How can I do a drawing of this shape؟

Question

HAMZA NASSAR 2018-8-17

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https://ww2.mathworks.cn/matlabcentral/answers/415208-how-can-i-do-a-drawing-of-this-shape

评论： HAMZA NASSAR 2019-12-1

image.png

%File of Result of stress by FEM v=0.35;% Poission ratio sv=66;% vertical stress sh=45;% min horizontal stress Pp=30;% pore pressure sH=75; % max horizontal stress a=0.08;

m=0:39; r=a+(5..*10.^-3+0.1.*m); n=linspace(1,120,40) theta= (3./2+3.*(n-1)) az=0; inc=0; b=((a.^2)./(r.^2)); N=40; % eqn 3.2.6 sx = sH.*(cos(az).* cos(r)).^2 + sh.*(sin(az).*cos(inc)).^2 + sv.*(-sin(inc).^2); sy = sH.*(-sin(az)).^2 + sh.*(cos(az)^2); sz = sH.*((cos(az).* sin(inc)).^2) + sh.*(sin(az).^2).*(sin(inc).^2)+ sv.*(cos(inc).^2); syz = (sH.*-sin(az).*sin(inc).*cos(az))+(sh.*cos(az).* sin(inc).*sin(az))+(sv.*cos(inc).*0); sxz = (sH.*cos(az).* cos(az).*sin(r).*sin(r))+(sh.*sin(az).*cos(inc).*sin(az).*sin(inc))+(sv.*- sin(r).*cos(inc)); sxy =(sH*cos(az)* cos(inc)*-sin(az))+ (sh*sin(az)*cos(inc)*cos(az))+(sv*-sin(inc)*0); sigma_tt = 0.5.*(sx + sy).*(1 +b) - 0.5.*(sx - sy).*(1 + 3.*b).*cos(2.*theta) - sxy.*(1 + 3.*b).*sin(2.*theta);