Collapse mechanism: continuous beam ABCD, BMD

Assumptions (the figure is not fully clear)

• Continuous beam A-B-C-D, prismatic, with the same plastic moment $M_p$ in all spans: A fixed; B, C and D simple supports. AB = 8 m with 100 kN at E, 4 m from A; BC = 5 m with 6 kN/m over the whole span and 75 kN at F, 3 m from B; CD = 4 m with 50 kN at G, 2 m from C. The given loads are the factored (collapse) loads, so $M_p$ is the value required.
• Degree of indeterminacy $= 6 - 3 = 3$ (A fixed: 3; B, C and D: 1 each). Each span can fail on its own, and the largest $M_p$ among the spans governs.

Mechanism 1: span AB (hinges at A, E and B)

AB is fixed at A and continuous at B, so it acts as a fixed-ended beam. With rotation $\theta$ at A and at B and $2\theta$ at E, the deflection under the load is $4\theta$.

$$M_p(\theta + 2\theta + \theta) = 100\times 4\theta \;\Rightarrow\; M_p = 100\ \text{kN·m}$$

This is the usual $PL/8 = 100\times 8/8$ for a fixed beam with a central load.

Mechanism 2: span BC (hinges at B, F and C)

BC is also treated as a fixed-ended span. Part BF (3 m) rotates $2\phi$ and part FC (2 m) rotates $3\phi$ (since $3\times2\phi = 2\times3\phi$), so the deflection under the point load is $6\phi$. Hinge rotations: B $2\phi$, F $5\phi$, C $3\phi$.

• Internal work $= M_p(2\phi + 5\phi + 3\phi) = 10M_p\phi$.
• External work $= 75\times6\phi + 6\times\tfrac12\times5\times6\phi = 540\,\phi$.

$$M_p = \frac{540}{10} = 54\ \text{kN·m}$$

Mechanism 3: span CD (hinges at C and G)

CD has a simple end at D, so two hinges are enough. Part CG (2 m) rotates $\theta$ about C, so the deflection under the load is $2\theta$; part GD (2 m) then rotates $\theta$ about D, and the hinge at G turns $2\theta$.

• Internal work $= M_p(\theta + 2\theta) = 3\,M_p\theta$.
• External work $= 50\times2\theta = 100\theta$.

$$M_p = \frac{100}{3} = 33.33\ \text{kN·m}$$

Collapse mechanism and required $M_p$

The largest requirement is for span AB, so the collapse mechanism is the failure of AB with hinges at A, E and B (a beam mechanism), and $M_p = 100$ kN·m. Spans BC and CD stay elastic and carry the moments below.

Bending moment diagram at collapse

The hinges give $M_A = -100$, $M_E = +100$ and $M_B = -100$ kN·m. The hinge at B holds the moment $-M_p$ while BC and CD remain elastic as a two-span beam, so the three-moment equation at C (with $M_B = -100$, $M_D = 0$) gives the hogging support moment

$$M_C = -39.31\ \text{kN·m}$$

The sagging moments are then $M_F = +44.42$ kN·m under the 75 kN load in BC and $M_G = +30.35$ kN·m under the 50 kN load in CD. All are below $M_p$, so no other hinge forms and the diagram is statically admissible.

Shape (sagging plotted above the line): in AB, a triangle with a peak of $+100$ at E and $-100$ at A and B; in BC, a curve starting at $-100$ at B, rising to about $+44.42$ at F and falling to $-39.31$ at C; in CD, straight lines from $-39.31$ at C through $+30.35$ at G to 0 at D.

Answer: $M_p = 100$ kN·m; the collapse mechanism is span AB, with plastic hinges at A, at E (under the 100 kN load) and at B.