Collapse load of frame with W and W/2 loads

Assumptions

• Frame ABCDE: A (0,0) hinged, B (0,6), C (4,6), D (8,6), E (8,3) fixed. Column AB $2M_p$ (6 m high), beam BCD $1.5M_p$ (BC = CD = 4 m), column DE $M_p$ (3 m high; the right base is 3 m above A's level, so the columns have different heights).
• Loads: W at C, and W/2 per metre on column AB acting horizontally to the right. $W_c$ is expressed in terms of $M_p$ (in kN·m, lengths in m).
• Method: virtual work, $\sum M_p\theta = \sum P\delta$.

Mechanisms

Mechanism 1: beam mechanism

Hinges at B, C and D (the weaker section at D is the column DE, $M_p$). Beam BC rotates $\theta/2$ at B, the two beam halves rotate $\theta/2$ each in opposite senses, so C deflects $4\times\theta/2 = 2\theta$.

Internal work $= \frac{11}{4}\,M_p\,\theta$.

Equation: $W_c\times 2\,\theta = \frac{11}{4}\,M_p\theta$, so $W_c = \frac{11}{8}\,M_p = 1.3750\,M_p$.

Mechanism 2: sway mechanism

The beam BD translates horizontally by $3\theta$ (column DE, 3 m high, rotates $\theta$ about E); column AB rotates $\theta/2$ about A. Hinges at B (beam side, $1.5M_p$), D (in DE) and E. W does no work.

Internal work $= \frac{11}{4}\,M_p\,\theta$.

Equation: $W_c\times \frac{9}{2}\,\theta = \frac{11}{4}\,M_p\theta$, so $W_c = \frac{11}{18}\,M_p = 0.6111\,M_p$.

Mechanism 3: combined mechanism

Combination of 1 and 2 removing the hinge at B: hinges at C, D and E.

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

Equation: $W_c\times \frac{13}{2}\,\theta = 4\,M_p\theta$, so $W_c = \frac{8}{13}\,M_p = 0.6154\,M_p$.

Refinement: sway with the hinge inside the column AB. For a UDL the plastic hinge in the loaded column does not form at B but at the section of maximum moment, at a height $y$ above A. Let the beam translate by $\Delta = 3\phi$ ($\phi$ = rotation of DE). The part AK rotates $\theta$ about A, so $y\theta = \Delta$, and the part KB translates with the beam. Hinges: K (rotation $\theta$, $2M_p$), D ($\phi$, $M_p$), E ($\phi$, $M_p$), where $\phi = y\theta/3$.

$$\begin{aligned} \text{Internal} &= M_p\theta\,(2 + \tfrac{2y}{3}) \\ \text{External} &= \tfrac W2\left[\tfrac{y^2\theta}{2} + (6-y)\,y\theta\right] = \tfrac W2\,\theta\,y\left(6 - \tfrac y2\right) \end{aligned}$$ $$W = \frac{M_p\,(4 + \tfrac{4y}{3})}{y\,(6 - \tfrac y2)}$$

Setting $dW/dy = 0$ gives $y^2 + 6y - 36 = 0$, so $y = 3(\sqrt5 - 1) = 3.708$ m, and

$$W_c = 0.5818\,M_p$$

(At $y = 6$, a hinge at the top of the column, this expression gives 0.667; mechanism 2, with the hinge in the beam, gives 0.611.)

Collapse load

The smallest value is the sway mechanism with the hinge in the column AB, and it is lower than the beam (1.375) and combined (0.615) mechanisms:

Answer: $W_c \approx 0.582\,M_p$ (in kN when $M_p$ is in kN·m and lengths are in m).