Introduction to Continuum Mechanics 3 Episode 6 docx

4.9, then as in Section 4.1, we obtain where n is the unit outward normal vector to the boundary, T is the stress tensor evaluated atthe boundary and t is the force vector per unit area

the characteristic equation has the form

Therefore A = 0 is an eigenvalue and its direction is obviously n = 63 The remaining values are

eigen-To find the corresponding eigenvectors, we set (7/j-A(5,y)n^ = 0 and obtain for either

Example 4.6.2

Do the previous example for the following state of stress: 7\2 = T 2 i = 1000 MPa, all other TIJ are zero,

Solution, From Eq (4.6.10), we have

Corresponding to the maximum normal stress T\ — 1000 MPa, Eq (4.6.11) gives

and corresponding to the minimum normal stress TI - -1000 MPa, (i.e., maximum

compres-sive stress),

'The maximum shearing stress is given by

which acts on the planes bisecting the planes of maximum and minimum normal stresses, i.e.,the ei-plane and the e2-plane in this problem

4.7 Equations of Motion - Principle of Linear Momentum

In this section, we derive the differential equations of motion for any continuum in motion.The basic postulate is that each particle of the continuum must satisfy Newton's law of motion.Fig 4.8 shows the stress vectors that are acting on the six faces of a small rectangular elementthat is isolated from the continuum in the neighborhood of the position designated by */

Let B = Bffi be the body force (such as weight) per unit mass, p be the mass density atjq

and a the acceleration of a particle currently at the position*/; then Newton's law of motiontakes the form, valid in rectangular Cartesian coordinate systems

Fig 4.8

Dividing by Ajt^A^A^ and letting A*/-» 0, we have

Since I,, = Te, = 7J/ e;-, therefore we have (noting that all e,- are of fixed directions in Cartesiancoordinates)

In invariant form, the above equation can be written

and in Cartesian component form

These are the equations that must be satisfied for any continuum in motion, whether it be solid

or fluid They are called Cauchy's equations of motion If the acceleration vanishes, then Eq (4.7.2) reduces to the equilibrium equations

or,

Example 4.7.1

In the absence of body forces, does the stress distribution

where vis a constant, satisfy the equations of equilibrium?

Solution Writing the first(/ =1) equilibrium equation, we have

Similarly, for i = 2, we have

Solution Substituting the given stress distribution in the first term on the left-hand side of

Eq (4.7.3b), we obtain

Therefore,

or,

4.8 Equations of Motion in Cylindrical and Spherical Coordinates

In Chapter 2, we presented the components of divT in cylindrical and in spherical dinates Using those formulas, we have the following equations of motion: [See alsoProb 4.34]

coor-Cylindrical coordinates

We note that for symmetric stress tensor, T^-TQ r =Q.

Example 4.8.1The stress field for the Kelvin's problem (an infinite elastic space loaded by a concentratedload at the origin) is given by the following stress components in cylindrical coordinates

where

and A is a constant Verify that the given state of stress is in equilibrium in the absence of body

forces

Solution From R = r + z , we obtain

Thus,

Thus, the left hand side of Eq, (4.8 la) becomes, with B r = 0

In other words, the r-equation of equilibrium is satisfied

Since T& = TQ Z = 0 and TQQ is independent of 0, therefore, with BQ — a@ — 0, the second

equation of equilibrium is also satisfied

The third equation of equilibrium Eq (4.8 Ic) with B z — a z = 0 can be similarly verified.[see Prob 4.35]

Spherical coordinates

Again, we note that for symmetric stress tensor, T^- T0 = 0 and7^- 7^=0.

4.0 Boundary Condition for the Stress Tensor

If on the boundary of some body there are applied distributive forces, we call them surfacetractions We wish to find the relation between the surface tractions and the stress field that

is defined within the body

If we consider an infinitesimal tetrahedron cut from the boundary of a body with its inclinedface coinciding with the boundary surface (Fig 4.9), then as in Section 4.1, we obtain

where n is the unit outward normal vector to the boundary, T is the stress tensor evaluated atthe boundary and t is the force vector per unit area on the boundary Equation (4.9.1) is calledthe stress boundary condition

Fig 4.9

Example 4.9.1Given that the stress field in a thick wall elastic cylinder is

where A and B are constants.

(a) Verify that the given state of stress satisfies the equations of equilibrium in the absence ofbody forces

(b) Find the stress vector on a cylindrical surface r = a.

(c) If the surface traction on the inner surface r = r/ is a uniform pressure /?/ and the outer surface r = r 0 is free of surface traction, find the constants^ and B.

Solution.

The above results, together with Tj$ = T n = 0, give a value of zero for the left hand side of

Eq (4.8 la) in the absence of a body force component Thus, the r-equation of equilibrium issatisfied Also, by inspection, one easily sees that Eq (4.8 Ib) and Eq (4.8 Ic) are satisfied

Example 4.9.2

It is known that the equilibrium stress field in an elastic spherical shell under the action of

external and internal pressure in the absence of body forces is of the form

(a) Verify that the stress field satisfies the equations of equilibrium in the absence of bodyforces

(b) Find the stress vector on spherical surface r=a.

(c) Determine A and B if the inner surface of the shell is subjected to a uniform pressure PJ

and the outer surface is free of surface traction

Solution.

(a)

Thus, Eq (4.8.2a) is satisfied when B r = a r = 0, Eqs (4.8.2b) and (4.8.2c) can be similarly

verified, [see Prob.4.38]

(b) The unit normal vector to the spherical surface is n = ep thus the stress vector is given by

i.e.,

(c)The boundary conditions are

Thus,

From Eqs (viii) and (ix), we obtain

Thus,

4.10 Piola Kirchhoff Stress Tensors

Let dA 0 be the differential material area with unit normal n0 at the reference time t 0 and

dA that at the current time t of the same material area with unit normal n We may refer to

dA 0 as the undeformed area and dA as the deformed area Let df be the force acting on the deformed area dAn In Section 4.1, we defined the Cauchy stress vector t and the associated Cauchy stress tensor T based on the deformed area dAn, that is

The stress vector ^ , defined by the above equation is a pseudo-stress vector in that, beingbased on the undeformed area, it does no describe the actual intensity of the force We notehowever, that t0 has the same direction as the Cauchy stress vector L

The first Piola-Kirchhoff stress tensor (also known as the Lagrangian Stress tensor) is a

linear transformation T0 such that

The relation between the first Piola-Kirchhoff stress tensor and the Cauchy stress tensorcan be obtained as follows:

Since

therefore

Using Eqs (4.10.2) and (4.10.4), Eq (ii) becomes

Using Eq (3.28.6), i.e.,

we have,

The above equation is to be true for all n 0 , therefore,

This is the desired relationship

In Cartesian components, Eq (4.10.6a) reads

From Eq (4.10.6a), we obtain

which in Cartesian components, reads

We note that when Cartesian coordinates are used for both the reference and the current

We also note that the first Piola-Kirchhoff stress tensor is in general not symmetric

(B) The Second Piola-Kirchhoff Stress Tensor

Let

where

In Eq (4.10.8b), df is the (pseudo) differential force which transforms, under the deformation gradient F into the (actual) differencial force df at the deformed position (one njay compare

the transformation equation df = Fdf with d\ - ¥dX); thus, the pseudo vector t is in general

in a different direction than that of the Cauchy stress vector t

The second Piola-Kirchhoff stress tensor is a linear transformation T such that

where we recall n 0 is the normal to the undeformed area From Eqs (4.10.8a) (4.10.8b) and(4.10.9) we have

We also have (see Eqs (4.10.3) and (4.10.4)

Comparing Eqs (i) and (ii), we have

Equation (4.10.10) gives the relationship between the first Piola-Kirchhoff stress tensor TQ and

the second Piola-Kirchhoff stress tensor T Now, from Eqs (4.10.6a) and (4.10.10), one easilyobtain the relationship between the second Piola-Kirchhoff stress tensor and the Cauchy stresstensor T as

We note that the second Piola-Kirchhoff stress tensor is always a symmetric tensor if theCauchy stress tensor is a symmetric one

Example 4.10.1The deformed configuration of a body is described by

If the Cauehy stress tensor for this body is

(a) What is the corresponding first Piola-Kirchhoff stress tensor

(b) What is the corresponding second Piola-Kirchhoff stress tensor

Solution From Eqs (i), we have

detF = 1Thus, from Eq (4.10.6a), we have, the first Piola-Kirchhoff stress tensor:

The second Piola-Kirchhoff stress tensor is, from Eq (4.10.11)

Example 4.10.2The equilibrium configuration of a body is described by

If the Cauchy stress tensor for this body is

(a) What is the corresponding first Piola-Kirchhoff stress tensor

(b) What is the corresponding second Piola-Kirchhoff stress tensor and

(c) calculate the pseudo stress vector associated with the first Piola-Kirchhoff stress tensor onthe 63 - plane in the deformed state

(d) calculate the pseudo-stress vector associated with the second Piola-Kirchhoff stress tensor

on the 63 - plane in the deformed state

Solution From Eqs (i), we have

Thus, from Eq (4.10.6), we have, the first Piola-Kirchhoff stress tensor:

The second Piola-Kirchhoff stress tensor is, from Eq (4.10.11)

(c) For a unit area in the deformed state in the €3 direction, its undeformed area dA 0 n 0 isgiven by Eq (3.28.6) That is

With detF = 1, and the matrix F given above, we obtain

Thus, n 0 = 62 and

gives

i.e, ^ = 2503 MPa We note that this vector is in the same direction as the Cauchy stress vector,

its magnitude is one fourth of that of the Cauchy stress vector, because the undeformed area

is 4 times that of the deformed area

(d) We have, from Eq (4.10.9)

Thus,

We see that this pseudo stress vector is in a different direction from that of the Cauchy stressvector (We note that the tensor F transforms e into the direction of e.)

4.11 Equations of Motion Written With Respect to the Reference Configuration,

In this section, we shall show that with respect to the reference configuration, the equations

of motion can be written as follows:

where (T 0 )ij are the Cartesian components of the first Piola-Kirchhoff stress tensor, p 0 is the

density in the reference configuration, Xj are the material coordinates and B/ and a/ are

body force per unit mass and the acceleration components respectively

From Eq (4.10.7b), we have

Thus,

j

Now,

and we can show that the last term of Eq (ii) is zero as follows:

t In Chapter 7, an alternate shorter proof will be given.

By using the following identity [see Prob 4.40] for any tensor A^^^s)

we obtain,

so that

Thus,

Substituting Eq (vii) in the Cauchy's Equation of motion [ Eq.(4.7.2b)], we get

Since dV = (detF>/K0 [See Eq (3.29.3)], therefore,

where p0is the initial density Thus, we have, in terms of the first Piola-Kirchhoff stress tensorand with respect to the material coordinates, the equations of motion take the following form

whereas in terms of the Cauchy stress tensor and with respect to the spatial coordinates, theequations of motion take the form

In invariant notation, Eq (4.11.4) reads

where Div denotes the divergence with respect to the material coordinates X and Eq (4.11.5)reads

where div denotes the divergence with respect to the spatial coordinates x

4.12 Stress Power

Referring to the infinitesimal rectangular parallelepiped of Fig 4.8 which is repeated herefor convenience, let us compute the rate at which work is done by the stress vectors and bodyforce on the particle as it moves and deforms

Fig 4.8 (repeated)

The rate at which work is done by the stress vectors t_Cj and ^ on the pair of faces having

—QI and ej as their respective normal is:

where we have used the fact that t^-v = Te^v/e/ = v/e/'Tej = v/7/i , and dV=dxidx^dx^

denotes the differential volume Similarly, the rate at which work is done by the stress vectors

Including the rate of work done by the body force (pBdV- v = pBiVjdV) the total rate of

work done on the particle is

Eq (4.12.1) takes the form

However, we have from the equations of motion

is known as the stress power It represents the rate at which work is done to change the volume

and shape of a particle of unit volume

For a symmetric stress tensor

where £>{y are the components of the rate of deformation tensor defined in Section 3.13.Equation (4.12.8a) can be written in the invariant form

Show that the stress power can be expressed in terms of the first Piola-Kirchhoff stresstensor T0 and the deformation gradient F as the following

Solution In Sect 3.12, we obtained [see Eq (3.12.4)],

Since dx - F dX [ see Eq (3.7.2)], therefore

Equation (i) is to be true for all dX, thus

or,

Using Eqs (4.12.7) and (4.12.10b), the stress power can be written

Since [see Eq (4.10.7)]

therefore,

Using the identity tr(ABCD) = tr(BCDA) = tr(CDAB) and the relation detF = ~~ Eq (iii)becomes

Example 4.12.2Showjthat the stress power can be expressed in terms of the second Piola-Kirehhoff stresstensor T and the Lagrange strain tensor E * as follows

Solution From Eq (3.13.6),

and Eq (3.7.2)

we obtain

From Eq (3.24.2), we obtain

so that

Compare Eq (i) with Eq (4.12.12), we obtain

Using Eq (4.10.11), that is

we have for the stress power

Making use of Eq (4.12.13), Eq (ii) becomes

4.13 Rate of Heat Flow Into an Element by Conduction

Let q be a vector whose magnitude gives the rate of heat flow across a unit area byconduction and whose direction gives the direction of heat flow, then the net heat flow by

conduction Q c into a differential element can be computed as follows:

Referring to the infinitesimal rectangular parallelepiped of Fig 4.10, the rate at which heatflows into the element across the face with e^ as its outward normal isK~*l"el)jc +dx,x ,X 3 ^ X 2^ X 3 and tnat across the face with -ej as its outward normal is[Orei)je jc ,* d x ?d x 3- Thus, the net rate of heat inflow across the pair effaces is given by

where #/ = q " e/ Similarly, the net rate of heat inflow across the other two pairs of faces is

so that the total net rate of heat inflow by conduction is

Example 4.13.1

Using the Fourier heat conduction law q = -/cV@, where V© is the temperature gradient

and K is the coefficient of thermal conductivity, find the equation governing the steady-state

distribution of temperature

Solution From Eq (4.13.1), we have, per unit volume, the net rate of heat inflow is given

by

Now, if the boundaries of the body are kept at fixed temperature, then when the steady-state

is reached, the net rate of heat flow into any element in the body must be zero Thus, thedesired equation is

For constant /c, this reduces to the Laplace equation

4.14 Energy Equation

Consider a particle with a differential volume dVat the position x at time t Let U denote its internal energy, KE the kinetic energy, Q c the net rate of heat flow by conduction into theparticle from its surroundings,^ the rate of heat input due to external sources (such as

radiation) and P the rate at which work is done on the particle by body forces and surface forces (i.e., P is the mechanical power input) Then, in the absence of other forms of energy

input, the fundamental postulate of conservation of energy states that

Now, using Eq (4.12.6) and Eq (4.13.1), we have

thus, Eq (4.14.1) becomes

If we let u be the internal energy per unit mass, then

In arriving at the above equation, we have used the conservation of mass principle

Thus, the energy equation (4.14.1) becomes