UNIT 5Torsion of Circular Shafts and Thin Cylinders

Syllabus: Torsion of Circular Shafts: Theory of pure torsion – Derivation of Torsion equations : T/J = q/r = Nθ/L – Assumptions made in the theory of pure torsion – Torsional moment of resistance – Polar section modulus – Power transmitted by shafts – Combined bending and torsion and end thrust – Design of shafts according to theories of failure.Thin Cylinders: Thin seamless cylindrical shells – Derivation of formula for longitudinal and circumferential stresses – hoop, longitudinal and volumetric strains – changes in dia, and volume of thin cylinders– Thin spherical shells.

PART – 1Torsion of Circular Shafts

INTRODUCTION: Shafts are widely used in engineering application to transmit power from one point to another,

from a motor to a machine tool, from an engine to the rear axle of an automobile or from a steam or hydraulic turbine to an electric generator. The shafts used for the purpose may be solid or hollow.

When a shaft is made to transmit power, a pure twisting couple or torque acts about its polar axis. Shear stresses are set up perpendicular to the radius at all transverse sections. As a result complementary shear stresses are developed on the longitudinal planes which cause a distortion of filaments.

As the cross-section across any transverse plane is symmetrical, the points remain at the same radial distance before and after the twist. The angle of twist is reckoned over a length of the shaft.

Torsional deformation of a circular shaft:

Torque is a moment that tends to twist a member about its longitudinal axis. Its effects is of primary concern in the design of axles or drive shafts used in vehicles and machinery. A small illustration can be made by using a shaft which is made of highly deformable material such as rubber, Fig. 5.1a. When the torque is applied, the circles and longitudinal grid lines originally marked on the shaft tend to distort into the pattern shown in Fig. 5.1b. One can note that twisting causes the circles to remain circles, and each longitudinal grid line deforms into a helix that intersects the circles at equal angles. Also, the cross sections from the ends along the shaft will remain flat –that is, they do not warp or bulge in or out – and radial lines remain straight during the deformation, Fig. 5.1b. From these observations we can assume that if the angle of twist is small, the length of the shaft and its radius will remain unchanged.

Table: 5.1 Analogy Between Axial Deformation and TorsionAxial Deformation TorsionAxial Force (F) Torque (T)Elongation (e) Twist angle(φ)Normal Stress (σ ) Shear stress(τ )

Extensional Strain (ϵ ) Shear strain (γ)Modulus of Elasticity(E) Shear Modulus(G)

Definition of Torsion: Consider a shaft rigidly clamped at one end and twisted at the other end by a torque T=F×d applied in a plane perpendicular to the axis of the bar such a shaft is said to be in torsion.

Effects of Torsion: The effects of a torsional load applied to a bar are (i) To impart an angular displacement of one end cross 1 section with respect to the other end. (ii) To setup shear stresses on any cross section of the bar perpendicular to its axis.

GENERATION OF SHEAR STRESSES

The physical understanding of the phenomena of setting up of shear stresses in a shaft subjected to torsion may be understood from the figure 1 to 3.

Fig 1: Here the cylindrical member or a shaft is in static equilibrium where T is the resultant external torque acting on the member. Let the member be imagined to be cut by some imaginary plane mn'.

Fig 2: When the plane 1mn' cuts remove the portion on L.H.S. and we get a fig 2. Now since the entire member is in equilibrium, therefore, each portion must be in equilibrium. Thus, the member is in equilibrium under the action of resultant external torque T and developed resisting Torque T r .

Fig 3: The Figure shows that how the resisting torque Tr by virtue of an infinites small shear forces acting on the plane perpendicular to the axis of the shaft. Obviously such shear forces would be developed by virtue of shear stresses. Therefore we can say that when a particular member (say shaft in this case) is subjected to a torque, the result would be that on any element there will be shear stresses acting. While on other faces the complementary shear forces come into picture. Thus, we can say that when a member is subjected to torque, an element of this member will be subjected to a state of pure shear.

Shaft: The shafts are the machine elements which are used to transmit power in machines.

Twisting Moment: The twisting moment for any section along the bar / shaft is defined to be the algebraic sum of the moments of the applied couples that lie to one side of the section under consideration. The choice of the side in any case is of course arbitrary.Shearing Strain: If a generator ab is marked on the surface of the unloaded bar, then after the twisting moment 'T' has been applied this line moves to ab'. The angle γ ' measured in radians, between the final and original positions of the generators is defined as the shearing strain at the surface of the bar or shaft. The same definition will hold at any interior point of the bar.

Modulus of Elasticity in shear: The ratio of the shear stress to the shear strain is called the modulus of elasticity in shear OR Modulus of Rigidity and in represented by the symbol G.

G= τγ

Angle of Twist: If a shaft of length L is subjected to a constant twisting moment T along its length, than the angle θ through which one end of the bar will twist relative to the other is known is the angle of twist.

For the purpose of designing a circular shaft to withstand a given torque, we must develop an equation giving the relation between twisting moment, maximum shear stress produced, and a quantity representing the size and shape of the cross-sectional area of the shaft.

DERIVATION OF TORSION EQUATION:When a circular shaft is subjected to torsion, shear stresses are set up in the material of the shaft. To determine the magnitude of shear stress at any point on the shaft, consider a shaft fixed at one end AA and free at the end BB as shown in fig.16.1 Let CD is any line on the outer surface of the shaft. Now let the shaft is subjected to a torque T , the shaft at the end BB will rotate clockwise and every cross-section of the shaft will be subjected to shear stresses.

The point D will shift to D’ and hence line CD will be deflected to CD’ as shown in fig.16.2a The line OD will be shifted to OD’ as shown in fig.16.2b

Let R =Radius of shaftL = Length of shaftT =Torque applied at the end BBτ = Shear stress induced at the surface of the shaft due to torque T

C/G = Modulus of rigidity of the material of the shaftφ =∠DCD ' also equal to shear strainθ =∠DOD ' and is also called angle of twist.

Now distortion at the outer surface due to torque T = DD’∴ Shear strain at outer surface ¿Distortion per unit length

¿ Distortionat the outer surfacelengthof shaft

¿ DD'

L

¿ tanφ

¿φ [if φ is very small]

∴ Shear strain at outer surface, φ=DD'

L ------------------- (i)

Now from Fig. 16.2 b Arc DD '=OD×θ=Rθ {∵OD=R=Radiusof shaft}

Substituting the value of DD ' in equation (i), we get Shear strain at outer surface

φ=R×θL ----------------- (ii)

Now the modulus of rigidity (G) of the material of the shaft is given as

G= Shear stress inducedShear strain produced

=Shear stress at the outer surfaceShear strainat outer surface

G= τ

R×θL

¿τ ×LR×θ

∴GθL

= τR ------------------- 5.1

∴ τ=R×C×θL

Now for a given shaft subjected to a given torque (T), the values of G,θ and L are constant. Hence shear stress produced is proportional to the radius R.

τ ∝R∨ τR

=constant --------------(iii)

RESISTING TORQUE:

From the point of view of equilibrium, the external torque T must be balanced by the resisting torque(Tr). The resisting torque is the sum of the moments of tangential shearing stress acting on any

transverse section. Fig. 21.5 shows a circular shaft subjected to an external torque T. Consider a ring radial distance r from the centre of the shaft and of thickness δr . The intensity of shear stress q is equal to fs,r/R, from equation

The elementary resisting couple set up this thin ring is given by δ T r=(q . δa ) . r

Where δa = area of the elementary ring ¿2πrδr Hence the resisting couple T r set up by the whole cross-section of the shaft is

T r=∑r=0

r=R

δT r=∫0

R f sRδa. r2=

f s

R∫0R

δa . r2

In the above expression, the term∫0

R

δa . r2 represents the polar moment of inertia (J) about its

longitudinal axis of the shaft.

Hence T r=f sR.J

T r

J=

f sR

Finally equating T r to T, and combining equations we obtain

qr=f s

R=T r

J=Gθ

L ------------ Torsion equation

The term J/R is known as torsional section modulus. The term G J is known as torsional rigidity may be looked as torque required to introduce unit angle of twist in unit length, and is called as torsional rigidity or stiffness of shaft. This term is analogous to the term flextural rigidity (EI) used in bending. NOTE:

1. The relative stiffness of two shafts is measured by the inverse ratio of the angles of twist in equal lengths of shafts when subjected to equal torques.

2. Maximum and Mean torque: if the torque fluctuates, the greatest value of the torque must be used in evaluating the maximum shear stress due to torsion. On the other hand for calculating power P, mean torque should be used.

3. J for solid shaft ¿π32

D4

4. J for Hollow shaft ¿π32

(D4−d4 )

POWER TRANSMITTED: Consider a shaft to a torque T and rotating at N revolutions per minute (rpm). Power is defined as the rate of doing work. Taking second as the unit of time then, angle through which torque moves is

ω= N60

×2π

ω=2πN60

Power=work done per second

¿T × 2 πN60

¿2πN T60

If T is taken in N – m, then unit of power is in N – m/sec i.e., Watt. Since a Watt is a small quantity in practice, it is expressed in kilo watt (kW).

Note that 1 metric H.P. = 0.75 kW.

STRAIN ENERGY IN TORSION: When a torque T is applied to a shaft, it gets twisted by an angle θ. Thus the twisting moment does the work on the shaft and this work done is stored as strain energy in the shaft. From the torsional formula the relationship between T and θ is linear as shown in fig. When T is applied gradually the angle of rotation increases linearly and reaches the value θ.

∴ strain energy U=work done ¿area under thecurve

¿ 12Tθ

¿ 12T ( TLGJ )

¿ 12T 2LGJ

i) For a solid shaft:

J= π32

D4 and T= τRJ

∴ strain energy for a solid shaft U=12T 2LGJ

¿ 12 ( τR J )

2 LGJ

¿ 12

τ2 J 2L

GJ (D2 )2

¿2 τ2JL

G D2

¿2τ2( π32 D4)L

G D2

¿ π16

τ2D2LG

¿( π4 D2

L) τ2

4G

¿ τ2

4G×volume

Strain energy stored per unit volume = τ2

4G

ii) For hollow shaft: J=π32

(D 4−d4 )

Area A=π4

(D2−d2 )

U=12T 2LGJ

¿ 12 ( τR J )

2

¿ 12

τ2 J 2L

GJ ( D2 )2

¿2 τ2 JL

G D2

¿2τ2 π32

(D4−d4 ) L

G D2

¿ π16

τ2 (D2−d2 ) (D2+d2 ) LGD2

¿ 14× τ2 (A ) (D2+d2 )L

G D2

¿ 14× τ2 (D2+d2 )

GD2 ×volume

Strain energy per unit volume ¿ 14× τ2 (D2+d2 )

GD2