Question

B-For the joint below 8.8 property class M12 bolts are used. a) How much shear force can hold this fastening without slippage, when friction “f” betwewn the connected members is 0.222 b) What are slipping and bolt safeties when shearing force is 2300 N? c) What are slipping and bolt safeties when shearing force is 10300 N? TE? F=?

Answer 1

au M12 bolt, tensile stress area is 84.30mm2 8.8 class bolt ; proof strength= GoodPa (Sp) Friction force (fo) PH Р Fax Axial force a bolt is given by Fax = 0.75*AX Sp Fax = 0.75% 8 4.30 x 600 where 37,935 N "Now, slip resistance for a bolt is given by = MK Nefax / 8 M = friction factor (0.22) k=1 for clearance holes Me = No. of contact surfaces (₂1) fax = Axial preload force 8=1l for service load, 1.25 for max load Sup resistance = 0.22*1 X 1 X 37,935 = 8 345.7N ()- bor 5 bolts = 8345-3x5 = 41+28.5 N for max shear load (8=1-25) .:P = 41728.5 = 33,382.8N On 33-3828 KN 1.25

Now, when P = 23oon 41428-5/1-25 Es = 23oo 14.51 [8= 1.25] When P= 10, 300 N f.s = 41,728.5/1.25 10,300 =3.24