Question

A nylon thread is subjected to a 8.5-N tension force. Knowing that E 5 3.3 GPa and that the length of the thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.

Answer 1

Answer:

a)diameter of the thread = 0.546mm

b) stress in the thread = 36.3MPa

Explanation:

Hooke's law

$E = \frac{Stress}{Strain}$

strain = 1.1% = 0.011

Stress = strain* Young’s modulus

= 3.3 × 10⁹ × 0.011

= 36.3 × 10⁶Pa

stress = 36.3MPa

stress in the thread = 36.3MPa

b) Now,

stress = Force / area

Force = 8.5N

Stress = 36.3 × 10⁶

Area = Force / stress

=  8.5 /  36.3 × 10⁶

= 0.234 × 10⁻⁶m²

Area =

$\frac{\pi }{4} d^2\\Thus,\\d^2 =0.234 \times 10^-^6. \frac{4}{\pi } \\d = \sqrt{0.298 \times 10^-^6 } \\= 5.46 \times 10^-^4m\\=0.546mm$

diameter of the thread = 0.546mm