Collapse load of frame with UDL on column

Assumptions

• Portal frame: A (0,0) hinged, B (0,5), C (6,5), D (6,0) fixed. Column AB has plastic moment $M_p$, beam BC and column CD have $2M_p$.
• The loads (two 20 kN point loads at 2 m and 4 m from B, and 10 kN/m on column CD) are applied proportionally with a load factor $\lambda$; the 10 kN/m acts horizontally towards the left on CD. The collapse load factor $\lambda_c$ is found in terms of $M_p$ (in kN·m).
• Method: kinematic (upper-bound) theorem with virtual work, $\sum M_p\theta = \lambda \sum P\delta$.

Possible hinge positions

B (in the weaker member AB, $M_p$), under each 20 kN load, C and D. Degree of indeterminacy = 5 - 3 = 2 (A hinged gives 2 reactions, D fixed gives 3), so with 5 critical sections the number of independent mechanisms is 5 - 2 = 3: two beam mechanisms (one with the hinge under each load) and the sway mechanism. Combining them gives further mechanisms.

Mechanism 1: beam mechanism

Joints B and C do not move. The beam BC forms hinges under the second load P2 (4 m from B) and at C; at B the hinge forms at the top of the weaker column AB ($M_p$). With rotation $\theta$ at P2, the left portion (4 m) rotates $\theta/3$ and the right portion (2 m) rotates $2\theta/3$ (rigid portions); the deflection under the second load is $4\times\theta/3$.

Internal work $= \frac{11}{3}\,M_p\,\theta$.

Equation: $\lambda\times40\,\theta = \frac{11}{3}\,M_p\theta$, so $\lambda = \frac{11}{120}M_p = 0.0917\,M_p$.

With the hinge under the first load instead (2 m from B), the portions rotate $2\theta/3$ and $\theta/3$: internal work $= 2M_p\theta + M_p\cdot\frac{2}{3}\theta\ (\text{B}) + 2M_p\cdot\frac{1}{3}\theta\ (\text{C}) = \frac{10}{3}M_p\theta$, and the external work is again $40\theta$, so $\lambda = \frac{1}{12}M_p = 0.0833\,M_p$ (the lower beam value).

Mechanism 2: sway mechanism

Columns rotate through $\theta$ about A and D, and the beam translates horizontally by $5\theta$. Hinges form at B (the weaker section is AB, $M_p$), C and D. The vertical loads do no work. (The column CD is moved to the left by the UDL.)

Internal work $= 5\,M_p\,\theta$.

Equation: $\lambda\times125\,\theta = 5\,M_p\theta$, so $\lambda = \frac{1}{25}M_p = 0.0400\,M_p$.

Mechanism 3: combined mechanism

Combine mechanisms 1 and 2 so that the hinge at C cancels. Both columns rotate $2\theta/3$ (sway to the left), the beam portion from B to the hinge rotates $\theta/3$ in the opposite sense (hinge at B $= 2\theta/3 + \theta/3 = \theta$), and the portion from the hinge to C rotates $2\theta/3$ with the column (hinge at the load $= \theta$, no hinge at C).

Internal work $= \frac{13}{3}\,M_p\,\theta$.

Equation: $\lambda\times\frac{370}{3}\,\theta = \frac{13}{3}\,M_p\theta$, so $\lambda = \frac{13}{370}M_p = 0.03514\,M_p$.

(The same combination with the hinge under the first load gives $\lambda = \frac{11}{245}M_p = 0.0449\,M_p$, which is higher.)

Collapse load

The smallest value governs (combined mechanism, lower than both the beam value 0.0833 and the sway value 0.04):

$$\lambda_c = \frac{13}{370}M_p = 0.0351\,M_p$$

So the collapse loads are: point loads $20\lambda_c = 0.703\,M_p$ kN each and UDL $10\lambda_c = 0.351\,M_p$ kN/m (with $M_p$ in kN·m). A check with a linear-programming search over all possible hinge sets gave the same value $0.03514\,M_p$.

Answer: collapse load factor $\lambda_c = 13M_p/370 \approx 0.0351\,M_p$ (combined mechanism with hinges at B, under the 20 kN load at 4 m, and at D).