Collapse value of P: portal frame with horizontal P/2

Assumptions

• Portal frame with fixed bases A and E, columns of height $\ell/2$ and a beam of span $\ell$ ($\ell/2 + \ell/2$) with mid-point C. All members have the plastic moment $M_P$.
• Loads: $P$ vertical at mid-span C and $P/2$ horizontal (to the right) at the top-left joint B. Take $\ell = 1$ in the tables.
• Fixed bases: 6 reactions, so the degree of indeterminacy $= 6 - 3 = 3$. Critical sections: A, B, C, D, E (5), so there are $5 - 3 = 2$ independent mechanisms, beam and sway; their combination is the third mechanism.

Mechanisms

Mechanism 1: beam mechanism

Taking $\ell = 1$. No sway; hinges at B, C and D. The hinge at C rotates $\theta$ and the hinges at B and D rotate $\theta/2$ each (the two halves of the beam turn $\theta/2$ in opposite senses).

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

Equation: $P\times \frac{1}{4}\,\theta = 2\,M_p\theta$, so $P = 8\,M_p/\ell$.

Mechanism 2: sway mechanism

The columns (height $\ell/2$) rotate $\theta$ and the beam moves $\ell\theta/2$. Hinges at A, B, D and E, each rotating $\theta$.

Internal work $= 4\,M_p\,\theta$.

Equation: $P\times \frac{1}{4}\,\theta = 4\,M_p\theta$, so $P = 16\,M_p/\ell$.

Mechanism 3: combined mechanism

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

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

Equation: $P\times \frac{3}{8}\,\theta = 3\,M_p\theta$, so $P = 8\,M_p/\ell$.

Collapse load

The beam and combined mechanisms give the same value, which is smaller than the sway value (16):

Answer: $P_c = 8M_P/\ell$ (and the horizontal load at collapse is $P_c/2 = 4M_P/\ell$).