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Hw3.pdf
Homework 3
Plasticity, Viscoplasticity and Creep
1
The following state of stress exists at a point in an aluminum alloy component: σx = 100
MPa, σy = 100 MPa, σz = 100 MPa, and τxy = 100 MPa. What would have to be the
value of the uniaxial yield stress of the alloy, σY in MPa, that would result in yielding of
the material, under the imposed multi-axial stress state, according to the Tresca or the von
Mises yield criteria? Comment.
2
Consider a single crystal of Nickel pulled in tension with the stress axis coincident with the
[1̄23] crystallographic direction. Determine the slip system (slip plane and direction) which
is most highly stressed under these conditions. If the load increases further, which one would
be the second operational slip system?
3
A simply supported steel beam has length L = 1, breadth b = 1 and height h = 0.1. The
elastic modulus is E = 2 × 1011 and Poisson’s ratio is ν = 0.3. A tension test on a sample
of the beam material, has produced the results in the following table:
1
e(−)
0.0
0.002
0.02
0.20
σ(P a)
0.0
400 × 106
401 × 106
420 × 106
A downward distributed force Q N/m2 is applied at the center of the uppwer surface of
the beam. Assume a relatively small but reasonable value of Q then use the finite element
method to determine approximate solutions to this problem for various values of the applied
load.
If the assumed force is unable to induce plasticity, increase it gradually until plastic
deformation is obtained. What is the threshold value of the load required to produce plastic
deformation in the most highly stressed areas of the beam?
4
The following expression has been found useful in describing the results of uniaxial tensile
tests at various strain rates and temperatures and well into the plastic domain of the alloy Ti6Al-4V , commonly used in aerospace (D. Lesuer, ”Experimental Investigations of Material
Models for Ti-6Al-4V and 2024-T4 Aluminum Alloys”, Final Report, DOT/FAA/AR-00/25,
September 2000).
σ = (A + Bn )(1 + C ln ˙∗ )(1 − T ∗m )
Here, σ, , ˙∗ are, respectively, the stress, strain and strain rate (dimensionless and
referred to ˙ = 1s−1 ). The dimensionless temperature T ∗ is given by
T∗ =
T − 298
Tm − 298
where T is the actual temperature (in Kelvins) and Tm is the melting point of the alloy.
Parameter and physical property values for Ti-6Al-4V are given in the following table
2
Parameter
Value
A
1098 M P a
B
1092 M P a
n
0.93 (-)
C
0.014 (-)
m
1.1 (-)
Tm
1878 K
ρ
4428 kg/m3
Cp
580 J/kgK
a) For ˙ = 10−4 s−1 and T = 298K, compute and draw the stress-strain curve, determine
the offset yield stress (corresponding to = 0.002 and determine also the strain and stress
at the onset of necking (i.e. when dσ/d = σ).
b) Repeat the calculation in (a) above, but now for T = 1298K and ˙ = 10−4 s−1 .
c) Repeat the calculation in (a) above, but now for ˙ = 104 s−1 and T = 298K.
d) Repeat the calculation in (a) above, but now for ˙ = 104 s−1 and T = 1298K.
e) Make 3D plots of σ − − ˙ at constant temperature T and comment.
f) Make 3D plots of σ − − T at constant strain rate ˙ and comment.
5
The following form of a generalized creep law, valid for both the primary and secondary
stages of creep has been proposed
1
1
σ = K M ˙ N
where σ is the applied stress, is the strain, ˙ is the strain rate and K, M, and N are
temperature dependent material constants.
Integration of the above with respect to time at constant stress and with zero strain at
time t = 0 yields upon rearrangement, the following form of Andrade’s law for primary creep
1 = [
M
N +M σ N M
( ) ] N +M × t N +M
M
K
For the secondary stage of creep alone, the power law can be used and this can be written
as
σ
˙2 = ( )n
λ
where λ and n are additional, temperature dependent material constants.
3
Finally, the value of strain marking the transition from primary to secondary creep at
any given temperature and stress, 12 , has the form
12 = At−α
12
where t12 is the time at which the transition occurs and A and α are material parameters
which can be shown to be related to the previous materials constants by
α=
1−
n
N
n( NN+M
)−1
M
and
A = [(
1
N + M 1− n K n 1−n(N +M
)/N M
) N( ) ]
M
λ
The following material property data values have been determined for IN100 exposed to
a stress of 180 MPa at 1000 degrees Celsius.
Property
λ
n
K
N
M
Value
860 M P a − s
9.7
1450 M P a − s
10.8
9.8
Use the information above to determine the strain for transition from primary to secondary creep under the given conditions. Then use the result to construct the creep curve
( − t) for both, primary and secondary stages of creep in this case.
6
A simply supported beam has length L = 3, breadth b = 1 and height h = 0.1. The beam
density is 8000 kg/m3 and it is maintained at a uniform temperature of 1250 degrees Celsius.
Because of the high temperature involved, it may be expected that the beam will sag in the
course of time due to its own weight.
Assume the Bailey-Norton law
c = Aσ n tm
4
is applicable in this case, where the stress is in Pa and time in seconds. The values of the
parameters in the law are (in SI units) as follows:
A = 1.523 × 10−17 exp(−28, 392/(T + 273))
n = 3.15 and m = 1. where T is the temperature in degrees Celsius.
Also assume the value of the elastic modulus at the given temperature is given as a
function of temperature (in Pa) by
E = 1.5 × 1010
Use the finite element method to determine approximate solutions to this problem and
determine the maximum deflection of the beam as a function of time.
7
Consider a long, pressurized, thick walled cylinder (inner radius a, outer radius b = 1.5a).
The internal pressure is p.
Rearrange the equations given in class in dimensionless form and then plot/tabulate the
ratios σr /p and σθ /p versus the ratio r/b for the following values of the power law parameter,
n = 1; n = 2; n = 5; n = 8. Include also for comparisons the results obtained assuming that
deformation in the tube is purely elastic.
5
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