Given: Four-25-mm-diameter rivets with an axial load of
T = 75 kN distributed equally as shown. The thickness of
the plates is 25 mm and width of plate is 150 mm.
find (a) the shear stress on the rivets
(b) the bearing stress between a plate and a rivet;
(c) the minimum average tensile stress in each plate.
(a) the shear stress on the rivets
Shear stress= Pd/4t
shear stress= (75kN*0.025)/(4*0.15)
shear stress= 1.875/0.6
shear stress= 3.125Pa
(b) the bearing stress between a plate and a rivet;
fbr=P/Dt
= 75kN/(0.025*0.15m)4
=75kN/0.015
=5,000 psi
(c) the minimum average tensile stress in each plate.
Minimum average tensile stress
Sut= A/Dm
Sut= 75/(25*015)
Sut= 46.2775kpsi
Comments
Leave a comment