Elastic Response of Anisotropic Materials 177

a. Calculate the strains induced by this stress tensor, assum­ing no change in temperature or moisture content (i.e., assume t.T= AM= 0).

b. Calculate the strains induced by this stress tensor and a tempera­ture increase of 100°C (assume t.M = 0).

c. Calculate the strains induced by this stress tensor and a 2% increase in moisture content (assume t.T = 0).

d. Calculate the strains induced by this stress tensor, a temperature increase of 100°C, and a 2"1<, increase in moisture content.

4.4. Consider the following strains:

£., = 1500 (.lm/m

y,. = 750 f.lrild

fw "' - 2000 ~tm/m

y, = - 500 J.lr•ld

r = - 1750(.lm/m

Y •. "' !l'l<l (.l rad

a. Calculate the stress tensor that caused these strains, assuming no change in temperature or moisture content (i.e., assuming t.T=AM =0).

b. Calculate the stress tensor that caused these strains, if these strains were caused by the simultaneous effects of stress and a temperature decrease of 100°C (assume AM= 0).

c. Calculate the stress tensor tha t caused these strains, if these strains were caused by the simultaneous effects of stress, a temperature decrease of 100°C, and n 2% increase in moisture content.

~ An orthotropic mnterinl with the following properties is considered

in problems 4.5 through 4.8:

l: 11 = 100 Gl'<1

V11 = 0.20

C1! = 60 Gl'a

a 11 = 1 ~tm/m - "C

IS 11 = 100 (.1111/111 - %M

E!! = 200 G l'n

VII =-0.25

G1.1 = 75 GP<l

a12 = 25 ~tm/m cc

~~! = 6!Xl ~t m/m - ".,M

E11 = 75GPa

v!J = 0.60

G!.I = 50GI'<l a 11 = 15(.lm/m -"C

I~" = HXJO J.lln I m - " .. M

~Calculate the compliance mntrix, S,,. ----------------''-4.6. Calculate the stiffness matrix, C,,.

4.Z Consider the following stress tensor:

[

(Jll <J;, = (J21

<J u

(J·~] [ 75 (J2.1 = 10

(Jll -25

10

-90 30

-25] 30 (MPn)

25

252 Structural Analysis of Polymeric Composite Materials

5.1. Calculate the reduced compliance matrix for the materials listed below, first by hand calculation and then using program UNIDIR. Use material properties listed in Table 3.1.

a. Glass/epoxy

b. Kevlar/epoxy

c. Graphite/epoxy

[s.2. Calculate the reduced stiffness matrix for the materials listed below, first by hand calculation and then using program UNIDIR. Use material properties listed in Table 3.1.

(a. Glass/epoxy

b. Kevlar/epoxy

c. Graphite/epoxy 5.3. Confirm that:

5.4. Referring to Example Problem 5.4, confirm that the predicted stresses and strains at points Band Care consistent with Equations 5.2 and 5.3.

5.5. A thin unidirectional glass/epoxy composite laminate is simultane­ously subjected to a uniform temperature change t:.T = -175°C, an increase in moisture content AM= 0.5%, and the following in-plane stresses:

5.6.

5.7.

[s.8.

crn = 350MPa

cr:u = 40MPa

Determine the resulting strains (£11, £221 and y1~, first by hand cal­culation and then using program UNIDIR. Use material properties listed in Table 3.1.

Repeat Problem 5.5 for a unidirectional Kevlar/epoxy composite laminate.

Repeat Problem 5.5 for a unidirectional graphite/epoxy composite laminate.

A thin unidirectional glass/epoxy composite laminate is simultane­ously subjected to a uniform temperature change t:.T = -275°F and

Unidirectional Composite Laminates Sttbject to Plane Stress 253

an unknown plane stress state. The following strains are measured as a result (moisture content remains constant):

£u = -1250Jlin./in.

£:u = 2000Jlin./in.

Determine the stresses (cr11,cr22, and td, first by hand calculation and then using program UNIDIR. Use material properties listed in Table 3.1.

5.9. Repeat Problem 5.8 for a unidirectional Kevlar/epoxy composite laminate.

5.10. Repeat Problem 5.8 for a unidirectional graphite/epoxy composite laminate.

5.11. A square unidirectional glass/epoxy composite laminate with dimensions 1 m x 1 m is clamped between an infinitely rigid frame, as shown in Figure 5.37. Material properties are listed in Table 3.1. Initially the clamped composite is stress-free, but the temperature is subsequently decreased by 100°C. The thermal expansion coef­ficient of the rigid walls is zero, so the rigid walls do not expand or contract. a. Calculate the stresses (cr11, cr22, td induced by this change in

temperature, first by hand calculation and then using program UNIDIR.

FIGURE 5.37

Infinitely rigid frame

Composite laminate

+X

Clamped composite laminate considered in Problems 5.10, 5.11, and 5.12.

Material Properties 117

material property will be given, suitable for use with anisotropic materials. That is, the properties of a composite material when referenced to a nonprin­cipal material coordinate system will be discussed first. These general defi­nitions will then be specialized to the principal material coordinate system, that is, they will be specialized for the case of orthotropic or transversely isotropic composites.

Typical values of the properties discussed in this chapter measured at room temperatures are listed for glass/epoxy, Kevlar/epoxy, and graphite/ epoxy in Table 3.1. These properties do not represent the properties of any specific commercial composite material system, but rather should be viewed as typical values. Due to ongoing research and development activities within

~ TABLE3.1

Typical Properties of Common Unidirectional Composites

Glass/ Kevlar/ Graphite/ Property Epoxy Epoxy Epoxy

En 55GPa JOOGPa 170GPa

(8.0 Msi) (15Msi) (25 Msi)

1:~~ I6GPa 6GPa 10 Gl'a

(2.3 Msi) (0.90 Msi) (1.5 Msi)

Vu 0.28 0.33 0.30

Gu 7.6GPa 2.1 GPa 13GPa

(1.1 Msi) (0.30 Msi) (1.9 Msi) (I'

au J050MPa 131!0MJ>a 1500 MPa

(ISO ksi) (200 ksi) (218 ksi)

a~ 690MPa 280MPa 1200 Ml'a

(100 ksi) (40 ksi) (175 ksi) (I'

au 45MPa 35MPa SOMPa

(5.8 ksi) (2.9 ksi) (7.25 ksi)

a~ 120MPa 105MPa IOOMPa

(16 ksi) (15 ksi) (14.5 ksi)

'tlz 40MPa 40MPa 90MPa

(4.4 ksi) (4.0 ksi) (13.1 ksi)

lln 6.7J.Lm/m-•c -3.6J.Lm/m-•c -0.9J.Lm/m-•c

(3.7J.Lin./in.-•F) (-2.0 J.Lin./in.-°F) (-0.5J.Lin./in.-"F)

!l:!z 25J.Lm/m-•c 58J.Lm/m-•c 27J.Lm/m-•c

(I4J.Lin./in.-"F) (32J.Lin./in.-•F) (ISJ.Lin./in.-"F)

llu 100 J.Lm/m-o/oM 175J.Lm/m-'YoM 50J.Lm/m-%M

(100 J.Lin./in.-%M) (175J.Lin./in.-%M) (50 J.Lin./ in.-%M)

~ 1200 J.Lm/m-%M 1700 J.Lm/m-%M 1200 J.Lm/m-%M

(1200 J.Lin./in.-%M) (1700 J.Lin./in.-'YoM) (1200 J.Lin./in.-o/oM)

Ply 0.125mm 0.125mm 0.125mm Thickness (0.005 in.) (0.005 in.) (0.005 in.)

392 Structural Analysis of Polymeric Composite Materials

a. Midplane strains and cun·atures

b. Strains (Ew Eyy Y~y) induced in ply 2

c. Stresses (crx., crYY' txy) induced in ply 2 d. Stresses (cr11, cr22, td induced in ply 2

6.18. A 10 em x 10 em square [0/±30/90], graphite/epoxy laminate is supported between three infinitely rigid walls and friction­less rollers, as shown in Figure 6.35. A load N ... = -7500 N/m is applied to the plate. Review Table 6.13, then (ignore the possibil­ity of buckling) use hand calculation to determine N.., N.,, N,., E~ .• , E~, and y;,.

6.19. A [20/65/-25], graphite/epoxy laminate is cured at 175'C and then cooled to room temperatures (20°C). Moisture content remains at 0%. The following loads are then applied:

N,.=30kN/m Nyy = -7kN/m Nxy = 0

Mxx = 10 N-m/m Myy = Mxy = 0

Using properties listed in Table 3.1 and assuming ply thicknesses of0.125 mm:

a. Determine all ply strains and stresses using program CLT

b. Prepare plots similar to Figure 6.13, showing the through­thickness variation of strains Ew Eyy' and Yxy

c. Prepare plots similar to Figure 6.14, showing the through­thickness variation of strains E11 , E22, and Y1~

d. Prepare plots similar to Figure 6.15, showing the through­thickness variation of stresses crxx, cr•Y' and t .fV

e. Prepare plots similar to Figure 6.16, showing the through­thickness variation of stresses crllt cr~2, and t 12

Laminate

FIGURE 6.35

Rigid walls

Frictionless rollers

A [0/±30/90). graphite/epoxy laminate supported between rigid walls and frictionless rollers.

Tltennomechanical Behavior of Multi angle Composite Laminates

~ 6.20. Repeat Problem 6.19 for a glass/epoxy laminate. 6.21. Repeat Problem 6.19 for a Kevlar/epoxy laminate.

393

6.22. A [20/65/-25]; "hybrid" laminate is cured at 175°C and then cooled to room temperatures (20'C). Plies 1 and 6 are graphite/ epoxy, plies 2 and 5 are glass/epoxy, and plies 3 and 4 are Kevlar/ epoxy. The following loads are then applied:

N,,. = 30 kN/m Nyy = -7 kN/m Nxy = 0

M .. = 10 N-m/m Myy = Mxy = 0

Moisture content remains constant at 0%. Using properties listed in Table 3.1 and assuming the graphite/epoxy, glass/epoxy, and Kevlar/epoxy plies have thicknesses of 0.125 mm, 0.200 mm, and 0.15 mm, respectively:

a. Determine all ply strains and stresses using program CLT b. Prepare plots similar to Figure 6.13, showing the through­

thickness variation of strains En , Eyyt and Yxy c. Prepare plots similar to Figure 6.14, showing the through­

thickness variation of strains E11, £u, and y12

d. Prepare plots similar to Figure 6.15, showing the through­thickness variation of stresses crn , crYY' and t xy

e. Prepare plots similar to Figure 6.16, showing the through­thickness variation of stresses cr11, cr22, and t 12

6.23. Referring to Figure 6.21, the ANSYS analysis described in Example Problem 6.9, and the values for the in-plane strains and stresses at points B and C listed in Table 6.11, use program UNIDIR to confirm that the stresses and strains predicted at points B and C for each ply are consistent.

6.24. Referring to Figure 6.21, the ANSYS analysis described in Example Problem 6.10, and the values for the effective stresses at points A and C listed in Table 6.12, use program CLT to confirm that the strains predicted at points A and C are consistent with Equation 6.38.

6.25. Referring to Figure 6.21, the ANSYS analysis described in Example Problem 6.9, and the values for the in-plane strains and stresses at points A, B, and C listed in Table 6.11, determine the stresses relative to the 1-2 coordinate system induced in each ply.

a. at point A b. at point B

c. at point C

Thermomecha11ical Belwvior of Multiangle Composite Laminates 285

(u) 0.6

e 0.4 .§. " 0.2 .: .g

0 ·;;; 0 0.. V> - 0.2 V> ... c:::

-'"' v - 0.4 :2 1-<

- 0.6 · 1----,.-----.------.----·--·---,----, - :!00 - 200 - 100 0 100 200

Normal slmin, r,_, (~tm/m)

(b) 0.6

e 0.'~ ---!'ry_s_---------------------.§. .. 0.2 c ~

0 ·;;; 0 0..

____ F~_z _______ _____________ _ - - - - - _P]t ~ - - - - - - - - - - - - - - - - - - - - - -

-------~l~------------------------------~ry~---- - -----------------::: - 0.2 ...

c::: -'"' v

- OA :2 1-<

--------- _!'!y_:~------------------ --------------~t~---------- - -----------

------------~r! _____________________ _ - 0.6 --------,,------ .-------·--·--

- 1750 - 1250 - 750 - 250

Nonm1l stmin, rrr (~tm/m)

(c) 0.6

·e 0.·~ ------------------ -~r ~ ------- --.§. -------------------~tz ____ _____ _ .. 0.2 .: -------- - ----------~~~--------- - -.g 0 ·;;;

0 0..

Ply 5

===================0~ ~============-V> 0.2 "' ... --------- - --------- ~~ ] _____________ _ c:::

-'"' v -0.'~

~ ----------- - -- - - ---~ry} ____ _____ _____ _ - ------------------ ~ry} _______________ _

- 0.6 . ----. 0 200 400 600 1100 1000 1200

~ FIGURE b.13 Through-th ickrwss strain plots i mpl i~d by tlw midpl.1 11l' slmins <1 nd nrr\'.llu res disnrssl·d in

xample l'robk•m 6.1. Str.1i ns rl'fl' rl'lll'l'd to the x- ymord inall' sysll' lll . (<1) Norn1.1l stra in r,.; (b) normal slr.1 in r.,.; (c) she.1r strain y, •.

Thermomechanical Behavior of Multiangle Composite Laminates

(a) 0.6 -,-------------------

~ \~~-

~ 0.4 -~------------------------= ~ ~ -_'\:.=-=-_~~---- =p(,.--:y 7=-

i ·: ----=:::::::~~~~y::::::!ji~::~,~:;:::-1---· ~ - 0.2 --- .c.__ ____ ,_Ply.:L_ ____ , _________ I----· J;i t-::_----------------------------------------- -rfy-2(\--:2 - 0.4 .----------------- -rh=-1-< --~----

-·0.6 ----.----,---.---.--.-.---,--,---. - 1500 - 1300 - 1100 - 900 - 700 - 500 - 300 -100 100 300

Normal strain, r 11 (~m/m)

- 1500 - 1:300 - 1100 - 900 -700 - 500 - 300 - 100 100

Nnrmalstrain, £n (~tm/m)

(c) 0.6 ·

c: ~

:2-M f-o

- 0.6 ·1----- 1000 - 500

--------.--.--0 51KI 1000 1500

Shear strain, y12 (~r.wtl

289

~FIGURE &.14 Through-thkknes~ s t ra in plots implied by thL• midplane :.trains .md curvaturL•s discussL•d in Example l'roblem 6.1. Strains referenced to the 1- 2 coordinate syslem. (.1) Norn1.1l str.1in r 11; (b) normal stra in ~2; (c) shear strain Y12•

Predicting Failure of a Multiangle Composite Laminate 421

stresses. This reflects how strains are usually measured in practice. That is, strains measured in practice (using strain gages, for example) are usually measured relative to the service condition (at room temperature, say), rather than at the cure temperature of the composite.

HOMEWORK PROBLEMS

Notes: Computer programs CLT, LAMFAIL, and PROGDAM are used in the follow­ing problems. These programs can be downloaded from the following website:

http://depts.wnshinglon.edu/nmtns/computcr.html

7.1. Assume a new room-temperature cure graphite-epoxy prepreg system with properties listed in Table 3.1 has been developed. A [0/+30), laminate is produced using this material, and is, therefore, initially stress-and-strain free at room temperature. The room­temperature laminate is then subjected to the biaxial tensile loads shown in Figure 7.11 (i.e., Nvv = N, . ./2 and Nrv = M rr = Mw = Mxv = 0). Use the following process to. determine the tensile loads N.n arid N,,v necessary to causL' first-ply failure, according to the maximum stress failure criteria:

a. Usc program CL.:f to determine the stresses (cr11 , cr~~, t 12)

induced in each ply by unit loads N" = 1, N,,,, = 1/2 (i.e., usc ~ = Q~ .

b. Determine which ply (or plies) is closest to the failure cond i­tion dictated by the maximum stress fa ilure criterion, using the failure strengths listed for graphite/epoxy in Tabll' :u.

c. Ca lculate the increase in unit loads that will e<lllse first-ply ply fa ilure.

d . Confirm the results of part (c) using program LAM FAIL.

N., - N,j 2

N_,_,

FIGURE 7.11 A (O/:t30(, ln m in.lll' Jn,Jdcd in binxi<ll ll•nsion, as dcscribl•d in l'robk•m 7.1.

422 Structural Analysis of Polymeric Composite Materials

~ 7.2. Repeat Problem 1 for a room-temperature cure glass/epoxy system.

7.3. Repeat Problem 1 for a room-temperature cure Kevlar/epoxy system.

7.4. Repeat Problem 1 for the following load condition:

N.,9 = SN.. M,., = Muv = M,. = 0

Z5. Repeat Problem 1, except assume the laminate is produced using a graphite-epoxy system cured at 175°C (350°F).

7.6. Repeat Problem 1, except assume the laminate is produced using a glass/epoxy system cured at 175°C (350"F).

7.Z Repeat Problem 1, except assume the laminate is produced using a Kevlar/epoxy system cured at 175°C (350°F).

Z8. Three different first-ply failure envelopes for a [0/45/90/-45]. graphite-epoxy laminate are shown in Figure Z7. The three points listed below lie on these failure envelopes. In each case, usc pro­gram CLT to determine which ply(ies) are predicted to fail, accord­ing to the maximum stress failure criterion:

a. N.u=32Z3 kN/m N1,1,=0 kN/m (this point lies on the AT= AM= 0 curve)

b. N.u=79.41 kN/m NY.~ =O kN/m (this point lies on the AT= -155°C, AM= 0 curve)

c. Nn = 159.1 kN/m Nw = 0 kN/m (this point lies on the AT= -155°C, AM = 1% ·curve)

Z9. Three different first-ply failure envelopes for a [0/45/90/-45).... graphite-epoxy laminate are shown in Figure Z7. The three points listed below lie on these failure envelopes. In each case, usc pro­gram CLT to determine which ply(ies) arc predicted to fail, accord­ing to the maximum stress failure criterion:

a. Nu = 0 kN/m Nw = 32Z3 kN/m (this point lies on the AT= AM= 0 curve)

b. N,., = 0 kN/m Nw = 79.41 kN/m (this point Jies on the AT= -155"C, AM=(} curve)

c. N .. = 0 kN/m Nw = 159.1 kN/m (this point lies on the AT= -155°C, AM =·i% curve)

7.10. Three different first-ply failure envelopes for a [0/45/90/-45). graphite-epoxy laminate are shown in Figure ZZ The three points listed below lie on these failure envelopes. In each case, usc pro­gram CLT to determine which ply(ics) arc predicted to fail, accord­ing to the maximum stress failure criterion: