Collapse value of P: continuous beam

Set-up

Beam ABC: A fixed, B and C supports; AB = 6 m with $P$ at mid-span J (3 m from A); BC = 6 m with $2P$ at K, 2 m from B and 4 m from C. $M_p$ is the same throughout. Degree of indeterminacy = 2; the possible mechanisms are the failure of each span alone and their combination.

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

Mid-span deflection $\delta = 3\theta$; hinge rotations $\theta$, $2\theta$, $\theta$.

$$P\times3\theta = M_p(\theta + 2\theta + \theta) = 4M_p\theta \;\Rightarrow\; P = \frac{4}{3}M_p = 1.333\,M_p$$

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

Part BK (2 m) rotates $\theta_1$ about B and part KC (4 m) rotates $\theta_2$ about C, with $2\theta_1 = 4\theta_2$, so $\theta_1 = 2\theta_2$. The deflection under the load is $\delta = 4\theta_2$.

• Hinge at B: $\theta_1 = 2\theta_2$; hinge at K: $\theta_1 + \theta_2 = 3\theta_2$.
• Internal work $= M_p(2\theta_2 + 3\theta_2) = 5M_p\theta_2$.
• External work $= 2P\times4\theta_2 = 8P\theta_2$.

$$8P\theta_2 = 5M_p\theta_2 \;\Rightarrow\; P = \frac{5}{8}M_p = 0.625\,M_p$$

Mechanism 3: combined (hinges at A, J and K; no hinge at B)

With no hinge at B, J-B-K rotates as one rigid piece about support B, so when K moves down J moves up. Let this piece rotate $\theta$: K moves down $2\theta$ and J moves up $3\theta$. Hinge rotations: A $\theta$, J $2\theta$, K $\theta + \dfrac{2\theta}{4} = 1.5\theta$.

• Internal work $= M_p(\theta + 2\theta + 1.5\theta) = 4.5M_p\theta$.
• External work $= 2P\times2\theta - P\times3\theta = P\theta$.

This gives $P = 4.5M_p$, much larger. (In the opposite sense the external work is negative, so no mechanism forms.)

Collapse load

The lowest value is for span BC:

Answer: $P_c = \dfrac{5}{8}M_p = 0.625\,M_p$ (hinges at B and under the $2P$ load; $2P_c = 1.25M_p$).

Check (equilibrium)

At collapse $M_B = -M_p$ and $M_K = +M_p$. For BC: $M_K = \dfrac{2P\times2\times4}{6} - \dfrac{4}{6}M_p = \dfrac{8P}{3} - \dfrac{2}{3}M_p$; setting this equal to $M_p$ gives $P = \dfrac{5}{8}M_p$. In AB, $M_A$ is not fixed by statics at collapse (span AB is still indeterminate). For example, with $M_A = -0.5M_p$ the moment under $P$ is $\dfrac{PL}{4} + \dfrac{M_A + M_B}{2} = 0.9375M_p - 0.75M_p = 0.19M_p$, well within $M_p$, so the yield condition is also satisfied.