Question

The figure illustrates the nonpermanent connection of a steel cylinder head to a grade 30 cast-iron pressure vessel using N bolts. A confined gasket seal has an effective sealing diameter D. The cylinder stores gas at a maximum pressure Pg For the specifications given in the table for the specific problem assigned, select a suitable bolt length, assuming bolts are available in increments of 5 mm. Then determine the yielding factor of safety np the load factor nl, and the joint separation factor no. Take the value of Efor steel as 207 GPa and for cast iron as 100 GPa. F E -D es Head thickness (A) Cylinder thickness (8) Cylinder ID Gasket sealing diameter (D) Bolt circle diameter (6) Head outer diameter ( Number of bolts (1) Cylinder pressure (Pg) Bolt grade (6) Bolt spec. 20 mm 20 mm 60 mm 100 mm 200 mm 300 mm 10 6.15 MPa ISO 9.8 M12 * 1.75

Please show all working clearly and box and label each answer. thank you

Answer 1

solution Fiest of all we need to calculete, grip length Lg = A+B Lg - 20+20 Lg= 40mm 'so, Height of Nut (H) foe size of get 12.000 (Hexagona) > H 10.8mm by using table. (L)) so, fastener length L> Lg tH L> 40+10.8 L> 50.8mm Te 55mm Ans so, the suitable length of bolt is 55 mm PAGE 1

Now, length Calculate the thead fe metric balt will be Lr ad +6 = 2X 12+ 6 30mm TL So, By using table, we have expression fos unthzended length(la) LOOLT (grip) La = 55-30 Ld 25mm Similarly (LA) toe threaded Length It = lg-ld (Grip) =40-25 1 Lt Lt = 15 mm Now, Calculate, Area of Unthreaded postion- Ad Id² IX 12 At : 113.1 mm? TPAGE 2 2

md for the eaded Area will be using table (from d=12mm) A84.5 mm Now, Bolt stiffnen will be (ko) ko Ad At Es Adat t Atld A we have given, E = 207 GPa (steel) E = 100 Pa (CT) . kb 113.1 x 843 X 207 113, 1X15 184.3725 kb 518.8 MN/m so, Member stiffness for steel Cylinder head will be (ka) 0.5774 E ki ( 1.155+++D-d)(D+d) (1.155t + D + dod) where, In t 20mm 0:5774 T007) x(12) (H.155 (29) +1% -12)(18+12) (1155 (20) +18719) (18-12) (Page 3 Toni

k, = 4470 M N/m // Now, Member stiffness for cast Iron ka - 0.5774 xTTXdxE Im [(-155 410-4)+d) (1.155 t++d)(0-1) ke 0:5774 X TTX12X100 In 1.155X 20+18-12)(18+12) (1.155X20 +10 +12)(18-12) 2159 MN/m ka 1 Woo, Member stiffness for entize frustum t L K2 - kmaluhoo + 2159 Ikm = 1456 MN/m // so Calculate. Joint constance is a Kb с — kbtkm 518.8 1456+ 5188 c - 0.263 [muehl //

Therefor, foe M12 + 1.75 Peo of strength th in table, 6 p 650 MPa. Preload will be Fi - 0.75 X At x6p 0.75 X 84.3 X6507753 fi = 41.1 KN Wow, total external Load perbolt is PT = Pg x Ag o bere Pg = 6.15 MPa given, Ag - Aeffective area of cylinder, 2. Ag = (dg)" - 7 x 100 =7853.98 mm² N = Number of bolts. N=10 PT0615 10°° x 7853.98 10 Py = 4.83 kN/bolt TPAGE 5

2) zielding factor of safety- • (tos) y = Np PTX ct fi 6. AL 4650X10-3 -3 X 84.3 0.263 X 4.83 +4111 np =1.293 - Ans (2> Overload factor of safety- Ni= 6p At - fi CPT 650x133x 84,3-411 0.263 X4.83 10.78 Ans. ni A i) Separation factor of safety Fi Pr (1-() 41.1 4,83 (1-0-263) no Ans Mo = 11:55 No = 11.55 11.551 PAGF6

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