Page 240 - Materials Science and Engineering An Introduction

212 • Chapter 6 / Mechanical Properties of Metals

6.40 A steel alloy to be used for a spring application must 6.48 For a brass alloy, the following engineering
have a modulus of resilience of at least 2.07 MPa stresses produce the corresponding plastic engi-
(300 psi). What must be its minimum yield strength? neering strains prior to necking:
6.41 Using data found in Appendix B, estimate the
modulus of resilience (in MPa) of cold-rolled Engineering Engineering
17-7PH stainless steel. Stress (MPa) Strain
315 0.105
True Stress and Strain 340 0.220
6.42 Show that Equations 6.18a and 6.18b are valid when
there is no volume change during deformation. On the basis of this information, compute the en-
gineering stress necessary to produce an engineer-
6.43 Demonstrate that Equation 6.16, the expression ing strain of 0.28.
defining true strain, may also be represented by
6.49 Find the toughness (or energy to cause fracture)
A 0 for a metal that experiences both elastic and plas-
P T = ln a b
A i tic deformation. Assume Equation 6.5 for elastic
when the specimen volume remains constant dur- deformation, that the modulus of elasticity is 103
ing deformation. Which of these two expressions GPa (15 * 10 6 psi), and that elastic deformation
is more valid during necking? Why? terminates at a strain of 0.007. For plastic de-
formation, assume that the relationship between
6.44 Using the data in Problem 6.30 and Equations stress and strain is described by Equation 6.19,
6.15, 6.16, and 6.18a, generate a true stress–true are 1520 MPa
strain plot for stainless steel. Equation 6.18a be- in which the values for K and n
(221,000 psi) and 0.15, respectively. Furthermore,
comes invalid past the point at which necking be- plastic deformation occurs between strain values
gins; therefore, measured diameters are given in of 0.007 and 0.60, at which point fracture occurs.
the following table for the last three data points,
which should be used in true stress computations. 6.50 For a tensile test, it can be demonstrated that
necking begins when
Load Length Diameter
N lb f mm in. mm in. ds T = s T (6.32)
dP T
159,500 35,850 54.864 2.160 12.22 0.481
Using Equation 6.19, determine an expression for
151,500 34,050 55.880 2.200 11.80 0.464
the value of the true strain at this onset of necking.
124,700 28,000 56.642 2.230 10.65 0.419
6.51 Taking the logarithm of both sides of Equation
6.45 A tensile test is performed on a metal specimen, 6.19 yields
and it is found that a true plastic strain of 0.16 is
produced when a true stress of 500 MPa (72,500 log s T = log K + n log P T (6.33)
psi) is applied; for the same metal, the value of in the plas-
K in Equation 6.19 is 825 MPa (120,000 psi). Thus, a plot of log s T versus log P T
tic region to the point of necking should yield a
Calculate the true strain that results from the ap- straight line having a slope of n and an intercept
plication of a true stress of 600 MPa (87,000 psi).
(at log s T = 0) of log K.
6.46 For some metal alloy, a true stress of 345 MPa Using the appropriate data tabulated in
(50,000 psi) produces a plastic true strain of 0.02. Problem 6.30, make a plot of log s T versus log P T
How much does a specimen of this material elongate and determine the values of n and K. It will be nec-
when a true stress of 415 MPa (60,000 psi) is applied essary to convert engineering stresses and strains
if the original length is 500 mm (20 in.)? Assume a to true stresses and strains using Equations 6.18a
value of 0.22 for the strain-hardening exponent, n. and 6.18b.

6.47 The following true stresses produce the corre- Elastic Recovery after Plastic Deformation
sponding true plastic strains for a brass alloy:
6.52 A cylindrical specimen of a brass alloy 10.0 mm
True Stress (psi) True Strain (0.39 in.) in diameter and 120.0 mm (4.72 in.)
long is pulled in tension with a force of 11,750 N
60,000 0.15
(2640 lb f ); the force is subsequently released.
70,000 0.25
(a) Compute the final length of the specimen at
What true stress is necessary to produce a true this time. The tensile stress–strain behavior for
plastic strain of 0.21? this alloy is shown in Figure 6.12.