MAT 33 – Fall 2008
Homework #7 – Solutions
Due:
10/27/08
1.)
For a brass alloy under tension, the following engineering stresses produce
the corresponding plastic engineering strains, prior to necking.
Engineering Stress (MPa)
Engineering Strain
315
0.105
340
0.220
(i)
Calculate the
true
stress,
σ
T
and
true
strain and
ε
T
for both of these
loading situations
(ii)
The true stress and true strain are related by
σ
T
= K
ε
T
n
.
Calculate
the strain hardening exponent, n, and the constant K for this
material.
(iii)
Calculate the
engineering stress
necessary to produce an
engineering
strain
of 0.32 in this material.
(i) Since
σ
T
=
σ
(1 +
ε
) then
σ
T
1
= (315 MPa)(1+ 0.105)= 348 MPa
σ
T
2
= (340 MPa)(1+ 0.220)= 415 MPa
Similarly for strains, since
ε
T
= ln(1 +
ε
) then
ε
T
1
= ln (1 + 0.105)= 0.09985
ε
T
2
= ln (1 + 0.220)= 0.19885
(ii)
Taking logarithms of
σ
T
= K
ε
T
n
we get
log
σ
T
= log K + n log
ε
T
which allows us to set up two simultaneous equations for the above pairs of true
stresses and true strains, with
K
and
n
as unknowns.
Thus
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log (348) = log K + n log(0.09985)
)
log (415) = log K + n log(0.19885)
)
Solving for these two expressions yields
K
= 628 MPa
and
n
= 0.256.
(iii)
Converting
ε
= 0.32 to true strain
0.278
=
0.32)
+
(1
ln
=
T
ε
The corresponding
σ
T
to give this value of
ε
T
[using
σ
T
= K
ε
T
n
] is just
MPa
453
=
)
MPa)(0.278
(628
=
K
=
0.256
n
T
T
ε
σ
Now converting this
σ
T
to an engineering stress
MPa
343
=
32
.
0
1
453
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 Spring '07
 keith
 Deformation, Tn, MPa

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