Assume that the bolt is perpendicular to the inclined plane. Then the force can be resolved into components that are normal and tangential to the plane. The two components are
Fn=Fsinθ,Ft=Fcosθ.
The area of the 8 mm bolt is A=4πd2=4π(8×10−3)2=16π×10−6m2
The normal and tangential stresses are
σn=Fn/A=16πFsinθMPa
σt=Ft/A=16πFcosθMPa The factors of safety are (F in N )
fn=σnσnallow=16πFsinθMPa500MPa=Fsinθ8000π
ft=σtσtallow=16πFcosθMPa300MPa=Fcosθ4800π Let F=8kN=8×103N,θ=50°. Then the factors of safety are
fn=8000sin50°8000π=4.1
fn=8000cos50°4800π=2.9
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