Collapse load of propped cantilever, 6 m span

Set-up

Propped cantilever of span $L = 10$ m (A fixed, B roller) with a point load $W$ at C, 6 m from the fixed end ($a = 6$ m, $b = 4$ m). The hinges form at A and under the load at C (the two sections of largest moment).

Mechanism and virtual work

Part AC rotates $\theta$ about A, so C deflects $\delta = a\theta = 6\theta$. Part CB rotates $\phi$ about B with $b\phi = \delta$, so $\phi = \dfrac{6\theta}{4} = 1.5\theta$.

• Rotation at A $= \theta$; rotation at C $= \theta + \phi = 2.5\theta$.
• Internal work $= M_p\theta + M_p(2.5\theta) = 3.5M_p\theta$.
• External work $= W\delta = 6W\theta$.

$$6W_c\theta = 3.5M_p\theta \;\Rightarrow\; W_c = \frac{3.5}{6}M_p = 0.5833\,M_p$$

(In general $W_c = \dfrac{M_p(2b + a)}{ab}$.)

Check by equilibrium

With $M_A = -M_p$ and $M_B = 0$: $M_C = \dfrac{Wab}{L} - M_p\dfrac{b}{L} = \dfrac{24W}{10} - 0.4M_p$. Setting $M_C = M_p$ gives $W = \dfrac{1.4M_p}{2.4} = 0.5833M_p$, and the moment nowhere exceeds $M_p$.

Answer: $W_c = 0.583\,M_p$ (kN, with $M_p$ in kN·m).