chapter  9
13 Pages

Torsion in circular shafts

D = 146 mm (Answer) The coupling bolts will therefore lie on a pitch circle of diameter 146 x 1·5 = 219 mm (say 220 mm) (see Fig. 9.5) The whole of the torque is transmitted through the coupling via the six bolts, so that the shear moment for each bolt will amount to one-sixth of the total torque. Let the shearing force developed in each bolt = P newtons Then the total torque transmitted = 6 X P X 110 Nmm

Fig. 9.5

Hence

= 0·66 P Nm

P 45757 0·66 = 69329 N

Since the allowable shearing stress in the bolts is 100 N/mm2: 69329

area of bolt required 100

= 693 mm2

EXERCISES CHAPTER 9

1. Calculate the torque exerted by a shaft of diameter 50 mm when the maximum shearing stress is 20 N/mm2 •

4. Determine the maximum shearing stress in a solid steel shaft of diameter 250 nun if the angle of twist is one degree over a length of 1 . 8 m and the modulus of rigidity is 80 kN Inun2 .