Question

A shaft is made of an aluminum alloy having an allowable shear stress of τallow = 100 MPa. If the diameter of the shaft is 100 mm, determine the maximum torque T that can be transmitted. What would be the maximum torque T′ if a 75-mm-diameter hole were bored through the shaft? Sketch the shear-stress distribution along a radial line in each case.

Answer 1

Answer:

(a) Maximum torque is 19.63 KN.m for a diameter of 100 mm.

(b) Maximum torque is 13.42 KN.m for a hollow rod of outer diameter 100 mm  and inner diameter 75 mm.

Explanation:

The maximum torque is given by the formula:

Tmax = τmax J/c

(a)

J = (π/2)r^4 = (π/2)(0.05 m)^4

J = 9.8174 x $10^{-6}m^{4}$

T = (100 $10^{6}$ Pa)(9.8174 x $10^{-6}m^{4}$)/0.05 m

T = 19.63 KN.m

(b)

J = (π/2)(r1^4 - r2^4) = (π/2)[(0.05 m)^4 - (0.0375 m)^4

J = 6.711 x $10^{-6}m^{4}$

for maximum torque, we will take c = external radius = 0.05 m

T = (100 $10^{6}$ Pa)(6.711 x $10^{-6}m^{4}$)/0.05 m

T = 13.42 KN.m

The shear stress distribution along a radial line in each case is given in the attachement.

Answer 2

If the diameter of the shaft is 100 mm, the maximum torque is equal to $19.63\;kNm$

Given the following data:

Shear stress = 100 MPa.

Diameter of shaft = 100 mm to m = 0.01 meter.

Radius = $\frac{0.1}{2}$ = 0.05 m.

How to calculate the maximum torque (T).

Mathematically, the maximum torque of a shaft is given by this formula:

$\tau_{max}=\frac{Tr}{J} \\\\\tau_{max}=\frac{Tr}{\frac{\pir^4}{2} }\\\\\tau_{max}=\frac{2Tr}{\pi r^4}\\\\T=\frac{\tau_{max}(\pi r^4)}{2r} \\\\T=\frac{100 \times 10^6 \times 3.142 \times ( 0.05^4) }{2 \times 0.05}\\\\T=19.63 \times 10^3\\\\T=19.63\;kNm$

For the hollow shaft, we have:

$\tau_{max}=\frac{T'r}{J} \\\\\tau_{max}=\frac{T'r}{\frac{\pi (r^4-R^4)}{2} }\\\\\tau_{max}=\frac{2T'r}{\pi (r^4-R^4)}\\\\T'=\frac{\tau_{max}(\pi (r^4-R^4))}{2r} \\\\T'=\frac{100 \times 10^6 \times 3.142 \times ( 0.05^4-0.0375^4) }{2 \times 0.05}\\\\T'=13.42 \times 10^3\\\\T'=13.42\;kNm$

For the shear stress at the inner surface, we have:

$\tau_{\rho}=\frac{T' \rho}{J} \\\\\tau_{\rho}=\frac{T' \rho}{\frac{\pi (r^4-R^4)}{2} }\\\\\tau_{\rho}=\frac{2T' \rho}{\pi (r^4-R^4)} \\\\\tau_{\rho}=\frac{2 \times 13.42 \times 10^3 \times 0.0375}{3.142 \times (0.05^4-0.0375^4)}\\\\\tau_{\rho}=75.0 \times 10^6\\\\\tau_{\rho}=75.0\;MPa$

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