Principles of Engineering Mechanics (2nd Edition) Episode 12 pps

addition we will assume that the material only undergoes small deformation and that this deformation is proportional to the applied 12.4 Elementary strain The longitudinal, or axial, s

216 Introduction to continuum mechanics

the device would give a reading for a specific point in space and does not follow a particular particle of fluid A flow velocity device would also

be attached to the pipe

A system of co-ordinates which relates prop- erties to a specific point in space is known as Eulerian Thus pressure p is a function of x and

time t

The particle velocity is here defined by

dr

dt

Consider first how measurements of deforma-

tion are made for the pipe itself TO determine note here that the partial derivative is not how much the material has been stretched we required, however the velocity will be a function could measure the relative movement of two of both x and t

marks, one at x = xA and the other at x = xB In summary, Lagrangian co-ordinates refer to a Note that if the Pipe moves as a rigid body the particular particle whilst Eulerian co-ordinates marks will move with the pipe SO the marks will refer to a particular point in space

not be at their original locations We must regard

XA and XB as being the ‘names’ of the marks: that 12.6 Ideal continuum

is they define the original positions of the marks An ideal solid is defined as one which is

In this context x does not vary, SO we must use a homogeneous and isotropic, by which we mean different symbol to denote the displacement of that the properties are uniform throughout the the marks from their original positions The region and SO not depend on orientation In symbols uA and uB will be used addition we will assume that the material only

undergoes small deformation and that this deformation is proportional to the applied

12.4 Elementary strain

The longitudinal, or axial, strain is defined to be

loading system This last statement is known as the change in length per unit length

Hooke’s law

thus the strain I = ~ (12-2) isotropic and the term is usually restricted to

incompressible, inviscid fluids This is clearly a

XB - X A

As the distance between the marks approaches to good approximation to the properties of water in

and the effects of viscosity are confined to a thin (12.3) layer close to a solid surface known as the

boundary layer For gases such as air, which are The partial differential is required since strain very compressible, it is found that the effects of could vary with time as well as with position compressibility in flow processes are not signi-

ficant until relative velocities approaching the

12.5 Particle velocity

The velocity of a particle at a given value of x, say ’peed Of sound are reached

x A , is simply

12.7 Simple tension

au

E = -

ax

v = - dU

Again the partial derivative is used to indicate

that x is held constant

The above co-ordinate system is known as

Lagrangian

If we are concerned with the fluid in the pipe

then a pressure measuring device would be fixed

to the pipe and, assuming that the pipe is rigid,

Figure 12.2 shows a straight uniform bar length L ,

cross-section area A and under the action of a

12.9 The control volume 217

tensile force F The state of tension along the bar

surface is F to the right and on a left-facing

d2U

is constant, this means that if a cut is made ( F + E & ) - F = ( P A & ) ,

at

anywhere along the bar the force on a right-facing

surface it is F to the left

It follows that at any point the tensile load

divided by the original cross-section area, F / A , is

constant and this quantity is cajled the stress (+ A

negative stress implies that the load is compres-

aF d2U

Since nominal o r engineering stress is defined as force/original area, then dividing both sides of equation 12.8 bY A gives

ax

sive If the extension under this load is 6 then the a a a2u

ax

By Hooke’s law 6 m F so E ~c a or

We have already shown that the strain

a U

property of the material known as Young’s ax

modulus

Re-arranging the above equations gives

so substituting 12.11 into 12.9 and using 12.10 we (12.7) finally obtain

FL

a=

A E

a2u a2u

a x 2 at2

A state of tension resulting in an extension is

regarded as being associated with a positive stress

and a positive strain (See Appendix 8 f o r a

discussion of material properties )

Figure 12.3 shows an element of a uniform bar

which has no external loads applied along its

length, the external loads or constraints occurring

only at the ends The material is homogeneous the solution of which is u = a + bx where a and b

with a density P l Young’s modulus E and a are constants depending on the boundary

conditions

constant cross-section area A

du

dx

This is a very common equation in applied physics and is known as the wave equation

In a statics case the right-hand side of equation 12.12 is zero so that u is a function of x only,

d2 u

- = o

dx2

Now strain

E = - = b = V I E = ( F / A ) / E

so if at x = 0 u = 0 then a = 0, u = Fx/(AE) At

x = L the displacement is equal to F L / ( A E ) , as

expected

12.9 The control volume

The equations of motion developed for rigid bodies and commonly used for a solid continuum refer t o a fixed amount of matter However for fluids it is usually more convenient to concentrate

on a fixed region of space with a volume V and a

surface S The properties of the fluid are expressed as functions of spatial position and of time, it being noted that different particles will occupy a given location at different times

The mass of an element of length dx is pAdx

and this is constant as these quantities refer to the

original values

Resolving the forces in the x direction and

equating the net force to the mass of the element

times its acceleration gives

218 Introduction to continuum mechanics

p by pv in the development of the continuity equation This is possible since p is the mass per unit volume and pv is the momentum per unit volume Thus the change in momentum in time At

is

a(pv) d V

AG = [I, pv ( v - &) + 1, at ]

AG Now force F = limAt-o-

At

= [ p ( v - d S ) + [ " Y d V (12.14)

5

A t time t the control volume is shown in

Fig 12.4 by the solid boundary A t time t + At the 12.12 Streamlines

position of the set of particles originally within the A streamline is a line drawn in space at a specific

control volume is indicated by the dashed time such that the velocity of the fluid at that

The velocity of the fluid at an elemental part of The distance along the streamline is s and, as in the surface is z, and the outward normal t o the path co-ordinates, e, is the unit tangent vector and

At the elemental surface area, dS, the increase

in mass in the time At is

p(dSvAtcosa) = pv-e,dSAt = pv-dSAt

Note that the area vector (dS = e n d s ) is defined

as having a magnitude equal to the elemental

surface area and a direction defined by the

outward normal unit vector, e,

Integrating over the whole surface we obtain

the net total mass gained by the original group

due to the velocity at the surface In addition to

this there is a further increase in mass due to the

density over the whole volume changing with

time

Thus the change in mass,

ds

dt Thus v = vet = - e,

Am = [ I s p v - d S + Iv$dV]At If the flow is steady, that is the velocity at any

point does not vary with time, a streamline is also

a path line

12.10 Continuity

Since the mass must remain constant

- = [ s p v - d S + [ ? d V = O (12.13) volume

this is known as the continuity equation a finite volume We now wish to obtain an

expression for an elemental volume correspond-

T o obtain the equations of motion we need to Figure 12.6 shows a rectangular element with consider the time rate of change in linear sides dx, dy and dz Considering the continuity

momentum This is achieved by simply replacing

Equation of motion for a fluid

equation we first evaluate the surface integral

12.14 Euler’s equation for fluid flow 219

component of velocity u since the streamlines at this surface may be diverging

A stream tube could have been used where the curved surface is composed of streamlines, but this means that the cross-section area would be a variable and the effect of pressure on this surface would have to be considered

\,pu-dS = [ pvx+-dx a 2 ] dydz-pv,dydz

[ a:1

+ pvy+-dy dzdx-pvydzdx

+ pv,+-dz dxdy-pv,dxdy

First we need to apply the continuity equation

- -

- [ apvx +- apvy + ””.I & dy dz so with reference to Fig 12.7

The vector operator V is defined, in Cartesian

(p+$ds)(v+E*)dA co-ordinates, to be

aP

at

V = i- + j - + k -

ax ay az

Neglecting second order terms

SO with pu = ipv, + jpv, + kpv,

l , p u d S = V-pudxdydz p-dV+ as v- d V + p u d S + - d V = dS at 0 (12.16)

where dV = dsdA

In applying the force equation we are going to include a body force, in this case gravity, in addition to the pressure difference Resolving

The operation V (pu) is said to be the divergence

of the pv field and is often written as div(pu)

Also [ v 2dV=-dxdydz at a t forces along the streamline

so the complete continuity equation is

aP

d F = p d A - p+-ds ( E ) dA

or

or

d F = pgcOSa dsdA (12.17)

In applying the momentum equation we shall

choose a small cylindrical element with its axis

surface, of area dS ’, there could be a small radial

(12.15)

aP

at

v*pu+- = o

equation 12.14, along a streamline However at the curved d G = ( p + g d s ) ( V + % d s ) av 2 dA-pvudA

220 Introduction to continuum mechanics

= 2pv-dsdA +v2-dsdrdA +puudS’

The right-hand side of this expression can be

simplified by subtracting v times equation 12.16 to

give

combining with 12.17 and dividing through by

dsdA

and finally re-arranging gives

1 ap av av -gcosa - = v- +-

p as as at (12.18) This is known as Euler’s equation for fluid flow

Since v = v(s, t ) ,

dv av ds av dt av av

dt as dt at dt as at

- - - - - + =-v+-

the right-hand side of 12.18 may be written as

dv

dt

-

If we consider the case for steady flow where the

velocity at a given point does not change with

time, Euler’s equation may be written

1 do dv

the partial differentials have been replaced by

total differentials because v is defined to be a

function of s only Multiplying through by ds and

integrating gives

-I gcosads- I - - - - +constant

now cos ads = dz thus

I f + + g z = constant

If p is a known function of p then the integral can

be determined but if we take p to be constant we have

+ ”’ + gz = constant

this is known as Bernoulli’s equation

This equation is strictly applicable to steady flow of a non-viscous, incompressible fluid; it is, however, often used in cases where the flow is changing slowly The effects of friction are usually accounted for by the inclusion of experimentally determined coefficients As has already been

mentioned, the effects of compressibility can often be neglected in flow cases where the relative speeds are small compared with the speed of sound in the fluid

SECTION B Two- and three-dimensional continua

We are now going to extend our study of solid continua to include more than one dimension In our treatment of one-dimensional tension o r compression we did not consider any changes in the lateral dimensions Although we are going to use three dimensions we shall restrict the analysis

to plane strain conditions By plane strain we mean that any group of particles which lie in a plane will, after deformation, remain in a plane

It is possible that the plane will be displaced from the original plane but will still be parallel to it

It is an experimental fact that a stress applied in one direction only will produce strain in that direction and also at right angles to the stress axis

If a specimen is strained within the x-y plane then, if the strain in the z direction is to be zero, there must be a stress in the z direction as well as

in the x and y directions Conversely, if stresses are applied in the x and y directions with a zero stress in the z direction, there will be a resulting strain in the z direction as well as those in the x

and y directions The two-dimensional analyses presented later are based on the latter case

12.17 Poisson’s ratio

If Hooke’s law is obeyed, then the transverse

strain produced in axial tension will also be

proportional to the applied load; thus it follows

that the lateral strain will be proportional to the

axial strain The ratio

transverse strain

axial strain

- - v

where v is known as Poisson’s ratio

If a uniform rectangular bar, as shown in Fig

12.8, is loaded along the x axis then

E, = u x / E

E~ = -vu,/E

and E, = - v c , / E

12.1 8 Pure shear

Figure 12.9 shows a rectangular element which is

deformed by a change in shape such that the

length of the sides remain unaltered The shear

strain yxy is defined as the change in angle

(measured in radians) of the right angle between

adjacent edges This is a small angle consistent

with our discussion of small strains

12.19 Plain strain 221

Figure 12.10

This shows the equivalence of the complementary shear stresses

Again by Hooke’s law, shear stress is proportional to the shear strain

where G is known as the Shear Modulus or as the Modulus of Rigidity

-

Referring to Fig 12.11 it is seen that the shear strain can be expressed in terms of partial differential coefficients as

Yxy = Y1+ Y2

auy au,

ax ay

yxy = - +-

12.19 Plane strain

The rectangular element, shown in Fig 12.12, has one face in the XY plane and is distorted such that

the corner Points A , B , c and D rnOve in the XY

plane only

The translation of point A is u and that of point

C is u + du For small displacements

( 12.23)

The loading applied to the element to produce

pure shear is as shown in Fig 12.10 This set of

forces is in equilibrium, SO by considering the sum

of the moments of the two couples in the xy plane

(12.m) F,dr-F,dy=O

rXy = F,/(dydz)

The shear stress is defined as

d u = -dx+-dy i+ 2dx+Ldy

Substitution into equation 12.20 gives or in matrix form

222 Introduction to continuum mechanics

au, au,

[:::I= k 4 [;] (12.24)

Let us now introduce the notation

Figure 12.13

1 au

xy 2 ( ax $)

(see Fig 12.13)

The 112 in the strain matrix spoils the simplicity

of the notation therefore it is common to replace

Figure 12.12

a u x -

aY

4Yxy by Exy

~ - u , , ~ etc

In this notation the strain in the x direction 12.20 Plane stress

The triangular elements shown in Fig 12.14 are in equilibrium under the action of forces which have components in the x and y directions but not in the z direction Note that the surface abcd has

area dydzi and area abef has an area dxdzj; these

are the vector components of the area e’f’c’d’

The sense of the stress component, shown on the diagram, is such that when multiplied by the area

vector it gives the force vector

Ex, = ux,x

similarly EYY = UY,Y

and the shear strain

and equation 12.24 becomes

yxy = uy,x + u , , ~

[::;

I = I [::: [;I :;::]

The square matrix can be written as the sum of a

symmetrical and an anti-symmetrical matrix By

this means the shear strain can be introduced

I

1

+ [ặx,y - uy,x) 0

u x , x I ( u x , y + uy,x 1

0 -%uy,x- UXJ [I::: :::I 7 [ l ( u x , y + uy,x)

[::;

UY 3 Y

therefore

Resolving in the x direction we obtain

+[lY -?I} [;I (12-25) where Oxy is the rigid body rotation in the xy

plane given by

F, = u,dydz+ rxydxdz

Fy = c,dxdz+ rxydydz

or, in matrix form,

dy dz

[ :] = [ rz :][ dxdz]

Letting dydz = S, and dxdz = S,

In many texts rXy is replaced by mxy

The values of the components of stress and strain

depend on the orientation of the reference axes

In Fig 12.15 the axes have been rotated by an

angle 8 about the z axis

12.22 Principal strain 223

they may now be transformed by use of the transformation matrix

12.22 Principal strain

Since (du) = [T](du’) and (dx) = [TI(&’) we can write

[TKdu’) = (1.1 + [aLIWl(d4

(du‘) = [TIT{[El + [filI[Tl(dx’)-

and pre-multiplying by [TIT we obtain

The rotation [a] is not affected by the change

in axes because they are rotated in the xy plane The transformed strain matrix is

[&’I = [TIT[&] IT1

- cos0 sin8 E,,

[-sin e cos e ] [ E,, 2::]

cos0 -sine

x [ sine cos@

= [I:; 3

-

1

where

E’ ,, = E,, cos2 e + E,, sin2 e

E’,, = E , ~ C O S ~ e + &,sin2 e

E’,, = ( E ~ - ~,,)sinBcose

( E y y - E x x )

x = x’cos8-y’sine

which, in matrix form, becomes

(x) = ITl(x‘>

+ E,, (cos’ e - sin2 e)

-

- sm28+ E , , C O S ~ ~ (12.30) also (E’,,+ E ’ ~ , ) = ( E , + E ~ ) (12.31) From equation 12.30 it is seen that it is possible

(12.27)

2

or, in abbreviated form

The matrix [T1 is a transformation matrix It is

inversion that the inverse of this matrix is the

same as its transpose

easily shown from the geometry or by matrix to choose a va1ue for 8 such that &’xy = 0- The

value of 8 is found from

(12.32)

The axes for which the shear strain is zero are Writing equations 12.25 and 12.26 in abbrevi- known as the principal strain axes Let us

therefore take our original axes as the principal axes, that is E,, = 0 The longitudinal strains are now the principal strains and will be denoted by E~

in the x direction and by E~ in the y

2Exy

( E , - Eyy 1

tan28 =

I

cose sin8 -sine cos6

[TI-’ = [TIT = [

ated form as

(du) = {[El + [a1 I (k )

and ( F ) = [u](S)

224 Introduction to continuum mechanics

From equations 12.28 and 12.29 we now have expressed in terms of rotated co-ordinates we

may write

(Ef,-&’ ) ( E 1 - E 2 ) ~ ~ ~ 2 e

f l =

thus [ T ] ( F ’ ) = [a][T](Sf)

so pre-multiplying by [TIT gives

and from equation 12.30

( E ~ - ~ ~ ) s i n 2 8

[of] = P I T [a1 [TI

- E l x y =

A simple geometric construction, known as

strains and the angle 8 Figure 12.16 shows a

axis, the ordinate being the negative shear strain

The location of the centre is given by the average

strain, and the radius of the circle is half the

difference between the principal strains It is seen

that this diagram satisfies the above equations

therefore Mohr’s circle, gives the relationship between the

circle plotted with its centre on the normal strain cos0 sin8 a,, aXy

[-sine wse][uxy uy,]

cos0 -sine

.[ sin8 cos0

-

-

I

where

a’ xx = a,, cos2 e + cry, sin2 e

urYy = uyy cos2 e + a,, sin2 e

dXy = (ayy - u,,) sin ecos e

+ aXy2cosBsin 8 (12.33)

- uxy2cosBsin8 (12.34)

+ oxy (cos2 e - sin2 e)

2

- (uyy - uxx) sin 28 + a,, cos20 (12.35)

also (a’,, + dYy) = (a, + uyy)

From equations 12.33 and 12.34 we now have

(ufxx - utyy) - (ul - u 2 ) ~ ~ ~ 2 e

-

and from equation 12.35

(ul -u2)sin28

2

-(+Ixy =

The form of these equations is the same as those for strain therefore a similar geometrical construction can be made, which is Mohr’s circle for stress as shown in Fig 12.17

Because we have taken the material to be isotropic it follows that the principal axes for stress coincide with those for strain This is because normal stresses cannot produce shear strain in a material which shows no preferred directions

Figure 12.16

It can be seen that when 6 = 7d4 the shear

strain is maximum and the normal strains are

equal If the circle has its centre at the origin then

for 0 = 7r/4 the normal strains are zero So for the

case of pure shear the principal strains are equal

and opposite with a magnitude E,, = y,,/2

In the case of uniaxial loading E~ = - vsl hence

the radius of the circle is ( E ~ + m1)/2 which also

equals the maximum shear strain at 8 = 7~14

so y x , = E , ( l + v ) = u l ( l + Z ) ) / E

Equation 12.26 can also be written in abbreviated

form as

( F ) = [ m )

and since the components of any vector can be

12.24 The elastic constants 225

Because of the symmetry b must be equal to c so

we can write

~1 = ( b + ( ~ - b ) ) ~ i + b ~ 2 + 6 ~ 3

or u1 = ~ ( E ~ + E ~ + E ~ ) + ( U - ~ ) E ~

Let b = A and (a - b ) = 2 p where A and p are the Lame constants, and introducing dilatation A, the

sum of the strains, we have

and again because of symmetry

12.24 The elastic constants

So far we have encountered three elastic

constants namely Young's modulus ( E ) , the shear

modulus ( G ) and Poisson's ratio (v) There are

three others which are of importance, the first of

which is the bulk modulus

For small strains the change in volume of a

rectangular element with sides dx, dy and dz is

(&xx &I dY dz+ (&yydY ) dzdx + (E==dZ) dxdY- Figure 12-18

The volumetric strain, also known as the

dilatation, is the ratio of the change in volume to

the original volume; thus the dilatation

A = E,, + eyy + E, Let us now consider the case of pure shear, see

Fig 12.18- We have already Seen that (+I = - ~ X Y 9

a 2 = r x y , s1 = - E , ~ and E~ = e,y so substituting into equations 12.37 and 12.38 we have

It should be remembered that shear strain has no

effect on the volume

The average stress

a,,, = ( a x , + c y , + a, 113 - r x y = h A + 2 p ( - ~ , ~ )

rxy = AA + 2 p X y

and the bulk modulus K is defined by and

Solving the last two equations shows that A = 0

and rxy = 2peXy giving

Now consider the case of pure tension, see

Fig- 12-19> such that u z = 0 and E z = - V E I

(12.36) (For fluids the average stress is the negative of the

pressure p )

The two other constants are the Lame

constants and they will be defined during the

following discussion

In general every component of stress depends

consider an element which is aligned with the

principal axes of stress and strain, then each

principal stress will be a function of each principal

f l a w = KA

2Exy Y x y

linearly on each component of strain If we Substitution into equations 12.38 and 12.39 gives

~1 = AA + 2 ~ ~ 1

0 = AA- ~ / A v E ~