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)4HW (
Example (1):
Givens:
= 230 GPa
= = = 4.6 GPa
= = = 92 GPa
= = = 1.769 GPa
= 0.25
= 0.3
= 0.6
= 0.4
,,,:Calculate
= . + . = (0.4*230) + (0.6*4.6) = 94.76 GPa
= = = 7.565 GPa
= . + . = (0.3*0.6) + (0.25*0.4) = 0.28
= = = 2.911 GPa
Example (2):
Calculate: , , , for a unidirectional lamina for the
following material combinations: = 0.45 and = 0.55
1. E-glass and polyester (isophthalic)
From table 2.1:
E-glass has the following properties:
= 72.345 GPa
= 0.22
= 3.45 GPa
From table 2.4:
Polyester has the following properties:
= 3.4 GPa
= 0.38
= 75.9 GPa
= . + . = (0.55*72.345) + (0.45*3.4) = 41.32
= = = 7.145 GPa
= . + . = (0.38*0.55) + (0.22*0.45) = 0.308
= = = 6.047 GPa
2. S-glass and epoxy9310:
From table 2.1:
S-glass has the following properties:
= 85 GPa
= 0.22
= 4.8 GPa
From table 2.4:
Epoxy9310 has the following properties:
= 3.12 GPa
= 0.38
= 75.8 GPa
= . + . = (0.55*85) + (0.45*3.12) = 48.154
= = = 6.635 GPa
= . + . = (0.38*0.55) + (0.22*0.45) = 0.308
= = = 8.297 GPa
3. Carbon T300 and vinyl esler:
From table 2.1:
Carbon T300 has the following properties:
= 230 GPa
= 0.2
= 3.53 GPa
From table 2.4:
Vinyl esler has the following properties:
= 3.4 GPa
= 0.38
= 82.7 GPa
= . + . = (0.55*230) + (0.45*3.4) = 128.03
= = = 7.421 GPa
= . + . = (0.38*0.55) + (0.2*0.45) = 0.299
= = = 6.202 GPa
Example (3):
Select a matrix, fiber and fiber volume fraction to obtain
material with and
for different fibers and matrices we find that toBy using matlab
get these conditions we use:
D-glass as fiber and Bismaleiraide as a matrix
Matlab code:
clear;clc;close all;
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 =

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Hw4

  • 1. )4HW ( Example (1): Givens: = 230 GPa = = = 4.6 GPa = = = 92 GPa = = = 1.769 GPa = 0.25 = 0.3 = 0.6 = 0.4 ,,,:Calculate = . + . = (0.4*230) + (0.6*4.6) = 94.76 GPa = = = 7.565 GPa = . + . = (0.3*0.6) + (0.25*0.4) = 0.28 = = = 2.911 GPa
  • 2. Example (2): Calculate: , , , for a unidirectional lamina for the following material combinations: = 0.45 and = 0.55 1. E-glass and polyester (isophthalic) From table 2.1: E-glass has the following properties: = 72.345 GPa = 0.22 = 3.45 GPa From table 2.4: Polyester has the following properties: = 3.4 GPa = 0.38 = 75.9 GPa = . + . = (0.55*72.345) + (0.45*3.4) = 41.32 = = = 7.145 GPa = . + . = (0.38*0.55) + (0.22*0.45) = 0.308 = = = 6.047 GPa
  • 3. 2. S-glass and epoxy9310: From table 2.1: S-glass has the following properties: = 85 GPa = 0.22 = 4.8 GPa From table 2.4: Epoxy9310 has the following properties: = 3.12 GPa = 0.38 = 75.8 GPa = . + . = (0.55*85) + (0.45*3.12) = 48.154 = = = 6.635 GPa = . + . = (0.38*0.55) + (0.22*0.45) = 0.308 = = = 8.297 GPa 3. Carbon T300 and vinyl esler: From table 2.1: Carbon T300 has the following properties: = 230 GPa = 0.2 = 3.53 GPa From table 2.4: Vinyl esler has the following properties: = 3.4 GPa = 0.38 = 82.7 GPa
  • 4. = . + . = (0.55*230) + (0.45*3.4) = 128.03 = = = 7.421 GPa = . + . = (0.38*0.55) + (0.2*0.45) = 0.299 = = = 6.202 GPa Example (3): Select a matrix, fiber and fiber volume fraction to obtain material with and for different fibers and matrices we find that toBy using matlab get these conditions we use: D-glass as fiber and Bismaleiraide as a matrix Matlab code: clear;clc;close all; 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 =