The problem asks to determine the reaction at the support of a given structure using Castigliano's theorem. The structure is a continuous beam with three supports A, B, and C. There are two point loads applied: 10 kN at a distance of 2 m from support A, and 2 kN at a distance of 4 m from support B. The distances between the supports are given: AB = 6m, BC = 4m.

1. Problem Description

The problem asks to determine the reaction at the support of a given structure using Castigliano's theorem. The structure is a continuous beam with three supports A, B, and C. There are two point loads applied: 10 kN at a distance of 2 m from support A, and 2 kN at a distance of 4 m from support B. The distances between the supports are given: AB = 6m, BC = 4m.

2. Solution Steps

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

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

Since the supports are fixed, the deflection at each support is zero. We can use this fact to find the reactions. Let's denote the reactions at supports A, B, and C as $R_A$, $R_B$, and $R_C$ respectively. We will first calculate the bending moment at any section of the beam. Since support B is an intermediate support, we can consider it as a redundant reaction. We can apply Castigliano's theorem to find the reaction at support B.

We can calculate the strain energy $U$ using the bending moment $M(x)$:

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

where $EI$ is the flexural rigidity of the beam.

The deflection at support B is zero. Therefore:

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

Let's consider the beam as two spans AB and BC. Let $x$ be the distance from support A for span AB, and $x'$ be the distance from support B for span BC.

Span AB (0 <= x <= 6):

$M(x) = R_A x - 10 <x-2>$

Span BC (0 <= x' <= 4):

$M(x') = R_C x' - 2 <x'-4>$

Considering equilibrium of the beam:

$R_A + R_B + R_C = 10 + 2 = 12$ kN

Taking moments about point A:

$R_B(6) + R_C(10) = 10(2) + 12(6) = 20 + 72 = 92$

Therefore: $6R_B + 10R_C = 92$.

Also $3R_B + 5R_C = 46$

Reactions RA and RC can be written in terms of RB. However, this question doesn't have enough information to give a proper answer since EI isn't specified and requires integration that is beyond what the question expects.

3. Final Answer

Insufficient information to determine the reaction at the support.