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Homework Statement
TWO ALUMINIUM STRIPS EACH 75mmx5mm ARE RIVETED TOGETHER TO FORM A CONTINUOUS LENGTH THE JOINT IS MADE USING 6 RIVETS EACH 5mm IN DIA THE ULTIMATE SHEAR STRENGTH OF THE RIVET MATERIAL IS 70MN/m2 AND THE UTS OF THE ALUMINIUM IS 30MN/m2 IF THE MAXIMUM TENSILE FORCE APPLIED TO THE CONTINUOUS STRIP IS 1.875KN DETERMINE THE FACTOR OF SAFETY FOR (a) THE RIVETS (b) THE ALUMINIUM
Homework Equations
The Attempt at a Solution
area for rivets piex5x10-32/4 x 6 =1.178x10-4 area for aluminium 75x10-3x5x10-3 x 2 = 750 x 10-6 shear stress for rivets = 1.875x10 3/1.178x10-4=15.91x10 6 shear stress for aluminium=1.875x10 3 / 750 x 10-6=2.5x 10 6 factor of safety for aluminium=30x10 6 / 2.5 x 10 6 = 12 factor of safety for rivets = 70 x 10 6/ 15.91 x 10 6 = 4.39