ch01-03 stress & strain & properties

03 Solutions 46060 5/7/10 8:45 AM Page 4
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3–5. A tension test was performed on a steel specimen
having an original diameter of 12.5 mm and gauge length
of 50 mm. Using the data listed in the table, plot the
stress–strain diagram, and determine approximately the
modulus of toughness. Use a scale of 20 mm = 50 MPa and
20 mm = 0.05 mm>mm.
Stress and Strain:
s = P e = dL A (MPa) L (mm/mm)
0 0
90.45 0.00035
259.9 0.00120
308.0 0.00204
333.3 0.00330
355.3 0.00498
435.1 0.02032
507.7 0.06096
525.6 0.12700
507.7 0.17780
479.1 0.23876
Modulus of Toughness: The modulus of toughness is equal to the
total area under the stress–strain diagram and can be
approximated by counting the number of squares. The total
number of squares is 187.
Load (kN)
0
11.1
31.9
37.8
40.9
43.6
53.4
62.3
64.5
62.3
58.8
Elongation (mm)
0
0.0175
0.0600
0.1020
0.1650
0.2490
1.0160
3.0480
6.3500
8.8900
11.9380
(u t ) approx = 187(25)A10 6 B ¢ N m ≤ 2 a0.025 m b = 117 MJ>m3
m
Ans.
4