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MAE 20
Winter 2011
Assignment 5
6.7 For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the
modulus of elasticity is 115 GPa (16.7 × 10
6 psi).
(a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm
2
(0.5 in.
2 ) without plastic deformation?
(b) If the original specimen length is 115 mm (4.5 in.), what is the maximum length to which it may be
stretched without causing plastic deformation?
Solution
(a) This portion of the problem calls for a determination of the maximum load that can be applied without
plastic deformation (F y
). Taking the yield strength to be 275 MPa, and employment of Equation 6.1 leads to
Fy =! y A 0 = ( 275 " 10
6 N/m
2
2
= 89,375 N (20,000 lb f
(b) The maximum length to which the sample may be deformed without plastic deformation is determined
from Equations 6.2 and 6.5 as
li = l 0 1 +
E
= (115 mm) 1 +
275 MPa
3 MPa
' =^ 115.28 mm^ (4.51^ in.)
6.22 Consider the brass alloy for which the stress-strain behavior is shown in Figure 6.12. A cylindrical
specimen of this material 6 mm (0.24 in.) in diameter and 50 mm (2 in.) long is pulled in tension with a force of
5000 N (1125 lbf). If it is known that this alloy has a Poisson's ratio of 0.30, compute: (a) the specimen elongation,
and (b) the reduction in specimen diameter.
Solution
(a) This portion of the problem asks that we compute the elongation of the brass specimen. The first
calculation necessary is that of the applied stress using Equation 6.1, as
F
A 0
F
d 0
2
5000 N
2
6 N/m
2 = 177 !MPa (25,000 psi)
From the stress-strain plot in Figure 6.12, this stress corresponds to a strain of about 2.0 × 10
of strain, Equation 6.
! l = " l 0
)(50 mm) = 0.10 mm (4 # 10
(b) In order to determine the reduction in diameter ∆d, it is necessary to use Equation 6.8 and the definition
of lateral strain (i.e., ε x
= ∆d/d 0
) as follows
! d = d 0
x
= # d 0
z
= # (6 mm)(0.30) (2.0 % 10
= – 3.6 × 10
6.29 A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of
2.000 in. (50.800 mm) is pulled in tension. Use the load–elongation characteristics tabulated below to complete
parts (a) through (f).
Load Length
N lbf mm in.
(b) The elastic modulus is the slope in the linear elastic region (Equation 6.10) as
E =
200 MPa $ 0 MPa
3 MPa = 62.5 GPa (9.1 % 10
6 psi)
(c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain curve at
approximately 285 MPa (41,000 psi ).
(d) The tensile strength is approximately 370 MPa (54,000 psi), corresponding to the maximum stress on
the complete stress-strain plot.
(e) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by one-hundred. The
total fracture strain at fracture is 0.165; subtracting out the elastic strain (which is about 0.005) leaves a plastic
strain of 0.160. Thus, the ductility is about 16%EL.
(f) From Equation 6.14, the modulus of resilience is just
U
r
y
2
2 E
which, using data computed above gives a value of
U
r
( 285 MPa)
2
3
MPa)
= 0.65 MN/m
2 = 0.65! 10
6 N/m
2 = 6.5! 10
5 J/m
3
(93.8 in.- lb
f
/in.
3
6.41 Using the data in Problem 6.28 and Equations 6.15, 6.16, and 6.18a, generate a true stress–true
strain plot for aluminum. Equation 6.18a becomes invalid past the point at which necking begins; therefore,
measured diameters are given below for the last four data points, which should be used in true stress computations.
Load Length Diameter
N lbf mm in. mm in.
Solution
These true stress-strain data are plotted below.
6.44 The following true stresses produce the corresponding true plastic strains for a brass alloy:
True Stress (psi) True Strain
