Finite Element Method

(DE ZG 513)
Lecture - Introduction to stress, traction, strain
BITS Pilani Dr. Amol,
Pilani Campus Dept. of Mechanical Engineering, BITS Hyderabad
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What is Mechanical Stress

Stress is an internal resistance offered by the material when subjected to external loads

F d F

Normal Stress =

The above expression is for the average stress induced across any transverse section perpendicular to the axis of
the bar and away from the region of loading

Since the geometry given above is uniform, the average stress remain same for all cross sections
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What is Mechanical Stress

In general, a structural member or a machine element will not possess uniform geometry of shape or size, and
the loads acting on it will also be complex

For example, an automobile crankshaft or an aircraft wing are subjected to loadings that are both complex as
well as dynamic in nature

In such cases, one will have to introduce the concept of the state of stress at a point
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Introduction to Body force, Surface force and Stress/Traction vector

Consider a body occupying a region of space referred to a rectangular coordinate system Oxyz, as shown in
Fig below

In general, the body will be subjected to two types of forces –

(1) Body force

(2) Surface force

The body forces act on each volume element of the body such as
gravitational force, the inertia force and the magnetic force

The surface forces act on the surface, at a point or area elements

of the body
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Introduction to Body force, Surface force and Stress vector

Wind force (Acts as a Surface force) Surface force

Weight of the bar (Act

as a Body force)

Weight of the Tree (Acts as a Body force)

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Introduction to Body force, Surface force and Traction

The surface force can either act at a point or acts over a certain area
Traction

In the figure force F acts at a point whereas ‘p’ is distributed over a Normal
certain area (n)
When the surface force acts over an area, it is often termed as surface
traction or traction vector. In the figure, ‘p’ is considered as traction

Important*- Surface traction need not act normal to the surface

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Assuming that under the action of both body forces and surface forces,
the body is in equilibrium.

CUTTING PLANE
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Introduction to Body force, Surface force and Stress vector

Let the body be cut into two parts C and D by a plane passing through
point P as shown in Figure

If we consider the free-body diagrams of C and D, then each part is in

equilibrium under the action of the externally applied forces and the
internally distributed forces across the interface

In part D, let ΔA be a small area surrounding the point P

In part C, the corresponding area at P’ is ∆A’

These two areas are distinguished by their outward drawn normals

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Introduction to Body force, Surface force and Stress vector

The action of part C on ∆A at point P can be represented
by the force vector and the action of part D on ΔA’ at
P’ can be represented by the force vector .

Traction vector or Stress Vector =

Traction vector and normal of the plane are in different directions

The value of traction vector depends upon the area as well as the normal
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Normal and shear stress

Normal stress Component

Resultant traction vector

Shear stress Component

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Rectangular stress Components

x- plane

x- axis

τxy -Stress acting on x plane in y direction

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Rectangular stress Components

Y-plane

X-plane
Above stress components are
Z-plane related with the Net traction forces
(T)
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Stress Components on an arbitrary plane

Consider a small tetrahedron at P with three of its faces normal to the

coordinate axes, and the inclined face ABC having its normal parallel to n

Let h be the perpendicular distance from P to the inclined face ABC

If the tetrahedron is isolated from the body and a free-body

diagram is drawn, then it will be in equilibrium under the action of
the surface forces and the body forces
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Stress Components on an arbitrary plane
Let be the resultant stress vector on face ABC. This can
be

resolved into components , , parallel to the three

axes x, y, and z
On the three faces, the rectangular stress components are
σx, τxy , τxz , σy, , τyz , τyx , σz, τzx , and τzy

If A is the area of the inclined face(ABC) then

Area of BPC = projection of area ABC on the yz plane = A nx

Area of CPA = projection of area ABC on the xz plane = A ny

Area of APB = projection of area ABC on the xy plane = A nz

Let the body force components in x, y and z directions be ϒx, ϒy and ϒz respectively, per unit volume.
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Stress Components on an arbitrary plane
The volume of the tetrahedron is equal to (1/3) A*h, where h is the
perpendicular distance from P to the inclined face ABC

For the equilibrium of the tetrahedron, the sum of the forces in x, y

and z directions must individually vanish. Thus, for force
equilibrium
in x direction

Cancelling A

Similarly, for equilibrium in y and z directions

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Stress Components on an arbitrary plane
In the limit as h tends to zero, the oblique plane ABC will
pass through point P,

=0

Consequently, one gets following equations

The above equations is known as Cauchy Stress Formula

This equation shows that the nine rectangular stress
components at P will enable one to determine the stress
components on any arbitrary plane passing through point P.
As the h reduced to zero, tetrahedron will become a point ‘P’
and therefore

Stress state at a point

Total 9 Stress Components - 3 Normal & 6 Shear components

Stress is a second order tensor as it needs both plane and direction σij , where i-plane, j-direction
Since, we have derived Cauchy’s Formula, we will compute net traction vector acting on a plane with normal ‘n’
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Stress Components on an arbitrary plane
If is the resultant stress vector on plane ABC, we have

If σn and τn are the normal and shear stress components

By Substituting
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Equality of cross shear
Consider an infinitesimal rectangular box surrounding a state of stress. Let the dimensions of the sides be ∆x, ∆y and
∆z
Taking moments about the Z axis, one gets

In the limit as ∆x, ∆y and ∆z tend to zero, the above equation gives

Similarly taking moments about other two axes gives

Thus, the cross shears are equal, and of the nine rectangular components, only SIX are
independent.
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Principal Stress
We have seen that the normal and shear stress components can be determined on any plane with normal n, using
Cauchy's formula given by

From the strength of materials perspective, following questions is

very important

Is there any plane passing through the given point

on
which the resultant stresses are wholly normal
?
Any component of ‘n’ lying in plane ABC will be zero value, hence
Plane ABC will be free of any Shear (τ)
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Principal Stress
Let us assume that there is a plane n with direction cosines nx , ny and nz on which the stress is wholly normal

Let σ be the magnitude of this stress vector. Then we have

The components of this along the x, y and z axes are

We also know that from the Cauchy’s formula

This equation can be written as

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So far, attention has been focused on the state of stress at a point.

But we observed that the state of stress in a body varies from point to point

Therefore, set of conditions need to be established to ensure equilibrium

Stress equilibrium equation

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Stress equilibrium equation

Consider a small rectangular element from its ∆z

parent body
∆y
∆x
The faces are marked as 1, 2, 3 etc. On face No. 1,
the stress components are σx, τxy and τxz.
τxz

τxy

On the right hand face, i.e. face No. 2, the stress components are

Following a similar procedure, the stress components on the six faces of the element can be determined
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Stress equilibrium equation

Let the body force components per unit volume in the x,y,z directions be ϒx, ϒy, and. ϒFor
z force equilibrium in X direction
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Stress equilibrium equation
Cancelling terms, and dividing by ∆x∆y∆z, we get
1

Similarly, equating forces in the y and z directions respectively to

zero, we get two more equations.

Equations 1,2,and 3 are called stress equilibrium equations (STATIC) and must be
satisfied at all points in the body
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Stress equilibrium equation
Let us consider a uniaxial loading (say in x direction) with NO body forces
In that case, σx will only be a non-zero value

This results in
𝑑σ𝑥
𝑜𝑟 =0,
𝑑𝑥
We know that, from Hook’s law σ𝑥 𝐸ε𝑥 ε − 𝑆𝑡𝑟𝑎𝑖𝑛
=

Substituting the above, we get 𝑑(𝐸ε𝑥) = 0

𝑑𝑥
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Concept of Strain
If the distance between any two arbitrary points in a body changes during or after application of the external stimuli,
then the body is considered to undergo strain

Body (state 1) Body (state 2)

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Concept of Strain
A strain can be either
(i) the extensional strain (Normal Strain ‘ε’)
(ii) the shear strain (ϒ) and (iii) Volumetric strain

Extensional (Normal) Strain = Change in length/Original length

Undeformed Deformed body (Strain in x direction) Deformed body (Strain in y direction)

body
strain in x direction strain in y direction
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Concept of Strain Deformed body

Shear
Strain

Θ1 equal to 90
Undeformed
Undeformed Deformed body body
body
Shear strain is Change in the orientation = θ1-
θ2

Shear strain
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Concept of Strain

In a general three-dimensional system, the state of strain at a point is given as

Given are displacements in x, y and z direction

Normal strain

Shear strain

From three displacements, we have computed six strain fields

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Equilibrium equation in terms of displacement

𝑑σ𝑥
1-D Stress Equillibrium 𝑜𝑟 =0,
equation 𝑑𝑥
We know that, from Hook’s law σ𝑥 𝐸ε𝑥
=

𝑑(𝐸ε𝑥)
Substituting the above, we get =0
𝑑𝑥

Substituting strain in terms of displacement , we finally obtain

𝑑 𝑑𝑢𝑥
𝐸 =0
𝑑𝑥 𝑑𝑥
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Relation between Stress and Strain
Relationship between Normal Stress and Normal Strain

The longitudinal stress σxx (or (σx )) is given as σxx = F/A

Considering cross sectional area as A,
From the Hook’s law,longitudinal stress (σxx) and longitudinal strain (εxx)are related as σxx = E εxx , E is modulus of elasticity
Stress (σxx) acting in X direction will cause elongation along X axis but contraction along Y axis

Hence there will be a strain along the y axis, which is called as a transverse strain (εyy)

The transverse strain (εyy) and the longitudinal strain (εxx) are related as εyy = -ν εxx , ν is Poisson’s Ratio

Finally, writing εxx and εyy in terms of stress σxx , we have εxx = σxx/E, and εyy = -ν εxx = -ν (σxx/E)
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Relation between Stress and Strain
Similarly considering the loading in Y direction,

We compute σyy = F/A

The stress and strain are related asσyy = E εyy

Stress (σyy) acting in Y direction will cause contraction along X axis (εxx)

Contraction Strain εxx and εyy are related as εxx = -ν εyy

Finally writing εxx and εyy in terms of stress σyy we

, have εyy = σyy/E, and εxx = -ν (σyy/E)
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Relation between Stress and Strain

Two-dimensional state of loading

From loading along X axis we computed
εxx = σxx/E 1
εyy = -ν (σxx/E) 2

From Loading along Y axis, we computed

εyy = σyy/E 3
εxx = -ν (σyy/E) 4

Since, Stress and strain are linearly related, we can use principle of superposition to calculate net strains in x and y
directions
Using principle of Superposition, Equation (1) and (4) will give Net strain in X direction εisxx = (1/E)*(σxx - νσyy ),

Equation (2) and (3) will give Net strain in Y direction isεyy = (1/E)*(σyy - νσxx )
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Relation between Stress and Strain
Relationship between Normal Stress and Normal Strain
Considering a general three-dimensional state of loading, we obtain
εxx = (1/E)*(σxx - νσyy - νσzz )
εyy = (1/E)*(σyy - νσxx - νσzz ONLY for ISOTROPIC
)
εzz = (1/E)*(σzz - νσxx - νσyy ) Materials
By mathematical manipulation, One can also write σxx , σxx , σzz in terms of εxx ,
εxx , εzz
Relationship between Shear Stress ( ) and Shear Strain ( )
ϒxy = τxy /G
ϒyz = τyz /G G = Shear Modulus
ϒzx = τzx /G
Relationship between Shear Modulus (G) and Young’s Modulus (E)
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Relation between Stress and Strain
Generalized Relationship between Stress and Strain

We know that , at a given point, we have 6 stress Components (=σxx, σyy, σzz, τxy, τyz , τzx)

We also know that , at a given point, we have 6 strain Components (εxx, εyy, εzz, ϒxy, ϒyz , ϒzx)

For an ANISOTROPIC
Material = Total 36
Constants
For an ISOTROPIC Material
= 2 Constants (E and ν)

6X1 6X6 6X1 For ORTHOTROPIC Material =

Stress Elasticity Matrix Strain
(36 Constants) ??
Explore Yourself
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Relation between Stress and Strain
Generalized Relationship between Stress and Strain

For an ISOTROPIC Material , which has 2 Constants (E and

ν)

‘D’ matrix (Symmetric)

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In general parts and components used in various engineering applications are subjected to complex loading

Stress analysis of three-dimensional parts/components subjected to complex loading is challenging and time
taking
Therefore, many times bodies under the state of three-dimensional stress state are analyzed under two-
dimensional state of loading

When a 3D body is analyzed as a 2D state of stress/strain, then it comes under plane stress or plane strain
analysis
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Plane Stress Condition
Plane stress theory is applied to the bodies bounded by two parallel planes separated by a distance that
is very small in comparison to the other dimensions of the body

A material is said to be under plane stress if the stress components are zero across a particular plane

σxx ≠ 0 σyy ≠ 0 τxy ≠ 0

GIVEN
σzz = 0 τyz = 0 τzx = 0
=0
εxx = (1/E)*(σxx - νσyy - νσzz εxx ≠ 0
) =0
εyy = (1/E)*(σyy - νσxx - νσzz ) εyy ≠ 0
=0
εzz = (1/E)*(σzz - νσxx - νσyy ) ≠ 0

ϒxy = τxy/G
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Plane Strain Condition
Plane strain theory is applied to the bodies bounded by two parallel planes separated by a distance that
is very large in comparison to the other dimensions of the body

A material is said to be under plane strain if all the strain components are zero across a particular plane

εzz = 0 ϒzx = 0 ϒzy = 0

GIVEN
σxx ≠ 0 σyy ≠ 0 τxy ≠ 0
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Plane Strain

εzz = (1/E)*(σzz - νσxx - νσyy ) = 0 σzz = ν (σxx +

σyy)
εxx = (1/E)*(σxx - νσyy - νσzz ) = (1/E)*( σxx - νσyy – ν(ν(σxx + σyy )) )

εxx = (1/E)*( σxx (1- ν2)- ν(1+ ν )σyy )

Similarly

εyy = (1/E)*( σyy (1- ν2)- ν(1+ ν )σxx )

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