Question

A simply supported beam is under a uniformly distributed load (UDL) along the full span. The mid-span deflection is measured as 24 mm. If the length and depth of the beam is doubled while keeping other parameters unchanged, the mid-span deflection is _______ mm.

Hint:

In beam deflection problems, remember that the deflection is proportional to the length raised to the fourth power and inversely proportional to the moment of inertia, which is proportional to the cube of the depth for a rectangular cross-section.

Answer 1

The deflection \(\delta\) of a simply supported beam under a uniformly distributed load is calculated using the formula: \[ \delta = \frac{5 w L^4}{384 E I} \] where \(w\) is the uniform load per unit length, \(L\) is the beam's length, \(E\) is the material's Young’s modulus, and \(I\) is the moment of inertia of the beam's cross-section.
Effect of Doubling Beam Length:
Deflection is proportional to \(L^4\). Consequently, doubling the length increases deflection by a factor of \(2^4 = 16\).
Effect of Doubling Beam Depth:
For a rectangular section, the moment of inertia \(I\) is proportional to the cube of the depth, \(I \propto d^3\). Doubling the depth results in an 8-fold increase in the moment of inertia (\(2^3 = 8\)), which reduces deflection by a factor of 8. 
Combined Effect:
A 16-fold increase in deflection due to doubled length is counteracted by an 8-fold decrease due to doubled depth. The net effect is an increase in deflection by a factor of: \[ \frac{16}{8} = 2 \] Therefore, the new deflection is: \[ \delta_{{new}} = 2 \times \delta_{{old}} = 2 \times 24 \, {mm} = 48 \, {mm} \]