
iterations adjustments
Posted:
Nov 23, 2012 4:01 PM


Hello, I really have to deal with the problem below. I am a beginner in matlab and my deadline is quite pressing. Thank you in advance! Well, I have to guess an initial value "n" in order to find the ultimate value "Pu". When the extracting Pu deviates from a specific value (Pe) more than 0.1% then I need the code to make use of the last value of "n" until the parameter "Pu"reach the Pe. Otherwise the parameter n is kept. herein is the code: function [Pu] = KN (c,B,d,rho,fck,fsy,Es,eta,fcube,h) % Starting Assumption c = 569; r1 = 130; r2 = 90; B = (130 * 90)^0.5; Ap = 0.1397; h = 43; rho = Ap / h; d = 0.5 * h; fck = 48.6; sigmap = 1.84; Fp = sigmap * h; fpk = 2576; fsy = fpk  (Fp / Ap); Es = 1.83 * 10^5; fcube = 0.8 * fck; eta=0.45; ddelta = 1; j = 0; delta = h; while abs(ddelta/delta) > 0.01 j = j + 1; if j > 300 break end Fc = 0.8 * 2/3 * fck * (h/2  delta/4); Ft = d * rho * fsy; Fbmax = Fc  Ft Mbmax = (Ft * (2 * d  h)  Fc * (d  13*h/16  3 * delta/32)); Fb = eta * Fbmax Mb = eta * Mbmax X = calX(c,B,d,Es,fsy,fcube,rho,Fb,Mb); y = caly(c,B,d,Es,fsy,fcube,rho,Fb,X); if B / d < 2 tasi = 0.0035 * (1  0.22*(B/d))*(1 + B/(2*y)); else tasi = 0.0019 * ( 1 + B/(2*y)); end delta1 = 1/2*tasi*(cB); ddelta = delta  delta1; delta = delta1; end TA = calTA (c,B,d,y,X); p1 = P1 (B,y,d,fcube,TA); p2 = P2 (c,B,y,d,Es,fsy,X,rho,Fb); function [P1] = P1 (B,y,d,fcube,TA) % B = diameter of the load area, given in [mm] % y = depth of the compression zone in KN model, given in [mm] % d = average effective depth of the slab, given in [mm] % fcube = compressive strength of concrete, measured on standard cubes, % given in [N/mm2] % TA = tan(alpha) % % calculate ft % if B/d < 2 ft = 825 * (0.35 + 0.3* (fcube/150))*(1  0.22 * (B / d)); else ft = 460 * (0.35 + 0.3 * (fcube/150)); end % % calculate falpha % falpha = TA * (1  TA)/(1 + TA * TA); % % calculate P1 P1 = pi * (B / d) * (y / d) * (B + 2 * y) / (B + y) * ft * falpha * d * d /1000; function [P2] = P2 (c,B,y,d,Es,fsy,X,rho,Fb) % c = diameter of the slab, given in [mm] % B = diameter of the load area, given in [mm] % y = depth of the compression zone in KN model, given in [mm] % d = average effective depth of the slab, given in [mm] % fsy = yield stress of reinforcing steel, given in [N/mm2] % Es = modulus of elastisity of reinforcing steel, given in [N/mm2] % X = boundary moment and normal force ratio % rho = reinforcement percentage % Fb = horizontal lateral normal force, given in [kN] % % Calculate kz ky = 3*(c  B)/(2*(3*d  y)); kz = ky  3 * X * c / (4 * (3 * d  y)); % Calculate R1 R2overBeta if B / d < 2 tasi = 0.0035 * (1  0.22*(B/d))*(1 + B/(2*y)); else tasi = 0.0019 * ( 1 + B/(2*y)); end rs = Es / fsy * tasi * (d  y); C0 = B/2 + 1.8 * d; if rs > C0 R1 = rho * fsy * d * ((rs  C0) + rs * log(c/(2 * rs)))/1000; R2overBeta = rho * fsy * d * C0/1000; else R1 = rho * fsy * d * rs * log(c/(2*C0))/1000; R2overBeta = rho * fsy * d * rs/1000; end % % Calculate P2 % P2 = 2 * pi / kz * (R1 + R2overBeta + Fb * (c/2/1000)); function [ta] = calTA (c,B,d,y,X) ky = 3*(c  B)/(2*(3*d  y)); kz = ky  3 * X * c / (4 * (3 * d  y)); A = 1/4.7*(1+y/B)*log(c/(B+2*y)); ta = ((kz + 1) sqrt((kz + 1)*(kz + 1)  4 * (kz + A)*(A + 1)))/(2*(kz+A)); function [y] = caly(c,B,d,Es,fsy,fcube,rho,Fb,X) dp = 100; j = 0; y = d; while abs(dp)>0.1 j = j+1; if j > 300 break end TA = calTA (c,B,d,y,X); p1 = P1(B,y,d,fcube,TA); p2 = P2(c,B,y,d,Es,fsy,X,rho,Fb); dp = p1  p2; y = y  dp/1000; end function [X] = calX(c,B,d,Es,fsy,fcube,rho,Fb,Mb) dX = 1; k = 0; X = 0; X1 = 1; while abs (dX) > 0.0000001 k = k + 1; if k > 300; break end y = caly(c,B,d,Es,fsy,fcube,rho,Fb,X); TA = calTA (c,B,d,y,X); p1 = P1 (B,y,d,fcube,TA); p2 = P2 (c,B,y,d,Es,fsy,X,rho,Fb); X1 = 4*pi*(Mb/((p1+p2)*1000/2)); dX = X1  X; X = X + dX/20; end
the code is a bit long but tahnk you again for your time and the advices!!!

