Homework answers / question archive / MECHANICAL ENGINEERING DEPARTMENT Consider the [90/+θ/-θ]S laminate with the lamina elastic properties and strengths given in Table 1 below and the lamina thermal expansion behavior listed in Table 2 below and plotted in Figure 1 below

MECHANICAL ENGINEERING DEPARTMENT Consider the [90/+θ/-θ]S laminate with the lamina elastic properties and strengths given in Table 1 below and the lamina thermal expansion behavior listed in Table 2 below and plotted in Figure 1 below

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MECHANICAL ENGINEERING DEPARTMENT

1. Consider the [90/+θ/-θ]S laminate with the lamina elastic properties and strengths given in
   Table 1 below and the lamina thermal expansion behavior listed in Table 2 below and
   plotted in Figure 1 below. A high temperature resin system is used to fabricate this
   laminate, with the processing temperature equal to 400oF, at which the laminate is
   considered stress-free. If the laminate is slowly cooled to 0oF, and temperatureindependent elastic response is assumed, determine the lamina stress vector for each of
   the three unique plies (90o, +θ and –θ) at 0oF. Assume a uniform temperature is achieved
   through the thickness of the laminate. Perform these calculations for the range of θ values
   equal to 10, 20, and 30 degrees.
   a) Calculate and then plot as a function of θ, for θ ranging from 10 to 30 degrees, the
   maximum lamina fiber axial stress, the maximum lamina transverse stress, and the
   maximum lamina shear stress.
   b) Using the Hashin Failure Criteria, determine which (if any) of the [90/+θ/-θ]S layups will
   survive cool down to 0oF, for the range of θ values equal to 10, 20, and 30 degrees.
   Table 1. Constituent material elastic, strength and physical properties of Problem #1
   intermediate modulus carbon/epoxy lamina.
2.  

Elastic Property	Value
E1	25 x 106 psi
E2	1.7 x 106 psi
G12	0.65 x 106 psi
ν12	0.30
ρ	0.056 #/in3
σAtu	110.0 x 103 psi
σAcu	110.0 x 103 psi
σTtu	9.0 x 103 psi
σTcu	20.0 x 103 psi
τAu	11.0 x 103 psi
vf	0.6
Ply thickness	0.0052 inch

ME 7502 – Reinforced Composite Structures Fall 2022 Mid-Term Examination
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Table 2. Thermal strain versus temperature data for Problem #1 intermediate modulus
carbon/epoxy lamina.

Temperature (oF)	(?L/L)AXIAL	(?L/L)TRANSVERSE
-100	8.00E-05	-2.00E-03
-80	6.40E-05	-1.60E-03
0	0.00E-06	0.00E-06
100	-8.00E-05	2.00E-03
160	-1.28E-04	2.20E-03
300	-2.40E-04	2.67E-03
400	-2.40E-04	3.00E-03

Figure 1. Thermal expansion behavior of intermediate modulus carbon/epoxy lamina.
-3.00E-03
-2.00E-03
-1.00E-03
0.00E+00
1.00E-03
2.00E-03
3.00E-03
4.00E-03
-200

-1

00

0

10

0

20

0

30

0

40

0

500
Thermal Strain (in/in)
Temperature (F)
Thermal Expansion Behavior of Intermediate Modulus
Carbon/Epoxy Lamina
(dL/L)axial (dL/L)transverse
ME 7502 – Reinforced Composite Structures Fall 2022 Mid-Term Examination
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2. A [0/+60/-60]S laminate with the ply properties listed in Table 4 below is to be subjected to a
temperature change from its initial temperature of 75oF. This temperature change can be
expressed as a nonlinear temperature change function through the thickness of the
laminate, by the equation
Δ?? = ????2 + ???? + ??
where z is the through thickness coordinate, and the constants a, b and c are given by
?? =
?100?3?
?2 ∗ ?????????2 = 308,185.4
?? = -2 ∗ (?) ∗ ?? = -9615.3845
?? = -150 + ?2 ∗ ?? = -75
with h = 3 * tply or half the thickness of the 6-ply laminate. Using this quadratic function for
?T, we can obtain the following temperature change distribution through the thickness of
the laminate, as detailed in Table 3 below.
Table 3. Temperature change distribution for [0/60/-60]S laminate of Problem #2.

i	zi	?T(zi)
0	-0.0156	150.000
1	-0.0104	58.333
2	-0.0052	-16.667
3	0.0000	-75.000
4	0.0052	-116.667
5	0.0104	-141.667
6	0.0156	-150.000

a) For the nonlinear temperature change distribution as detailed in Table 3 above, determine
the stresses in the lamina coordinate system at both the top and bottom in each of the 0o,
+60o and -60o plies.
b) Given the lamina strengths in Table 4 below, determine if the laminate subjected to this
nonlinear temperature change distribution could be expected to survive with no excessive
lamina stresses and therefore with no damage to the laminate.
c) Assuming the same initial stress-free temperature of 75oF and by subjecting this same
[0/+60/-60]S laminate to a simple uniform temperature of -75oF, answer the question “Is the
nonlinear through thickness temperature gradient more stressing on the laminate than the
uniform through thickness temperature of -75oF?”
ME 7502 – Reinforced Composite Structures Fall 2022 Mid-Term Examination
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Table 4. Constituent material elastic, strength and physical properties of Problem #2 intermediate
modulus carbon/epoxy lamina.

Property	Lamina Value
E1	25 x 106 psi
E2	1.7 x 106 psi
G12	1.3 x 106 psi
ν12	0.3
α1 (-200oF to 200oF)	-0.3 x 10-6 in/in/oF
α2 (-200oF to 200oF)	19.5 x 10-6 in/in/oF
σLtu	110 x 103 psi
σTtu	4.0 x 103 psi
τLu	9.0 x 103 psi
σLcu	110 x 103 psi
σTcu	20 x 103 psi
Ply thickness	0.0052 inch

3. Consider the quasi-isotropic [0/45/90/-45]S laminate containing lamina with the mechanical
properties listed in Table 5 below.
Table 5. Lamina material properties for [0/45/90/-45]S laminate of Problem #3.

Elastic Property	Value
E1	25 x 106 psi
E2	1.7 x 106 psi
G12	0.65 x 106 psi
ν12	0.30
ρ	0.056 #/in3
σAtu	110.0 x 103 psi
σAcu	110.0 x 103 psi
σTtu	7.0 x 103 psi
σTcu	7.0 x 103 psi
τAu	9.0 x 103 psi
vf	0.6
Ply thickness	0.0052 inch

The laminate is fabricated at the (stress-free) temperature of 250oF, and for all temperature
changes of interest from 250oF, in this problem the lamina coefficients of thermal expansion
are
α11 = -0.653x10-6 in/in/oF; α22 = 7.5x10-6 in/in/oF.
The laminate is subjected to multiple cyclic thermal loading blocks, each of which contains
ME 7502 – Reinforced Composite Structures Fall 2022 Mid-Term Examination
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the three (3) thermal cycle load steps shown in Table 6 below. Using the appropriate S-N
curves from Figures 2 below and utilizing a cumulative damage approach, perform the
calculations to estimate the maximum allowable number of cyclic loading blocks (NB) to
which the laminate may be subjected prior to the occurrence of any ply fatigue failures.
Table 6. A single thermal cyclic loading block with three (3) thermal cycle load steps for quasiisotropic [0/45/90/-45]S laminate.

Thermal Cycle Load Step	TMIN (oF)	TMAX (oF)	No. of Cycles/Step ni
1	-230	10	100
2	-190	140	100
3	-110	214	100

Hint: For each thermal cycle load step, compute the maximum and minimum stresses
corresponding to TMIN and TMAX respectively (laminate is stress-free at 250oF). From these
compute Ri and using the appropriate curves from Figure 2, find Ni.
Note 1: Find the total number of cycles to failure by using the simple Palmgren-Miner
approach
? ??????????(???? ??) = ???? ∗ {????11 + ????22 + ????33} = 1
where NB = the number of cyclic loading blocks, ni = the number of cycles at stress
amplitude σI and Ni = the lifetime under constant amplitude σI as defined by the SN curve.
Note 2: The data points corresponding to all of the Figure 2 fatigue S-N curves are provided
in the “#2 - Data” worksheet of the spreadsheet file “ME 7502 Fall 2022 Mid-Term
Exam Fatigue Problem.xls.”
ME 7502 – Reinforced Composite Structures Fall 2022 Mid-Term Examination
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Figure 2a. Fatigue S-N curves at R = 0.1, 0.25 and 0.5 for C/Ep unidirectional in
longitudinal [0] tension and compression.
0
20
40
60
80
100
120
1.0E+00 1.0E+01 1.0E+02 1.0E+03 1.0E+04 1.0E+05 1.0E+06 1.0E+07
Stress, ksi
Cycles, N
C/Ep, [0] Tension & Compression Fatigue S-N Curves
R = 0.1
R = 0.25
R = 0.5
ME 7502 – Reinforced Composite Structures Fall 2022 Mid-Term Examination
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Figure 2b. Fatigue S-N curves at R = 0.1, 0.25 and 0.5 for C/Ep unidirectional in transverse [90]
tension and compression.
8 7 6 5 4 3 2 1 0
1.0E+00 1.0E+01 1.0E+02 1.0E+03 1.0E+04 1.0E+05 1.0E+06 1.0E+07
Stress, ksi
Cycles, N
C/Ep, [90] Tension & Compression Fatigue S-N Curves
R = 0.1
R = 0.25
R = 0.5
ME 7502 – Reinforced Composite Structures Fall 2022 Mid-Term Examination
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Figure 2c. Fatigue S-N curves at R = 0.1, 0.25 and 0.5 for C/Ep unidirectional in longitudinal [0]
shear.
9 8 7 6 5 4 3 2 1 0
10
1.0E+00 1.0E+01 1.0E+02 1.0E+03 1.0E+04 1.0E+05 1.0E+06 1.0E+07
Stress, ksi
Cycles, N
C/Ep, [0] Shear Fatigue S-N Curves
R = 0.1
R = 0.25
R = 0.5
ME 7502 – Reinforced Composite Structures Fall 2022 Mid-Term Examination
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4. Consider the uniformly loaded simply supported beam of rectangular cross-section shown
in Figure 3 below. The beam is composed of unidirectional composite material, whose
properties are listed in Table 7 below. The fiber direction of the unidirectional composite is
aligned with the beam X axis. For this beam, b = 0.5 inch and h = 4 inch.
a. Generate the maximum uniform load (wmax) versus beam length L curves for this
beam, considering that failure can occur by excessive shear stresses, excessive
axial stresses due to bending, or by lateral instability.
b. Plot the maximum deflection δmax as a function of beam length L for this beam,
including the effects of shear deformation.
Figure 3. Sketch of uniformly loaded simply supported beam of rectangular cross-section.
Note: Lateral instability of a uniformly loaded simply supported orthotropic beam of
rectangular cross-section is governed by the expression
CIE
L
wL
CR 2 XX Z
85.12
)( =
where the torsional rigidity C is given by the equation
63.01( )
3
3
hG
hbG bG
C
XY
XZ XZ
⋅ ⋅
= -
L
h
b

w (#/inch)

X

Z
ME 7502 – Reinforced Composite Structures Fall 2022 Mid-Term Examination
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Table 7. Lamina material properties for unidirectional composite material of Problem #4.

Elastic Property	Value
E1	25 x 106 psi
E2	1.7 x 106 psi
G12	0.65 x 106 psi
ν12	0.30
ρ	0.056 #/in3
σAtu	110.0 x 103 psi
σAcu	110.0 x 103 psi
σTtu	7.0 x 103 psi
σTcu	20.0 x 103 psi
τAu	9.0 x 103 psi
vf	0.6
Ply thickness	0.0052 inch