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 given by the formula: \[ \delta = \frac{5 w L^4}{384 E I} \] where:
\(w\) is the uniform load per unit length,
\(L\) is the length of the beam,
\(E\) is the Young’s modulus of the material,
\(I\) is the moment of inertia of the beam's cross-section.
When the length of the beam is doubled:
The deflection is proportional to \(L^4\), so doubling the length will increase the deflection by a factor of \(2^4 = 16\). 
When the depth of the beam is doubled:
The moment of inertia \(I\) for a rectangular section is proportional to the cube of the depth, \(I \propto d^3\). Doubling the depth increases the moment of inertia by a factor of \(2^3 = 8\), which decreases the deflection by a factor of 8. 
Overall effect:
Doubling the length increases the deflection by a factor of 16. 
Doubling the depth decreases the deflection by a factor of 8. Thus, the total effect is an increase in deflection by a factor of: \[ \frac{16}{8} = 2 \] Therefore, the new deflection is: \[ \delta_{{new}} = 2 \times \delta_{{old}} = 2 \times 24 = 48 \, {mm} \]