Question

A solid shaft is subjected to an axial load P = 200 kN and a torque T = 1.5 kN.m. a) Determine the diameter of the shaft if the maximum shear stress should not exceed 100 Mpa. b) Using the diameter, determine the maximum normal stress.

Answer 1

Answer:

a) 42 mm

b) 144.4 MPa

Explanation:

Load P = 200 kN = 200 x 10^3 N

Torque T = 1.5 kN-m = 1.5 x 10^3 N-m

maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa

diameter of shaft d = ?

From T = τ * $\frac{\pi }{16}$ * $d^{3}$

substituting values, we have

1.5 x 10^3 = 100 x 10^6 x $\frac{3.142 }{16}$ x $d^{3}$

$d^{3}$ = 7.638 x 10^-5

d = $\sqrt[3]{7.638 * 10^-5}$ = 0.042 m = 42 mm

b) Normal stress = P/A

where A is the area

A = $\frac{\pi d^{2} }{4}$ = $\frac{3.142*0.042^{2} }{4}$ = 1.385 x 10^-3

Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = 144.4 MPa