1. torsion - limba.wil.pk.edu.pllimba.wil.pk.edu.pl/~az/cwicz/class_1.pdf · torsion definitions...

Adam Paweł Zaborski

1. Torsion

Definitions

torsion angle – the angle between the final and initial radius vectors of a point, Fig. 1.1; please note, that

the angle is defined in the cross-section, its unit is [rad] or []; this is the rotation angle of one cross-

section with respect to another

Fig. 1.1 Definition of the torsion angle

unit torsion angle – the torsion angle per meter

torsion moment, twisting couple, torque – the moment the vector of which is parallel to the bar axis; it may

be shown in different ways, Fig. 1.2

Fig. 1.2 Different symbols used to present a torque

The most common application of torsion is provided by transmission shafts, which can be either solid or

hollow.

Solution of BVP

de Saint-Venant’s assumptions

1. The projection of the cross-section onto its initial plane remains unchanged (as the rigid membrane),

Fig.1.3

2. The cross-section distortion consists in the movement “out-of-plane” only and is the same for all cross-

sections.

w

v

A

A'

Fig. 1.3 de S-V’s assumptions

The displacement components can be written:

),,(

,,

,cos,sin

,1cos,sin,'

zyu

xywxzv

wv

,

where:

– torsion angle,


– unit torsion angle,

– distortion function; it does not depend on x,

We calculate the strain matrix:

,

0

00

021

21

yz

Tzy

and the stress matrix:

0

00

0 yGzG

Tzy

From the first Navier’s equation we get:

02 (1)

the remaining equations are identically fulfilled. We see that the distortion function is harmonic.

The static boundary conditions are:

on the side surface:

0

nymz

zy, (2)

the remaining conditions are identically fulfilled,

on the bottoms:

1,1

yGqzGq

zvzyvy , (3)

(1st equation is identically fulfilled).

From the kinematic conditions of fixing at the point O, we have:

.0,0,00,0,0

zy (4)

The Laplace equation (1), with zero boundary conditions with the partial derivatives (2), makes so-called

Neumann’s problem, which has a solution accurate up to a constant. For the cross-section with one axis of

symmetry we have:

0

zy.

Because we can do nothing with the remaining conditions on the bottoms (3), we define the loading in the

same form as for the shear stresses:

., yGzGzvzyvy

(5)

Calculating the torque, we have:

F F

syzvyvzs JGdFzzGyyGdFzqyqM (6)

where the torsion inertia moment is defined as:

dFzyzyJF

yzs

22 (7)

In the general case the torsion inertia depends on the distortion function and indirectly on the BVP’s

solution.

Eventually we get the relationship between the unit torsion angle, torsion moment and the torsion rigidity

GIs:

s

s

GJ

M (8)

The unit of the unit torsion angle is [1/m = rad/m].


Torsion of bars with axisymmetric cross-section

Basic formulae

In the particular case of axial symmetry (a circle, a ring), the distortion function is identically equal to zero

and the torsion inertia moment reduces to the polar inertia moment:

00),( JJzy s

It means that the plane cross-section remains a plane after the deformation, it doesn’t deplane.

The stresses, (5), become: ., yGzG vzvy

z

y

xy

xz

Fig. 1.4. Shear stress distribution

The resultant shear stress is:

rJ

MrGyzG s

xzxy0

2222 , [MPa].

The shearing stress varies linearly with the distance from the axis of the shaft, Fig. 1.4.

The torsion angle is calculated from the unit torsion angle provided that the torsion moment is constant:

00 GJ

lM

GJ

M

dx

d ss

l

s

dx [1 = rad]

This is the angle through which one section rotates with respect to the other. Note, that the torsion angle is

dimensionless, so the angle unit in the above formula is a radian, not a degree.

When the interval is loaded by a continuous twist moment of intensity )(xm , the formulae become:

x

dmxM0

)()( , and

x

dGJ

Mx

0 0

)()(

Design conditions

There are two main requirements:

the requirement of the strength and

the requirement of usability.

The first requirement means that every structure has to sustain applied load. The second one consists in

several demands of durability, rigidity, resistance to severe weather conditions and so on. From the point of

view of the strength of materials course two principal design conditions should be listed:

the ultimate limit state

the serviceability limit state, usually the stiffness of the structure.

In the particular case of the torsion, the ultimate limit state is checked by:

tR)max(

where tR is the shear strength of a given material. The suitable value can be found in the standards.

Because the maximum value of the stress is attained at the side surface for Rr , we introduce so-called

cross-section twist factor or modulus 0W :


R

JW

def0

0 ,

so:

0

0

0

0)max(W

MR

J

M

The stiffness condition means that the unit torsion angle or the torsion angle doesn’t exceed acceptable

(permissible) values:

acceptable , or acceptable

Transmission shafts

The principal specifications to be met in the design of a transmission shaft are the power to be transmitted

and the speed of rotation of the shaft. The role of the designer is to select the material and the dimensions

of the cross-section of the shaft, so that the maximum shearing stress allowable in the material will not be

exceeded when the shaft is transmitting the required power at the specified speed.

The power P associated with the rotation of a rigid body subjected to a torque M0 is:

0MP ,

where is the angular velocity of the body expressed in radians per second. But, f 2 , where f is the

frequency of the rotation, i.e., the number of revolutions per second. We write:

fMP 20 .

Hollow shaft

The linear repartition of the shearing stresses in the twisted circular bar signifies that the outer fibers are the

most useful and the central ones are almost not used. Therefore, the use of hollow shafts makes sense,

especially in the case where the weight is more important than the price of the element.

The torsion inertia moment for the hollow shaft with outer diameter D and inner diameter d is:

4444

0323232

dDdD

J

,

and the cross-section torsion factor is:

4400

162 dD

DD

JW

.

Examples

Example 1.1

The shaft with the diameter of 4 cm and length of 2 m is fixed at one end and loaded at another end by such

torque that the point A on the side surface is displaced to the point A’. The arc AA’ length is 1 mm.

Determine the torsion angle, the unit torsion angle, the torque, the maximum shear stress and the angular

strain on the side surface. G = 80 GPa.

Solution

From the definition of the twist angle we calculate:

05.04

2.02

1.0

d[rad].

For constant torsion moment we have:

025.02

05.0

l [rad/m].

Using the formula of the unit torsion angle we get:

500025.032

04.01080

49 ss JGM [Nm]


(it’s quite significant value, the average value for the VW golf is about 150 Nm). The maximum stress

value is:

69

0

0

0

0 104002.0025.010802

d

GRJ

GJ

W

Mmax [Pa] 40 [MPa].

From the Hooke’s equation we determine

0005.002.0025.022

22 max

d

G.

Example 1.2

Determine the rotation angle between section A and section E of the shaft in Fig. 1.5, 41 d cm, 32 d cm,

G = 80 GPa.

A

B

C D E

30 Nm 40 70

d1 d2

0.3 0.5 m 0.4 0.4

Fig. 1.5 Twisted shaft

Solution

The formula of the twist angle:

0

0

GJ

lM

is valid only in the case of all parameters constant, so we have to cut the shaft into the intervals with

constant cross-section inertia moment and constant torque: AB, BC, CD and DE.

We have:

AB: 4.0ABl m, 60704030 ABM Nm, 741 1051.2

32

d

J AB m4, 00120.0

1051.21080

4.06079

AB rad

BC: 4.0BCl m, 104030 BCM Nm, 741 1051.2

32

d

J BC m4, 000199.0

1051.21080

4.01079

BC rad

CD: 3.0CDl m, 104030 CDM Nm, 842 1095.7

32

d

JCD m4, 000472.0

1095.71080

3.01089

CD rad

DE: 5.0DEl m, 30DEM Nm, 842 1095.7

32

d

J DE m4, 00236.0

1095.71080

5.03089

DE rad

and the total angle of the twist is:

00289.000236.0000472.0000199.000120.0 DECDBCABAE rad.

Fig. 1.6 presents the diagram of the twist angle.


Fig. 1.6 Diagram of the twist angle

Example 1.3

Determine the diameter of the shaft in Fig. 1.7, if the acceptable values of (a) the shear stress is 150tR

MPa, (b) unit twist angle is 05.0acc [rad/m] and (c) the twist angle between the sections B and D is 0.02

[rad]. The Kirchhoff module value is G = 80 GPa, 12 2.1 dd .

A B

C

D

0.3 1.2 m 0.7 m

300 Nm 800 Nm

Fig. 1.7 Shaft under torsion

Solution

the 1st condition will be fulfilled when:

1500

0max tR

W

MMPa

hence the following equations:

30

3

0

0

0 1616

tt

R

MdR

d

M

W

M

and

0217.010150

300163

6

ABd m, 0181.0

10150

30016

2.1

13

6

BCd m, 0214.0

10150

50016

2.1

13

6

CDd m

so

0217.0,,max CDBCAB dddd m

the 2nd

condition will be fulfilled when

05.00

0 accGJ

M

hence, we have:

49

0

4

0

05.01080

3205.0

32

Md

dG

M

angle

0

0,0005

0,001

0,0015

0,002

0,0025

0,003

0 0,5 1 1,5


so

0296.005.01080

300324

9

ABd m, 0246.0

05.01080

30032

2.1

14

9

BCd m, 0280.0

05.01080

50032

2.1

14

9

CDd m

so 0296.0,,max CDBCAB dddd

the 3rd

condition will be fulfilled when

02.0 accBD rad

hence, we have:

02.0

2.11080

322.1500

2.11080

323.03004949

00

ddGJ

lM

GJ

lM

CD

CDCD

BC

BCBCCDBCBD

so

0354.03.03002.150002.01080

32

2.1

14

9

d m

Taking into account the results obtained we finally assume:

cm6.3m036.0 d

Example 1.4

For the shaft in Fig. 1.8 determine the function of the twist moment and draw the unit torsion angle

diagram. Determine the values and draw the diagram of the torsion angle. 31 d cm, 1.32 d cm, 80G

GPa.

d1 d2

0.2 0.5 1.2 m 0.6

150 Nm/m

Fig. 1.8 Shaft with loading

Solution

In order to solve the redundant problem we need to go back to the BVP and the complete set of its

equations. As we know, there are:

the statics equations of the equilibrium of the forces (Navier’s equations with the SBC),

the geometric equations of compatibility (Cauchy’s equations with KBC), and

the constitutive equations describing behavior of the material.

For the particular problem of the shaft twist we have:

one statics equation 00 M (the resultant torque should be zero for static equilibrium)

one geometric equation of compatibility 0 EAAE (the twist angle between one end and another,

both fixed)

the constitutive equations 0

0

GJ

lM

We release the shaft end from constraints:


B A

MA ME

0.2 0.5 1.2 m 0.6

150 Nm/m

D C E

Fig. 1.9 Shaft under loading

We have two unknowns EA MM , and two equations, the static one and the kinematic one:

07.0150 EA MM

0 DECDBCAB

The second equation can be written as:

02.17.01505.05.05.01502.01502.02.05.01506.0

02020101

GJ

M

GJ

M

GJ

M

GJ

M AAAA

Multiplying both sides by 01GJ , with 8771.00201 JJ we get:

05.11005.16.295.08771.032.06.0 AAAA MMMM ,

hence we obtain:

47.62AM Nm, 53.42EM Nm.

Fig. 1.10, 1.11 and 1.12 present the diagrams of the twist moment, unit twist angle and twist angle.

Fig. 1.10 Twist moment diagram

Moment

-60

-40

-20

0

20

40

60

80

0 0,5 1 1,5 2 2,5


Fig. 1.11 Unit twist angle diagram

Fig. 1.12 Twist angle diagram

theta

-0,008

-0,006

-0,004

-0,002

0

0,002

0,004

0,006

0,008

0,01

0,012

0 0,5 1 1,5 2 2,5

alpha

0

0,001

0,002

0,003

0,004

0,005

0,006

0,007

0,008

0 0,5 1 1,5 2 2,5


Torsion of noncircular members

Membrane analogy

It has been proved by de Saint-Venant that the distortion function vanishes for some particular cross-

section shapes only, e.g. axisymmetrical, triangular, elliptical etc. In general, if the distortion function is

non-zero, the BVP should be solved.

By use of some kind of parametrization (of so-called Prandtl’s functions), the BVP may be written in

another form of Poisson’s equation:

G22

where the function is the Prandtl’s function:

yz

xzxy

, .

The distribution of shearing stresses in a noncircular member may be visualized more easily by using the

membrane analogy. An elastic membrane subjected to a uniform pressure constitutes an analog of the bar

in torsion. The same BVP describes the deformation of the membrane and the shearing stresses in the

twisted bar. The shearing stress at some point will have the same direction as the horizontal tangent to the

membrane and its magnitude will be proportional to the maximum slope of the membrane. Moreover, the

torque will be proportional to the volume between the membrane and its initial plane.

Rectangular cross-section

In the case of rectangular cross-section the steepest slope occurs at the midpoint of the larger side of the

rectangle, Fig.1.13

Fig. 1.13 Prandtl’s function for rectangular section

Furthermore, the shearing stress is zero at the corners and at the center of the section, Fig. 1.14.


=0

=0

max

ma

x

Fig. 1.14 Shearing stress in rectangular cross-section

The solution for a rectangular section may be obtained by expansion of the warping function in series:

0

).()(,n

nn zgyfyzzy .

The solution can be presented as formulae:

max3

2max ,,

sbh

s

bh

s hbJhb

M,

where s is the shearing stress at the middle of the shorter side and the coefficients , and are stated in

the table 1.1.

h/b 1 1.25 1.5 2 3 5 10 ∞

0.2082 0.2212 0.2310 0.2459 0.2672 0.2915 0.3123 0.3333

0.1406 0.1717 0.1958 0.2287 0.2633 0.2913 0.3123 0.3333

1.0 0.9159 0.8591 0.7958 0.7533 0.7429 0.7423 0.7423

Table 1.1 Coefficients for maximum shear stress, torsion inertia moment and additional shearing stress

Thin-walled cross-section

A thin-walled cross-section member is a bar the cross-section of which is made from profile, for example

from bent sheet of steel. The exact solution of such twisted profiles is difficult and usually not needed in

practice. There are simplified methods used instead.

From the point of view of practical calculation there are three types of these cross-sections, Fig. 1.15:

(a) the open thin-walled profiles folded (which can be uncurled and so-called middle-line doesn’t fork),

(b) the open profiles unfolded (which cannot be uncurled and the middle-line branches),

(c) the closed profiles (the middle-line is a closed line).

a) b) c)

Fig. 1.15 Thin-walled profiles with middle-line (a), (b) and (c)

Profiles with middle-line not branched

Some of the thin-walled cross-sections may be substituted by one rectangle with the same area and the

longer side length is equal to the middle-line of the profile.

For example, a C profile with the web 300x10 mm and the flanges 90x16 mm may be substituted by a

rectangle with middle-line length l = 2x90 + 300 = 480 mm and average width b = 12.25 mm.

Hydrodynamic analogy

There is an analogy between the distribution of the shearing stresses in the transverse section of a shaft and

the distribution of flow lines and velocities in water flowing through a closed channel of unit depth and

variable width, Fig. 1.16. In the last case, the water flow as well as the shear flow are constant, const .


v = const

Fig. 1.16 Hydrodynamic analogy: flow lines and flow velocities in hollow shaft

Profiles with middle-line branched

The thin-walled cross-sections with the branched middle-line can be solved applying the hydrodynamic

analogy. The section is cut into rectangles and it is assumed that:

each rectangle works independently, transmitting a part of the applied torque, siM ,

the unit torsion angle of each rectangle is the same, idemi .

With the static equilibrium equation, we get a set of n equations for n component rectangles:

i is

is

si

sisis ni

GJ

M

GJ

MMM 1,,1,,

)1(

)1( ,

hence we find siM and the unit torsion angle of the whole cross-section..

The shorter cut lines are, the better the approximation is.

Hollow profiles

The shearing stress in closed thin-walled cross-sections may be determined by means of the hydrodynamic

analogy. The flow of the stress through every section is constant: const2211 . The shearing stress,

running around the cross-section and acting on a different lever, constitutes the torque as a sum of the

shearing stress actions. Assuming that the stress repartition is constant in each section of the closed

channel, the torque may be written as:

c c

s AdsssdsM 2)()( ,

where F is the area inside the middle-line of the profile, Fig. 1.17.

1

ds2

Fig. 1.17 Shearing stress circulation in closed profile

Hence, the extreme shearing stress will be attained in the section of the channel with the smallest width:

min

max

A

M s

2.

The above formula is known as Bredt’s first formula.

To determine the deformation caused by the torsion we use the formula:

GJ

M s ,


where GJ is the torsional stiffness, and the inertia moment is (Bredt’s second formula):

c

AJ

ds

AJ

c

22 4,

4const .

Bredt’s theory has been developed on the basis of a simplifying assumption for the stress distribution

which gets closer to the actual distribution when the wall becomes thinner.

Examples

Example 1.5

Determine the maximum shearing stress and the unit twist angle of the cross-section in Fig. 1.18. Compare

the solution for the profile with not branched middle-line with the Bredt’s solution.

10 300

100

16

16

Fig. 1.18 Twisted cross-section

Solution

a) We consider the cross-section as a rectangle with the length equal to the length of the middle-line:

4745100216300 l mm

To obtain the same area we take:

41.12474

588058803230010210016 A mm

We have:

3

12.38

41.12

474

l

The maximum shearing stress is:

ss

s

s MM

W

M 5

2

31max 1011.4

47441.12

G

M

G

M

GJ

M ss

s

s 6

3

31

1031.347441.12

1

b) We divide the cross-section into 3 rectangles, two flanges 16100 , 25.6

l and one web 10268 ,

8.26

l. The parameters and for the web are

31 and for the flanges (from linear interpolation)

are 2976.0 and 2977.0 . Using the Bredt’s assumptions, we have:

– the static equation

sMMM wfl2

– the compatibility equation

flwwfl

w

w

fl

flwfl MM

MM

GJ

M

GJ

M7326.0

268103333.0100162977.0 33

and, finally:


ss M

M 5

2

31

1000.326810

2681.0

w , ss M

M 5

21080.4

100162976.0

366.0

fl , sM51080.4max

G

M

G

M ss 6

31000.3

100162977.0

366.0

Conclusions:

Both methods give different answers, the relative errors are: for maximum shearing stress 17% and for the

unit twist angle about 10%.

Example 1.6

Determine the shearing stress in the T-beam (the web 30x5 cm and the flange 6x20 cm), twisted by the

torque 20 kNm,. while l/ = 6: = 0.299, = 0.299, h/b = 3.33: = 0.272, = 0.269.

MPaMPa

kNm,kNm

flw 0.52,8.43

18.10,82.91020

1021620269.01012530299.0

0036.02.0272.010180

0025.03.299.09820

213

21

8

2

8

121

MMMM

G

M

G

M

Example 1.7

To get an idea on applicability of Bredt’s formulae let’s analyze a particular case of twisted ring of the

outer diameter R and inner diameter r, where both exact and approximate solutions can be found. For the

sake of analysis we will use a parameter which is the ratio of the cross-section thickness to the middle-

line radius: .

a) Bredt’s solution

2max

2

1

2

ssB M

A

M

32 2

1

2

2

G

M

GA

M ssB

b) exact solution

22max4

2

s

s

s M

W

Mexact

224

2

G

M

GJ

M s

s

s exact

The ratio of Bredt’ solution to the exact solution is:

4

4,

24

4 22

exact

Bredt

exact

Bredt .

The above formulae as functions of parameter are presented in Fig. 1.19.

The maximum value of shearing stress is underestimated and quickly starts to be significant (from 1.0 )

and error associated to the unit twist angle remains small up to the larger values of the parameter (up to

0.5).


Fig. 1.19 Comparison of Bredt’s solution with exact solution

Review problems

Problem 1.1

An engine is connected to a generator by a large hollow shaft with inner and outer diameters of 100 mm

and 150 mm, respectively. Knowing that the allowable shearing stress is 85 MPa, determine the maximum

torque that can be transmitted: (a) by the shaft as designed, (b) by a solid shaft of the same weight, (c) by a

hollow shaft of the same weight and of 200 mm outer diameter.

Ans.: a) 45.2 kNm, b) 23.3 kNm, c) 70.5 kNm

Problem 1.2

Knowing that each of the shafts AB, BC and CD consists of solid circular rods, Fig. 1.20, determine the

shaft in which the maximum shearing stress occurs and the magnitude of that stress.

201 d mm, 252 d mm, 303 d mm.

Ans.: 58.7 MPa in the 2nd

shaft.

D

C

B

A

270 Nm

110 Nm

90 Nm

Fig. 1.20 Shafts with loads


Problem 1.3

A gear, Fig. 1.21, has two cogs: a greater one and a smaller one, with the diameters of 20 mm and 8 mm,

respectively. Knowing that the allowable shearing stress is 85 MPa, determine the diameters of the shafts if

the torque 2201 M Nm.

M1 M2

Fig. 1.21 Gear with two cogs and shafts

Ans.: 0174.0,0236.0 21 dd m.

Problem 1.4

Knowing that the allowable shearing stress of a brass shaft with both ends fixed, Fig. 1.22, is 55 MPa,

determine the diameter of the shaft. Using the calculated diameter and knowing that the Kirchhoff’s

modulus is 37 GPa, determine the rotation angle of the section where the torque is applied.

0.6 m 1 m

170 Nm

Fig. 1.22 Brass shaft with load

Ans.: 0214.0d m, 79.4rad0837.0

Problem 1.5

Determine the diameter of the steel shaft, Fig. 1.23, knowing that the allowable shear stress is 80 MPa, and

the maximum value of the rotation angle should not exceed the value of 2 degrees, G = 80 GPa.

100 Nm

0.6 m 0.7 m

170 Nm

Fig. 1.23 Shaft with loading

Ans.: 0198.0d m (cf. the twist angle diagram).


Problem 1.6

Using an allowable shearing stress of 40 MPa, design a solid steel shaft of a VW engine to transmit 100

kW at the speed of 5000 rpm

Ans.: 0290.0d m

Problem 1.7

Determine the maximum shearing stress and the unit twist angle for the cross-sections stated in Fig. 1.24.

Fig. 1.24 Different profiles

Addendum

Glossary

torsion angle, twist angle – kąt skręcenia

unit torsion angle – jednostkowy kąt skręcenia

torsion moment – moment skręcający (jako siła przekrojowa)

torque – moment skręcający (jako obciążenie)

wheel nut torque - moment dokręcenia śruby

torsion inertia moment – moment bezwładności na skręcanie

torsion modulus, torsion/twist factor – wskaźnik na skręcanie

distortion, warping – spaczenie, deplanacja

torsion rigidity – sztywność skręcania

shaft – wał, wałek

gear – przekładnia

cog – koło zębate, ząb (koła zębatego)

brass – mosiądz

rpm (rotations per minute) – (ilość) obrotów na minutę

middle-line – linia środkowa

web – środnik

flange – półka

profile – profil

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