Castigliano: propped cantilever, moment and rotation

The beam AB (span 10 m) is fixed at A and has a roller at B. Loads: 30 kN/m over the whole span and 50 kN at the centre. EI is constant. Take the roller reaction $R_B$ as the redundant.

        50 kN
          v      30 kN/m
 vvvvvvvvvvvvvvvvvvvvvvvvvv
 ||==========+===========o
 A (fixed)  5 m          B (roller, R_B)
 |<------------ 10 m ---->|

Bending moment (x measured from B)

$$M = R_B\,x - 15x^2 - 50\langle x-5\rangle, \qquad \frac{\partial M}{\partial R_B} = x$$

Redundant from Castigliano

The deflection at B is zero:

$$\frac{\partial U}{\partial R_B} = \frac{1}{EI}\int_0^{10} M\,x\,dx = 0$$ $$\int_0^{10}\left(R_B x^2 - 15x^3\right)dx - 50\int_5^{10}(x-5)x\,dx = 0$$ $$\frac{1000}{3}R_B - 37500 - 5208.33 = 0 \ \Rightarrow\ R_B = 128.125\ \text{kN}$$

(Check: UDL alone gives $3wL/8 = 112.5$ kN and the central load gives $5P/16 = 15.625$ kN; their sum is 128.125 kN.)

Moment at the fixed support

$$V_A = 350 - 128.125 = 221.875\ \text{kN}$$ $$M_A = R_B(10) - 30\cdot\frac{10^2}{2} - 50(5) = 128.125(10) - 1500 - 250 = -468.75\ \text{kNm}$$

The fixed-end moment is 468.75 kNm (hogging).

Rotation at the roller B

Apply a dummy couple $M'$ at B. Then $\partial M/\partial M' = 1$ at the sections (the extra term caused by the change of $R_B$ integrates to zero because $\int M\,x\,dx = 0$), so

$$\theta_B = \frac{\partial U}{\partial M'}\bigg|_{M'=0} = \frac{1}{EI}\int_0^{10} M\,dx$$ $$\int_0^{10} M\,dx = R_B\frac{10^2}{2} - 15\frac{10^3}{3} - 50\frac{5^2}{2} = 128.125(50) - 5000 - 625 = 781.25$$ $$\theta_B = \frac{781.25}{EI}\ \text{rad (anticlockwise)}$$

Answer: fixed-end moment $M_A = 468.75$ kNm (hogging); rotation at the roller $\theta_B = \dfrac{781.25}{EI}$ rad.