A rod with a cross -sectional area of A and loaded in tension with an axial force of P= 2000 Ibf undergoes a stress of sigma =P/A Using a material strength of 24 kpsi and a design factor of 3 .0. determine the minimum diameter of a solid circular rod. Using Table A-17 , select a preferred fractional diameter and determine the rod's factor of safety. Since A=pi d^2/4,sigma =P/A and from Eqcdot (1-3),sigma =S/n_(d) . then sigma =(P)/(A)=(P)/(pi d^2)/4=(S)/(n_(d)) Solving for d yields d=((4Pn_(d))/(pi S))^1/2=((4(2000)3)/(pi (24000)))^1/2=0.564in From Table A -17, the next higher preferred size is (5)/(8)in=0.625 in. Thus, when n_(d) is replaced with n in the equation developed above, the factor of safety n is n=(pi Sd^2)/(4P)=(pi (24000)0.625^2)/(4(2000))=3.68 Thus rounding the diameter has increased the actual design factor.