Contmech Imp

Surface force. Surface traction

Consider, next, forces acting inside a continuous body. Suppose that a part of the body B occupies a

region R which has the surface S.

δS

t( )nS

R

n

B

P

O

x

Ass. Point P ∈ S, δS is a small element of Saround P , and n ⊥ δS at P with nini = 1.

Ass. Material outside R exerts a force

δp = t(n)

δS

on the material inside the surface through the ele-

ment δS.

δp is called the surface force (pintavoima) and t(n)

the mean surface traction (jannitysvektori, traktio)

across dS from the outside to the inside of R.

A force −δp and a traction −t(n) are transmitted across dS from the inside to the outside of R.

Clearly, δp depends on the position of P (or its position vector x) and the orientation of n, but as

δS → 0, t(n) is assumed to tend to a limit independent of the shape of δS. Henceforth, denote

t(n)

(x) = limδS→0

δp

δS

and call t(n) the (surface) traction at the point P on the surface with normal n. In general, n ∦ t(n).

– Typeset by FoilTEX – 1

Components of stress

At P , there is a traction t(n) associated with each direction n

through P . In particular, there is a traction associated with all base

vectors ei of a given coordinate system. Denote these by

ti := t(ei).

Thus, e.g., t1 is the force per unit area exerted on the negative side

of a surface normal to the x1 axis by the material on the positive

side of this surface.

Resolve the vectors ti in components:

ti = Tijej (i = 1, 2, 3; summation over j).

⇒ ti · ek = Tijej · ek = Tijδjk = Tik ⇒ Tij = ti · ej

Quantities Tij are the stress components (jannityskomponentit).

e1

t 1

x1

e1

e2

T11

T12

T13

t 1

T31

T21

t 2

T22

T23

T32

T33

e3

t 3

If T11 > 0, material on the positive side of a surface with n = e1 is pulling the material on the

negative side of the surface and vice versa (tension), and if T11 < 0, material on the positive side is

pushing the material on the negative side and vice versa (compression).

Stress components Tij with i = j are called the normal (or direct) stress components (normaalijan-

nitykset) and those with i 6= j are called shearing stress components (leikkausjannitykset).


Traction on any surface. Cauchy’s law

Q2

Q3

Q1

e3−

t 3− δ S3

t 1− δ S1

e1−

t( )δSn

t 2− δ S2

e2−

e3

e1

e2

n

P

Consider forces acting on the tetrahedron

PQ1Q2Q3 ⊂ B.

The unit normals of, tractions on, areas of and surface

forces on the four faces are

face normal traction area force

PQ2Q3 −e1 −t1 δS1 −t1δS1



Q1Q2Q3 n t(n) δS t(n)δS

Areas δSi are projections of δS to the coordinate planes.

⇒ δSi = n · eiδS = niδS.

A body force (tilavuusvoima) may also be present, with mean value b / unit mass inside the tetrahedron.

Ass. For any part of a body, the rate of change of momentum is proportional to the resultant force

acting. This is known as Cauchy’s law of motion, generalization of NII. For PQ1Q2Q3, in particular

⇒ − t1δS1 − t2δS2 − t3δS3 + t(n)

δS + ρb δV = ρf δV

⇒ t(n)

= niti + ρδV

δS(f − b)

ass. δS→0−→ niti = niTijej

⇒ t(n)j = niTij components of t

(n)


Transformation of stress components. Stress tensor

Traction t(n) is given by

t(n) = niti = niTijej

in the base ei. In particular, choosing n = ei, we have ti = Tijej.

The same results must hold in another base, i.e., if n = ep = Mpiei, then ni = Mpi and

tp := t(ep) = MpiTijej = MpiTijMqjeq = MpiMqjTijeq

But we should define the stress components in the base eq analogously:

tp = Tpqeq.

Thus,

Tpq = MpiMqjTij i.e., T = MTMT

with T = (Tij)

and the stress components form a second-order tensor,

T = Tijei ⊗ ej

called Cauchy stress tensor. It completely describes the state of stress of a body. Hereafter, we will

call it simply the stress tensor (jannitystensori).

Since n is a vector and T is a tensor, we may write the equation t(n)j = niTij in form

t(n) = n · T


Example

Problem. The state of tress throughout a continuum is given wrt. the Cartesian coordinate system

(O, ei) by the matrix

T =

0

@

3x1x2 4x22 0

4x22 0 2x3x2

0 2x3x2 0

1

A

Determine the normal stress component on the cylindrical surface φ := x22 + x2

3 − R2 = 0.

Solution. First, n ‖ ∇φ = 2x2e2 + 2x3e3. Normalize magnitude to unity:

n =2x2e2 + 2x3e3

(4x22 + 4x2

3)1/2

=x2e2 + x3e3

(x22 + x2

3)1/2

=x2e2 + x3e3

R.

Thus,

t(n)

= n · T =x2e2 + x3e3

R·

· (3x1x2e1 ⊗ e1 + 4x22e1 ⊗ e2 + 4x

22e2 ⊗ e1 + 2x3x2e2 ⊗ e3 + 2x3x2e3 ⊗ e2)

= R−1[4x32e1 + 2x3x

22e3 + 2x2

3x2e2]

⇒ Tnn = n · t(n) = R−2(2x2

3x22 + 2x2

3x22) = 4R−2x2

3x22 = 4R2 sin2 ϕ cos2 ϕ = R2 sin2 2ϕ

where x2 = R cos ϕ and x3 = R sin ϕ have been used. Thus, the body is in tension (Tnn ≥ 0).


Equations of equilibrium

Consider, next, the body B in equilibrium.

δS

t( )nS

R

n

B

P

O

x

Ass. R is an arbitrary region of B and S is its

surface with unit normal n.

Ass. Two kinds of forces act on the material: surface

forces acting across S and body forces.

Ass. In equilibrium, the resultant force and the resul-

tant torque about O acting on the material is zero:

ZZ

St(n)

dS +

ZZZ

Rρb dV = 0

ZZ

Sx × t

(n)dS +

ZZZ

Rρx × b dV = 0.

In terms of components

0 =

ZZ

SniTijdS +

ZZZ

Rρbj dV =

ZZZ

R

„

∂Tij

∂xi

+ ρbj

«

dV

0 =

ZZ

SejpqxpnrTrq dS +

ZZZ

Rejpqρxpbq dV =

ZZZ

Rejpq

„

∂(xpTrq)

∂xr

+ ρxpbq

«

dV,

where the divergence theorem,RR

niTij dS =RRR

∂iTij dV , has been used.


But since R is arbitrary

∂Tij

∂xi

+ ρbj = 0, ejpq

„

∂(xpTrq)

∂xr

+ ρxpbq

«

= 0.

However, as ∂xp/∂xr = δpr, we get

0 = ejpq

„

xp

∂Trq

∂xr

+ δprTrq + ρxpbq

«

= ejpqxp

„

∂Trq

∂xr

+ ρbq

«

+ ejpqTpq = ejpqTpq

Thus,

j = 1 : 0 = e123=+1

T23 + e132=−1

T32 = T23 − T32

j = 2 : 0 = e213=−1

T13 + e231=+1

T31 = T31 − T13

j = 3 : 0 = e312=+1

T12 + e321=−1

T21 = T12 − T21

and, therefore, Tij = Tji. Thus, in equilibrium, the stress tensor is symmetric and

∇ · T + ρb = 0.

We will later prove that the symmetry of T prevails also for bodies in motion. Thus, T will be treated

as a symmetric tensor hereafter.


Boundary conditions in equilibrium

Consider a finite body B, occupying the region R, in equilibrium.

Let S be the surface of the area R, and nS be the outward unit vector and dS an element of that

surface. The traction at the surface is, thus,

t(nS)

= nS · T

and the surface force caused by this traction on the element dS is t(nS) dS. In equilibrium, this has

to be balanced by an external surface force, p dS. Thus, the boundary condition at the surface is

p dS = t(nS) dS ⇒ p = nS · T .

In particular, if the body is in vacuum, we must require

nS · T = 0

at its surface. On the other hand, if the body B1 is immersed in a medium constituting another body,

B2, the boundary condition at the separating surface is

nS · T(1) = nS · T

(2),

where T (1) and T (2) are the stress tensors measured inside the bodies 1 and 2, respectively.


Example: the law of Archimedes

Consider a body B1 with constant density ρ1 in a homogeneous gravitational field g = −ge3 immersed

in an incompressible fluid (body B2; density ρ2) with

T(2)

= −p(x) I,

where p(x) is the hydrostatic pressure of the fluid. The equilibrium

equation for the fluid reads

ρ2ge3 = ∇ · T(2)

= −∇p

⇒ ∂3p = −ρ2g; ∂jp = 0, j = 1, 2

∴ p = p0 − ρ2gx3,

where p0 is constant. Thus, surface traction exerted by the fluid on the

surface of B1 reads

n · T(2)

= −(p0 − ρ2gx3)n.

ρ1 g

2

ρ2

1B

B

Assuming that also T (1) is isotropic, T (1) = −P (x)I, and the interior of B1 to be in equilibrium,

P = P0 − ρ1gx3 and n · T (1) = −(P0 − ρ1gx3)n. Thus, for an equilibrium to exist requires

P0 − ρ1gx3 = p0 − ρ2gx3 ∀ x ∈ S1

and this is possible only if ρ1 = ρ2 and P0 = p0.


If ρ2 6= ρ1, there is a non-vanishing resultant force F acting on the body B1

F =

ZZ

S1

n · T(2) dS +

ZZZ

R1

ρ1g dV =

ZZZ

R1

(∇ · T(2) + ρ1g)dV

= −ZZZ

R1

(∇p + gρ1e3)dV =

ZZZ

R1

g(ρ2 − ρ1)e3 dV = gV1(ρ2 − ρ1)e3

Note: the tensor field T (2) is defined only in the region outside the boundary of the body B1. However,

the mathematical evaluation of the surface integral can be done by continuing the field to the region

inside the body B1, i.e., by taking the derived form p(x) = p0−gρ2x3 to apply also inside the region.

This is denoted by adding the tilde in

T(2)

= −pI

p(x) = p0 − gρ2x3 ∀ (x1, x2, x3) ∈ R1 ∪ R2.

The eq. for F is the familiar law of Archimedes, giving the buoyancy (noste) as

b = F − ρ1V1g = gV1(ρ2 − ρ1)e3 − (−ρ1V1ge3) = ρ2V1ge3.

Note also that the shape of the body B2 does not affect the buoyancy.


Principal stress components. Principal axes of stress

In general, t(n) ∦ n; the traction has a tangential component to a surface (with normal n) and as well

as a normal component. However, for certain special directions n we may have t(n) ‖ n. Then

n · T = t(n)

= TnT symm.=⇒ T · n = Tn

This clearly occurs, when T is one of the three principal components of T , denoted by T1, T2, and T3,

and n is the unit vector determining the corresponding principal axis. We call the principal components

Ti as the principal stress components and the principal axes ni as the principal axes of stress.

The principal stress components are the roots of the equation

det(T − T I) = 0.

Usually, we denote the principal stress components so that T1 ≥ T2 ≥ T3.

If the orthogonal principal axes, ni, of T are taken as the base vectors, then the matrix of T is diagonal,

T = diag(T1, T2, T3).

Thus,

T = T1n1 ⊗ n1 + T2n2 ⊗ n2 + T3n3 ⊗ n3.

Note, however, in general the stress components vary from point to point, and the principal components

Ti and axes ni are functions of position.


Example 1

Problem. Let the matrix of T in some coordinate system (O, ei) and in appropriate units be

T =

0

@

1 2 3

2 4 6

3 6 1

1

A

Find the principal components and pricipal axes of stress.

Solution. The principal stress components are obtained from

0 =

˛

˛

˛

˛

˛

˛

1 − T 2 3

2 4 − T 6

3 6 1 − T

˛

˛

˛

˛

˛

˛

= (1 − T )[(4 − T )(1 − T ) − 36] − 2[2(1 − T ) − 18] + 3[12 − 3(4 − T )]

= (1 − T )(−32 − 5T + T 2) − 2(−16 − 2T ) + 9T

= −T3+ 6T

2+ 40T = −T (T

2 − 6T − 40) = −T (T − 10)(T + 4)

Thus, T1 = 10, T2 = 0, T3 = −4.


The corresponding eigenvector components are obtained from Tn(i) = Tin

(i) (no summation), i.e.,

8

>

<

>

:

n(i)1 + 2n

(i)2 + 3n

(i)3 = Tin

(i)1

2n(i)1 + 4n

(i)2 + 6n

(i)3 = Tin

(i)2

3n(i)1 + 6n

(i)2 + n

(i)3 = Tin

(i)3

(i = 1, 2, 3; no summation)

i = 1, T1 = 10 : n(i)2 = 2n

(i)1 ∧ 3n

(i)3 = 5n

(i)1

i = 2, T2 = 0 : n(2)1 = −2n

(2)2 ∧ n

(2)3 = 0

i = 3, T3 = −4 : n(i)2 = 2n

(i)1 ∧ n

(i)3 = −3n

(i)1

Thus, as ni = n(i)j ej and ‖ni‖ = 1 ∀ i = 1, 2, 3,

n1 =e1 + 2e2 + 5

3e3q

1 + 4 + 259

=3e1 + 6e2 + 5e3√

9 + 36 + 25=

3e1 + 6e2 + 5e3√70

n2 =−2e1 + e2√

1 + 4=

−2e1 + e2√5

n3 =e1 + 2e2 − 3e3√

1 + 4 + 9=

e1 + 2e2 − 3e3√14


Example 2

Ass. The stress tensor T has distinct principal components T1, T2 and T3 (in descending order).

Problem. Show that T1 is the maximum and T3 is the minimum of the normal component of stress,

n · T · n, for any orientation of n.

Solution. We have

T = n · T · n = niTijnj.

Choosing the pricipal axes as the base vectors,

T = T1e1 ⊗ e1 + T2e2 ⊗ e2 + T3e3 ⊗ e3,

and, thus,

T (n) = (n · e1)2T1 + (n · e2)

2T2 + (n · e3)

2T3

:= n21T1 + n2

2T2 + n23T3

where ni are the components of n = niei. We, therefore, need to find the extrema of T (ni) subject

to the constraint f(ni) := nini − 1 = 0. This can be done using an undetermined Lagrangian

multiplier σ:∂T

∂ni

+ σ∂f

∂ni

= 0, f(ni) = 0

giving Tini + σni = 0 (no summation) for the first three equations.


Thus, solve

Tini + σni = 0 (no summation)

nini − 1 = 0

Multiply the upper one with ni and sum over i to get

0 =X

i

(niTini + σnini) = T + σ

Thus,

Tini = Tni (no summation)

at the three extrema. Thus, either T = Ti or ni = 0, and if the principal stresses are all distinct,

T = T1, n1 = ±1, n2 = n3 = 0

T = T2, n2 = ±1, n1 = n3 = 0

T = T3, n3 = ±1, n1 = n2 = 0

are the only possible solutions for the extremal values of T (ni) and the corresponding surface normals

ni. Thus, the normal stress component relative to any orientation of the surface normal n varies between

T3 and T1.


Stress invariants. Stress deviator

As T is a symmetric second-order tensor, it has three independent invariants. These are denonted by

J1, J2 and J3, called stress invariants, and defined as

J1 = T1 + T2 + T3 = tr T = Tii

J2 = −(T1T2 + T1T3 + T2T3) = 12{tr T

2 − (tr T )2} = 1

2(TijTji − TiiTjj)

J3 = T1T2T3 = det T .

Note that the sign of J2 is opposite of the previously introduced tensor invariant I2.

Earlier, we considered a simple spherical stress tensor of form,

T = −pI,

where p is the hydrostatic pressure of the fluid. If the stress tensor is of this form, it is said that the

body is in a pure hydrostatic state of stress. Clearly, p = −13tr T .

Even if T is not isotropic, it is often convenient to decompose in an isotropic and deviatoric part:

T = S + 13(tr T )I = S + 1

3J1I = S − pI; p := −13tr T

The deviatoric (Sii = 0) part S is called the stress deviator tensor. It has the components

Sij = Tij − 13Tkkδij.


The principal axes of S are those of T . (This is clear since the principal axes of −pI are arbitrary.)

Let S1, S2 and S3 denote the principal components of S. Thus,

S1 + S2 + S3 = tr S = 0,

and, because

Sini = S · ni = {T − 13(tr T )I} · ni = (Ti − 1

3tr T )ni (no summation)

for all principal axes ni, we have

S1 = 13(2T1 − T2 − T3), S2 = 1

3(2T2 − T1 − T3), S3 = 13(2T3 − T1 − T2)

Because S is deviatoric, it has only two basic invariants. These are taken to be

J′2 = −(S2S3 + S3S1 + S1S2) = 1

2tr S2

J ′3 = S1S2S3 = det S = 1

3tr S3

It is sometimes convenient to adopt J1, J ′2 and J ′

3 as the basic invariants of T .


Shear stress

The normal stress component on a surface normal to the x1 axis is T11. The shear stress is the

resultant of the other two components of traction, i.e.,

T 12e 2

T 13e 3

+T 11e 1

T12e2

e3

t1

e1e2

T 13e 3

T12e2 + T13e3,

which has a magnitude ofq

T 212 + T 2

13.

More generally, i.e., relative to the surface normal to n, the

normal stress component is n · t(n) = n · T · n. Thus,

the shear stress is

t(n) − (n · t

(n))n = n · T − (n · T · n)n

= n · T − (n · T ) · (n ⊗ n)

= (n · T ) · {I − n ⊗ n}= nrTrs(δsj − nsnj)ej

Let T1 ≥ T2 ≥ T3 be the principal stress components and n1, n2 and n3 be the corresponding

principal axis directions. It can be shown that as n is varied, the magnitude of the shear stress attains

a maximum of 12(T1 − T3) along a direction which is aligned with 1√

2(n1 ± n3).

Note: 12(T1 − T3) = 1

2(S1 − S3) and that shear stress vanishes on any surface for pure hydrostatic

state of stress.


Some simple states of stress

(a) Hydrostatic state of stress

T = −pI,

where the hydrostatic pressure is generally a function of position. Leads to the equilibrium

equation

−∇p + ρb = 0,

where b(x) is the body force per unit mass at position x.

Regarding the body forces to be negligible, the equilibrium equation reads

0 = ∇ · T , i.e., ∂iTij = 0 ∀j = 1, 2, 3.

As there are six independent stress components and only three scalar equilibrium equations, so there’s

no unique solution for the state of stress in equilibrium. Some possible cases fulfilling the equilibrium

equations for b = 0 are considered below.

Homogeneous stress. Stress with all its components spatially constant. Special cases include

(b) Uniform tension or compression in the x1 direction is given by

T11 = σ, T22 = T33 = T12 = T13 = T23 = 0

Example: a uniform cylindrical bar parallel to the x1 axis with uniform forces applied on the

planar ends of the bar and no forces on the cylindrical surface. If σ > 0 the bar is in tension and

if σ < 0 it is in compression. Principal axes: e1 and two arbitrary orthogonal vectors ⊥ e1.


(c) Uniform shear stress in the x1 direction on planes x2 = const. arises if

T21 = τ, T11 = T22 = T33 = T13 = T23 = 0

Example: laminar shear flow of viscous fluid with v = v(x2)e1. Principal axes of stress: e3

and 1√2(e1 ± e2).

Inhomogeneous equilibrium states of stress include

(d) Pure bending. Let

T11 = cx2, T22 = T33 = T12 = T13 = T23 = 0,

where c is a constant. Approximates the

state of stress in a prismatic beam parallel

to x1 axis with couples about axes parallel

to x3 acting on its end faces (see Fig.).

Plane x2 = 0 chosen so that the resul-

tant on each face = 0. Principal axes as

in (b).

x1

x3

x2

(e) Plane stress.

T11 = T11(x1, x2), T22 = T22(x1, x2), T12 = T12(x1, x2), T33 = T13 = T23 = 0

Here, the equilibrium eqs. read ∂jTij = 0, where i, j = 1, 2. One principal axis e3 and the

others in the x1x2 plane.


(f) Pure torsion. Suppose that

T13 = x2g(r), T23 = −x1g(r), T11 = T22 = T33 = T12 = 0,

where r =q

x21 + x2

2 and g(r) is an arbitrary function. Gives the state of stress in a cylindrical

bar, whose axis of symmetry coincides with x3 axis and which is twisted by couples acting about

the axis and applied to the ends of the cylinder.

Principal directions: radial direction er = 1r(x1e1+x2e2) and the two bisectors 1√

2(eϕ±e3)

of the tangential and axial directions, eϕ = 1r(−x2e1 + x1e2) and e3.

x3x2

x1


Date post:	12-Jan-2016
Category:	Documents
Upload:	trupti
View:	228 times
Download:	1 times

Contmech Imp

Documents