Question #208642

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 


1
Expert's answer
2021-06-21T05:52:35-0400

Rim rotation =600r/min= 600 r/min

Allowable shear stress =63MPa= 63 MPa

Density of Materials =7800kgm3=7800 \frac{kg}{m^3}

velocity =πDN60=π×1×60060=31.41msec= \frac {\pi DN}{60}= \frac {\pi \times 1\times 600}{60} = 31.41 \frac{m}{sec}

Hoop Stress σh=ρ×v2=7800×31.412=7.69Mpa\sigma_h = \rho \times v^2 =7800 \times 31.41^2 = 7.69 Mpa

Now Force due to above stress

Fs=σh×(2×t×l)=7.69×106×0.012×0.1)F_s = \sigma_h \times (2\times t \times l)= 7.69 \times 10^6\times 0.012 \times 0.1) =18475.985N=18475.985 N

According to shear stress theory,

Fs=12×π4×d2×τsF_s = 12 \times \frac{\pi}{4}\times d^2\times \tau_s

d2=Fs12π4×τ=18475.985×412×π×63×106d^2 = \frac{F_s}{12\frac{\pi}{4}\times \tau} = \frac{18475.985\times 4}{12\times \pi\times 63\times 10^6}

d=5.6mmd = 5.6 mm


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!
LATEST TUTORIALS
APPROVED BY CLIENTS