Question 8.18: A brake band attached to the hinge by means of a riveted joi...

\text { Given } P=10 kN \quad t=3 mm \quad \sigma_{t}=80 N / mm ^{2} .

\tau=60 N / mm ^{2} \quad \sigma_{c}=120 N / mm ^{2} .

Step I Diameter of rivets
There are four rivets in the lap joint, which are subjected to single shear. From shear consideration.

4\left[\frac{\pi}{4} d^{2} \tau\right]=P \quad \text { or } \quad 4\left[\frac{\pi}{4} d^{2}(60)\right]=10 \times 10^{3} .

∴        d = 7.28 or 8 mm           (a).

From crushing consideration,

4 d t \sigma_{c}=P \text { or } 4 d(3)(120)=10 \times 10^{3} .

∴          d = 6.94 or 7 mm          (b).

From (a) and (b), it is observed that shearing becomes the criterion of design. Therefore,
d = 8 mm
Step II Width of band
Considering tensile strength of plate along the section XX,

(w-2 \times 8)(3)(80)=10 \times 10^{3} .

∴          w = 57.67 or 60 mm
Step III Pitch of rivets

m = 1.5d = 1.5(8) = 12 or 15 mm
p + 2m = w or p + 2(15) = 60
p = 30 mm

p_{t}=p=30 mm .