Materials 101 Problem Set #4
1.) The tensile stress-strain curve for a cylindrical specimen of brass is reproduced in Figure 1.
A. Find the following:
a) The work done to fracture per unit volume
The work done to fracture per unit volume is the area underneath the Nominal Stress vs. Nominal
Strain curve. If we had an equation to describe this curve, we could integrate the equation from zero
to the strain at fracture. Since we don’t have this equation available, estimate the work done by
estimating the area beneath the curve. Work done to fracture per unit volume ~ 20.5 boxes * 10
J/m^3/box = 205 J/m^3.
b) The Young's modulus of brass
The Young’s modulus is found from the slope of the Nominal Stress vs. Nominal Strain curve in in the
linear regime where nominal strain is very small. In this case you should estimate the slope of the
curve. E ~100 MPa/.0025 = 40,000 MPa.
c)The ultimate tensile stress
The ultimate tensile stress (UTS) occurs where the Nominal Stress vs. Nominal Strain curve shows a
maximum. In this case the UTS ~530 MPa.
d) The 0.2% offset yield stress
To find the 0.2% offset yield stress, extrapolate a line with the same slope as the linear regime (which
is equal to the Young’s modulus) from the x axis of Nominal Stress vs. Nominal Strain curve upwards
until it intersects with the curve. The stress at this intersection is the 0.2% offset yield stress. Here it
is ~ 150 MPa.
e)The proportional limit
The proportional limit occurs where the stress and strain stop being linear or proportional. This is
sometimes used as a measure the yield stress (but not in this class) and is ~ 95 MPa.
B. Suppose a similar specimen is given a tensile prestrain of 0.2. On subsequent testing this specimen you
would find an approximate yield stress of
492 MPa
?
When the specimen is loaded again, the stress would be proportional to the strain, starting at the 0.20
strain, until it hits the normal stress-strain curve. Applying this to the first σn-εn plot given, a nominal
yield stress based on the original cross-sectional area Ao would be 410 MPa. However if as usual the
cross-sectional area Ao1 is measured before the second test we need to find this area in order to get the
correct stress from the value we read off the original stress-strain curve. We know this stress (the true
stress) = 410 Ao/Ao1 , where Ao/Ao1 = Lo1/Lo = 1+εn = 1.20. The yield stress will be thus 492 MPa.
Another way to look at this problem is as follows: The sample will not start to yield again (undergo
permanent plastic deformation) until it reaches a stress equivalent to that stress it was experiencing the
last time it was yielding. In this case yielding last occurred at a prestrain of 0.2. We know that
permanent plastic deformation occurred during the prestrain, so we know that the area of the sample
changed from Ao to Ao1.
Yield stress (σy) is the point where elastic deformation stops and permanent plastic deformation begins.
You can calculate the yield stress after the prestrain, by converting the nominal stress of the prestrain of
0.2 (this is the strain at which the sample last yielded) to what it would be with the new area Ao1. (As I
stated earlier, the yield stress will occur at the a stress equivalent to the stress value it last yielded at.
This calculation finds that equivalent stress.) However, as stated earlier, the sample has a new area, Ao1.
So σy = F/Ao*( Ao/Ao1)= σn*( Ao/Ao1). Note, this is the same thing as calculating the true stress at a
nominal strain of 0.2.
What is the true strain and true stress at a point of a curve in Figure 1 that corresponds to 0.35 nominal strain?
At the nominal strain of 0.35 we can find the true strain by using the fact that true strain = ln(L/Lo) =
ln(1+ εn ) = 0.30. To find the true stress you need to multiply the nominal stress 520 MPa by Ao/A =
1+ εn = 1.35. This gives a true stress of 702 MPa. Why can't you compute the real ε and σ at the nominal
strain of 0.45 from this curve? This is because at nominal strain of 0.40,where the nominal stress-nominal
strain curve is a maximum, necking starts. Hence at 0.45 nominal strain, necking has already occurred.
Because the curve is obtained by monitoring the variation of nominal strain from changes in the length,
beyond 0.40 strain the data cannot be used to calculate the true strain.
1
600
Nominal Stress (MPa)
500
400
X
Brass (25 °C)
300
Ultimate Tensile Stress
200
100
0
0
0.1
0.2
0.3
0.4
0.5
0.6
Nominal Strain
250
Nominal Stress (MPa)
200
Low strain region of the above curve
150
0.2% offset yield stress
100
The proportional limit
50
0
0.000
0.002
0.005
0.010
0.015
0.020
0.025
0.030
0.035
Nominal Strain
2
equivalent
Problem 5
5.) The true stress-strain curve for the steel in question 4 is given by
.
σt = 897 (εt) 0.3 ( ε t) 0.01 (stress in MPa)
i) What is the true strain at the ultimate tensile tensile stress? Assume τ=0 for
part i).
ii) What is the ultimate tensile stress for the steel at a εt = 1 sec -1 ?
Answer:
i)
The ultimate tensile stress occurs at the maximum in the force F vs.
true strain curve. Since F = σt A, we can find the maximum with
true strain from dF/dεt = 0 = A(dσt / dε ) + σt(dA/dεt) where A =
Aoexp(-εt). Taking the derivatives we find that εt UTS = 0.3 the
strain hardening coeffficient.
Additionally using the following relationship where σt = Kεm,
ε = m at necking since the UTS corresponds to the point at which
the deformation starts to localize, forming a neck.
ii)
From part i) (σt) at the UTS(strain) = 897(0.3)0.3(1)0.01 = 625 MPa
=σt. To get the ultimate tensile stress, which is the nominal stress,
we need to multiply σt by A/Ao, which comes from the following
equation. σt=σ * (Ao/A) . So we can solve for the nominal stress
(σ)UTS = 625 *exp(-εt) = 625(.714) = 463 MPa.
Materials 101 Problem Set #4
6.) The hotel balcony designed in Problem 4 collapses, killing 200 people. It is suspected
that the supporting rod was the cause of (or at least a contributing factor in) the failure.
You examine a failed connecting rod and find the dimensions in the sketch below. There
is no evidence of any torsional strain, implying τ during the failure was zero.
0.95 inches in diameter
0.50 inches in diameter
i) What was the local true strain at fracture?
ii) Is the appearance of the failed supporting rod consistent with its being constructed
with the steel in part b above? Give quantitative reasoning.
Problem Set #4 Question #6
Part i)
Ao = A (1+En) where En is the nominal strain.
Solve for nominal strain and convert to true strain according to:
E(true) = ln (1+En).
En=Ao/A-1=.95/.5-1= 0.9 Ao/A=(1/0.5)^2 = 4
Et=ln(1+0.9)=0.64
ln(Ao/A) = 1.38
Part i)
This unnecked diameter, which is reduced from the
original 1 inch diameter, is the key. This means that the true strain at
the UTS (the last strain at which the sample was deforming uniformly) was
-2ln(0.95) = 0.102. But the strain hardening exponent given in question 3
(which should equal the strain at the UTS) is given as 0.3. SO it looks as
if the wrong steel was used.
1