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EML 4230 Introduction to Composite Materials
Chapter 3 Micromechanical Analysis of a Lamina
Elastic Moduli
Dr. Autar KawDepartment of Mechanical Engineering
University of South Florida, Tampa, FL 33620
Courtesy of the TextbookMechanics of Composite Materials by Kaw
Strength of Materials Approach
h, t = A ff
and h, t = A mm
ht = A cc
Strength of Materials Approach
t
t =
A
A = V
c
f
c
ff
V - 1 =t
t =
A
A = V
f
c
m
c
mm
Strength of Materials Approach
1
2 3
h
tc
tm/2 tf
tm/2
h
tc
Lc
FIGURE 3.3Representative volume element of a unidirectional lamina.
Strength of Materials Approach
h
tc
tm/2 tf
tm/2
c c
FIGURE 3.4A longitudinal stress applied to the representative volume element to calculate the longitudinal Young’s modulus for a unidirectional lamina.
Longitudinal Young’s Modulus
,A = F ccc and ,A = F fff
A = F mmm
, E = c1c and , E = fff
mmm E =
F + F = F mfc
Longitudinal Young’s Modulus
A E + A E = A E mmmfffcc1
: then),( If mfc = =
A
A E + A
A E = E
c
mm
c
ff1 V E + V E = E mmff1
Longitudinal Young’s Modulus
V E + V E = E mmff1
V E
E = FF
f1
f
c
f
,A = F ccc and ,A = F fff
A = F mmm
, E = c1c and , E = fff
mmm E =
Longitudinal Young’s Modulus
FIGURE 3.5Fraction of load of composite carried by fibers as a function of fiber volume fraction for constant fiber to matrix moduli ratio.
Example
Example 3.3
Find the longitudinal elastic modulus of a unidirectional Glass/Epoxy lamina with a 70% fiber volume fraction. Use the properties of glass and epoxy from Tables 3.1 and 3.2, respectively. Also, find the ratio of the load taken by the fibers to that of the composite.
Example
Example 3.3
Ef = 85 Gpa
Em = 3.4 GPa
GPa 60.52 =
0.3 3.4 + 0.7 85 = E1 )()()()(
Example
Example 3.3
FIGURE 3.6Longitudinal Young’s modulus as function of fiber volume fraction and comparison with experimental data points for a typical glass/polyester lamina.
ExampleExample 3.3
0.9831 = 0.7 60.52
85 =
F
F
c
f )(
Transverse Young’s Modulus
h
tc
tm/2 tf
tm/2 c
c
FIGURE 3.7A transverse stress applied to a representative volume element used to calculate transverse Young’s modulus of a unidirectional lamina.
Transverse Young’s Modulus
mfc = =
mfc + =
, t = ccc
and , t = fff
mmm t =
,
E = c
c2
and ,E
= f
ff
E =
m
mm
Transverse Young’s Modulus
andt
t E
1 +
t
t
E
1 =
E
1
c
m
mc
f
f
,2
E
V + E
V =
E
1
m
m
f
f
2
ExampleExample 3.4
Find the transverse Young's modulus of a Glass/Epoxy lamina with a fiber volume fraction of 70%. Use the properties of glass and epoxy from Tables 3.1 and 3.2, respectively.
Example
Example 3.4
= 85 GPaE f
Em = 3.4 GPa
GPa 10.37 = E
3.4
0.3 +
85
0.7 =
E
1
2
2
Transverse Young’s Modulus
FIGURE 3.8Transverse Young’s modulus as a function of fiber volume fraction for constant fiber to matrix moduli ratio.
Transverse Young’s Modulus
V4
= s
d f2
1
V32
= s
d f2
1
Transverse Young’s Modulus
s
d
(b)
(a)
s
d
FIGURE 3.9Fiber to fiber spacing in (a) square packing geometry and (b) hexagonal packing geometry.
Transverse Young’s Modulus
FIGURE 3.10 Theoretical values of transverse Young’s modulus as a function of fiber volume fraction for a boron/epoxy unidirectional lamina (Ef = 414 GPa, vf = 0.2, Em = 4.14 GPa, vm = 0.35) and comparison with experimental values. Figure (b) zooms figure (a) for fiber volume fraction between 0.45 and 0.75. (Experimental data from Hashin, Z., NASA tech. rep. contract no. NAS1-8818, November 1970.)
Transverse Young’s Modulus
h
tc
tm/2 tf
tm/2
1 1
(b)
(a)
tc
tm/2
tf
tm/2
tc + cT
Lc
tf + fT
h
tc
tm/2 tf
tm/2
1 1
(b)
(a)
tc
tm/2
tf
tm/2
tc + cT
Lc
tf + fT
FIGURE 3.11A longitudinal stress applied to a representative volume element to calculate Poisson’s ratio of unidirectional lamina.
Major Poisson’s Ratio
Tm
Tf
Tc + =
,t
= f
TfT
f
and ,t
= m
TmT
m
t =
c
TcT
c
Tmm
Tff
Tcc t + t = t
Major Poisson’s Ratio
,- =
Lf
Tf
f
and ,- = Lm
Tm
m
L
c
Tc
12 - =
Lmmm
Lfff
Lc12c t - t- = t-
Major Poisson’s Ratio
Lmmm
Lfff
Lc12c t - t- = t-
: then, If Lm
Lf
Lc = =
mmff12c t + t = t
t
t + t
t =
c
mm
c
ff12 V + V = mmff12
ExampleExample 3.5
Find the Major and Minor Poisson's ratio 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.
Example
Example 3.5
0.2 = f
0.3 = m
0.230 =
0.3 0.3 + 0.7 0.2 = 12 )()()()(
ExampleExample 3.5
E1 = 60.52 Gpa
E2 = 10.37 GPa
ExampleExample 3.5
0.03941 = 60.52
10.37 0.230 =
E
E = 1
21221
In-Plane Shear Modulus
h
tc
tm/2 tf
tm/2 c
c
FIGURE 3.12An in-plane shear stress applied to a representative volume element for finding in-plane shear modulus of a unidirectional lamina.
In-Plane Shear Modulus
mfc + =
,t = ccc
and ,t = fff
mmm t =
In-Plane Shear Modulus
,
G =
12
cc
and ,G
= f
ff
G =
m
mm
,t = ccc
and ,t = fff
mmm t =
mfc + =
tG
+ t G
= t G
mm
mf
f
fc
12
c
In-Plane Shear Modulus
tG
+ t G
= t G
mm
mf
f
fc
12
c
: then, If mfc = =
t
t G
1 +
t
t
G
1 =
G
1
c
m
mc
f
f12
G
V + G
V =
G
1
m
m
f
f
12
ExampleExample 3.6
Find the in-plane shear modulus of a Glass/Epoxy lamina with a 70% fiber volume fraction. Use properties of glass and epoxy from Tables 3.1 and 3.2, respectively.
Example
Example 3.6
GPa 85 = E f 0.2 = f
GPa 35.42 =
0.2 + 1 2
85 =
+ 1 2E = G
f
ff
)(
)(
ExampleExample 3.6
GPa 3.4 = Em 0.3 = m
GPa 1.308 =
0.3 + 1 2
3.40 =
+ 1 2
E = G
m
mm
)(
)(
ExampleExample 3.6
GPa 4.014 = G
1.308
0.30 +
35.42
0.70 =
G
1
12
12
In-Plane Shear Modulus
FIGURE 3.13 Theoretical values of in-plane shear modulus as a function of fiber volume fraction and comparison with experimental values for a unidirectional glass/epoxy lamina (Gf = 30.19 GPa, Gm = 1.83 GPa). Figure (b) zooms figure (a) for fiber volume fraction between 0.45 and 0.75. (Experimental data from Hashin, Z., NASA tech. rep. contract no. NAS1-8818, November 1970.)
Longitudinal Young’s Modulus
V E + V E = E mmff1
V - 1V + 1
= E
E
f
f
m
2
+ E / E
1- E / E = mf
mf
)(
)(
ExampleExample 3.7
Find the transverse Young's modulus 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. Use Halphin-Tsai equations for a circular fiber in a square array packing geometry.
ExampleExample 3.7
a
b
2
2
FIGURE 3.14Concept of direction of loading for calculation of transverse Young’s modulus by Halphin–Tsai equations.
ExampleExample 3.7
2 = Ef = 85 GPa Em = 3.4 GPa
0.8889 =
2 + 85/3.4
1 - 85/3.4 =
)(
)(
ExampleExample 3.7
GPa 20.20 = E
0.70.8889 1
0.70.88892+ 1 =
3.4E
2
2
))((
))((
Transverse Young’s Modulus
FIGURE 3.15 Theoretical values of transverse Young’s modulus as a function of fiber volume fraction and comparison with experimental values for boron/epoxy unidirectional lamina (Ef = 414 GPa, νf = 0.2, Em = 4.14 GPa, νm = 0.35). Figure (b) zooms figure (a) for fiber volume fraction between 0.45 and 0.75. (Experimental data from Hashin, Z., NASA tech. rep. contract no. NAS1-8818, November 1970.)
Transverse Young’s Modulus
Ef/Em = 1 implies = 0, (homogeneous medium)
Ef/Em implies = 1, (rigid inclusions)
Ef/Em implies (voids)
0
1
- =
Transverse Young’s Modulus
a
b
12 12
FIGURE 3.16Concept of direction of loading to calculate in-plane shear modulus by Halphin–Tsai equations.
Transverse Young’s Modulus
V + V = mmff12
V - 1V + 1
= G
G
f
f
m
12
+ G / G
1 - G / G =
mf
mf
)(
)(
V40 + 1 = 10f
Example
Example 3.8
Using Halphin-Tsai equations, find the shear modulus of a Glass/Epoxy composite with a 70% fiber volume fraction. Use the properties of glass and epoxy from Tables 3.1 and 3.2, respectively. Assume the fibers are circular and are packed in a square array. Also get the value of the shear modulus by using Hewitt and Malherbe’s8 formula for the reinforcing factor.
Example
Example 3.8
1 = GPa 35.42 = G fGPa 1.308 = Gm
0.9288 =
1 + 835.42/1.30
1 - 835.42/1.30 =
)(
)(
Example
Example 3.8
GPa 6.169 = G
0.7 0.9288 1
0.7 0.9288 1 + 1 =
1.308G
12
12
)()(
)()()(
Example
Example 3.8
2.130 =
0.7 40 + 1 =
V 40 + 1 = 10
10f
)(
Example
Example 3.8
0.8928 =
2.130 + 1.308 / 35.42
1 - 1.308 / 35.42 =
)(
)(
Example
Example 3.8
GPa 8.130 = G
0.7 0.8928 - 1
0.7 0.8928 2.130 + 1 =
1.308G
12
12
)()(
)()()(
END
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