Answer to Question #156664 in Math for ali

Question #156664

The material for the bolt shown in the angled joint above has an ultimate tensile strength of 500 MPa and a shear strength of 300 MPa. The diameter of the bolt is 8 mm. Determine the factor of safety in operation.


1
Expert's answer
2021-01-20T17:46:05-0500


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.F_n=F\sin\theta, F_t=F\cos\theta.

The area of the 8 mm bolt is A=πd24=π(8×103)24=16π×106m2A=\dfrac{\pi d^2}{4}=\dfrac{\pi (8\times10^{-3})^2}{4}=16\pi\times10^{-6}m^2

The normal and tangential stresses are 


σn=Fn/A=Fsinθ16πMPa\sigma_n=F_n/A=\dfrac{F\sin\theta}{16\pi}MPa

σt=Ft/A=Fcosθ16πMPa\sigma_t=F_t/A=\dfrac{F\cos\theta}{16\pi}MPa

The factors of safety are (FF in NN )


fn=σnallowσn=500MPaFsinθ16πMPa=8000πFsinθf_n=\dfrac{\sigma_n^{allow}}{\sigma_n}=\dfrac{500MPa}{\dfrac{F\sin\theta}{16\pi}MPa}=\dfrac{8000\pi}{F\sin\theta}

ft=σtallowσt=300MPaFcosθ16πMPa=4800πFcosθf_t=\dfrac{\sigma_t^{allow}}{\sigma_t}=\dfrac{300MPa}{\dfrac{F\cos\theta}{16\pi}MPa}=\dfrac{4800\pi}{F\cos\theta}

Let F=8kN=8×103N,θ=50°.F=8kN=8\times10^3N, \theta=50\degree. Then the factors of safety are


fn=8000π8000sin50°=4.1f_n=\dfrac{8000\pi}{8000\sin50\degree}=4.1

fn=4800π8000cos50°=2.9f_n=\dfrac{4800\pi}{8000\cos50\degree}=2.9


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog