Table of Contents

— David Wagner 2009/10/18 12:39

√ME 5463-001 Fracture Mechanics, Homework #11

Figure E5.1

5.2 Shown in Figure E5.1* is a dark-field isochromatic pattern of a large plate with a deep crack approaching the free edge. The model was made from 0.200 inch thick polycarbonate (fsigma = 40 psi/in/FR), and the scribed grid lines are 1.0 inch apart. Estimate the stress intensity factor from this pattern, using both the apogee and differencing methods. Compare the two methods as applied to this example.

* This figure is available in high-resolution digital form at http://www.wam.umd.edu/~rsanford http://terpconnect.umd.edu/~sanford/exercises.html.

Annotated Drawing

  • f_sigma = 40
  • N_1 = 2
  • r_1 = 194.5 182.2 px / 600 px/in.
  • theta_1 = 75.11 deg.
  • N_2 = 3
  • r_2 = 93.8 87.7 px / 600 px/in.
  • theta_2 = 82.65 deg.
  • a = 1.2 in

[It's difficult to see the apogee markers on the drawing at right.]

Apogee Method

% apogee_sigma_0x.m
 
function sigma_0x=apogee_sigma_0x(K,r_m,theta_m)
sigma_0x = K./sqrt(2*pi*r_m) .* sin(theta_m)*cos(theta_m)./...
   (cos(theta_m)*sin(3*theta_m/2) + 1.5*sin(theta_m)*cos(3*theta_m/2));
return
end
% apogee_0.m
%
% This function returns zero when K is correct.
 
function makezero = apogee_0(K,r,theta,N,f_sigma,t)
 
makezero=(K./sqrt(2*pi*r).*sin(theta) - apogee_sigma_0x(K,r,theta)*sin(3*theta/2)).^2 ...
  + (apogee_sigma_0x(K,r,theta)*cos(3*theta/2)).^2 ...
  - (N*f_sigma/t)^2;
 
return
end
% hw_11a.m
 
     K = [320:1:329];
    r_1 = 182.2/600;
theta_1 = 75.11*pi/180.;
    N_1 = 2;
f_sigma = 40.;
      t = 0.2;
 
apogee_results = [K;apogee_0(K,r_1,theta_1,N_1,f_sigma,t)]
apogee_results =

    320.000    321.000    322.000    323.000    324.000    325.000    326.000    327.000    328.000    329.000
  -3908.247  -2931.150  -1951.003   -967.808     18.436   1007.728   2000.069   2995.459   3993.898   4995.385

The function is zero between K=323 and 324, and K = 324 is a reasonable estimate.

Calculating K by the apogee method using N_2, r_2, and theta_2 results in a K of about 504.

Differencing Method

% diff_f.m
 
function f=diff_f(r_a,theta)
 
f = sqrt(sin(theta)^2 + 2*sqrt(2*r_a)*sin(theta)*sin(3*theta/2)+2*r_a  );
 
return
end
% diff_k.m
 
function K=diff_k(N_1,r_1,theta_1, N_2,r_2,theta_2,a,t,f_sigma)
 
K = sqrt(2*pi)*(N_2-N_1)*sqrt(r_1*r_2)*f_sigma/((diff_f(r_2/a,theta_2)*sqrt(r_1) - diff_f(r_1/a,theta_1)*sqrt(r_2))*t);
 
return
end
% hw_11b.m
 
    r_1 = 182.2/600;
theta_1 = 75.11*pi/180.;
    N_1 = 2;
    r_2 = 87.7/600;
theta_2 = 82.65*pi/180.;
    N_2 = 3;
      a = 1.2;
f_sigma = 40.;
      t = 0.2;
 
K = diff_k(N_1,r_1,theta_1,N_2,r_2,theta_2,a,t,f_sigma)
K =  669.95

Comparison

None of these results are consistent.

The results could likely be improved by using image processing techniques to find the apogee points.


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