Question

A nylon thread is to be subjected to a 10-N tension. Knowing that E 5 3.2 GPa, that the maximum allowable normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the required diameter of the thread.

Answer 1

0.631 mm

Step-by-step explanation:

Strain=
`\frac {\triangle L}{L}=1%` hence 0.01

Stress=strain* Young’s modulus hence

Stress=
`0.01*3.2 Gpa=32 N/mm^(2)`

But stress,
`\sigma=\frac {P}{A}=\frac {P}{0.25\pi d^(2)}`

Making d the subject then

`d=\sqrt {\frac {P}{0.25\pi \sigma}}`

`d=\sqrt {\frac {10}{0.25\pi*32}}\approx 0.631 mm`