Collapse load of two-span beam

Set-up

Beam ABC: A fixed, B and C supports (C is a simple end support). AB = 6 m with a 20 kN load at mid-span J; BC = 6 m with a UDL of 10 kN/m. The loads are increased proportionally by a load factor $\lambda$ (the collapse loads are $20\lambda$ kN and $10\lambda$ kN/m); $M_p$ is the same for both spans. Reactions: A (3) + B (1) + C (1) = 5, so the degree of indeterminacy is $5 - 3 = 2$. A single-span mechanism needs 3 hinges for a fixed-ended span, or 2 hinges for a span with a simple end.

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

AB behaves as a fixed-ended beam. Mid-span deflection $\delta = 3\theta$; rotations are $\theta$ at A, $2\theta$ at J and $\theta$ at B.

• Internal work $= M_p(\theta + 2\theta + \theta) = 4M_p\theta$
• External work $= 20\lambda\times 3\theta = 60\lambda\theta$

$$60\lambda\theta = 4M_p\theta \;\Rightarrow\; \lambda = \frac{4}{60}M_p = 0.0667\,M_p$$

Mechanism 2: span BC (hinges at B and K in the span)

BC has a hinge at B and a hinge K at distance $x$ from the simple end C. Part BK rotates $\theta$ about B and part KC rotates $\phi = \dfrac{(6-x)}{x}\theta$ about C (both give the same deflection at K, $\delta_K = (6-x)\theta$).

• Rotation at B $= \theta$; at K $= \theta + \phi = \dfrac{6\theta}{x}$.
• Internal work $= M_p\theta\left(1 + \dfrac{6}{x}\right)$.
• External work $=$ UDL $\times$ area of the deflected triangle $= 10\lambda\times\dfrac{6\delta_K}{2} = 30\lambda(6-x)\theta$.

$$\lambda = \frac{M_p\,(x+6)}{30\,x\,(6-x)}$$

Minimising with respect to $x$: $\dfrac{d\lambda}{dx} = 0 \Rightarrow x^2 + 12x - 36 = 0 \Rightarrow x = 6(\sqrt2 - 1) = 2.485$ m from C (3.515 m from B). Then

$$\lambda = \frac{M_p\times8.485}{30\times2.485\times3.515} = 0.03238\,M_p$$

Mechanism 3: combined (hinges at A, J and in BC)

With no hinge at B, the parts J-B and B-K form one rigid piece that rotates about support B, so one of the two loads always moves against its own direction. If J moves down, the part of BC next to B lifts, the UDL does negative work and $\lambda > M_p/15$ (higher than mechanism 1). If the BC side moves down, J lifts and the 20 kN load does negative work; the best hinge position (about 4.5 m from B) gives $\lambda \approx 0.093M_p$. Neither is critical.

Collapse load

The smallest value governs, which is the failure of span BC:

$$\lambda_c = 0.0324\,M_p$$

So the collapse loads are $20\lambda_c = 0.648\,M_p$ kN at J and $10\lambda_c = 0.324\,M_p$ kN/m on BC ($M_p$ in kN·m).

Answer: collapse load factor $\lambda_c = 0.0324\,M_p$ (collapse of span BC, with hinges at B and 3.52 m from B).