Answer:
a)diameter of the thread = 0.546mm
b) stress in the thread = 36.3MPa
Explanation:
Hooke's law
[tex]E = \frac{Stress}{Strain} \\[/tex]
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 =
[tex]\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[/tex]
diameter of the thread = 0.546mm
