ANALYSIS OF STRESS AND STRAIN I

When dealing with a new design or design improvement

problem the design engineer knows that the design must

(1) function according to some prescribed requirements and

(2) have an acceptable level of structural integrity. Stress

analysis is that part of the design process which strives to

ensure that each element of a given system will not fail to

meet the structural requirements of the design throughout

the specified life of the system.

Stress analysis tools range from the countless theoretical

techniques employed in elementary and advanced mechanics

of materials and the mathematical theory of elasticity, to

computer-based numerical procedures such as the finite

element method and to the wide variety of available

experimental techniques.

Practical stress analysis problems range from problems

which closely resemble simple models which have known

closed form solutions to complex problems which are not

easily adaptable to the various classical techniques. For the

complex problem, the analysit must either obtain an

approximation from a theoretical analysis, seek a numerical

solution, perform an experimental analysis or carry out a

combination of these approaches. In order to be competent,

the analysit should have a good working knowledge of the

theoretical, numerical and experimental procedures

employed in the field of stress analysis.

Experimental stress and strain analysis requires thorough

understanding of stresses and strains developed in any

structural member. Therefore, relationships between

deformations, strains, and stresses developed in a body

must be well understood. Further, generalized Hooke’s law,

equilibrium equations and compatibility equations, and Airy’s

stress functions for determining stresses at any point in a

member subjected to known loads/stresses on its boundary

are fundamental concepts. Plane stress state, plane strain

state, and three-dimensional displacement field along with

strain matrix should also be discussed.

DEFINITION AND COMPONENTS OF STRESS

Stress is simply an internally distributed force within a body. Let’s consider a body in

equilibrium subject to a system of external forces, Figure 1. Under the action of

these forces, internal forces are developed within the body. To examine the stresses

at some interior point Q, we use an imaginary plane to cut the body at a section a-a

through Q, dividing the body into two parts. As the forces acting on the entire body

are in equilibrium, the forces acting on one part alone must also be in equilibrium;

this requires the presence of forces on plane a-a. These internal forces, applied to

both parts, are distributed continuously over the cut surface. This process, referred

to as the Method of Sections, is relied on as a first step in solving all problems

involving the investigation of internal forces. The resulting figure is the Free Body

Diagram.

Figure 1 Method of sections: (a) Sectioning of a loaded body; (b) free body with external and internal forces; (c) enlarged area DA with components of the force D F.

An element of area DA, located at point Q on the cut surface, is acted on by force DF.

In general DF will not lie along x, y or z axis. Decomposing into components parallel

to x, y and z, we define the normal stress sx and the shearing stresses txy and txz.

Figure 1 Method of sections: (a) Sectioning of a loaded body; (b) free body with external and internal forces; (c) enlarged area DA with components of the force D F.

dA

dF

A

F

dA

dF

A

F

dA

dF

A

F

zz

Axz

yy

Axy

xx

Ax

D

D

D

D

D

D

DD

D

00

0

limlim

lim

tt

s

Written in integral form

A

xzz

A

xyy

A

xx dAFdAFdAF tts ,,

On a given surface only one normal stress and one shear stress can exist (Figure 2). The net

tangential force acting on the surface will be equal to ; the net shear stress

will then be equal to

22

zy FF DD

22

xzxynetx ttt

Figure 2 Stress components acting on a differential element Net shear stress

Stress components

To describe the state of stress at point Q completely it would be necessary to examine

other surfaces by making different planar slices. Since different planar slices would

necessitate different coordinates and different free body diagrams, the stresses on each

slice would be, in general, quite different. So, to understand the complete state of stress

at point Q, every possible surface intersecting point Q should be examined. However,

this would require an infinite number of slices surrounding point Q. That is, if point Q

were completely isolated from the body, it would be described by an infinitesimal

sphere. Naturally, this would be impossible to do, so a coordinate transformation is

necessary for this.

In order to enable the determination of the stresses on an infinite number of planes

passing through a point Q, thus defining the stresses at that point, we need only specify

the stress components on three mutually perpendicular planes passing through the point

(Figure 3). These three planes, perpendicular to the coordinate axes, contain three hidden

sides of an infinitesimal cube. It should be emphasized that when moving from point Q to

point Q the values of stress will, in general, change. Also, body forces, such as weight of a

body acting through the whole volume of the body can exist. These stresses lead to the

stress tensor.

Figure 3 Element subjected to three- dimensional stress. All stresses have positive sense.

STRESS TENSOR

All stresses acting on a cube of infinitesimally small dimensions, i.e. ∆x = ∆y = ∆z → 0 are

identified by the diagram of a cube. The first subscript on normal stress σ or shear stress τ

associates the stress with the plane of the stress, i.e. subscript defines the direction of the

normal to the plane, and the second subscript identifies the direction of stress as

txx = normal stress on yz plane in x-direction

txy = shear stress on yz plane in y-direction

txz = shear stress on yz plane in z-direction.

Similarly the stresses on xz plane and xy planes are identified as shown in Figure 1. Stress

tensor τij can be written as

Figure 4 Stresses on a cube element

zzzyzx

yzyxyx

xzxyxx

ij

ttt

ttt

ttt

t

But, generally the normal stress is designated by σ.

So, txx = σxx = σx , tyy = σyy = σy , tzz = σzz = σz

Stress tensor can be written as

Stress tensor is symmetric, i.e. tij = tji

Complementary shear stresses are txy and tyx; tzy and tyz ; tzx and txz

Complementary shear stresses meet at diametrically opposite corners of an element to satisfy the equilibrium conditions.

zzyzx

yzyyx

xzxyx

ij

stt

tst

tts

t

NORMAL STRAIN

Stresses on a rectangular parallelepiped DxDyDz element cause a change of the element both

dimensionally and in shape. The normal stresses cause the element to expand and/or shrink in the

x, y and z directions so that the element remains a rectangular parallelepiped. The shear stresses do

not basically cause dimensional changes, but shear causes the element to change shape from a

rectangular parallelepiped to a rhombohedron.

Initally let’s consider only sx applied to the element as shown in Figure 5. The element increases in

length in x direction and decreases in length in y and z directions. The dimensionless rate of

increase in length is defined as the normal strain where

ex, ey and ez represent the normal strains in x, y

and z directions. Thus the new length in any

direction is equal to its original length plus the

rate of increase (normal strain) times its

original length.

Figure 5 Deformations due to sx

zzz

yyy

xxx

z

y

x

DDD

DDD

DDD

e

e

e

By using Hooke’s law,

As the element elongates in x direction, it contracts in y and z directions. If the material is linear,

these contractions are also directly proportional to the normal stress. It is common practice to

express the contractions in terms of the primary normal strain, in this case ex using the Poisson’s

ratio n. If the material is isotropic,

Ex

x

se

Ex

xzy

snneee

The normal strains caused by sy and sz are

similar to the strains caused by sx:

E

E

E

E

zzyx

zz

y

yzx

y

y

snneee

se

snneee

se

For an element undergoing sx, sy and sz simultaneously, the effect of each stress can be added

using the concept of linear superposition. Thus the general stress-strain relationship for a linear,

homogeneous, isotropic material is

yxzz

zxyy

zyxx

E

E

E

ssnse

ssnse

ssnse

1

1

1

If the strains are known, stresses in terms of strains can be expressed as

yxzz

zxyy

zyxx

E

E

E

eenennn

s

eenennn

s

eenennn

s

1211

1211

1211

SHEAR STRAIN

The change in shape of the element caused by shear stresses can be illustrated first by examining

the effect of txy alone, Figure 6. The shear strain gxy is a measure of the deviation of the stresses

element from a rectangular parallelepiped. It is defined as

DABBADBADxyDg

Figure 6 Pure shear

where gxy is in dimensionless radians. For a linear, homogeneous,

isotropic material the shear strain is directly related to the shear

stress by

G

xy

xy

tg

Similarly, the remaining shear strans are related to the corresponding

shear stresses.

GGzx

zx

yz

yz

tg

tg ,

n

12E

G

It can be shown that for a linear, homogeneous, isotropic material the

shear modulus is related to the modulus of elasticity and Poisson’s

ratio by

Strain tensor in a biaxial case will be (Figure 7)

Figure 7 Shear stresses and shear strains on a differential element

y

xy

xy

x

eg

ge

2

2

STRAIN TENSOR

Three-dimensional strain tensor will be

z

zyzx

yz

y

yx

yzxy

x

egg

ge

g

gge

22

22

22

PLANE STRESS CONDITION

Under plane stress condition, stresses σx, σy, τxy are represented in one plane, i.e. central plane of a

thin element in yx plane, as shown shaded in Figure 8.

Figure 8 Plane stress state

In this case stresses σz = txz = tyz = 0.

If E and v are Young’s modulus and

Poisson’s ratio of the material, respectively,

then strains are

yxz

xyy

yxx

E

E

E

ssne

nsse

nsse

1

1

Shear strain

But shear strain gxy is equally divided about x and y axes both.

Strain tensor for plane stress condition will be as follows:

For a particular set of three orthogonal planes, where shear stresses are zero and normal stresses on

these planes are termed as principal stresses σ1, σ2, and σ3, stress tensor will be

G

xy

xy

tg

z

y

xy

xy

x

e

eg

ge

00

02

02

3

2

1

00

00

00

ss

s

PLANE STRAIN CONDITION

For strain condition shown in Figure 9 the strain tensor is

Figure 9 Plane strain state

In this, strain

Therefore,

To satisfy the condition of plane strain

Moreover,

y

yx

xy

x

eg

ge

2

2

0

0

yxz

z

z

EEssns

e

e

yxz ssns

0 zxyz gg

Example

A 100 mm cube of steel is subjected to a uniform pressure of 100 MPa

acting on all faces. Determine the change in dimensions between parallel

faces of cube if E = 200 GPa, v = 0.3.

DEFORMATIONS IN CARTESIAN COORDINATES

There are two types of strains resulting from deformation of a body, i.e. (a) linear or extensional

strains and (b) shear strains, resulting in change of shape. Let us consider an element of dimensions

∆x, ∆y, with change in x- and y-direction as ∆u and ∆v as shown in Figure 10. A body of dimensions

∆x and ∆y, in x y plane, represented by ABCD, deforms to AB′C′D as shown in Figure 10 (a).

Strain in x-direction,

Again it is deformed to A′B′CD as shown in Figure 10 (b), strain in y-direction,

.

Now the body in x-y plane is deformed to A′B′C′D, as shown in Figure 10 (c).

Figure 10 (a) Small element (b) Normal strain (c) Shear strains

xu

x DD

e

yv

y DD

e

Shear strain,

Moreover, linear strains are on both side of coordinate axes

In order to study the deformation, or change in shape, one has to consider a displacement field(s) for

a point in a body subjected to deformation.

A point P is located at position vector OP = r and displaced to position vector OP′ = r′. The

displacement vector S = r′ - r. The displacement vector (Figure 11) a function of x, y, z, has

components u, v, w in x-, y-, z-directions, such that

where S is a function of initial coordinates of point P.

u = u (x, y, z)

v = v (x, y, z)

w = w (x, y, z)

function of x, y, and z.

Figure 11 Displacement vectors

D C

D' C'

Dx

Dx+(du/dx) Dx

(dv/dx) Dx

(du/dy) Dy

y

yyu

x

xxv

D

D

D

D

/tan,

/tan 21

or,

kwjviuS

These strains at point P are defined by

and shear strains are

In x-y plane, or in a two-dimensional case, strains are

Example

The displacement field for a given point of a body is

at point P (1, −2, 3) determine displacement components in x-, y-, z-directions.

What is the deformed position?

4322 10624332 kzxjyxiyxS

Example 1.1

The displacement field for a body is given by

write down the strain components and strain matrix at point (2,1,2).

422 103242 kyxjziyxS

Figure 12 Stresses and elastic displacements in cylindrical coordinates

DEFORMATIONS IN CYLINDRICAL COORDINATES

In many problems the geometry of the component may not be suitable for the use of a rectangular

(Cartesian) coordinate system, and is more practical to use a different coordinate system. Problems

like thin or thick-walled pressure vessels, circular rings, curved beams, and half plane problems are

more suitable to a cylindrical coordinate system.

In figure 12 (a), an infinitesimal element is

constructed using r coordnates and the

corresponding normal and shear stresses are

shown. The depth of the element in the z

direction is Dz. It is assumed that both Dr and D

are approaching zero.

The stress-strain relations are the same as in

rectangular coordinates:

ssnse

ssnse

ssnse

rzz

zr

zrr

E

E

E

1

1

1

Figure 12 Stresses and elastic displacements in cylindrical coordinates

and

zrzr

zz

r

G

r

E

E

E

tn

g

tn

g

tn

g

12

12

12

The strain-displacement relations in cylindrical

coordinates are determined similar to Cartesian

coordinates but due to the complexity of the

deformed element, the deformation due to radial

and tangential displacements are viewed

separately, Figure 9 (b) and (c).

The initial and deformed elements are indicated

by dotted and solid lines, respectively. The net

radial strain is due to the radial displacement

only and is given by

r

u

r

urruu rrrrr

D

D

/e

The net tangential strain is

where the first term is due to the radial displacement and

the second to tangential displacement. Simplifying gives

e

D

D

D

DD

r

uuu

r

rur r /

e

u

rr

ur 1

The shear strain gr is equal to a + b. Thus

r

rrrurruu

r

uuu rrrr D

DD

D

D

///

g

Simplifying

r

u

r

uu

rr

r

g

1

The strains in z and zr planes are developed in a similar manner.

z

uzz

e

r

u

z

u

z

uu

rzr

zrz

z

g

g

1

A special case is encountered for problems symmetric with respect to the longitudinal z axis, which

are referred to as axisymmetric problems (such as pressurized circular disks), where vartiations

with respect to are zero and due to symmetry u = 0 everywhere. Thus, for axisymmetric

problems we can write

r

urer

urr

e

z

uzz

e

r

u

z

u zrzrzr

ggg 00

GENERALIZED HOOKE’S LAW

For a simple prismatic bar subjected to axial stress σx and axial strain εx (Figure 13), Hooke’s

law states that stress ∝ strain are related as

where E is the proportionality constant and Young’s modulus of elasticity of the material.

Figure 13 Simple bar subjected to axial stress and axial strain

xx

xx

E es

es

xx es ,

But for an elastic body subjected to six stress components σx, σy, σz, tyx, tyz, tzx and six strain

components, i.e. εx, εy, εz, gxy, gyz, gzx, the generalized Hooke’s law can be expressed as

σx = A11 εx + A12 εy + A13 εz + A14 gxy + A15 gyz + A16 gzx

σy = A21 εx + A22 εy + A23 εz + A24 gxy + A25 gyz + A26 gzx

σz = A31 εx + A32 εy + A33 εz + A34 gxy + A35 gyz + A36 gzx

τxy = A41 εx + A42 εy + A43 εz + A44 gxy + A45 gyz + A46 gzx

τyz = A51 εx + A52 εy + A53 εz + A54 gxy + A55 gyz + A56 gzx

τxz = A61 εx + A62 εy + A63 εz + A64 gxy + A65 gyz + A66 gzx

where A11, A12, …, A65, A66 are 36 elastic constants for a given material.

For homogeneous linearly elastic material, above noted six equations are known as

generalized Hooke’s law. Similarly, strains can be expressed in terms of stresses as follows:

For a fully anisotropic material there are 21 constants

In generalized Hooke’s law and for orthotropic materials there are nine constants in Hooke’s law.

Say for an isotropic material having the same elastic properties in all directions and which has no

directionally varying property, there are three principal stresses σ1, σ2, σ3 and three principal

strains ε1, ε2 and ε3, then Hooke’s law can be written as

where A, B, and C are elastic constants.

zxyzxyzyxzx

zxyzxyzyxx

BBBBBB

BBBBBB

tttsssg

tttssse

666564636261

161514131211

3211 eees CA

In the above equation ε1 is the longitudinal strain along σ1 but ε2 and ε2 are lateral strains so,

constants B = C, therefore

σ1 = A ε1 + B ε2 + C ε3

= A ε1 + B ε1 − B ε1 + B ε2 + B ε3

= (A − B)ε1 + B(ε1 + ε2 + ε3)

where ε1 + ε2 + ε3 = ∆, a cubical dilatation = first invariant of strain σ1 = (A − B)ε1 + ΔB

Let us denote (A − B) by 2 μ and B by λ, then

σ1 = λ Δ + 2με1

Similarly σ2 = λ Δ + 2με2

σ3 = λ Δ + 2με3

λ and μ are called Lame’s coefficients.

σ1 + σ2 + σ3 = 3λ Δ + 2μ(ε1 + ε2 + ε3)

= 3λ Δ + 2μ Δ

= (3λ + 2μ) Δ

Adding s1, s2 and s3

Principal stress

Solving for ε1, we get

sss

23321

D

13211

11

223

2

esss

s

es

D

3211 223

ss

s

e

From elementary strength of materials

Therefore, Young’s modulus

Poisson’s ratio,

3211

1 ssnse E

23E

n

2

Example

For steel, Young’s modulus E = 208000 MPa and Poisson’s ratio,

n = 0.3. Find Lame’s coefficients λ and μ.

ELASTIC CONSTANTS K AND G From the previous part we know that

Putting the value into above equation of n, we get for E

From the elementary strength of materials we know that

Therefore Lame’s coefficient μ = G, shear modulus.

If σ1 = σ2 = σ3 = p, hydrostatic stress or volumetric stress

Therefore, Bulk modulus is,

Shear modulus, , Lame’s coefficient.

2

23EorE

n

n

2

2or

n

n

12

22E

n 12GE

ModulusBulkKp

p

,3

23

23

3

D

D

3

23 K

G

Example 1.2

For aluminium, E = 67000 MPa, Poisson’s ratio, v = 0.33. Determine

the Bulk modulus and shear modulus of aluminium.

GENERAL THREE DIMENSIONAL STRESS TRANSFORMATIONS

As stated earlier, a state of stress not only depends on position within a structure, but it also

depends on the orientation of the surface containing the stresses. The analysis of stress

variation with respect to orientation can be carried out by using coordinate transformation. This

relates an arbitrary stress surface to a set of known stresses on three mutually orthogonal

surfaces defined by the stress matrix

zzyzx

yzyyx

xzxyx

stt

tst

tts

s

Let us consider the state of stress at a point, Figure 14. If

an arbitrary oblique plane is passes through the solid so

that the plane intersects the three mutually perpendicular

reference planes, a tetrahedral element about the point will

be isolated as shown in Figure 15 (a).

Figure 14 General state of stress

Let us consider the x' axis to be normal to the oblique plane and y '

and z ' axes to be tangent to the plane. The orientation of the normal

to the oblique surface can be established by the angles x ' x, x ' y and

x ' z which relate the axis to the x, y and z axes, Figure 15 (b).

Given the coordinates of the vertices A (a, 0, 0), B (0, b, 0) and

C (0, 0, c), the equation for the oblique plane is

Figure 15 Stresses on oblique surface

01cz

b

y

ax

The direction cosines of the surface normal are proportional to

the gradients (derivatives ) of the surface equation with respect

to the x, y and z directions, respectively. For the equation of a

plane the derivatives are simply the coefficients of the x, y and

z terms and are called directional numbers. The direction

cosines are thus,

cK

bK

aK zxyxxx

1cos

1cos

1cos

Where K is the proportional constant. The sum of the squares of

the directional cosines is unity.

1222

cK

bK

aK

Solving for K yields 222 )()()( abacbc

abcK

Thus x ' x is 222 )()()(

cosabacbc

bcaK

xx

From vector mechanics, the oblique area A0 is equal to the

cross product ACAB21

222

0 )()()(21

abacbcA

Since bcAx 21

xxxx

xx AAA

A

cos

2

2cos 0

0

Similarly zxzyxy AAAA cos,cos 00

Direction cosines can shortly be denoted as

zxzxyxyxxxxx nnn coscoscos

So zxzyxyxxx nAAnAAnAA 000

To determine the normal stress sx' acting on the oblique surface,

a summation of forces in the x' direction is necessary. The force

due to each stress is obtained by multiplying the stress by the

area over which the stress acts. Next, the component of the

force in the x ' direction is determined. For example, the force

due to sx is . The component of this force in the

x ' direction is . Summing all forces in x' direction

for equilibrium

xxxxx nAA 0ss

xxxxx nnA 0s

0000

000

0000

yxzxyzxxzxzxzxyxyz

xxyxxyzxxxzxyxxxxy

zxzxzyxyxyxxxxxx

nnAnnAnnA

nnAnnAnnA

nnAnnAnnAA

ttt

ttt

ssss

Simplifying results in

For a complete transformation of stresses with respect to the arbitrary

oblique surface, it is necessary to determine the shear stresses t x'y'

and t x‘z' .If we define the direction cosines of the y' axis relative to the

xyz coordinate system as ny‘x , ny‘y and ny‘z, a summation of forces in

the y' direction for equilibrium yields

xxzxzxzxyxyzyxxxxyzxzyxyxxxx nnnnnnnnn tttssss 222222

00000

000000

yyzxyzxyzxzxzyyxyzxyyxxy

zyxxzxyyxxxyzyzxzyyyxyxyxxxyx

nnAnnAnnAnnA

nnAnnAnnAnnAnnAA

tttt

ttssst

Simplifying results in

xyzxzyxxzx

yyzxzyyxyzxyyxyyxxxyzyzxzyyyxyxyxxxyx

nnnn

nnnnnnnnnnnnnn

t

ttssst

The final shear stress t z’x‘ on the oblique surface is found in a similar manner. Defining the direction

cosines of the z' axis relative to the xyz coordinate system as nz‘x , nz‘y and nz‘z, equilibrium of forces in

the z' direction yields

xzzxzzxxzx

yzzxzzyxyzxzyxyzxxxyzzzxzyzyxyxzxxxxz

nnnn

nnnnnnnnnnnnnn

t

ttssst

This set of equations are completely sufficient for the determination of the state of stress on any internal

surface in which an arbtrarily selected tangential set of coordinates is used (in this case y' z' coordinates).

xyzyzxzyyyyzyyxyxyzyzyyyxyxy nnnnnnnnn tttssss 222222

For a complete transformation of the stress element to that of a rectangular element oriented by the x' y' z'

coordinate system, the six stresses on the two surfaces with normals in the y' and z' directions must also

be determined (sy‘ , t y’z‘ , t y’x‘ , sz‘ , t z’y‘ , t z’x‘ ). Since it is assumed that no body force acts on the body the

cross shear will be equal to each other (t y’x‘ = t x’y‘, t y’z‘ = t z’y‘, t z’x‘ = t x’z‘). Thus with cross shears being

equal, it is only necessary to evaluate sy‘ , sz‘ and t y’z‘ .

xzzzzxzzyzyzyzxzxyzzzyzyxzxz nnnnnnnnn tttssss 222222

xzzyzzxyzx

yzzyzzyyyzxzyyyzxyxyzzzyzyzyyyxzxyxzy

nnnn

nnnnnnnnnnnnnn

t

ttssst

Example

The state of stress at a point relative to an xyz coordinate sytem is given by

σx = -8 MPa, σy = 4 MPa, σz = -5 MPa, τxy = 6 MPa, τyz =2 MPa, τzx = -2 MPa.

Determine the state of stress on an element that is oriented by first rotating the

xyz axes 45° about the z axis and then rotate the resulting axes 30 ° about the

new x axis.

Example 1.3

The state of stress at a point relative to an xyz coordinate sytem is given by

Determine the state matrix relative to a coordinate system defined by first rotating

the xyz axes 45° about the x axis and then rotate the resulting axes -45 ° about

the new z axis.

MPa

101525

154030

25300

s

PLANE STRESS TRANSFORMATIONS

For the state of plane stress shown in Figure 16, sz t yz = t zx =0. Plane stress

transformations are normally performed in the xy plane as shown in Figure 16

(b). The angles relating the x'y'z' axes are

09090

9090

9090

zzyzxz

zyyyxy

zxyxxx

Figure 16 Plane stress transformations

Thus,

100

0cossin

0sincos

zzyzxz

zyyyxy

zxyxxx

nnn

nnn

nnn

Stresses are obtained as

tsst

tsss

tsss

22

22

22

sincoscossin

sincos2cossin

sincos2sincos

xyyxyx

xyyxy

xyyxx