4.1 Torsion of Circular Shaft | unit 4 torsion of circular shaftsand principal stresses and strains

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Unit - 4

Torsion of Circular Shaft and Principal Stresses and Strain

4.1 Torsion of Circular Shaft

1. Theory of torsion:

A member which subjected to couple which produce rotation about its longitudinal axis is known as Twisting moment or Torsion.

Due to Torque applied on shaft, the shaft produces shear stress and shear strain in the material.

In practice, torque is produced in shaft of motors or power equipment’s e.g. Fans, axles of vehicles, pumps, etc.

Shaft are circular in section only; hollow shaft is preferable for design.

2. Assumption made in the theory of torsion:

The shaft should be perfectly straight and uniform in cross-section.

Material of shaft is homogeneous and isotropic.

Circular shaft remains circular after twisting.

Plane section of shaft remain plane before and after twisting.

Twist is uniform along the length of shaft.

Maximum shear stress induced in shaft does not exceed elastic limit.

Torque applied on the shaft in the plane perpendicular to the axis of shaft.

3. Derivation of torsion formula:

Consider a shaft of radius R and length L is subjected to torque T on the free end and other end is fixed.

Initially the line of the shaft is horizontal AB before twisting and after twisting it takes position AB’ as shown in fig.

Shear strain <BAB’ = and Angle of twist <BOB’ =

For longitudinal section,

tan = tan =

BB’=L ………………..(i)

From cross-section,

Length of arc = Radius Angle tan

BB’=r …………….(ii)

From equation I and ii

L = r ………….(iii)

By definition shear modulus,

G== ……..(iv)

Equation for (iii) and (iv),

= …………….(v)

Maximum shear stress occurs at radius R.

Now consider elemental ring of area dA at radius r from centre O.

Shear force on elemental ring = Stress Area

dF = dA

Torsional moment on ring,

dT = dF r = rdA)

dT = dA

Now torsional moment on whole sectional,

= dA

But we know, dA = polar moment of inertia = J

T = J

From equation (vi) and (vii), we get,

= =

4.2 Stresses, Strain and Deformation in Determinate and Indeterminate Shaft of Hollow, Solid, Homogeneous Cross-Section Subjected Twisting Moments.

A metal bar 15mm diameter subjected to pull of 40KN elongated by 0.5 mm over a gauge length of 500 mm. In a torsion test on the same material, maximum shear stress of 45Mpa was measured on a bar of 50mm diameter and angle of twist over a length of 300 mm was measured 0.4 degree. Determine Poisson ratio for the material.

Ans:

Given:

Tensile test: D = 15mm P = 40KN

= 0.5 mm L = 500 mm

= 45Mpa D = 50 mm

= 0.4 L = 300mm

To find:

(i) Modulus of elasticity from tension test.

E = = = 226.35Mpa

(ii) Shear modulus from torsion test:

G = = 77.35 Mpa

(iii) Passions ratio

E = 2G(1+)

= 1.463-1 = 0.463

2. A hallow circular of 200mm external diameter, thickness of metal 20mm is rotating at 180rpm. The angle of twist of 3m length was found to be 0.7. Calculate the power transmitted and maximum shear stress induced in the material. Take G = 80 Mpa.

Ans:

Given

D = 200 mm

t = 20 mm

N = 180 rpm

L = 3m = 3000 mm

= 0.01222 radian

Internal diameter d= D – 2t = 200 - 220 = 160 mm

T = 30.2208 N.M

Power p = = 569.65 KW

4.3 Power Transmitted by Shaft:

It is the average torque and corresponding angle turned per unit duration of time.

Power = Average torque Angle of rotation/sec.

P = T

P =

Where, T = Average (mean) torque in N-m.

N = No. of revolution per minute(rpm)

4.4 Principal Stress and Strain:

We have studied the effect of simple stresses which were either normal or tangential acting on particular plane only.

Most of cases on actual member, more complex condition of stresses are developed.

In this chapter, we study the analytical and graphical method to find the stresses acting on an inclined plane of the member subjected combined stresses.

The plane on which the normal stress or the shear stress reaches their maximum intensity have particularly significant on the materials.

Concept Principal Plane:

The plane on which only normal stresses acts and no shear stress is called principal plane.

The plane AB, BC, CD and AD as shown in fig carries only normal stress and no shear stress. These planes are called as Principal planes.

2. Principal stress:

The magnitude of normal stress acting on the principal plane is called as principal stress.

3. Major Principal stress:

It is maximum value of normal stress acting on the principal plane.

It is the maximum intensity of direct stress.

4. Minor Principal Stress:

It is minimum value of normal stress acting on the principal plane.

It is the minimum intensity of direct stress.

5. Major Principal Plane:

A plane which carries major principal stresses is called as major principal plane.

6. Minor Principal Plane:

A plane which carries the minor principal stress is called minor principal plane.

4.5 Normal and Shear Stresses on Oblique Plane:

The resultant of the normal stress and the tangential stress is called as resultant stress. It is denoted by .

Where,

= Tangential stress

Angle of obliquity:

The angle made by the resultant stress with the normal stress is called as angle of obliquity.

It is denoted by

tan =

2. Stresses on oblique section of body

Following are the various cases, for determination of stresses on an oblique section of a body:

A member subjected to axial stress.

A member subjected to normal stresses in two mutually perpendicular direction.

A member subjected to shear stress.

A member subjected to normal stresses in two mutually perpendicular direction accompanied by a simple shear stress.

4.6 Magnitude and Orientation of Principal Stresses and Maximum Shear Stresses:

Member subjected to normal stresses in two mutually perpendicular direction accompanied by a simple shear stress:

Consider a rectangular element ABCD of uniform cross-sectional area and unit thickness as shown in fig.

It is subjected to two normal stresses and and a shear stress .

Consider a plane BE inclined at an angle to face BC as shown in fig.

Horizontal force on face BC = Stress on BE Area = (BC 1)

Vertical force on CE = (CE 1)

Vertical force on BC due to,

Shear Stress = BC 1

Horizontal force on CE due to shear stress = CE 1

From fig.

Resultant normal forces,

= (BC) + (CE 1) + (BC + (CE 1) …………(i)

= + + +

= + + sin2 ……………(ii)

Resultant tangential force on plane BE, along x-axis.

= = - - +

= sin2 -

Resultant stress, =

Angle of obliquity =

2. Member subjected to direct stresses on two mutually perpendicular plane accompanied with shear stress.

Select the co-ordinate system representing direct stress on x-axis and shear stress on y-axis with origin at O.

Locate point P and Q, OP = and OQ =

Locate point R ( and S()

Joint RS intersecting the x-axis at point C

CP = CQ =

Radius of circle, R =

The circle intersect abscissa at point A and B shear stress is zero.

References:

Mechanics of structure vol. I & II by S.B. Junnarkar and Dr. H. J. Shah, Twenty second edition, Charotar publishing house pvt. Ltd.

Strength of materials by R. Subramanian, Oxford University Press.

Structural Analysis: A Matrix Approach by Pundit and Gupta, McGraw Hills.

Structural Analysis by Hibbler, Pearson Education.

Structural Analysis: M. M. Das, B. M. Das---PHI Learning Pvt. Ltd. Delhi.

Fundamentals of Structural Analysis: 2nd ---West---Wiley.

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