For this hinged-base portal frame (indeterminacy 1, critical sections B, C, D), how many independent mechanisms are there?

2 (beam and sway); the combined one is derived from them

3 (beam, sway, combined all independent)

In the beam mechanism, why is the external work of the 70 kN horizontal load zero?

Because there is no sway, so B has no horizontal displacement

Because the load is smaller than the vertical load

Because the hinge at B cancels it

Because horizontal loads never do work

The mechanisms give Mp = 157.5, 140 and 227.5 kN·m. Which is taken as the collapse moment and why?

227.5, the combined mechanism, because the largest required Mp governs

140, the smallest, as it fails first

2069 Asar · 8 marks

Collapse moment of portal frame, load factor 1.75

Short answer

Factored loads: 70 kN horizontal, 210 kN vertical. Beam, sway and combined mechanisms require Mp of 157.5, 140 and 227.5 kN·m. The combined mechanism governs; collapse moment is 227.5 kN·m.

Calculate the collapse moment after establishing possible failure mechanisms for the portal frame shown in figure below. Use load factor 1.75. [Figure: portal frame, both bases hinged, height 4 m, span 6 m (3 m + 3 m); 40 kN horizontal at the top-left joint; 120 kN vertical at mid-span; all members $M_P$ . Printed as the alternative (OR) to the continuous beam question above.]

Given and factored loads

Portal frame with both bases hinged, height 4 m, span 6 m, all members $M_p$ . Working loads: 40 kN horizontal at the top-left joint B and 120 kN vertical at mid-span C. With the load factor 1.75, the factored loads are $40\times1.75 = 70$ kN and $120\times1.75 = 210$ kN. The frame must just collapse under these loads, so each mechanism gives a required $M_p$ and the largest one governs.

Degree of indeterminacy $= 4 - 3 = 1$ ; the critical sections B, C and D give $3 - 1 = 2$ independent mechanisms (beam and sway), plus their combination.

Mechanisms

Mechanism 1: beam mechanism

No sway; hinges at B, C and D. Each half-span rotates $\theta/2$ (in opposite senses), so the hinge at C turns $\theta$ and C deflects $3\times\theta/2 = \frac{3}{2}\theta$ .

Hinge	Rotation	Plastic moment	Internal work
C (mid-span)	$1\theta$	$M_p$	$1\,M_p\theta$
B	$\frac{1}{2}\theta$	$M_p$	$\frac{1}{2}\,M_p\theta$
D	$\frac{1}{2}\theta$	$M_p$	$\frac{1}{2}\,M_p\theta$

Internal work $= 2\,M_p\,\theta$ .

Load	Displacement along the load	External work
70 kN horizontal at B	$0\theta$	$0\,\text{kN·m}\,\theta$
210 kN at C	$\frac{3}{2}\theta$	$315\,\text{kN·m}\,\theta$

Equation: $M_p\times 2\,\theta = 315\,\theta$ (kN·m), so $M_p = \frac{315}{2} = 157.50$ kN·m.

Mechanism 2: sway mechanism

The columns (4 m) rotate $\theta$ about the hinged bases and the beam moves $4\theta$ ; hinges at B and D only.

Hinge	Rotation	Plastic moment	Internal work
B	$1\theta$	$M_p$	$1\,M_p\theta$
D	$1\theta$	$M_p$	$1\,M_p\theta$

Internal work $= 2\,M_p\,\theta$ .

Load	Displacement along the load	External work
70 kN horizontal at B	$4\theta$	$280\,\text{kN·m}\,\theta$
210 kN at C	$0\theta$	$0\,\text{kN·m}\,\theta$

Equation: $M_p\times 2\,\theta = 280\,\theta$ (kN·m), so $M_p = 140$ kN·m.

Mechanism 3: combined mechanism

Combine 1 and 2 so that the hinge at B cancels; hinges at C and D.

Hinge	Rotation	Plastic moment	Internal work
C (mid-span)	$1\theta$	$M_p$	$1\,M_p\theta$
D	$1\theta$	$M_p$	$1\,M_p\theta$

Internal work $= 2\,M_p\,\theta$ .

Load	Displacement along the load	External work
70 kN horizontal at B	$2\theta$	$140\,\text{kN·m}\,\theta$
210 kN at C	$\frac{3}{2}\theta$	$315\,\text{kN·m}\,\theta$

Equation: $M_p\times 2\,\theta = 455\,\theta$ (kN·m), so $M_p = \frac{455}{2} = 227.50$ kN·m.

Collapse moment

The combined mechanism needs the largest $M_p$ :

Answer: $M_p = 227.5$ kN·m (the collapse moment of the frame).

✓ Answer checked by Claude Opus 5.5 (high effort)

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Collapse moment of portal frame, load factor 1.75

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