Question

An eyebolt is to be made out of steel having an ultimate tensile strength of 500 N/mm². If the load to be lifted using this eyebolt is 500 kgf then using a factor of safety of 2 find the core diameter (minor diameter) of the thread for eyebolt.

1. \(4\sqrt{\frac{5}{\pi}}\)mm
2. \(5\sqrt{\frac{5}{\pi}}\) mm
3. \(15\sqrt{\frac{5}{\pi}}\) mm
4. \(2\sqrt{\frac{5}{\pi}}\)mm

Answer 1

Correct Answer - Option 1 : \(4\sqrt{\frac{5}{\pi}}\)mm

Concept:

\(W = \frac{σ_u}{nA_c}\)

where W = load, σ_u = ultimate tensile strengh, n = factor of safty, A_c = core diameter

Calculation:

Given:

σ_u = 500 N/mm², n = 2, W = 500 kgf = 500 × 10 = 5000 N  (∵ 1 N = 1 kgf × g)

Let "d" is the core diamter of bolt then, \(A_c= \frac{\pi d^2}{4}\)

So we have,

\(W=\frac{\sigma_u\times \pi d^2}{4\times n}\)

\(d^2 = \frac{4\times2\times 5000}{500\times\pi}\)

\(d=4\sqrt{\frac{5}{\pi}}\)

Hence the core diameter of eyebolt is \(4\sqrt{\frac{5}{\pi}}\).