CONSERVATION OF LINEAR MOMENTUM FOR A …fc.civil.tamu.edu/resources/en/engr214/Chapter3.pdf48 CHAPTER 3. CONSERVATION OF LINEAR MOMENTUM FOR A CONTINUUM fluid bo undary fluid bo undary

Chapter 3

CONSERVATION OF LINEARMOMENTUM FOR ACONTINUUM

Figure 3.1:

45

46 CHAPTER 3. CONSERVATION OF LINEAR MOMENTUM FOR A CONTINUUM

3.1 Macroscopic vs. Microscopic View of Forces

This chapter will consider the conservation of linear momentum for a continuum. Before we beginthis study, consider a brief review of conservation of linear momentum for some simple cases wherethe concept of a continuum is not utilized. In physics and ENGR 211, we learned that a systemhaving velocity possessed linear momentum, and an analysis of such a system required that we applythe conservation of linear momentum. Consider mass flowing through a pipe as shown below: If the

flow in

flow outpipe

Figure 3.2: Flow in a Curved Pipe

flow is assumed to be at steady state, then the equation for the conservation of linear momentumreduces to

0 =∑

(mv)in −∑

(mv)out +∑

Fext, (3.1)

If we consider the system to be the pipe with the fluid flowing through it, then the system hasmomentum entering and leaving as a result of the velocity of the fluid flowing into and out of thepipe. In addition, the supports provide reactions on the pipe and these reactions result in externalforces on the system boundary which in effect transfer linear momentum into the system. Notice thatwe have taken a macroscopic view by considering only the linear momentum entering and leavingthe system. Such a macroscopic view accounts for the change in linear momentum from entrance toexit but does not provide any information on how the linear momentum may vary inside the systembetween the entrance point and the exit point. Further, any variation of fluid velocity over the crosssection of the pipe has been completely neglected.

Consider next the example of the frame member shown below:

x

y

10 kN

2 m

5 m

8 kN

Figure 3.3: Cantilevered Frame Member

In this case, the structure is static. Therefore, the conservation of linear momentum reduces to

0 =∑

Fext, (3.2)

3.1. MACROSCOPIC VS. MICROSCOPIC VIEW OF FORCES 47

For the static case, the external forces on the system (applied loads and reactions at the supports)give the complete picture of conservation of linear momentum. Using the conservation of linearmomentum, one can obtain the reaction forces at the support as shown below. Note that theconservation of angular momentum is required for calculation of the reaction moment, which hasbeen covered in ENGR 211 for a system and will be discussed in the next chapter for a continuum.

Reactions at supportF = 10 kNV = 8 kNM = -20 kN-m

10 kN

MF

V

2 m

5 m

8 kN

Figure 3.4: Reactions for Cantilevered Frame Member

Applying conservation of linear and angular momentum for the structure taken as a systemallows us to determine the reaction forces and moments at the support. However, it provides noinformation about how these forces are distributed over the cross section of the member.

In the examples considered above, the forces on the system boundary were idealized as pointforces. While we may think of point loads as a convenient idealization, point forces do not actuallyexist. Instead, forces are always applied over some finite area. Consider again the cantilevered framemember in Figure 3.4. As was briefly discussed in Chapter 1 (and which will be demonstrated indetail in Chapter 13), the concentrated (point) force and moment reactions are the resultant offorces that are distributed over the cross-section in some manner schematically shown below:

=F

M

Actual forcedistribution

Idealized forceand momentresultants

V

Figure 3.5: Actual Force Distribution and Force and Moment Resultants on Cross Section

For the pipe flow problem in Figure 3.2, consider the system to be only the fluid. For such asystem, the pipe boundary must exert forces on the fluid boundary. These boundary forces on thefluid are what make the fluid turn as it flows through the pipe.

Note that the forces are distributed along the pipe in some fashion that has yet to be determined.Because of viscous effects, it can be shown that the pipe will also exert shear (or drag type) forceson the fluid which tend to slow the fluid near the pipe boundary.

It is important to note that the external forces applied to the system boundary are in factdistributed over a finite area and that they vary with position. This implies that the amount oflinear momentum transferred to the fluid by the boundary forces will change with position and willdepend on the area over which the forces are applied. This realization naturally introduces theconcept of the traction vector, which will be discussed in detail.


fluid boundary

fluid boundary

forces on fluidboundary

fluid flow

Figure 3.6: Forces exerted on Fluid boundary by Pipe

3.2 Traction Vectors

In order to construct the statement of conservation of linear momentum, it is first necessary to intro-duce some new terminology that will make the description convenient. Consider a three-dimensionalcontinuum body occupying volume V at some time, and suppose that a cutting plane A with unitouter normal n is passed through the system. Due to the external forces applied to the body (andperhaps due to its motion), internal forces are generated in the body, which appear on the free bodydiagrams of the two pieces of the body as distributed forces.

z

x

y

-n

∆AA

1F

4F

1F

2F

3F

2F

3F4F

n ∆F

−∆F

Figure 3.7: Definition of Traction Vector

The traction vector, t(n), is defined as the intensity of the distributed forces as follows:

t(n) = lim∆A→0

∆F∆A

, (3.3)

where ∆F is the force acting on the infinitesimal area ∆A. Note that the notation t(n) impliesthat the force is acting on a plane defined by unit outward normal n and is not necessarily in thedirection of n. Also notice that the traction will have units of force per area.

As shown below, opposite faces at the cut boundary have normals in opposite directions, anddue to Newton’s third law, action forces are in the opposite direction of reaction forces. Therefore,the following definition can be given to the traction vector t(−n):

t(−n) = lim∆A→0

−∆F∆A

= − lim∆A→0

∆F∆A

= −t(n)

3.2. TRACTION VECTORS 49

( )nt

n

( )-nt

-n∆A

∆A

t(−n) ≡ lim −∆F∆A

= − lim ∆F∆A

= −t(−n)

Figure 3.8: Cauchy’s Lemma

We can therefore state that at any point in a body for a cut with given direction n, the tractionsare equal in magnitude and opposite in direction on the two faces of the cut so that:

t(−n) = −t(n) Cauchy’s Lemma (3.4)

The above is essentially a statement of Newton’s third law written in terms of tractions ratherthan forces. Thus, two objects that are in contact with each other exert equal and opposite tractionson each other at every point of contact.

One point that must be stressed is that the traction vector at a point also depends onthe orientation of the cut. Therefore, one can define infinitely many traction vectors at a point,since there are infinitely many ways to introduce a cut through the material at that point. As anexample, assume that the forces are distributed uniformly on the cross section of a truss memberunder tension, shown in the Figure below.

i -i

( )it ( )it

FFx

Figure 3.9: Cross section of a truss member

The traction vectors t(i) in the cut with normal i, or t(−i) in the cut with direction −i are shownin the Figure above. Now assume that at the same point a cut is introduced horizontally instead ofvertically. A schematic view of that situation is shown below

Since no forces are applied in the j direction, the traction vector t(j) applied on the cut withnormal j will be zero.

Another example that gives a physical understanding of the traction vector is when one standson both feet, the traction vector applied by the ground to one’s feet is half of the traction vectorapplied when they stand on only one foot (assuming a uniform distribution of contact forces). Themagnitude of the traction vector is inversely proportional to the surface area of the foot one standson. Therefore, a flat foot will have tractions with much smaller magnitudes than those of a ballerinaon her toes, as shown in the figure below.

Finally, think about the pyramids. Imagine what the traction vector on the foundation wouldhave been had they been built to point downward!

To further consolidate the notion of the traction vector the following examples are discussedbelow:


j

FF

( ) =jt 0

x

y

( )it (- )it

Figure 3.10: Horizontal cut of truss member

Figure 3.11: Relationship between traction vector and cross-sectional area

Example 3-1

Uniaxial Tension in the x-direction

x( ) 5t =i ii-i

-5t = i(-i)

Figure 3.12:

Find the three components of the t(i) traction vector on the right and left faces perpendicularto the x-direction of a body under uniform tension. Assume units of kN

m2 for the magnitude of thetraction vector.

Solution

3.2. TRACTION VECTORS 51

On the x face:

t(i)x= i · t(i)

= i · 5i =⇒ t(i)x= 5

kNm2

t(i)y= j · t(i)

= j · 5i =⇒ t(i)y= 0

t(i)z= k · t(i)

= k · 5i =⇒ t(i)z= 0

=⇒[t(i)

]=

[5 0 0

] kNm2

On the −x face:

t(−i)x= i · t(−i)

= i · (−5i) =⇒ t(−i)x= −5

kNm2

t(−i)y= j · t(−i)

= j · (−5i) =⇒ t(−i)y= 0

t(−i)z= k · t(−i)

= k · (−5i) =⇒ t(−i)z= 0

Due to the fact that the bar is under static equilibrium (COLM),

t(i)x= −t(−i)x

t(i)y= −t(−i)y

t(i)z= −t(−i)z

Example 3-2

Uniaxial Compression in the y-direction

( ) 3t j j j

-j( ) 3t =

-jj

y

x

= -

Figure 3.13:

Find the three components of the t(j) vector on the top and bottom faces perpendicular to they-direction for a body subjected to uniform compression (units for magnitude of traction are kN

m2 ).

Solution


On the y face:

t(j)x= i · t(j)

= i · (−3j) =⇒ t(j)x= 0

t(j)y= j · t(j)

= j · (−3j) =⇒ t(j)y= −3

kNm2

t(j)z= k · t(j)

= k · (−3j) =⇒ t(j)z= 0

=⇒[t(j)

]=

[0 −3 0

] kNm2

On the −y face:

t(−j)x= i · t(−j)

= i · (3j) =⇒ t(−j)x= 0

t(−j)y= j · t(−j)

= j · (3j) =⇒ t(−j)y= 3

kNm2

t(−j)z= k · t(−j)

= k · (3j) =⇒ t(−j)z= 0

Due to the fact that the bar is under static equilibrium,

t(j)x= −t(−j)x

t(j)y= −t(−j)y

t(j)z= −t(−j)z

Example 3-3

Pure uniform shear in the xy-plane of a body with a square cross section.

y

x

-j

j

i

-i

t(−j) = 3i

t(j) = −3i

t(i) = −3jt(−i) = 3j

Figure 3.14:

Find the three components of the traction vector on each face for a body subjected to uniformshear (units for the magnitude of traction are kN

m2 ).

Solution

3.3. CONSERVATION OF LINEAR MOMENTUM 53

On the x face:

t(i)x= i · t(i)

= i · (−3j) =⇒ t(i)x= 0

t(i)y= j · t(i)

= j · (−3j) =⇒ t(i)y= −3

kNm2

t(i)z= k · t(i)

= k · (−3j) =⇒ t(i)z= 0

=⇒[t(i)

]=

[0 −3 0

] kNm2

On the −x face:

t(−i)x= i · t(−i)

= i · (3j) =⇒ t(−i)x= 0

t(−i)y= j · t(−i)

= j · (3j) =⇒ t(−i)y= 3

kNm2

t(−i)z= k · t(−i)

= k · (3j) =⇒ t(−i)z= 0

On the y face:

t(j)x= i · t(j)

= i · (−3i) =⇒ t(j)x= −3

kNm2

t(j)y= j · t(j)

= j · (−3i) =⇒ t(j)y= 0

t(j)z= k · t(j)

= k · (−3i) =⇒ t(j)z= 0

=⇒[t(j)

]=

[−3 0 0

] kNm2

On the −y face:

t(−j)x= i · t(−j)

= i · (3i) =⇒ t(−j)x= 3

kNm2

t(−j)y= j · t(−j)

= j · (3i) =⇒ t(−j)y= 0

t(−j)z= k · t(−j)

= k · (3i) =⇒ t(−j)z= 0

3.3 Conservation of Linear Momentum

We now turn our attention to the construction of the conservation of linear momentum statement fora continuum. Linear momentum must be conserved. In a differential volume element, momentumchanges are caused by mass flowing into and out of the system and by forces acting upon the system.When considering continua, we divide the forces into two groups:


1) body forces, which are long-range forces (i.e., the force affects the object, but doesn’t touchthe body; e.g., gravity) and

2) contact forces, which are short-range forces (i.e., forces that arise from contact with thebody; e.g., boundary tractions caused by friction).

It is useful to construct some verbal descriptions of the amount of linear momentum entering orleaving a system considering the following points:

• Linear momentum may enter or leave the system with mass.

• Linear momentum may also enter or leave the system at a certain rate.

• When mass enters a system traveling at a velocity, it adds momentum to the system.

• The rate at which mass enters the system is min (where min = dmin

dt )

• Therefore, the rate at which linear momentum enters the system is given by (mass rate in) ×(velocity of mass flowing in) = (mv)in.

• Similarly, the rate at which linear momentum leaves the system is (mv)out.

We will now begin the derivation of the conservation of linear momentum for continuous mediain Cartesian coordinates. For clarity, this will be done in three steps beginning with some one-dimensional problems and finally for the full three-dimensional case.

The verbal and mathematical statements of conservation of linear momentum (COLM) may bewritten as:

Accumulation oflinear momentumwithin systemduring time period ∆t

=

Linear momentumentering systemduring time period ∆t

−

Linear momentumleaving systemduring time period ∆t

or

Psys|t+∆t − Psys|t =∑

(mv)in∆t −∑

(mv)out∆t +∑

Fext∆t (3.5)

where P is the amount of linear momentum in the system, or at a specific instant of time t1 and as∆t → 0

dPdt

=∑

(mv)in −∑

(mv)out +∑

Fext

Recall from discussions and derivations in ENGR 211, the external forces on the system (the last termin the above equations) are responsible for transferring linear momentum associated with boundaryforces to linear momentum within the system. To clearly understand that the term

∑Fext∆t is

a momentum term, consider a dimensional analysis. Recalling that force has units of mass timesacceleration, one can write F∆t →

(kgm

s2

)s = kgm

s → mv. For a continuum, we will be consideringhow tractions (instead of forces) applied to the boundary transfer linear momentum to the system.In fact, we can think of tractions as measuring linear momentum flux on the boundary at which theyact. Flux is the term used to define a property like mass or linear momentum entering or actingupon a surface area during a certain time period. For mass flowing through an area at a certain rate,the mass flux would have units such as kg

m2s . To see that a boundary traction is a linear momentumflux, consider a dimensional analysis, and write the units for a traction:

traction =forcearea

→(kg

ms2

) 1m2

=[kg

ms

] 1m2

1s→ momentum

(area)(time)

→ momentum per unit area per unit time → momentum flux


3.3.1 Conservation of Linear Momentum in 1-D

Consider a one dimensional (1-D) continuum body that interacts with the environment only inthe x-direction and that is subject to zero body forces. By 1-D, we mean that the importantproperties like density, velocity and tractions are functions of one coordinate only (x), while theactual continuum body is placed in 3-D physical space. In addition, for the simplicity of derivations,we will assume that all vector quantities have only one non-zero component in the x-direction.

fluid flow

fluid flowx

y

z

Area=Ax

Figure 3.15: 1-D Fluid Flow

An example of 1-D flow could be the flow between two parallel flat plates or a straight cylindricalpipe where the fluid velocity varies only in the x direction (there is no flow in the y or z directions),and a pressure gradient exists only in the x direction. A truss member would be an example of a 1-Dsolid where tractions exist only in one coordinate direction. For these 1-D problems, we assume thatthe region of concern has cross-sectional area Ax at x as was previously assumed in the derivationof conservation of mass.

We schematically sketch a 1-D system as shown below. The cross-hatched areas are meant toindicate that no normal components of velocity or traction exist on these boundaries (there are onlynon-zero x-components of all vector quantities). In addition, all pertinent quantities are functionsof x only. We define a system (control volume) of length ∆x located at position x as shown.

-i +i

i ∆x

system

)( x x xv vρ )( x x x x

v vρ +∆

( )tx x x+∆ i( )t

x x -i

x x + ∆x

Figure 3.16: 1-D System with Mass Flow and Tractions (no body forces)

Note that the system is drawn in plan form and the unit vector i indicates the orientation of thecross section at x + ∆x. The term t(−i)x

∣∣x

is defined to be the x-component of the traction vectoron the −i cross section of the system (control volume) at point x (the normal to the surface of thesystem is always defined to be the outward unit normal vector–in this case −i), while t(i)x

∣∣x+∆x

isthe x-component of the traction vector on the i cross-section of the system at point x + ∆x. Notethat while the tractions are shown on the sketch as vectors located at a point, they represent forcesper unit area that are applied to the entire cross-sectional area Ax. The terms ρvxvx representmomentum flux crossing the system boundary Ax. To define the linear momentum crossing thesystem boundary, consider the following word description. Momentum flux is momentum per unitarea per unit time. The momentum flux in the x direction is the mass flux in the x direction (ρvx)


times the velocity in the x direction (vx) or ρvxvx. The total momentum entering a boundaryduring some time interval is given by the product of momentum flux, area, and time interval or(ρvxvx) × (Ax) × (∆t). The momentum of the system at any given time is given by the product ofthe mass of the system and velocity = [(mass density) × (volume)] × (velocity) = ρ(Ax∆x)vx. Inconsidering the traction terms we note two important matters:

1) The tractions must be multiplied by the area upon which they act in order to obtain theresultant force and

2) a traction is defined as the intensity of the contact forces applied by the environment or thesystem.

A component of the traction vector is said to be positive when it acts in the positive coordinatedirection. With these basic components of the terms necessary for linear momentum, we can nowdevelop the conservation of linear momentum:

Psys|t+∆t − Psys|t =∑

(mv)in∆t −∑

(mv)out∆t +∑

Fext∆t (3.6)

or

(ρAx∆xvx)|t+∆t − (ρAx∆xvx)|t = (ρvxvxAx)|x ∆t − (ρvxvxAx)|x+∆x ∆t (3.7)

+ (t(−i)xAx)

∣∣x

∆t + (t(i)xAx)

∣∣x+∆x

∆t

Divide by Ax∆x∆t, assume Ax to be constant, and use Cauchy’s Lemma (t(−i)x

∣∣x

= t(i)x

∣∣x) to

obtain

(ρvx)|t+∆t − (ρvx)|t∆t

= −(ρvxvx)|x+∆x − (ρvxvx)|x

∆x+

t(i)x

∣∣x+∆x

− t(i)x

∣∣x

∆x(3.8)

Take the limit as ∆x → 0, ∆t → 0 to obtain

∂(ρvx)∂t

= −∂(ρvxvx)∂x

+∂t(i)x

∂x(3.9)

The above is the conservation of linear momentum in 1-D. It is a partial differential equation inx and t. Note that the three unknown quantities are all functions of space (x) and time (t) (i.e.,ρ = ρ( x, t ), vx = vx(x, t ), t(i)x

= t(i)x(x, t )). It is also interesting to note that although we have

added a new equation, we have also added a new unknown, t(i)x(remember that only ρ and vx were

unknown quantities in the conservation of mass in 1-D). Equation (3.9) is often called the equationof motion since it defines the motion of the fluid (or solid) in terms of the velocity.

In the preceeding development, no distinction has been made as to whether the“system” is asolid or a fluid. In fact, the conservation of linear momentum developed here applies to both solidand fluid bodies. For a fluid, the tractions that have been included are the result of pressures orpressure gradients within the fluid. Such fluid pressure will cause tractions that act normal to thesurface of a fluid element or a fluid boundary. As an example, consider fluid under static pressure p,with vx = 0. Equation (3.9) will reduce to ∂t(i)x

∂x = 0, or after integration, t(i)x= constant. Based

on the definition of pressure, we observe that t(i)x= −P , as schematically shown below. Notice that

t(−i)x= −t(i)x

= P .Note that for a solid body under static equilibrium, the conservation of linear momentum reduces

to 0 = ∂t(i)x

∂x also. Integrating this equation with respect to x gives t(i)x= constant; that is, the

traction in the x-direction is independent of position x. Note that if we had a solid member ofconstant cross-sectional area Ax, then the force in the member would be constant and given byF = t(i)x

Ax assuming the traction to be uniform over the cross section. A 1-D solid with tractionsin one direction only (x-direction in this case) is precisely the two-force truss member that was


( ) - Ptx= i( ) Pt

x- =i

Figure 3.17: Fluid under static pressure

considered in ENGR 211. For the two-force member, it was observed that the force was constantalong the length (i.e., the same on both ends) and was directed along the length of the member.

All the essential terminology and steps for developing conservation of linear momentum in 2-Dand 3-D have now been developed. Before going further in the development of conservationof linear momentum for 2-D and 3-D bodies, it is imperative that the student thoroughlyunderstand all the steps taken in this section .

3.3.2 Conservation of Linear Momentum in 2-D (Rectangular CartesianCoordinates)

Consider now a 2-D material system formed by an infinitesimal rectangular domain, which containsno component of velocity or traction or body force in the z-direction. Here we will consider bodyforces. The linear momentum entering and leaving the system as well as tractions acting on thesystem must all be defined in order to develop the conservation of linear momentum in 2-D.

j

-j

i-i

systemlinearmomentum in

linearmomentum out

linearmomentum out

linearmomentum in

traction on+j surface

traction on-j surface

traction on+i surface

traction on-i surface

bodyforce

xj

i

(x, y) (x + ∆x, y)

(x, y + ∆y) (x + ∆x, y + ∆y)

Figure 3.18: Schematic of Conservation of Linear Momentum for 2-D Region

Linear Momentum Flux. Linear momentum flux is defined to be positive when directed inthe positive coordinate direction, i.e., when mass is being carried with a velocity in the positivecoordinate direction. To define the linear momentum entering the −i face (located at position x),we again turn to our word statement for linear momentum flux.

(linear momentum flux into boundary) = (mass flux into boundary) × (velocity of mass at boundary)


The velocity of the mass at the −i boundary (point x) is given by v|x. The mass flux into the −iboundary (point x) is that due only to the velocity component normal to the surface (i.e., vx) andis given by (ρvx)|x. Thus the linear momentum flux on the left boundary is given by (ρvxv)|x. Thelinear momentum flux entering and leaving the system is given by:

(ρvxv)|x = linear momentum flux entering the −i face at point x

(ρvxv)|x+∆x = linear momentum flux leaving the i face at point x + ∆x

(ρvyv)|y = linear momentum flux entering the −j face at point y

(ρvyv)|y+∆y = linear momentum flux leaving the j face at point y + ∆y

The linear momentum in the system at time t is given (msysv)|t, or one can write the linearmomentum per unit volume in the system at time t as (ρv)|t.

Tractions. Tractions are defined to be positive when they act outward on the system boundary.Note that this is consistent with the notion that tensile forces on a truss member are positive asdiscussed in ENGR 211. The tractions acting on the system are given by:

t(−i) = traction on the surface having outward normal −i, also called the −i or −x facet(i) = traction on the surface having outward normal i, also called the +i or +x face

t(−j) = traction on the surface having outward normal −j, also called the −j or −y facet(j) = traction on the surface having outward normal j, also called the +j or +y face

Note that the traction t(−i) is positive when it acts in the −i direction, and t(i) is positive when itacts in the +i direction, i.e., positive tractions are always directed outward.

As was stated previously, boundary tractions can be considered as linear momentum flux on theboundary where they act. We showed this with a units analysis which is repeated here for clarity.Simply writing the units for a traction:

traction =forcearea

→(kg

ms2

) 1m2

=[kg

ms

] 1m2

1s→ momentum

(area)(time)

→ momentum per unit area per unit time → momentum flux

Body Forces. In many cases body forces such as that due to gravity may act on the body. Definea body force per unit mass as g, hence the body force per unit volume is given by ρg.

With these definitions, we now show another sketch of the system (differential control volumeelement) with the appropriate linear momentum, traction and body forces.

Recall that the total momentum entering a boundary during some time interval is given by theproduct of momentum flux, area, and time interval. Applying the conservation of linear momentumprinciple to the 2-D differential system gives the following:

(ρv)|t+∆t − (ρv)|t

∆x∆y∆z =

(ρvxv)|x − (ρvxv)|x+∆x

∆y∆z∆t

+

(ρvyv)|y − (ρvyv)|y+∆y

∆x∆z∆t (3.10)

+

t(−i)

∣∣x

+ t(i)

∣∣x+∆x

∆y∆z∆t

+

t(−j)

∣∣y

+ t(j)

∣∣y+∆y

∆x∆z∆t + ρg∆x∆y∆z∆t

As it was discussed earlier in the conservation of mass chapter, ∆z can have any value. We useCauchy’s Lemma in the above so that t(−i) = −t(i) and t(−j) = −t(j), divide by (∆x∆y∆z∆t) and


j

-j

i-i ρ g

system

y

( )x xvρ v

( )x x xvρ

+ v

( )y y yvρ

+ v

( )y yvρ v

xj

i

( ) y y+ jt

( ) x -it

( ) y -jt

( ) x x+ it

(x, y) (x + ∆x, y)

(x, y + ∆y) (x + ∆x, y + ∆y)

Figure 3.19: 2-D Differential Element with Linear Momentum, Tractions and Body Force

take the limit as ∆x,∆y, ∆t → 0. The conservation of linear momentum in 2-D becomes (in vectorform):

∂(ρv)∂t

= −

∂(ρvxv)∂x

+∂(ρvyv)

∂y

+

∂t(i)

∂x+

∂t(j)

∂y+ ρg (3.11)

We now have two equations for the conservation of linear momentum in 2-D. The components ofthe above vector equation are given in Cartesian Coordinates by:

∂(ρvx)∂t

= −

∂(ρvxvx)∂x

+∂(ρvyvx)

∂y

+

∂t(i)x

∂x+

∂t(j)x

∂y+ ρgx

∂(ρvy)∂t

= −

∂(ρvxvy)∂x

+∂(ρvyvy)

∂y

+

∂t(i)y

∂x+

∂t(j)y

∂y+ ρgy

Notice that there are nine dependant variables, i.e., ρ( x, y, t ), vx( x, y, t ), vy( x, y, t ), t(i)x( x, y, t ),

t(i)y(x, y, t ), t(j)x

(x, y, t ), t(j)y(x, y, t ), gx(x, y, t ), gy(x, y, t ) with one scalar quantity, ρ, and eight

components of the four vector quantities (v = vxi + vyj, t(i) = t(i)xi + t(i)y

j, t(j) = t(j)xi + t(j)y

j,and g = gxi + gyj). For these nine quantities usually only the body force is known (g), while theremaining seven are unknown and, of course, thus far there are only three equations at our disposal(one for conservation of mass and two for conservation of linear momentum).

3.3.3 Conservation of Linear Momentum in 3-D (Rectangular CartesianCoordinates)

A full three dimensional control volume element is represented here by an infinitesimal parallelepipedhaving dimensions ∆x, ∆y, and ∆z.

We again define tractions, momentum flux and body forces in a manner consistent with that inthe previous section

Now sum the linear momentum entering and leaving the system and set this equal to the change


x

y

z

ρ g( )x x

vρ v

( )x x xvρ

+∆ v

( )y y yvρ

+∆ v

( )y yvρ v

x

y

z

( )z zvρ v

( )z z zvρ

+∆ v

j

ik

( )ty y+∆ j

( )tz− k

( )tx x+∆ i

( )ty− j

( )tz z+∆ k

( )tx− i

Figure 3.20: 3-D Differential Element with Linear Momentum, Tractions and Body Force

in momentum of the system during an infinitesimal time interval ∆t:(ρv)|t+∆t − (ρv)|t

∆x∆y∆z =

(ρvxv)|x − (ρvxv)|x+∆x

∆y∆z∆t

+

(ρvyv)|y − (ρvyv)|y+∆y

∆x∆z∆t

+

(ρvzv)|z − (ρvzv)|z+∆z

∆x∆y∆t (3.12)

+

t(−i)

∣∣x

+ t(i)

∣∣x+∆x

∆y∆z∆t

+

t(−j)

∣∣y

+ t(j)

∣∣y+∆y

∆x∆z∆t

+

t(−k)

∣∣z

+ t(k)

∣∣z+∆z

∆x∆y∆t + ρg∆x∆y∆z∆t

Substituting Cauchy’s lemma (t(i) = −t(−i), t(j) = −t(−j), t(k) = −t(−k)) into (3.12), dividing bothsides of the equation by ∆x∆y∆z∆t, and taking the limit ∆x,∆y, ∆z → 0 and ∆t → 0, we obtain:

∂(ρv)∂t

= −

∂(ρvxv)∂x

+∂(ρvyv)

∂y+

∂(ρvzv)∂z

+

∂t(i)

∂x+

∂t(j)

∂y+

∂t(k)

∂z+ ρg (3.13)

The above represents three equations, with sixteen dependant variables: the scalar mass density, ρ,and the components of each vector:

v = vxi + vyj + vzk

t(i) = t(i)xi + t(i)y

j + t(i)zk

t(j) = t(j)xi + t(j)y

j + t(j)zk

t(k) = t(k)xi + t(k)y

j + t(k)zk

g = gxi + gyj + gzk


The total number of equations available to us currently is only four, one from COM and three fromCOLM.

We will now simplify the COLM equations using the COM. By expanding the partial derivativesof ∂(ρv)

∂t and −

∂(ρvxv)∂x + ∂(ρvyv)

∂y + ∂(ρvzv)∂z

we can re-write equation (3.13) as:

v∂ρ

∂t+

∂v∂t

= −v

∂(ρvx)∂x

+∂(ρvy)

∂y+

∂(ρvz)∂z

−

ρvx∂v∂x

+ ρvy∂v∂y

+ ρvz∂v∂z

(3.14)

+∂t(i)

∂x+

∂t(j)

∂y+

∂t(k)

∂z+ ρg

Conservation of mass can now be used to simplify the last result. Note that in Chapter 2 theconservation of mass is given by:

∂ρ

∂t= −

(∂(ρvx)

∂x+

∂(ρvy)∂y

+∂(ρvz)

∂z

)(3.15)

Multiplying the conservation of mass equation (3.15) by v, we see that the first term on the left-handside of the conservation of linear momentum equation cancels the first term on the right-hand side,so that the vector form of the conservation of linear momentum equation reduces to:

Conservation of Linear Momentum/Equations of Motion

ρ∂v∂t

+ ρ

(vx

∂v∂x

+ vy∂v∂y

+ vz∂v∂z

)= ρg +

∂t(i)

∂x+

∂t(j)

∂y+

∂t(k)

∂z(3.16)

Noticing that the second term on the left hand side of the above equation can be written as the dotproduct of the del operator, ∇, with the velocity v, the COLM can finally be written in vector formas follows:

Vector Form of Conservation of Linear Momentum

ρ∂v∂t

+ ρ (v · ∇)v = ρg +∂t(i)

∂x+

∂t(j)

∂y+

∂t(k)

∂z(3.17)

The Cartesian components of the above vector form of COLM are given by:

ρ∂vx

∂t+ ρ

(vx

∂vx

∂x+ vy

∂vx

∂y+ vz

∂vx

∂z

)= ρgx +

∂t(i)x

∂x+

∂t(j)x

∂y+

∂t(k)x

∂z

ρ∂vy

∂t+ ρ

(vx

∂vy

∂x+ vy

∂vy

∂y+ vz

∂vy

∂z

)= ρgy +

∂t(i)y

∂x+

∂t(j)y

∂y+

∂t(k)y

∂z(3.18)

ρ∂vz

∂t+ ρ

(vx

∂vz

∂x+ vy

∂vz

∂y+ vz

∂vz

∂z

)= ρgz +

∂t(i)z

∂x+

∂t(j)z

∂y+

∂t(k)z

∂z

Equations (3.16), or alternately equations (3.17) or (3.18), are called the equations of motionsince they relate the motion of the fluid (or solid) to the velocity components.

3.3.4 Relationship of Tractions to Stresses: Cauchy’s Formula

Tractions play an important role in the development of the Conservation of Linear Momentum as ithas been determined in the previous section. Consequently, we need to develop our understandingof tractions in a formal manner. In particular, we need to find a way of dealing with the infinitely


many ways we can define the traction vector at a point, if we want to have any hope of being ableto solve the equations of motion. For simplicity, we limit our discussion to a two-dimensional planesolid body of constant thickness that lies in the x-y plane with all forces applied in the x-y plane aswell. Later, we will extend the two-dimensional results to three-dimensional bodies. Assume a bodyhas various external loads (body forces are assumed to be zero) and supports as shown below:

f

b = thickness

x

y

z

1F

2F

Figure 3.21: Two-Dimensional Body with Applied Loads

Consider a system (free-body) obtained by passing a cutting plane with a unit normal n throughthe body as shown below. If we remove the right portion of the body, we must place tractions t(n)

on the cut surface as shown to the right in the figure below. The notation t(n) signifies the tractionat any point on the surface whose outward unit normal is given by n. It does NOT mean that thetraction is normal to the surface.

Note that the magnitude and direction of t(n) varies with position along the cut surface and, fora fixed point, varies with the orientation of the cut at that point. We can now apply cutting planesnormal to the x and y axes to form a new free body as shown in Figures 3.23 and 3.24:

Assume that the cutting planes are selected such that a very small (differential size) free-bodyis obtained, as shown in Figure 3.24. In fact the differential element is so small that the tractionvectors are assumed to be constant along the faces n, i, and j of the differential element.

As before, we must place tractions on those surfaces where the cuts were made and a portionof the body was removed. Note that the unit outward normal on the right (x) face is given by i.Hence, on the right face, we will denote the traction as t(i). Similarly, on the top (y) face the unitoutward normal is given by j and the traction there is denoted by t(j). The unit outward normal onthe inclined face can be written as

n = nxi + nyj = (cos θ)i + (sin θ)j (3.19)= [cos(β + π)]i + [sin(β + π)]j = −(cos β)i − (sinβ)j

The tractions on each face can be written in terms of their Cartesian components as shown below:Thus, we obtain:

t(n) = t(n)xi + t(n)y

j (3.20)

t(i) = t(i)xi + t(i)y

j (3.21)t(j) = t(j)x

i + t(j)yj

The area over which the traction t(n) acts may be defined as A(n). Note that we assume that A(n) isa differential area so that it can be assumed that the traction is a constant over the area upon which


-n

f

x

y

z

f

b

Original Body

Free Body #1 Free Body #2

2F

2F

1F

1F

2R

1R

3R

n

( )nt( )− nt

Figure 3.22: Two-Dimensional Body Showing Tractions of Cut Surface

f

x

yn

( )nt

Figure 3.23: Two-Dimensional Body with Two Cutting Planes Shown


x

y

n β

i

( )nt

j

( )it

( )jt

β

( )A j

( )A i( )A n

θ

Figure 3.24: 2-D Free-Body Differential Volume Element with Tractions Applied

β

x

y

( )A j

( )A i

( )A n

( )tyj

( )txj

( )txi

( )tyi

( )tyn

( )txn

Figure 3.25: 2-D Free-Body with Tractions in x-y Coordinate System

it acts. From geometry, the areas of the right and top faces are then given by A(i) = A(n) cos β andA(j) = A(n) sinβ, respectively, as shown in the Figure above.

We are now in a position to apply conservation of linear momentum to the free-body. For astatic body, COLM reduces to

0 =∑

Fext

The z-component of COLM is identically satisfied and the x- and y-components become (recall thatforce is equal to traction times area):

x-component: 0 = t(n)xA(n) + t(i)x

A(i) + t(j)xA(j) (3.22)

y-component: 0 = t(n)yA(n) + t(i)y

A(i) + t(j)yA(j)

Substituting the area relations into (3.26) gives:

x-component: 0 = t(n)xA(n) + t(i)x

A(n) cos β + t(j)xA(n) sinβ (3.23)

y-component: 0 = t(n)yA(n) + t(i)y

A(n) cos β + t(j)yA(n) sinβ

The area can be cancelled out to obtain the following

t(n)x= −t(i)x

cos β − t(j)xsinβ (3.24)

t(n)y= −t(i)y

cos β − t(j)ysinβ


Recalling the definition of the unit normal vector on the inclined face (n = nxi + nyj = (cos θ)i +(sin θ)j = −(cos β)i − (sinβ)j), we see that the above traction relations (3.24) can be written as

t(n)x= nxt(i)x

+ nyt(j)x(3.25)

t(n)y= nxt(i)y

+ nyt(j)y

The above two equations provide the relationships between the traction components that act on asurface with unit outward normal n and the x and y components of tractions required to keep thefree-body in equilibrium, i.e., satisfy conservation of linear momentum. The above formula is valid inthe presence of body forces and for non-zero velocities as well. It conveys the very important resultthat the components of the traction vector on the plane with orientation n are a linear combinationof the components of the traction vectors on planes with orientation i and j. It is called the Cauchyformula.

In order to simplify the writing and understanding of the traction components on the right sideof the equations, we turn to a different notation by defining the traction components on each faceto be Cauchy stress components. The traction term t(i)x

represents the x direction componentof the force per unit area that acts on a face whose normal is i (x face). This traction componentwill be redefined as the normal stress σxx component. Likewise the component of traction term t(j)x

represents the x direction component of the force per unit area that acts on a face whose normal isin the y direction and hence it will be denoted as the shear stress component σyx. Thus, we definethe stress components as follows:

σxx = t(i)x, σyx = t(j)x

, σxy = t(i)y, σyy = t(j)y

(3.26)

Substituting these stress component definitions into the Cauchy formula (3.25) gives

t(n)x= nxσxx + nyσyx (3.27)

t(n)y= nxσxy + nyσyy

It becomes convenient to express the above two relations (3.27) in matrix notation. We have alreadydefined the traction and normal vectors in 2-D as:

t(n) = t(n)xi + t(n)y

j and n = nxi + nyj

The traction and normal vector can be written in matrix notation as[t(n)

]=

[t(n)x

t(n)y

](3.28)

[n] =[

nx ny

](3.29)

We now define the Cauchy stress tensor (in 2-D) as

[σ] =[

σxx σxy

σyx σyy

]=

[t(i)x

t(i)y

t(j)xt(j)y

](3.30)

With the above vector and matrix notations and the two traction equations, equation (3.27) cannow be written in vector form as

Cauchy’s Formula in Vector Form

t(n) = n · σ (3.31)

or, in matrix form as

Cauchy’s Formula in Matrix Form


x

y

z

n

( )nt

( )tyyyσ = j

( )tyxyσ = i

( )txyxσ = j

( )txxxσ = i

Figure 3.26: 2-D Free-Body with Tractions and Stress Components

[t(n)] = [n][σ] (3.32)

We note that through the re-definition of the tractions in terms of stresses, the free-body is nowdrawn as:

The vector meaning of the equation t(n) = n · σ is as follows: the quantity n · σ provides theprojection of the stress tensor σ expressed in Cartesian coordinates, onto a plane which has anoutward unit normal vector n. The left side is the traction vector t(n) that exists on this plane.Recall that the usual vector dot product is applied to two vectors and results in a scalar. In the abovematrix notation, the row vector [n] (1×2) is dotted into each column vector (2×1) of the [σ] matrix(2 × 2) and results in a row vector [t(n)] (1 × 2). Thus, when a vector is multiplied with a tensor(represented by a square matrix), one obtains a vector. The stress tensor σ is therefore an objectthat is formed by the two traction vectors t(i) and t(j) in a way that, when the normal to a planeis multiplied with the matrix of σ, the result is the traction vector applied on that plane. Given amaterial point, the stress σ is defined for that point regardless of any orientation (in contrast, t(n)

depends on the orientation n).Note that it turns out that the normal stresses (σxx, σyy ) are positive when the traction vector

acts in the direction of the outward unit normal to the surface the traction acts upon. Also, thenormal components of stress are positive when the material is under tension and negative undercompression. The shear stress σxy is positive when the traction vector acting on a surface with unitnormal in the direction of x-axis is in the direction of y-axis. On a face with a unit normal in the −xdirection (call this the negative x face), the positive direction of σxy is in the −y direction. Similarlyσyx is positive when the traction vector acting on a surface with unit normal in the direction ofthe y-axis is in the direction of the x-axis. On the face with a unit normal in the −y direction,the positive direction of σyx is in the −x direction. This is a direct result of the definition of thecomponents of stress with respect to the components of the traction vector given by equation (3.26).

Figures 3.27 and 3.28 below summarize the relationship between traction vectors and stresscomponents for the 2-D and 3-D case. Note in Figure 3.27, that the tractions and stresses are shownon the negative faces.

The expression [t(n)] = [n][σ] (3.32) is quite general and can be shown to also apply for three-dimensional traction vectors and stress states shown below:

In order to derive the three-dimensional form of

t(n) = n · σ (3.33)


x

y

z

θ − n

( )− nt

( )tx− n

( )ty− n

( )tyyyσ − = − j

( )txxxσ − = − i

( )tyxyσ − = − i

( )tx

yxσ − = − j

Figure 3.27: 2-D Free-Body with Tractions and Stress Components (negative faces)

Cauchy’s tetrahedron in termsof tractions

y

x

Cauchy’s tetrahedron in termsof stresses

( )t j

( )t i

( )t k

( )t n

ni

j

k

nxxσ xyσ

xzσ

yyσ

yxσ

yzσ

zxσ

zyσ

zzσ

z

z

x

y

( )nt

Figure 3.28: Cauchy’s Tetrahedron

and

[t(n)] = [n][σ], (3.34)

we simply carry out the steps followed above but in three dimensions. Figure 3.29 shows thecomponents of the three traction vectors on the x, y, and z faces of a differential volume element.

Extending the definitions for components of the Cauchy stress in 3-D, we can easily deriveCauchy’s formula for 3-D.

t(n)x= σxxnx + σyxny + σzxnz

t(n)y= σxynx + σyyny + σzynz (3.35)

t(n)z= σxznx + σyzny + σzznz

Equation (3.35) was obtained by Augustine Cauchy in 1822. It was a breakthrough in the devel-opment of continuum mechanics, since it demonstrated for the first time that the components of


x

yt(j)

z

t(i)

0i

k

jt(k)

yyσ

yxσ xyσ

xxσ

zzσ

zyσ yzσ

xzσ

zxσ

Figure 3.29: Differential Volume Element with Stresses and Tractions (+ faces only)

the traction vector at some point are linear functions of the orientation of the plane on which thetraction is applied.

Cauchy’s formula may also be written in the following convenient matrix notation:

[t(n)x

t(n)yt(n)z

]=

[nx ny nz

] σxx σxy σxz

σyx σyy σyz

σzx σzy σzz

(3.36)

[t(n)

]= [n] [σ]

The Cartesian components of the stress tensor in 3-D are therefore given by

[σ] =

σxx σxy σxz

σyx σyy σyz

σzx σzy σzz

=

t(i)x

t(i)yt(i)z

t(j)xt(j)y

t(j)z

t(k)xt(k)y

t(k)z

(3.37)

In vector form we have t(n) = n · σ.It can be seen from Cauchy’s formula that if the stress tensor, σ, is known at a given point in

an object, the traction vector, t(n), can be found on any plane at that point simply by knowing theunit normal, n, to that plane and performing the above matrix multiplication. Therefore, if the ninecomponents of the stress tensor, σ, are known at a point, they uniquely define all traction vectorsin any orientation at that point.

As an example consider the following problem for which [σ] =

5 0 0

0 −5 00 0 0

MPa (note:

1 MPa = 1000 kNm2 ):

The traction vector on a plane with unit normal n = −√

22 i +

√2

2 j is given by

[t(n)] =[−

√2

2

√2

2

] [5 00 −5

]=

[− 5

√2

25√

22

]


5xx MPaσ =

5yy MPaσ =

5yy MPaσ =

5xx MPaσ =

x

y

ij

Figure 3.30:

( )tyn

( )nty

x

n( )t

xn

( )t 5MPayyy− = =j

( )t 5MPaxxx= =i

Figure 3.31:

Note that as a special case, one has from equation (3.33).

t(i) = i · σ = σxxi + σxyj + σxzk,[t(i)

]=

[σxx σxy σxz

]t(j) = j · σ = σyxi + σyyj + σyzk,

[t(j)

]=

[σyx σyy σyz

](3.38)

t(k) = k · σ = σzxi + σzyj + σzzk,[t(k)

]=

[σzx σzy σzz

]Revisiting the examples presented earlier in Section 3.2 may help clarify the relationship betweenstress and the traction vector (all units for the components of traction and stresses are in MPa).

Example 3-4

Uniaxial Tension in the x-directionOn the x face: n = i

t(i) = i · σ = σxxi + σxyj + σxzk = 5i =⇒

σxx = t(i)x= 5

σxy = 0σxz = 0

On the −x face: n = −i

t(−i) = (−i) · σ = −σxxi − σxyj − σxzk = −5i =⇒

σxx = −t(−i)x= 5

σxy = 0σxz = 0

Example 3-5


x( ) 5t − = − i i ( ) 5t =i ii-i

x( )σ xx x= it( )σ xx x = − it

Figure 3.32:

( ) 3t = j j j

-j( ) 3t − =j j

y

x

( )σ yy y− = jt

( )σ yy y= jt

Figure 3.33:

Uniaxial Compression in the y-direction (units are in MPa)On the y face: n = j

t(j) = j · σ = σyxi + σyyj + σyzk = −3j =⇒

σyx = 0σyy = t(j)y

= 5σyz = 0

On the −y face: n = −j

t(−j) = (−j) · σ = −σyxi − σyyj − σyzk = −5j =⇒

σyx = 0σyy = −t(−j)y

= 5σyz = 0

Example 3-6

Pure uniform shear in the xy plane (units are in MPa).On the x face: n = i

t(i) = i · σ = σxxi + σxyj + σxzk =⇒

σxx = 0t(i) = −3j =⇒ σxy = −3

σxz = 0

On the −x face: n = −i

t(−i) = (−i) · σ = −σxxi − σxyj − σxzk =⇒

σxx = 0t(−i) = 3j =⇒ σxy = −3

σxz = 0


y

x-j

j

i

-i

t(−j) = 3i

t(j) = −3i

t(i) = −3jt(−i) = 3j

σxy = −t(−i)yσxy = t(i)y

σyx = t(j)x

σyx = −t(−j)x

Figure 3.34:

On the y face: n = j

t(j) = j · σ = σyxi + σyyj + σyzk =⇒

t(j) = −3i =⇒ σyx = −3σyy = 0σyz = 0

On the −y face: n = −j

t(−j) = (−j) · σ = −σyxi − σyyj − σyzk =⇒

t(−j) = 3i =⇒ σyx = −3σyy = 0σyz = 0

Note that on the negative x and y faces, the components of stress are equal to the negativecomponents of the traction vector, as the Figure above indicates.

The following examples should help illustrate the importance of Cauchy’s formula and the sig-nificance of the stress tensor.

Example 3-7

Given: The following stress tensor in Cartesian coordinates.

[σ] =

5 −2 0

−2 −1 00 0 −3

MPa

Required :

a) Sketch the stress cube representation

b) Determine the traction vectors: t(i), t(j), t(k)

c) Make a sketch of t(i), t(j), t(k) acting on the appropriate plane(s)

Solution

b)[t(n)

]= [n] [σ] [i] =

[1 0 0

], [j] =

[0 1 0

], [k] =

[0 0 1

]


5

-1

3

2

x

z

y

1

5

2

a)

Figure 3.35:

[t(i)

]=

[1 0 0

] 5 −2 0

−2 −1 00 0 −3

= 5i − 2j MPa

[t(j)

]=

[0 1 0

] 5 −2 0

−2 −1 00 0 −3

= −2i − j MPa

[t(k)

]=

[0 0 1

] 5 −2 0

−2 −1 00 0 −3

= −3k MPa

tj

tk

tix

z

y

c)

Figure 3.36:

Example 3-8

If the components of a stress tensor in a rectangular Cartesian coordinate system (RCCS) are

[σ] =

2 −2 0

−2 −5 00 0 3

MPa

then find the traction vector acting on a plane whose unit outward normal is

n = −13i +

23j +

23k


Solution

t(n) = n · [σ]

=(−1

3i +

23j +

23k)·

2 −2 0

−2 −5 00 0 3

=[−1

3(2) +

23(−2)

]i +

[−1

3(−2) +

23(−5)

]j +

[23(3)

]k

=(−2 − 4

3

)i +

(2 − 10

3

)j +

63k

= −63i − 8

3j +

63k MPa

This next example is simply linear algebra, but should help orient the components of the tractionon different planes. Also, later we will discuss how materials yield and this will become extremelyimportant.

Example 3-9

For the traction vector in Example 3-8, determine

a) its magnitude,

b) its component parallel to the unit normal (i.e., normal to plane),

c) its component perpendicular to the normal (i.e., parallel to the plane) (hint: the tractionvector is resolved into a vector normal to the plane and a vector parallel to the plane),

d) the angle between t(n) and n.

SolutionFrom Example 3-8, n = − 1

3 i + 23 j + 2

3k and t(n) = − 63 i − 8

3 j + 63k.

a)

magnitude of t(n) =

[(−6

3

)2

+(−8

3

)+

(63

)2] 1

2

=[36 + 64 + 36

9

] 12

=

√1369

=2√

343

MPa

b)

t‖n = (t(n) · n)n

=(−2i − 8

3j + 2k

)·(−1

3i +

23j +

23k)

n

=(

69− 16

9+

129

)n

=29n

=29

(−1

3i +

23j +

23k)

MPa


c)

t⊥n = t(n) − t‖n

=(−2i − 8

3j + 2k

)− 2

9

(−1

3i +

23j +

23k)

=(−6

3+

227

)i +

(−8

3− 4

27

)j +

(63− 4

27

)k

=(−54 + 2

27

)i +

(−72 − 427

)j +

(54 − 4

27

)k

= −5227

i − 7627

j +5027

k MPa

d)

t(n) · n =∣∣t(n)

∣∣ |n| cos θ =⇒ θ = cos−1

[t(n) · n∣∣t(n)

∣∣ |n|]

= cos−1

2

9(2√

343

)(1)

= cos−1

(1

3√

34

)= 86.7

3.4 Conservation of Linear Momentum in Terms of Stresses

Using the definition of the traction vector in terms of Cauchy stress, equation (3.32) or equation(3.38), the conservation of linear momentum equation, (3.17), reduces to

ρ∂v∂t

+ ρ (v · ∇)v = ρg +(i

∂

∂x+ j

∂

∂y+ k

∂

∂z

)· σ (3.39)

or

Conservation of Linear Momentum in terms of Cauchy Stresses

ρ∂v∂t

+ ρ (v · ∇)v = ρg + ∇ · σ (3.40)

In matrix representation, the above conservation equations of linear momentum can be written inthe following form:

[ρ∂vx

∂t ρ∂vy

∂t ρ∂vz

∂t

]+ ρ

(vx

∂

∂x+ vy

∂

∂y+ vz

∂

∂z

) [vx vy vz

]

=[

ρgx ρgy ρgz

]+

[∂∂x

∂∂y

∂∂z

] σxx σxy σxz

σyx σyy σyz

σzx σzy σzz

(3.41)

The vector form of the conservation of linear momentum equation gives three component equations.The components of equation (3.41) are:

Conservation of Linear Momentum in terms of Cauchy Stresses

3.4. CONSERVATION OF LINEAR MOMENTUM IN TERMS OF STRESSES 75

x-component: ρ

[∂vx

∂t+ vx

∂vx

∂x+ vy

∂vx

∂y+ vz

∂vx

∂z

]= ρgx +

∂σxx

∂x+

∂σyx

∂y+

∂σzx

∂z

y-component: ρ

[∂vy

∂t+ vx

∂vy

∂x+ vy

∂vy

∂y+ vz

∂vy

∂z

]= ρgy +

∂σxy

∂x+

∂σyy

∂y+

∂σzy

∂z(3.42)

z-component: ρ

[∂vz

∂t+ vx

∂vz

∂x+ vy

∂vz

∂y+ vz

∂vz

∂z

]= ρgz +

∂σxz

∂x+

∂σyz

∂y+

∂σzz

∂z

Note that the number of dependant variables in the above partial differential equations is still 16(ρ, vx, vy, vz, gx, gy, gz, σxx, σxy, σxz, σyx, σyy, σyz, σzx, σzy, σzz) all of which may be functionsof x, y, z and t. With respect to the previous form of Conservation of Linear Momentum equations(3.16) the components of the three traction vectors t(i), t(j) and t(k) have been replaced by the ninecomponents of the Cauchy stress tensor. In many cases, particularly fluid applications, it becomesuseful to consider a state of stress consisting of only normal stresses (i.e., all shear stresses are zero).

The stress tensor reduces to [σ] =

σxx 0 0

0 σyy 00 0 σzz

. For a fluid, the hydrostatic stress state is

one where fluid pressure P is the same in all directions and thus the normal stresses are equal to

−P . The hydrostatic state would correspond to a stress tensor given by [σ] =

−P 0 0

0 −P 00 0 −P

.

The negative sign is due to hydrostatic pressure acting inward on a body (causing compression).Equating the trace (sum of the diagonal terms) of the two stress tensors gives σxx+σyy +σzz = −3P .Consequently we define the hydrostatic pressure P as

P = hydrostatic pressure = − (σxx + σyy + σzz)3

(3.43)

In the analysis of fluids and inelastic solids, it is useful to consider a stress tensor where the averagehydrostatic pressure is removed or subtracted from the Cauchy stress tensor. The motivation forthis is that hydrostatic stress does not cause motion of a fluid element and generally does not causefailure of a solid. Such a stress definition is called the deviatoric stress tensor S and is definedby

S = deviatoric stress tensor = σ − 13

(σxx + σyy + σzz) I = σ + P I (3.44)

where I is the identity tensor (it is represented by the identity matrix, [I] =

1 0 0

0 1 00 0 1

. In

expanded form, this becomes Sxx Sxy Sxz

Syx Syy Syz

Szx Szy Szz

=

(σxx + P ) σxy σxz

σyx (σyy + P ) σyz

σzx σzy (σzz + P )

(3.45)

The Cauchy stress tensor can now be written in terms of the deviatoric stress tensor as

σ = S − P I (3.46)

Utilizing the last result, conservation of linear momentum can now be written in terms of deviatoricstress as:

Conservation of Linear Momentum in terms of Deviatoric Stress and Pressure

ρ∂v∂t

+ ρ (v · ∇)v = ρg + ∇ · S −∇P (3.47)

The last result can be expanded into component form to obtain:


Conservation of Linear Momentum in terms of Deviatoric Stress and Pressure

x-component: ρ

[∂vx

∂t+ vx

∂vx

∂x+ vy

∂vx

∂y+ vz

∂vx

∂z

]= ρgx +

∂Sxx

∂x+

∂Syx

∂y+

∂Szx

∂z− ∂P

∂x

y-component: ρ

[∂vy

∂t+ vx

∂vy

∂x+ vy

∂vy

∂y+ vz

∂vy

∂z

]= ρgy +

∂Sxy

∂x+

∂Syy

∂y+

∂Szy

∂z− ∂P

∂y(3.48)

z-component: ρ

[∂vz

∂t+ vx

∂vz

∂x+ vy

∂vz

∂y+ vz

∂vz

∂z

]= ρgz +

∂Sxz

∂x+

∂Syz

∂y+

∂Szz

∂z− ∂P

∂z

Note that the nine components of σ have been replaced by the eight independent components of S(because Sxx + Syy + Szz = 0) and the pressure P . It is useful to view the above linear momentumequations in the following manner. For any physical problem, there will be a set of inputs thatcauses outputs. In the above, we can think of the velocity components as outputs and the gravity,deviatoric stress and pressure gradients as inputs. Notice that inputs such as a pressure gradient ∂P

∂xor gravity component gx will cause the fluid to move and produce a velocity such as vx (the output).

We finally note that the conservation of linear momentum equation involves velocity components.Because velocities describe the motion of the continuum, the conservation of linear momentumequations are frequently referred to as the “equations of motion.”

3.4.1 Applications in Fluids

At this point we would like to introduce some “solvable” problems, but in order to do this, we mustpresent the concept of material behavior. That is the essence of this section and it will allow us toanalyze some interesting problems in the examples to follow.

Constitutive Equations for Incompressible Newtonian FluidsFor fluids we define the rate of strain in terms of the gradients of velocity. Therefore, we define

the deformation rate tensor, D, by:

D =12

[∇v + ∇vT

](3.49)

Experimentally is has been found that many fluids develop stresses proportional to the deformationrate, with the constants of proportionality independent of position and the deformation rate atconstant temperature. For isotropic fluids, which is true for most cases with fluids, the deviatoricstress, S, will depend on only two material parameters λ and µ, i.e.

S = λtr(D)I + 2µD (3.50)

But since incompressibility (from the COM) requires that tr(D) = ∇v = 0 the final form of theconstitutive equation for incompressible Newtonian fluids reduces to

S = 2µD (3.51)

or, in component form:

Sxx = 2µ∂vx

∂xSxy = Syx = µ

(∂vx

∂y+

∂vy

∂x

)

Syy = 2µ∂vy

∂ySyz = Szy = µ

(∂vy

∂z+

∂vz

∂y

)(3.52)

Szz = 2µ∂vz

∂zSzx = Sxz = µ

(∂vz

∂x+

∂vx

∂z

)

Consequently, for an incompressible Newtonian fluid, only one material parameter (µ) is requiredto describe it’s constitutive behavior.


In order to evaluate the material property µ, one can design many experiments. Consider the“Couette flow” example (discussed below) for a fluid flowing in the x-z plane between two parallelplates (walls) such that the only non-zero velocity component is vz. We assume that the shear stressin the x-z plane is related to the velocity gradient in the x direction by the viscosity coefficient µ:

Sxz = Szx = µ∂vz

∂x(3.53)

This last assumption is called a constitutive equation that must be verified by experiment. In theexperiment, we measure the shear force on the wall (to obtain Sxz) and the fluid velocity gradient(∂vz

∂x ). From experiment, we observe that the traction (shear stress) on a wall is proportional to thefluid velocity gradient normal to wall (∂vz

∂x ) as shown in the following Figure:

viscosity

coefficientslopeµ = =

Sxz = µ∂vz

∂x∂vz

∂x

Sxz

Figure 3.37:

The material property µ is called the viscosity coefficient for an incompressible Newtonian fluid.The development of constitutive equations for a solid will be discussed in Chapter 9. For the timebeing, we simply state that it is necessary to establish this material relationship in order to completethe solution to the following examples.

The conservation of linear momentum equation (3.47) can be further reduced after substitutionof the constitutive equations into the following form:

ρ

[∂v∂t

+ (v · ∇)v]

= ρg + ∇ · (2µD) −∇P

= ρg + µ[∇ ·

(∇v + ∇vT

)]−∇P (3.54)

= ρg + ∇ · (2µD) −∇P

The above equations are called the Navier-Stokes equations for incompressible Newtonian fluids.There are three equations in components along x, y, z directions in Cartesian coordinates, given by:

ρ

[∂vx

∂t+ vx

∂vx

∂x+ vy

∂vx

∂y+ vz

∂vx

∂z

]= ρgx + µ

∂2vx

∂x2− ∂P

∂x

ρ

[∂vy

∂t+ vx

∂vy

∂x+ vy

∂vy

∂y+ vz

∂vy

∂z

]= ρgy + µ

∂2vy

∂y2− ∂P

∂y(3.55)

ρ

[∂vz

∂t+ vx

∂vz

∂x+ vy

∂vz

∂y+ vz

∂vz

∂z

]= ρgz + µ

∂2vz

∂z2− ∂P

∂z

The above three equations are supplemented by the COM equation for incompressible fluids whichis given in component form by:

∂vx

∂x+

∂vy

∂y+

∂vz

∂z= 0 (3.56)


Equations (3.55) and (3.56) are four partial differential equations for the unknown functions vx( x, y, z, t ),vy( x, y, z, t ), vz( x, y, z, t ), and P (x, y, z, t ) assuming that gx, gy, gz, µ, and ρ are known. Note thatsince ρ is assumed to be constant and known, the pressure P has become an unknown that must bedetermined.

Example 3-10Couette Flow

Couette flow is the case of fluid flow between two parallel plates separated by a distance d, wherethe bottom plate is stationary and the top plate moves horizontally at a velocity uo. There is nopressure differential (gradient) along the length of the plate. The driving force is the movement ofthe top plate. Assume the flow is steady and incompressible.

z

x

L

d

uo

flow

(wall velocity)

Figure 3.38:

Assume the flow is steady state and incompressible:

steady state =⇒ ∂

∂t= 0

incompressible =⇒ ρ = constant, or 0 = ∇ · v (see equation 2.18)

and that body forces are zero:

body force (gravity) is zero =⇒ gx = gy = gz = 0

From the statement of the problem, we assume that there is no fluid flow normal to the plate (xdirection) or in the y direction. Consequently, the kinematic boundary conditions are written:

no flow in x or y direction =⇒ vx = vy = 0

a) Conservation of mass (continuity equation) is given by

∂ρ

∂t= −

[∂(ρvx)

∂x+

∂(ρvy)∂y

+∂(ρvz)

∂z

]

and reduces to

∂(ρvz)∂z

= 0 or∂vz

∂z= 0 =⇒ vz = f(x, y ) + C1

For plane motion, the fluid velocity vz is assumed to be independent of the y direction, i.e.,vz = vz(x). Thus, the final solution for vz from conservation of mass is given by:

vz = vz(x) (3.57)


b) Conservation of linear momentum equations are given by:

ρ

(∂vx

∂t+ vx

∂vx

∂x+ vy

∂vx

∂y+ vz

∂vx

∂z

)= ρgx +

∂Sxx

∂x+

∂Syx

∂y+

∂Szx

∂z− ∂P

∂x

ρ

(∂vy

∂t+ vx

∂vy

∂x+ vy

∂vy

∂y+ vz

∂vy

∂z

)= ρgy +

∂Sxy

∂x+

∂Syy

∂y+

∂Szy

∂z− ∂P

∂y

ρ

(∂vz

∂t+ vx

∂vz

∂x+ vy

∂vz

∂y+ vz

∂vz

∂z

)= ρgz +

∂Sxz

∂x+

∂Syz

∂y+

∂Szz

∂z− ∂P

∂z

and reduce to (recall the assumption of zero pressure gradients):

x-component of linear momentum: 0 =∂Sxx

∂x+

∂Syx

∂y+

∂Szx

∂z

y-component of linear momentum: 0 =∂Sxy

∂x+

∂Syy

∂y+

∂Szy

∂z(3.58)

z-component of linear momentum: 0 =∂Sxz

∂x+

∂Syz

∂y+

∂Szz

∂z

c) At this point conservation of mass and linear momentum has been reduced to four equations:(3.57) and (3.58) with 7 unknowns: vz, Sxx, Syy, Szz, Sxy = Syx, Syz = Szy and Sxz =Szx. Obviously, we can not obtain a solution unless we make further assumptions or haveadditional equations. Assume the flow produces no normal stresses (a good assumption basedon experimental observation). Since flow is in the x-z plane, there is no shear stress in the x-yor y-z plane (shear stress only in x-z plane). Hence, we write

Sxx = Syy = Szz = Sxy = Syx = Syz = Szy = 0

We assume that the shear stress in the x-z plane is related to the velocity gradient in the xdirection by the viscosity coefficient µ:

Sxz = Szx = µ∂vz

∂x(3.59)

d) With these assumptions on stress, the 3 linear momentum equations reduce to:

x-component: 0 = 0y-component: 0 = 0 (3.60)

z-component: 0 = µ∂2vz

∂x2

Equation (3.60) can now be solved for the velocity. We assume a “no slip” boundary conditionfor the fluid at each wall. Since the lower wall is stationary, then vz(x = 0) = 0. The upperwall moves at a velocity uo so that vz(x = d) = uo. Integrating the third momentum equation,and assuming µ is a constant, we obtain vz = C1x+C2. Substituting boundary conditions forvelocity yields

vz(x) = uo

(x

d

)(3.61)

Thus, the velocity profile is a straight line. The shear stress is given by

Sxz = µ∂vz

∂x= µ

(uo

d

)(3.62)

We note that the shear stress is constant from bottom to top. Thus, as uo increases, theshear stress increases. As d increases, the shear stress decreases. At the wall, the shear stress(traction) acting on the wall must be equal and opposite to the shear stress (traction) actingon the fluid!


z

x

L

d

uo

( ) xz o dv x u=

(wall velocity)

flow

Figure 3.39:

Some real world examples of Couette flow:

1 Wing moving through calm air at speed uo. At some distance far away from the wing (normalto direction wing is moving), the air is motionless - think of this point as a fixed boundarywhere the fluid velocity is zero. At the surface of the wing, the fluid velocity is uo if we assumea no-slip condition - think of this as the moving boundary. So, the flow looks just like Couetteflow.

2 Piston moving up and down in the cylinder of an engine. Between the piston and cylinder wallis lubrication oil with a thickness of d. The cylinder wall is the fixed boundary and the pistonwall is the moving boundary.

Some viscosity coefficient values:

Air at standard sea level conditions: µ = 1.79 × 10−5 kgm sec

Water: µ = 1.005 × 10−3 kgm sec at 20 C

Motor oil: µ = 1.07 kgm sec at 20 C

Note: 1 centipoise = 10−3 kgm sec = 6.72 × 10−4 lbm

ft sec

Example 3-11Poiseulle Flow

Poiseulle flow is the case of fluid flow between two fixed parallel plates separated by a distanced with a pressure gradient in the z direction. The driving force is the pressure differential from leftto right. Assume the flow is steady and incompressible.

Assume the flow is steady state and incompressible:

steady state =⇒ ∂

∂t= 0

incompressible =⇒ ρ = constant, or 0 = ∇ · v (see equation 2.18)

and that the only non-zero body force is in the x direction:

body force (gravity) is zero =⇒ gx = gy = gz = 0

From the statement of the problem, we assume that there is no fluid flow normal to the plate (xdirection) or in the y direction. Consequently, the kinematic boundary conditions are written:

no flow in x or y direction =⇒ vx = vy = 0


z

x

L

d flow

Figure 3.40:

a) Conservation of mass (continuity equation) is given by

∂ρ

∂t= −

[∂(ρvx)

∂x+

∂(ρvy)∂y

+∂(ρvz)

∂z

]

and reduces to∂(ρvz)

∂z= 0 or

∂vz

∂z= 0 =⇒ vz = f(x, y ) + C1

For plane motion, the fluid velocity vz is assumed to be independent of the y direction, i.e.,vz = vz(y). Thus, the final solution for vz from conservation of mass is given by:

vz = vz(x) (3.63)

b) Conservation of linear momentum equations are given by:

ρ

(∂vx

∂t+ vx

∂vx

∂x+ vy

∂vx

∂y+ vz

∂vx

∂z

)= ρgx +

∂Sxx

∂x+

∂Syx

∂y+

∂Szx

∂z− ∂P

∂x

ρ

(∂vy

∂t+ vx

∂vy

∂x+ vy

∂vy

∂y+ vz

∂vy

∂z

)= ρgy +

∂Sxy

∂x+

∂Syy

∂y+

∂Szy

∂z− ∂P

∂y

ρ

(∂vz

∂t+ vx

∂vz

∂x+ vy

∂vz

∂y+ vz

∂vz

∂z

)= ρgz +

∂Sxz

∂x+

∂Syz

∂y+

∂Szz

∂z− ∂P

∂z

and reduce to:

x-component of linear momentum:∂P

∂x= ρgx +

∂Sxx

∂x+

∂Syx

∂y+

∂Szx

∂z

y-component of linear momentum:∂P

∂y=

∂Sxy

∂x+

∂Syy

∂y+

∂Szy

∂z(3.64)

z-component of linear momentum:∂P

∂x=

∂Sxz

∂x+

∂Syz

∂y+

∂Szz

∂z

c) At this point conservation of mass and linear momentum has been reduced to four equations:(3.63) and (3.64) with ten unknowns: vz, P , ρ, gx, Sxx, Syy, Szz, Sxy = Syx, Syz = Szy andSxz = Szx. Obviously, we can not obtain a solution unless we make further assumptions orhave additional equations. We will assume that the body force gx and density ρ are specified(known). Assume the flow produces no normal stresses (a good assumption based on experi-mental observation). Since flow is in the x-z plane, there is no shear stress in the x-y or y-zplane (shear stress only in x-z plane). Hence, we write

Sxx = Syy = Szz = Sxy = Syx = Syz = Szy = 0


We assume that the shear stress in the x-z plane is related to the velocity gradient in the xdirection by the viscosity coefficient µ:

Sxz = Szx = µ∂vz

∂x

d) With the assumption on stress, the 3 linear momentum equations reduce to:

x-component:∂P

∂x= ρgx

y-component:∂P

∂y= 0 (3.65)

z-component:∂P

∂z= µ

∂2vz

∂x2

Equations (3.65) can now be solved for the pressure P and velocity vz. We start with thethird linear momentum equation. We assume a “no slip” Boundary Condition for the fluidat each wall. Since both walls are stationary, then vz(x = 0) = 0 and vz(x = d) = 0. Sincethe problem stated that the pressure gradient in the z direction was the driving force, we takethe pressure gradient in the z direction dP

dz to be a given value (a specified input). Integratingthe third (z) momentum equation with respect to z and assuming µ is a constant, we obtainvz =

(12µ

dPdz

)x2 + C1x + C2. Substituting boundary conditions for the velocity yields:

vz(x) =12µ

dP

dz(x2 − xd) =

12µ

dP

dzx(x − d) (3.66)

z

xflow

L

dvz (x) =

1

2µ

dP

dzx (x − d)

Figure 3.41:

Thus, the velocity profile for vz is a quadratic in x. The shear stress is given by

Sxz(x) = µdvz

dx=

dP

dz

(x − d

2

)(3.67)

which is a maximum at either wall, and is zero at the center. Interestingly, the shear stressdoes NOT depend on the viscosity coefficient, µ, for this particular problem.

A photograph of velocity profiles of fluid starting from rest and flowing from left to right isshown below. The left vertical line is the fluid at rest. As the fluid flows to the right andbecomes steady-state, the velocity profiles become parabolic in shape (right-most profiles).Note that the parabolic profiles elongate indicating that the fluid is “sticking” to the wall dueto viscosity effects.

Example 3-12Flow in a Tube


Figure 3.42:

Using cylindrical coordinates, one can easily describe the flow in a tube with only one dimension.As seen below, the only component of velocity which is non-zero would be the flow along the axisof the tube.

COM

for ρ = constant =⇒ ∂ρ

∂t+ ∇ · (ρv) = 0 =⇒ ∇ · v = 0

or1r

∂(rvr)∂r

+1r

∂vθ

∂θ+

∂vz

∂z= 0 =⇒ ∂vz

∂z= 0 =⇒ vz = vz(r)

z

r

− dP

dz

R2

4µ v vr = =θ 0

Figure 3.43:

COLM

=⇒ 0 = −∂P∂r

0 = − 1r

∂P∂θ

=⇒ P = P (z)

0 = −∂P

∂z+ µ

d2vz

dr2+ µ

1r

dvz

dr=⇒ µ

(d2vz

dr2+

1r

dvz

dr

)=

dP

dz= constant

=⇒ vz =dP

dz

1µ

r2

4+ C1 ln r + C2

Boundary Conditions

C1 = 0 for finite vz for r = 0

C2 = −dP

dz

1µ

R2

4for vz|r=R = 0

∴ vz =dP

dz

1µ

R2

4

[r2

R2− 1

]vmax = −dP

dz

R2

4µ

Srz = µdvz

dr= µ

dP

dz

r

2µ=

dP

dz

r

2


Flow rate: Q =∫A

vz dA =∫ π

0vz2πr dr = −dP

dzπR4

8µ vavg = QA = vmax

2

1-D steady state incompressible flow in a cylindrical annulus (Annular Flow)

The flow is now constrained by the surfaces of two concentric cylinders. Again using cylindricalcoordinates, it has the same functional form for the velocity profile as the flow in a tube, but withdifferent values for the constants.

R1

R2

Figure 3.44:

vz =dP

dz

1µ

r2

4+ C1 ln r + C2

vz(R1) = 0 =dP

dz

1µ

R21

4+ C1 lnR1 + C2

vz(R2) = 0 =dP

dz

1µ

R22

4+ C1 lnR2 + C2

C1 = −dP

dz

1µ

R22 − R2

1

41

ln(

R2R1

)C2 = −dP

dz

1µ

R21

4− C1 lnR1

= −dP

dz

1µ

R21

4+

dP

dz

1µ

R22 − R2

1

4lnR1

ln(

R2R1

)

vz =dP

dz

1µ

r2

4− dP

dz

1µ

R22 − R2

1

4ln r

ln(

R2R1

) − dP

dz

1µ

R21

4+

dP

dz

1µ

R22 − R2

1

4lnR1

ln(

R2R1

)

=dP

dz

1µ

r2 − R2

1

4− R2

2 − R21

4

ln(

rR1

)ln

(R2R1

)

vz = −dP

dz

R22

4µ

(1 −

(R1

R2

)2)

ln(

rR1

)ln

(R2R1

) −(

r

R2

)2

+(

R1

R2

)2

vz = −dP

dz

R21

4µ

1 −

(1 −

(R2

2

R21

− 1)) ln

(r

R1

)ln

(R2R1

) −(

r

R1

)2

+ 1

vz = −dP

dz

R21

4µ

[1 −

(r

R1

)2

+K2 − 1lnK

ln(

r

R1

)]


K =R2

R1

Example 3-13Couette Viscometer

This final example is actually one way viscosity is measured experimentally. In a couette vis-cometer fluid is placed between the outside wall of a cup and the sides of a bob. A constant angularvelocity is then applied to the bob. This constant velocity and the resulting, measured torque canbe used to determine the viscosity of the fluid.

Ri

RoΩ

y

z

d

Figure 3.45:

For plane couette flow with angular velocity, Ω, C.L.M. yields

dP

dz= 0

= µ

d2vz

dy2=⇒ vz(y) = C1y + C2

vz(0) = 0 =⇒ C2 = 0vz(d) = ΩRo = C1(Ro − Ri)

∴ vz =ΩyRo

Ro − Ri= Ω

r − Ri

Ro − RiRo

y = r − Ri

vz = ΩRo∆r

∆R

Syz = µdvz

dy= µC1 =

ΩRo

∆Rµ

M = 2πR2i Lµ

ΩRo

∆R

µ =M(∆R)

2πR2i LΩRo

∴ a measurement of M (torque) yields viscosity, µ.


Ω L

M

Figure 3.46:

3.5 Closing Notes on Stresses and Conservation of LinearMomentum

We note that equations (3.42) in terms of Cauchy stress, or equations (3.48) in terms of deviatoricstress, form a set of coupled partial differential equations which are difficult to solve. Later in thischapter some simple 1-D examples are considered where the equations can be reduced to a singleordinary differential equation and solved for the velocity distribution.

For the static case, equation (3.42) reduces to:

Conservation of Linear Momentum for a Static Case

0 = ρgx +∂σxx

∂x+

∂σyx

∂y+

∂σzx

∂z

0 = ρgy +∂σxy

∂x+

∂σyy

∂y+

∂σzy

∂z(3.68)

0 = ρgz +∂σxz

∂x+

∂σyz

∂y+

∂σzz

∂z

For a solid, these equations are often referred to as the equilibrium equations since they providea statement of equilibrium that the internal forces body (which produce internal stresses) withina continuum must satisfy. We note again that equation (3.68) provides 3 equilibrium equationswhich are in terms of nine unknown stress components σxx, σyy, σzz, σxy, σyx, σxz, σzx, σyz,and σzy. Consequently, the differential equations of equilibrium alone cannot be used to solve forthe unknown stresses; in fact, additional equations will be required to extract a solution. Theseadditional equations are the kinematic equations and constitutive equations. They will be studiedin detail in later chapters. Note that for a 1-D case (x-component only) with no body forces,conservation of linear momentum reduces to 0 = ∂σxx

∂x (actually an ordinary differential equation0 = dσxx

dx ). Integration gives σxx = constant which is precisely the case of a truss member whereinthe axial force, traction and normal stress are constant in a member!

In this text, we have not considered conservation of linear momentum in any coordinate sys-tem except Cartesian coordinates. For cylindrical coordinates (or any other coordinate system) thederivation is easily obtained in vector notation by using the del operator in the appropriate coor-dinate system. Figure 3.47 shows a sketch of a differential element in cylindrical coordinates withthe stresses applied. Also shown on Figure 3.47 are the stress components in 2-D and 3-D Cartesiancoordinates. Note the similarity in labeling the stresses (the subscripts) in the two different coordi-nate systems. The notation for stresses can be easily remembered by noting the following: the firstsubscript refers to the face or direction of the outward unit normal of the plane the stress acts on,and the second subscript refers to the direction of the stress.

3.5. CLOSING NOTES ON STRESSES AND CONSERVATION OF LINEAR MOMENTUM 87

Stress Components in Cartesian Coordinates

y

x

z

Stress Components in Cylindrical Coordinates

x

y

z

r

r∆

z∆

∆θ

θ σrθ = σθr

σrz = σzr

σzθ = σθz

σrr

σθθσzz

σxy

σyx

σxx

σyy

σθr

σθz

σrz σrθ

σzθ

σzr

σxx

σxy

σyy

σyx

σxy = σyx

σxz = σzx

σyz = σzy

σyy

σyy

σxx

σxx

σzz

σzz

σxy

σxy

σxz

σxz

σyz

σyz

σyx

σyx

σzx

σzx

σzy σzy

Figure 3.47: Stress Components in Cartesian and Cylindrical Coordinates


Deep Thought

I discovered today that conservation of linear momentum is not everything there is in life ;

there is also stress.

3.6. QUESTIONS 89

3.6 Questions

3.1 In your own words, describe the meaning of the conservation of linear momentum.

3.2 Explain the physical significance (not the mathematical definition) of each term in the Con-servation of Linear Momentum Equation. Refer to the global terms of the equation (e.g. ρgaccounts for the body forces due to gravity acting on the volume element). Write the equationin Cartesian coordinates.

3.3 State the reduced expression of the Conservation of Linear Momentum Equation for a staticsystem. Explain every simplification that you make.

3.4 (a) Write the conservation of linear momentum in vector form.(b) Write the equation of equilibrium in the x-direction.

3.5 Is a traction vector that is associated with a particular surface normal to, parallel to, or at anangle to that surface?

3.6 The stress tensor [σ] can be a function of position, and describes the state of stress at a specificpoint in a material. How is the stress tensor related to the traction vector?

3.7 Considering only the direction in which gravity acts, what is the 1-D equation of motion for astatic, incompressible fluid?

3.8 In your own words, describe the meaning of the terms body force and contact force. Give twoexamples of each.

3.9 Explain in your own words how the momentum of a volume element may be changed if thevolume element is fixed in an inertial (non-accelerating) coordinate system.

3.10 Write the vector form of the equation of continuity for an incompressible fluid. Write theequation using cylindrical coordinates.

3.11 In terms of the extra stress [S], write the vector form of the equation of motion (i.e. conser-vation of linear momentum) for steady-state fluid flow. Write it in Cartesian coordinates.

3.7 Problems

3.12 Derive the Conservation of Linear Momentum in 2-D.

3.13 GIVEN : A fluid flowing in a pipe with radius b is in steady state with negligible body force.

REQUIRED :

(a) Write out the conservation of mass and linear momentum in as simplified a form aspossible, accounting for symmetry and physics.

(b) Enumerate the unknowns and what they depend on (independent variables).(c) Determine the number of additional equations required to obtain a solution, and guess

what these equations might be.(d) Construct a set of admissible boundary conditions.

3.14 For the steady state, 1-D flow of an incompressible plane Couette flow, we find the followingequations that satisfy both the conservation of mass and the conservation of linear momentum:

∂P

∂x= c1,

∂P

∂y= 0,

∂P

∂z= c2,

Solve these equations for P , the pressure distribution in the fluid.


3.15 The state of stress at a certain point of a body is given by

[σ] =

2 −1 3

−1 4 03 0 −1

MPa

(a) Find the traction vector on the plane whose normal is in the direction

2i + 2j + k

(b) Find the traction vector on the plane whose normal is in the direction

i − 3j + 2k

3.16 If the components of a stress tensor [σ] in a rectangular Cartesian coordinate system are

[σ] =

3 −1 0

−1 −8 00 0 5

MPa

then find the traction vector acting on a plane whose unit outward normal is

n =23i − 1

3j +

53k

3.17 For the traction vector in 3.14, determine

(a) its magnitude,

(b) its component normal to the plane

(c) its component parallel to the plane (hint: the traction vector is resolved into a vectornormal to the plane and a vector parallel to the plane),

(d) the angle between t(n) and n.

3.18 GIVEN : The following stress tensor in Cartesian coordinates

[σ] =

7 −3 0

−3 −2 00 0 −4

ksi

REQUIRED :

(a) Sketch the stress cube representation

(b) Determine the traction vectors t(i), t(j), t(k):

(c) Make a sketch of t(i), t(j), t(k) acting on the appropriate plane(s)

3.19 GIVEN : The following stress tensor in Cartesian coordinates,

[σ] =

2 −1 0

−1 3 40 4 0

ksi

REQUIRED :

(a) Find the traction vector t(n) acting on a plane given by the following outward unit normal:

n = − 1√3i +

1√3j +

1√3k

3.7. PROBLEMS 91

(b) Determine∣∣t(n)

∣∣(c) Determine the components of t(n) parallel and normal to the surface given by n.

(d) Determine the angle between t(n) and n.

3.20 For (a) and (b), find the traction vector acting on a plane whose unit outward normal is givenby n if the components of the stress tensor are given by [σ].

(a) [σ] =

2 0 −5

0 3 0−5 0 1

MPa n = 2

3 i + 23 j − 1

3k

(b) [σ] =

1 5 1

5 4 01 0 −2

ksi n = − 3

5 i + 45 j

3.21 For the traction vector in problem 3.18(a), determine

(a) its component normal to the plane;

(b) its component parallel to the plane;

(c) its magnitude and the angle between t(n) and n.

3.22 For a rectangular Cartesian coordinate system, the stress tensor [σ] is given by

[σ] =

σxx σxy σxz

σyx σyy σyz

σzx σzy σzz

MPa

In terms of the components of this tensor (σxx, σxy, etc.), what is the traction vector actingon a surface that has unit outward normal given below? For each case, also sketch the planewith the specified unit normal n and show the components of t(n) (in terms of σxx, σxy, etc.)that act on the plane.

(a) i

(b) j

(c) k

(d) (i+j)√2

3.23 Consider this 2-D stress tensor in the x-y plane:

[σ] =

3 2 0

2 −8 00 0 0

MPa

(a) Draw a stress block in the x-y plane, and place the components of stress on each face.

(b) Using the summation of forces, determine the traction vector t(n) on a plane

having a unit normal vector with a 30 rotation.

(c) Resolve the traction into a normal and a shear component of the face described by theunit normal.

(d) Repeat b) and c) for a unit normal rotated 90 counter-clockwise from the given normal.

(e) How do these computed traction components fit into the new stress tensor?


3.24 In the absence of body forces, do the stress components given by

σxx = a[y2 + b(x2 − y2)], σyy = a[x2 + b(y2 − x2)], σyy = ab(x2 + y2)

σxy = −2abxy, σxz = 0, σyz = 0

satisfy the equations of equilibrium (conservation of linear momentum equations with zerovelocity components)? In the above expressions, a and b are constants.

3.25 Solve for the values of a and b in the stress components given by

σxx = a[y2 + b(x2 − y2)] + 5(b − 1)x, σyy = a[x2 + b(y2 − x2)] − 7(a + 3)y, σyy = ab(x2 + y2)

σxy = −2abxy + 2bx − ay, σxz = 0, σyz = 0

3.26 Using Cauchy’s Formula t(n) = n · σ, for the following stress tensor

[σ] =

10 5 0

5 0 00 0 1

MPa

n = cos(θ)i + sin(θ)j + 0k

find:

(a) The plane at which the shear component of the traction vector equals zero (i.e., find theangle θ).

(b) The normal component of the traction vector found in part (a). What is this normalcomponent of the traction vector called?

(c) What is the angle between the maximum shear stress plane and the principal stress plane?

(d) What is the value of the maximum shear stress for the stress tensor given above?

Hints:

– Find the components of t(n) that are normal and parallel (shear) to the plane describedby n, again in terms of θ.

– Solve for θ iteratively; initially choose 20 < θ < 35.

3.27 A local stress state is given in the figure below.

(a) Find the Cauchy stress tensor that represents the local stress state.

(b) Find the traction vector on an inclined plane forming an angle of 120 with the x-axisusing the Cauchy formula.

(c) Calculate the components of the vector that are normal and parallel to the plane.

3.28 A local stress state is shown in the figure below.

(a) Find the Cauchy stress tensor which represents the local stress state;

(b) Find the traction vector on an inclined plane with unit normal n shown in the figureusing the Cauchy formula;

(c) Calculate the components of the vector that are normal and parallel to the plane;

3.7. PROBLEMS 93

x

y

z

10 MPa

20 MPa

600

15 MPa

Problem 3.27

z

45o

n

a

a

x

y

Problem 3.28

(d) Decompose the Cauchy stress into pressure P and deviatoric (extra) stress S.

3.29 GIVEN : A slender bar with a uniaxial load applied along its axis, resulting in a traction vectort(i) = 10i kN

m2 , which is constant along x. The other two traction vectors t(j) = t(k) = 0 alongthe rod.

REQUIRED :

(a) Find the components of the stress tensor σ in the rod.

(b) What is the traction vector t(n) applied on the plane with normal n that forms an angleθ = 30 with the x-axis and a 90 angle with the z-axis?

3.30 A flat-bottomed barge with a wetted surface area of 24 ft. by 96 ft. moves through the water.Assume the boat does not “push” any water in front of it. How much force (in pounds) isrequired to propel the boat at 15 mph in water that is 6 ft. deep?

3.31 A piston with a diameter of 3 in. moves in a cylinder of diameter 3.01 in. The piston has alength of 2.5 in. Oil with a viscosity of 1,000 centipoise fills the gap between the piston andcylinder. How much force is required to move the cylinder if its velocity is 4 ft

sec?


y

x

z z

y

x

x

tt

∆ x = 10m∆ y = 5m∆ z = 5m

n(-i)

(i)θ

Problem 3.29

3.32 GIVEN : The Couette viscometer shown consists of a rotating cup with a viscous fluid insideof it. A torque spins the cup with an angular velocity Ω.

Ω = 10radsec

δ = 0.5 inRo = 2.75 inRi = 2.5 in (Radius of inner cylinder)µ = 2.5 × 10−3 centipoiseL = 12 in

REQUIRED : Find torque and Srθ.

Ω

Rotating Cup Fluid

Problem 3.32

3.33 Two layers of liquids with viscosities µ1 and µ2 and densities ρ1 and ρ2 and with equal depthsb flow between two fixed horizontal parallel plates. Find the velocity distribution for steadystate, 1-D, incompressible flow. Verify that the velocity found satisfies conservation of massand conservation of linear momentum.

3.34 Consider the velocity gradients shown below for an incompressible Newtonian fluid. Find thestress tensor [σ] in terms of pressure P and viscosity µ for cases (a) and (b).

(a) ∇v =

−8 0 0

4 2 00 0 3

(b) ∇v =

2 3 0

0 −5 00 0 8

3.7. PROBLEMS 95

3.35 GIVEN : The following velocity field for a incompressible Newtonian fluid with viscosity µ =0.96 Pa · s:

vx = 2x2 − y2, vy = −4xy, vz = 0

REQUIRED :

a) Components of [S] and [σ] at the point ( 2, 1, 1 )

b) Shear and normal traction vector components on a plane with normal

n =1√2

(i + j)

3.36 Substitute for σxx, σyy, σzz, σxy, σyx, σxz, σzx, σyz, and σzy in terms of velocity gradients,viscosity, and pressure into the Equations of Motion (Conservation of Linear Momentum). Dothis in component form (all three directions), and then re-express the result in vector form.

3.37 GIVEN : Plane Couette flow between two plates, where the top plate is moving at a constantvelocity v, and the bottom plate is stationary. REQUIRED : Derive the velocity distributionvz(y).

3.38 Obtain the steady-state, incompressible, 1-D flow of an incompressible viscous fluid layer ofuniform depth d flowing down an inclined plane which makes and angle θ with the horizontal.

dx

y vx = V

Problem 3.39

3.39

1. Write the Conservation of Linear Momentum for incompressible Newtonian Fluids (Navier-Stokes Equation)

2. Find the velocity of the fluid vx as a function of position y, given vx(d) = V

3. Find the shear stress Sxy at y = 0 and at y = d.

3.40 Please the velocity profile given the same properties as in problem 3.37, but with the upperboundary condition changed from known velocity to known shear stress.

dx

y Sxy

Problem 3.40


V

V

h

y

x

fluid between plates

Problem 3.39

GIVEN : Two parallel plates contain a Newtonian viscous fluid and they are moving in oppositedirections at a constant velocity V as shown in the figure. Assume no body forces, no externalpressure and steady state.

REQUIRED :

1. Find the velocity of the fluid vx(y), assuming V and h are known.

2. Draw the velocity profile and find the value of the fluid velocity Vx at y = 0, y = h4 ,

y = −h4 , for V = 10 m

s and h = 0.1 m

3. Does the velocity profile depend on the viscosity of the fluid? (Explain your answer)

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