Site pages
Current course
Participants
General
20 February  26 February
27 February  5 March
6 March  12 March
13 March  19 March
20 March  26 March
27 March  2 April
3 April  9 April
10 April  16 April
17 April  23 April
24 April  30 April
Lesson 15. SOLVED EXAMPLES
Module 3. Stress Lesson 15
SOLVED EXAMPLES 15.1 Problem A solid shaft subjected to a torque of 1500Nm. Find the necessary diameter of the shaft, if the allowable shear stress is 600N/cm^{2}. The allowable twist is 1^{0} for every 20 diameter length of the shaft. Take G=8*10^{5} N/cm^{2} Solution: Given Torque T=1500 Nm =1500*10^{2} Ncm Allowable shear stress ז_{s}=600N/cm^{2} Angle of Twist ө=1^{0}=π/180 radians Length of shaft, l=20D G=8*10^{5} N/cm^{2} 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) ז_{s}D^{3} 1500*10^{2}= (π/16)*600*D^{3} On solving D=10.8cm. ii. For stiffness Using the relation T/J=ז_{s}/R=G ө/L Where T=Torque J=Polar moment of inertia in cm^{4}= (π/32)*D^{4} R=Radius of curvature in cm G=Rigidity modulus in N/cm^{2} 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/cm^{2} Solution: Given. Diameter of shaft, D = 25cm Max. Shear stress, fs = 460 kg/cm^{2} Let T = Torque transmitted by the shaft Using the relation, T= (π/16) x fs x D^{3} with the usual notations T= (π/16) x 460 x 25^{3} = 1411300 kg/cm 14113 kg/m
15.3 Problem A solid steel shaft is to transmit a torque of 10000 kgm. If the shearing stress is not to exceed 450 kg/cm^{2}, find the minimum diameter of the shaft.
Solution: Given Torque, T= 10,000 kgm = 1000000 kgcm Max, shear stress, fs= 450 kg/cm^{2} Let D= Diameter of the shaft. Using the relation, Or T= (π/16) x fs x D^{3} with the usual notations 1000000 = (π/16) x 450 x D^{3}
D^{3} = (1000000 x 16) / (π x 450) = 11320 cm D = 22.45 cm
