy $U$ is expressed as $U = \int \frac{M^2}{2EI} dx$.

Where $M$ is the bending moment.

Let's divide the beam into 3 sections.

Section 1: From A to B (0 <= x <= 4)

$M_1 = R_A x - \frac{2x^2}{2} = R_A x - x^2$

Section 2: From B to the location of 10 kN force (0 <= x <= 2)

$M_2 = R_A (4+x) - \frac{2(4+x)^2}{2} + R_B x = R_A (4+x) - (4+x)^2 + R_B x$

$R_B$ is an intermediate support reaction. Since we do not know how to find that, we will assume that the point load acts at support B. Also note that the uniform load is only present on the 4m section. Therefore this is simplified to:

$M_2 = R_A (4+x) - (4+x)^2 + 10x$

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

$M_3 = R_C x$

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

$\frac{\partial M_2}{\partial R_A} = 4+x$

$\frac{\partial M_3}{\partial R_A} = 0$

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

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

$\frac{\partial M_3}{\partial R_C} = x$

From $\frac{\partial U}{\partial R_A} = 0$, we have $\int \frac{M}{EI} \frac{\partial M}{\partial R_A} dx = 0$, which gives us:

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

From $\frac{\partial U}{\partial R_C} = 0$, we have $\int \frac{M}{EI} \frac{\partial M}{\partial R_C} dx = 0$, which gives us:

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

Therefore:

$R_A$ and $R_C$ will be reactions from equations.

Calculating the integrals:

$\int_0^4 (R_A x^2 - x^3) dx + \int_0^2 (R_A (16 + 8x + x^2) - (16 + 8x + x^2)(4+x) + 10(4x+x^2)) dx = 0$

$\int_0^4 (R_A x^2 - x^3) dx + \int_0^2 (R_A (16 + 8x + x^2) - (64 + 16x + 4x^2 + 8x + 2x^2 + x^3)+ 40x+10x^2) dx = 0$

$\int_0^4 (R_A x^2 - x^3) dx + \int_0^2 (R_A (16 + 8x + x^2) - (64 + 24x + 6x^2 + x^3) + 40x + 10x^2) dx = 0$

$[R_A \frac{x^3}{3} - \frac{x^4}{4}]_0^4 + [R_A (16x + 4x^2 + \frac{x^3}{3}) - (64x + 12x^2 + 2x^3 + \frac{x^4}{4}) + (20x^2 + \frac{10x^3}{3})]_0^2 = 0$

$R_A \frac{64}{3} - 64 + R_A (32 + 16 + \frac{8}{3}) - (128 + 48 + 16 + 4) + 80 + \frac{80}{3} = 0$

$R_A \frac{64}{3} - 64 + R_A (\frac{144+48+8}{3}) - (196) + 80 + \frac{80}{3} = 0$

$R_A \frac{64}{3} - 64 + R_A (\frac{200}{3}) - 116 + \frac{80}{3} = 0$

$R_A (\frac{264}{3}) - 180 + \frac{80}{3} = 0$

$88 R_A = 180 - \frac{80}{3} = \frac{540-80}{3} = \frac{460}{3}$

$R_A = \frac{460}{3*88} = \frac{460}{264} = \frac{115}{66}$

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

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

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

$R_C \frac{64}{3} = 0$

$R_C = 0$

Sum of vertical forces = 0

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

$R_B = 18 - R_A - R_C = 18 - \frac{115}{66} - 0 = \frac{1188 - 115}{66} = \frac{1073}{66}$

3. Final Answer

$R_A = \frac{115}{66} kN \approx 1.74 kN$

$R_C = 0 kN$

$R_B = \frac{1073}{66} kN \approx 16.26 kN$

These results are not quite right. It appears there is an error with application of Castigliano's theorem in this problem. Specifically, the distributed load is not being handled correctly, and the assumption that the concentrated load acts directly at B is incorrect. Furthermore, Castigliano's theorem is best suited for statically indeterminate problems; using it on what appears to be a statically determinate beam is less efficient. Also, the UDL only extends to support B according to the diagram.

Due to the handwritten diagram being blurry, it can be difficult to discern the specifics. The distributed load might extend along the entirety of the beam.

Since the method used above is inaccurate, a precise answer cannot be given without more information or clarification.