Answer to Question #156664 in Math for ali

Question

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.

Answer 1

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

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

The area of the 8 mm bolt is $A=\dfrac{\pi d^2}{4}=\dfrac{\pi (8\times10^{-3})^2}{4}=16\pi\times10^{-6}m^2$

The normal and tangential stresses are 

$$\sigma_n=F_n/A=\dfrac{F\sin\theta}{16\pi}MPa$$

$$\sigma_t=F_t/A=\dfrac{F\cos\theta}{16\pi}MPa$$

The factors of safety are ($F$ in $N$ )

$$f_n=\dfrac{\sigma_n^{allow}}{\sigma_n}=\dfrac{500MPa}{\dfrac{F\sin\theta}{16\pi}MPa}=\dfrac{8000\pi}{F\sin\theta}$$

$$f_t=\dfrac{\sigma_t^{allow}}{\sigma_t}=\dfrac{300MPa}{\dfrac{F\cos\theta}{16\pi}MPa}=\dfrac{4800\pi}{F\cos\theta}$$

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

$$f_n=\dfrac{8000\pi}{8000\sin50\degree}=4.1$$

$$f_n=\dfrac{4800\pi}{8000\cos50\degree}=2.9$$