The data shown in the following table are from a tensile test of high-strength steel. TENSILE-TEST DATA Load (lb) Elongation (in.) 1.000 2.000 6.000 10.000 12.000 12.900 13.400 13.600 13.800 14.000 14,400 15.200 16.800 18.400 20.000 22,400 22.600 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 ultimate stress percent elongation in 2.00 in percent reduction in area 0.0336 0.0507 0.1108 Fracture The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure). Gage length @ At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel. (Submit a file with a maximum size of 1 MB.) Choose File No fie chosen This answer has not been graded yet ultimate stress, percer percent elongation in 2.00 in. (at fracture), and percent reduction in area (at fracture). (Enter the proportional limit, modulus of elasticity, yield stress, and ultimate stress in psi.) Determine the proportional limit, modulus of elasticity (the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, proportional limit 59.91 x modulus of elasticity yield stress 30046 Xp 68.897 X P 112.8307539 X psi 30.83%
The data shown in the following table are from a tensile test of high-strength steel. TENSILE-TEST DATA Load (lb) Elongation (in.) 1.000 2.000 6.000 10.000 12.000 12.900 13.400 13.600 13.800 14.000 14,400 15.200 16.800 18.400 20.000 22,400 22.600 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 ultimate stress percent elongation in 2.00 in percent reduction in area 0.0336 0.0507 0.1108 Fracture The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure). Gage length @ At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel. (Submit a file with a maximum size of 1 MB.) Choose File No fie chosen This answer has not been graded yet ultimate stress, percer percent elongation in 2.00 in. (at fracture), and percent reduction in area (at fracture). (Enter the proportional limit, modulus of elasticity, yield stress, and ultimate stress in psi.) Determine the proportional limit, modulus of elasticity (the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, proportional limit 59.91 x modulus of elasticity yield stress 30046 Xp 68.897 X P 112.8307539 X psi 30.83%
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:The data shown in the following table are from a tensile test of high-strength steel.
TENSILE-TEST DATA
Load (lb) Elongation (in.)
1,000
2,000
6,000
10,000
12,000
12,900
13,400
13,600
13,800
14,000
14,400
15,200
16,800
18,400
20,000
22,400
22,600
0.0002
0.0006
0.0019
0.0033
0.0039
0.0043
0.0047
0.0054
0.0063
0.0090
0.0102
0.0130
0.0230
0.0336
0.0507
0.1108
Fracture
The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure).
Gage
length
Ⓡ
At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in.
Plot the conventional stress-strain curve for the steel. (Submit a file with maximum size of 1 MB.)
Choose File No file chosen
This answer has not been graded yet.
Determine the proportional limit, modulus of elasticity (the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in. (at fracture), and percent reduction in area (at fracture). (Enter the proportional limit, modulus of elasticity, yield stress, and ultimate stress in psi.)
proportional limit
59.91 X psi
modulus of elasticity
30046 X
psi
68.897 X psi
yield stress
ultimate stress
112.8307539 X psi
percent elongation in 2.00 in.
percent reduction in area
30.83✔ %
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