Strain energy deflection at A, spring support

Given and assumptions

Beam ABC, $EI$ constant: overhang $AB = 1.5$ m, span $BC = 3$ m. Load $P = 20$ kN (down) at the free end A, hinge at B, and C supported on a spring of stiffness $k = EI/9$ (kN/m, with $EI$ in kNm$^2$). The deflection of A is found by Castigliano's theorem, $\delta_A = \partial U/\partial P$, using the total strain energy of beam bending plus the spring.

Reactions

Take moments about B:

$$R_C\times3 = P\times1.5 \;\Rightarrow\; R_C = 0.5P = 10\ \text{kN (downward, spring in tension)}$$ $$R_B = P + 0.5P = 1.5P = 30\ \text{kN (upward)}$$

Bending moments (x from A, y from C)

• $AB$: $M = Px$, $0 \le x \le 1.5$
• $BC$: $M = R_C\,y = 0.5Py$, $0 \le y \le 3$

Strain energy

$$U = \int_0^{1.5}\frac{(Px)^2}{2EI}dx + \int_0^{3}\frac{(0.5Py)^2}{2EI}dy + \frac{R_C^2}{2k}$$ $$U = \frac{P^2}{2EI}\left[\frac{1.5^3}{3} + 0.25\times\frac{3^3}{3}\right] + \frac{(0.5P)^2\times 9}{2EI} = \frac{P^2}{2EI}\left[1.125 + 2.25\right] + \frac{1.125P^2}{EI}$$ $$U = \frac{1.6875P^2 + 1.125P^2}{EI} = \frac{2.8125\,P^2}{EI}$$

Deflection at A

$$\delta_A = \frac{\partial U}{\partial P} = \frac{5.625\,P}{EI} = \frac{5.625\times20}{EI}$$

Parts: beam bending $= 67.5/EI$; spring (rotation of the beam about B caused by the spring's extension) $= 45/EI$.

Answer: $\delta_A = \dfrac{112.5}{EI}$ downward (in metres when $EI$ is in kNm$^2$).