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2.66 A bar 1 m long is bent and then stress re-
lieved. The radius of curvature to the neutral
axis is 0.50 m. The bar is 30 mm thick and
is made of an elastic, perfectly plastic material
e=(0.030)
2(0.50) = 0.03
Since Y= 600 MPa and E= 200 GPa, we find
e= (2)(0.003) = 0.006
2.67 Assume that a material with a uniaxial yield
stress Yyields under a stress state of principal
(σ1−σ2)2+ (σ2−σ3)2+ (σ3−σ1)2= 2Y2
Now consider a new stress state where the prin-
cipal stresses are
+ [(σ2+p)−(σ3+p)]2
+ [(σ3+p)−(σ1+p)]2
2.68 Give two different and specific examples
in which the maximum-shear-stress and the
tension, simple compression, equal biaxial ten-
sion, and equal biaxial compression. Thus, ac-
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2.69 A thin-walled spherical shell with a yield stress
Yis subjected to an internal pressure p. With
criteria will give the same results. We will now
demonstrate this more rigorously. The princi-
find that
σ1−0 = Y
2.70 Show that, according to the distortion-energy
1.15Ywhere Yis the uniaxial yield stress of the
that σ2=σ1/2. Substituting these into the
distortion-energy criterion given by Eq. (2.37)
2.71 What would be the answer to Problem 2.70 if
the maximum-shear-stress criterion were used?
2.72 A closed-end, thin-walled cylinder of original
length l, thickness t, and internal radius ris
subjected to an internal pressure p. Using the
σ1=pr
2t;σ2=pr
t;σ3= 0
=pr
in order to produce a tensile membrane stress,
2.73 A round, thin-walled tube is subjected to ten-
sion in the elastic range. Show that both the
thickness and the diameter of the tube decrease
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2.74 Take a long cylindrical balloon and, with a thin
longitudinal direction, or (4) an ellipse? Per-
form this experiment and, based on your obser-
couraged to assign the students the task of pre-
dicting the shape numerically; an example of a
and l3. Assuming that the material is rigid and
perfectly plastic, show that volume constancy
requires that the following expression be satis-
From the definition of true strain given by
Eq. (2.9) on p. 35, ln l1
2.76 What is the diameter of an originally 30-mm-
diameter solid steel ball when it is subjected to
a hydrostatic pressure of 5 GPa?
Table 2.1 on p. 32 we take values for steel of
ν= 0.3 and E= 200 GPa, so that
2.77 Determine the effective stress and effective
strain in plane-strain compression according to
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2.78 (a) Calculate the work done in expanding a 2-
mm-thick spherical shell from a diameter of 100
mm to 140 mm, where the shell is made of a ma-
terial for which σ= 200+50ǫ0.5MPa. (b) Does
your answer depend on the particular yield cri-
terion used? Explain.
by
σ1=σ2=pt
2t
u= 2σ1ǫ1= 2Yln rf
ro
W= 8πY r2
otoZrf
ro
dr
r= 8πY r2
otoln rf
ro
which is the same expression obtained earlier.
To obtain a numerical answer to this prob-
lem, note that Yshould be replaced with an
average value ¯
Y. Also note that ǫ1=ǫ2=
ln(140/100) = 0.336. Thus,
1.5= 206 MPa
Hence the work done is
2.79 A cylindrical slug that has a diameter of 1
in. and is 1 in. high is placed at the center of
a 2-in.-diameter cavity in a rigid die. (See the
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Compressible
matrix
F
d
1"
1"
2"
Obtain an expression for the force Fversus pis-
ton travel dup to d= 0.5 in.
The total force, F, on the piston will be
F=Fw+Fm,
where the subscript wdenotes the workpiece
and mthe matrix. As dincreases, the matrix
pressure increases, thus subjecting the slug to
transverse compressive stresses on its circum-
ference. Hence the slug will be subjected to tri-
axial compressive stresses, with σ2=σ3. Using
the maximum shear-stress criterion for simplic-
ity, we have
σ1=σ+σ2
where σ1is the required compressive stress on
the slug, σis the flow stress of the slug mate-
rial corresponding to a given strain, and given
as σ= 15,000ǫ0.4, and σ2is the compressive
stress due to matrix pressure. Lets now deter-
mine the matrix pressure in terms of d.
volume of the cavity when d= 0 is π. Hence
the original volume of the matrix is Vom =3
4π.
The volume of the matrix at any value of dis
σ2, is now given by
σ2=4(40,000)
3d=160,000
3d(psi)
The absolute value of the true strain in the slug
is given by
ǫ= ln 1
1−d,
with which we can determine the value of σfor
any d. The cross-sectional area of the workpiece
at any dis
Aw=π
4(1 −d)in2
and that of the matrix is
Am=π−π
4(1 −d)in2
The required compressive stress on the slug is
σ1=σ+σ2=σ+160,000
3d.
We may now write the total force on the piston
as
F=Awσ+160,000
3d+Am
160,000
3dlb.
The following data gives some numerical re-
sults:
d Awǫ σ F
(in.) (in2) (psi) (lb)
0.1 0.872 0.105 6089 22,070
0.2 0.98 0.223 8230 41,590
0.4 1.31 0.510 11,460 82,030
0.5 1.571 0.692 12,950 104,200
0
Displacement (in.)
0 0.1 0.2 0.3 0.4 0.5
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2.80 A specimen in the shape of a cube 20 mm on
each side is being compressed without friction
in a die cavity, as shown in Fig. 2.35d, where the
width of the groove is 15 mm. Assume that the
(20)(20)(20) = (h)(x)(x)
Fig. 2.35d on p. 67). The absolute value of the
true strain is
ǫ= ln 20
The cross-sectional area on which the force is
acting is
have σ1= 1.15Yf, or
2.81 Obtain expressions for the specific energy for
a material for each of the stress-strain curves
shown in Fig. 2.7, similar to those shown in
Section 2.12.
(a) For a perfectly-elastic material as shown in
Fig 2.7a on p. 40, this expression becomes
1
this is identical to an elastic material for
ǫ1< Y/E, and for ǫ1> Y /E it is
2Y
E2
E
=Y2
2E+Y ǫ1−Y2
E=Yǫ1−Y
2E
(e) For an elastic, linear strain hardening ma-
terial, the specific energy is identical to
an elastic material for ǫ1< Y/E and for
=Y1−Ep
Eǫ1+Epǫ2
1
2
jected to three principal (normal) stresses of σ1,
σ2= 0, and σ3=−σ1/2. What is the value of
σ1when the metal yields according to the von
Mises criterion? What if σ2=σ1/3?
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2.83 A steel plate has the dimensions 100 mm ×100
mm ×5 mm thick. It is subjected to biaxial
tension of σ1=σ2, with the stress in the thick-
ness direction of σ3= 0. What is the largest
possible change in volume at yielding, using the
von Mises criterion? What would this change
in volume be if the plate were made of copper?
2.85 An aluminum alloy yields at a stress of 50 MPa
in uniaxial tension. If this material is subjected
to the stresses σ1= 25 MPa, σ2= 15 MPa and
σ3=−26 MPa, will it yield? Explain.
According to the maximum shear-stress crite-
rion, the effective stress is given by Eq. (2.51)
on p. 69 as:
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V=πd2h
4=π(1)2(1)
4= 0.785 in3
Heat = Work = V u =VKǫn+1
1.5
(0.785)(30,000) = 4.20
Therefore, ǫ= 2.60. Using absolute values, we
2.87 A solid cylindrical specimen 100-mm high is
compressed to a final height of 40 mm in two
In the first step, we note that ho= 100 mm and
h1= 70 mm, so that from Eq. (2.1) on p. 30,
Note that if the operation were conducted in
one step, the following would result:
2.88 Assume that the specimen in Problem 2.87 has
for each step.
d1=dorho
h1
= 80r100
70 = 95.6 mm
4d2
A2=π
p. 37) so that Eq. (2.11) on p. 35 yields
σ1= 180(0.357)0.20 = 146.5 MPa
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2.89 Determine the specific energy and actual energy
expended for the entire process described in the
0.916, K= 180 MPa and n= 0.20, we have
2.90 A metal has a strain hardening exponent of
0.22. At a true strain of 0.2, the true stress
is 20,000 psi. (a) Determine the stress-strain
relationship for this material. (b) Determine
σ=Kǫn→20,000 = K(0.20)0.22
or K= 28,500 psi. Therefore, the stress-strain
relationship for this material is
ln Ao
Aneck =n= 0.22
Consequently,
Aneck =Aoe−0.22
and the maximum load is
P=σA =σultAneck.
2.91 The area of each face of a metal cube is 400 m2,
and the metal has a shear yield stress, k, of 140
load applied to the z-direction to cause yield-
ing according to the Tresca criterion? Assume
stresses in the x- and y- directions are
σx=−40,000
400 =−100 MPa
σmax −σmin =Y= 2k= 280 MPa
It is stated that σ3is compressive, and is there-
2.92 A tensile force of 9 kN is applied to the ends of
a solid bar of 6.35 mm diameter. Under load,
the diameter reduces to 5.00 mm. Assuming
uniform deformation and volume constancy, (a)
determine the engineering stress and strain, (b)
determine the true stress and strain, (c) if the
original bar had been subjected to a true stress
of 345 MPa and the resulting diameter was 5.60
mm, what are the engineering stress and engi-
neering strain for this condition?
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from Eq. (2.1) on p. 30 as:
A=9000
4(5.00)2= 458 MPa
(c) If the final diameter is df= 5.60 mm, then
the final area is Af=π
4d2
f= 24.63 mm2.
Ao
π
4(6.35)2= 268 MPa
2.93 Two identical specimens 10-mm in diameter
and with test sections 25 mm long are made
of 1112 steel. One is in the as-received condi-
tion and the other is annealed. What will be
the true strain when necking begins, and what
will be the elongation of these samples at that
instant? What is the ultimate tensile strength
for these samples?
n= 0.19. At necking, ǫ=n, so that the strain
will be ǫ= 0.08 for the cold-rolled steel and
ǫ= 0.19 for the annealed steel. For the cold-
l=enlo=e0.08(25) = 27.08 mm
or 8.32 %. To calculate the ultimate strength,
we can write, for the cold-rolled steel,
UTS = P
Ao
=UTStrueAoe−n
Ao
= UTStruee−n
stress state σ1,σ2=σ1/3, σ3= 0. Sketch the
Mohr’s circle diagram for this stress state. De-
termine the stress σ1necessary to cause yielding
by the maximum shear stress and the von Mises
criteria.
For the stress state of σ1,σ1/3, 0 the following
figure the three-dimensional Mohr’s circle:
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1
2
3
32
3−02
9σ2
9σ2
9σ2
Solving for σ1gives σ1= 125 MPa. According
to the Tresca criterion, Eq. (2.36) on p. 64 on
2.95 Estimate the depth of penetration in a Brinell
hardness test using 500-kg load, when the sam-
worked aluminum with a yield stress of 200
kg/mm2. From Fig. 2.22 on p. 52, we can esti-
expression:
HB = 2P
(πD)(D−√D2−d2)
Because the radius is 5 mm and one-half the
penetration diameter is 1.5 mm, we can obtain
αas
α= sin−11.5
5= 17.5◦
The depth of penetration, t, can be obtained
from
t= 5 −5 cos α= 5 −5 cos 17.5◦= 0.23 mm
2.96 The following data are taken from a stainless
2500 0.02
3600 0.20
4500 0.60
4600 (max) 0.86
4586 (fracture) 0.98
(2.9), (2.10), and (2.8) on pp. 33-35:
0.02 2.02 0.00995 0.0554 45.1
0.2 2.2 0.0953 0.0509 70.7
0.6 2.6 0.262 0.0431 104
0.98 2.98 0.399 0.0376 120
The true stress-true strain curve is then plotted
25
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2.97 A metal is yielding plastically under the stress
state shown in the accompanying figure.
50 MPa
20 MPa
(a) Label the principal axes according to their
proper numerical convention (1, 2, 3).
(b) What is the yield stress using the Tresca
criterion?
(c) What if the von Mises criterion is used?
(d) The stress state causes measured strains
of ǫ1= 0.4 and ǫ2= 0.2, with ǫ3not being
measured. What is the value of ǫ3?
or
2Y2= 12,600
2.98 It has been proposed to modify the von Mises
yield criterion as:
(σ1−σ2)a+ (σ2−σ3)a+ (σ3−σ1)a=C
Fig. 2.36 on p. 67).
For plane stress, one of the stresses, say σ3, is
zero, and the other stresses are σAand σB. The
yield criterion is then
(σA−σB)a+ (σB)a+ (σA)a=C
For uniaxial tension, σA=Yand σB= 0 so
that C= 2Ya. These equations are difficult
to solve by hand; the following solution was
obtained using a mathematical programming
package:
Y
B
von Mises
a=12
a=4
2.99 Assume that you are asked to give a quiz to stu-
dents on the contents of this chapter. Prepare
three quantitative problems and three qualita-
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