[Foundations of Engineering Mechanics] Statics of Rods || Rods Interacting with Liquid or Air Flows

6. Rods Interacting with Liquid or Air Flows

This chapter is devoted to the problems of interaction between rods and external or internal flows of liquid or air.

6.1 Introduction

The general linear and nonlinear equilibrium equations valid for all types of space-curved rods were derived in Chapters 1, 2, and 3. The only restriction was that the stress-strain state caused by the external loads was governed by the generalized Hooke's law. As it was mentioned above, this book as well as its continuation (Svetlitsky (1987)) deal with problems that are linear in physical sense.

In most applied problems, external loads (q, p(j), IL, and T(v)) are known a priori. For example, dead forces (see Figs. 4.14 and 4.19a), follower forces (see Figs. 4.18 and 4.26), electromagnetic attracting forces (see Fig. 4.16), etc.

Forces may be either dependent on rod deformations (more precisely, dependent on the displacements Uj and the angles of rotation of the attached axes (}j as shown in Figs. 4.15 and 4.16) or independent of rod deformations (see Figs. 4.14 and 4.19). Forces that depend on Uj and {}j are termed feedback forces. For most feedback forces, the relationship between the forces and moments on one hand and the displacements Uj and the angles {}j on the other hand can be explicitly obtained. Hence, no difficulties arise in integration of the equilibrium equations and the equations of motion. The only possible difficulty is the choice of the most efficient numerical method. Serious difficulties indeed appear when we cannot determine the forces as functions of displacements of the axial points and angles of rotation. For example, this is the case in the problem of interaction between a drilling bit and a metal (see Fig. 0.20) and in the problem of interaction between a space-curved rod and a liquid or air flow (see Fig. 0.18).

Experimental data on interaction between rods and liquid or air flows, as a rule, can be obtained for a straight rod when the rod axis is perpendicular to the vector of the flow velocity. For other angles between the vectors Vo and el (the angles 'Pa in Fig. 6.1), there is no comprehensive experimental data. Needless to say that it is a hard problem to obtain these expressions for curvilinear rods for an arbitrary angle 'Pa. However, lacking the explicit

V. A. Svetlitsky, Statics of Rods© Springer-Verlag Berlin Heidelberg 2000

272 6. Rods Interacting with Liquid or Air Flows

Fig. 6.1.

relations for interaction forces, we cannot determine the stress-strain state of a curvilinear rod and we cannot study its static and dynamic stability. Thus, a practically important problem of determination of approximate formulas for aerohydrodynamic forces arises. Of course, for particular cases 'Pa = 0 and 'Pa = 7r /2, the approximate formulas must be identical to those obtained from the experimental investigations for straight rods. In what follows, aerohydrodynamic forces and moments are called aerodynamic loads.

The prime objective of this chapter is the derivation of the components of aerodynamic loads in the Cartesian and in the attached coordinate systems (Sects. 6.4 and 6.6). It is assumed that the rod is placed into a uniform flow. Rods of circular cross section are discussed in Sect. 6.4. For these rods, the center of gravity of each cross section coincides with its stiffness center. Rods of noncircular cross section (when the center of gravity and the stiffness center do not coincide) are considered in Sect. 6.6. Basic concepts and notions of aerohydrodynamics used in this chapter and experimentally obtained formulas for aerodynamic loads are collected at the beginning of Sects. 6.2 and 6.3.

A rod element of circular cross section in an air flow is shown in Fig. 6.l. The vector of the flow velocity is denoted by Vo. The element is subjected to a distributed aerodynamic force qa = qn + ql' A rod placed into a flow may take a shape that differs much from its original shape. This original shape corresponds to the equilibrium configuration of the rod when the medium is at rest. The aerodynamic loads depend on the shape of the rod axis. More precisely, the forces depend on the angle 'Pa between the tangent to the rod axis (the vector el) and the vector of the flow velocity Vo as shown in Fig. 6.1.

It is worth noting that the prime difficulty which appears in the interaction problems is that the aerodynamic forces depend on the shape of the rod and on the orientation of the axial line in the flow.

For large values of flow velocity, a rod of small bending stiffnesses may take an equilibrium configuration that differs dramatically from the natural configuration. Hence, nonlinear problems of statics of rods in a flow occur. Usually, it is assumed that the flow around the rod is steady and vortex-

6.2 Basic Concepts of Aerohydrodynamics 273

free. The latter is true only for a bounded range of the flow velocity values and for rods of comparatively smooth profile (e.g. for rods of circular cross section). For a rod of triangular or rectangular cross section, even a small rotation of the rod about its axial line may cause a considerable change of the aerodynamic forces applied to the rod, say, qn and ql as shown in Fig. 6.1.

It should be noted that for large values of the Reynolds number, the flow is unsteady, and, because of the Karman forces, the rod may break into oscillation with respect to its equilibrium configuration. Strictly speaking, this problem is not the problem of statics. On the other hand, the problem can be treated by methods of statics because the shape of the rod axis in this equilibrium configuration can be determined from the equilibrium equations of the rod in which the aerodynamic forces qn and ql are replaced by their averaged, constant values.

6.2 Basic Concepts of Aerohydrodynamics

6.2.1 Eulerian and Lagrangian Representations

A motion of a liquid or a gas can be either steady or unsteady. Consider a fixed point in space. If pressure, density, and the vector of the flow velocity at this point are time-independent parameters, then we can say that a steady motion of the medium takes place. If some of these parameters change in time, then the motion is termed unsteady.

A motion of a liquid or a gas can be described by use of the Eulerian variables or the Lagrangian ones. The Lagrangian representation of motion implies that we trace an individual particle of medium and determine the parameters (velocity, density, pressure, etc.) of this particle. Within the Eulerian representation, we observe the parameters of medium at a fixed point and pay no attention to the behavior of individual particles.

Xz Z'PftJ (Xi tt); t)

u(t)

UO

P (Xio;t,)

Fig. 6.2.


Lagrangian Variables Within the framework of the Lagrangian representation, we are to study the motion of individual particles of medium (see Fig. 6.2). This approach comprises the analysis of the following problems:

(1) The determination of the vector parameters (velocity v and acceleration VI) and the scalar parameter (density p) of a material particle as functions of time t.

(2) The derivation of the relationship between these parameters for different material particles.

First, we are to learn to distinguish different material particles. This can be done as follows: we ascribe to each material particle the coordinates of the point that this particle occupied at time to, e.g. the Cartesian coordinates XiO as shown in Fig. 6.2. The Cartesian coordinates of the material particle at any time tare

(6.1)

Similarly, the other characteristics of motion can be expressed through XiO and t. Thus, XiO and t are called Lagrangian variables.

The components of the velocity vector v and of the acceleration vector V, density and pressure are related to the Lagrangian variables as follows:

. _ aXi(XjO , t) . Xi - at '

" a2Xi(XjO, t) Xi = at2 ;

p = p (X jO , t) ;

p=p(XjO,t).

(6.2)

(6.3)

(6.4)

(6.5)

It should be noted that unlike the notion point accepted in mathematics, a material particle should be thought of as a nonzero volume of medium. This volume is small as compared to the total volume that the medium occupies. The volume of a material particle may change during motion of the medium while its mass remains the same. Hence, the density p of a particle may also change.

The Lagrangian representation is convenient when displacements of material particles are small, e.g. in the case of small oscillations of a medium.

Eulerian Variables Let us investigate the change of the parameters (velocity, pressure, etc.) of a medium at a fixed point. In the Lagrangian representation, an observer moves along with an individual material particle and measures its characteristics while in the Eulerian representation, an observer watches a fixed spatial place and measures characteristics of the medium at this point. Within the framework of the Eulerian approach, the following problems arise:

1 The upper dot signifies the differentiation with respect to time t.


(1) The determination of vector and scalar characteristics at a fixed point as functions of time t;

(2) The derivation of relationships between these characteristics of neighboring particles.

The Eulerian variables are the current coordinates Xi of material particles and time t (see Fig. 6.2). We have

V=V(Xi(t),t); P = P(Xi(t) , t); p = P (Xi(t) , t). (6.6)

The Eulerian approach enables us to determine the vector and scalar fields, i.e. the velocity v, the acceleration v, density p, and pressure p.

Using the Eulerian variables, we can express the total derivative of the velocity v with respect to time t (so-called material or substantial derivative) as follows:

dv av 3 av cit = at + L OX' Xi·

i=l '

(6.7)

The same rule of differentiation applies in the case of any other vector field (not necessarily v).

The physical meaning for the terms in the right-hand side of (6.7) is as

follows: ~~ is a partial derivative which indicates the rate of change of v at

a fixed point; ~v , i = 1,2,3, are partial derivatives which indicate the rate UXi

of change of v as the coordinates Xi vary while time t is held constant. In terms of the Lagrangian variables, we have

dv av dt at

because XjO do not depend on t. Information on the behavior of individual material particles is of little

interest in most applied problems. Consequently, the Eulerian approach has some advantages over the Lagrangian one.

A set of scalar functions of coordinates Xl, X2, and X3 may be treated as one vector-function. This vector-function assigns a vector to a point with the coordinates (Xl, X2, X3). Hence, we obtain a set of vectors that is called a vector field. To illustrate a vector field, one should draw the lines of the field. A line of a vector field (field line) is a curve that is tangent to this field (see Fig. 6.3). For any point A, there exists a unique field line passing through A. The field lines of a velocity field are called streamlines. If the velocity at a point A is equal to zero, then this point is termed a critical point of the velocity field. The field at such a point has a singularity. Thus, the name critical.

Streamlines provide insight into an instantaneous state of a liquid. It is especially advantageous in the problems concerned with bodies interacting


{~ X,

Fig. 6.3.

Xl

x, Fig. 6.4.

with flows (see Fig. 6.4). For a steady motion of a medium, the streamlines and the path lines coincide (see Fig. 6.3). A stream tube is the collection of streamlines that intersect a closed curve as illustrated in Fig. 6.5. Since the material particles move in the direction of the velocity field, they cannot cross the boundary of a stream tube. In the case of a steady motion, representing the flow as a set of stream tubes, we can understand many subtle points and reduce the analysis of the spatial flow to the analysis of a one-dimensional flow. Then, using the law of conservation of mass and the law of conservation of energy for each tube, we can examine the problem of interaction.

6.2.2 Basic Principles of Aerodynamics

Equation of Continuity Let us consider a steady motion of a fluid. According to the law of conservation of mass, the mass of liquid or gas carried through each cross section of a stream tube is the same for equal time intervals, that is,

pv(s) F(s) = const, (6.8)

where p is the density of liquid, v( s) is the magnitude of the average velocity in the tube, and F( s) is the cross-sectional area of the tube. For incompressible fiuids, we have p = const, hence,


Fig. 6.5.

vF = const. (6.9)

The relations (6.8) and (6.9) are valid if applied to the motion of ideal liquid in a pipeline of varying cross section F(s), i.e. the condition (6.8) is not local but integral one because it contains the finite area F.

Let us derive the equation of continuity in terms of the Eulerian variables. Consider a liquid element of mass 15m = pJv and volume Jv. During motion, the mass of the elementary volume remains unchanged, that is,

pJv = Po Jvo = const,

where Po and Jvo are the volume and the density at time to. Considering the total derivative of (6.10), we get

dp d (Jv) _ 0 dt + p Jvdt - .

The term

d (Jv) Jv dt

(6.10)

(6.11)

measures a rate of change of the volume at a fixed point. In some textbooks on hydrodynamics, this ratio is shown to be equal to the divergence of the vector field v at this point,

3

d (Jv) = divv = L 8vXi •

Jv dt 8x; ;=1

(6.12)

In terms of the Eulerian variables, we have

3 dp 8p "" 8p dt = 8t + ~ 8x. Xi ,

i=l '

(6.13)

in view of (6.13), we obtain the following equation of continuity:


(6.14)

For an incompressible liquid (p = const), the relation (6.14) yields

OVX1 + OVX2 + OVX3 = O. OXI OX2 OX3

(6.15)

Bernoulli's Equation An elementary volume of liquid moving with a velocity v is shown in Fig. 6.5. Since the liquid is ideal, there is no friction forces and thus the volume is subjected to the gravity force and difference of pressure multiplied by the area of the tube dF. We have

dv op p ds dF - = - -;::;- ds dF + pg ds dF

dt uS

or, by use of the Eulerian variables,

Op = - Os + pg, (6.16)

where

ds -=V' dt '

g = -gi2 ;

here el is a unit vector tangent to the axial line of the stream tube. Using local derivatives, we can rearrange (6.16) as follows:

[ 8v 8v 1 8p . - + w x v + - v + (a: xv) v = - - - a: x p - pg12 . ot os os

(6.17)

For steady motions, the tube keeps its spatial position and the vector and scalar parameters of the flow (velocity v, density p, and pressure p) do not depend on time t, i.e. ov/ot = 0 and w = O. Thus, from (6.17) it follows that

ov 2 op . P os vel + pv CB3e2 = - Os el - CB3pe2 - pg12 .

Multiplying both sides of (6.18) by el, we get

1 0 2 op OX2 2' p os (v ) + os + pg Os = 0,

where

ox·) OS- = (i2 . eIJ·

Integrating (6.19), we obtain Bernoulli's equation

(6.18)

(6.19)


pv2

2 + P + pgX2 = const

or

pv2 pv2 -t + PI + pgx2(0) = 2 + P + pgX2, (6.20)

where VI, PI, and X2 (0) are the velocity, the pressure, and the coordinate X2 at cross section 1 (see Fig. 6.5). From (6.20) it follows that in the case of a

2

steady motion of an incompressible liquid, the sum of the kinetic energy, y, and the potential energy, p+ pgX2, for a material volume remains unchanged.

For an unsteady motion, we have ~~ 1 o. Anyway, whenever the tube

occupies the same spatial position during motion (w = 0), we have

PV 2 is 8v - + P + pgX2 + -a ds = const . 2 0 t

(6.21)

This equation is valid for any tube provided the velocity is a function of time t alone (the velocity does not depend on the coordinate s). This is true for incompressible liquids when the cross-sectional area of the tube remains the same. In this case, using (6.21), we get

PV 2 av 2 + P + pgX2 + at = const . (6.22)

Equation (6.22) can be used for analysis of nonstationary flows in pipelines.

Xz Po A

Fig. 6.6.

Consider some examples which illustrate an application of Bernoulli's equation. A vessel supplied with a pipeline is shown in Fig. 6.6. The liquid runs out of the vessel through the pipeline. Our purpose is to determine the outflow velocity V and the pressure as a function of the coordinate X2(S). All the stream tubes take their origin at the free surface A. The initial velocity of the liquid in each tube is equal to zero, and the original pressure is


the atmospheric pressure Po. One such a tube is illustrated in Fig. 6.6. The outflow velocity of liquid particles is denoted by v (the cross section B). The pressure at the output cross section B is equal to Po.

For the cross section B, Bernoulli's equation gives

pv 2 2=pg(hA -hB ). (6.23)

This equation can be solved for v. Since the liquid is incompressible and the pipeline is of constant cross section, we see that the velocity v is the same for any cross sections of the pipeline. For an arbitrary cross section, Bernoulli's equation takes the form

pv2

Po + pghA = pgX2(S) + 2 + P (s). (6.24)

Combining (6.23) and (6.24), we arrive at

Note that in this example, the flow is steady provided the level of liquid in the vessel does not change (hA = const). If this condition is not satisfied, then the flow is unsteady and the functions v(t) and p(s, t) are to be determined from (6.22).

Fig. 6.7.

As the second example, let us consider a rod of noncircular cross section in a fluid flow (Fig. 6.7). There exists a streamline that ends at the point B. Since the point B is stationary, t.he velocity of the liquid particle that is carried to the point B must equal zero. Assuming that the flow tube ending at the point B is a horizontal one, we determine the increment of pressure at the point B as follows:

pV6 6.p = PB - Po = 2 ;

here Vo and Po are the velocity and the pressure of the flow. Let JF be a small part of the rod surface containing the point B. The

flow acts on the rod element with a force JX (so-called drag force) given by the relation

<5X = <5Fpv5 2

6.3 Experimental Results 281

Consequently, we can conclude that the aerodynamic force that acts on a rod is linear in density and quadratic in flow velocity. This fact is in excellent agreement with the experimental results.

In this example, Bernoulli's equation enables us to determine the pressure only at the stationary point B. To determine the pressure at the other points of the rod surface, one should use the general equations of aeroelasticity. Unfortunately, it seems to be impossible to derive the explicit formulas for aerodynamic forces. Consequently, experimental studies are of great importance.

6.3 Experimental Results

A cross section of a rod immersed into a plane homogeneous liquid or air flow is shown in Fig. 6.8. The cross section is subjected to a distributed pressure p. If the flow velocity equals zero, then the pressure is constant and the resultant force and the resultant moment acting on an element of unit length equal zero. For a nonzero velocity, the pressure is no longer constant, i.e. elements of equal length may experience different resultant loads. Hence, the cross section is subjected to a nonzero moment /11 and a force with projections qX 2 and qX 3 on the reference axes X 20X3.

_.

Fig. 6.8.

We assume that the aerodynamic forces QXl and Q X 2 and the moment /11 are applied to the center of gravity of the cross section (the point 0) as illustrated in Fig. 6.8. The aerodynamic forces Q:C2 and QX3 are termed a lifting force and a drag force, respectively. The distribution of pressure over the boundary is governed by the Reynolds number Re = vll v, where v is a


coefficient of kinematic viscosity of the liquid or air and l is a characteristic of the boundary size (e.g. the diameter of a circular cross section).

For a rod of an arbitrary cross section, the explicit formulas for the forces qXi and the moment f.J.I can hardly be obtained theoretically. However, using the theory of dimension and similarity, we are able to derive approximate formulas.

First, as the most simple case, let us consider incompressible ideal liquid. A motion of the liquid can be characterized by the following parameters: velocity v, density p, and characteristic length l. Assume that the force qX 3

is defined by the relation

(6.25)

where cnl is a non dimensional coefficient, a, b, and C are unknown constants, and Vn is the projection of the vector Vo on a plane that is perpendicular to the rod axis. The left-hand side of (6.25) has the dimension of distributed force, hence, the right-hand side must be of the same dimension:

(6.26)

here L, M, and T have the dimension of length, mass, and time, respectively. The relation (6.26) implies the following equations in the unknowns a, b,

and c:

0= a - 3b + c; b = 1; a = 2.

As a result, we have c = 1 and

1 .) QX 3 = "2 cn1Pv:;,l. (6.27)

Consider the projections of the rod cross sections on a plane that is orthogonal to the velocity vector v = -ViI (see Fig. 6.8). We can take the square of the least area of these projections as the characteristic length l. For example, for a rod of circular cross section, we have I = d, where d is the diameter of a cross section. For the rod shown in Fig. 6.8, we can put l = b. Finally, we arrive at

pv2 Q - c b n

X3 - nl 2' (6.28)

where cnl is the nondimensional coefficient of head resistance. Similar reasoning gives us the formulas for the distributed force QX2 and

the moment f.J.l'

pv.~b Q:C2 = CI -2- ; (6.29)

pv;, bl 1 f.J.! = em --2-' (6.30)

6.4 Aerodynamic Forces Acting on Rods of Circular Cross Section 283

y

Fig. 6.9.

where II is the characteristic size of the rod cross section. For symmetrical cross sections, the aerodynamic coefficients Cn !, Cl, and

Cm in the right-hand sides of (6.28)-(6.30) are functions of the Reynolds number and the angle of attack aa (see Figs. 6.8 and 6.9). These coefficients are to be determined experimentally.

To account for the fact that the moment depends on the sign of the angle aa, we write

(6.31)

In the case of a rod of circular cross section, the aerodynamic moment /-Ll is equal to zero. Moreover, if the Reynolds number varies within a certain interval, the aerodynamic coefficients Cn and Cl remain the same (Grafskii, Kazakevich (1983), Devnin (1975), and Kazakevich (1977)).

6.4 Aerodynamic Forces Acting on Rods of Circular Cross Section

Determine the projections of the aerodynamic forces when the flow velocity vector Vo is parallel to the plane X l OX3 as illustrated in Fig. 6.10. We have

Vo = Vo cos a . i l + Vo sin a . Is ,

where a is the angle between the vectors i l and e v (Svetlitsky (1982)). An element of a rod of circular cross section and the distributed aerody

namic forces acting on this element are shown in Fig. 6.10. Consider rods of circular cross sections (see Fig. 6.11a,b). Let the center of gravity of each cross section (the point 0) coincide with its stiffness center (the point Od. The case when these centers do not coincide is considered in Sect. 6.6.

Let 'Pa be the angle between the tangent vector el and the vector Vo. Assume that the magnitudes of the aerodynamic forces are as follows:

1 2 Iqnl = '2cnpvnd, Vn = Ivolsin'Pa; (6.32)

1 2 Iqll = '2ClPVl d, VI = Ivol cos 'Pa ; (6.33)


VIz .. .. .. ... ...

i,

Fig. 6.10.

(a)

"ua (c)

-~ ~

(a) (b)

.a-I .

(c) (d) Fig. 6.11.

x,


here en and CI are aerodynamic coefficients, v n is the normal component of the vector vo, and VI is the tangent component of Vo. The angle CPa satisfies the relation

(Vo . er) I I .

coscpa= Ivol =xlcosa+x3sma.

Determine now the projections of the drag force vector qn on the immovable coordinate axes. The following equalities must hold:

3

el = Lxjij; j=l

the unit vector e v is parallel to the flow velocity vector Vo.

(6.34)

(6.35)

From (6.35) it follows that el, ev , and qn are coplanar vectors. The reason for this is that the vectors qn and v n equal in direction.

For a rod of circular cross section, the aerodynamic force qn is orthogonal to the rod axis and directed along the vector V n , while the lifting force ql

vanishes. Equation (6.32) can be written as follows:

(6.36)

here qnxi are the vector components in the immovable frame. In components, the relations (6.34) and (6.35) read

dXI dX2 dX3 qnxl ds + qnx2 ds + qnx3 ds = 0 ; (6.37)

(dX3 dX I .) - cosa - -- sma ds ds

(6.38)

Equations (6.36)-(6.38) can be solved for the three components qnxi as follows:

(6.39)

(6.40)

(6.41 )

Let us obtain the components of qn in the attached coordinate system. We have

3

qn = Lqnx)j. j=l


Using the transformation matrix OJ), we can write

3 • ~[(l) Ij = ~ kj ek,

j==l

therefore,

or

3

ek = Lqnkek,

k==l

where qnk are the components of the vector qn in the attached basis,

Recall that

[ (1) , 11 = Xl ; [ (1)-X'·

13 - 3'

v=1,2,3.

j==l

(6.42)

Let us show that qnj = O. In the attached frame, the vector qn is orthogonal to the base vector el of the attached coordinate system. Substituting the expressions for qnx J into (6.43) and setting v = 1, we arrive at

. [ , , 2 qnl = qnOsm<pa cosax l - (Xl) COS<pa

- (x;)2 cos <Pa + x~ sin a - (x~)2 cos <Pa].

Since

cos ax~ + sin ax~ = cos <Pa ,

we get

. { [' )2 , 2 , 2 } _ qnl=qnOsm<pa cos<pa- (Xl + (X 2 ) + (X 3 ) ]COS<pa =0.

The formulas for the components of the drag force qn can be obtained without resort to the expressions for qnxj. In the attached axes, the relation (6.34) becomes an identity because the vector el is orthogonal to the vectors e2 and e3. In the Cartesian basis, the vector e v can be written as follows:

e v = cos ail + sinab;

hence, in the attached basis, we get


(1) (1).) ( (1) (1).) ev = (lll cosa + 113 sma e1 + 12l cosa + 123 sma e2

( (1) (1).) + 131 cosa + 133 sma e3·

Therefore, (6.35) and (6.36) give two equations in the unknowns qn2

and qn3:

(I (1) 1(1)·) (I (1) 1(1).) - O· qn2 31 cos a + 33 sm a - qn3 21 cos a + 23 sm a = , 2 2 2· 4

qn2 + qn3 = qnO sm t.pa·

On rearrangement, we finally obtain

. (I (1) I (1). ) qn2 = qnO sm t.pa 21 cos a + 23 sm a

. (1(1) 1(1).) qn3 = qnO sm t.pa 31 cos a + 33 sm a (6.43)

Thus we can use the expressions for the components of the aerodynamic forces either in the form (6.42) (referred to the Cartesian basis) or in the form (6.43) (referred to the attached basis).

Determine now the projections of the tangent force vector q1 on the Cartesian axes. We have

j=1,2,3. (6.44)

In the attached coordinate system, we have

(6.45)

where

In view of (6.44), we arrive at

2 dX1 q1xj = qlO cos t.pa ds sgn( cos t.pa) ;

2 dX2 q1X2 = qlO cos t.pa ds sgn(cos t.pa) ;

2 dX3 q1x3 = qlO cos t.pa ds sgn( cos t.pa) , (6.46)

where

C1 2 q10=2 PVod .

The factor sgn (cos t.pa) is introduced to account for the direction of the vector Q1, that is,


if sgn(cos'Pa) > 0, then ql = Iqllel;

if sgn(cos'Pa) <0, then ql=-Iqllel'

Thus, the total aerodynamic force acting on a rod element of unit length has the following components in the Cartesian coordinate system:

(6.4 7)

(6.48)

(6.49)

For further analysis, we introduce the nondimensional counterparts of the components of the aerodynamic forces (qnxi' qn2' qn3' qlxi' and qd:

[3

iinj = qnj A33 (0) ; [3

iiI = ql A33 (0) ; [3

iiaxi = qaxi A33 (0) .

The nondimensional aerodynamic force qa has the following components (here the tilde is dropped out):

_ (1), 2 . ( , ) . qaxl - qlO Xl cos 'Pa + qnO sm 'Pa cos a - Xl cos 'Pa ,

_ (1), 2 1 ,. 2 . qax2 - qlO x 2 cos 'Pa - "2 qnO x 2 sm 'Pa,

_ (1), 2 . ( . , ) . qax3 - qlO X3 cos 'Pa + qnO sm 'Pa sm a - X3 cos 'Pa ,

here

x,

Fig. 6.12.

(6.50)

(6.51 )

(6.52)

(6.53)

Consider an example. Let us determine the aerodynamic forces qlxj' qnxj' ql, qn2' and qn3 acting on a circular rod of circular cross section (see Fig. 6.12). The vector of flow velocity lies in the plane Xl OX3' The aerodynamic forces depend on the direction cosines xj of the vector el (see (6.39)-(6.41) and


(6.43)-(6.45)). In the case under consideration, the direction cosines depend on the shape of the rod axis. Let us assume that the displacements of the points of the axial line are small. Hence, we can determine xj using the method given in Appendix 5. The form of the axial line in the natural configuration is assumed to be known. In this example, we have

1 Ro =-,

1f (6.54)

where Xi and Ro are non dimensional quantities. Differentiating (6.54) with respect to 'T/, we get

(6.55)

1./J t--t--t---IH~~+--=~~4--l--+----1

• q,/t'ID r---'---r---r-;;:-;;--.-'-..-~--'-"----' I

a6t--t--h~~~~~--+--+--~~~~

(b)

Fig. 6.13.

Since the rod axis is a plane curve, we have

2' 2' Xl + x 2 = 1. (6.56)

Using (6.54) and eliminating X2 from (6.55), we arrive at

(Xl - Ro) x~ + V R~ - (Xl - Ro)2 x~ = O. (6.57)


Solving (6.57) for x~ and thus eliminating x~ from (6.56), we get

0.7 t--+~~'-+--+--I--+-~~=----t--1

J 'Inx/o ,----,----,------,----=-----,

0.0 r---t-----t----t-1.~~~_i 0.3 t---+---+--~JIC.~-+-~~-I

Ot--~~-~~~-h~-~~~~ 0.8

-a6~~~~~~--t---~--~ -0.9 L......::~....::::::..-=:--.....:~ __ ...l.-__ ..L-__ ...J

a (b) q '10' r-__ -=j0.=2 __ ..::0T-. t,.:...-._~a.~6-___:~O~.8~----1.,'l

flXJ

-O.B F---+----::;;i..--4.=---+---:-.;;.====I

(e)

Fig. 6.14.

(6.58)

Integration of (6.58) gives us the relationships Xl = Xl (7]) and x~ = x~ (7]). From (6.56) it follows that


The problem under consideration was solved for the following numerical values: Cn = 1, C1 = 0.1, l = 2.0 m, p = 1.0 kg· m-3 , d = 0.1 m, E = 2.1011 N· m- 2 , and A33(0) = 4.10-2 N· m2 .

For some values of the angle 0: and Ivol = 103 em· S-l, the plots of Q1xj

versus 7] are shown in Fig. 6.13a,b. For the same angles, the plots of qj versus 7] are shown in Fig. 6.14a-c. The relationships between the components of the aerodynamic force qj in the attached axes and the coordinate 7] are illustrated in Fig. 6.15 a-c. The plot of the absolute value of the total aerodynamic force acting on the rod versus the nondimensional coordinate 7] is shown in Fig. 6.16. The expressions obtained for the aerodynamic forces enable us to determine Q j, M j , {) j, and Uj from the equilibrium equations of the rod.

'1,w"r---....,.---.-=;::--,---"7T----, ~5~--_r--~--_w~~-~--~

1.01---+--J~-+-=-~~~+---~

o 0.4 (a.) 0.6 'I

-1.B ~---'-__ --' ___ -'-__ ---L_--..:3I~

fj . lOs rO---,:-=-----'=r----=r=-----=-r=----=L,

-0.4 """""~==+::::::::::.---+---+---=~:=:=~

(c)

Fig. 6.15.


o~--~~~~--~~--~~~ 0.2 0.4- 0.6 0.8 ?

Fig. 6.16.

6.5 Stress-Strain State of a Rod Interacting with an Air Flow

First, let us consider a special case when deformation of a rod caused by aerodynamic forces is small (see Fig. 6.12), that is, we assume that the shape of the rod under loading differs little from its natural shape. In this case, the equations of the zeroth approximation (1.116)-(1.119) may be used. We have

dQiO) (0) dry - <£30Q2 = -ql ,

(6.59)

6.5 Stress-Strain State of a Rod Interacting with an Air Flow 293

du~O) (0) _ . dry - CE30U2 - 0,

d (0) ~ + "" u(O) - .0(0) - O. d17 =30 1 'U3 - ,

(0) dU3 .0(0) - 0 + 'U2 - .

d17

-0.4 I----~~~tr____i

-0.6 '-----"-------' J (a)

Q;II~10'.--------_._------_....

o.gr-------4---~L-~

~61----~~---___i

O'Jr--7~~~--~~

'I H~mr~--------.-------,

Fig,6.17.

q;'!10S

0.6

0.3

0

-0.3

0.3

0.2

0,1

0

11s~m,S

0.1

0.05

(6.60)

'I

(b)

0.5 'I (of)

Equations (6.59) are to be integrated for the zero initial data. The plots of QjO) and M;o) versus 17 for some values of the angle 0: are shown in Fig. 6.17a-f Substituting a new variable 171 = 1 - 17 into (6.60), we integrate it for the zero initial data. The plots of'l9jO) and ujO) are illustrated in Fig. 6.18.


~:dm+rO ________ ~o.~5 ______ ~~ r-~------~--------~

Or-----~_+~------~

0 (a) 0.5 u/-10·

(b) 9-J .IO+

Uz {15°;

-O.Z 0.3 uzf*j'

uJ (f5';

-0.* 'I

-O.B (el

Fig. 6.18.

Consider now the most general case when the stress-strain state of a rod is governed by nonlinear equilibrium equations. A rigid hosepipe in a liquid or air flow is shown in Fig. 6.19. The aerodynamic forces acting on the hosepipe result in displacement of its axial points and rotation of the attached axes. Thus, the nonlinear equilibrium equations need be used.

Hosepipes designed for liquid transfer are widely used in engineering, e.g. various fueling systems (see Figs. 0.13 and 0.18), systems for lifting concretions from the sea bed (nodule pipes), systems of oil transfer (see Fig. 0.14). Traditionally, a hosepipe is considered as an absolutely flexible rod. Of course, this is just approximately so. The matter is that the bending and torsional stiffnesses of a real hosepipe are actually small but still different from zero. That is why the notion rigid hosepipe has appeared. Thus, from the mathematical point of view, there is no difference between a rod and a rigid hosepipe.

If the liquid is at rest, then the axial line of the hosepipe is a curve in the plane Xl OX2. The fluid flow results in the distributed aerodynamic forces acting on the hosepipe. These forces depend on the magnitude of the vector of flow velocity as well as on its direction. Assume that the flow is steady and vortex-free. This assumption is introduced for simplicity but it allows us to determine the stress-strain state of the hosepipe for large deflection of the rod axis.

The axial line of the loaded hosepipe is shown in Fig. 6.19 as a dot-anddash line.


Q+dQ

Fig. 6.19.

The nonlinear equations governing the stress-strain state of the hosepipe are as follows (see Sect. 1.3):

dQ "dr] + ce x Q + ql + qn + g = 0 ;

dM dry + ce x M + el x Q = 0 ;

d'!9 +L-1L ce(l) -L-1A-1M=O' dry 1 2 0 1 ,

du ((1)) (1) (1) _ . dry + ce x u + III - 1 el + 121 e2 + 131 e3 - 0,

M = A (ce - ce~l))

or

(6.61)

here ql and qn are the distributed aerodynamic forces and

3

g = - L 9 (12 . ei) ei i=1

is the distributed gravity force acting on the hosepipe.


As already proved, the components of the vectors ql and qn in the attached basis satisfy the relations (6.42), (6.43), and (6.45). The variables [Xl in (6.61) are the elements of the matrix L(I) (see (A.57)). Equations (6.61) are solved numerically by the method of step-by-step loading (see Sect. 2.3). Whatever numerical method is used, we must first define the initial approximation to the sought variables.

As an initial approximation of the state vector,

ZeD) = (Q(O) M(O) '19(0) u(O»)T , , , ,

let us take the one that corresponds to an absolutely flexible rod subjected only to the gravity force (vo = 0). As a result, we obtain the zeroth ap-

proximations QiO) and xjO)' (recall that £;0)', j = 1, 2 , 3, appear in the expressions for the aerodynamic forces).

o -1.104

5 10 15 20 X,

(a)

,/ \9; --V ""'"

.-:;

--

10

5

o -5

4.5 0

(b)

./ -03.---~

o

-1.104

o 0 5 10 15 20 X,

4.5/<£1J3 I .. X3

(c)

~ M2 -.-

I --

~ M1 ---'...J

1/ \

o 0.2 0.4 0.6 0.8 YJ 0 0.2 0.4 0.6 0.8

(d) (e)

Fig. 6.20.

Now we can solve the nonlinear equations by the method of iterations. If

xjO)', j = 1, 2, 3, are found, then, specifying the components of the vector

of flow velocity V6j ) , j = 1, 2, ... , n, we can evaluate the corresponding

aerodynamic forces qiO) and q~O). Thus, having found V6 j ) , we obtain the first approximation of the state vector Z{l) as follows:


Z(l)1 + f l (Q(1), M(l), '19(1), u(l») = f 2(qiO), q~O), g(O»). (6.62)

Once 19(1) and u(1) are found we determine X(l) q(l) q(l) and g(l) and J J ' J'l,n, ,

therefore, we obtain the second approximation

Z(2)1 + f l (Q(2), M(2), '19(2), u(2») = f 2(qil ) , q~l), g(1»),

and so on.

1 0 ~-I--+-I-

5 1---f--f-"7'+--It---l

Ol'k::::--+~+-:~'---;----f

- 5 '----'-_...L.---'-_-'----'

o 5 10 15 20 XI

(a)

N/ - - v< / I

~ O2

X J

y t'-\

O2 \

~ o

_1.104

o 0.2 0.4 0.6 0.8 11

(d)

Fig. 6.21.

X2

10

5

0

- 5 4.5 0

(b)

Mj

0 _5.103

-1.104

-1.5.104

-2.104

o 0 5 10 15 20 XI

45~tjjLI j XJ

(c)

~ ./

/ \ 1\ '\ I M1 IL-.: ~3 \

lM2 \

o 0.2 0.4 0.6 0.8 11 (e)

This example illustrates application of the explicit relations for the aerodynamic forces obtained in Sect. 6.4 for the case of arbitrary angles between the vectors Vo and el. Nowadays there exist software packages for solution of two-point boundary-value problems. These packages may be useful in engineering design provided the explicit expressions for the aerodynamic forces are given.

The above problem was solved for the following data: l = 30 m, Xlk = 21 m, X2k = 12 m, A1l = 7.36· 104 N . m2 , and A22 = A33 = 2.2· 105 N . m2 . The flow velocity Vo was taken 30m·s- l (air flow) and 3m·s-1 (liquid flow). The solution was evaluated for the angles a = 45° and a = 90°.

The numerical results are illustrated in Figs. 6.20 and 6.21. The shape of the rod axis for Vo = 0 is shown in Figs. 6.20a and 6.21a as a dash-and-dot


line. The projections of the axial line on the coordinate planes for a = 45° and Vo = 30 m . S-1 (air flow) are illustrated in Fig. 6.20a-c as a solid line and for Vo = 3 m . S-1 (liquid flow) as a dashed line.

It is seen from the plots that the hosepipe in the air flow does not deflect much out of the vertical plane (the projections of the maximum deflection does not exceed 5% of the hosepipe length). In the case of the liquid flow (for Vo = 3m· S-1), this deflection makes about 15%. The plots for the case of a = 90° are shown in Fig. 6.21a-c. In this case, the out-of-plane deflections are greater than those for a = 45°. For the hosepipe in the liquid flow, the projections of the maximum deflections make 22.5% of the hosepipe length.

The plots of Qj and M j versus 71 for a = 45° and a = 90° (vo = 30 m 'S-1)

are shown in Figs. 6.20d,e and 6.21d,e. For the hosepipe in the liquid flow, the relationship between the axial force Q1 and 7] are shown as dashed lines.

6.6 Aerodynamic Forces Acting on Rods of Noncircular Cross Section

An element of a rod of noncircular cross section is shown in Fig. 6.22. The center of gravity of the cross section (the point 0) coincides with its stiffness center (the point Od.

Fig. 6.22.

Vo ------------

In general, these two points are different as indicated in Fig. 6.23. A rod of circular cross section (Fig. 6.1) is subjected to the drag force

qnl and the axial force q1. A rod of noncircular cross section (Fig. 6.22) is

6.6 Aerodynamic Forces Acting on Rods of Noncircular Cross Section 299

e(1) X2 2

Vn -

Fig. 6.23.

additionally subjected to the lifting force q), which is orthogonal to the vector qn" and the aerodynamic moment /Ll (see Sect. 6.3). According to the sign convention, the cross section in Fig. 6.23 is rotated counterclockwise by an angle aa > O. Hence, the lifting force is directed downwards.

-----vn

Fig. 6.24.

The aerodynamic forces acting on a rod are the drag force qn, and the lifting force q). The vectors of these forces lie in the plane perpendicular to the rod axis. These forses depend on the normal component of velocity v n (see Figs. 6.23 and 6.24). The vectors qn, and Vn are codirectional vectors. The vector q) is perpendicular to the vector qn, . Since the direction of the force q) depends on the sign of the angle of attack aa, we write q) = IqJie) sgnaa; here e) is a unit vector collinear to the vector q).

Experimental investigations show that the aerodynamic coefficients Cn ,

and c) depend on the angle of attack aa. The nondimensional absolute values of the aerodynamic forces are as follows:

in view of the formula Vn = Va sin i.pa, we have

_I 1- (1)· 2 . gn, - qn, - Cn, gn, Sin i.pa,

1 1 (1) . 2

q) = q) = c)q) 8m i.pa,

where

q (l) = q(l) n, )

1 pV6bl3 2' A33(O) .

(6.63)


When determining the forces qnl and ql experimentally, we assume that these forces are applied to the center of gravity (the point 0) of a cross section.

6.6.1 Components of qnl and ql in the Cartesian Coordinate System

The drag force qnl satisfies the following relations:

qnl . el = 0;

qnl . (e v x ed = 0 ;

pV5bl3 sin2 <Pa

Iqnll = Cnl (O:a) 2A33(0) . (6.64)

These relations are identical to (6.34) and (6.35). Recall that we used (6.34) and (6.35) to derive the expressions for the components of qn in the Cartesian basis. In view of (6.30)-(6.41), we obtain the components of qnl in this basis as follows:

qnlXI = Cnlq~ll) sin <Pa(coso: - x~ cos<Pa);

_ (I)' , . qnlx2 - -Cnl qnl sm <PaXl cos <Pa ,

qn lX3 = Cnlq~~) sin <pa(sin 0: - x~ cos<pa). (6.65)

These relations are identical to those for qn, except that the aerodynamic coefficient Cn for rods of circular cross section does not depend on the angle of attack O:a, while the coefficient cnl depends on this angle. The lifting force ql which acts on a rod of non circular cross section is orthogonal to the drag force qnl' Hence, the vector ql must satisfy the following relations:

or

ql' el = 0;

ql·en=O;

qlXl x~ + qlx2X~ + qlx3x~ = 0;

qlXI e XI + qlx2ex2 + qlx3ex3 = 0;

qrxi + ql2x2 + ql2x3 = (clql sin2 <Pa)2 .

(6.66)

(6.67)

(6.68)

(6.69)

The unit vector en in (6.67) and the vector Vn are collinear. Let us now obtain the components ex; of the vector en in the Cartesian frame. We have


In a more expanded notation, we have the following equations in the three unknowns ex;:

(6.70)

Equations (6.70) are similar to (6.34)-(6.36), thus, on rearrangement, we get

eX1 = (cos a - x~ cos <Pa) _._1_ ; sm<Pa

, 1 eX2 = -x2 cos <Pa -.-- ;

sm<pa

( . , ) 1 eX3 = sm a - x3 cos <Pa -.--. sm<pa

Solving (6.69) for qlx; and multiplying the result by sgnaa, we arrive at

qlx1 = -q?) Cl ~1 sgn aa ;

qlx2 = -ql(l) Cl ~2 sgn aa ;

_ (1) A qlX3 - -ql Cl LJ.3 sgn aa ,

where

Substituting ~j and ex; into (6.71), we get

(1). . , Q1X1 = ql Cl sm <Pa sm aX2 sgn aa ;

(1) . ( ., , ) Qlx2 = ql C! sm <Pa - sm aX1 + cos aX3 sgn aa ;

_ (1). , Qlx3 - -ql Cl sm <Pa cos aX2 sgn aa .

(6.71)

(6.72)

The coefficient cl(aa) in (6.72) depends on the angle of attack and on the shape of a rod cross section. As noted above, the relationships between these coefficients and the angle of attack aa can be obtained experimentally. The experimental data for the aerodynamic coefficients en1 , el, and em versus aa

are given in Kazakevich (1977). Knowing these relationships, we are able to solve the equilibrium equations numerically. If a rod deforms under the action of aerodynamic forces, then the angle of attack reads O'a = O'aO + aa1, where O'aO is the initial (given a priori) angle of attack and O'a1 is an additional angle caused by deformation of the rod. The additional angle can be determined from the equilibrium equations. For small displacements of the axial points and small angles of rotation of the attached axes, the expression for the additional angle will be obtained later (see (6.89)).


The vectors qn, and ql must satisfy the orthogonality condition, that is, qn, . ql = 0 or, in the expanded notation,

From (6.65) and (6.72) we see that the orthogonality condition for the vectors qn, and ql holds.

To perform a numerical analysis of the equilibrium equations for rods of noncircular cross section, one must know the relationship between the aerodynamic coefficients cn, and CI and the angle of attack aa (see (6.65) and (6.72)). The sign of the angle aa must also be known.

Let us obtain the explicit expressions for the angle aa. For any location of the basis ej relative to the vector Vo, the angle aa is the angle between the normal component of the velocity Vn and the base vector e3. Then,

where

Vo = vo(cosai1 + sinah).

Since

we obtain

where

The scalar products take the form

(Vo· e2) = vo(cosaZii) + sinaZii));

(vo· e3) = vo(cosaZii) + sin aiR)) .

Therefore,

1 (1) 1(1) . 21 cos a + 23 sm a .

sin aa = i===========;:===============::O,

( I (1) cos a + 1 (1) sin a) 2 + (I (1) cos a + z (1) sin a) 2 21 23 31 23

Z(l) cosa + 1(1) sina cosaa = /=======3=1===;:====3=3============

( z (1) cos a + z (1) sin a) 2 + (Z (1) cos a + z (1) sin a) 2 21 23 31 23

Using the properties of the matrix L(1) (see (A.49)), we have

(6.73)

(6.74)


Z (1)2 + Z (1)2 + Z (1)2 - 1 . 11 21 31 - , 222

Z(1) +Z(1) +Z(1) -1' 13 23 33 - ,

Z(1)Z(1) +Z(1)Z(1) +Z(1)Z(1)-O 11 13 21 23 31 33 - .

The expression under the radical sign in (6.74) can be rearranged as follows:

? 2 ( (1) . Z(I))- ( Z(I) . Z(I)) _ 2 _. 2 cos aZ21 + sm a 23 + cos a 31 + sm a 33 - I-cos 'Pa - sm 'Pa·

Finally, we get

Z(1) cosa + Z(1) sina sinaa = 21 . 23

Z (1) cos a + Z (1) sin a cosaa = 31 . 33

sm 'Pa sm'Pa

6.6.2 Components of qnl and ql in the Attached Coordinate System

Recall that

3

qn, = L qn,xjij, j=l

3

q, = L q'Xj sgn (aa)ij ; j=1

after the transformation from the basis {i j } to the basis {ej}, we get

qn, = t (t qnlxJk~)) eK; K=1 j=1

q, ~ 1;, (t. q", ,gn a.l J:,') eK

From (6.75) and (6.76) it follows that

3

q -"'q Z(I). n 1 K - ~ nl~Cj K j ,

j=1

(6.75)

(6.76)

(6.77)

It can be shown that qnl and ql1 are equal to zero. The nonzero aerodynamic forces qn, and q, in the attached frame are as follows:

- (1)' (Z (1) Z (1). ) . qn12 - -Cn, qn! sm 'Pa 21 cos a + 23 sm a ,

(1)' (z(l) Z(I).) qn!3 = Cn! qn, sm 'Pa 31 cos a + 33 sm a ; (6.78) (1). (1) Z(I).)

q'2 = -c,q, sm 'Pa (l31 cos a + 33 sm a sgn aa ; _ (1). (1) Z (1). ) q'3 - -c,q, sm 'Pa (l21 cos a + 23 sm a sgn aa . (6.79)


Let us consider a rod of noncircular symmetrical cross section with one axis of symmetry. In the attached frame, we have

ql = qlO cos2 'Pa sgn (cos 'Pa) ;

q2 = qnj 2 + q]2 ;

q3 = qn j 3 + q]3 ;

. 2 f.J,1 = f.J,1O sm 'Pa sgn aa , (6.80)

Recall that the forces are assumed to be applied to the center of gravity of the cross section.

Up to now, the aerodynamic force have been considered as the sum

where qnj is the drag force and q] is the lifting force. For a numerical analysis, it may be convenient to deal with the projections of the aerodynamic force on the principal axes of a cross section (it is especially so for axisymmetrical cross sections) (see Fig. 6.24). The aerodynamic coefficients C2 and C3 in the expression for the forces q2 and q3 are related to the coefficients enj and c] as follows:

C2 = c] cosaa + cnj sinaa ;

C3 = -c] sin aa + cnj cos aa . (6.81)

Let us obtain the expressions for the aerodynamic forces and moments. The forces are assumed to be applied to the stiffness center 0 1 . As noted above, in experimental studies the aerodynamic forces and the moment are assumed to be reduced to the center of gravity (the point 0 in Fig. 6.23). The principal axes were used in the derivation of the equilibrium equations for rods of solid cross section, i.e. the forces were assumed to be applied to the center of gravity O. This point is also the stiffness center of the cross section (see Fig. 6.8).

A force that is directed along one of the principal axes results in a displacement of the point 0 in this direction. Likewise, a moment whose vector is directed along a principal axis of a cross section results in rotation of the cross section about this axis. For example, a force P that is collinear with the vector e2 as indicated in Fig. 6.24 results in a displacement of the center of gravity such that the displacement vector lies in the coordinate plane (e2, e3). Moreover, under the action of P, the attached axes rotate relative to their original position. Now let a force P be directed along the base vector e~ 1) referred to the center of stiffness 0 1 . In this case, the only effect pro

duced by this force is a displacement along the vector e~I). Suppose that a moment J.Ll is perpendicular to the plane of drawing and its vector passes through the point O. As a result, the cross section rotates remaining in the

6.7 Small Displacements of Axial Points 305

plane of drawing while the point 0 is at rest. Recall the relationship between the vector M and the curvatures (the vector ce) obtained on the assumption that each component of the vector M acts independently from the other components:

M j = Ajj (rej - rejo) ;

here Ajj are non dimensional stiffness parameters related to the axes passing through the center of stiffness.

Now let the load be applied to the center of gravity; we have

(1) (1) qn, = qn,2e 2 + qn,3e3 ;

_ (1) (1) . q] - q]2e2 - q]3e3 ,

_ _ (l (1) (1) + 1(1) (1) + l (1) (1») g - go 12 e 1 22 e 2 32 e 3 , (6.82)

where

is the magnitude of the nondimensional gravity force. The twisting moment, which performs rotation about the axis e~I), (the vector ei1) is perpendicular to the plane of drawing, see Fig. 6.24) can be written as follows:

(6.83)

here

. 2 (1) J-Ll = PlO sm 'Pa sgn O'ael ;

J-L - aq e(l) . lqn, - - n,2 1 ,

J-Llql = -aq]2 sgn O'ae il) ; _ l (1) (1) (J) (1)

J-Lg - -ago 22 e 1 + agol 12 e2 .

The explicit formulas for aerodynamic forces and moments derived in this section are valid for arbitrary deflections of the rod axis provided the stress-strain state of the rod is governed by the generalized Hooke's law.

6.7 Increments of Aerodynamic Forces at Small Displacements of Axial Points

In the preceding sections, the expressions for aerodynamic forces and moments valid for any displacements of axial points were obtained. The aerodynamic forces depend on the direction cosines of the vector el. We have


where xjo are the direction cosines, which are characteristics of the shape of the axial line in the natural configuration, and U Xj are finite changes of direction cosines caused by the aerodynamic forces.

Consider small displacements of the axial points. Let us establish the relationship between increments of the projections of the aerodynamic forces and small displacements of the axial points. Assuming that these increments are also small, we get

q1Xj = q1xjO + D.q1Xj ;

qnxj = qnxjO + D.qnxj ;

qlxj = qlxjO + D.qlxj ,

where D.q1xj' D.qnxj' and D.qlxj are small quantities.

(6.84)

For a rod of circular cross section, we have C1 = const, Cn = const, Cl = 0, and Cm = 0; consequently, the relations (6.39)~(6.41) and (6.46) give

D.q1x, = ~ (8q1Xj 8cosi.pa + 8qlx j )/ u'. L 8cosi.p 8x' 8x' XK' K=l a K K X~(=XKO

, U:rJ\

Let the vector Vo be parallel to the plane Xl OX3. We have

D.q1XI = qi~) COSi.paO(2X lO cosa + COSi.paO) . U~, (1)2 ."

+ q10 cos i.pao sm aX10ux3 ;

" 2 (1) ,,(1) 2 , w.q1x2 = q10 cos i.paO cos ax20 u x + q10 cos i.paO Ux2

2 (1) ." + q10 cos i.paO sm aX20 u X3 ;

" 2 (1) I , W.qlx3 = qlO cos i.paO cos aX30 u XI

(1) ( 2" ) , + qlO cos i.paO cos i.paO + sm ax30 U X3 ;

D. _ (x~o cos a cos 2i.pao - cos i.paO cos2 a sin 2i.pao) , qnxI - qnO . - 2 U XI sm i.pao

. ( x~o cos 2i.pao - cos i.pao cos a) , + qnO sm a . U X3 ; sm i.paO .

, cos 2i.pao ,. , D.qnx2 = qnO x 20 cos a. UXI - qnO sm i.pao cos i.paoux2

sm i.pao

, . cos 2i.pao , + qnOx20 sm a. U x ; sm i.paO 3

" cos a ( , . , W.qnx3 = qnO -.-- X30 cos 2i.pao - sm i.paO cos i.paO) U x sm i.paO 1

[ sin a(x~o cos 2i.pao - sin a cos i.paO) sin 2i.pao] , +qno . - 2 U X3 ' sm i.pao

(6.85)

(6.86)


where

COS <{JaO = x~o cos a + x;o sin a .

Equations (6.85) and (6.86) can be represented in vector form as follows:

ilq1 =A(1)u'· x x' ilq = A (2)U' .

nx x' (6.87)

here the elements of the matrices A (1) and A (2) can be easily determined from (6.85) and (6.86).

6.7.1 Rods of Noncircular Cross Section

In the case of rods of noncircular symmetrical cross section, the coefficients cn , ) c" and Cm depend on the angle of attack aa. If the angle of attack varies little, we get

Let us obtain the relationship between ilaa and small angles of rotation of . {}' . { } (1) (1) (1) (1) the basIs ej relatIve to the basIs ejO . The elements l21 , l23 , l31 , and l33

of the matrix L(1) in (6.79) are defined by (A.57). Recall that L(1) = LL(O); here L(O) is the matrix of transformation from the basis {i j } to the basis {e jO}

(see (A.55)), L is the matrix of transformation from the basis {ejo} to the basis {ej}, and the basis {ejo} is associated with the natural configuration

of the rod. The elements li~O) of the matrix L(O) are assumed to be known. For small angles of rotation of the attached axes, the matrix L is as follows (see (A.46)):

Hence, we have

(1) _ (0) (0) (O){) . l21 - 121 + 131 {)1 - 111 3,

1(1) -1(0) + I (O){) _1(0){) . 23 - 23 33 1 13 3,

1 (1) - 1(0) _ l (O){) + I (O){) .. 31 - 31 21 1 11 2,

1 (1) - 1(0) _ l (O){) + I (O){). 33 - 33 23 1 12 2· (6.88)

Expanding sinaa defined by (6.79) into a series and using (6.88), we get (only linear terms are retained)


where

a1 aO(a1 - bd aOb2 aOa2 . / . C1 D D2 C2 = - D2; C3 = D' D = V a6 + b6 ;

(0) l (0) . b l (0) l (0) . ao = l21 cos a + 23 sm a ; 0 = 31 cos a + 33 sm a ;

a1 = li~) cos a + li~) sin a;

a2 = l1(~) cos a + l1(~) sin a;

(0) (0) . b1 = l21 cosa + l23 sma;

(0) (0) . b2 = III cosa + l12 sma.

Since sin aa = sin aaO + cos aaO .6.aa, we get

.6.aa = C1 '!9 1 + c2'!92 + c3'!93 cos aaO

bo cosaao = D' (6.89)

The increments of the projections of the aerodynamic forces qnjxj and q]Xj are related to the increments of the angle of attack .6.aa as follows:

On rearrangement, we have


A _ 0 8c, Cl 191 + C2 192 + C3 193. . , uq'Xl - q, -8 sm <PaO sm ax20

aa cos <PaO o o· . , + q, CIO sm <PaO sm aUX2

00· (' 2· ')' - q, CIO sm a XlO cos a + sm a cos aX30 UX1 00· ,(,.2. ')' - q, CIO sm ax20 x30 sm a + sm a cos ax lO UX3 ;

A _ 0 8c, Cl 191 + C2 192 + C3 193 uq'X2 - ql -8 sin<pao(sinax~o - cosax~o)

aa cos <PaO

+ qpcPo[ (- sin aX~OAl + cos aX~OAl + sin a sin <PaO) U~,

+ (- sin aX~OA2 + cos aX~OA2 - cos a sin <p~o) U~3 J ;

A _ 0 8c, cl 191 + C2 192 + C3 193 . , uq'X3 - q, -8 sm <PaO cos ax20

aa cos <PaO

o 0(' 'A' 'A' , ) + q, CIO sm <PaO cos aUX2 - 1 x 20 cos aUX1 - 2x 20 cos aUX3 '

where

_ 1 d(l) 2. qn,OO - '2 p Vo , cPo = CIO sgn aaO ;

(6.90)

(6.91)

A ,. 2· , 2 = x 30 sm a + sm a cos axlO .

In vector notation, we have

(6.92)

The increment of the moment 6.J.Ll can be expressed as follows:

(6.93)

here

For small Uj and 19 j, the stress-strain state of a rod interacting with an air or liquid flow may be determined by use of the expressions for increments of the aerodynamic forces (6.90) and (6.92) and for an increment of the moment (6.93).


6.8 Rods Containing Internal Liquid Flows

There exists a great number of publications devoted to the problem of statics of rods containing an internal liquid flow. The extensive list of reference on the subject is available in the monograph Svetlitsky (1982).

In most publications, only rods of circular cross section (pipelines) are treated (see Fig. 6.25a). However, real constructions include hollow rods of noncircular cross section (see Fig. 6.25b-d) when the cross-sectional bending stiffnesses are different. For example, a tool for deep-hole drilling has a cross section shown in Fig. 6.25c. Such a shape of the cross section improves chip removal and provides efficient cooling of the tool during drilling.

(a) (b)

(c) (d) Fig. 6.25.

A liquid flow may result in additional forces applied to the rod. In the case of straight rods, these forces may lead to loss of static stability.

Fig. 6.26.

6.8 Rods Containing Internal Liquid Flows 311

An element of a rod and an elementary volume of the liquid enclosed into this element as well as the forces applied to the rod and the liquid are shown in Fig. 6.26a,b. The elementary volume of the liquid moves with velocity Wo = WOel, where Wo is the velocity of liquid particles averaged over the cross section. According to the d'Alembert principle, we have

O(POel) dwo - - f + m2g - m2 -- = 0 as dt'

Po = PoF, (6.94)

where F is the cross-sectional area of the rod element, Po is the liquid pressure, m2 is the mass of liquid enclosed into a rod element of unit length, and f is the distributed force of interaction between the liquid and the walls of the rod. In the case of ideal liquid, the vector f lies in the plane defined by the vectors e2 and e3, i.e. f is orthogonal to the vector el. (This fact is valid both for statics and dynamics.) The flow velocity w is a function of t and s. Thus, using the Eulerian variables, we get the total derivative

dwo oWo oWo cit = at + as Wo . (6.95)

If the rod is in equilibrium and the absolute value of the velocity Wo remains unchanged (the motion of the liquid is steady), then ow%t = O. For F = const, from (6.109) it follows that

dwo 2 oel) 2 2 cit = Wo as = w o re30 e 2 - wO re20 e 3 . (6.96)

Taking into consideration the force of gravity acting both on the rod and the liquid, we can represent the equilibrium equation as follows:

(6.97)

Eliminating the vector f from (6.94) and (6.97), we arrive at

(6.98)

or

where

Summing the moments acting on the element, we get


dM P ds +(el XQ)+I-L+ LT(v)J(s-sv)=O;

v=l

(6.100)

this equation is identical in form to (1.9). Since el x Q = el x Q(1), we can rewrite (6.100) as follows:

dM P ds + (el x Q(1)) + I-L + LT(v) J (s - sv) = 0;

v=l

(6.101)

here I-L is a distributed moment applied to the rod, say, the distributed aerodynamic moment.

Fig. 6.27.

Concentrated forces p(i) applied to a rod may be divided into two groups: forces produced by the liquid flow and other external forces. This is a key feature of problems of statics and dynamics of a rod that contains a liquid flow. One such a rod is shown in Fig. 6.27.

Fig. 6.28.

At the points where two tubes are joined together such that one tube makes an angle 2f3 with the other, concentrated forces p(2) and p(:l) appear (see Fig. 6.28). Using the principle of balance of moment of momentum for fluids, we get the following expression for the absolute value of the concentrated force (see Fig. 6.28):


IP(C)I = 2Po cos (3 + 2m2w5 cos (3 = 2 (Po + m2w5) sin ~;

here Po is the pressure in the liquid. The direction of the force p(c) is shown in Fig. 6.28. The displacement vector u and the vectors ae and iJ satisfy the relations

(1.61) and (1.60), respectively. To obtain nondimensional equilibrium equations, we should put (see Sect. 1.1)

s 'T/ = l ; Wo = wo[Po ; Q = QA33(0) .

[2 ' ae = ~[;

M = MA33(0) . [ ,

i A33(0) 'Y = [3 ;

- (i) p(i) = P A33(0).

[2 '

qA33 (0) q = [3

iJ,A33 (0) f.L = [2 ;

A . -~ . R _ .POA33 (0) . " - A33 (0) , 0 - [2 '

- (v) T(v) = T A 33 (O).

[ , Po =

On rearrangement, the following nondimensional equilibrium equations for a rod of constant cross section interacting with an internal flow can be obtained (as before, a tilde is dropped out in the notation of nondimensional quantities) :

here

(6.102)

(6.103)

(6.104)

(6.105)

(6.106)

m2 nl=--

ml +m2 (6.107)

Equations (6.102)~(6.106) differ from (1.57)~(1.61) in that (6.102)~(6.107) contain the vector Q(1) and the concentrated forces p(i). The first component of the vector Q(1) is a function of the flow velocity wo and the pressure Po (see (6.107)). Owing to similarity of these systems of equations, we see that the problem under consideration can be solved by the numerical methods


discussed in Chap. 2; however, the boundary conditions must be satisfied by the component QP) = Q1 - (Po + n1 w6) instead of the component Q1.

In the attached coordinate system, the equilibrium equations take the form

dQ(1) n -d- + <E x Q(l) + , + q + L p(i) <5 (T) - T)i) = 0 ;

T) ~1 (6.108)

d~ P d + <E X ~ + e1 X Q(l) + JL + LT(v) <5 (T) - T)v) = 0;

T) v=l

(6.109)

d'!9 (1) L - + L<E - <E = 0 . 1 dT) 0 , (6.110)

du dT) + <E X U + (lll - 1) e1 + l21 e2 + l31 e3 = 0 ; (6.111)

~ = A (<E - <E61)) . (6.112)

The expanded form of (6.108) is as follows:

dQ n ci+<EXQ-<EX [(Po+n1 w6) erj+,+q+ L p(i) <5 (T)-T)i) = 0; (6.113)

T) i=l

here

<E X [(Po + n1w6) ell = <E3 (Po + n1w6) e2 - <E2 (Po + n1w6) e3·

In view of the boundary conditions, from (6.108)-(6.112) it follows that

Q Q(1) P, 2 = 1 + 0 + n1 wo· (6.114)

Let us consider some examples to illustrate how an internal liquid flow affects the stress-strain state and the static stability of a rod.

w, p ~ <.~.-.>:> p ~X2 W 1 • ~-----=' . ~ s. ~ ====---_____ ,:;::::V X

1

X3

Fig. 6.29.

Consider a compressed-twisted straight rod whose axial line is not perfectly straight due to imperfections (Fig. 6.29). When the liquid is at rest, the shape of the axial line of the rod is characterized by the small curvatures


~re20 and ~re30. Taking into account the initial twist relO, we can represent the curvature vector as follows:

re = (relO + ~red el + (~re2 + ~re20) e2 + (~re3 + ~re30) e3 . (6.115)

Assuming that ~rej and ~rejO are small and using (6.108)-(6.112), we obtain the linear equilibrium equations

MlO = T, MlO

relO = --; All

(6.116)

these equations are valid in the general case when the bending stiffnesses of the rod are different, that is, A22 ::f A33 .

In vector form, the system of equations (6.116) reads

Z' +AZ = ~b, (6.117)

where


A=

0 -reID 0 Qi1)

0 0 0 0 A33

reID 0 Qi1)

0 0 0 0 0 A22

0 -1 0 MlO

0 0 0 0 -- - reID

MlO A33

1 0 reID - -- 0 0 0 0 0 A22

1 0 0

A22 0 0 -relO 0 0

1 0 0 0

A22 reID 0 0 0

0 0 0 0 0 -1 0 -relO

0 0 0 0 1 0 reID 0

Qi1) = Q10 - (Po + n1 W6) , Q10 = -P.

P

Y]2

p

Fig. 6.30.

To account for the influence of the gravity force acting both on the rod and the liquid, one should treat Q10 and Po as functions of TJ. Equations (6.117) allow us to analyze the relationship between the stress-strain state of the rod


and the small initial curvatures ~Ge20 and ~Ge30' As the liquid runs through curved segments of the rod (~Ge20 :j:. 0 and ~Ge30 :j:. 0), centrifugal forces occur. These forces result in an additional deformation of the rod.

An oil derrick is shown in Fig. 6.30. A flow of a special liquid brings the turbo drill into rotation and lifts the rock. The cross-sectional area of the clearance between the turbodrill and the conduit is large as compared to the cross-sectional area of the hole in the turbodrill. Hence, for incompressible liquids, the velocity of the liquid in the turbodrill Wo exceeds considerably the velocity of the liquid in the clearance Wi. Thus, the flow in the clearance may be neglected. Consequently, if the turbodrill has no additional supports, then the rod shown in Fig. 6.29 may be treated as a simplified model of such a turbodrill. Using (6.117), we can determine the limiting values of Wo and Po (for specified values of ~Ge20 and ~Ge30) for which the magnitude of the displacement vector lui satisfies the inequality

where ~ is the radial clearance between the drilling bit and the conduit.

p

:1;-

Fig. 6.31.

Let us examine the influence of liquid flows on static stability of compressed-twisted rods. First, consider a simple in-plane problem of stability of a straight rod which interacts with a liquid flow (see Fig. 6.31). Assume that the rod is subjected to a tensile force P. Setting ~Ge20 = ~Ge30 = 0 in (6.116), on rearrangement, we arrive at

(6.118)

Setting Q10 = P, we have

u~v +(-P+Po+niw5)u~=0. (6.119)

Solving this equation, we get

(6.120)

where


k = J -P + Po + nlw5· In the case of simply supported ends, we have Cl = C3 = C4 = 0 and

C2k2 sin k = O. Since k f:. 0 and C2 f:. 0, it follows that sin k = O. Thus k = 7fn and

T

112

Fig. 6.32.

The critical values P, Po, and nl w5 correspond to n critical tensile force is

(6.121)

1. Hence, the

(6.122)

Thus we see that a tensile force may result in loss of stability of the rod containing an internal liquid flow.

P

90 w= 10

80

70

60

50 0 1 2 T

Fig. 6.33.

As another example, consider a compressed-twisted straight rod containing an internal liquid flow (see Fig. 6.32). To account for reaction forces,


which appear at the intermediate supports at loss of stability, and determine the critical loads P and T, one can use (4.245). The term Qh of the matrix A in (4.245) should be replaced by Qi:) such that

Qi:) = QlO - (Po + n1 w5) .

Now the critical loads can be determined by the method discussed in Sect. 4.6. The only difference is the boundary condition at "1 = 1. In this example (Fig. 6.32), the rod is simply supported at "1 = "11, "1 = "12, and "1 = 1 such that only axial displacements of these points are allowed. Hence, the follower moment must satisfy the following boundary-condition equations:

Zs(l) = O.

Solving the problem numerically, we obtain the plots of the first two eigenvalues p.1 and p.2 versus T: (1) for three values of Wo and for "1 = 0.3 and "1 = 0.8 (see Fig. 6.33); (2) for three values of "1 and for w = 5 (see Fig. 6.34). These results were obtained with regard to the initial twist reh '10.

h1 = 0.2, h2 = 0.8 I

t--+--+---+---th1 = 0.1, h2 = 0.8

30 L-_...L...._---'--_-----'

o 1 2 T Fig. 6.34.

It can be seen from the plots that the critical force p.1 decreases as the flow velocity increases. For All = 0.577, A22 = 0.664, and A33 = 1, the critical force P*l at Wo = 10 is less than the critical force at Wo = 0 by 9%. Consequently, the critical values of the force essentially depend on the flow velocity.

[Foundations of Engineering Mechanics] Statics of Rods || Rods Interacting with Liquid or Air Flows

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