Lecture Notes of Mechanics of Solids, Chapter 4 1

Chapter 4 Torsion of Circular Shafts In addition to the bars/rods under axial loads as discussed in Chapters 1 to 3, there are other loading cases in engineering practice. In this chapter we will discuss the effects of applying a torsional loading to a long straight circular member such as a shaft or tube, as extracted from the machine showing in Fig. 4.1. We are going to show how to determine both the

• Shear strain and shear stress • The angle of twist

TurbineMachine Generator

A BShaft

Transmit mechanical

power

Transmit electrical

power

Wires

Driven Torque TD Resistant Torque TR

F.B.D.

Fig. 4.1 Engineering example of torsional shaft

4.1 SHEAR STRESS/STRAIN RELATIONSHIP (4th: 69-70,106-107; 5th:69-70,106-107)

Shear Stress Let’s recall the definition of shear stress in Chapter 2. When parts of a deformable body try to slip past another part, a shear stress is set up.

P

∆F∆Fn

∆Ft

Cross section ∆A

Shearing in torsion

T

T

Shear force

τ

Fig. 4.2 Definition of shear stress and shearing in torsion

The shear stress equation was defined in section 2.1 as:

AF

lim t

A ∆∆

=τ→∆ 0

(4.1)

which is a shear force intensity that acts parallel to the material cross sectional plane as shown in Fig. 4.2. It is worth pointing out that the shear stress in an element always comes with pairs to maintain equilibrium as shown in Fig. 4.3.

Lecture Notes of Mechanics of Solids, Chapter 4 2

Shear Deformation We call the deformation created by shear stress as Shear Strain, given the symbol γ (gamma). It is defined as the change in angle of the element, it is a non-dimensional quantity. Unit of shear strain is radian

γ

τ

ττ

τ Fig. 4.3 Element of material with applied shear stress τ and shear strain γ

Hooke’s Law for Shear By conducting a similar material testing to normal stress (Section 2.4), there is a linear relationship (for most engineering materials) between the shear stress and shear strain, as shown in Fig. 4.4.

τ

γ1

Gradient = G

Fig. 4.4 Relationship of shear stress τ – shear strain γ for linear elastic material

This relationship is called Hooke's law for Shear and is represented by equation Eq. (4.2).

γ=τ G (4.2) where: G = Shear Modulus of Elasticity (for short, Shear Modulus) or Modulus of Rigidity. As the shear modulus is a material property (determined by material shear testing), it is related to the Young's Modulus E and Poison's ratio v by the following equation (we are going to prove this late),

( )vEG+

=12

(4.3)

Now that we have these relationships we can examine the effect of an applied torque to the shaft of circular cross section.

4.2 TORSION OF CIRCULAR SECTIONS (4th: 177-188;5th:177-188) Assumptions • This analysis can only be applied to solid or hollow circular sections • The material must be homogeneous • Torque is constant and transmitted along bar by each section trying to shear over its neighbor. • Transverse planes remain parallel to each other. • For small angle of rotation, the length of shaft and its radius remain unchanged.

Lecture Notes of Mechanics of Solids, Chapter 4 3

Shear Strain/Stress Distribution Examine the deformation of a length dx between two transverse planes of a shaft with an applied torque T. For this differential element, assume the left end is fixed and the right end rotates by dφ due to the applied torque T, where dφ is termed as Angle of Twist of the element

dx

γ

dφFixEnd

TwistedEnd

xda T

ρ

Fig. 4.5 Small transverse element with applied torque T rotated by an amount dφ

The surface of radius “ρ” rotates through angle γ, which is shear strain. The arc is defined as length da, which is equal to:

dxdda γ=ϕρ= which gives that:

dxdϕ

ρ=γ (4.4)

where: dxdϕ = Rate of Twist (4.5)

which is constant for the cross-sectional plane. Eq. (4.4) states that the magnitude of shear strain for any of these elements varies only with its radial distance ρ. By using Hooke’s law:

γ=τ G (4.2) and by substituting for shear strain γ, Eq. (4.4), Eq. (4.2) becomes that:

dxdG ϕ

ρ=τ (4.6)

which relates the shear stress linearly to the distance ρ away from the centre of the section. As a result, the shear stress distribution then looks as Fig. 4.6.

T

Distribution of shear stress

Fig. 4.6 Shear stress distribution in circular section with applied torque T

Torque T and Rate of Twist We now equate the applied torque T to the torque generated in the section by the shear stress distribution. To do this, look at a small circumferential section dA, as shown in Fig. 4.7.

Lecture Notes of Mechanics of Solids, Chapter 4 4

Elementaldistribution of

shear stress

ρ

dρ

dA

Fig. 4.7 Shear stress distribution in elemental section dρ of cross section

The elemental torque of a thin circular strip of thickness dρ is given by

( )dAdT τ⋅ρ= (4.7) Integrating over the area of the circular beam gives

( )∫∫ ρρπρρτ=ρτ= ddAT

A2

Substituting for shear stress as Eq. (4.6):

∫ ρ ρρ

ϕρπρ= d

dxdGT 2 (4.8)

Since the rate of twist dxdϕ is constant through the section, it is not a function of radius ρ. If we assume a homogeneous material, G is also constant, so:

JdxdGd

dxdGT ϕ

=ρπρϕ

= ∫ ρ32 (4.9a)

or GJT

dxd

=ϕ (4.9b)

Polar Moment of Inertia J We represent the integral term as the geometric rigidity of the cross section. We call this term the Polar (Second) Moment of Inertia, J.

∫ ρ ρπρ= dJ 32 (4.10)

This term indicates the cross sectional properties to withstand the applied torque. Since this applies to circular bars, the standard terms for J are:

• Solid Shaft of radius R, diameter D :

3222

44

03 DRdJ

R π=

π=ρπρ= ∫ (4.11)

• Hollow Shaft with Inner Radius Ri and Outer radius Ro:

( ) ( )

3222

44443 ioioR

R

DDRRdJ o

i

−π=

−π=ρπρ= ∫ (4.12)

R

D

Ro

Do

Ri

Di

Lecture Notes of Mechanics of Solids, Chapter 4 5

• Thin Walled Tube with t < R/10:

Rm = mean radius of the thin walled tube iom RRR ≈≈ and io RRt −=

From Eq. (4.12),

( )( )( ) ( )( ) tRtRRRRRRRRJ mmmioioio3222 222

22π=

π=−++

π=

tRJ m32π= (4.13)

Engineer's Theory of Torsion ( ETT ) When we equate Eq. (4.9) with the shear stress term, Eq. (4.6), gives that:

ρτ

=ϕ

=dxdG

JT (4.14)

or ρ=τJT (4.15)

where T = the internal torque at the analyzed cross-section; J = the shaft’s polar moment of inertia; G = shear modulus of elasticity for the material ρ = radial distance from the axis (centre). This is called Engineer's Theory of Torsion ( ETT ). The Maximum Shear Stress The maximum shear stress can be computed as

oRJT

max =τ (4.16)

where Ro is the radius of the outer surface of circular shaft. Example 4.1 Compare the weight of equal lengths of hollow and solid shafts to transmit a torque T for the same maximum shear stress. For hollow shaft, the inner and outer diameters have relationship Di = 2/3 Do = 2/3 DH. From ETT (Eq. 4.15):

DJJttanconsT 2

=ρ

==τ

For Solid Shaft:

32

4S

SolidD

Jπ

=

For Hollow Shaft:

44

4

8165

3232

32 HHHHollow DDDJ ×π

=

−

π=

If we then equate the RHS of the above equation (due to the same T and τ), we get:

HollowHSolidS DJ

DJ

=

22 , i.e. H

Hollow

S

Solid

DJ

DJ 22

=

Substituting for the J's we get:

Rm

t

Lecture Notes of Mechanics of Solids, Chapter 4 6

33

8165

1616 HS D

D×

π=

π

which gives that:

07516581

3 .DD

S

H == or DH = 1.075 DS, an increase in size by 7.5%

Comparing the weight ratio:

6420

4

32

4

2

22

.D

DD

AA

VV

S

HH

S

H

S

H =π

−

π

==

which is a reduction in weight of 35.8 % if the hollowed shaft is used!

4.3 ANGLE OF TWIST (4th: 198-212;5th:198-212) The maximum shear stress is one of major design constraints in relation to strength of shaft. However, sometime the design may depend on restricting the amount of rotation or twist when the shaft is subjected to a torque. Angle of Twist for General Cases In this section, we will develop a formula for determining the angle of twist φ (phi) of one end of a shaft with respect to its other end as shown in Fig. 4.8.

From Eq. (4.4), we have ρ

γ=ϕdxd

x

y

z

J(x)

L

BC

x dx

T1T2T3

φ

Fig. 4.8 Rotational shaft under general loading conditions

According to Hooke’s law, Gτ=γ , and substituting Eq. (4.15) i.e. ( ) ( )xJxT ρ=τ , we have

( )( )dxxGJ

xTd =ϕ

Integrating over the entire length L of the shaft, we obtain the angle of twist for the whole shaft as

( )( )∫=ϕ

L

odx

xGJxT (4.17)

where

Lecture Notes of Mechanics of Solids, Chapter 4 7

φ = the angle of twist of one end with respect to other end, measured in radian T(x) = the internal torque at arbitrary position x, found from the method of sections and

equation of moment equilibrium J(x) = the shaft’s polar moment of inertia expressed as a function of position x. G = shear modulus of elasticity for the material Single Constant Torque and Uniform Cross-Section Area

T

T

L

φGJ

Fig. 4.9 Uniform shaft under a constant torque T

Usually in engineering practice, the material is homogeneous and the shaft’s cross-sectional area and applied torque are constant as shown in Fig. 4.9. Eq. (4.17) becomes

GJTL

=ϕ (4.18)

Multiple Torque and Cross-Section Areas If the shaft is subjected to several different torques, or consists of a number of different the cross-sectional areas or shear moduli, Eq. (4.18) can be applied to each segment of the shaft where these quantities are all constant.

∑=ϕi ii

ii

JGLT

(4.19)

Sign Convention of Internal Torque

+T

+T

-T

-T

+φ

-φ

Fig. 4.10 Sign conventions for torque and angle of twist

In order to apply the above equation (Eq. (4.18)), we must develop a sign convention for internal torque and angle of twist of one end with respect to the other end. To do this, we will use the right-hand rule, whereby both the torque and angle will be positive, provided the thumb is directed outward from the shaft when the fingers curl to give the tendency for rotation, as illustrated in Fig. 4.10.

Lecture Notes of Mechanics of Solids, Chapter 4 8

Example 4.2: a) Determine the maximum shear stress and rate of twist of the given shaft if a T = 10kNm torque is applied to it; b) if the length of the shaft is 15 m, how much would it rotate by? c) if restrict the maximum shear stress level within τallow = 60MPa, what is the maximum torque that the shaft can transmit? Let G = 81GPa, D = 75 mm.

T=10kNm

L=15m

φD=75mm

a) The maximum shear stress:

( ) 4644

101063320750

32m..DJ −×=

π=

π=

( ) ( ) ( )( ) MPa.

./.

J/DT

JTRo

max 7120101063

20750101026

3=

×

××===τ

−

The rate of twist: ( )

( ) ( ) m/rad..GJ

Tdxd 039740

101063310811010

69

3=

×××

×==

ϕ−

which equates to : ( ) m/..

dxd o2772039740180

=π

×=

ϕ

b) If the shaft is 15 m long, the angle of rotation at the free end is

°=×°=ϕ

=ϕ 15534152772 ..Ldxd

or directly from Eq. (4.18) ( )

( ) ( ) ( )π×=°==×××

××==ϕ

−/..rad.

.GJTL 1805960155345960

10106331081151010

69

3

c) The maximum shear stress must be less than the allowable stress; allowo

max JTR

τ≤=τ , i.e.

( ) ( ) mkN..

.R

JT

o

allowmax ⋅=

×××=

τ≤

−974

037501060101063 66

4.4 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS (4th: 213-220; 5th:213-220)

Compound Shafts A compound shaft is one made from more than one material property. The aim here is to determine how much of the applied torque is carried by each material. Equilibrium:

00 21 =−−==∑ TTTM x (4.19) Compatibility:

21 ϕ=ϕ=ϕ (4.20)

++

Lecture Notes of Mechanics of Solids, Chapter 4 9

J1

J2

T1 T2

T

Distribution of Shear Stress (G1<G2)

Distribution of Shear Strain

G1

G2Fully

bonded

y

z

Substituting Eq. (4.18) into Eq. (4.20) and equating with Eq. (4.19), we can find T1 and T2, hence the rate of twist and shear stresses carried by each material.

22

2

11

1

JGLT

JGLT

==ϕ , i.e 222

111 T

JGJGT =

TJGJG

JGT2211

222 += ; T

JGJGJGT

2211

111 += (4.21)

( )TJGJGG

2211

22 +

ρ=τ ; ( )TJGJG

G

2211

11 +

ρ=τ (4.22)

Indeterminate Shafts

L

LAC

A

BC

T

LBC

T

TB

TA RoA

CB

Global Equilibrium (for ground reactions):

00 =−−==∑ BAx TTTM (4.23) Since only one equilibrium equation is relevant and there are two unknowns, this problem is statically indeterminate. However, the angle of twist of one end of the shaft with respect to other end is zero. We can give compatibility condition as

Compatibility: 0==ϕ ∑i ii

iiBA JG

LT

and note that the internal torque in segment CB is negative by using the right-hand rule. ( )

0=−

+JG

LTJGLT BCBACA (4.24)

From Eqs. (4.22) and Eq. (4.23), we have

TL

LT BC

A

= ; T

LL

T ACB

= (4.25)

Therefore,

TLJ

RLBCmaxA

o=τ ; TLJ

RLACmaxB

o=τ (4.26)

++

Lecture Notes of Mechanics of Solids, Chapter 4 10

COMPARISON OF TORSIONAL SHAFTS WITH AXIALLY LOADED BARS/RODS Axial Loaded Bar/Rod

(Chapters 1, 2&3) Torsional Shaft

(Chapters 4) Load Type Force F (N) Torque T (Nm) Sign Convention

Tension or Compression

+

-

+F

+F

-F

-F

Tension

Compression

+T

+T

-T

-T

+φ

-φ

+

-

Right-Hand Rule

Geometric Property A – Area (m2) J – Polar Moment of Inertia(m4) Material Property E – Young’s Modulus (Pa) G – Shear Modulus (Pa) Stress

Normal stress: AF

avg =σ (Pa)

UniformDistribution

Shear stress: ρ=τJT (Pa)

T

Distribution of shear stress

Strain Normal strain:

LL∆

=ε (ms) Shear strain γ - radian

Hooke’s Law ε=σ E γ=τ G Deformation

Deflection (m) (Elongation/Contraction)

General: ( )( ) ( )dx

xAxExFL

∫=δ0

Single uniform: EAFL

=δ

Multi-segments: ∑=δi ii

ii

AELF

Angle of Twist (Radian)

General: ( )( ) ( )dx

xJxGxTL

∫=ϕ0

Single uniform: GJTL

=ϕ

Multi-segments: ∑=ϕi ii

ii

JGLT

Work Strain Energy

pPW ∆21

=

∑=i ii

ii

AELF

U2

2

(for a Truss)

TTW ϕ=21

∑=i ii

ii

JGLT

U2

2

(Multi-segments)

Work-Strain Energy Method P

UP

2=∆

TU

T2

=ϕ

Castigilinao’s Method

( )

∂∂

=∆ ∑ii

i

i

iiP AE

LPF

F

By introducing a virtual force Q:

( )0=

∂∂

= ∑Qii

i

i

iiQ AE

LQF

F∆

( )

∂∂

= ∑ii

i

i

iiT JG

LTT

Tϕ

By introducing a virtual torque S:

( )0=

∂∂

= ∑Sii

i

i

iiS AE

LST

Tϕ

Lecture Notes of Mechanics of Solids, Chapter 4 11