A key having a square cross-section of side d/4 and length l is used to transmit torque T from the shaft of diameter d to the hub of a pulley. Assuming the length of the key to be equal to the thickness of the pulley, the average shear stress developed in the key is given by
Key Shear Stress Calculation
When a key is used to transmit torque from a shaft to a pulley or hub, it experiences a shearing force. This shearing force causes shear stress within the key. To determine the average shear stress developed in the key, we need to consider the torque transmitted, the dimensions of the shaft, and the dimensions of the key.
Torque Transmission Principle
The torque (\(T\)) transmitted by the shaft is resisted by a tangential force (\(F\)) acting at the periphery of the shaft where the key is seated. This force acts on the key, causing it to shear. The relationship between torque, force, and the shaft's diameter (\(d\)) is:
\[T = F \times \text{radius of shaft}\]
Since the radius of the shaft is \(d/2\), we can write:
\[T = F \times \frac{d}{2}\]
From this, the tangential force (\(F\)) acting on the key can be expressed as:
\[F = \frac{T}{d/2} = \frac{2T}{d}\]
Key Dimensions and Shear Area
The problem states that the key has a square cross-section with a side of \(d/4\). The length of the key is given as \(l\). It is also mentioned that the length of the key is equal to the thickness of the pulley, which confirms that \(l\) is the effective shearing length.
When the key transmits torque, it tends to shear along its length. The area resisting this shear force is the product of the key's length and its width (which is the side of its square cross-section).
- Side of the square key (\(s\)) = \(d/4\)
- Length of the key (\(l_k\)) = \(l\)
Therefore, the shear area (\(A\)) of the key is:
\[A = \text{length of key} \times \text{side of key}\]
\[A = l \times \frac{d}{4}\]
Average Shear Stress Calculation
The average shear stress (\(\tau\)) developed in the key is defined as the shear force divided by the shear area.
\[\tau = \frac{\text{Shear Force} (F)}{\text{Shear Area} (A)}\]
Substituting the expressions for \(F\) and \(A\) derived above:
\[\tau = \frac{\frac{2T}{d}}{l \times \frac{d}{4}}\]
Now, we simplify the expression:
\[\tau = \frac{2T}{d} \times \frac{4}{l d}\]
\[\tau = \frac{8T}{l d^2}\]
This formula gives the average shear stress developed in the key based on the given parameters: torque (\(T\)), key length (\(l\)), and shaft diameter (\(d\)).
Important Questions from Keys
The forces experience by a Key used in a gear train for power transmission is
A key of 14 mm width, 9 mm height and 100 mm length is mounted on a shaft of 50 mm diameter. If the allowable shear stress for the key material is 50 MPa, what is the maximum torque that can be transmitted?
- A key of width and height of 6 mm each is used to fix a gear on a shaft of 20 mm diameter. The shaft is used to transmit 10 kW power at 600 rpm to the gear. Permissible shear stress in the key is 80 N/mm$^2$, while compressive stress in the key is neglected. The minimum length of the key, in mm, is _____(round off to 2 decimal places).
- Which of the following key transmits power through frictional resistance only?
Feather keys are generally _______
