Torsion 6th Chapter

7/31/2019 Torsion 6th Chapter

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4. Torsional Loading4.0 Calculation of Torque in Shafts

Torque is a moment about the axis of a member (for example a powertransmission shaft) and may be denoted by a curved arrow, or a double-headed

arrow according to the right-hand rule. In calculating internal torques in a shaftit is important to follow a sign convention. One of the following may be used:

This may be simply shown in 2-D as in the following figure.

Either convention would be acceptable. Convention 1 will be used in thefollowing examples, unless stated otherwise. Also, it is not necessary to show

both the double headed arrows and the curved arrows. Either one would besufficient. For the shaft in Figure 4.3, determine the torque and sketch its

variation along the axis, as a torque diagram.

Case 1: Shaft subject to several concentrated torques

Figure 4.1

Convention 1 Convention 2

Figure 4.2

100 Nm

260 Nm

90 Nm20 Nm

Figure 4.3


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The reaction may be found by writing the overall equation of equilibrium.

For the overall freebody in Figure 4.4,

T = 0 gives:-TA -20 - 90 +260 -100 = 0

TA = 50 Nm.

Internal torquesmay be found

by consideringthe equilibrium

of segments ofthe shaft. For

example,making an

imaginary cutbetween C and D and considering the right-handside freebody gives

T = TCD -260 + 100 =0TCD = 160 Nm

The same result would be obtained by taking a free-body left of the cut.Similarly the torque at various sections may be calculated using the method of

sections.

This results in the torque diagram in Figure 4.6.

Figure 4.4

100 Nm

260 Nm

90 Nm20 Nm

A B C D E

TA

Figure 4.5

100 Nm

260 Nm

20 Nm

A B C D E

TA

TCD

TCD90 Nm

Figure 4.6-100 Nm

160 Nm

70 Nm50 Nm

A B C D E


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Case 2: Shaft subject to distributed torsional loading

Sometimes, a shaft may be subject to a distributed loading. Examples include,

drill-pipes, and beams subject to uniform eccentric loading. Consider the

drilling shaft shown in Figure 4.7 which is subject to a torsional loading of 200Nm/m between B and C and is driven at end A.

For equilibrium, TA = 200 (N/m) (3 m) = 600 Nm.The torque distribution may be found my applying the method section.

Making a cut at distance x from A where x > 5 m yields the free-body diagramin Figure 4.8

T = -TBC+ = 0TBC =At x = 5 m, TBC=

At x = 8 m, TBC=

200 Nm/m

A

3 m

TA

Figure 4.7

B C

5 m

20 Nm/m

(8-x) m

TBC

C

600 Nm

Figure 4.9 Torque Diagram

Figure 4.8


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4.1 Torsion of circular shafts

Shafts are skeletal structural members subject to torsional loadings. It isimportant to determine the stress distribution and angle of twist in designing

shafts. Rotating shafts are made of circular cross section (either hollow, orsolid) and this is the category of section we will first consider.

Assumptions:

During twisting, radii remain straight and plane sections remain plane. Stresses and strains are within the elastic, and proportional limit. Material is homogeneous, and

isotropic.

Consider an infinitesimal shaft segmentsubject to an induced torque T (seechapter 1 for calculation of inducedtorque). Let us now cut out a typical disk

element of radius r from this element.

Since the distortions are small, thecircumferential lengthAA" =

Denoting distance OA by r, AA" =Using these equations, = (/x) rAs the distances areinfinitesimal, this

may be written as:

= (1)

Recalling the

assumption that the

radii remainstraight, the ratio

(d/dx) is aconstant. This implies that the shear strain increases linearlywith the radius. At the centre, the strain is zero, and it is

maximum at the surface of the shaft.From Hooke's law in shear,

= (2)Combining equations (1) and (2) we get, = (3)

This means the shear stress and shear strain vary linearly with r.

R

r

x

O

A

A"r

A'

A''

A

O

B

B'

r

B''

x

xA'

A

A"

B'B"

B


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m, m

A

F

The maximum shear strain and stress occur near the

surface (when r = R), and may be denoted by m and mrespectively.

Therefore, m = G m (4)

Infinitsimal force F = ( )

Infinitsimal torque T = F ( )

Substituting for dF we get, T = ( ) r

Using equation (3), T =G(d/dx)r2( ) (5)

From this, T = (6)

This integral is referred to as the

____________________________________

and is denoted by the letter _________

Hence equation (6) may be written as, T = GJdx

d(7)

where J = r2 dA (7a)

Eliminatingdx

dfrom equations (3) and (7) and rearranging to make the

subject gives:

=J

Tr(8)

From equation (7),dx

d=

Integrating gives: 2 - 1 = dxGJ

T21

x

x (9a)

For uniform shaft segment made of homogeneous material, subject to constant

induced torque, this reduces to: 2 - 1 =GJ

)xx(T 12 (9b)


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Example:The power loading on a uniform shaft of solid circular cross section is shown inthe figure below. The speed of the shaft is 900 rpm. The shear modulus (G) of

steel may be taken as 80 GPa. If the allowable shear stress is 60 MPa, find theminimum shaft diameter. For this diameter, what would be the angle of twist

between A and D?

Shaft speed = 900 rpm = 900/60 =15 Hz = 30 rad/s = 94.25 rad/sHence the torque loading on the shaft would be as follows:At A, TA = 5000/94.25 = 53.1 Nm;

Similarly, At C, TC =74.3 Nm, and at D, TD =21.2 NmThe torque at B (input) may be found from the overall equilibrium equation:

TB 53.1-21.2-74.3=0 gives TB = 148.6 NmTorque diagram may be obtained by using method of sections. For this case, the

torque diagram is given below. Note this would correspond to the signconvention given below. The opposite convention would give opposite results. In

the past exam solutions, the opposite convention has been used.

Maximum shear stress = 95.5(R)/J = 2(95.5)/( R3) 60106which gives R 10.04 mm. Use R = 11 mm. or diameter = 22 mm.This gives J = (11)4/2 mm4 =22.998 10-9 m4, and G = 80109 Pa.

D-A = (D-C)+ (C-B)+(B-A)=)10998.22)(1080(

)8.0)(2.21()9.0)(5.95()2.1)(1.53(99

++

=-0.021 radians = -1.22(negative sign indicates D rotates in a clockwise direction relative to A).

2 kW5 kW

AB

C

1.2 m 0.9 m 0.8 m

7 kW

MOTOR

D

TDTCTBTA

53.1 Nm

-21.2 Nm

-95.5 Nm

Torque DiagramSign

convention


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Shear Stress due to torque in thin-walled tubes

Shear Flow

Longitudinal force on AB =

Longitudinal force on DC =

For longitudinal equilibrium, these forces mustbe equal and opposite giving:

Shear flow due to torque is constant.

Shear flow - torque relationship

We have, q = ()(t) = a constant ....(1)

Now consider the torque due to shear flow acting

over an infinitesimal length L as shown. If the

thickness of the wall is t, and the shear stress is

, then the torque about an arbitrary point O is

given by:

T == .....(2)

where a the perpendicular distance between O and the line of action of the

shear stress.Using equation (1), this equation may be written in terms of shear flow q as:

T == q(L)(a) .....(3)

This may be expressed in terms of the area bounded by the

triangle OAA' which has a base length L and height a.

Since OAA' = (1/2)(a)(L)

Thus, infinitesimal torque T = 2 q (OAA') ....(4)

By integration, T =2q( ) ..(5)

Therefore =t

q=

tA2

T..(6)

The torque-twist relationship may be shown to be given by:

=

t

ds

GA4

T

dx

d2

where the integral 1/t is computed along the centerline of the

wall for the full length of the perimeter. For a uniform shaft subject to constanttorque this gives:

=

t

ds

GA4

)xx(T)(

2

1212 ..(7)

B

D

B

A

C

L

x

A

C

D

1

12

2

O

AA

a


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Torsion of Non-Circular Members/Rectangular Sections:

The determination of shear stress and

strain distribution in non-circular sectionssubject to torsion is beyond the scope of

this course except for some remarks onthe specific case of rectangular sections

for which results are readily available. Fora rectangular bar of length L and cross

sectional dimensions a, b, the maximumshear stress is given by the formula

2

1

maxabc

T=

and the angle of twist between the two ends is given by:3

2

12

abGc

TL=

in which the coefficients c1 and c2 depend on the aspect ratio of the cross

section (a/b) and are tabulated in text books.

a/b c1 c21.0 0.208 0.1406

1.5 0.231 0.1661

2.0 0.246 0.229

3.0 0.267 0.263

4.0 0.282 0.281

5.0 (1-0.63 b/a)/3 (1-0.63 b/a)/3 1/3 1/3

Note that thin walled

open sections may betreated as rectangular

members with an aspectratio of infinity to find

the maximum stress and

angle of twist (see figureon the right)

In such cases the maximum stress would be given by2max

ab

T3=

and the relative angle of twist would be given by:312

Gab

TL3=

For example the maximum stress in a rectangular tube with and without a crackmay be found as follows:

Without a crack:

abt2

T=

and with a crack2max t)ba(2

T3

+=

a

b

L

T

T

b

b

a

a

b b

a


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Examples:

Calculate the shear stress distribution in the following thin walled tubes subject

to a torque T. Take the wall thickness t as uniform. All shapes have a totalcircumferencial length L, thus making use of same amount of material. Which

shape is most efficient, and which is the worst?


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Torsional Loading

1. For the torsional loading examples in Chapter 1, determine the minimum

shaft diameter for a solid shaft if the allowable stress in shear is 60 MPa.Calculate the angle of twist between the two ends if the shear modulus is 80

GPa.

2. Determine the maximum torque that can be inducedin the following thin-walled section, if the allowable

shear stress is 50 MPa. The wall thickness is 4 mm.The dimensions given are measured along the

centreline of the wall. For this torque, what is theangle of twist between the ends of a segment having

5 m length? Take G as 70 GPa.

3. The torque diagram for a 2 m long shaft of circular hollow cross section isshown below. It is subject to a constant torque for a distance of 0.8 m, and alinearly varying torque for the remainder. If its outer diameter is 40 mm

and inner diameteris 35 mm, what is

the maximum stressinduced? What is

the angle of twist

between the twoends if G = 80 GPa.

4. A uniform shaft ofsolid circular cross

section is fixed at bothends, and is subject to atorque of 180 Nm applied

at one third of its span. Ifthe maximum shear stress

in the shorter segment is 30 MPa, determine the diameter of the shaft.

5. A compound shaft consists of a 20 mm diameter solid steel shaft and analuminium tubing of 25 mm inner diameter and 30 mm outer diameter. One

end of this compound shaft is fixed and the other end is subject to a torquethrough a pin. Find the maximum torque that may be applied to this shaft if

the allowable shear stresses in steel andaluminium are 110 MPa and 80 MPa

respectively. The shear moduli are, 80

GPa for steel and 27 GPa for aluminium.

39 mm

15 mm

39 mm

40 Nm

- 60 Nm

-200 Nm

Torque Diagram

2L/3L/3

180 Nm


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Solution to Tutorials in TorsionPlease check carefully and report any errors or steps that are not clear.

1. Case 1: |Tmax | = 160 Nm (from the torque diagram).

allowable = 60 MPa

64 1060)2/R(

R160

gives, R 0.0119 m

Let R = 12 mm

J=R4/2 = 32572 mm4 = 33.57210-9 m4G= 80 GPa = 80109 PaGJ = 2606 Nm2Lengths of the members are:

LAB = 0.5 m; LAB = 0.5 m; LBC = 1.0m; LCD = 1.2 m; LDE = 0.8 m;

2606

)8.0100()2.1160()0.170()5.050(AE

+++= radians = 0.079 radians =

4.55Case 2: |Tmax | = 600 Nm (from the torque diagram).allowable = 60 MPa

6

41060

)2/R(

R600

gives, R 0.0185 m

Let R = 19 mm

J=R4/2 = 102354 mm4 = 102.410-9 m4G= 80 GPa = 80109 PaGJ = 8188 Nm2

8188

))5600(AB

= radians

8188

))3300(

8188

)3(Tdx

)3)(8188(

)x8(600dx

GJ

T )average(CB8

5

8

5

CB

BC

==

== radians =

C - A = (B - A) + (C - A) = 0.476 radians = 27.3

2. From Pythagoras' theorem,

h = )1539( 22 =36 mm; Area of the triangle =(1/2)(h)(30) = 540 mm2 and the area of the semicircle is

(1/2)(15)2 = 353.4 mm2Total enclosed area A = 893.4 mm2.If the allowable shear stress is 50 MPa and the wallthickness is 4 mm,

)104)(104.893(2

T36

= 50106 gives: T 357 Nm

tds

=(239+15)/4=31.28;

28.311070)104.893(4

5357)(

92612

=

radians = 0.25 radians

39 mm

15 mm

39 mm

h


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|Tmax| = 200 Nm

J =2

)RR( 4i

4

O

= 104103 mm4= 10410-9 m4

J

RT Omax

max

=

Pa10104

)1020)(200(9

3

= = 38.5 MPa.

Since G = 80 GPa, GJ = 8320 Nm2

0149.08320

)2.1)(130(

8320

)8.0)(40()()( ABBcAc =

+=+= radians = -0.85

C turns in a clockwise direction relative to A.

3. Let the reactions at theends A and C be TA and TC

respectively. For overallequilibrium,180 - TA - TC = 0 (1)

This is the only equationfrom statics, and sincethere are two unknowns the

problem is staticallyindeterminate. This may

be done in two ways.Method 1:The internal torques may

be found by method ofsections, in terms of either

of the two unknownreactions. Working in terms of TC, the internal torques are:TAB = 180-TC .(2)

and TBC = -TC .(3)

GJ3

L)T3180(

GJ

)3/L)(T180(

GJ

)3/L2)(T()()( CCCABBCAC

=

+

=+=

.(4)

But since both ends are fixed, for compatibility, C - A = 0 .(5)

Substituting equation (4) into equation (5) gives, 180-3TC = 0 giving TC = 60 Nm.From equation (1) TA = 120 Nm. (this is not actually required)

The induced torques may be found by putting TC = 60 Nm into equations (2) and (3)which give the following induced torques: TAB = 120 Nm and TBC = -60 Nm. Shearstress in the shorter segment (AB) is given as 30 MPa.

Therefore 64AB

1030)2/R(

)R)(120(=

= Pa. Solving for R we get, R = 13.66 mm, d = 27.3

mm.

Method 2: Releasing the constraint at one end and reapplying it to enforce compatibilitygives the same results.

40 Nm

- 60 Nm

-200 Nm

Torque Diagram

A BC

2L/3L/3180 Nm

A B C

TA TC180 Nm

-TC

180-TC


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4. Let the torques induced in the steel and aluminium shafts be TSt and TAl respectively.For equilibrium,TSt +TAl = T (1)

For compatibility the angle of twist between thetwo ends must be the same in both shafts.

St = Al (2)where St is the relative angle of twist betweenthe two ends.The constitutive equations are:

StSt

St

StJG

LT= (3a) and

AltAl

Al t

AlJG

LT= (3b)

Substituting these into equation (2) gives:

StSt

AlAl

StAl

JG

JGTT = (4)

JAl = (154-12.54)/2 = 41172 mm4 and JSt = (104)/2 = 15708 mm4GSt = 80 GPa and GAl = 27 GPa.Substituting these into equation (4) gives: TAl = 0.8846 TSt (5)Using (1) and (5) we get,

TSt = 0.531 T (6a); and TAl = 0.469 T (6b)

For ensuring that the stress in the steel shaft remains allowable, St 110 MPaUsing the stress-torque relationship,

6

St

StSt 10110J

RT Pa.

TSt (1570810-12)(110106)/(1010-3) = 172.8 NmFrom equation (6a) T 172.8/0.531 = 325.4 Nm (7a)Similarly for ensuring that the stress in the aluminium pipe remains allowable, Al 80MPaUsing the stress-torque relationship,

6

Alt

Al,OAl1080

J

RT Pa.

TAl (4117210-12)(80106)/(1510-3) = 219.6 NmFrom equation (6b) T 1219.6/0.469 = 468.2 Nm (7b)Condition (7a) is more restrictive.

Therefore T 325.4 Nm.

T

TAlTSt

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Torsion 6th Chapter