Question
Download Solution PDFA 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
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
The Force of the shaft circumference is given by
\(\rm{P=\frac{P}{d/2}} =\frac{2P}{d}\)
Shearing area = width × length
\(\rm{\frac{d}{4}\times l=\frac{ld}{4}}\)
\(\rm{Average ~shear~ stress = \frac{P} {Shearing ~area}} \)
\(\rm{Average ~shear~ stress =\frac{\frac{2T}{d}}{\frac{ld}{4}}}=\frac{8T}{ld^2}\)
Last updated on May 30, 2025
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