Carlos Sarria
MANE6940
Final Exam
Problem1
a. Use the above expressions and the data in the table to calculate the distributions of stresses
in the disk. Then use Hooke’s law to obtain the strains and the radial displacement u® in the
disk. Write down the calculated values of all these quantities at r=a and r=b.
The given equations (see below), allow us to find the stress distributions in the radial and tangential
directions.
Substituting the given values into these equations the following results are obtained:
r=a
r=b
Exact solution
Sr Sp
er
ep
ez
u( r )
0 1.62E+07
-0.00012
0.0041
-0.00012 4.05E-05
0 5.12E+06 -3.80E-05 0.00013 -3.80E-05 3.84E-05
Using Hooke’s Law (see below) we can calculate the strain tensor components:
𝜀𝑟 =
1
∗ (𝜎𝑟 − 𝜈 ∗ 𝜎𝜑 )
𝐸
𝜀𝜑 =
1
∗ (𝜎𝜑 − 𝜈 ∗ 𝜎𝑟 )
𝐸
𝜀𝑧 = −
𝜈
∗ (𝜎𝑟 + 𝜎𝜑 )
𝐸
The expression below was obtained by integrating the εr equation above with respect to r:
b. Build a finite element model of the system, solve and compare the results obtained with the
exact solution.
Using COMSOL, the following results were obtained at r=a:
FEA Solution
Sr
Sp
er
ep
ez
u( r )
r=a
1055 1.62E+07 -1.20E-04
4.10E-04 -1.20E-04
4.10E-05
r=b
0.81 5.06E+06 -3.80E-05
1.30E-04 3.90E-05
3.80E-05
The results are very close with the exception of the radial stress at r=a. This is probably due to the
proximity to the constrained edge.
c. Use the stress equations to determine the rotation speed N that will first produce yielding in
the disk. Write down the calculated value.
To determine the angular velocity required to cause yielding of the disk we need to set the σφ equation
defined above equal to the yield strength of the material, 60MPa. Then we solve for omega and obtain:
ω=513rad/s or 4899RPM.
d. Extend the finite element model by adding plasticity and use it to determine the rotation
speed at which the elastic-plastic boundary reached the middle of the disk. Write down the
calculated value.
This was done using a trial and error gradually increasing the rotating speed until the plastic stress
reached the middle of the beam. The figure below shows that the beam sees stresses right at the yield
point until about 0.11m, which is the middle of the beam. The velocity that was input into COMSOL was
680rad/s or 6493RPM.
Problem 2
a. Write down the primary slip system to be activated and the value of the resolved shear
stress.
To determine the primary slip plane, we can calculate the Schmidt Factors of all 12 possible slip systems
to determine the largest one. The Schmidt Factor is defined as: Ms=cos(phi)cos(lambda). The table
below summarizes all the values of factors. It can be seen that the highest factor are obtained with the
system: {1 -1 1} <0 1 1> with a Schmidt Factor of 0.47
Slip Plane Indices
h1
k1
1
1
l1
1
Slip Direction Indices
h3
k3
l3
0
-1
1
Tensile Stress Axis
h2
k2
2
1
l2
3
1
1
1
1
0
-1
2
1
3
1
1
1
-1
1
0
2
1
3
-1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
0
-1
1
0
1
0
-1
-1
1
1
0
1
2
2
2
2
1
1
1
1
3
3
3
3
-1
1
1
-1
0
-1
2
1
3
-1
1
1
1
1
0
2
1
3
1
-1
1
0
1
1
2
1
3
1
-1
1
1
0
-1
2
1
3
1
-1
1
-1
-1
0
2
1
3
Schmidt
Factor
0.34992
711
0.17496
355
0.17496
355
0
0
0
0.11664
237
0.29160
592
0.17496
355
0.46656
947
0.11664
237
0.34992
711
To calculate the residual value of the shear stress we find the ratio of tau/sigma=cos(lambda)cos(phi).
Tau/sigma=0.47. This is the Schmidt factor.
b. As the crystal deforms, the crystal rotates trying to align the slip direction with the stress
axis. Write down which will be the secondary slip system to be activated.
The secondary slip system will be the one resulting in the second highest Schmidt factor. The table
above shows that the secondary systems would be: {1 1 1} <0 -1 1>, with an Ms of 0.35, and {-1 -1 1} <-1
1 0> also with an Ms of 3.5.
Problem 3
a. Assume T=800C and calculate the deflection u(x,t) for 0 <=t<=3600s. What is the maximum
deflection at t=3600s?
After inputting all the given data and carrying out the double integral to calculate IN we obtain an
expression for u(x,t):
U(x,t)=
By setting T=800C, x=L and t=3600s we obtain a maximum deflection of -4.5e-4m
b. Assume T=900C and determine the maximum deflection at t=3600s.
Following the same procedure as above we obtained a maximum deflection of -4.3e-3m.
Problem 4
a. Use the above to determine the fracture stress of the sample. From your result, will the
sample fracture before yielding, or yield before fracturing?
After inputting all the given information we can calculate the fracture stress of the sample, σf, which is
4.69e8Pa. This is lower than the material’s yield strength of 9e8Pa meaning that the material will
fracture before it yields.
b. Build a finite element model, compute and comment on your result.
The COMSOL model predicts a maximum stress of 2e10Pa at the tip of the crack. In reality, the
specimen will never reach this value because the crack will start propagating. If we want to calculate
the stress concentration, Kt, we need to first calculate the nominal stress given by the equation:
𝜎=
𝑃
𝐴
Where A is the cross sectional area: 0.1*0.03=0.003m^2. Therefore the nominal stress is 4e8Pa. The
stress concentration is defined by:
𝐾𝑡 =
𝜎𝑚𝑎𝑥
𝜎𝑛𝑜𝑚
Substituting the COMSOL max stress we find a Kt of 50.