At a hinge at joint B, which plastic moment is used for the internal work in the beam mechanism?

The weaker member meeting there: 2Mp of column BA

In the sway mechanism, why does the hinge at C in column CD rotate 3theta?

Because CD is 1 m high while AB is 3 m, so rotating about hinge D it must turn 3theta for the same beam translation 3theta

In the combined mechanism, what is Mp and what is the internal work?

Internal 6Mp theta, external 35 theta, Mp = 5.83 kN·m

Internal 4.5Mp theta, Mp = 4.44

Internal 6Mp theta, external 20 theta, Mp = 3.33

Mechanisms give 4.44, 4.29 and 5.83 kN·m. Which governs?

5.83, the largest, from the combined mechanism

2066 Jestha · 8 marks

Plastic moment capacity Mp of frame

Short answer

The combined mechanism governs: internal work 6Mp theta equals external work 35 theta, giving Mp = 35/6 = 5.83 kNm. Hence member capacities are AB 11.67, BC 17.5, CD 5.83 kNm.

Determine the value of plastic moment capacity Mp for the frame loaded as follows. [Figure: frame; beam B-C (3Mp) of 4 m (2 m from B to the 20 kN load) with a 10 kN load/moment at the right end; left column BA (2Mp) with A fixed; column from the beam to a hinged support D (Mp); heights 3 m and 1 m marked.]

Assumptions (the figure is not fully clear)

Frame: A (0,0) fixed, B (0,3), C (4,3), D hinged 1 m below C. Beam BC is 4 m with the 20 kN at mid-span J (2 m from B); the 10 kN at the right end C is taken as a horizontal force to the right.
Plastic moments: column AB $2M_p$ , beam BC $3M_p$ , column CD $M_p$ . The loads are ultimate loads and the required $M_p$ is the largest value given by the mechanisms.
Degree of indeterminacy $= 5 - 3 = 2$ ; critical sections A, B, J, C: $4 - 2 = 2$ independent mechanisms (beam and sway) plus the combination.

Mechanisms

Mechanism 1: beam mechanism

No sway; hinges at B (in the column BA, $2M_p$ is weaker than the beam $3M_p$ ), J (2 m from B, mid-span) and C (in the column CD, $M_p$ ). B and C rotate $\theta/2$ and J rotates $\theta$ .

Hinge	Rotation	Plastic moment	Internal work
J (under 20 kN)	$1\theta$	$3M_p$	$3\,M_p\theta$
B (column BA)	$\frac{1}{2}\theta$	$2M_p$	$1\,M_p\theta$
C (column CD)	$\frac{1}{2}\theta$	$M_p$	$\frac{1}{2}\,M_p\theta$

Internal work $= \frac{9}{2}\,M_p\,\theta$ .

Load	Displacement along the load	External work
20 kN at J	$1\theta$	$20\,\text{kN·m}\,\theta$
10 kN at C	$0\theta$	$0\,\text{kN·m}\,\theta$

Equation: $M_p\times \frac{9}{2}\,\theta = 20\,\theta$ (kN·m), so $M_p = \frac{40}{9} = 4.44$ kN·m.

Mechanism 2: sway mechanism

Column AB (3 m) rotates $\theta$ about the fixed base A and the beam moves $3\theta$ ; column CD (1 m high) rotates $3\theta$ about the hinge D. Hinges at A, B (in BA) and C (in CD).

Hinge	Rotation	Plastic moment	Internal work
C (column CD)	$3\theta$	$M_p$	$3\,M_p\theta$
A (fixed base)	$1\theta$	$2M_p$	$2\,M_p\theta$
B (column BA)	$1\theta$	$2M_p$	$2\,M_p\theta$

Internal work $= 7\,M_p\,\theta$ .

Load	Displacement along the load	External work
20 kN at J	$0\theta$	$0\,\text{kN·m}\,\theta$
10 kN at C	$3\theta$	$30\,\text{kN·m}\,\theta$

Equation: $M_p\times 7\,\theta = 30\,\theta$ (kN·m), so $M_p = \frac{30}{7} = 4.29$ kN·m.

Mechanism 3: combined mechanism

Combine 1 and 2 so that the hinge at B cancels; hinges at A, J and C.

Hinge	Rotation	Plastic moment	Internal work
C (column CD)	$2\theta$	$M_p$	$2\,M_p\theta$
J (under 20 kN)	$1\theta$	$3M_p$	$3\,M_p\theta$
A (fixed base)	$\frac{1}{2}\theta$	$2M_p$	$1\,M_p\theta$

Internal work $= 6\,M_p\,\theta$ .

Load	Displacement along the load	External work
20 kN at J	$1\theta$	$20\,\text{kN·m}\,\theta$
10 kN at C	$\frac{3}{2}\theta$	$15\,\text{kN·m}\,\theta$

Equation: $M_p\times 6\,\theta = 35\,\theta$ (kN·m), so $M_p = \frac{35}{6} = 5.83$ kN·m.

Required plastic moment

The combined mechanism requires the largest $M_p$ :

Answer: $M_p = \dfrac{35}{6} = 5.83$ kN·m (so AB $= 11.67$ , BC $= 17.5$ and CD $= 5.83$ kN·m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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Plastic moment capacity Mp of frame

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