The bonded area of stapler pin, A = 2 x (0.3 X 0.05) + 0.5 x 0.05 = 0.055 in2
(a) Force F = 1.2 #
Shear stress = F/A = 1.2 / 0.055 = 21.81 psi
(b) Force F = 4.8 #
Shear stress = F / A = 4.8/0.055 = 87.27 psi
A stapler works by "shearing" one staple off of a row of staples. Assume that the...
A stapler works by "shearing" one staple off of a row of staples. Assume that the staple has a bonded thickness of 0.05" along its edge. a. If F #, determine the shear stress in the bond along the edge of the staple. b. If we choose to require--# of force to shear one of the staples, how strong (in psi) should we make the bond between the staples? 6. 0.5 in. 0.3 in. A stapler works by "shearing" one...
1. The following is a one-layer truss structure fabricated from aluminum tubing with outside diameter and wall thickness as 0.4 inch and 0.05 inch, respectively. The Young's modulus and Poisson's ratio for aluminum are E= 10 x 10°psi, v=0.33, respectively. The applied load P is 60 Ib. Please answer = 60° B Ra all Cled URb The internal force for member AC is: -30 lb 34.6 lb 30 lb -34.6 lb Question 2 1 points Save Answer 1. The following...