The problem asks to determine the reaction at support B of a continuous beam ABC using Castigliano's theorem. The beam has supports at A, B, and C. A downward force of 10 kN acts 2 meters from support A. A downward force of 2 kN acts 4 meters from support B (which is 6 meters from support A). The distance between A and B is 4 meters, and the distance between B and C is 4 meters. Therefore, the total length of the beam is 8 meters.

1. Problem Description

The problem asks to determine the reaction at support B of a continuous beam ABC using Castigliano's theorem. The beam has supports at A, B, and C. A downward force of 10 kN acts 2 meters from support A. A downward force of 2 kN acts 4 meters from support B (which is 6 meters from support A). The distance between A and B is 4 meters, and the distance between B and C is 4 meters. Therefore, the total length of the beam is 8 meters.

2. Solution Steps

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

$\delta = \frac{\partial U}{\partial P}$

For support reactions, since there is no displacement at the support, we have $\frac{\partial U}{\partial R_B} = 0$, where $R_B$ is the reaction force at support B.

The total strain energy due to bending is given by:

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

where $M$ is the bending moment, $E$ is the modulus of elasticity, and $I$ is the moment of inertia.

Therefore, we have:

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

We need to find the bending moment $M$ in terms of $x$ and $R_B$.

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

We have $R_A + R_B + R_C = 10 + 2 = 12 kN$.

Taking moments about A:

$4R_B + 8R_C = 10(2) + 2(6) = 20 + 12 = 32$

$R_B + 2R_C = 8$, which implies $R_C = (8-R_B)/2 = 4 - R_B/2$

$R_A = 12 - R_B - R_C = 12 - R_B - (4-R_B/2) = 8 - R_B/2$

Now we need to divide the beam into three sections:

Section 1: A to the 10 kN force (0 <= x <= 2)

$M_1 = R_A x = (8 - R_B/2)x$

$\frac{\partial M_1}{\partial R_B} = -\frac{x}{2}$

Section 2: 10 kN force to B (2 <= x <= 4)

$M_2 = R_A x - 10(x-2) = (8 - R_B/2)x - 10x + 20 = -2x - (R_B/2)x + 20$

$\frac{\partial M_2}{\partial R_B} = -\frac{x}{2}$

Section 3: B to 2 kN force (4 <= x <= 6) (x is from A)

$M_3 = R_A x - 10(x-2) + R_B(x-4) = (8 - R_B/2)x - 10x + 20 + R_B x - 4 R_B = -2x + (R_B/2) x - 4R_B + 20$

$\frac{\partial M_3}{\partial R_B} = \frac{x}{2} - 4$

Section 4: 2kN force to C (6 <= x <= 8) (x is from A)

$M_4 = R_A x - 10(x-2) + R_B(x-4) - 2(x-6) = (8-R_B/2)x - 10x + 20 + R_B x - 4R_B - 2x + 12 = -4x + (R_B/2)x - 4R_B + 32$

$\frac{\partial M_4}{\partial R_B} = \frac{x}{2} - 4$

$\int_0^2 M_1 \frac{\partial M_1}{\partial R_B} dx + \int_2^4 M_2 \frac{\partial M_2}{\partial R_B} dx + \int_4^6 M_3 \frac{\partial M_3}{\partial R_B} dx + \int_6^8 M_4 \frac{\partial M_4}{\partial R_B} dx = 0$

$\int_0^2 (8 - R_B/2)x (-\frac{x}{2}) dx + \int_2^4 (-2x - (R_B/2)x + 20)(-\frac{x}{2}) dx + \int_4^6 (-2x + (R_B/2) x - 4R_B + 20)(\frac{x}{2} - 4) dx + \int_6^8 (-4x + (R_B/2)x - 4R_B + 32)(\frac{x}{2} - 4) dx = 0$

Evaluating the integrals (omitted for brevity):

$- \frac{4}{3}(8 - R_B/2) + \frac{1}{6}(16 + 6 R_B - 240) + \frac{1}{12}(-240 + 44 R_B) + \frac{1}{12}(-576 + 92 R_B) = 0$

Simplifying:

$-32/3 + 2R_B/3 -37.33 + R_B = -20 - R_B + 80/3 + (-20 +11/3 R_B)+(-48+23/3 R_B)=0$

$-32/3 + 20 + 48/3$

$-37.33-3.33(3*4)+105(2)-40*2 = R_B$

After integrating and simplifying, we get:

$R_B (20/3 + 44/12 + 92/12) = 32/3 + 104/3 + 240/12 + 576/12$

$R_B (\frac{80+44+92}{12}) = (\frac{128+416+240+576}{12})$

$216R_B = 1360$

$R_B = 1360/216 = 6.296$

3. Final Answer

The reaction at support B is 6.30 kN.