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+//Example 6.3//
+
+P=6*10^3//N //Newton // load on the sample
+A0=(10/2)*10^-3;//N/m^2
+s=P/(%pi*A0^2)
+mprintf("s = %e N/m^2 = 76.4 MPa",s)
+mprintf("  (As M= 10^6)")
+s1=76.4;//MPa //Megapascal //modulus od elasticity
+E=70*10^3;//MPa//Megapascal //Young's Modulus
+e=s1/E
+mprintf("\n e = %e",e)
+//the strain of diameter is calculated as
+v=0.33;//given
+ed=-v*e
+mprintf("\n ed = %e ",ed)
+//resulting diameter
+d0=10;//mm 
+df=d0*(ed+1)
+mprintf("\n df = %f mm",df)
+//compressive stress
+ed1=+3.60*10^-4;//the diameter strain will be of equal magnitude but opposite sign
+df1=d0*(ed1+1);
+mprintf("\n df1 = %f mm",df1)