Page 239 - Materials Science and Engineering An Introduction

Questions and Problems • 211

6.34 For the tempered steel alloy whose stress–strain
Load Length behavior can be observed in the Tensile Tests mod-
N in. mm ule of VMSE, determine the following:
lb f
0 0 2.500 63.50 (a) the approximate yield strength (0.002 strain
310 1380 2.501 63.53 offset)
625 2780 2.502 63.56 (b) the tensile strength
1265 5630 2.505 63.62 (c) the approximate ductility, in percent elongation
1670 7430 2.508 63.70 How do these values compare with those for the
1830 8140 2.510 63.75 oil-quenched and tempered 4140 and 4340 steel
2220 9870 2.525 64.14 alloys presented in Table B.4 of Appendix B?
For the aluminum alloy whose stress–strain be-
2890 12,850 2.575 65.41 6.35 havior can be observed in the Tensile Tests module
3170 14,100 2.625 66.68 of VMSE, determine the following:
3225 14,340 2.675 67.95 (a) the approximate yield strength (0.002 strain
3110 13,830 2.725 69.22 offset)
2810 12,500 2.775 70.49 (b) the tensile strength
Fracture (c) the approximate ductility, in percent elongation
How do these values compare with those for the
(a) Plot the data as engineering stress versus en-
gineering strain. 2024 aluminum alloy (T351 temper) presented in
Table B.4 of Appendix B?
(b) Compute the modulus of elasticity.
6.36 For the (plain) carbon steel alloy whose stress–
(c) Determine the yield strength at a strain offset strain behavior can be observed in the Tensile
of 0.002. Tests module of VMSE, determine the following:
(d) Determine the tensile strength of this alloy. (a) the approximate yield strength
(e) Compute the modulus of resilience. (b) the tensile strength
(f) What is the ductility, in percent elongation? (c) the approximate ductility, in percent elongation
6.32 A cylindrical metal specimen 15.00 mm in di- 6.37 A cylindrical metal specimen having an original
ameter and 120 mm long is to be subjected to a diameter of 12.8 mm (0.505 in.) and gauge length
tensile force of 15,000 N. of 50.80 mm (2.000 in.) is pulled in tension until
(a) If this metal must not experience any plastic fracture occurs. The diameter at the point of frac-
deformation, which of aluminum, copper, brass, ture is 8.13 mm (0.320 in.), and the fractured gauge
nickel, steel, and titanium (Table 6.2) are suitable length is 74.17 mm (2.920 in.). Calculate the ductility
candidates? Why? in terms of percent reduction in area and percent
elongation.
(b) If, in addition, the specimen must elongate no
more than 0.070 mm, which of the metals that sat- 6.38 Calculate the moduli of resilience for the materi-
isfy the criterion in part (a) are suitable candidates? als having the stress–strain behaviors shown in
Why? Base your choices on data found in Table 6.1. Figures 6.12 and 6.22.
6.33 For the titanium alloy whose stress–strain be- 6.39 Determine the modulus of resilience for each of
havior can be observed in the Tensile Tests mod- the following alloys:
ule of Virtual Materials Science and Engineering
(VMSE), determine the following: Yield Strength
(a) the approximate yield strength (0.002 strain Material MPa psi
offset) Steel alloy 830 120,000
(b) the tensile strength Brass alloy 380 55,000
(c) the approximate ductility, in percent elongation Aluminum alloy 275 40,000
How do these values compare with those for the Titanium alloy 690 100,000
two Ti-6Al-4V alloys presented in Table B.4 of
Appendix B? Use the modulus of elasticity values in Table 6.1.