Answer:
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Explanation:
Given that:
5-hp single-cylinder engine is used. At most, the belt transmits 60 percent of this power. The driving sheave has a diameter of 6.2 in. and the driven, 12.0 in. The belt selected should be as close to 92 in pitch length as possible. The engine speed is governor-controlled to a maximum of 3100 rev/min. Select a satisfactory belt, and specify it using the standard designation.
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..........23÷357=0.0644257703........
Answer:
Max shear = 8.15 x 10^7 N/m2
Explanation:
In order to find the maximum stress for a solid shaft having radius r, we will be applying the Torsion formula which can be written as;
Allowable Shear Stress = Torque x Radius / pi/2 x radius^4
Putting the values we have;
T = 2000 N/m
Radius = Diameter/2 = 0.05 / 2 = 0.025 m
Putting values in formula;
Max shear = 2000 x 0.025 / 3.14/2 x (0.025)^4
Max shear = 8.15 x 10^7 N/m2