A load of 160200 N (36010 lbf) is applied to a cylindrical specimen of a steel a
ID: 1817742 • Letter: A
Question
A load of 160200 N (36010 lbf) is applied to a cylindrical specimen of a steel alloy that has a cross-sectional diameter of 10.7 mm (0.4213 in.). If the original specimen length is 498 mm (19.61 in.), how much will it increase in length when this load is applied?
A load of 160200 N (36010 lbf) is applied to a cylindrical specimen of a steel alloy that has a cross-sectional diameter of 10.7 mm (0.4213 in.). If the original specimen length is 498 mm (19.61 in.), how much will it increase in length when this load is applied?Explanation / Answer
Stress is defined as force/area = F/A
Here the force is 160200N and the area is *(d/2)² = 89.92mm²
The stress on the graph is measured in MPa which is equal to N/mm²
=>Stress = 160200N/89.92N/mm² = 1781.5 MPa
On the graph is seen that this stress results in a strain of approximately 0.02.
Strain is the difference in length per original length = L/L, and L is what we look for and L is given 498mm.
=> 0.02 = L/498mm
=> 0.02*498mm = L
=> L = 9.96mm
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