Answer to Question #208642 in Mechanical Engineering for Mkhuseli Dlamini

Question

A thin rim is made by joining two halves with rivets as shown in figure 3.8. Determine the required diameter of the twelve rivets if the rim must rotate at 600 r/min and the allowable shear stress in the rivets is 63 MPa. Ignore the effect of the cover plates. The density of the material used 7800 kg/m3 

Answer 1

Rim rotation $= 600 r/min$

Allowable shear stress $= 63 MPa$

Density of Materials $=7800 \frac{kg}{m^3}$

velocity $= \frac {\pi DN}{60}= \frac {\pi \times 1\times 600}{60} = 31.41 \frac{m}{sec}$

Hoop Stress $\sigma_h = \rho \times v^2 =7800 \times 31.41^2 = 7.69 Mpa$

Now Force due to above stress

$F_s = \sigma_h \times (2\times t \times l)= 7.69 \times 10^6\times 0.012 \times 0.1)$ $=18475.985 N$

According to shear stress theory,

$F_s = 12 \times \frac{\pi}{4}\times d^2\times \tau_s$

$d^2 = \frac{F_s}{12\frac{\pi}{4}\times \tau} = \frac{18475.985\times 4}{12\times \pi\times 63\times 10^6}$

$d = 5.6 mm$