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arXiv:physics/0211004v1 [physics.ed-ph] 1 Nov 2002

Stress in Rotating Disks and Cylinders

Thomas B. Bahder

U. S. Army Research Laboratory

2800 Powder Mill Road

Adelphi, Maryland, USA 20783-1197

(Dated: February 2, 2008)

The solution of the classic problem of stress in a rotating elastic disk or cylinder, as solved in

standard texts on elasticity theory, has two features: dynamical equations are used that are valid

only in an inertial frame of reference, and quadratic terms are dropped in displacement gradient

in the definition of the strain. I show that, in an inertial frame of reference where the dynamical

equations are valid, it is incorrect to drop the quadratic terms because they are as large as the

linear terms that are kept. I provide an alternate formulation of the problem by transforming the

dynamical equations to a corotating frame of reference of the disk/cylinder, where dropping the

quadratic terms in displacement gradient is justified. The analysis shows that the classic textbook

derivation of stress and strain must be interpreted as being carried out in the corotating frame of

the medium.

I.INTRODUCTION

The problem of stresses in rotating disks and cylinders is important in practical applications to rotating machinery,

such as turbines and generators, and wherever large rotational speeds are used. The textbook problem of stresses

in elastic rotating disks and cylinders, using the assumption of plane strain or plane stress, is published in classic

texts, such as Love [1], Landau and Lifshitz [2], Nadai [3], Sechler [4], Timoshenko and Goodier [5], and Volterra and

Gaines [6]. The standard approach presented in these texts has two characteristic features:

1. Newton’s second law of motion is applied in an inertial frame of reference to derive dynamical equations for the

continuum (see Eq. (1) below), and

2. quadratic terms in displacement gradient are dropped in the definition of the strain tensor (see Eq.(20) below).

In this paper, I show that, for a rotating elastic body, the second feature of the solution is inconsistent with the first:

dropping the quadratic terms in the displacement gradient is an unjustified approximation in an inertial frame of

reference. In what follows, I refer to the method that is employed in Ref. [1, 2, 3, 4, 5, 6] as ‘the standard method’,

and for brevity, I will refer to a cylinder as a generalization of both a disk and a cylinder.

The classic problem of stress in an elastic rotating cylinder is complex because the undeformed reference state

of the body is the non-rotating state. The deformed state is one of steady-state rotation. The analysis of the

problem must connect the non-rotating reference state to the rotating stressed/strained state. These two states are

typically connected by large angles of rotation. When large angles of rotation are present, the quadratic terms in the

displacement gradient cannot be dropped (in an inertial frame of reference) in the definition of the strain [7, 8, 9]. The

problem of stress analysis when large-angle rotations are present is well known and has been discussed by a number

of authors in general contexts, see for example [7, 8, 9]. However, large-angle rotations in the problem of a rotating

elastic cylinder have not been dealt with in a technically correct manner, because quadratic strain gradient terms are

incorrectly dropped in the ‘standard method’ [1, 2, 3, 4, 5, 6].

In this work, I formulate the elastic problem of a rotating cylinder in a frame of reference that is corotating with

the material. In this corotating frame, the quadratic terms in the displacement gradient can be dropped, and the

resulting differential equations are linear and can be solved.

In section II, I review the ‘standard method’ of solution used in Ref. [1, 2, 3, 4, 5, 6] and show that for a rotating

cylinder the displacement gradient in an inertial frame of reference is of order unity, and therefore quadratic terms

(in strain tensor definition) cannot be dropped when compared to the linear terms. Section III contains the bulk

of the analysis. I describe the corotating systems of coordinates and the transformation of the velocity field to the

corotating frame. I use the velocity transformation rules to transform the dynamical Eq. (1) from the inertial frame

to the corotating frame (see Eq. (61) or (62)), where extra terms arise known as the centrifugal acceleration and the

coriolis acceleration. In section IV, I write the explicit component equations for stress (in cylindrical coordinates)

for the rotating elastic cylinder in its corotating frame. To display the resulting solution concretely, I derive the

well-known formula for the stress in the rotating cylinder for the case of plane stress, as computed in the corotating

frame. Stress is an objective tensor, i.e., stress is independent of observer motion [10, 11], so the physical meaning

of stress in the corotating frame is the same as in the inertial frame. Therefore, the stress field components in the

corotating frame are equal to the stress field components in the inertial frame, see Eq. (36).

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II.STANDARD SOLUTION METHOD

In the ‘standard method’ [1, 2, 3, 4, 5, 6], the stress analysis of elastic rotating cylinders starts with the dynamical

equations, which, in generalized curvilinear coordinates are given by [2, 10, 11]

σkj

;j+ ρfk= ρak

(1)

where σkjare the contravariant components of the stress tensor, fkis the vector body force, and akis the acceleration

vector. In Eq. (1), repeated indices are summed and the semicolon indicates covariant differentiation with respect to

the coordinates. Expressed in terms of the velocity field in spatial coordinates, the acceleration is given by [10, 11]

ak=∂vk

∂t

+ vjvk

;j

(2)

where vais the velocity field, and the semicolon indicates covariant differentiation with respect to the coordinates,

and vjvk

time dependent. Equation (1) is derived by applying Newton’s second law of motion to an element of the medium.

Newton’s second law is valid only in an inertial frame of reference, and consequently the validity of Eq. (1) is limited

to inertial frames of reference.

In the ‘standard method’ of solution, Eq. (1) is applied by invoking an “effective body force”, of magnitude equal to

the centrifugal force in the rotating frame. In the inertial frame, there is actually no effective force (such as Coriolis

or cetrifugal force). For the case of a body rotating about its principle axis, a more careful determination of the terms

fk− akin Eq. (1) comes from setting the body force to zero (or setting equal to some applied force) and computing

the material acceleration akfor a given body motion. For a rigid body, or a uniform density elastic cylinder that is

rotating about its axis of symmetry at a constant angular velocity ωo, the Cartesian velocity field components are:

v1= −ωoy, v2= ωox, and v3= 0, where superscripts 1,2,3 indicate components on the Cartesian basis vectors

associated with the x,y,z-axes (in the inertial frame). Corresponding to this velocity field, the cylindrical components

of the acceleration field are given by

;jis called the convective term. In Eq. (1), the stress σkj, acceleration ak, and body force fk, are generally

¯ ak=∂¯ vk

∂t

+ ¯ vb¯ vk

;b=?−rω2

o,0,0?

(3)

where I have chosen the z-axis as the symmetry axis and the bar over the components indicates that they are in the

inertial frame of reference in cylindrical coordinates. For the case where there are no body forces, with the acceleration

in Eq. (3), Eq. (1) in cylindrical coordinates leads to the three equations

¯ σ11

,1+ ¯ σ12

,2+ ¯ σ13

,3+¯ σ11

r

,3+3

− r¯ σ22= −ρrω2

o

(4)

¯ σ12

,1+ ¯ σ22

,2+ ¯ σ23

r¯ σ12= 0

,3+¯ σ13

r

(5)

¯ σ13

,1+ ¯ σ23

,2+ ¯ σ33

= 0(6)

where the superscripts 1,2,3 enumerate tensor components on the r,φ,z coordinate basis vectors respectively, in

cylindrical coordinates and the commas indicate partial differentiation with respect to these coordinates.

For steady rotation at a uniform angular velocity ωo, and assuming the absence of elastic waves, there is rotational

symmetry about the z-axis so the stress components do not depend on azimuthal angle φ. Therefore, all derivatives

with respect to φ are zero, leading to the equations:

¯ σ11

,1+ ¯ σ13

,3+¯ σ11

r

,3+3

− r¯ σ22= −ρrω2

o

(7)

¯ σ12

,1+ ¯ σ23

r¯ σ12= 0

,3+¯ σ13

r

(8)

¯ σ13

,1+ ¯ σ33

= 0(9)

I introduce physical components of stress, σrr, σφφ,σzz, σrφ, σrz, and σφz, with units of force per unit area and

which are related to the tensor components ¯ σ11, ¯ σ22, ¯ σ33, ¯ σ12, ¯ σ13, and ¯ σ23, by [10, 11]

¯ σrr= ¯ σ11,(10)

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¯ σφφ= r2¯ σ22

¯ σzz= ¯ σ33

¯ σrφ= r¯ σ12

¯ σrz= ¯ σ13

¯ σφz= r¯ σ23

(11)

(12)

(13)

(14)

(15)

Expressing Eq. (7)–(9) in terms of the physical components, I obtain the well-known equations valid in an inertial

frame of reference [1, 2, 3, 4, 5, 6],

∂¯ σrr

∂r

?1

+∂¯ σrz

∂z

+¯ σrr− ¯ σφφ

r

= −ρrω2

o

(16)

∂

∂rr¯ σrφ

?

+1

r

∂¯ σφz

∂z

+∂¯ σzz

+3

r2¯ σrφ= 0

+¯ σrz

r

(17)

∂¯ σrz

∂r ∂z

= 0(18)

Note that Eqs. (16)–(18) have been derived using Newtons’s second law, and so they are valid only in an inertial

frame of reference. In particular, Eqs. (16)–(18) are not valid in a rotating frame of reference.

When a rotating disk or cylinder is analyzed, the assumption of plane stress or plane strain is often made. In

both cases, stresses must be related to strains by constitutive equations. For the simplest case of a homogeneous,

isotropic, perfectly elastic body, the constitutive equations in curvilinear coordinates in an inertial frame can be

written as [2, 10, 11]

σik= λegik+ 2µeik

(19)

where λ and µ are the Lam´ e material constants, eikare the contravariant strain tensor components, e = ea

contraction of the strain tensor, and gikare the contravariant metric tensor components.

The Eulerian strain tensor eikis related to the displacement field uiby [10, 11]

a

is the

eik=1

2

?ui;k+ uk;i+ um;ium

;k

?

(20)

In the ‘standard method’ of solution [1, 2, 3, 4, 5, 6], the quadratic terms um;ium

equations that can be solved (for example, by using the Airy stress function [12]).

However, dropping the quadratic terms in Eq. (20) is not justified for a rotating body because these (dimensionless)

terms um;i are of order unity. To prove this assertion, it is sufficient to consider the limiting case of a rigid body

in steady-state rotation at constant angular speed ωo. The deformation mapping function gives the coordinates zk

(here taken to be Cartesian) of a particle at time t in terms of the particle’s coordinates Zkin some reference state

(configuration) at time t = to:

;kare dropped, which leads to linear

zk= zk(Zm,t)(21)

so that zk(Zm,to) = Zk. The deformation mapping function has an inverse, which I quote here for later reference

Zm= Zm(zk,t)(22)

Both zkand Zmrefer to the same Cartesian coordinate system. The coordinates of a particle initially at Zkat

t = to= 0 rotating about the z-axis are given by the deformation mapping function

zk= Rk

mZm

(23)

where the orthogonal matrix Rkmis given by

Rk

m=

cosωot

−sinωot cosωot 0

0

sinωot 0

01

(24)

The displacement vector field for this deformation mapping function is given by [13]

u = umIm= (zm− Zm)Im= (δm

k−¯Rm

k)zkIm

(25)

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where Imare the unit Cartesian basis vectors and¯Rm

kis the transpose matrix that satisfies

¯Rm

kRk

l= δm

l

(26)

where δm

appearing in Eq (20) are of order unity and therefore the quadratic terms um;ium

not small. More specifically, the dropped terms in Eq. (20) vary in time between -2 and 0 (in Cartesian components):

l

= +1 if m = l and 0 if m ?= l. Therefore, from Eq. (25), it is clear that gradients of displacement um;k

;kcannot be dropped because they are

1

2um;ium

;k=

−1 + cosωot

0

0

00

−1 + cosωot 0

00

(27)

The above calculation was done for a rigid body, but clearly, a similar error is introduced for elastic bodies. Therefore,

in general, for a rotating elastic body, the quadratic terms in displacement gradients in Eq. (20) cannot be dropped [14].

In cylindrical components, the relation between the physical components (see Ref. [10, 11]) of strain, ¯ errand ¯ eφφ,

and physical components of the displacement field, (ur,uφ,uz), is given by

¯ err =

∂ur

∂r

−1

2

??∂ur

∂r

?2

+

?∂uφ

∂r

?2

+

?∂uz

∂r

?2?

(28)

¯ eφφ =

ur

r

−

1

2r2

?(ur)2+ (uφ)2?

(29)

(I use a bar over ¯ errand ¯ eφφto indicate that these are cylindrical components, and indices rr and φφ (as distict from

11 and 22) to indicate that these are physical components and not tensor components. See the Appendix and Table

I and II for notation conventions.) In Eq. (28) and (29), I have assumed that there is no dependence on φ and z, so I

have set derivatives with respect to these variables to zero.

In the ‘standard method’ of solving for the stress in a rotating cylinder [1, 2, 3, 4, 5, 6], the quadratic terms in

Eq. (28) and (29) are incorrectly dropped.

The straight forward approach to correctly studying the stresses in a rotating disk or cylinder, involves keeping the

quadratic terms in displacement gradient in Eq. (20). However, this approach does not appear promising because it

leads to insoluble nonlinear differential equations. In the next section, I approach the problem by using a transforma-

tion to a corotating frame of reference, in which dropping the quadratic terms can be justified for moderate angular

velocity of rotation ωo.

III.TRANSFORMATION TO THE ROTATING FRAME

As discussed in the introduction, the problem of an elastic rotating cylinder is complicated because the un-

stressed/unstrained reference state is the non-rotating state, while the stressed (strained) state is rotating, and these

two states are typically related by a large (time-dependent) angle. The analysis of the rotating disk or cylinder must

relate the stresses in the rotating state to the reference configuration, which I take to be the non-rotating state. I

define a transformation from an inertial frame of reference to the corotating frame of reference of the cylinder. This

transformation provides a relation between the rotating stressed state and the non-rotating reference configuration.

A.Coordinate Systems

Starting from an inertial frame of reference, S, defined by the Cartesian coordinates zk= (x,y,z), I make a

transformation to a rotating frame of reference S′. The rotating frame will be corotating with the cylinder so that

in this frame S′the azimuthal velocity field will be zero at all times. The transformation from an inertial system of

coordinates to a rotating system of coordinates is most simply done using Cartesian coordinates. On the other hand,

the assumed cylindrical symmetry of the problem begs for use of cylindrical coordinates. Hence I will make use of four

systems of coordinates. In the inertial frame S, I have two systems of coordinates: a Cartesian system of coordinates

zk= zk = (x,y,z), and a cylindrical system of coordinates xi= (r,φ,z). In the corotating frame of reference,

S′, I have a Cartesian system of coordinates z′k= (x′,y′,z′) and a cylindrical coordinate system x′i= (r′,φ′,z′).

These coordinates are summarized in Table I and the Appendix. I also introduce notation for tensor components

in each of the four coordinate systems, see Table II . The Cartesian components of the stress tensor in the inertial

frame S will be denoted by σik. In the same inertial frame S, the cylindrical components of stress will be ¯ σik. The

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Cartesian components of stress in the corotating frame S′will have a prime, σ′ik. In this same corotating frame, S′,

the cylindrical components of stress will be denoted by using a tilde, ˜ σik.

From the vantage point of an inertial frame of reference, S, with Cartesian coordinates zk, consider a cylinder

whose symmetry axis is aligned and colocated with the coordinate z-axis. At time t = −∞, take the cylinder to

be non-rotating. Now assume that in the distant past, around the time t ∼ −T, the cylinder begins a slow angular

acceleration lasting a long time, on the order of 1/ǫ, where 1/ǫ << T. An example of such an angular acceleration

function is

ω(t) =1

2ωo[1 + tanh(ǫ(t + T))](30)

where I assume that τ << 1/ǫ << T and τ is the longest time constant in the problem. This inequality states that

the acceleration occurs slowly, τ << 1/ǫ, slower than any time scale in the problem, and that this acceleration occurs

in the distant past, 1/ǫ << T, so that at t = 0, I have a steady-state situation of a cylinder rotating at constant

angular speed ωo. By slowly accelerating the cylinder, I avoid introducing modes of vibration. As the cylinder’s

angular velocity increases from t = −∞, each particle comprising the cylinder moves along a spiral trajectory (with

increasing radius). From the point of view of the inertial frame S, the stresses on a given element of the medium

(particle) are such that they cause the particle to experience an acceleration, moving along the spiral path. At t = 0,

the cylinder has achieved its maximum angular velocity ωo. Due to the assumption of a perfectly elastic medium, at

t = 0 the velocity field has zero radial component; all particles of the cylinder are moving azimuthally (in a plane

perpendicular to the z-axis with zero radial component). The velocity field is that of a rigid body and the acceleration

field is given by Eq. (3).

Now I introduce the corotating frame of reference, S′, with the Cartesian coordinates z′

rotation is equal to that of the cylinder at all times. The coordinates zk(≡ zk) and z′

k, whose angular velocity of

k(≡ z′k) are related by

z′

i= Aik(t) zk

(31)

where the time dependent matrix Aik(t) is given by

Aik(t) =

cos(θ − θo)

−sin(θ − θo) cos(θ − θo) 0

0

sin(θ − θo) 0

01

(32)

where θ is a function of time given by the integral of ω(t):

θ(t) =ωo

2ǫ[ǫ(t + T) + logcosh(ǫt + ǫT) + log2] (33)

and θ(0) = θo.

By construction, in the corotating frame S′the particles comprising the material are not rotating about the z′-axis;

there is zero azimuthal component of the velocity field at all times. As the angular velocity ω(t) increases from

t = −∞, each particle comprising the cylinder experiences an increasing effective centrifugal force that displaces the

particle to a larger radius. In this rotating frame S′, there will (in general) also be a Coriolis force. However, in S′,

for moderate angular speed ωo, the strain eik will be small, and the gradients of the displacement field will also be

small. Consequently, dropping the quadratic terms um;ium

Note that the transformation that relates cylindrical components in inertial frame S and rotating frame S′is given

by the identity matrix

;kin Eq. (20) will provide a good approximation to eik.

∂xi

∂x′ k=

∂xi

∂za

∂za

∂z′m

∂z′m

∂x′k= δi

k

(34)

Furthermore, the time dependent transformation between inertial cylindrical coordinates xi= (r,φ,z) in S and

corotating cylindrical coordinates x′ i= (r′,φ′,z′) in S′, is given by

r′= r

φ′= φ − (θ(t) − θo)

z′= z

(35)

where θ(t) is given by Eq. (33). The relation between cylindrical stress components ¯ σikin the inertial frame S and

cylindrical stress components ˜ σikin the corotating frame S′is

¯ σik(xn) =

∂xi

∂x′ a

∂xk

∂x′b˜ σab(x′ n) = ˜ σik(x′ n)(36)