The problem is to determine the reaction at the support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a downward point load of 10 kN at B. There is also a uniformly distributed load of 2 kN/m between B and C. The length between A and B is 4 meters, and the length between B and C is 4 meters.

1. Problem Description

The problem is to determine the reaction at the support B of a continuous beam using Castigliano's theorem. The beam is supported at A, B, and C. There is a downward point load of 10 kN at B. There is also a uniformly distributed load of 2 kN/m between B and C. The length between A and B is 4 meters, and the length between B and C is 4 meters.

2. Solution Steps

Castigliano's second theorem states that the partial derivative of the total strain energy $U$ with respect to a force $R$ is equal to the displacement at the point of application of that force in the direction of the force. Mathematically, $\frac{\partial U}{\partial R} = \Delta$, where $\Delta$ is the displacement.

Since support B is a rigid support, the displacement at B is zero. Thus, $\frac{\partial U}{\partial R_B} = 0$. To solve this, we express the bending moment $M$ as a function of $R_B$, and then compute the strain energy $U = \int \frac{M^2}{2EI} dx$. We then differentiate $U$ with respect to $R_B$ and set it equal to zero.

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

We have the total length of beam is 8 m.

Take section at distance x from A for segment AB and distance y from C for segment BC.

For the segment AB (0 <= x <= 4):

$M_1 = R_A x$

For the segment BC (0 <= y <= 4):

$M_2 = R_C y + 2y(y/2) = R_C y + y^2$

Equilibrium equations:

$\sum F_y = 0 \implies R_A + R_B + R_C - 10 - 2*4 = 0 \implies R_A + R_B + R_C = 18$

$\sum M_A = 0 \implies R_B*4 + R_C*8 - 10*4 - 2*4*6 = 0 \implies 4R_B + 8R_C = 40 + 48 = 88 \implies R_B + 2R_C = 22$

Consider $R_B$ as redundant reaction. Remove support B and find deflection at B.

$R_A + R_C = 18 - R_B$

$R_C = 11 - R_B/2$

$R_A = 18 - R_B - (11 - R_B/2) = 7 - R_B/2$

Then the moments are:

$M_1 = (7-R_B/2)x$ for AB

$M_2 = (11-R_B/2)y + y^2$ for BC

Castigliano's theorem:

$\frac{\partial U}{\partial R_B} = \int_0^4 \frac{M_1}{EI} \frac{\partial M_1}{\partial R_B} dx + \int_0^4 \frac{M_2}{EI} \frac{\partial M_2}{\partial R_B} dy = 0$

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

$\frac{\partial M_2}{\partial R_B} = -y/2$

$\int_0^4 (7-R_B/2)x * (-x/2) dx + \int_0^4 ((11-R_B/2)y+y^2) * (-y/2) dy = 0$

$\int_0^4 (R_B/2 - 7)(x^2/2) dx + \int_0^4 (R_B/2 - 11) (y^2/2) - y^3/2 dy = 0$

$(R_B/2 - 7) [x^3/6]_0^4 + (R_B/2 - 11) [y^3/6]_0^4 - [y^4/8]_0^4 = 0$

$(R_B/2 - 7)(64/6) + (R_B/2 - 11)(64/6) - (256/8) = 0$

$(R_B/2 - 7)(32/3) + (R_B/2 - 11)(32/3) - 32 = 0$

$(32/3)(R_B - 18) = 32$

$R_B - 18 = 3$

$R_B = 21$

3. Final Answer

21 kN