Question

A 6061-T6 aluminum pulley transmits 75 hp at 1,750 rpm. The shaft is 1% inches in diameter and is made from cold-drawn 1040 steel. The key is to be made from cold-drawn 1020 steel. For a safety fac- tor of 3, compute the required square key length for both shear and compression

Answer 1

Given: Transmitted power = 75hp = 55.92Kw

Speed = 1750Rpm

Diameter of shaft = 1.75inch

Here we have:

$p=\frac{2\pi NT}{60} \Rightarrow T=\frac{P*60}{2\pi N}$

$T= \frac{ 55.92 * 60 * 10^3 }{ 2 \pi *1750}=305.14Nm$

For Square Key:

$\tau =\frac{2T}{dbl}...........(1)$

$\sigma _c = \frac{ 4 T }{ d h l}$

For cold drawn 1040 steel:

Ultimate yield strength = 530*10⁶ N/m²

Ultimate shear strength = 0.57 * 530 * 10⁶ = 302.1 * 10⁶ N/m²

Allowable shear strength = (302.1 * 10⁶)/factor of safety

= 302.1 * 10⁶/3

= 100.7 * 10⁶ N/m²

width b = 0.009525

Then putting values in eq (1)

Length of Key l = 0.691(shear)

l = 0.345m(comp)

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