We are asked to determine the reaction at support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a 10 kN load acting downward and a 2 kN load acting downward. The distances between the supports and loads are given in the figure: AB = 2 m, BC = 4 m. The support at A is 4 m to the left of the 10kN load, the 10kN is 2m to the left of support B, and support C is 4m to the right of support B.

1. Problem Description

We are asked to determine the reaction at support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a 10 kN load acting downward and a 2 kN load acting downward. The distances between the supports and loads are given in the figure: AB = 2 m, BC = 4 m. The support at A is 4 m to the left of the 10kN load, the 10kN is 2m to the left of support B, and support C is 4m to the right of support B.

2. Solution Steps

Castigliano's second theorem states that the partial derivative of the total strain energy (U) with respect to a force (P) applied at a point is equal to the displacement (delta) at that point in the direction of the force.

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

Since support B is a fixed support, the vertical deflection at B is zero. Therefore, we can apply a vertical reaction force $R_B$ at B, calculate the strain energy in the beam as a function of $R_B$, and then set the derivative of the strain energy with respect to $R_B$ equal to zero.

$\frac{\partial U}{\partial R_B} = 0$

First, we determine the reactions at supports A and C in terms of $R_B$.

Take the sum of moments about point A equal to zero:

$\sum M_A = 0$

$R_B \cdot 6 + R_C \cdot 6 - 10 \cdot 4 - 2 \cdot 10 = 0$

$6R_B + 6R_C = 40 + 20 = 60$

$R_B + R_C = 10$

$R_C = 10 - R_B$

Take the sum of forces in the vertical direction equal to zero:

$\sum F_y = 0$

$R_A + R_B + R_C - 10 - 2 = 0$

$R_A + R_B + (10 - R_B) = 12$

$R_A + 10 = 12$

$R_A = 2$ kN

Now, consider the bending moments in each section of the beam.

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

$M_1(x) = R_A x = 2x$ (0 <= x <= 4)

$M_1(x) = R_A x - 10(x-4) = 2x - 10x + 40 = -8x + 40$ (4 <= x <= 6)

Section BC: (0 <= x <= 6)

$M_2(x) = R_C x = (10-R_B)x$ (0 <= x <= 4)

$M_2(x) = R_C x - 2(x-4) = (10-R_B)x - 2x + 8 = (8-R_B)x + 8$ (4 <= x <= 6)

The total strain energy due to bending is:

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

Where E is Young's modulus, I is the second moment of area of the cross-section, and M is the bending moment.

We are only concerned about bending moment here, so:

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

Since the deflection at point B is zero:

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

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

$\frac{\partial M_2}{\partial R_B} = -x$ (0 <= x <= 4)

$\frac{\partial M_2}{\partial R_B} = -x$ (4 <= x <= 6)

$\int_{0}^{6} M_1 * 0 + \int_{0}^{4} (10-R_B)x * (-x) dx + \int_{4}^{6} ((8-R_B)x + 8) * (-x) dx = 0$

$\int_{0}^{4} (10-R_B)x^2 dx + \int_{4}^{6} ((8-R_B)x^2 + 8x) dx = 0$

$(10-R_B)[\frac{x^3}{3}]_0^4 + (8-R_B)[\frac{x^3}{3}]_4^6 + 8[\frac{x^2}{2}]_4^6 = 0$

$(10-R_B)(\frac{64}{3}) + (8-R_B)(\frac{216-64}{3}) + 8(\frac{36-16}{2}) = 0$

$(10-R_B)(\frac{64}{3}) + (8-R_B)(\frac{152}{3}) + 8(10) = 0$

$640 - 64R_B + 1216 - 152R_B + 240 = 0$

$2096 + 640 - 216R_B = 0$

$216R_B = 2856$

$R_B = \frac{2856}{216} = 13.222$

3. Final Answer

The reaction at support B is 13.22 kN.