You don't need all that load and elongation data to answer the question. At fracture, the diameter was reduced by a factor 9.6/15 = 0.64. The area was reduced by the square of that factor, 0.4096
The area at that time was 40.96% of the initial area.
The following data was obtained from a tensile test of steel. The test specimen was 15mm in diameter and 50 mm in length.
Load(kN)/elongation mm)
5 /0.005
10 /0.015
30 /0.048
50 /0.084
60 /0.102
64.5 /0.109
67 /0.119
68 /0.137
69 /0.16
70 /0.229
72 /0.3
76 /0.424
84 /0.668
92 /0.965
100 /1.288
112 /2.814
127 /fracture
Given that at fracture, the minimum diameter was 9.6mm. calculate the reduction in cross-sectional area. Expressing the answer as a percentage of the original cross-sectional area.
1 answer