Castigliano: reaction at B, 20 kN/m beam

Reading of the figure: a continuous beam ABC with $AB = 6$ m and $BC = 10$ m carrying 20 kN/m over the whole length, A hinged, B and C on rollers (a beam with only A and B supported would be determinate, so C is taken as a roller). EI is constant. Take $R_B$ as the redundant.

   20 kN/m
 vvvvvvvvvvvvvvvvvvvvvvvvvv
 A o--------o B-----------o C
 |<-- 6 -->|<---- 10 ---->|

Reactions in terms of $R_B$

Moments about C: $R_A(16) + R_B(10) = 20(16)(8)$

$$R_A = 160 - 0.625\,R_B$$

Bending moments (x from A)

$$\begin{aligned} 0 \le x \le 6:\quad M &= R_A x - 10x^2, & \frac{\partial M}{\partial R_B} &= -0.625x \\ 6 \le x \le 16:\quad M &= R_A x + R_B(x-6) - 10x^2, & \frac{\partial M}{\partial R_B} &= 0.375x - 6 \end{aligned}$$

Castigliano's theorem

The support B does not settle, so $\partial U/\partial R_B = 0$:

$$\int_0^6 M\,\frac{\partial M}{\partial R_B}\,dx + \int_6^{16} M\,\frac{\partial M}{\partial R_B}\,dx = 0$$

Substituting $R_A$ and integrating gives

$$75.000\,R_B - 15800.00 = 0$$ $$R_B = \frac{15800.00}{75.000} = 210.67\ \text{kN}$$

Other reactions

$$R_A = 160 - 0.625(210.67) = 28.33\ \text{kN}, \qquad R_C = 320 - 210.67 - 28.33 = 81.00\ \text{kN}$$

Check: $M_B = 28.33(6) - 20(6)(3) = -190$ kNm, which agrees with the three moment equation $2M_B(16) = -\tfrac{20}{4}(6^3 + 10^3)$.

Answer: $R_B = 210.67$ kN (upward).