Strain energy method for frame displacement

Given data and assumptions

Frame fixed at joint 0, all members of solid circular section $d = 100$ mm, $E = 2\times10^5$ N/mm$^2$ ($= 2\times10^8$ kN/m$^2$). Taking the figure as a cantilever frame with horizontal member 0-1 (3 m), vertical member 1-2 (1 m) and horizontal member 2-3 (2 m), the 500 kN load $P$ acts downward at joint 3. Shear energy is neglected (small for slender members; $G$ is not given). Vertical displacement of joint 3 is found by Castigliano's theorem: $\delta_3 = \partial U/\partial P$.

$$I = \frac{\pi d^4}{64} = 4.9087e-06\ \text{m}^4,\quad EI = 981.7\ \text{kNm}^2,\quad A = 0.00785\ \text{m}^2,\quad AE = 1.5708e+06\ \text{kN}$$

Internal forces (s measured from joint 3)

Strain energy and displacement

$$U = \int\frac{M^2}{2EI}dx + \int\frac{N^2}{2AE}dx$$ $$\delta_3 = \frac{\partial U}{\partial P} = \frac{P}{EI}\left[\int_0^2 s^2ds + (2)^2(1) + \int_0^3 (2+s)^2 ds\right] + \frac{PL_{2-1}}{AE}$$ $$\int_0^2 s^2ds = 2.667,\qquad 2^2\times1 = 4,\qquad \int_0^3(2+s)^2ds = \frac{5^3-2^3}{3} = 39$$ $$\delta_{bending} = \frac{500\times(2.667+4+39)}{981.7} = \frac{500\times45.667}{981.7} = 23.258\ \text{m}$$ $$\delta_{axial} = \frac{500\times1}{1.5708e+06} = 0.318\ \text{mm}$$

Answer: vertical displacement of the loaded joint $\approx 23.26$ m downward (bending governs; axial part is only 0.32 mm). The very large value shows that the given load is extremely heavy for a 100 mm rod; the small-deflection theory is used as asked.