Consider a tensile test of a unidirectional laminate. For the matrix material, a

Question

Consider a tensile test of a unidirectional laminate. For the matrix material, assume that the modulus, Em = 3.0 GPa, and that major Poisson's ratio I'm-03. Use MATLAB or equivalent to generate plots of the following quantities: Pe a) Plot "7P vs Vy for IM-7 carbon fiber (Ef = 301 GPa) E-glass (Er = 72.4 GPa) S-glass (Ef -86.9 GPa) Kevlar (Ef = 141 GPa) · . . b) At what values of do the fibers carry 75% of the load? Plot% vs.%m for a general unidirectional laminate in tension on a log-log scale. Plot this curve for Vf = 0.1 , 0.25, 0.75, and 0.9. On this same log-log plot, draw vertical lines corresponding c) to the ratio of %m for the four materials in part a).

Explanation / Answer

I have wrote two different programs: 1st one isfor part (a) and (b) and second one is for part (c).

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clc;clear all;
E_m=3; %Gpa
v_m=0.3;
E_f_cf=301;
E_f_eg=72.4;
E_f_sg=86.9;
E_f_k=141;
% Variation of volume fraction of fibre
Vf=0.01:0.01:0.9;
% equivalent Youngs modulus;
E_eq_cf=E_m*(1-Vf)+Vf*E_f_cf;
% Load ratio will be the ratio of Youngs modulus
P_fc_cf=E_f_cf./E_eq_cf;
% equivalent Youngs modulus with E glass;
E_eq_eg=E_m*(1-Vf)+Vf*E_f_eg;
P_fc_eg=E_f_eg./E_eq_eg;
% equivalent Youngs modulus; S glass
E_eq_sg=E_m*(1-Vf)+Vf*E_f_sg;
P_fc_sg=E_f_sg./E_eq_sg;
% equivalent Youngs modulus; Kevlar
E_eq_k=E_m*(1-Vf)+Vf*E_f_k;
P_fc_k=E_f_k./E_eq_k;
% plotting the result
figure(1)
plot(Vf,P_fc_cf,Vf,P_fc_eg,Vf,P_fc_sg,Vf,P_fc_k,'LineWidth',1.5);
legend('IM7 carbon fiber','E Glass', 'S Glass','Kevlar')
ylabel('P_f/P_c');xlabel('Volume fraction of fibre')
hold on;
% putting text boxes for the value of the 75% fibre load
ycf=0.75*max(P_fc_cf);
text(0.02,ycf,'leftarrow 75% of load carried by IM7 carbon fibre fibre')

yeg=0.75*max(P_fc_eg);
text(0.03,yeg,'leftarrow 75% of load carried by E Glass fibre')

ysg=0.75*max(P_fc_sg);
text(0.03,ysg,'leftarrow 75% of load carried by S Glass fibre')

ycf=0.75*max(P_fc_k);
text(0.025,ycf,'leftarrow 75% of load carried by Kevlar fibre')

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for part (c), Since the E_f/E_m will be a constant value so, there will not be any varaition with P_f/P_c. But if we want to plot for the E_f/E_eq then we can get the varaition because E_f and E_m both don't vary with volume fraction. Although, I have written the code You can modify according to your requirement.

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clc;clear all;
E_m=3; %Gpa
v_m=0.3;
E_f_cf=301;
E_f_eg=72.4;
E_f_sg=86.9;
E_f_k=141;
% Variation of volume fraction of fibre
Vf=[0.1,0.25,0.75,0.9];
% equivalent Youngs modulus;
E_eq_cf=E_m*(1-Vf)+Vf*E_f_cf;
% Load ratio will be the ratio of Youngs modulus
P_fc_cf=E_f_cf./E_eq_cf;
% equivalent Youngs modulus with E glass;
E_eq_eg=E_m*(1-Vf)+Vf*E_f_eg;
P_fc_eg=E_f_eg./E_eq_eg;
% equivalent Youngs modulus; S glass
E_eq_sg=E_m*(1-Vf)+Vf*E_f_sg;
P_fc_sg=E_f_sg./E_eq_sg;
% equivalent Youngs modulus; Kevlar
E_eq_k=E_m*(1-Vf)+Vf*E_f_k;
P_fc_k=E_f_k./E_eq_k;
% plotting the result

E_fm_cf=E_f_cf./E_m;
E_fm_eg=E_f_eg./E_m;
E_fm_sg=E_f_sg./E_m;
E_fm_k=E_f_k./E_m;
loglog(E_fm_cf,P_fc_cf,E_fm_eg,P_fc_eg,E_fm_sg,P_fc_sg,E_fm_k,P_fc_k,'LineWidth',1.5);
ylabel('P_f/P_c');xlabel('E_f/E_m')

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