Collapse load of portal frame with W/2

Assumptions

• Portal frame: A (0,0) hinged, B (0,L), C (L/2, L), D (L, L), E fixed at the base of the right column, which is $2L/3$ high (so E is at height $L/3$ above A). Column AB $M_p$, beam BD $3M_p$ (span L), column DE $M_p$.
• Loads: $W/2$ horizontal (to the right) at B and $W$ vertical at mid-span C. $W_c$ in terms of $M_p$ and $L$.
• Degree of indeterminacy $= 5 - 3 = 2$; critical sections B, C, D, E: $4 - 2 = 2$ independent mechanisms (beam and sway) and their combination.

Mechanisms (tables with $L = 1$)

Mechanism 1: beam mechanism

Lengths in units of $L$ (take $L = 1$ in the tables). No sway; hinges at B (in the weaker column AB), at C and at D (in the column DE, $M_p$). Rotations $\theta/2$ at B and D, $\theta$ at C.

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

Equation: $W_c\times \frac{1}{4}\,\theta = 4\,M_p\theta$, so $W_c = 16\,M_p$ per $L$.

Mechanism 2: sway mechanism

The right column DE (height $2L/3$) rotates $\theta$ about the fixed base E, so the beam moves $2L\theta/3$; the left column AB (height $L$, hinged base) rotates $2\theta/3$. Hinges at B (in AB), D (in DE) and E.

Internal work $= \frac{8}{3}\,M_p\,\theta$.

Equation: $W_c\times \frac{1}{3}\,\theta = \frac{8}{3}\,M_p\theta$, so $W_c = 8\,M_p$ 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 $= 5\,M_p\,\theta$.

Equation: $W_c\times \frac{1}{2}\,\theta = 5\,M_p\theta$, so $W_c = 10\,M_p$ per $L$.

Collapse load

The sway mechanism governs (beam 16 and combined 10 are higher):

Answer: $W_c = 8M_p/L$ (hinges at B in column AB, D in column DE and E).