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Plastic moment capacity of frame, 60 kN

Short answer

Three mechanisms analyzed: beam mechanism gives M_p = 24 kN·m, sway mechanism gives M_p = 60 kN·m, combined gives M_p = 48 kN·m. The sway mechanism governs, requiring M_p = 60 kN·m.

Find the plastic moment capacity of the frame shown in figure below during collapse. [Figure: portal frame; A (left base) hinged, F (right base) fixed; left column AB with 3Mp, 20 kN/m UDL horizontally; beam B-C-D 2Mp with 60 kN vertical at C (4 m from B, total 6 m); right column D-E-F with 40 kN horizontal at E, Mp; heights 2 m and 4 m marked.]

Assumptions

  • Portal frame: A (left base) hinged, B (0,4), C (4,4), D (6,4), F (right base, 6,0) fixed, E at 2 m above F. Columns AB and DF are 4 m high; beam BCD is 6 m with the 60 kN at C (4 m from B).
  • Plastic moments: AB 3Mp3M_p, beam BCD 2Mp2M_p, DEF MpM_p. MpM_p is the value to be found.
  • Loads taken as ultimate: 20 kN/m on AB (horizontal, to the right), 60 kN vertical at C and 40 kN horizontal at E (to the right).
  • Degree of indeterminacy =5−3=2= 5 - 3 = 2; critical sections B, C, D, E, F (A is hinged): 5 - 2 = 3 independent mechanisms. The required MpM_p is the largest from all mechanisms.

Mechanisms

Mechanism 1: beam mechanism

No sway; hinges at B (beam end, 2Mp2M_p, weaker than the column 3Mp3M_p), at C (4 m from B, 2Mp2M_p) and at D, where the hinge forms in the weaker column DF (MpM_p). The left part of the beam (4 m, B to C) rotates θ/3\theta/3 and the right part (2 m, C to D) rotates 2θ/32\theta/3, so the hinge rotations are θ/3\theta/3 at B, θ\theta at C and 2θ/32\theta/3 at D.

HingeRotationPlastic momentInternal work
C (under 60 kN)1θ1\theta2Mp2M_p2 Mpθ2\,M_p\theta
D (top of column DF)23θ\frac{2}{3}\thetaMpM_p23 Mpθ\frac{2}{3}\,M_p\theta
B (beam end)13θ\frac{1}{3}\theta2Mp2M_p23 Mpθ\frac{2}{3}\,M_p\theta

Internal work =103 Mp θ= \frac{10}{3}\,M_p\,\theta.

LoadDisplacement along the loadExternal work
60 kN at C43θ\frac{4}{3}\theta80 kN⋅m θ80\,\text{kN·m}\,\theta
40 kN at E0θ0\theta0 kN⋅m θ0\,\text{kN·m}\,\theta
20 kN/m on AB-0 kN⋅m θ0\,\text{kN·m}\,\theta

Equation: Mp×103 θ=80 θM_p\times \frac{10}{3}\,\theta = 80\,\theta (kN·m), so Mp=24M_p = 24 kN·m.

Mechanism 2: sway mechanism

The columns (4 m high) rotate θ\theta about A and F, and the beam translates 4θ4\theta. Hinges at B (beam end, 2Mp2M_p is smaller than the column 3Mp3M_p), D (in the right column, MpM_p) and F.

HingeRotationPlastic momentInternal work
B (beam end)1θ1\theta2Mp2M_p2 Mpθ2\,M_p\theta
D (column)1θ1\thetaMpM_p1 Mpθ1\,M_p\theta
F (fixed base)1θ1\thetaMpM_p1 Mpθ1\,M_p\theta

Internal work =4 Mp θ= 4\,M_p\,\theta.

LoadDisplacement along the loadExternal work
60 kN at C0θ0\theta0 kN⋅m θ0\,\text{kN·m}\,\theta
40 kN at E2θ2\theta80 kN⋅m θ80\,\text{kN·m}\,\theta
20 kN/m on AB-160 kN⋅m θ160\,\text{kN·m}\,\theta

Equation: Mp×4 θ=240 θM_p\times 4\,\theta = 240\,\theta (kN·m), so Mp=60M_p = 60 kN·m.

Mechanism 3: combined mechanism

Combine 1 and 2 so that the hinge at B cancels; hinges under the 60 kN load, at D and at F.

HingeRotationPlastic momentInternal work
C (under 60 kN)1θ1\theta2Mp2M_p2 Mpθ2\,M_p\theta
D (column)1θ1\thetaMpM_p1 Mpθ1\,M_p\theta
F (fixed base)13θ\frac{1}{3}\thetaMpM_p13 Mpθ\frac{1}{3}\,M_p\theta

Internal work =103 Mp θ= \frac{10}{3}\,M_p\,\theta.

LoadDisplacement along the loadExternal work
60 kN at C43θ\frac{4}{3}\theta80 kN⋅m θ80\,\text{kN·m}\,\theta
40 kN at E23θ\frac{2}{3}\theta803 kN⋅m θ\frac{80}{3}\,\text{kN·m}\,\theta
20 kN/m on AB-1603 kN⋅m θ\frac{160}{3}\,\text{kN·m}\,\theta

Equation: Mp×103 θ=160 θM_p\times \frac{10}{3}\,\theta = 160\,\theta (kN·m), so Mp=48M_p = 48 kN·m.

A sway mechanism with hinges at B, under the 40 kN load at E, and at F (value 53.3 kN·m) and other mechanisms were also checked; they all require less.

Required plastic moment

The largest requirement comes from the sway mechanism:

Answer: Mp=60M_p = 60 kN·m (so AB =180= 180, beam =120= 120 and column DF =60= 60 kN·m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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