Question 9.20:
Radius, r = d/2 = 3 X 10-3 m
Maximum shearing stress = 6.9 × 107 Pa
Maximum stress = Manimum load or force / Area
Maximum force = Maximum stress × Area
= 6.9 × 107 × π × (r) 2
= 6.9 × 107 × π × (3 ×10–3)2
= 1949.94 N
Each rivet carries one quarter of the load.
∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N
Two strips of metal are riveted together
at their ends by four rivets, each of diameter 6.0 mm. What is the
maximum tension that can be exerted by the riveted strip if the
shearing stress on the rivet is not to exceed 6.9 × 107
Pa? Assume that each rivet is to carry one
quarter of the load.
Solution:
Diameter of the metal
strip, d = 6.0 mm = 6.0 × 10–3 mRadius, r = d/2 = 3 X 10-3 m
Maximum shearing stress = 6.9 × 107 Pa
Maximum stress = Manimum load or force / Area
Maximum force = Maximum stress × Area
= 6.9 × 107 × π × (r) 2
= 6.9 × 107 × π × (3 ×10–3)2
= 1949.94 N
Each rivet carries one quarter of the load.
∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N
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