A 2.50-in-diameter shaft carrying a chain sprocket has one end milled in the form of a square to permit the use of a hand crank. The square is 1.75 in on a side. Compute the maximum shear stress on the square part of the shaft when a torque of 15 000 lb . in is applied.

Also, if the length of the square part is 8.00 in, compute the angle of twist over this part. The shaft material is steel with $G = 11.5 \times 10^6$ psi.

Question

A 2.50-in-diameter shaft carrying a chain sprocket has one end milled in the form of a square to permit the use of a hand crank. The square is 1.75 in on a side. Compute the maximum shear stress on the square part of the shaft when a torque of 15 000 lb . in is applied.

Also, if the length of the square part is 8.00 in, compute the angle of twist over this part. The shaft material is steel with $G = 11.5 \times 10^6$ psi.

Also, if the length of the square part is 8.00 in, compute the angle of twist over this part. The shaft material is steel with [latex]G = 11.5 × 10^6[/latex] psi.

Accepted Answer

Objective:

Compute the maximum shear stress and the angle of twist in the shaft.
Given:

Torque = [latex]T[/latex] = 15 000 lb . in; length = [latex]L[/latex] = 8.00 in.
The shaft is square; thus, a = 1.75 in.
[latex]G = 11.5 × 10^6[/latex] psi.

Analysis:

Figure 3–10 shows the methods for calculating the values for [latex]Q[/latex] and [latex]K[/latex] for use in Equations (3–12) and (3–13).

[latex] au_{\max }=T / Q[/latex] (3.12)

[latex] heta=T L / G K[/latex] (3.13)

Results:
[latex]\begin{aligned} Q &=0.208 a^{3}=(0.208)(1.75 \mathrm{in})^{3}=1.115 \mathrm{in}^{3} \ K &=0.141a^{4}=(0.141)(1.75 \mathrm{in})^{4}=1.322 \mathrm{in}^{4} \end{aligned}[/latex]
Now the stress and the deflection can be computed.
[latex]\begin{gathered} au_{\max }=\frac{T}{Q}=\frac{15000 \mathrm{lb} \cdot \mathrm{in}}{\left(1.115\mathrm{in}^{3} ight)}=13460 \mathrm{psi} \ heta=\frac{T L}{G K}=\frac{(15000\mathrm{lb} \cdot \mathrm{in})(8.00 \mathrm{in})}{\left(11.5 imes 10^{6}\mathrm{lb}/ \mathrm{in}^{2} ight)\left(1.322 \mathrm{in}^{4} ight)}=0.0079 \mathrm{rad}\end{gathered}[/latex]
Convert the angle of twist to degrees:
[latex] heta=(0.0079 \mathrm{rad})\left(180^{\circ} \pi \mathrm{rad} ight)=0.452^{\circ}[/latex]
Comments:

Over the length of 8.00 in, the square part of the shaft twists [latex]0.452^{\circ}[/latex]. The maximum shear stress is 13460 psi, and it occurs at the midpoint of each side as shown in Figure 3-10.

Question 3.8: A 2.50-in-diameter shaft carrying a chain sprocket has one e...

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