torsion of circular shafts

Torsion of Circular Cross-Section Shafts

47

9. TORSION OF CIRCULAR CROSS-SECTION SHAFTS

9.1 Introduction 9.1.1 Torsion in Structures

Engineering structures are frequently loaded by twisting couples (known as Torque). Typical structures that undergo such loading are power transmission shafts such as the prop-shaft of an automobile and twisting tools such as screwdrivers. In general, any rotating shaft must have a torque applied to create its rotation. Shafts may be solid (as in the case of the screwdriver) or hollow (as for the prop-shaft).

In this course only shafts with circular cross-sections will be considered as they allow a number of simplifying assumptions to be made so that the stresses and deformations due to torsion can be evaluated.

9.1.2 Vector representation of torque

As torque is equivalent to a couple (i.e. a moment) it is a vector quantity and has both a magnitude and a sense of direction. Torque is frequently represented by either a curved arrow around the axis of rotation or by a vector arrow along the axis of rotation (in these notes, when using the vector arrow notation, a double-headed arrow will represent torque to differentiate it from an axial load). The rotation of the torque may be determined from the arrow vector using the right hand screw rule or right hand thumb rule.

When considering the equilibrium of a structure the torque contributes to the moments about the axis of rotation and the sum of the moments about this axis must equal zero. For a structure only loaded in torsion (like a prop-shaft) this means that in equilibrium the structure must have opposing torques of equal magnitude and opposite sense as shown in Figure 9.1

TB

TA

TB

TA

Torque applied at a point X = TX

Torque in a length XY = TXY

TAB

Figure 9.1

To help differentiate between torque applied at a point and torque in a length of shaft (not necessarily equal to each other) the following nomenclature will be used:

• TX = An applied torque at point X

• TXY = The net torque in length XY

The net torque in a length of shaft is the torque that causes the shaft to twist (in the same way that the net axial force in a shaft is the force that causes the shaft to lengthen or shorten) and can only reliably be determined using a FBD of the shaft with the shaft cut at some point in the length XY.

Transformations of Stress and Strain

48

9.1.3 Example: Variation of torque along a shaft

Calculate and graph the variation of torque along the stepped shaft as shown in Figure 9.2.

A B C D 3m 2m 2m

16kNm 2kNm 12kNm

8kNm

1m E

10kNm

20kNm

30kNm

Figure 9.2

Step 1: FBD of entire shaft to determine resisting torque applied at support A

Step 2: Separate shaft into two FBDs with a cut in length AB to determine TAB

Step 3: Repeat step 2 for lengths BC, CD, and DE

Remember that either left or right hand side FBDs can be used to determine the torque in the length under investigation

9.1.4 Units

As Torque is a moment it has the units of Force (MLT-2)× Length (L), i.e generalised units of ML2T-2. In the SI system this is commonly Newton metres (Nm).


49

9.2 Stress due to Torsion 9.2.1 Static equilibrium considerations

dV TAB

r

dA

A

B

Figure 9.3

Consider a length AB of solid shaft as shown in Figure 9.3 with a torque TAB along its length. For equilibrium the sum of the Forces and Moments must equal zero. The only forces capable of resisting the applied Torque are shear forces on the cross-sectional faces. For an elementary (very small) area dA, denote the shear force occurring on that area as dV. The moment that this shear force creates is equal to dV multiplied by its distance from the rotational axis of the shaft (r). The sum of all these elementary shear moments must then equal TAB for equilibrium. This sum is represented as the integral over an area and is denoted by

∫A (an Area Integral) in the equation below.

∫

∑ ∫

=⇒

=−=

A

AB

A

ABaxisshaft

dV.rT

0dV.rTM

As τ = V/A, dV = τ .dA this may be written as:

∫=A

AB dArT τ (9.1)

As r is not constant over the area of integration it must remain within the integral. Recall that for the case of axial loading the distribution of normal stress across a cross-section was assumed to be constant. This however, is NOT a valid assumption for shear stress due to torsion and τ will vary across the cross-section and must also remain within the Area integral. This means that equation (9.1) is insufficient to solve for τ . As consideration of the sum of the forces applied to the structure does not provide any extra information from which to solve for τ (as all the elementary forces dV are matched by equal and opposite forces on the opposite side of the shaft axis) this problem is statically indeterminate and the deformation of the structure must be considered (i.e. geometric compatibility).

9.2.2 Axial shear stress

Previously it has been shown that shear stresses cannot exist in a single plane. Consider a small element of the shaft as shown in Figure 9.3 with a shear stress τ acting on the cross sectional face. For equilibrium there must be complementary shear stresses acting on the axial faces of the same magnitude.


50

The existence of these axial shear stresses can be shown by considering the relative motion of two edges of a paper tube when subjected to a twisting motion. Whilst in the paper roll sliding occurs in an engineering structure the sliding is resisted by the continuity of material and a stress is generated.

9.2.3 Geometric Compatibility

Now consider the deflection of the shaft of radius R, fixed to a rigid support at the end A with a Torque (TAB) acting along its length - as shown in Figure 9.4(a). Under the influence of the Torque the free end will rotate through an angle φ AB called the angle of twist. Consider also a surface element ABCD located within the shaft at a distance r from the axis of revolution. This element can be used to obtain the generalised relationship between the angle of twist and the shear stress generated in the shaft.

TAB

(a)

LAB

A D

B C

C’

φAB

B’

r

O

A D

B C’ B’ C τ γAB

(b)

R

Figure 9.4

At this point the importance of considering a circular cross-section is highlighted. Under torsion a circular cross-section remains plane and undistorted. This can be shown by considering the point B, located at radius r from the axis of revolution, moving around the shaft axis by an angle φ AB to a new point B’. For a circular cross-section, both B and B’ (and C and C’) are at the same distance from the shaft axis and are both in the same plane as the original un-deformed cross-section. This does not occur for instance in square cross-sections (see Figure 9.5(b)) where more complex analysis methods are required to assess the effect of torsion on stress distributions.

Figure 9.51

1 From: Beer, Johnston and Dewolf, Mechanics of Materials, 2002


51

As with axial loading where the generalised (uniform) stress condition breaks down near the point of loading, the assumption of true planar rotation of the cross-sections also breaks down near the point of loading. Saint Venant’s principle however, allows the performance away from the point of loading to be modelled by assuming that the torque is applied to the shaft via rigid endplates.

Using these assumptions and considering only small deflections in the elastic region of the material, the thin elementary surface ABCD is deformed under the action of the applied Torque to new shape AB’C’D as shown in Figure 9.4(b). The angle of twist and the shear strains generated in surface element AB’C’D can be related as follows:

{ }

{ }ϕϕγ

γγγ

rBBasL

r

smallisasL

BB

AB

ABAB

ABABAB

≈=

=≈

'

'tan

This equation shows that the shear strain on an elementary surface of a circular cross-sectional shaft varies with the distance from the axis of revolution. This means that the maximum shear strain is found on the outer surface of the shaft (when r = R).

In the elastic region τ = Gγ therefore the equation above may be re-written in terms of the shear stress (ignoring subscripts) to give:

L

Grϕτ ==== (9.2)

As G, φ and L are constants for a given cross-section shear stress is seen to vary linearly across the cross section with a minimum of zero at the centre (provided the shaft is solid) and a maximum when r = R.

9.2.4 Angle of Twist

Equations (9.1) and (9.2) may now be combined to relate the torque and the angle of twist in a length (AB) of shaft.

========

====

====

====

∫∫∫∫

∫∫∫∫

∫∫∫∫

∫∫∫∫

A

ABAB

ABABAB

AAB

ABAB

A AB

ABAB

A

AB

dArJwhereL

JG

dArL

G

dAL

rGr

dArT

2

2

ϕ

ϕ

ϕ

τ

J is defined as the Polar Second Moment of Area of the cross-sectional Area. Methods to determine J will be discussed in later lectures. Rearranging the equation above gives the general form of the torsion equation for a length of shaft (ignoring subscripts):

GJ

TL=ϕ (9.3)

The product GJ is referred to as the Torsional Rigidity of a shaft and is an assessment of how resistant a shaft is to twisting.


52

9.2.5 Example: Twist in a oil well drill shaft

When an oil well is being drilled at a depth of 1800m, the typical torque required to commence drilling is approximately 60kNm. Determine the number of rotations of the top of the drill shaft (shown as drill string in Figure 9.6) at the point when drilling commences. Consider the shaft to be made from hollow steel (G = 79GPa) tubes with a polar second moment of area of J=1.074×10-4m4.

Figure 9.6

The angle of twist is calculated using equation (9.3):

srevolution

radians

.G

TLGJ

TL

========

====

××××××××====

====

−−−−4100741

ϕ

9.2.6 Shear Stress

Equations (9.2) and (9.3) may also be used to determine the variation of shear stress through a shaft:

GJ

TL.

L

GrL

Gr

====

==== ϕτ

J

Tr====τ (9.4)


53

The distribution of shear stress on a cross-section may now be shown graphically in Figure 9.7 as varying linearly with distance from the axis of rotation of the shaft with the maximum stress occurring when r = Ro.

r

τ τmax = TRo/J

Ro

r

τ τmax = TRo/J

Ro Ri

τmin = TRi/J

(a) (b)

Figure 9.7

9.2.7 Example: Maximum shear stress in a solid shaft

For the 50mm diameter solid shaft in Figure 9.12 calculate the maximum shear stress. Use J = 6.135×10-7

m4.

2.5kNm

2.5kNm

50mm

Figure 9.8

Maximum shear stress in solid shaft is found using equation (9.4):

MPa

.

.

TRJ

TR

o

omax

====

××××====

××××====

====

−−−−

−−−−

7

7

101356

101356

τ


54

9.3 Multiple Length Torsion Members 9.3.1 Stress at a cross-section

Equation (9.4) was derived for a simple shaft with torques applied to each end. However, it may also be used to determine the stress distribution on a cross-section for shafts of variable cross-section or where torques are applied at a location between the ends of the shaft. All that is required is knowledge of the torque applied at the cross-section of interest, which may be obtained from consideration of a Free Body Diagram with one end of the FBD being at the section of interest.

D

C

B

A

TA

TB

TC

TD

B

A

TA

TB

TBC

L AB

L’

(a) (b)

0<L’<L BC

Figure 9.9

Consider the shaft ABCD shown in Figure 9.9 with torques TA, TB, TC and TD applied at points A, B, C and D respectively. To determine the cross-sectional stress distribution in the shaft in length BC divide the shaft into appropriate FBDs at the section of interest (in this case a distance of L’ from point B) and solve for the torque (TBC) in length BC.

BABC

BCBAaxisshaft

TTT

TTTM

−=⇒

=−−=∑ 0

Knowing TBC, equation (9.4) may now be used to determine the stress distribution at any cross-section in length BC.

9.3.2 Twist in a multi-length shaft

Equation (9.3) for the angle of twist in a length of shaft may only be used if the length of shaft considered is:

• of constant G (i.e. homogeneous),

• has a uniform cross-section (i.e. is of constant J) and

• has no internal point torsion loads (i.e. has constant torque along the length of shaft).

If as in Figure 9.9 the shaft is subjected to torques at locations other than the ends or has changes in cross-section, then the shaft must be divided up into lengths (IJ) that meet the above criteria and the angle of twist for each length determined separately. The sum of the separate angles of twist will equal the total twist of the shaft such that the angle of twist of a multi-length shaft may be written as:

∑=n

IJIJ

IJIJ

JG

LTϕ where n is the number of IJ lengths (9.5)

In the example of Figure 9.9, the angle of twist at the cross-section of interest, relative to section A, is the sum of the twist in length LAB plus the twist in the length L’ (only two lengths are considered as the as the point of application of torque TB and the change in cross-section coincide). Thus the twist (relative to point A) at a distance L’ from B may be written as:

BCBC

BC

ABAB

ABAB

JG

'LT

JG

LT +=ϕ


55

9.3.3 Example: Variation of twist in a stepped shaft

The stepped steel shaft of Figure 9.10 is rigidly fixed at A and has torques applied at B and D. Calculate and graph the angle of twist along the shaft relative to section A.

A B C D 3m 2m 2m

J = 2m4 J = 1m4

6kNm 2kNm 12kNm 8kNm

1m E

Figure 9.10


56

9.4 Polar Second Moment of Area 9.4.1 Simplified method for Circular Cross-Sections

The Polar second moment of area of an Area is defined as:

dArJA∫= 2 (9.6)

For a circular cross-section the simplest elemental area to consider is a very thin disk as shown in Figure 9.11.

dA=2 r.dr

Ro

Ri

r dr

Figure 9.11

In Figure 9.11, dA = 2π r.dr, and the total area will be integrated across if r is considered to vary from the inner radius Ri to the outer radius Ro. Therefore J may be written as:

bar solid afor 2

e thick tubafor )(2

2

4

443

o

R

R

io

R

RRdrrJo

i

π

ππ

=

−== ∫ (9.7)

9.4.2 Units

Equation (9.6) shows that J will have general units of Length4 (L4). In the SI system this will commonly be m4 or mm4.


57

9.4.3 Example: Maximum shear stress in a solid and hollow shaft

Calculate the inner radius (Ri) for the 80mm diameter hollow shaft so that the maximum stress in the hollow shaft is identical to that in a solid shaft of outer diameter 50mm. Also plot the variation of shear stress (τ ) with radial position (r) for both shafts.

2.5kNm

2.5kNm

50mm

2.5kNm

2.5kNm 80mm 2Ri

(a) (b)

Figure 9.12

Maximum shear stress in solid shaft is found using equations (9.4) and (9.7):

====

====

======== 42 o

oomax R

TR

J

TRπ

τ

The inner radius of the hollow shaft is found using the same equations (but different version of equation (9.7) to account for difference in J):

=

=

=

=

=J

TRomaxτ

====

====

====

====

====J

TRiminτ

Ro

Ro Ri


58

9.4.4 Example - Electric motor driving a gear shaft

The electric motor of Figure 9.13 provides 2.4kNm of Torque to a gear shaft. The gears resist the applied torque as shown. With the dimensions of the solid shafts as shown, calculate the maximum shear stress in shaft lengths AB and BC.

B

0.8kNm

0.4kNm

1.2kNm

36mm

C

D

44mm

44mm 54mm

2.4kNm

A

Figure 9.13

FBD for length AB

2.4kNm

TAB

54mm

====

====

========∑∑∑∑

J

TR

M

omax

axis

τ

0

FBD for length BC

1.2kNm

44mm

2.4kNm

TBC

====

====

========∑∑∑∑

J

TR

M

omax

axis

τ

0


59

9.4.5 Example - Twist in a stepped shaft

The steel (G = 79GPa) circular cross-section stepped shaft in Figure 9.14 is fixed to a rigid support at A and has a Torque of 2kNm applied at D. The shaft is hollow between A and B and solid for the remainder. Calculate the angle of twist at D.

A B C D 0.05m 0.03m 0.04m

0.025m 0.035m 0.025m

2kNm

Figure 9.14

The twist in a stepped shaft is calculated from equation (9.5) considering lengths of constant G, constant cross-section and constant torque. As G is constant throughout and there are no torques applied between A and D only have to break the shaft down into lengths of constant cross-section. Therefore from equation (9.5):

{{{{ }}}}

(((( )))) (((( )))) (((( ))))°°°°========

++++++++−−−−××××××××

××××====

++++++++

−−−−====

++++++++====++++++++====

====∑∑∑∑

47204320

01250

040

01750

030

0125001750

050

1079

200022444494444

.rads.

.

.

.

.

..

.

R

L

R

L

RR

L

G

T

constantGandTasJ

L

J

L

J

L

G

T

JG

LT

JG

LT

JG

LT

JG

LT

CDo

CD

BCo

BC

ABio

AB

CD

CD

BC

BC

AB

AB

CDCD

CDCD

BCBC

BCBC

ABAB

ABAB

n

IJIJ

IJIJ

ππ

ϕ


60

9.5 Design of Power Transmission Shafts 9.5.1 Relationship between Power and Torque

A power transmission shaft is normally specified by the amount of power it needs to transfer and the rotational speed at which that power is to be transferred. Power is related to torque by the angular velocity of the shaft (ω ) by the equation.

)2.( fT

TP

πω

==

(9.8)

Which also relates power to the rotational frequency of the shaft (rotations per second). Therefore, if the power and speed requirements of a shaft are known, the diameter of the shaft and wall thickness may be determined by combining Equations (9.3), (9.4) and (9.8).

9.5.2 Example - Design of Formula 1 car power transmission shaft

Design the lightest power transmission shaft for driving a Formula 1 car wheel. Consider the full engine power capability as being transferred to the wheels, i.e. 350hp per wheel. The shaft must be capable of 1700rpm. The chassis designer advises you that the largest diameter the shaft may be is 35mm. A titanium shaft may be used with an allowable shear stress of 500MPa.

Convert to SI units:

1hp = 0.7457kW 700hp = 522kW

1rpm = 2π /60 rad/sec 1700rpm = 178.0 rad/sec

Calculate Torque to be transmitted:

kNmP

T 932.2178

522000===ω

Calculate outer radius for a solid shaft:

mm.T

R

R

TR

J

TR

maxo

o

oomax

51510500

2932223

63

42

====××××××××

××××========

========

ππτ

τπ

Therefore cross-sectional area of solid tube is π .(15.5)2 = 756.0 mm2

Calculate the inner radius for a hollow shaft with an outer diameter of 35mm:

(((( )))) (((( ))))

mm..

.TR

RR

TRRR

RR

TR

J

TR

max

ooi

max

oio

io

oomax

01310500

017502932201750

2

2

46

44

4

4444

2

====××××××××

××××××××−−−−====−−−−====

====−−−−⇒⇒⇒⇒−−−−

========

ππτ

πττ

π

Therefore cross-sectional area of hollow tube is π .(17.52 – 13.02) = 431.2 mm2. As mass = density × volume = density × length × area, the lightest shaft will be the shaft with the smallest cross-sectional area which is the hollow shaft with external diameter of 35 mm and internal diameter of 26mm.

9.6 Torsional statically indeterminate problems 9.6.1 Introduction

As with the normal and shear stress statically indeterminate problems discussed earlier static indeterminacy can also be encountered in torsional problems. In such cases the distribution of torque through a combination of shafts cannot be determined by simply using a FBD alone (statics). In general the method of solution is to look for geometric compatibility in the various shafts under consideration


61

(i.e. identify how the twist in the various shafts are related). This will allow a series of simultaneous equations to be generated which will then allow solution of the problem.

9.6.2 Example: Torsional Static Indeterminacy

A steel shaft (G = 77GPa) and an aluminium tube (G = 27GPa) are connected to a fixed support at A and to a rigid end plate at B as shown in Figure 9.15. If the initial stresses in both the tube and the shaft are zero and the maximum allowable shear stresses are 120MPa for steel and 70MPa for aluminium, determine the maximum torque that can be applied at B.

A B 0.5m

0.05m 0.076m 0.06m TB

Figure 9.15

Step 1: Draw FBD of the end plate at B to get an expression relating the torque in the aluminium tube (TAB-Al ), the torque in the steel shaft (TAB–St) and the applied torque TB

Step 2: Relate the deflections of the tube and the shaft to gain a second expression relating TAB-Al and TAB–

St

Step 3: Calculate the maximum allowable TAB-Al and TAB–St

Step 4: Solve for TB


62

10. TRANSFORMATIONS OF STRESS AND STRAIN

10.1 Equivalent Shear state for Torsion 10.1.1 Pure shear = bi-axial tension/compression

Previous sections have shown that a state of pure shear stress (which is the state of stress due to torsion) is equivalent to two equal and opposite direct stresses at right angles on planes at 450 to the shear stresses. Hence in torsion, the pure state of shear stress shown in element a of Figure 10.1, may also be considered as shown in element b (with σ = ±τ).

T T a b 45º

τ τ τ

τ

σ = - τ

σ = τ σ = - τ σ = τ

Figure 10.1

The tensile stress at 45º to the axis explains why brittle materials in torsion fracture on a 45º helix (this occurs when σ (=τ ) > σ uts). Figure 10.2(a) shows a ductile failure dominated by shear and Figure 10.2(b) a brittle failure dominated by tension.

(b)

(a)

Figure 10.2


63

10.1.2 Example: Tensile stress in a helically wound pipe

A pipe of 200mm outside diameter is made by wrapping a 10mm thick steel sheet at 45º around a central mandrel and then welding the sheet along its edges (as shown in Figure 10.3). Knowing that the maximum allowable tensile stress in a weld is 85MPa, calculate the maximum torque that may be applied to the pipe.

T T 45º

Figure 10.3

As torsion is a state of pure shear, the maximum tensile stress on the weld will be equal in magnitude to the maximum shear stress in the pipe as shown in Figure 10.1 (note the different orientation of the Torques reverses the directions of the shears and hence the tensile stresses).

Therefore:

( )

( )

kNm

R

RRT

J

RT

o

io

o

9.451.02

09.01.01085

2446

44max

max

maxmaxmax

=×

−×××=

−=

==

π

πτ

στ

10.2 General Stress Systems in 2-Dimensions

A τxy

τxy σ

x σx σ

y

σy

B

C D y

x

Figure 10.4

The stresses on an element in a component subjected to combined 2D loading (assuming no through thickness stresses, i.e. plane stress) are schematically shown in Figure 10.4. The element ABCD shown is taken of unit thickness and stresses are not to vary in the thickness direction. The reference system of co-ordinate axes are as shown. The normal stresses acting on the x and y faces are respectively σx and σy. The


64

shear stresses are denoted by τxy. Observe that the sign convention uses a shear stress on a positive face to be positive in the direction of the co-ordinate system. For such a stress element there are two points of importance:

i. What are the maximum stresses and on what plane (in what orientation) do they act? ii. What is the stress state on a chosen plane of interest?

10.3 Stresses on a general oblique plane Consider rotating the element ABCD by an angle θ to the x-axis so that it now has axes of x’ and y’ orientated at angle θ to the x and y axes. To determine the new stresses σ x’ , σ y’ and τ x’y’ on the element in terms of the original stresses consider the free body diagram of a prismatic element ADE and the stresses acting on it are as shown in Figure 10.5. The normal stress σx’ and shear stress τx’y’ act on the plane AE and maintain the equilibrium of the prismatic element.

y'

x'

A τx’y’

τx’y’ σ

x’ σx’ σ

y’

σy’

B

C

D

x

y

θ

A

E D

θ

y' x' θ

Area = A

Area = Asinθ

Area = Acosθ τ

xy

τx’y’ σ

x σy

σx’ θ

Figure 10.5

The stresses σx’ and τx’y’ are obtained by resolution of forces in the respective directions.

0sin.sin.cos.sin.cos.cos.sin.cos.

0cos.sin.sin.sin.sin.cos.cos.cos..

'''

''

=+−−+=

=−−−−=

∑∑

θθτθθσθθτθθστ

θθτθθσθθτθθσσ

AAAAAF

AAAAAF

xyyxyxyxy

xyyxyxxx

These expressions reduce to:

( ) ( )θθτθθσστθτθσθσσ

22''

'

sincoscossin

sinsincos

−+−−=

++

xyyxyx

xy2

y2

xx 2 =

These equation simplifies, after the use of some trigonometric identities to:

θτθσσσσ

σ 2sin2cos22' xy

yxyxx +

−+

+= (10.1)

θτθσσ

τ 2cos2sin2'' xy

yxyx +

−−= (10.2)

The stress in the y’ direction may be determined by replacing θ with (θ + π /2) to give:

θτθσσσσ

σ 2sin2cos22' xy

yxyxy −

−−

+= (10.3)


65

10.4 Principal stresses and principal planes Equations 10.4 and 10.5 may be rearranged to give:

( ) ( )

( ) ( )

2

2

2''

2

'

2

22

2''

2

222

'

22

2cos2sin.2cos2sin2

2cos2sin2

2sin2sin.2cos2cos2

2sin2cos22

xyyx

yxyx

x

xyxyyxyx

xyyx

yx

xyxyyxyx

xyyxyx

x

τσσ

τσσ

σ

θτθτθσσθσσ

θτθσσ

τ

θτθτθσσθσσ

θτθσσσσ

σ

+

−=+

+−⇒

+−−

−=

+

−−=

+−+

−=

+

−=

+−

Define two new variables σ ave and R:

τ+σ-σ

R

σ+σ=σ

xyyx

yxave

2

2

2

2

= (10.4)

The equations above may be simplified to:

( ) 22''

2' Rτσσ yxavex =+− (10.5)

Which is the equation for a circle of radius R centred at σ x’ = σ ave and τ x’y’ = 0 as shown in Figure 10.6

σmax

σave

σmin

R

τx'y’

σx'

Figure 10.6 τ+σ-σ

- σ+σ

R=σxy

yxyxave

2

2

min22

=−σ τ+σ-σ

σ+σ

R=σ xyyxyx

ave2

2

max22

+=+σ

σσσσmax and σσσσmin are called Principal stresses and planes on which these stresses act are called Principal planes. To find out what angle θ p the element ABCD has to be rotated to develop these principal stresses on its edges observe that at the point when the principal stresses are developed the shear stress is equal to zero. Therefore using equation 10.6 and setting τ x’y’ = 0 gives:


66

yx

xyp σσ

τθ

−=

22tan (10.6)

The maximum shear stress that can be developed on the element is equal to the magnitude of the radius of the circle: τ+

σ-σR xy

yxyx

2

2

max'' 2

±=±=τ (10.7)

when this maximum shear stress is achieved the axial stresses on the element are equal to σ ave so equations 10.4 or 10.6 may be used to determine the rotation of the element to create this stress state:

xy

yxs τ

σσθ

22tan

−−= (10.8)

These equations may then be used to determine the stress on an element as it is rotated through a stress field. Importantly it allows the maximum (principal stresses to be determined and the orientation at which they occur. This helps an engineer to determine along which plane a structure is likely to fail.

10.5 Graphical representation of stresses in 2-D. Given the stresses σx, σy and τxy acting on the x and y planes at a point (either on the surface or inside) on a structure, equations 10.4 and 10.5 can be used to calculate the normal and shear stresses on any other plane at that point. The planes on which maximum normal and shear stresses act and the magnitudes of these stresses can also be found. A simple graphical procedure (developed by German engineer Otto Mohr) can be followed to obtain the same results whilst avoiding lengthy and repetitive calculations.

Mohr’s procedure is based upon the observation that for any plane at that position (i.e. for any value of θ ),

if the stresses are plotted on a graph of σ versus all of the points obtained will lie on a circle. Knowing the normal and shear stresses on planes normal to x and y (arbitrary) axes at any point in a structure the following elemental diagrams and Mohr’s circle may be drawn.

y

x τxy

τxy σ

x σ

x σy

σy

σ1 σ

1 σ2

σ2

θp

σ1

σave

σ2

τ σ

Pt B (σy ,

τxy)

Pt A (σx ,

τxy)

2θ

p τxy τ

2yx σσ −

Figure 10.7


67

10.5.1 Method of Construction

i. Determine stresses on an element with local axes x and y (σ x, σ y & τ xy)

ii. Chose an appropriate origin for the σ and τ co-ordinate axes. Select a suitable scale, common for both the σσσσ and ττττ axes. Use of graph paper is advantageous. The positive τ axis is considered to be a shear stress that wants to rotate the element clockwise. Remember the jingle “As in the kitchen the clock is above the counter.”

iii. Locate points A (σ x, τ xy) and B (σ y, τ xy) to represent the elemental stresses on the x and y planes. Note σ x or σ y could be positive, zero or negative but Mohr’s method still works.

iv. Join AB. The intersection of line AB with the σ axis is the centre for the Mohr’s circle with a co-ordinate of (σ ave, 0).

v. Construct the Mohr’s circle with AB as diameter. The intercepts of the circle with σ axis (at 1 and 2) having stresses σ1 and σ2 are the principal stresses (maximum and minimum stresses).

vi. Recognise that from equation 10.11, the angle (2θ p) is the angle between the x and σ1 axes. So to orientate the element with the maximum principal stress (σ1) occuring on the x face the element must be rotated θ p in the same sense as point A must be rotated to move point A to the point of maximum stress.

vii. When rotating point A by θ p, point B moves to a position of minimum stress so the element is under pure bi-axial stress with the principal stresses applied (i.e. there are no shear stresses on the element in this orientation). To orientated the element so that the maximum stress occurs on the y face the element must be rotated 90º + θ p.

viii. To find the stresses on an element at any particular angle θ to the x axis simply rotate the line AB an angle corresponding to twice the element rotation (in the same sense) and calculate new values for the axial and shear stresses (see Figure 10.8).

y

x τxy

τxy σ

x σ

x σy

σy

θ

τ σ

Pt B (σy ,

τxy)

Pt A (σx ,

τxy)

2θ

τ

Pt A’ ( σx’ ,

τx’y’ )

Pt B’ (σy’ ,

τx’y’ )

σx’

σx’

σy’

σy’ τ

x’y’

Figure 10.8


68

10.6 Example An element of an engineering structure experiences a state of plane stress with σ x = 50MPa, σ y = -10MPa and τ xy = 40MPa. (a) Determine the principal planes and principal stresses, (b) determine the plane and stress components exerted on the element when it is subject to maximum shear stress and (c) if the material from which the element is made has maximum allowable tensile, compressive and shear stresses of 100MPa, -110MPa and 50MPa respectively predict whether the structure is likely to fail in this stress state.

MPa+τ+σ-σ

R

MPaσ+σ

=σ xy

yx

yxave

50402

1050

2

202

1050

2

22

2

2

=

+=

=

=−=

Mohr’s circle for this stress state is shown in Figure 10.9. From this diagram it is obvious that:

MPaR

MPaR

MPaR

ave

ave

50

305020

705020

max

min

max

==−=−=−=

=+=+=

τσσσσ

Also

( )

( ) °=⇒−=−=

°=⇒−

=−

=

4.1840

20502tan

6.262050

402tan

sxy

avexs

pavex

xyp

θτ

σσθ

θσσ

τθ

So the principal stresses are 70MPa and –30 MPa and are located on faces of an element rotated 26.6º from the x axis. The maximum shear stress of 50MPa occurs on an element rotated –18.4º from the x axis with axial stresses of 20MPa occurring simultaneously.

As the maximum allowable tensile and compressive stresses are greater than the principal stresses the structure will not fail in compression or tension. But as the maximum allowable shear stress is 50MPa any slight increase in the radius of the Mohr’s circle (by increasing any of the loading stresses) would cause the structure fail in shear along planes orientated at (approximately) 18.4º to the x-y axes.

σmax σ

max σmin

σmin τ

σ

Pt B (-10, 40)

Pt A (50, 40)

2θ

p

τ

σave = 20

2θ

s τmax

σmin

σmax

y

x

40

50

-10 θp θ

s σave

σave σ

ave

σave τ

max

Figure 10.9

Tutorial 2: Torsion and Transformation of Stress & Strain

69

TUTORIAL 2: TORSION AND TRANSFORMATIONS OF STRESS & STRAIN 2.1 Determine the torque in each section of the stepped shaft shown:

A B C D 1.2m 0.6m 1.4m

16kNm 20kNm 6kNm 2kNm

1.8m E

8kNm 10kNm

F 1.4m 1.6m 1.0m

Answers (26kNm, 20kNm, 0kNm, 8kNm, 6kNm, -10kNm)

2.2 Determine the torque required to cause an angle of twist of 5º in an aluminium tube of length 4ft with a polar second moment of area of 0.4in4 (Use G = 3.9×106psi). How much longer would the tube have to be to see an end-to-end twist of 7.5º? If the outer diameter of the tube is 0.75in and the inner diameter 0.5in, what are the mean and minimum shear stresses in the tube?

Answers (2.84kip.in, 2ft, 4431psi, 3545psi)

2.3 A 6ft long aluminium rod (G = 3.9×106psi) is bonded to a 4ft long brass rod (G = 5.6×106psi). Both rods are 0.5in in diameter and the free end of the brass rod is fixed to a rigid support. A 300lb.in torque is applied to the free end of the aluminium rod so that the maximum stress in the brass/aluminium composite rod is 12.225ksi. Determine the angle of twist at (a) the aluminium/brass joint and (b) the free end of the aluminium rod.

Answers (24º, 75.7º)

2.4 A pulley shaft has three pulleys loaded as shown. The length of shaft between end A and pulley B is 0.6m, 0.8m between pulleys B and C, 1.0m between pulleys C and D and 0.5m between pulley D and end E. The shaft is made of steel (G = 27GPa). Determine (a) the angle of twist between C and B and (b) D and B.

B

900Nm

500Nm

400Nm

36mm C

D

36mm

30mm 30mm

A

E Answers (8.54º, 2.11º)

2.5 A composite torsion bar is made from a 0.5m length of solid circular cross-section aluminium bonded end-to-end with a 0.25m length of solid circular cross-section brass. If a torque of 1250Nm is applied determine the required diameters of the two sections of the torsion bar. The maximum allowable stress in the brass rod is 50MPa and 25MPa in the aluminium.

Answers (Al 63.4mm, B 50.3mm)


70

2.6 A torque T is applied to a solid tapered shaft AB that is fixed at end B. Show by integration that the angle of twist at A is φ = 7TL/(12π Gr4).

2r

r A

B

L

T

2.7 A thin tube of outer diameter 60 mm is required to carry a torque of 150,000 Nmm. The shear stress must not exceed 27 N/mm2. a. What is the required wall thickness of the tube?

Answer (1.03mm)

2.8 A hollow shaft is 50mm outside diameter and 30mm internal diameter. An applied torque of 1.6kNm is found to produce an angular twist of 0.4º, measured on a length of 0.2m of the shaft. Calculate: a. The modulus of rigidity of the shaft material, and b. the maximum power which could be transmitted by the shaft at 2000 rpm if the maximum allowable shearing stress is 65 MN/m2.

Answers (86 GN/m2, 2897 kW)

2.9 A hollow cylindrical steel (Modulus of Rigidity = 12×106 lbf/in2) shaft, 5 in. outside diameter and 3 in. inside diameter, has a maximum shear stress of 8000 lbf/in2: a. What power can be delivered at 200 rpm? b. What is the maximum angle of twist in a 50 ft. length of the shaft? c. What percentage stronger would the shaft be if it were solid having the same outside diameter?

Answers (404.9kW, 9.17º, 14.90%)

2.10 A solid circular shaft of diameter D has to transit power P at a given speed of m rpm. A hollow tubular shaft, with an inside diameter equal to 2/3 of its outside diameter, transmits the same power at the same speed and the maximum permissible shear stress is the same for each shaft. Find the ratio of the weights of the two shafts.

Answer (Solid/Hollow = 1.55)


71

2.11 A circular shaft (shown below) is fixed to rigid supports at both ends (A and C). The shaft is solid from A to B and hollow from B to C and a 100Nm torque is applied at the mid-section (B). Determine: a. The torque exerted on the ends of the shaft by the supports b. The angle of twist at point B if the material used has a modulus of rigidity of 4GPa.

300mm

40mm 20mm A

150mm

100Nm

B

C

Answers (51.61Nm, 48.39Nm, 0.441º)

2.12 A composite shaft with a steel core and an aluminium outer (cross-section shown below) is 100mm long and is rigidly clamped at both ends. A torque of 122.5 Nm is applied 40mm from one end. Find a. the torque transmitted to each end, b. the rotation of the plane of application of the torque and c. the maximum shear stress in the bar.

rs

ra

rs = 1cm

ra = 2cm

Es = 196 GPa

Ea = 73.5 GPa νs = 0.30 νa = 0.25

Answers (73.5 Nm and 49 Nm, 3.62×10-4 rads, 6.762 MPa)

2.13 A horizontal shaft, securely fixed at each end has a free length of 10m. Viewed from one of the shaft ends, axial couples of 30kNm clockwise and 40kNm counter-clockwise act on the shaft at distances of 4m and 7m respectively from the viewed end. Determine: a. the end fixing couples in magnitude and direction and b. the diameter of the solid shaft for a maximum shearing stress of 60 MN/m. c. Draw a diagram to show how a line, originally parallel to the axis on the outer surface of the shaft will appear after the application of the couples and find the position along the (in metres) where the shaft suffers no angular twist.

Answers (6 kNm and 16 kNm, d = 0.127m, x = 5m)


72

2.14 A composite shaft of circular cross-section is 19.5in. long is rigidly fixed at each end. A 12in. length of the shaft is 2in. diameter and is made of bronze (Modulus of Rigidity = 6×106 lbf/in2) to which is joined a 7.5in. long, 1in. diameter shaft of steel (Modulus of Rigidity = 12×106 lbf/in2). If the shear stress in the steel is limited to 8000 lbf/in2, find: a. the maximum torque that can be applied at the joint, and b. the maximum shear stress in the bronze?

Answers (9426 lbf.in, 5000 lbf/in2)

2.15 A cylindrical pressure vessel with an internal diameter of 300mm and wall thickness of 10 mm has a gauge pressure of 400kPa. Calculate the axial and hoop stresses in the pressure vessel.

Answers: (6MPa, 3MPa)

2.16 Calculate the wall stresses in a spherical pressure vessel of 250mm internal diameter, with wall thickness of 6mm pressurised to pressure of 6atm.

Answer: (5.277MPa)

2.17 A flat bar is subjected to a longitudinal stress of 110MPa. Determine the maximum shear stress in the bar and the plane on which it occurs.

Answers: (55MPa, 45º)

2.18 A cylindrical pressure vessel subjected to internal pressure has an axial stress of 40MPa. Calculate the maximum shear stresses (in the plane of the material) and the inclination to the axis of the planes on which they occur – don’t forget the hoop stresses. If the vessel is made using a helically wound cylinder with the helix making an angle of 30º to the longitudinal axis, determine the stresses along the seam.

Answers: (20MPa, 45º, 70MPa, 17.3MPa)

2.19 In a structural member the stresses are found to be σ x,= 20MPa, σ y = -35MPa and τ xy = -28MPa. Determine (a) the shear and normal stresses on a plane making an angle 60º with the x axis (b) the principal stresses and show the planes on which each is acting and (c) the maximum shear stresses and the planes on which these act.

Answers: ((a) -9.8MPa, -45.5MPa (b) 31.7MPa @ 22.75º CW, -46.7MPa @ 67.5º CCW

(c) -39.24MPa @ 22.25ºCCW and 112.25ºCCW)

2.20 A shaft 1.25 inches in diameter transmits 100hp at 1800rpm. Determine the maximum tensile, compressive and shear stresses developed in the shaft (away from end regions). Indicate on a sketch the planes on which each acts.

Answers: (9125psi, 9125psi, 9125psi)

torsion of circular shafts

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