Collapse load of propped cantilever, midspan load

Set-up

Propped cantilever AC of span $L$: A fixed, C a roller, uniform $M_p$, point load $P$ at mid-span B. The beam is indeterminate to the first degree, so $1 + 1 = 2$ hinges cause collapse. The largest moments occur at the fixed end A and under the load B, so the plastic hinges form at A and B.

Mechanism and virtual work

Part AB rotates $\theta$ about A. Part BC rotates about C so that B has the same deflection $\delta = \dfrac{L}{2}\theta$. Since BC has the same length, it rotates $\theta$ the other way.

• Rotation at A $= \theta$; rotation at B $= \theta + \theta = 2\theta$.
• Internal work $= M_p\theta + M_p(2\theta) = 3M_p\theta$.
• External work $= P\delta = P\,\dfrac{L}{2}\theta$.

$$P_c\,\frac{L}{2}\theta = 3M_p\theta \;\Rightarrow\; P_c = \frac{6M_p}{L}$$

Check by equilibrium

With $M_A = -M_p$ (hogging) and $M_C = 0$, the moment under the load is $M_B = \dfrac{PL}{4} + \dfrac{M_A + M_C}{2} = \dfrac{PL}{4} - \dfrac{M_p}{2}$. Setting $M_B = M_p$ gives $P = \dfrac{6M_p}{L}$. The moment diagram is nowhere above $M_p$, so the static and kinematic values agree and this is the true collapse load.

Answer: $P_c = 6M_p/L$.