Collapse load of rectangular portal frame

Assumptions

• Portal frame with fixed bases. Beam B-C-D, span 6 m (3 m + 3 m), $M_p$. Left column AB, 4 m high, $M_p$; right column DE, 6 m high, $2M_p$. The beam is horizontal, so the left base A is 2 m above the right base E.
• Loads: $2P$ horizontal (to the right) at the top-left joint B and $3P$ vertical at mid-span C. $P$ is increased until collapse; the collapse load is the value $P_c$ (the actual loads are $2P_c$ and $3P_c$), given in terms of $M_p$ (kN·m, lengths in m).
• Degree of indeterminacy $= 6 - 3 = 3$, critical sections A, B, C, D, E (5): number of independent mechanisms $5 - 3 = 2$ (beam, sway), plus the combined mechanism.

Mechanisms

Mechanism 1: beam mechanism

No sway. Hinges at B, mid-span C and D, all in the beam ($M_p$ is less than the column $2M_p$ at D). The two beam halves rotate $\theta/2$ each, with $\theta$ under the load.

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

Equation: $P_c\times \frac{9}{2}\,\theta = 2\,M_p\theta$, so $P_c = \frac{4}{9}\,M_p = 0.4444\,M_p$.

Mechanism 2: sway mechanism

The left column (4 m) rotates $\theta$ about A, so the beam moves $4\theta$ to the right; the right column (6 m) rotates $2\theta/3$. Hinges at A, B, D (beam end) and E.

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

Equation: $P_c\times 8\,\theta = 4\,M_p\theta$, so $P_c = \frac{1}{2}\,M_p = 0.5000\,M_p$.

Mechanism 3: combined mechanism

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

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

Equation: $P_c\times \frac{17}{2}\,\theta = 3\,M_p\theta$, so $P_c = \frac{6}{17}\,M_p = 0.3529\,M_p$.

Collapse load

The lowest value is the combined mechanism (beam 0.444 and sway 0.5 are higher), and the LP check over all hinge combinations gave the same:

Answer: $P_c = \dfrac{6}{17}M_p \approx 0.353\,M_p$, i.e. collapse loads $W_p = 2P_c = 0.706\,M_p$ horizontal and $3P_c = 1.059\,M_p$ vertical (kN, with $M_p$ in kN·m).