## Lesson 15. SOLVED EXAMPLES

 Module 3. Stress Lesson 15SOLVED EXAMPLES 15.1 Problem A solid shaft subjected to a torque of 1500N-m. Find the necessary diameter of the shaft, if the allowable shear stress is 600N/cm2. The allowable twist is 10 for every 20 diameter length of the shaft. Take G=8*105 N/cm2 Solution: Given Torque T=1500 N-m =1500*102 N-cm Allowable shear stress זs=600N/cm2 Angle of Twist ө=10=π/180 radians Length of shaft, l=20D G=8*105 N/cm2 D=diameter of the shaft First we shall find out the diameter of the shaft for its strength and stiffness i. For strength Using the relation T=( π/16) זsD3 1500*102= (π/16)*600*D3 On solving D=10.8cm. ii. For stiffness Using the relation T/J=זs/R=G ө/L Where T=Torque J=Polar moment of inertia in cm4= (π/32)*D4 R=Radius of curvature in cm G=Rigidity modulus in N/cm2 L=Length of shaft in cm Ө=Angular Twist On solving D=13cm Therefore we shall provide diameter of 13 cm i.e. greater of two values For the same problem we can also find out the power P by using the power transmission relation that is P= 2πNT/60 15.2 Problem Find the torque which a shaft of 25 cm diameter can safely transmit, if the shear is not to exceed 460 kg/cm2 Solution: Given. Diameter of shaft, D = 25cm Max. Shear stress, fs = 460 kg/cm2 Let T = Torque transmitted by the shaft Using the relation, T= (π/16) x fs x D3 with the usual notations T= (π/16) x 460 x 253 = 1411300 kg/cm 14113 kg/m 15.3 Problem A solid steel shaft is to transmit a torque of 10000 kg-m. If the shearing stress is not to exceed 450 kg/cm2, find the minimum diameter of the shaft. Solution: Given Torque, T= 10,000 kg-m = 1000000 kg-cm Max, shear stress, fs= 450 kg/cm2 Let D= Diameter of the shaft. Using the relation, Or T= (π/16) x fs x D3 with the usual notations 1000000 = (π/16) x 450 x D3 D3 = (1000000 x 16) / (π x 450) = 11320 cm D = 22.45 cm