- #1
chris78
- 6
- 0
A load p of 5kn is applied to the tensile member shown and carried at the joint by a single 20mm diameter rivet.The angle of the shear joint is 60 degrees to the axis of the load.
calculate the tensile stress in the rivet
calculate the shear stress in the rivet
given that the ultimate tensile strength for the rivet is 80MN/m. What is the safety factor for this joint.
Homework Equations
The Attempt at a Solution
shear force=5sin30
shear force=2.5kn
Tensile force=5cos30
Tensile force=4.330127knCSA of rivet=pie x 0.01^2
CSA of rivet=3.14159 x 10^-4
Shear stress=force/area
Shear stress=2500N/3.14159 x 10^-4
Shear stress=7957753.876 NM/2
Tensile stress=4330.127N/3.14159 x 10^-4
Tensile stress=13783233.97 NM/2
Factor of safety= UTS/working stress
80MN/m=80 x 10^6 NM/2
Shear=80x10^6/7957753.976=10.053
Tensile=80x10^6/13783233.97=5.804
Factor of safety for joint=5.8
I posted a similar question yesterday and thank all for there help...i just wanted to make sure that the above is correct and I am resolving my vectors correctly to work out the shear and tensile forces.
Thanks again
Chris