Collapse load P: portal frame with UDL 2P/l

Assumptions

• Portal frame: A and F hinged, columns AB and EF of height $l$, beam BE of span $l$; the vertical loads $P$ act at C and D (the third points). A UDL of $2P/l$ acts horizontally to the right on column AB. All members have the same plastic moment $M_p$. Take $l = 1$ in the tables.
• Hinged bases: degree of indeterminacy $= 4 - 3 = 1$.

Mechanisms

Mechanism 1: beam mechanism

Taking $l = 1$. No sway; hinges at B, C and E. Part BC ($l/3$) rotates $2\theta/3$ and CE ($2l/3$) rotates $\theta/3$, so the hinge at C turns $\theta$.

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

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

Mechanism 2: sway mechanism (hinges at B and E)

The columns (height $l$) rotate $\theta$ about the hinged bases and the beam moves $l\theta$. The hinges are at B and E, at the beam ends.

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

Equation: $P\times 1\,\theta = 2\,M_p\theta$, so $P = 2\,M_p/l$.

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 $P = 2\,M_p/l$.

Mechanism 4: sway with the hinge inside the loaded column. With a UDL on AB, the maximum moment in the column is not at B but within the column, at height $y$ above A. Let the column part AK rotate $\theta$ about A; the upper part of the column and the beam translate $y\theta$, and the right column rotates $\phi = y\theta/l = y\theta$ (with $l = 1$). Hinges: K ($\theta$) and E ($\phi$).

$$\begin{aligned} \text{Internal} &= M_p\theta(1 + y) \\ \text{External} &= \tfrac{2P}{l}\left[\tfrac{y^2\theta}{2} + (1-y)\,y\theta\right] = 2P\theta\,y\left(1 - \tfrac y2\right) \end{aligned}$$ $$P = \frac{M_p(1+y)}{2y - y^2}$$

Minimising: $y^2 + 2y - 2 = 0$, so $y = \sqrt3 - 1 = 0.732\,l$, and

$$P = \frac{\sqrt3\,M_p}{4\sqrt3 - 6} = \left(1 + \frac{\sqrt3}{2}\right)\frac{M_p}{l} = 1.866\,\frac{M_p}{l}$$

Collapse load

This is lower than the sway (2.0), combined (2.0) and beam (6) values, and a check of the other hinge positions (for example, adding a hinge at C to mechanism 4) gives nothing lower:

Answer: $P_c = \left(1 + \dfrac{\sqrt3}{2}\right)\dfrac{M_p}{l} = 1.866\,\dfrac{M_p}{l}$.