Part B,G, and H.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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Part B,G, and H. Please provide proper units

A cold drawn 1144 Stressproof steel specimen with a diameter of 0.874 in. and a 1.0 in. gage length was tested to fracture. Load and
deformation data obtained during the tension test are given in the accompanying table. Determine
(a) the modulus of elasticity.
(b) the proportional limit.
(c) the yield strength (0.20% offset).
(d) the ultimate strength.
(e) the fracture stress.
(f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.664 in.
(g) the percent elongation at fracture.
(h) the percent reduction in area.
Tension Test of 1144 Steel
Load (lb)
0
5497
10686
15637
20599
25113
30253
34892
39589
45405
50410
55583
61290
64432
66808
69762
72845
75285
76700
76938
76333
74049
70631
66441
63531
Answer:
(a) E= i 21837.6
(b) OPL=
102.2
(c) 00.2% =
120
(d) OUTS=
(e) Ofr=105.95
(f) o
183.47
(g) % Elong - i
(h) % Reduction = i
Otr=
Change in Length
(in.)
0
i
128.307
0.00031
0.00062
0.00094
0.00125
0.00156
0.00187
0.00218
0.00250
0.00291
0.00333
0.00385
0.00468
0.00728
0.01248
0.02080
0.03120
0.04420
0.06240
0.08060
0.10140
0.11960
0.13780
0.15600
0.16640
ksi
ksi
ksi
ksi
ksi
ksi
%
%
Transcribed Image Text:A cold drawn 1144 Stressproof steel specimen with a diameter of 0.874 in. and a 1.0 in. gage length was tested to fracture. Load and deformation data obtained during the tension test are given in the accompanying table. Determine (a) the modulus of elasticity. (b) the proportional limit. (c) the yield strength (0.20% offset). (d) the ultimate strength. (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.664 in. (g) the percent elongation at fracture. (h) the percent reduction in area. Tension Test of 1144 Steel Load (lb) 0 5497 10686 15637 20599 25113 30253 34892 39589 45405 50410 55583 61290 64432 66808 69762 72845 75285 76700 76938 76333 74049 70631 66441 63531 Answer: (a) E= i 21837.6 (b) OPL= 102.2 (c) 00.2% = 120 (d) OUTS= (e) Ofr=105.95 (f) o 183.47 (g) % Elong - i (h) % Reduction = i Otr= Change in Length (in.) 0 i 128.307 0.00031 0.00062 0.00094 0.00125 0.00156 0.00187 0.00218 0.00250 0.00291 0.00333 0.00385 0.00468 0.00728 0.01248 0.02080 0.03120 0.04420 0.06240 0.08060 0.10140 0.11960 0.13780 0.15600 0.16640 ksi ksi ksi ksi ksi ksi % %
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