PRINCIPLES OF DAIRY MACHINE DESIGN

Module 3. Stress

Lesson 15
SOLVED 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

ז_s=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) ז_sD³

1500*10²= (π/16)*600*D³

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 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