Mathcad - ME 7502 Mid-Term Exam Problem #1...

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:


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


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


1 2 12E1E2

1

1

0

C2_Tmin

6 103

6 103

0

Tmax 280

TC2_max Tmax Tref


1 2 12E1E2

1

1

0

C2_Tmax

1.5 103

1.5 103

0


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


1 2 12E1E2

1

1

0

C3_Tmin

6 103

6 103

0

Tmax 352

TC3_max Tmax Tref


1 2 12E1E2

1

1

0

C3_Tmax

600.004

600.004

0


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.

Mathcad - ME 7502 Mid-Term Exam Problem #1...

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