Castigliano's second theorem: slope at A

Castigliano's second theorem: the rotation at a point equals the partial derivative of the strain energy with respect to a couple applied there:

$$\theta_A=\frac{\partial U}{\partial M_0}=\int \frac{M}{EI}\frac{\partial M}{\partial M_0}\,dx$$

Since no couple acts at A, apply a fictitious clockwise couple $M_0$ at A, write $M$ in terms of $M_0$, differentiate, then put $M_0=0$.

Reactions (with $M_0$)

Taking moments about B (AB = 5 m, BC = 2 m):

$$5R_A + M_0 = 50(2.5) - 10(2) = 105 \;\Rightarrow\; R_A = 21-\frac{M_0}{5}\ \text{kN},\qquad R_B = 39+\frac{M_0}{5}\ \text{kN}$$

(Check: $R_A+R_B=60=50+10$.)

Bending moments (x from A, sagging +)

• Portion A to mid-span ($0\le x\le2.5$): $M=R_Ax+M_0=\left(21-\frac{M_0}{5}\right)x+M_0$, so $\dfrac{\partial M}{\partial M_0}=1-\dfrac{x}{5}$
• Mid-span to B ($2.5\le x\le5$): $M=R_Ax+M_0-50(x-2.5)$, same derivative $1-\dfrac{x}{5}$
• Overhang BC: $M=-10x'$ (x' from C), independent of $M_0$, so it contributes nothing.

Slope at A (put $M_0=0$)

$$EI\,\theta_A=\int_0^{2.5}21x\left(1-\frac{x}{5}\right)dx+\int_{2.5}^{5}\big[21x-50(x-2.5)\big]\left(1-\frac{x}{5}\right)dx$$ $$= 175/4 + 425/24 = 1475/24 = 61.46\ \text{kN m}^2$$ $$\theta_A=\frac{61.46}{EI}\ \text{rad (clockwise, same sense as }M_0)$$

Check: $50\times5^2/16=78.13$ (central load) minus $20\times5/6=16.67$ (hogging moment from the overhang) = 61.46. ✓

Answer: $\theta_A = 61.46/EI$ rad (with EI in kNm²), clockwise.