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+// exa 3.2 Pg 63
+
+clc;clear;close;
+
+// Given Data
+P=6;// kN
+alfa=30;// degree
+Sut=250;// MPa
+n=2.5;// factor of safety
+
+sigma_w=Sut/n;// MPa (Working stress for the link)
+PH=P*10**3*cosd(alfa);// kN
+PV=P*10**3*sind(alfa);// kN
+
+t=poly(0,'t');// thickness of link
+A=2*t*t;// mm.sq. 
+sigma_d=PH/A;// N/mm.sq.
+M=PH*100+PV*250;// N.mm
+I=t*(2*t)**3/12;// mm^4 (Moment of Inertia)
+sigma_t=M*t/I;// N/mm.sq.
+//maximum tensile stress at the top fibres = sigma_d+sigma_t=sigma_w  ...eqn(1)
+expr=sigma_d+sigma_t-sigma_w ;// expression of polynomial from above eqn.
+t=roots(numer(expr));// solving the equation (as denominator will me be multiplied by zero on R.H.S)
+t=t(1);// mm // discarding -ve roots
+printf('dimension of cross section of link, t=%.f mm.',t)