Collapse load P of portal frame

Assumptions

• Portal frame ABCDEF: A and F hinged, columns AB and EF of height $L$, beam BE of span $L$ with C and D at the third points. All members have plastic moment $M_p$.
• Loads: $P$ horizontal (to the right) at B, and $P$ vertical at C and at D.
• Both bases are hinged, so the degree of indeterminacy is $4 - 3 = 1$; critical sections B, C, D, E: $4 - 1 = 3$ independent mechanisms (two beam mechanisms, one sway). The hinge position at a joint is in the beam end (same $M_p$).

Mechanisms (tables with $L = 1$)

Mechanism 1: beam mechanism

Taking $L = 1$. No sway; hinges at B, C and E. The beam part BC (length $L/3$) rotates $2\theta/3$ and CE ($2L/3$) rotates $\theta/3$ the other way, so C has $\theta$ and C deflects $2\theta/9$.

Internal work $= 2\,M_p\,\theta$.

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

Mechanism 2: sway mechanism

Both columns rotate $\theta$ about the hinged bases; the beam translates by $L\theta$. Hinges at B and E only (the beam stays straight).

Internal work $= 2\,M_p\,\theta$.

Equation: $P\times 1\,\theta = 2\,M_p\theta$, so $P = 2\,M_p$ per $L$.

Mechanism 3: combined mechanism

Combine 1 and 2 so that the hinge at B cancels; hinges at C and E.

Internal work $= 2\,M_p\,\theta$.

Equation: $P\times 1\,\theta = 2\,M_p\theta$, so $P = 2\,M_p$ per $L$.

The beam mechanism with the hinge under the load at D, and the combined mechanism with the hinge under D (value 3), are higher.

Collapse load

The sway and the combined mechanism give the same minimum, and the beam mechanism is much higher, so

Answer: $P_c = \dfrac{2M_p}{L}$.