Collapse load of fixed beam with UDL

Set-up

A beam of span $L$ fixed at both ends carries a UDL of $W$ kN/m. The plastic moment of the section is $M_p$. The beam is indeterminate to the second degree, so $2+1 = 3$ hinges are needed for collapse. By symmetry the hinges form at the two fixed ends A and B and at mid-span C (where the free moment is largest).

Mechanism and virtual work

Let the mid-span deflection be $\delta = \dfrac{L}{2}\theta$.

• Rotation at A: $\theta$; at B: $\theta$; at C: $2\theta$.
• Internal work $= M_p\theta + 2M_p\theta + M_p\theta = 4M_p\theta$.
• External work $=$ load $\times$ average deflection $= W L \times \dfrac{\delta}{2} = W L\times\dfrac{L\theta}{4} = \dfrac{WL^2}{4}\theta$.

Equating:

$$\frac{W_cL^2}{4}\theta = 4M_p\theta \;\Rightarrow\; W_c = \frac{16M_p}{L^2}$$

Check by equilibrium (static method)

At collapse the end moments are $-M_p$ (hogging) and the mid-span moment is $+M_p$. The free bending moment at mid-span is $WL^2/8$, and the moment diagram is lowered by $M_p$, so $\dfrac{WL^2}{8} = M_p + M_p = 2M_p$, giving $W = \dfrac{16M_p}{L^2}$, the same value. Hence the upper and lower bounds coincide and this is the true collapse load.

Answer: $W_c = 16M_p/L^2$ kN/m (total load $16M_p/L$).