Fourth Year Composite

materials

Report: Solution to Homework IV

Report No: IV Date: 15/4/2013

Submitted to: Dr. Mohammad Tawfik

Name

Mohammad Tawfik Eraky

محمد توفيق أحمد عراقي

2013/2014

Pb 4.5

Compute, 𝑬𝟏, 𝑬𝟐 , 𝑮𝟏𝟐 and , 𝝂𝟏𝟐 given that EF=230Gpa, Em =EF/50,

GF=EF/2.5, GM=EM/2.6, 𝝂𝒇 = 𝟎. 𝟐𝟓 ,𝝂𝒎 = 𝟎. 𝟑, and Vf = 40%.the fibers

have circular cross section .Assume there is no voids

Solution

Using MATLAB code project (1) appendix (A)

We get the following results

𝑬𝟏 = 𝟗𝟒. 𝟕𝟔 𝐆𝐩𝐚

𝑬𝟐 = 𝟕. 𝟓𝟔𝟓𝟖 𝐆𝐩𝐚

𝑮𝟏𝟐 = 𝟐. 𝟗𝟏𝟏𝟒 𝑮𝒑𝒂

𝝂𝟏𝟐 = 𝟎. 𝟐𝟖

Pb 4.6

Compute all the elastic properties ( 𝑬𝟏, 𝑬𝟐 , 𝑮𝟏𝟐 , 𝑮𝟐𝟑and , 𝝂𝟏𝟐 )for

unidirectional lamina for the following material combinations:-

(a) E-glass-polyester(Isophthalic)

(b)S-glass Epoxy (9310)

(c) carbon (T300)-vinylester

Given Vf = 0.55 , 𝝂𝒇 = 𝟎. 𝟐𝟐 and 𝝂𝒎 = 𝟎. 𝟑𝟖 compare the results

using (ROM) formulas except for G23

Solution

From table 2.1, 2.4 in Barbero introduction to composite materials

We get the following properties we need to solve that problem

Composite material Modulus of elasticity (Gpa)

E-glass 72.345

polyester(Isophthalic) 3.4

s-glass 85

Epoxy(9310) 3.12 Carbon(T300) 230

vinylester 3.4

Using MATLAB code (appendix B) we get the following results

(a) E-glass-polyester(Isophthalic)

𝑬𝟏 = 𝟒𝟏. 𝟑𝟏𝟗𝟖 𝐆𝐩𝐚

𝑬𝟐 = 𝟕. 𝟏𝟒𝟓𝟏 𝐆𝐩𝐚

𝑮𝟏𝟐 = 𝟐. 𝟔𝟎𝟓𝟐 𝑮𝒑𝒂

𝑮𝟐𝟑 = 3.5779 Gpa 𝝂𝟏𝟐 = 𝟎. 𝟐𝟗𝟐𝟎

(b)S-glass Epoxy (9310)

𝑬𝟏 = 𝟒𝟖. 𝟏𝟗𝟎𝟎 𝐆𝐩𝐚

𝑬𝟐 = 𝟔. 𝟕𝟗𝟖𝟑 𝐆𝐩𝐚

𝑮𝟏𝟐 = 𝟐. 𝟒𝟕𝟓𝟖𝑮𝒑𝒂

𝑮𝟐𝟑 = 3.4378 Gpa 𝝂𝟏𝟐 = 𝟎. 𝟐𝟗𝟐𝟎 is constant as it is a function in representative

volume and Poisson ratio which given as constant

(c) Carbon (T300)-vinylester

𝑬𝟏 = 𝟏𝟐𝟖. 𝟎𝟑𝐆𝐩𝐚

𝑬𝟐 = 𝟕. 𝟒𝟐𝟏𝟓 𝐆𝐩𝐚

𝑮𝟏𝟐 = 𝟐. 𝟔𝟗𝟒𝟓𝑮𝒑𝒂

𝑮𝟐𝟑 = 3.8509 Gpa 𝝂𝟏𝟐 = 𝟎. 𝟐𝟗𝟐𝟎

Pb 4.11

Select materials (fibers, matrix) with fiber volume fraction to

obtain

𝑬𝟏 > 𝟑𝟎𝑮𝒑𝒂 , 𝑬𝟏

𝑬𝟐< 𝟑. 𝟓

Solution

We choose from table 2.1, 2.4

D-glass as fibers with 𝑬𝐟 = 𝟓𝟓𝑮𝒑𝒂

Polyester Isophthalic with 𝑬𝐦 = 𝟑. 𝟒 𝑮𝒑𝒂

By performing many iteration on MATLAB (Appendix C) we get the

following material with the representative volume to satisfy the

required condition.

As we see from the figure the volume fraction of the fibers

Vf = 0.78 hence Vm=0.22 to satisfy the condition.

Appendix A

clc;clear all;close all; clc;clear all;close all; %given the following properities e_f=230 ; % modlus of elasticty of Fibers [input in GPa] e_m= 230/50 ; % Modlus of elasticty of matrix [input in GPa] v_m= 0.6; % Representative volume of the matrix [input] v_f=0.4; % Reprenstative volume of the fibers [input] g_m=e_m/2.6; % modlus of regidity of the matrix [input in Gpa] g_f= 230/2.5; % modlus of regidity of the fibers[input in Gpa] alpha_m= 60/1000000; % Thermal coefficient of the matrix [input /c] alpha_f=5.4/1000000 ; % Thermal coeffient of the fibers [input /c] new_m=0.3; new_f=0.25; %outputs

e_1=v_m*e_m+v_f*e_f; % Modlus of elastisity in fiber direction e_2=(e_m*e_f)/(v_f*e_m+v_m*e_f); % modlus of elasticty in direction

transverse to the fibers alpha_1=(v_f*e_f*alpha_f+v_m*e_m*alpha_m)/(v_f*e_f+v_m*e_m); % thermal

coeffiecient in direction 1 alpha_2=alpha_f*v_f+alpha_m*v_m ; % thermal coeffient in direction 2 new_12=new_f*v_f+new_m*v_m ; new_21=(e_2*new_12)/e_1; g12=(g_m*g_f)/(v_m*g_f+v_f*g_m);% inplane shear modlus

%display outputs

display(e_1); display(e_2); display(alpha_1); display(alpha_2); display(g12); display(new_12); display(new_21);

Appendix B %given the following properities e_f=72.345 ; % modlus of elasticty of Fibers [input in GPa] e_m= 3.4 ; % Modlus of elasticty of matrix [input in GPa] v_m= 0.45; % Representative volume of the matrix [input] v_f=0.55; % Reprenstative volume of the fibers [input]

alpha_m= 60/1000000; % Thermal coefficient of the matrix [input /c] alpha_f=5.4/1000000 ; % Thermal coeffient of the fibers [input /c] new_m=0.38; new_f=0.22; g_m=e_m/(2*(1+new_m)); % modlus of regidity of the matrix

[input in Gpa] g_f= e_f/(2*(1+new_f)); % modlus of regidity of the

fibers[input in Gpa] %outputs

e_1=v_m*e_m+v_f*e_f; %

Modlus of elastisity in fiber direction e_2=(e_m*e_f)/(v_f*e_m+v_m*e_f); %

modlus of elasticty in direction transverse to the fibers alpha_1=(v_f*e_f*alpha_f+v_m*e_m*alpha_m)/(v_f*e_f+v_m*e_m); %

thermal coeffiecient in direction 1 alpha_2=alpha_f*v_f+alpha_m*v_m ; %

thermal coeffient in direction 2 new_12=new_f*v_f+new_m*v_m ; new_21=(e_2*new_12)/e_1; %

inplane shear modlus g12=(g_m*g_f)/(v_m*g_f+v_f*g_m); eta_23=(3-4*v_m+(g_m/g_f))/(4*(1-v_m)); g_23=g_m*((v_f+eta_23*(1-v_f))/(eta_23*(1-v_f)+(v_f*g_m/g_f)));

%display outputs

display(e_1); display(e_2); display(alpha_1); display(alpha_2); display(g12); display(new_12); display(new_21); display(g_23);

Appendix C clc;clear all;close all; %given the following properities e_f=55; % modlus of elasticty of Fibers [input in GPa] e_m= 3.4 ; % Modlus of elasticty of matrix [input in GPa] v_m= 0:0.01:1; % Representative volume of the matrix [input]

for i=1:length(v_m) v_f(i)=1-v_m(i); e_1(i)=v_m(i)*e_m+v_f(i)*e_f; e_2(i)=(e_m*e_f)/(v_f(i)*e_m+v_m(i)*e_f); e_ratio(i)=(e_1(i))/(e_2(i)); e_ratioReq(i)=3.4; end plot(v_m,e_ratio,'b','linewidth',1.5);hold on plot(v_m,e_ratioReq,'r','linewidth',2); grid on xlabel('V_f'); ylabel('e_ratio')