Question

O-ring

The figure shows a pressure vessel in which an O-ring is used as a seal. The pressure vessel cap is made of steel, and the vessel itself is made of cast iron. The pressure inside the vessel varies between 0 MPa and 20 MPa. The dimensions shown in the figure are A = 20, B = 20, C = 100, D = 150, E = 200, y F = 300, all in millimeters. ISO class 8.8 bolts with a a diameter of 10 mm are used to make up the joint between the vessel cap and its cylindrical body. The bolts are installed at regular angular increments around the vessel. The manufacturer of the O-ring requires that, to avoid leaks, it must not expand more than 0.1 mm How many bolts are needed to maintain the O-ring seal and to prevent failure of the bolts against all the static failure modes (yielding, overload and separation) and for them to9 have infinite life against fatigue? In al cases, the minimum safety factor must be 1.15.

Answer 1

Detailed answer of above asked question is solved as below.

cap Free body diagrom. $ 300mm Given = P = 2ompa. & Ø 200mm FOS= 1.15 T 0150mm given material is te=20mm. Dyessel exlinder a te romm, vessel cast iron is used, properties are 100mm Ef = 54. apa d=010mm 2) vessel cap steel is used, properties are Ec=200 GP a 3) Bolt, iso. class 818 6014. Clow carbon steel is used syt = 640 mpa. E6= 220 GP a' Stiffness of bolt 16= tette = 20+20= 40mm ko Ab Eb = (Fg do?) Eb = ( x102) 22.10 481.8689x10 N/mm Stiffness of connected member Am = 1 / 1 (3d 2-d ²) = 271 do 2 = 21TX 102 = 628.8185mm? koof- Am Ef te 628.3185 x 54 x 10 = 1.69 64x100 N/mm 20 Kme= Imm Am Ec - 628.31854 200x100 - 6.083x10 Nm 20 I= km 1 km + 1 kmc = 3.89550106 6.2881106 km = 1:33578108 N/mm.

- Allowable stress for bolt bon ball b = Sydxo5 - 640xor 2 = 278~260 N/mm2 fos 15 - Resultant load on each bolt. fo= A b ball be . (1 x dp 2) 6 all b = (1 x 102) x 278.26 Fo= 21.858103 N Fo= ( kbp+Fi Fi s preloo Fiz Preload in each bold. Ckht km 21.95X10 = 431.96x103 431.96x103 +1:33 x106 → P = 89.12X103N Tutal force due to internal pressure Pi F = I d, 2xP; = I x 10024 20 = 157.07X100 N No. of bult required PoE N= 157:07*100 = 1.76 bolt → 89.12X103. N = 1.76 ²2 bolt required