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API has not established rules for “rounding off”.
ITAC API 653 Summary, 2007 Page 1- 11Minimum Thickness Calculation forWelded Tank Shell(API 653 Section 4.3.3.1)tmin= 2.6 (H-1)DGSEtmin=the minimum acceptable thickness, in inches, for each course ascalculated from the above formula; however, tminshall not be lessthan 0.1 inch for any tank course.D =nominal diameter of tank, in feet.H= height from the bottom of the shell course under consideration to themaximum liquid level when evaluating an entire shell course, in ft; or=height, from the bottom of the length L (see 4.3.2.1) from the lowest pointof the bottom of L of the locally thinned area to the maximum liquid level,in ft; or=height from the lowest point within any location of interest to themaximum liquid level, in ft.G =Highest specific gravity of the contents.S =Maximum allowable stress in pounds per square inch; use thesmaller of 0.80Y or 0.429T for bottom and second course; use thesmaller of 0.88Y or 0.472T for all other courses. Allowable shell stressesare shown in Table 4-1 for materials listed in the current and previouseditions of API Std. 12C and Std. 650. Note:For reconstructed tanks,S shall be per the current applicable standard.Y =Specified minimum yield strength of the plate; use 30,000 psi if notknown.T =The smaller of the specified minimum tensile strength of the plateor 80,000 psi; use 55,000 psi if not known.E =Original joint efficiency for the tank. Use Table 4-2if original E isunknown. E=1.0 when evaluating the retirement thickness in acorroded plate, when away from welds or joints by at least thegreater of one inch or twice the plate thickness.
ITAC API 653 Summary, 2007 Page 1- 12FOR UNKNOWN MATERIALMaximum Allowable Stress(in PSI)Use the smaller of:NOTE:The Third Edition of API 653 has added Table 4-1, Maximum AllowableShell Stresses (not for use for reconstructed tanks). This will make stresscalculations much easier.Sample problem for minimum thickness of a welded tank shell.An inspection of a welded, 138 foot diameter tank, 50 feet tall, 48 feet fill height showssome scattered pitting in the first course. What is the minimum shell thickness requiredfor this tank, if the specific gravity of the product is 0.9?tmin= 2.6 (H-1) DGSEtmin= ?D = 138'H = 48'G = .9S = 23,600 (from Table 4-1)E = 1tmin= 2.6 (48-1)((138)(.9)23,600(1)15,177.2423,600tmin= .643"First or Second Course(Yield).80Y = .80 X 30,000 = 24,000or(Tensile).429T = .429 X 55,000 = 23,595Other Courses(Yield).88Y = .88 X 30,000 = 26,400or(Tensile).472T = .472 X 55,000 = 25,960
ITAC API 653 Summary, 2007 Page 1- 13Practice Problemtmin= 2.6 (H-1) DGSEA 190' diameter tank has a pit that measures 5/16" deep in the first course, what is themin t, if the fill is 42 feet and the specific gravity is 0.6? (The pit is not in a weld seam orHAZ.) The material is unknown.S =Table 4-1E =Original joint efficiency for the tank. Use Table 4-2 if original E is unknown.
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Fall '19
Arc welding, Tank Bottom, Haward Technology Middle East