CHAPTER 4

STRESS AND STRAINBITS PILANI GOA

Objectives

STRESS• To know state of stress at a point

• To solve for plane stress condition applications

STRAIN• To know state of stress at a point

• To solve for plane stress condition applications

STRESS

Let’s considered that the body is cut in two halves by a plane passing through point ‘o’

Stress vector can be defined as

A

FT

A

n

0

)(

lim

Here T is force intensity or stress acting on a plane whose normal is ‘n’ at the point O.

Characteristics of stress

1. The physical dimensions of stress are force per unit area.

2. Stress is defined at a point upon an imaginary plane, which divides the element or material into two parts.

3. Stress is a vector equivalent to the action of one part of the material upon another.

4. The direction of the stress vector is not restricted.

Stress vector may be written in terms of its components with respect to the coordinate axes in the form:

kTjTiTT z

n

y

n

x

nn )()()()(

Body cut by a plane ‘mm’ passing through point ‘O’ and parallel to y-z plane

Rectangular components of the force vector F acting on the small area centered on point O

Stress components on positive x face at point O.

3-Dimentional State of Stress OR Triaxial State of Stress

x xy xz

yx y yz

zx zy z

A knowledge of the nine stress components is necessary in order to determine the components of the stress vector T acting on an arbitrary plane with normal n.

Definition of positive and negative faces:

Positive face of given section

If the outward normal points in a positive coordinate direction then that face is called as positive face

If the outward normal points in a negative coordinate direction then that face is called as Negative face

Negative face of given section

Stress components acting on the six sides of a parallelepiped.

PLANE STRESS CONDITION

PLANE STRESS CONDITIONFor example: Thin sheet which is being pulled by forces in the plane of the sheet.

X

Y

Z

The state of stress at a given point will only depend upon the four stress components.

x xy

yx y

PLANE STRESS CONDITION

Stress components in the z direction has a very small value compared to the other two directions and moreover they do not vary throughout the thickness.

Plane Stress Condition

If take the xy plane to be the plane of the sheet, then x , ’x , y , ’y , xy , ’xy , yx and ’yx will be the only stress components acting on the element, which is under observation.

x ’x

xy

’xy

yx

’yx

’y

y

EQUILIBRIUM OF A DIFFERENTIAL ELEMENT IN

PLANE STRESS

EQUILIBRIUM OF A DIFFERENTIAL ELEMENT IN PLANE STRESS

Stress components in plane stress expressed in terms of partial derivatives.

ΣM = 0 is satisfied by taking moments about the center of the element

xy = yx

Equality of Cross Shears

This Equation says that in a body in plane stress the shear-stress components on perpendicular faces must be equal in magnitude.

It could be shownzy = yz and xz = zx

Definition of positive and negative xy

xy

xy xy

xy

Positive Shear Negative Shear

ΣF = 0 is satisfied by taking moments about the center of the element

Three-dimensional equations in the form of indicial notation

STRESS COMPONENTS

ASSOCIATED WITH ARBITRARILY ORIENTED FACES IN

PLANE STRESS CONDITION

STRESS COMPONENTS ASSOCIATED WITH ARBITRARILY ORIENTED FACES IN PLANE STRESS

0FResolve forces in normal and along the oblique plane

x

yx

y

A

B C

0cos BCXY

AC cosABX sinABXY

sinBCY

AB = AC cos

BC = AC sin

x

xy

y

A

B C

y

yx

x

cossin2sincos 22XYYX

2

cos2-1sin and

2

12coscos 22

2sin2cos22 XY

YXYX

0sin BCXY

AC sinABX cosABXY

sinBCY)sin(coscossin)( 22 XYXY

2cos2sin2 XY

XY

MOHR'S CIRCLE REPRESENTATION

OF PLANE STRESS

xy

xy

x

OC= σx + σy / 2

R = [(σx - σy / 2)2 + xy2]1/2y

x’ (σ, )

y’ (σ+90, )

σ1σ2

90

22tan

12

1

yx

xy

222,1 )

2(

2 xyyxyx

max

max

smax

smax

2cos2sin2 XY

XY

]2cos2sin2

[0

XYXY

d

d

d

d

xy

xys

22tan max

22max )

2( xy

yx

Example:

For given plane stress state find out normal stress and shear stress at a plane 450 to x- plane. Also find position of principal planes, principal stresses and maximum shear stress. Draw the mohr’s circle and represent all the stresses.

x x

xy

xy

y

y

y= 50 MN/m2

x=110 MN/m2

xy= 40 MN/m2

= 450

= 120 MN/m2 +90= 40 MN/m2

= -30 MN/m2

1

1

2

2

1

1= 130 MN/m2 2= 30 MN/m2

1 = 26.560 2 = 116.560

max= 50 MN/m2

smax1 = 71.560 smax2 = 161.560

’x

’xmax

’y

’ymax

smax1

σ1σ2

max

max

x

Y

σoσo

x

Y

σoσo

x

Y

σo

σo

x

Y

σo

σo

x

Yσo

σo

σo

σo

x

Y

σoσo

σo

σo

x

Yσo

σo

σo

σo

x

Y

σoσo

σo

σo

x

Y

ζo

ζo

x

Y

ζo

ζo

Addition of Two States of stress

x

Yσoσo x

Yσoσo

Addition of Two States of stress

x

Yσoσo x

Yσoσo

ab2ζ o

2ζ o

x

Y

ζo

ζo

Find the principal stress directions if the stress at a point is sum of the two states of stresses as illustrated

30o

Find the principal stress directions if the stress at a point is sum of the two states of stresses as illustrated

x

Y2σo

2σoab

3σ o

3σ o

45o

3-D State of StressGeneral State of Stress

x xy xz

yx y yz

zx zy z

x xy

y yz

zx z

9 stress components 6 stress components

Mohr’s Circle for 3-D state of stress

σ2

σ1

σ3

Analysis of Deformation

Strain Analysis

ANALYSIS OF DEFORMATION

Rigid-body translation. Rigid-body rotation about c.

Displacement Components

Deformation without rigid-body motion.

Sum of all the displacements

Displacement Components

Uniform strain

Non Uniform strain

STRAIN

Normal (ε)

Shear (γ)

Definition of Normal Strain

x

Y

Δx

Δu

Δy

x

Y

Δx

Δv

Δy

Definition of Shear Strain

x

Y

Δx

Δu

Δy

Δv

x

v

1tan

y

u

1tan

Plane strain (Biaxial Strain)

RELATION BETWEEN STRAIN AND DISPLACEMENT IN PLANE STRAIN

PLAIN STRAIN CONDITION

STRAIN COMPONENTS ASSOCIATED WITH ARBITRARY SETS OF AXES

It could be possible to express these displacement components either as functions of x' and y' or as functions of x and y.

By Chain Rule of Partial Derivatives

Stress Formulae cossin2sincos 22

' XYYXx

2sin2cos22' XY

YXYXx

2cos2sin2'' XY

XYyx

2sin2cos22' XY

YXYXy

MOHR'S CIRCLE REPRESENTATION OF PLANE STRAIN (Center & Radius)

M’Circle for Principal &

Max. Shear Plane

M’Circle for any arbitrary Plane

Example

A sheet of metal is deformed uniformly in its own plane so that the strain components related to a set of axes xy arex = -200 X 10-6

y = 1000X 10-6

xy = 900 X 10-6

find the strain components associated with a set of axes x'y‘ inclined at an angle of 30° clockwise to the xy set, as shown in the fig. Also to find the principal strains and the direction of the axes on which they exist.

Location of x', y' axes:

Orientation of Principal Planes (I & II)

x

y

III

θ1

Orientation of Max Shear Planes (D & E)

x

y D

Eθ1+ 45

Orientation of Arbitrary Planes (x’ & y’)

x

y

x’y’

θ

Ex. 4.21 General State of Strain

• At a point in a body, the principal strains are

• What is the maximum shear strain component at the point? What is the orientation of the plane, which experiences the maximum shear plane?

εI = 700 x 10-6

εII = 300 x 10-6

εIII = -300 x 10-6

Electrical Resistance Strain Guage for uniaxial loading

x

ac

θa

b

Strain Rosettes for General Loading

θc θb

y

45o Strain Rosette

x

y

O

εa = 100 x 10-6

εb = 200 x 10-6

εc = 900 x 10-6

Ex. 4.22

Find the principal strains in the plane of rosettes

εa

εbεc

Answers to 4.22

εx = 100 x 10-6

εy = 900 x 10-6

γxy = -600 x 10-6

εI = 1000 x 10-6

εII = 0

γmax = 1000 x 10-6