Castigliano: reactions of two-span beam

Method: the two-span beam has 3 vertical reactions and 2 equilibrium equations, so it is indeterminate to the 1st degree. Take $R_B$ (upward) as the redundant. Since B does not settle, Castigliano's theorem gives $\partial U/\partial R_B=0$.

Reactions in terms of $R_B$

$\sum M_C=0$ with the UDL of total load $wL$ acting at $L/2$ from A (that is $3L/2$ from C):

$$R_A(2L)+R_B(L)=wL\cdot\frac{3L}{2}\ \Rightarrow\ R_A=\frac{3wL}{4}-\frac{R_B}{2}$$

Bending moments (x from A)

• $0\le x\le L$: $M=R_Ax-\dfrac{wx^2}{2}$, $\;\dfrac{\partial M}{\partial R_B}=-\dfrac{x}{2}$
• $L\le x\le2L$: $M=R_Ax-wL\left(x-\dfrac{L}{2}\right)+R_B(x-L)$, $\;\dfrac{\partial M}{\partial R_B}=-\dfrac{x}{2}+(x-L)=\dfrac{x}{2}-L$

Condition $\partial U/\partial R_B=0$

$$\int_0^L M_1\left(-\frac{x}{2}\right)dx+\int_L^{2L}M_2\left(\frac{x}{2}-L\right)dx=0$$

Evaluating the integrals (with $R_A$ substituted) gives

$$\frac{L^{3} \left(- 5 L w + 8 R_{B}\right)}{48}=0\ \Rightarrow\ R_B=\frac{5wL}{8}$$

Other reactions

$$R_A=\frac{3wL}{4}-\frac{5wL}{16}=\frac{7wL}{16}\ \uparrow,\qquad R_C=wL-R_A-R_B=-\frac{wL}{16}$$

$R_C$ is negative, so the support at C pulls the beam downward with $wL/16$.

Check: $R_A+R_B+R_C=\frac{7+10-1}{16}wL=wL$ ✓.

Answer: $R_A=\dfrac{7wL}{16}$ (up), $R_B=\dfrac{5wL}{8}$ (up), $R_C=\dfrac{wL}{16}$ (down).