Creating a function to calculate yield strength of a metal?

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James
James 2018-5-22
评论： Ameer Hamza 2018-5-22
I am working on the following problem.
This is the code i have so far.
if true
% code
grain_size = 0.1;
strength_vector = zeros(1,20);
n = 1;
while (grain_size < 2.0)
A = HallPetch(12000, 9600, grain_size);
strength_vector(n) = A;
n = n + 1;
grain_size = grain_size + 0.1;
end
M = (0.1: 0.1: 2.0);
M = M';
strength_vector = strength_vector';
fprintf('Grain Size (mm) Yield Strength(psi) \n');
fprintf('%5.1f %25.3f \n', [M, strength_vector]');
function [yield_strength] = HallPetch(sigma_nought, k, d)
yield_strength = sigma_nought + k*(d^(-1/2));
end
end
It seems to work up until the grain size equals 2.0, then the result it gives me is zero. Any ideas why?

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回答（1 个）

Ameer Hamza
Ameer Hamza 2018-5-22
编辑：Ameer Hamza 2018-5-22
Since you want to get the value up to 2 you need to increase the limit of while loop a little bit
while (grain_size < 2.01)
Now it will run for grain_size=2
Grain Size (mm) Yield Strength(psi)
0.1 42357.866
0.2 33466.253
0.3 29527.122
0.4 27178.933
0.5 25576.450
0.6 24393.547
0.7 23474.195
0.8 22733.126
0.9 22119.289
1.0 21600.000
1.1 21153.241
1.2 20763.561
1.3 20419.757
1.4 20113.481
1.5 19838.367
1.6 19589.466
1.7 19362.864
1.8 19155.418
1.9 18964.572
2.0 18788.225
Note you cannot use
while (grain_size <= 2)
because of finite precision of floating point.

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