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/ 0.160
70 /0.229
72 /0.300
76 /0.424
84 /0.668
92 /0.965
100 /1.288
112 /2.814
127 /fracture
(a) Using the data,plot a stress-strain curve, marking the proportional limit and calculate the ultimate strength
(b) Given that , at fracture , the minimum diameter was 9.5 mm, calculate the reduction in cross-sectional area, expressing the answer as a percentage of the original cross-sectional area.
To plot the stress-strain curve, we need to calculate stress and strain values from the given data.
(a) To calculate stress, we use the formula:
Stress = Load / Cross-sectional Area
Given that the test specimen has a diameter of 15mm, the cross-sectional area can be calculated as follows:
Cross-sectional Area = π * (Diameter/2)^2
= π * (15/2)^2
= 176.71 mm^2
Now, we can calculate the stress values by dividing the load values by the cross-sectional area.
Load (kN) Cross-sectional Area (mm^2) Stress (MPa)
----------------------------------------------------------
5 176.71 0.028
10 176.71 0.056
30 176.71 0.169
50 176.71 0.283
60 176.71 0.339
64.5 176.71 0.365
67 176.71 0.379
68 176.71 0.385
69 176.71 0.390
70 176.71 0.395
72 176.71 0.408
76 176.71 0.430
84 176.71 0.475
92 176.71 0.521
100 176.71 0.566
112 176.71 0.634
127 176.71 0.719
Now, we need to calculate the strain values by dividing the elongation values by the gauge length. The gauge length in this case is 50mm.
Elongation (mm) Strain
------------------------------
0.005 0.0001
0.015 0.0003
0.048 0.0010
0.084 0.0017
0.102 0.0020
0.109 0.0022
0.119 0.0024
0.137 0.0027
0.160 0.0032
0.229 0.0046
0.300 0.0060
0.424 0.0085
0.668 0.0134
0.965 0.0193
1.288 0.0258
2.814 0.0563
Fracture N/A
Now, we can plot the stress-strain curve using the stress and strain values. The proportional limit is the point where the curve deviates from a linear relationship. From the table, the proportional limit can be estimated to occur at a stress of approximately 0.39 MPa and a strain of approximately 0.004.
To calculate the ultimate strength, we find the maximum stress value in the table, which is 0.719 MPa.
(b) To calculate the reduction in cross-sectional area at fracture, we use the formula:
Reduction in Cross-sectional Area = (Original Cross-sectional Area - Minimum Cross-sectional Area) / Original Cross-sectional Area * 100
Given that the original cross-sectional area is 176.71 mm^2 and the minimum diameter at fracture is 9.5 mm, we can calculate the minimum cross-sectional area as follows:
Minimum Cross-sectional Area = π * (Minimum Diameter/2)^2
= π * (9.5/2)^2
= 70.88 mm^2
Now, we can calculate the reduction in cross-sectional area:
Reduction in Cross-sectional Area = (176.71 - 70.88) / 176.71 * 100
= 60.00%
Therefore, the reduction in cross-sectional area at fracture is 60% of the original cross-sectional area.