Chapter 3 Micromechanical Analysis of a LaminaUltimate Strengths of a Unidirectional Lamina

Dr. Autar KawDepartment of Mechanical Engineering

University of South Florida, Tampa, FL 33620

Courtesy of the TextbookMechanics of Composite Materials by Kaw

Fiber

Composite

Matrix

Strain, ε

Stress, σ

( σm) ult

( σf) ult

( εf ) ult ( εm ) ult

FIGURE 3.24Stress–strain curve for a unidirectional composite under uniaxial tensile load along fibers.

and

)()( ,

E =

f

ultfultf

σε

E =

m

ultmultm

)()( σ

ε

)()()()( V-1 E + V = fmultffultfultT1 εσσ

])([)()()(])([)( minimumfmultfminimumfultfminimumfultm V-1E + V >V-1 εσσ

)()()()()(

)(ultmultfmultf

ultfmultmminimumf + E -

E - < Vσεσ

εσ

])()()()()( criticalfmultfcriticalfultfultm V-[1E + V > εσσ

)()()()(

)(ultfmultf

ultfmultmcriticalf

E - E - < V

εσεσ

Example 3.13

Find the ultimate tensile strength for a Glass/Epoxy lamina with a 70% fiber volume fraction. Use the properties for glass and epoxy from Tables 3.1 and 3.2, respectively. Also, find the minimum and critical fiber volume fractions.

Example 3.13

and GPa, 85 = E f MPa1550 = ultf )(σ

10182301085101550)(

1

9

6

-

ultf

. =

= ε

×××

Example 3.13

and GPa, 3.4 = Em MPa72 = ultm )(σ

102117010431072

1

9

6

-

ultm

. =

.

= )ε(

×××

Example 3.13

MP1104 =0.7-1103.4100.1823 + 0.7101550 = 9-16

ultT1 ))()(())(()( ×××σ

Example 3.13

0.6422% =

100.6422 =

1072+100.1823103.4 - 101550100.1823103.4 - 1072 = V

2-

61-96

-196

minimumf

×

×××××××

))(())(()(

Example 3.13

%.=

.=

).)(. - ().)(. - ( = )V(

-

-

-

criticalf

67320

1067320

10182301043101550101823010431072

2

196

196

×

××××××

FIGURE 3.25Tensile coupon mounted in the test frame for finding the tensile strengths of a unidirectional lamina. (Photo courtesy of Dr. R.Y. Kim, University of Dayton Research Institute, Dayton, OH.)

T t

w

60o

Tab Length = 38mm

Gage Length = 153mm

Length = 229mm

Tab Thickness = 3.2mm

FIGURE 3.26Geometry of a longitudinal tensile strength specimen.

FIGURE 3.27Stress–strain curve for a [0]8 laminate under a longitudinal tensile load. (Data courtesy of Dr. R.Y. Kim, University of Dayton Research Institute, Dayton, OH).

Example 3.13

GPa, 187.5 = E1

and)( MPa,2896 = ult

T1σ

1.560% = ultT1 )(ε

(a) (b) (c)

FIGURE 3.28Modes of failure of unidirectional lamina under a longitudinal tensile load.

E|| = ||

1

11

σε

E|| = ||

1

1122

σνε

(a) Fiber Microbuckling Extensional Mode

Fiber Microbuckling Shear Mode (b)

(c) Transverse Tensile Failure of Matrix

Shear Failure

(d)

FIGURE 3.29Modes of failure of a unidirectional lamina under a longitudinal compressive load.

νε

σ12

ultT21

ultc1

E = )()(

)()()( 3/1fult

Tmult

T2 V - 1 = εε

1 + 1 -

EE

sd =

f

mult

Tmult

T2 )()()( εε

where]min[)( ,S ,S = c

2c1ult

c1σ

and

)(])([ ,

V - 13E E V

EE V - 1 + V2 = S

f

fmf

f

mff

c1

V - 1

G = Sf

mc2

V + V = multmfultfult12 )()()( τττ

])()[()( V + V 2 = multmfultfultc1 ττσ

Material

Experimental Strength

Equation 3.78(a)

Equation (3.78b)

MPa

MPa

MPa

Glass/Polyester

Type I Carbon/Epoxy

Kevlar 49/Epoxy

600-1000

700-900

240-290

8700

22800

13200

2200

2900

2900

Table 3.6. Comparison of experimental and predicted values of longitudinal compressivestrength of unidirectional laminae, Vf =0.50. (Source: Table 7.2 in Introduction to Composite Materials by D. Hull, 1981, 8 Cambridge University Press, 1981, Reprinted with the permission of Cambridge University Press)

Example 3.14

Find the longitudinal compressive strength of a Glass/Epoxy lamina with a 70% fiber volume fraction. Use the properties of glass and epoxy from Tables 3.1 and 3.2, respectively. Assume fibers are circular and are in a square array.

Example 3.14.

GPa, 85 = E f

3.1 Table From

0.20, = fν

and)( MPa,1550 = ultfσ

MPa35 = ultf )(τ

GPa, 3.4 = Em

3.2 Table From

0.30, = mν

and)( MPa,72 = ultmσ

MPa34 = ultm )(τ

Example 3.14.

GPa 60.52 = E1 0.23 = 12ν

Example 3.14.

0.9441 =

4(0.7) = sd 2

1

π

Example 3.14.

100.2117 =103.40

1072 =

1-

9

6

ultm

×××)(ε

Example 3.14.

100.1983 =

1 + 1 - 1085103.40.9441100.2117 =

2-

9

91-

ultT2

×

××

×)(ε

Example 3.14.

100.2373 =

- 1 100.2117 =

2-

-1ult

T2

×

× )7.0()()( 3/1ε

Example 3.14.

MPa521.8 =

0.23100.1983 1060.52 =

-29

ultC1

)()()( ××σ

Example 3.14.

MPa21349 =

0.7 - 131085103.40.7

1085103.4 0.7 - 1 + 0.7 2 = S

99

9

9C1 )(

))()(()( ××

××

Example 3.14.

GPa 1.308 = Gm

MPa4360 =

0.7 - 1101.308 = S

9C2

×

Example 3.14.

MPa4360 = 4360 21349, = ultC1 )min()(σ

Example 3.14.

MPa69.4 =0.31034 + 0.71035 2 = 66

ultC1 )])(())([()( ××σ

Example 3.14.

FIGURE 3.30IITRI fixture mounted in a test frame for finding the compressive strengths of a lamina. (Data reprinted with permission from Experimental Characterization of Advanced Composites, Carlsson, L.A. and Pipes, R.B., Technomic Publishing Co., Inc., 1987, p. 76. Copyright CRC Press, BocaRaton, FL.)

Example 3.14.

GPa, 199 = Ec1

and)( MPa,1908 = ult

c1σ

0.9550% = ultc1 )(ε

L1 L2

w

Strain Gage

Specimen Dimensions

L1, mm L2, mm w*, mm

12.7 ± 1 127 ± 1.5 12.7 ± 0.1 or 6.4 ± 0.1

FIGURE 3.31Geometry of a longitudinal compressive strength specimen. (Data reprinted with permission from Experimental Characterization of Advanced Composites, Carlsson, L.A. and Pipes, R.B., TechnomicPublishing Co., Inc., 1987, p. 76. Copyright CRC Press, Boca Raton, FL.)

FIGURE 3.32Stress–strain curve for a [0]24 graphite/epoxy laminate under a longitudinal compressive load. (Data courtesy of Dr. R.Y. Kim, University of Dayton Research Institute, Dayton, OH.)

Fiber

Matrix

Fiber

d

d

s

Matrix

Matrix

σ2

σ2

FIGURE 3.33Representative volume element to calculate transverse tensile strength of a unidirectional lamina.

,εδ cc s=

and,εδ ff d =

εδ mm d-s = )(

:whereδδδ mfc + =

εεε mfc sd - 1 +

sd =

εε mmff E = E εε mf

mc s

d - 1 + EE

sd =

)()( ultTm

f

mult

T2 s

d - 1 + EE

sd = εε

)()( ultT22ult

T2 E = εσ

Example 3.15

Find the ultimate transverse tensile strength for a unidirectional Glass/Epoxy lamina with a 70% fiber volume fraction. Use properties of glass and epoxy from Table 3.1 and Table 3.2, respectively. Assume the fibers are circular and are arranged in a square array.

Example 3.15

100.1983 = -2ult

T2 ×)(ε GPa 10.37 = E2

MPa20.56 =

100.19831010.37 = -29ult

T2 ))(()( ××σ

Example 3.15

GPa, 9.963 = E2

and)( MPa,53.28 = ultT2σ

0.5355% = ultT2 )(ε

Example 3.15

FIGURE 3.34Stress–strain curve for a [90]16 graphite/epoxy laminate under a transverse tensile load. (Data courtesy of Dr. R.Y. Kim, University of Dayton Research Institute, Dayton, OH.)

where)()( , E = ultC22ult

C2 εσ

)(1)( ultCm

f

mult

C2

sd

EE

sd = εε

−+

Example 3.16

Find the ultimate transverse compressive strength of a Glass/Epoxy lamina with 70% fiber volume fraction. Use the properties of glass and epoxy from Table 3.1 and Table 3.2, respectively. Assume the fibers are circular and are packed in a square array.

Example 3.16

,GPa 85 = E f

,GPa 3.4 = Em

,)( MPa102 = ultCmσ

and,GPa 10.37 = E2

0.9441 = sd

Example 3.16

0.0300 =

103.410102 = 9

6

ultCm ×

×)(ε

Example 3.16

100.2810 =

0.03 0.9441 - 1 + 1085103.40.9441 =

2-

9

9

ultC2

×

×× )()()(ε

Example 3.16

MPa29.14 = 100.28101010.37 = -29ult

C2 ))(()( ××σ

Example 3.16

FIGURE 3.35Stress–strain curve for a [90]40 graphite/epoxy laminate under a transverse compressive load perpendicular to the fibers.

Example 3.16

GPa, 3 = Ec2 9

and)( MPa,198 = ultc2σ

2.7% = ultc2 )(ε

:where∆∆∆ mfc + =

, s= c12c )(γ∆

and )( ,d = f12f γ∆

))(( m12m d - s = γ∆

)()()( m12f12c12 sd - 1 +

sd = γγγ

G = G ff12mm12 )()( γγ

)()( m12f

mc12

sd - 1 +

GG

sd = γγ

)()( ult m12f

mult12

sd - 1 +

GG

sd = γγ

)(

)()(

ult m12f

m12

ult1212ult12

sd - 1 +

GG

sd G =

G =

γ

γτ

Example 3.17

Find the ultimate shear strength for a Glass/Epoxy lamina with 70% fiber volume fraction. Use properties for glass and epoxy from Tables 3.1 and 3.2, respectively. Assume the fibers are circular and are arranged in a square array.

Example 3.17

GPa 35.42 = G f

GPa 1.308 = Gm

GPa 4.014 = G12

0.9441 = sd

MPa34 = ult m12 )(τ

Example 3.17

100.2599 =

101.3081034 =

1-

9

6

ult m12

×

××)(γ

Example 3.17

1 2

3

45o

45o

σx

σx

45o

45o

Strain-Gage Rosette

FIGURE 3.36Schematic of a [±45]2S laminate shear test.

Example 3.17

MPa9.469 =

100.2599 0.9441 - 1 + 1035.42101.3080.9441 104.014= 1-

9

99

ult12 )()()()( ×

××

×τ

Example 3.17

and )()( ,

2 = ultx

ult12σ

τ

)()()( ultyultxult12 || + || = εεγ

Example 3.17

GPa, 5.566 = G12

and )( MPa,87.57 = ult12τ

2.619% = ult12 )(γ

Example 3.17

FIGURE 3.37Shear stress–shear strain curve obtained from a [±45]2S graphite/epoxy laminate under a tensile load.