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Collapse load of portal frame with 1.5W

Short answer

Using virtual work, the beam, sway and combined mechanisms give 1.833Mp, 1.111Mp and Mp; the lowest is the combined mechanism with hinges at A, C and D. Therefore Wc = Mp, with Mp in kN·m and lengths in m.

Determine collapse load in the portal frame shown in figure below. [Figure: portal frame; horizontal 1.5W at the top-left joint; vertical W at mid-span of the beam (1.5Mp, spans 3 m + 3 m); left column 2Mp, 3 m + 2 m high, fixed base; right column Mp hinged at its base.]

Assumptions

  • Portal frame: A (0,0) fixed, B (0,3), C (3,3) at mid-span, D (6,3), E hinged at the base of the right column. The figure gives "3 m + 2 m": the left column AB is taken as 3 m high and the right column DE as 2 m high (so E is 1 m above the level of A). Beam BD is 6 m (1.5Mp1.5M_p), AB is 2Mp2M_p and DE is MpM_p.
  • Loads: 1.5W horizontal (to the right) at B and W vertical at C. WcW_c is found in terms of MpM_p (kN·m, lengths in m).
  • Method: virtual work, ∑Mpθ=∑Pδ\sum M_p\theta = \sum P\delta. Reactions: 3 at A and 2 at E, degree of indeterminacy =5−3=2= 5 - 3 = 2; critical sections A, B, C, D give 4 - 2 = 2 independent mechanisms (beam and sway), plus the combination.

Mechanisms

Mechanism 1: beam mechanism

No sway. Hinges at B (beam end, 1.5Mp1.5M_p is smaller than the column 2Mp2M_p), under W (mid-span) and at D (in the weaker column DE, MpM_p). The rotations are θ/2\theta/2 at B and D and θ\theta at C.

HingeRotationPlastic momentInternal work
C (under W)1θ1\theta32Mp\frac{3}{2}M_p32 Mpθ\frac{3}{2}\,M_p\theta
B (beam end)12θ\frac{1}{2}\theta32Mp\frac{3}{2}M_p34 Mpθ\frac{3}{4}\,M_p\theta
D (column DE)12θ\frac{1}{2}\thetaMpM_p12 Mpθ\frac{1}{2}\,M_p\theta

Internal work =114 Mp θ= \frac{11}{4}\,M_p\,\theta.

LoadDisplacement along the loadExternal work
1.5W horizontal at B0θ0\theta0W θ0W\,\theta
W at C32θ\frac{3}{2}\theta32W θ\frac{3}{2}W\,\theta

Equation: Wc×32 θ=114 MpθW_c\times \frac{3}{2}\,\theta = \frac{11}{4}\,M_p\theta, so Wc=116 Mp=1.8333 MpW_c = \frac{11}{6}\,M_p = 1.8333\,M_p.

Mechanism 2: sway mechanism

Left column AB (3 m) rotates θ\theta about A; the beam moves 3θ3\theta to the right; the right column DE (2 m high) rotates 3θ/23\theta/2 about E. Hinges at A, B (beam end) and D (in DE).

HingeRotationPlastic momentInternal work
D (column DE)32θ\frac{3}{2}\thetaMpM_p32 Mpθ\frac{3}{2}\,M_p\theta
A (fixed base)1θ1\theta2Mp2M_p2 Mpθ2\,M_p\theta
B (beam end)1θ1\theta32Mp\frac{3}{2}M_p32 Mpθ\frac{3}{2}\,M_p\theta

Internal work =5 Mp θ= 5\,M_p\,\theta.

LoadDisplacement along the loadExternal work
1.5W horizontal at B3θ3\theta92W θ\frac{9}{2}W\,\theta
W at C0θ0\theta0W θ0W\,\theta

Equation: Wc×92 θ=5 MpθW_c\times \frac{9}{2}\,\theta = 5\,M_p\theta, so Wc=109 Mp=1.1111 MpW_c = \frac{10}{9}\,M_p = 1.1111\,M_p.

Mechanism 3: combined mechanism

Mechanisms 1 and 2 are combined so that the hinge at B cancels. Hinges at A, C and D.

HingeRotationPlastic momentInternal work
D (column DE)54θ\frac{5}{4}\thetaMpM_p54 Mpθ\frac{5}{4}\,M_p\theta
C (under W)1θ1\theta32Mp\frac{3}{2}M_p32 Mpθ\frac{3}{2}\,M_p\theta
A (fixed base)12θ\frac{1}{2}\theta2Mp2M_p1 Mpθ1\,M_p\theta

Internal work =154 Mp θ= \frac{15}{4}\,M_p\,\theta.

LoadDisplacement along the loadExternal work
1.5W horizontal at B32θ\frac{3}{2}\theta94W θ\frac{9}{4}W\,\theta
W at C32θ\frac{3}{2}\theta32W θ\frac{3}{2}W\,\theta

Equation: Wc×154 θ=154 MpθW_c\times \frac{15}{4}\,\theta = \frac{15}{4}\,M_p\theta, so Wc=1 MpW_c = 1\,M_p.

Collapse load

The combined mechanism is the lowest (the sway mechanism gives 1.111Mp1.111M_p and the beam mechanism 1.833Mp1.833M_p), and a search over all hinge positions confirmed it:

Answer: Wc=MpW_c = M_p (hinges at A, C and D; the loads at collapse are 1.5MpM_p horizontal and MpM_p vertical, with MpM_p in kN·m and lengths in m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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