ME 7502 Fall 2012 Mid-Term Exam Problem 1 Solution
Unit definitions: ksi103lbf
in2 Msi 1000 ksi
The lamina elastic constants and thickness for this C/Epoxy lamina are:
E1 25 Msi E2 1.7 Msi G12 0.65 Msi 12 0.3
21E2
E112 21 0.02 layer thickness t 0.0052 in
let [Qt] = Q0 for Mathcad calculations,
Tensor [Qt] matrix:
V 1 12 21 intermediate calculation for [Qt] matrix:
Q0
E1
V
12E2
V
0
21E1
V
E2
V
0
0
0
2 G12
Q0
25.154
0.513
0
0.513
1.71
0
0
0
1.3
Msi
For transformation matrix, let [T] for a given angle be represented by T
also let c cos deg s sin deg
T c 2
s 2
s c
s 2
c 2
s c
2 s c
2 s c
c 2 s 2
Let the off-axis stiffness matrix [Qxt] be represented in Mathcad by Qbar(), where isthe angle of the layer:
Qbar T Q0 T 1
T 0( )
1
0
0
0
1
0
0
0
1
T 90( )
0
1
0
1
0
0
0
0
1
T 45( )
0.5
0.5
0.5
0.5
0.5
0.5
1
1
0
T 45( )
0.5
0.5
0.5
0.5
0.5
0.5
1
1
0
Qbar 0( )
25.154
0.513
0
0.513
1.71
0
0
0
1.3
Msi Qbar 45( )
7.623
6.323
5.861
6.323
7.623
5.861
11.722
11.722
12.919
Msi
Qbar 45( )
7.623
6.323
5.861
6.323
7.623
5.861
11.722
11.722
12.919
Msi Qbar 90( )
1.71
0.513
0
0.513
25.154
1.429 10 15
0
2.858 10 15
1.3
Msi
Since [At] matrix is
1
N
n
Qbar_n tn
, the [At] matrix for [0/+45/-45/90]
S is:
A0_p45_m45_90_s Qbar 0( ) t Qbar 45( ) t Qbar 45( ) t Qbar 90( ) t 2
A0_p45_m45_90_s
0.438
0.142
0
0.142
0.438
0
0
0
0.296
Msi in
Next we construct the stress resultant vector for our unit load vector:
Unit_N
1
1
0.5
psi in
Next we calculate the global strain vector:
_global_Nx A0_p45_m45_90_s1 Unit_N
_global_Nx
1.724 10 6
1.724 10 6
1.691 10 6
The lamina strain vectors are:
eps0_Nx T 0( ) 1_global_Nx epsp45_Nx T 45( ) 1
_global_Nx
eps90_Nx T 90( ) 1_global_Nx epsm45_Nx T 45( ) 1
_global_Nx
a) The lamina stress vectors are:
0_Nx Q0 eps0_Nx 0_Nx
44.244
3.833
2.198
psi
p45_Nx Q0 epsp45_Nx p45_Nx
85.901
1.809
0
psi
m45_Nx Q0 epsm45_Nx m45_Nx
2.587
5.857
0
psi
90_Nx Q0 eps90_Nx90_Nx
44.244
3.833
2.198
psi
(i) Using the maximum stress criterion, we first tabulate maximum lamina level stresses for the fiber axial direction, the fiber transverse directionand the shear direction. These results are as follows:
The maximum fiber axial direction stress is 85.73 psi in the +45o pliesThe maximum fiber transverse direction stress is 5.857 psi in the -45o pliesThe maximum shear stress is 2.198 psi in both the 0 o and 90o plies.
Computing load factors to produce initial failure,
(1) by fiber axial direction failure:
Xfa110 ksi
85.901 psi Xfa 1.281 103
(2) by fiber transverse direction failure:
Xft4 ksi
5.857 psi Xft 682.943
(3) by shear:
X9 ksi
2.198 psi X 4.095 103
Since Xft < Xfa < X , using the maximum stress criterion, the initial failure will be by a matrix mode transverse tensile failure in the -45 o plies,occuring at the load vector:
Nfailure_MS Xft Unit_N Nfailure_MS
682.943
682.943
341.472
psi in _global _global_Nx Xft_global
1.177 10 3
1.177 10 3
1.155 10 3
(ii) Using the Hashin criterion, for the 0o and 90o plies, for the fiber tensile mode, the relevant ply failure equation is
Xf2 {(tu)2 + (A
u)2}= 1. Using this equation, we find
Xf_0_901
44.244 psi
110 ksi
2 2.198 psi
9 ksi
2
Xf_0_90 2.125 103
For the 0o and 90o plies, the tensile matrix mode, the relevant ply failure equation is Xf2 {(T)2 + (12/u)2} = 1. Using this equation, we
find
Xm_0_901
3.833 psi
4 ksi
2 2.198 psi
9 ksi
2
Xm_0_90 1.011 103
For the +45o plies, for the fiber tensile failure mode,
Xf_p451
85.901 psi
110 ksi
2 0 psi
9 ksi
2
Xf_p45 1.281 103
For the -45o plies, for the fiber transverse or matrix tensile mode,
Xm_m451
5.857 psi
4 ksi
2 0 psi
9 ksi
2
Xm_m45 682.943
Since the Xm_m45 factor has the lowest value, by the Hashin criterion as well, the initial laminate failure will be by matrix transverse tension in
the -45o plies, occuring at the load:
Nfailure_Hashin Xm_m45 Unit_N Nfailure_Hashin
682.943
682.943
341.472
psi in
b) Now we remove the -45o plies, re-compute the [A] matrix, recompute the incremental strain and incremental stress vectors, and then determinethe load factor causing the next failure along with the corresponding failure mode:
Since [At] matrix is
1
N
n
Qbar_n tn
, the [At] matrix for [0/+45/90]
S is:
A0_p45_90_s Qbar 0( ) t Qbar 45( ) t Qbar 90( ) t 2
A0_p45_90_s
0.359
0.076
0.061
0.076
0.359
0.061
0.122
0.122
0.161
Msi in
Next we calculate the incremental global strain vector:
_incr_global_Nx A0_p45_90_s1 Unit_N
_incr_global_Nx
1.814 10 6
1.814 10 6
1.728 10 6
The incremental lamina strain vectors are:
eps0_incr_Nx T 0( ) 1_incr_global_Nx epsp45_incr_Nx T 45( ) 1
_incr_global_Nx eps90_incr_Nx T 90( ) 1_incr_global_Nx
The incremental lamina stress vectors are:
0_incr_Nx Q0 eps0_incr_Nx 0_incr_Nx
46.568
4.034
2.246
psi
p45_incr_Nx Q0 epsp45_incr_Nx p45_incr_Nx
89.136
1.966
1.049 10 15
psi
90_incr_Nx Q0 eps90_incr_Nx90_incr_Nx
46.568
4.034
2.246
psi
The lamina stress vectors prior to the addition of the incremental stresses are:
0 0_Nx Xft 0
3.0216 104
2.6177 103
1.5009 103
psi
p45 p45_Nx Xft p45
5.8666 104
1.2353 103
6.1528 10 13
psi
90 90_Nx Xft 90
3.0216 104
2.6177 103
1.5009 103
psi
The largest fiber direction stress will be in the +45o plies; the factor causing fiber direction failure will be determined from the equation:
Xfa110000 58666
89.136 Xfa 575.906
The largest transverse fiber direction stress will be in the 0o and 90o plies; the factor causing transverse direction matrix failure will be foundfrom the equation:
Xft4000 2617.7
4.034 Xft 342.662
The largest shear stress will also be in the 0o and 90o plies; the factor causing matrix shear failure will be found using the equation:
Xf9000 1500.9
2.246 Xf 3.339 103
Since Xft is the lowest factor, the second ply failure is due to matrix transverse tensile stress in both the 0 o and 90o plies.
The total strain vector is now:
_total_global _global Xft _incr_global_Nx _total_global
1.799 10 3
1.799 10 3
1.747 10 3
At this point, only the +45o plies remain to carry all of the load, which is now at the level:
N2 Unit_N 682.943 Xft( ) N2
1.026 103
1.026 103
512.803
psi in
Since [At] matrix is
1
N
n
Qbar_n tn
, the [At] matrix for [+45]
S is:
Ap45_s Qbar 45( ) t 2
Ap45_s
0.079
0.066
0.061
0.066
0.079
0.061
0.122
0.122
0.134
Msi in
Next we calculate the new global strain vector:
_new_global_Nx Ap45_s1 N2
_new_global_Nx
0.016
0.016
0.011
The new 45o lamina strain vector is:
epsp45_new_Nx T 45( ) 1_new_global_Nx
The new 45o lamina stress vector is:
p45_new_Nx Q0 epsp45_new_Nx p45_new_Nx
1.479 105
4.931 104
1.488 10 11
psi
Therefore, since the +45o ply is unable to carry the loads after all the other plies have experienced matrix failures, the second plyfailure automatically leads to fiber failure in the +45o ply. The three points defining the load factor versus laminate level XX curve are:
Load_factor
0
682.943
682.943 Xft
eps
0
_global0
_total_global0
0 2 10 4 4 10 4 6 10 4 8 10 4 0.001 0.0012 0.0014 0.0016 0.00180
200
400
600
800
1000
1200
Load_factor
eps
ME 7502 Mid-Term Exam Solutions
Problem #2 Solution
Lamina Properties:
MPaE1 25 106 E2 1.7 106
12 .30 G12 0.65 106 21 12
E2E1 21 0.02
1 0.653 10 6 2 7.5 10 6
Tref 400
layer thickness t .0052
With 11 = -0.653 x 10-6 in/in/oF and 22 = 7.5 x 10-6 in/in/oF, and 400oF as the stress freetemperature, since the laminate is quasi-isotropic, we may use the closed from expression forthe stresses within each lamina
01
1
/21)(
}{221112
112211
EE
TEl
(1) Considering the first temperature cycle, for T_min = -80oF, and T_max = 160oF,
Tmin 80
TC1_min Tmin Tref
C1_TminE1 2 1 TC1_min
1 2 12E1E2
1
1
0
C1_Tmin
6 103
6 103
0
Tmax 160
TC1_max Tmax Tref
C1_TmaxE1 2 1 TC1_max
1 2 12E1E2
1
1
0
C1_Tmax
3 103
3 103
0
For each stress component, the ratio of the minimum to maximum stress is:
RC1_Tmax1
C1_Tmin1
R 0.5
By comparing the maximum stresses to the fatigue limit stress in Figures 3a, 3b and 3c, itis apparent that the number of cycles until the laminate experiences first ply failure will begoverned by transverse tensile stress, since this is the lowest fatigue limit stress of all threelamina stresses. Therefore, using Figure 3b with a maximum transverse tensile stress of 6ksi and R = 0.5, we obtain the number of cycles to failure N1 = 100,000 cycles.
N1 100000
(2) Considering the second temperature cycle, for T_min = -80oF, and T_max = 280oF,
Tmin 80
TC2_min Tmin Tref
C2_TminE1 2 1 TC2_min
1 2 12E1E2
1
1
0
C2_Tmin
6 103
6 103
0
Tmax 280
TC2_max Tmax Tref
C2_TmaxE1 2 1 TC2_max
1 2 12E1E2
1
1
0
C2_Tmax
1.5 103
1.5 103
0
For each stress component, the ratio of the minimum to maximum stress is:
RC2_Tmax1
C2_Tmin1
R 0.25
Therefore, using Figure 3b with a maximum transverse tensile stress of 6 ksi and R = 0.25, weobtain the number of cycles to failure N2 using the formula (see Mid-Term Exam Excel sheet):
N2 10
6 7.1
0.3 N2 4.642 103
(3) Considering the third temperature cycle, for T_min = -80oF, and T_max = 352oF,
Tmin 80
TC3_min Tmin Tref
C3_TminE1 2 1 TC3_min
1 2 12E1E2
1
1
0
C3_Tmin
6 103
6 103
0
Tmax 352
TC3_max Tmax Tref
C3_TmaxE1 2 1 TC3_max
1 2 12E1E2
1
1
0
C3_Tmax
600.004
600.004
0
For each stress component, the ratio of the minimum to maximum stress is:
RC3_Tmax1
C3_Tmin1
R 0.1
Therefore, using Figure 3b with a maximum transverse tensile stress of 6 ksi and R = 0.1, weobtain the number of cycles to failure N3 using the formula (see Mid-Term Exam Excel sheet):
N3 10
6 7.2
0.4 N3 1 103
n1 5 104 n2 2000
Finally, using the Palmgren-Miner rule, the data for n1 = 50,000 and n2 = 2000 provided inTable 3 and the computed fatigue lifetimes Ni, we have the expression:
13
3
2
2
1
1 Nn
Nn
Nn
from which we can compute the number of cycles n3 until fatigue failure during the last thermalcycle:
n3 1n2N2
n1N1
N3 n3 69.113
Therefore it will require approximately 69 cycles of the 3rd thermal cycle to cause fatigue failureby transverse tension in each of the plies of the laminate.