hw4

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HW (4) Example (1) : Givens : E m = E f 50 = 4.6 Gpa E f = 230 GPa G m = E m 2.6 = 1.769 GPa G f = E f 2.5 = 92 GPa Req : E 1 , E 2 , G 12 , ν 12 Sol : E 1 = V f . E f + V . E m ( = 0.4 * 230 ( + ) 0.6 * 4.6 = ) 94.7GPa E 2 = E f .E m V f .E m +V m .E f = 2304.6 ( 0.44.6 )+( 2300.6 ) = 7.55GPa ν 12 = ν m . V m + ν f . V f ( = 0.3 * 0.6 ( + ) 0.25 * 0.4 = ) 0.26

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Page 1: Hw4

HW (4)

Example (1):

Givens:

Em = Ef50

= 4.6 Gpa

E f = 230 GPa

Gm = Em2.6

= 1.769 GPa

Gf = Ef2.5

= 92 GPa

Req:

E1 ,E2 , G12 , ν12

Sol:

E1 = V f.E f + V.Em) = 0.4*230) + (0.6*4.6 = (94.7GPa

E2 = Ef . Em

V f . Em+V m . Ef =

230∗4.6(0.4∗4.6)+(230∗0.6) = 7.55GPa

ν12 = νm.V m + ν f.V f) = 0.3*0.6) + (0.25*0.4 = (0.26

G12 = Gm .G f

V f .Gm+V m .G f =

1.769∗92(0.4∗1.769)+(0.6∗92) = 2.9 GPa

Page 2: Hw4

Example (2):

Req:: E1 , E2 , G12 , ν12 for the following material combinations:V m = 0.45 and V f = 0.55

a))E-glass and polyester(isophthalic)

From table 2.1: E-glass E f = 72.345GPaν f = 0.22Gf = 3.45 GPa

From table 2.4:Polyester Em = 3.4GPaνm = 0.38Gm = 75.9 GPa

E1 = V f .E f + V m.Em = (0.55*72.345) + (0.45*3.4) = 41.32

E2 = Ef . Em

V f . Em+V m . Ef =

72.345∗3.4(0.55∗3.4)+(72.345∗0.45) = 7.145GPa

ν12 = νm.V m + ν f.V f = (0.38*0.55) + (0.22*0.45) = 0.308

G12 = Gm .G f

V f .Gm+V m .G f =

75.9∗3.45(0.55∗75.9)+(0.45∗3.45) = 6.047GPa

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b))S-glass and epoxy9310:

From table 2.1: S-glass E f = 85GPaν f = 0.22Gf = 4.8 GPa

From table 2.4:Epoxy9310 Em = 3.12GPaνm = 0.38Gm = 75.8 GPa

E1 = V f .E f + V m.Em = (0.55*85) + (0.45*3.12) = 48.154

E2 = Ef . Em

V f . Em+V m . Ef =

85∗3.12(0.55∗3.12)+(85∗0.45) = 6.635GPa

ν12 = νm.V m + ν f.V f = (0.38*0.55) + (0.22*0.45) = 0.308

G12 = Gm .G f

V f .Gm+V m .G f =

75.8∗4.8(0.55∗75.8)+(0.45∗4.8) = 8.297GPa

c))Carbon T300 and vinyl esler:

From table 2.1: Carbon T300 E f = 230GPaν f = 0.2Gf = 3.53 GPa

From table 2.4:Vinyl esler Em = 3.4GPa

Page 4: Hw4

νm = 0.38Gm = 82.7 GPaE1 = V f .E f + V m.Em = (0.55*230) + (0.45*3.4) = 128.03

E2 = Ef . Em

V f . Em+V m . Ef =

230∗3.4(0.55∗3.4)+(230∗0.45) = 7.421GPa

ν12 = νm.V m + ν f.V f = (0.38*0.55) + (0.2*0.45) = 0.299

G12 = Gm .G f

V f .Gm+V m .G f =

82.7∗3.53(0.55∗82.7)+(0.45∗3.53) = 6.202GPa

Example (3)

Select a matrix, to obtain material with E1>30Gpa and E1

E2

<3.5

Using matlab for different fibers and matrices we find that to get these conditions we use:

D-glass as fiber and Bismaleiraide as a matrix

Matlab code:

clear;clc;closeall;Vm = 0.24;Vf = 1-Vm;Ef = 55;Em = 3.583;E1 = (Vf*Ef)+(Vm*Em)E2 = (Em*Ef)/((Vf*Em)+(Vm*Ef))F = E1/E2

Output:

E1 = 42.6599

E2 =12.3761

F =3.4470