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snow_tiger [21]
4 years ago
8

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.
Engineering
1 answer:
Rom4ik [11]4 years ago
7 0

Answer:

<em>a) 42 mm</em>

<em>b) 144.4 MPa</em>

<em></em>

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 = <em>42 mm</em>

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 = <em>144.4 MPa</em>

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