Collapse load of portal frame, height 2L

Assumptions

• Portal frame: span $2L$ ($L + L$), height $2L$. A hinged, E fixed (right base). Columns $M_p$, beam $2M_p$.
• Loads: $3W$ horizontal (to the right) at the top-left joint B and $2W$ vertical at mid-span C. Take $L = 1$ in the tables.
• Reactions $2 + 3 = 5$, degree of indeterminacy $= 2$; critical sections B, C, D, E: two independent mechanisms (beam, sway) plus the combined mechanism.

Mechanisms

Mechanism 1: beam mechanism

Taking $L = 1$. No sway; hinges at B (in column AB, $M_p$), at mid-span C ($2M_p$) and at D (in column DE, $M_p$). Rotations $\theta/2$ at B and D, $\theta$ at C.

Internal work $= 3\,M_p\,\theta$.

Equation: $W_c\times 1\,\theta = 3\,M_p\theta$, so $W_c = 3\,M_p$ per $L$.

Mechanism 2: sway mechanism

Both columns are $2L$ high. They rotate $\theta$ (A is hinged), the beam moves $2L\theta$ to the right. Hinges at B (in AB), D (in DE) and E.

Internal work $= 3\,M_p\,\theta$.

Equation: $W_c\times 6\,\theta = 3\,M_p\theta$, so $W_c = \frac{1}{2}\,M_p = 0.5000\,M_p$ per $L$.

Mechanism 3: combined mechanism

Combine 1 and 2 so that the hinge at B cancels. Hinges at C, D and E.

Internal work $= \frac{7}{2}\,M_p\,\theta$.

Equation: $W_c\times 4\,\theta = \frac{7}{2}\,M_p\theta$, so $W_c = \frac{7}{8}\,M_p = 0.8750\,M_p$ per $L$.

Collapse load

The sway mechanism governs:

Answer: $W_c = \dfrac{M_p}{2L} = 0.5\,\dfrac{M_p}{L}$.