Asked by rajesh jain
A mild steel rod is 150 cm long. The portion 80 cm long is 22 mm diameter and the remaining length is 20 mm diameter. If the total change in length is 0.072 cm, Determine the changes of each part in 1) Length, 2) Diameter, 3) Volume. Take poissions ratio= 0.3
Answers
Answered by
drwls
Consider the length change as a result of stress.
Stress x area is constant along the length.
Stress is (22/20)^2 times higher (or 21% higher) in the narrower part. The ratio of length change in the 22 and 20 mm portions is
80*1 :: 70*1.21 or 80::84.7.
A). 48.6% of the length change (0.035 cm) is in the 22 mm wide portion. 51.4% of the length change (0.037 cm) is in the 20 mm wide portion.
B) The relative diameter changes are 30% of the
relative length change.
C) Let relative length change = e
Relative volume change = (1+e)*(1-0.3e)^2 -1
Stress x area is constant along the length.
Stress is (22/20)^2 times higher (or 21% higher) in the narrower part. The ratio of length change in the 22 and 20 mm portions is
80*1 :: 70*1.21 or 80::84.7.
A). 48.6% of the length change (0.035 cm) is in the 22 mm wide portion. 51.4% of the length change (0.037 cm) is in the 20 mm wide portion.
B) The relative diameter changes are 30% of the
relative length change.
C) Let relative length change = e
Relative volume change = (1+e)*(1-0.3e)^2 -1
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