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F = 2PdgR2 T2 = 762 Ib/in <br /> 3) Cone unit force, T1, the Meridional unit force. See Figure No. A4 <br /> —D <br /> PA1 <br /> T1 T1 <br /> A A <br /> a Id <br /> 1 <br /> C <br /> Wt = weight of liquid in cone <br /> Figure No. A4 <br /> Al = PI/4* D2 <br /> Al = 22167.08 square inches 153.938 square feet <br /> V, Volume of Cone = (PI*D/2*D/2*d)/3 <br /> V= 622.1662 ft3 <br /> Wt = SG* 62.4 Ib/ft3 * V <br /> Wt = 38,823 Ib <br /> Sum Fy= 0 <br /> T1Cos(a)PI*D - Wt-PA, =0 <br /> Ti = (Wt + PA1)/Cos(a)PI*D <br /> Ti = 466 Ib/in <br /> 4) Knuckle Region Reinforcement Check per API 620 3.12 <br /> The measured cone bottom thickness is defined as th here, th = 0.240 inches <br /> The measured shell thickness is defined as tc here, tc= 0.236 inches <br /> R2, radius of curvature of cone at A= 96.99356 inches <br /> Rc, radius of curvature shell = 84 inches <br /> wh = 0.6* sgrt(R2 *th) wc= 0.6* sgrt(Rc*tc) <br /> wn = 2.894865 inches WC= 2.671449 inches <br /> Available reinforcing area, Aa= wh*th +wc*tc <br /> Aa= 1.32523 sq <br />