Talk:Hosford yield criterion

Hosford Yield Criterion and Tresca
It was me that changed the exponent n from 1 to ∞ to make the statement about Tresca correct. I think you shouldn't undo this change, because your initial statement is wrong (try filling in n = 1 in the Hosford yield equation and find that you end up with zero on the left side). Notice that von Mises and Tresca are the bounds of the Hosford yield criterion. The minimal value of n equals 2, which corresponds with von Mises and gives the outer bound. When n is increased, a Hosford yield surface is found that lies inside the von Mises yield surface and ultimatelly, the Hosford yield surface will correspond with the Tresca yield surface. I didn't know better than to change it in this way and thought maybe it would get you to think about your statement. Note that I didn't change the figures because I don't know how to do that. MWoldman (talk) 07:35, 9 September 2009 (UTC)


 * I still think you should change the exponent from 1 to ∞ to make the statement about Tresca correct. Try increasing the exponent and find that the higher the value for n, the more the yield surface will look like the Tresca yield surface. And isn't it counterintuitive that your yield surface for n = 10 lies between the yield surfaces for n = 1 and n =2? MWoldman (talk) 10:03, 10 September 2009 (UTC)


 * For Tresca the exponent has to be ∞. Take s1, s2, s3 are 50, 40, 10. n=1 gives you 25. This is WRONG. n=50 gives you 29.6 (don't forget the n-th root). For n going to infinity you get some kind of maximum norm. The value converges to max(...) - ergo, Tresca. -- Wbetz (talk) 21:04, 16 February 2011 (UTC)
 * Agreed and updated. Note that the figure is still accurate for plane stress (s3 = 0); the n = 1 and n = infty lines overlap.
 * A matlab script for plotting the 2D surface is given below. Bbanerje (talk) 22:39, 16 February 2011 (UTC)
 * Also, here's how the $$n \rightarrow \infty$$ limit is derived

\sigma_y = \lim_{n \rightarrow \infty} \|x\|_n=\lim_{n \rightarrow \infty}\left(|x_1|^n+|x_2|^n+\cdots+|x_p|^n\right)^{1/n} \,. $$
 * If the $$x_i$$ are arranged in ascending order, $$x_1 < x_2 < \dots < x_p$$, we can write

\sigma_y = \lim_{n \rightarrow \infty} \left[|x_p|^n\left\{\left(\frac{|x_1|}{|x_p|}\right)^n+\left(\frac{|x_2|}{|x_p|}\right)^n+\cdots+1\right\}\right]^{1/n} \,. $$
 * Taking the limit, we have,

\sigma_y = \left(|x_p|^n\right)^{1/n} = |x_p| = \max(|x_i|)\,. $$
 * Bbanerje (talk) 03:24, 17 February 2011 (UTC)


 * However, I am sorry that I have to correct myself. Besides n going to infinity, n=1 is also a valid solution (My example given above was wrong.).

\sigma_y = \left(\tfrac{1}{2}|\sigma_2-\sigma_3| + \tfrac{1}{2}|\sigma_3-\sigma_1| + \tfrac{1}{2}|\sigma_1-\sigma_2|\right) = \tfrac{1}{2} \cdot 2 \cdot \max(|\sigma_1 - \sigma_2|, |\sigma_2 - \sigma_3| , |\sigma_3 - \sigma_1| ) \,. $$
 * Which is also equal to Tresca. Therefore, two solutions exist.
 * Wbetz (talk) 08:39, 17 February 2011 (UTC)

Plotting the plane Hosford criterion
function plotHosford

s1 = -1.5:0.01:1.5; s2 = -1.5:0.01:1.5; s3 = 0;

for ii=1:length(s1) for jj=1:length(s2) sigy_n1 = hosford(s1(ii), s2(jj), s3, 1); sigy_n2 = hosford(s1(ii), s2(jj), s3, 2); sigy_n50 = hosford(s1(ii), s2(jj), s3, 50); s1_n1 = s1(ii)/sigy_n1; s2_n1 = s2(jj)/sigy_n1; s1_n2 = s1(ii)/sigy_n2; s2_n2 = s2(jj)/sigy_n2; s1_n50 = s1(ii)/sigy_n50; s2_n50 = s2(jj)/sigy_n50; plot(s1_n1, s2_n1, 'r.'); hold on; plot(s1_n2, s2_n2, 'b.'); hold on; plot(s1_n50, s2_n50, 'k.'); hold on; end end axis equal

function [sigy] = hosford(s1, s2, s3, n) s23 = abs(s2-s3); s31 = abs(s3-s1); s12 = abs(s1-s2); sigy = (0.5*(s23^n + s31^n + s12^n))^(1/n);