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Homework 4
MANE 7100
Chet VanGaasbeek
1
A thin metal plate (half-height = 1, half width = 0.2, thickness = 0.01 in meters) is loaded
in uniaxial tension parallel to its height that produces a nominal stress σnom = 107 Pa.
The plate has an elliptical hole (major axis a, minor axis b = 0.01) at its center with the
major axis perpendicular tothe loading direction. Assume isotropic linear elastic behavior
(E = 1011 Pa and ν = 0.3. Develop a finite element model of the loaded plate and use
it to determine stress concentration factor due to the hull as a function of the ratio a/b.
Consider values 1¡a/b¡10. Compare to Roark’s.
Using a shell traction load of 100 Pa, the resulting S22 (y directional stresses) are
plotted and shown below for ratios a/b = 1, a/b = 5, a/b = 10, in that respective order.
S, S22
SNEG, (fraction = −1.0)
(Avg: 75%)
+2.945e+04
+2.695e+04
+2.445e+04
+2.195e+04
+1.946e+04
+1.696e+04
+1.446e+04
+1.196e+04
+9.461e+03
+6.963e+03
+4.464e+03
+1.965e+03
−5.332e+02
ODB: Job−1.odb
Y
Z
X
Abaqus/Standard 6.12−1
Sun Dec 09 19:28:23 Eastern Standard Time 2012
Step: Step−1
Increment
1: Step Time = 1.000
Primary Var: S, S22
Deformed Var: U Deformation Scale Factor: +9.988e+05
1
S, S22
SNEG, (fraction = −1.0)
(Avg: 75%)
+4.706e+04
+4.312e+04
+3.917e+04
+3.523e+04
+3.128e+04
+2.733e+04
+2.339e+04
+1.944e+04
+1.550e+04
+1.155e+04
+7.604e+03
+3.658e+03
−2.880e+02
ODB: Job−1.odb
Y
Z
X
Abaqus/Standard 6.12−1
Sun Dec 09 19:26:16 Eastern Standard Time 2012
Step: Step−1
Increment
1: Step Time = 1.000
Primary Var: S, S22
Deformed Var: U Deformation Scale Factor: +9.990e+05
S, S22
SNEG, (fraction = −1.0)
(Avg: 75%)
+8.114e+04
+7.436e+04
+6.759e+04
+6.082e+04
+5.405e+04
+4.728e+04
+4.051e+04
+3.374e+04
+2.696e+04
+2.019e+04
+1.342e+04
+6.650e+03
−1.209e+02
ODB: Job−1.odb
Y
Z
X
Abaqus/Standard 6.12−1
Sun Dec 09 19:22:26 Eastern Standard Time 2012
Step: Step−1
Increment
1: Step Time = 1.000
Primary Var: S, S22
Deformed Var: U Deformation Scale Factor: +9.992e+05
2
A large steel sheet has a central crack of length 2a = 0.04 m and undergoes catastrophic
fracture at a fracture stress σF = 480 MPa. A second large sheet of the same steel has a
2
central crack of length 2a = 0.1m. Use the fracture stress equation in the form:
C
σF = √
a
where C is a constant and calculate its fracture stress.
The first plate has:
√
√
√
C = σF a = 480 × 106 0.02 = 6.788 × 107 Pa m
thus in the second plate:
C
6.788 × 107
√
= 303.5 MPa
σF = √ =
a
0.05
3
A forensic examination of a fatigue specimen showed that an initial crack of size 0.003m
grew to a final size of 0.008 m by fast fracture during testing. Use the Paris equation in the
form:
da
= C(∆K)m = C(∆σ)m (πa)m/2
dN
with m = 3 to obtain by integration an expression relating the initial and final crack
sizes with the stress amplitude and number of cycles to failure. Then, use the resulting
expression to determine the increase in fatigue life that can be expected if the initial crack
size is reduced to 0.001m without changing the final crack size. Give the increased life at
the ratio of the number of cycles to failure for the specimen with the smaller initial crack
to that of the original specimen.
Using the Paris equation and integrating between N = 0 and N = Nf for a between
0.003 and 0.008 yields:
2
1−m/2
1−m/2
− a0
) = −2(0.008−1/2 − 0.003−1/2 ) = 14.15 = C(∆σ)3 (π)3/2 Nf
(a
2−m f
With a0 = 0.001:
2
1−m/2
1−m/2
) = −2(0.008−1/2 − 0.001−1/2 ) = 40.88 = C(∆σ)3 (π)3/2 Nf
− a0
(a
2−m f
with Nf the final number of cycles at failure. The relative increase is the ratio between the
two numbers of cycles:
C(∆σ)3 (π)3/2 Nf′
C(∆σ)3 (π)3/2 Nf
=
Nf′
= 40.88/14.15 = 2.89
Nf
or an increase in fatigue life of 189%.
3