Question

2.A driver of mass 50kg at the end of cantilever diving board of 1m length. The diver oscillates at a frequency of 2 Hz. What is the flexural rigidity EI of the diving board? MCQ question

Answer 1

Concept:

The formula for deflection at the end of the cantilever beam will be:

δ = $\frac{WL^{3} }{3EI}$

where EI is the flexural rigidity and W= weight acting on the cantilever

Flexural Rigidity:

The force couple needed to bend a stationary, non-rigid structure by one unit of curvature is known as flexural rigidity, or the resistance a structure provides when being bent.

The formula for frequency in terms of deflection is:

f= $\frac{.4985}{\sqrt{δ} }$

Given:

Frequency =2 Hz

Mass = 50 kg

Find:

Flexural rigidity EI

Solution:

Putting the value given in the above formula for frequency:

2= $\frac{.4985}{\sqrt{δ} }$

> δ= .0621 m

Therefore, now putting the value in the formula for deflection:

.0621=$\frac{500}{3EI}$

> EI= 2683.84 N- sqm.

Hence, the flexural rigidity is 2683.84 N-m.

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