Page 217 - Materials Science and Engineering An Introduction

6.7 True Stress and Strain • 189

Table 6.4
K
The n and K Values
(Equation 6.19) for Material n MPa psi
Several Alloys Low-carbon steel (annealed) 0.21 600 87,000
4340 steel alloy (tempered @ 315 C) 0.12 2650 385,000
304 stainless steel (annealed) 0.44 1400 205,000
Copper (annealed) 0.44 530 76,500
Naval brass (annealed) 0.21 585 85,000
2024 aluminum alloy (heat-treated—T3) 0.17 780 113,000
AZ-31B magnesium alloy (annealed) 0.16 450 66,000

EXAMPLE PROBLEM 6.4

Ductility and True-Stress-at-Fracture Computations
A cylindrical specimen of steel having an original diameter of 12.8 mm (0.505 in.) is tensile-
tested to fracture and found to have an engineering fracture strength s f of 460 MPa (67,000 psi).
If its cross-sectional diameter at fracture is 10.7 mm (0.422 in.), determine
(a) The ductility in terms of percentage reduction in area
(b) The true stress at fracture

Solution
(a) Ductility is computed using Equation 6.12, as

12.8 mm 2 10.7 mm 2
a b p - a b p
2 2
% RA = * 100
12.8 mm 2
a b p
2
2
128.7 mm - 89.9 mm 2
= 2 * 100 = 30%
128.7 mm
(b) True stress is defined by Equation 6.15, where, in this case, the area is taken as the fracture
area A f . However, the load at fracture must first be computed from the fracture strength as

1 m 2
6 2 2
F = s f A 0 = (460 * 10 N/m )(128.7 mm )a 6 2 b = 59,200 N
10 mm
Thus, the true stress is calculated as

F 59,200 N
s T = = 2
A f 2 1 m
(89.9 mm )a 6 2 b
10 mm
8
2
= 6.6 * 10 N/m = 660 MPa (95,700 psi)