Figure 3-7 shows a shaft carrying two sheaves that are keyed to the shaft. Part (b) shows that a force F is transmitted from the shaft to the hub of the sheave through a square key. The shaft is 2.25 inches in diameter and transmits a torque of 14 063 Ib.in. The key has a square cross

Question

Figure 3-7 shows a shaft carrying two sheaves that are keyed to the shaft. Part (b) shows that a force F is transmitted from the shaft to the hub of the sheave through a square key. The shaft is 2.25 inches in diameter and transmits a torque of 14 063 Ib.in. The key has a square cross section, 0.50 in on a side, and a length of 1.75 in. Compute the force on the key and the shear stress caused by this force.

Accepted Answer

Objective Compute the force on the key and the shear stress.

Given Layout of shaft, key, and hub shown in Figure 3-7.
Torque = T = 14 063 lb-in; key dimensions = 0.5 × 0.5 × 1.75 in.
Shaft diameter - D = 2.25 in; radius = R = D/2 = 1.125 in.

Analysis Torque T = force F × radius R, Then F = T/R.
Use equation (3-4) to compute shearing stress: τ = F/A,.

[latex] au =sharing force / area in shear = F/A_{s}[/latex] (3-4)
Shear area is the cross section of the key at the interface between the shaft and the hub: [latex]A_s = bL.[/latex]

Results F = T/R = (14 063 lb.in)/( 1.125 in) = 12 5001b
[latex]A_s = bL[/latex] = (0.50 in)( 1.75 in) = 0.875 [latex]in^2[/latex]
τ = F/A = (12 5001b)/(0.875 [latex]in^2[/latex]) - 14 300 [latex]in^2[/latex]

Comment This level of shearing stress will be uniform on all parts of the cross section of the key.

Question 3.3: Figure 3-7 shows a shaft carrying two sheaves that are keyed...

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