Question

A simply supported beam of span \(6\,\mathrm{m}\) carries a central point load. If the plastic moment capacity is \(90\,\mathrm{kN\!-\!m}\), the collapse load is

• A. \(45\,\mathrm{kN}\)
• B. \(60\,\mathrm{kN}\)
• C. \(75\,\mathrm{kN}\)
• D. \(90\,\mathrm{kN}\)

Hint:

For a simply supported beam with a central point load, \[ \boxed{ W=\frac{4M_p}{L}. } \]

Answer 1

The Correct Option is B

Step 1: Picture the collapse mechanism.
At collapse a single plastic hinge forms right under the point load at midspan, and the two beam halves rotate about their end supports like a pair of rigid links hinged at the centre.
Step 2: Write the virtual work equation.
Give the beam a small virtual central deflection $\delta$. Each half rotates by $\theta = \dfrac{2\delta}{L}$, so the hinge at the centre opens up by $2\theta$. Equating external work done by the load to internal work absorbed by the hinge: \[ W\delta = M_p (2\theta) = M_p\left(\frac{4\delta}{L}\right). \]
Step 3: Solve for the collapse load.
\[ W = \frac{4M_p}{L} = \frac{4 \times 90}{6} = 60\,\mathrm{kN}. \]
\[ \boxed{60\,\mathrm{kN}} \]