MECHANICS OF SOLIDS (ME F211) Chaudhari...Mechanics of Solids Chapter-4 Stress and Strain Vikas Chaudhari BITS Pilani, K K Birla Goa Campus Stress and Strain Objectives Stress To know

BITS Pilani K K Birla Goa Campus VIKAS CHAUDHARI

MECHANICS OF SOLIDS

(ME F211)

Vikas Chaudhari BITS Pilani, K K Birla Goa Campus

Mechanics of Solids

Chapter-4

Stress and Strain


Stress and Strain

Objectives

Stress

To know state of stress at a point

To solve for plane stress condition applications

Strain

To know state of strain at a point

To solve for plane strain condition applications


Stress and Strain

Stress

Stress vector can be defined as

A

FT

A

n

0

)(

limT is force intensity or stress acting on a plane

whose normal is ‘n’ at the point O.


Stress and Strain

Characteristics of stress

The physical dimensions of stress are force per unit area.

Stress is defined at a point upon an imaginary plane, which

divides the element or material into two parts.

Stress is a vector equivalent to the action of one part of the

material upon another.

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 zn

y

n

x

nn )()()()(


Stress and Strain

Body cut by a plane ‘mm’ passing through point

‘O’ and parallel to y-z plane and

Consider the free body of the left part of the

plane ‘mm’

Divide plane mm in large number of small areas,

i.e. y z

A force F is acting on the small area A (A is

centered on the point O).

F is inclined to the surface mm at some

arbitrary angle.

Figure shows the rectangular components of

the force vector F acting on the small area

centered on point O.


Stress and Strain

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

Negative face of given section

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

Face Positive Negative

x face 1-2-3-4 5-6-7-8

y face 3-4-5-6 1-2-7-8

z face 1-4-5-8 2-3-6-7


Stress and Strain

Stress components on positive x face

Normal Stress

Shear Stress

Similarly on y face y, yx & yz and on z face z, zx & zy

stress components will exist.

0lim

x

xx

Ax

F

A

0

0

lim

lim

x

x

y

xyA

x

zxz

Ax

F

A

F

A


Stress and Strain

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.

Stress components acting on the six sides of a parallelepiped.


Stress and Strain

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.

For example: Thin sheet which is being pulled by forces in the plane of the sheet.

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

x xy

yx y

x

y

z


Stress and Strain

Plane stress condition (Stress components in z direction are zero)

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


Stress and Strain

Equilibrium of a Differential Element in Plane Stress

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

Following figure must satisfy equilibrium conditions i.e. M = 0 and F = 0


Stress and Strain


M = 0 is satisfied by taking moments about the center of the element After simplification, we obtain In the limit as x and y go to zero

02 2 2 2

xy yx

xy xy yx yx

x x y yM y z x y z x z y x z k

x x

02 2

xy yx

xy yx

x y

x x

xy yx


Stress and Strain

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 can also be shown : yz = zy and xz = zx

Definition of positive and negative xy

xy

xy

Positive Shear Negative Shear

xy

xy


Stress and Strain


F = 0 is satisfied by following two conditions After simplification, we obtain

0

0

yxxx x yx x yx

y xy

y y xy y xy

F x y z y x z y z x zx y

F y x z x y z x z y zy x

0

0

xyx

xy y

x y

x y


Stress and Strain

Stress Components Associated with Arbitrarily Oriented Faces in Plane Stress

x’ x’y’

x

xy y

M

P N

MP = MN cos

NP = MN sin


Stress and Strain

Resolve forces in normal and along the oblique plane i.e. along x’ and y’

' ' cos sin sin cos 0x x x xy y xyF MN MP MP NP NP

2 2

'cos sin 2 sin cos

x x y xy

2

cos2-1sin and

2

12coscos 22

' cos2 sin 22 2

x y x y

x xy


Stress and Strain

Resolve forces in normal and along the oblique plane i.e. along x’ and y’

' ' ' sin cos sin sin 0y x y x xy y xyF MN MP MP NP NP

2 2

' ' ( )sin cos (cos sin )x y y x xy

' ' sin 2 cos22

y x

x y xy

' cos2 sin 22 2

x y x y

y xy

Similarly


Stress and Strain

Mohr's Circle Representation of Plane Stress

122

2

2

x y

xyR

2

x yOC

x

y

O C

x

y

xy

xy

,

,

x xy

y xy

x

y

xy

xy y

y

x x y

x


Stress and Strain


x

y

O C

x

y

xy

xy

,

,

x xy

y xy

x

y

xy

xy y

y

x x y

x

x

x’

y’

x’y’

x’y’

y’

x’

' ' '

' ' '

' ,

' ,

x x y

y x y

x

y

2


Stress and Strain


90

22tan

12

1

yx

xy

22

2,1 )2

(2

xy

yxyx

22

max )2

( xyyx

x

y

O C

x

y

xy

xy 1 2

max

max

21


Stress and Strain

Problem:

Draw the mohr’s circle for following state of stresses


Stress and Strain

Problem:

Addition of Two States of stress


Stress and Strain

Problem:

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


Stress and Strain

Problem:

Find the principal stress and the orientation of the principal axes of stress for the following cases of plane stress.

a. b. c.

x = 40MPa x = 140MPa x = -120MPa

y = 0 y = 20MPa y = 50MPa

xy = 80MPa xy = -60MPa xy = 100MPa


Stress and Strain

Problem:

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


Stress and Strain

Problem:

If the minimum principal stress is -7MPa, find x and the angle which the principal axes make with the xy axes for the case of plane stress illustrated


Stress and Strain

Problem: A rectangular plate is under a uniform state of plane stress in the xy plane. It is known that the maximum tensile stress acting on any face (whose normal lies in the xy plane) is 75MPa. It is also known that on a face perpendicular to the x axis there is acting a compressive stress of 15MPa and no shear stress. No explicit information is available as to the values of the normal stress y, and shear stress xy acting on the face perpendicular to the y axis. Find the stress components acting on the face perpendicular to the a and b axes which are located as shown in the lower sketch. Report your results in an unambiguous sketch.

x

y 15MPa 15MPa

y

y

yx

yx

x

b

a

30o


Stress and Strain

Analysis of Deformation

n n n nu u i u j u k


Stress and Strain

Displacement of Continuous Body

a. Rigid-body translation

b. Rigid-body rotation about c

c. Deformation without rigid-

body motion

d. Sum of all the displacements


Stress and Strain

Definition of Strain Components

(b) Uniform Strain (c) Non-uniform Strain


Stress and Strain

Strain

Normal Strain

Shear Strain


Stress and Strain

Plane Strain (stain components in z direction are zero)

0

' 'limxx

O C OC

OC

0

' 'limyy

O E OE

OE

0 00 0

lim ' ' ' lim ' ' '2

xyx xy y

COE C O E C O E


Stress and Strain

Relation between Strain and Displacement in Plane Strain

The x and y components of the displacement of point O are indicated by u and v.

u and v must be continuous functions of x and y to ensure that the displacement be geometrically compatible i.e. no hole or void are created by the displacement.

Using the concept of partial derivatives, the displacement of point E and C can be expressed as shown in figure.


Stress and Strain

Relation between Strain and Displacement in Plane Strain

; ; x y xyu v v u

x y x y

x xy

yx y

xy yx

0 0

' 'lim limxx x

x u x x xO C OC u

OC x x

0 0

' 'lim limyy y

y v y y yO E OE v

OE y y

0 00 0

lim ' ' ' lim2 2 2

xyx xy y

v x x u y y v uC O E

x y x y

Plain strain condition where


Stress and Strain

Strain Components Associated with Arbitrary Sets of Axes

' ' ' '

' ' ' '; ;

' ' ' 'x y x y

u v v u

x y x y

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


Stress and Strain


'

'

' '

' ' '

' ' '

' ' '

' ' '

' ' ' ' ' '

' ' ' ' ' '

x

y

x y

u u x u y

x x x y x

v v x v y

y x y y y

v u v x v y u x u y

x y x x y x x y y y

'cos 'sin

'sin 'cos

' cos sin

' sin cos

x x y

y x y

u u v

v u v

By Chain Rule of Partial Derivatives

From geometry, following relations are obtained


Stress and Strain


'

2 2

'

2 2

'

cos sin cos cos sin sin

cos sin sin cos

cos sin sin cos

x

x

x x y xy

u v u v

x x y y

u v v u

x y x y

'

'

' '

cos 2 sin 22 2 2

cos 2 sin 22 2 2

sin 2 cos 22 2 2

x y x y xy

x

x y x y xy

y

x y y x xy


Stress and Strain

Mohr's Circle Representation of Plane strain

x

y

O C

x

y

,2

,2

xy

x

xy

y

x

y

x xy

xy y

2

2

xy

2

xy


Stress and Strain

Mohr's Circle Representation for any arbitrary Plane


Stress and Strain

Problem:

A sheet of metal is deformed uniformly in its own plane so that the

strain components related to a set of axes xy are

x = -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. Also to find the principal

strains and the direction of the axes on which they exist.


Stress and Strain

References

1. Introduction to Mechanics of Solids by S. H. Crandall et al

(In SI units), McGraw-Hill

Documents

MECHANICS OF SOLIDS (ME F211) Chaudhari...Mechanics of Solids Chapter-4 Stress and Strain Vikas Chaudhari BITS Pilani, K K Birla Goa Campus Stress and Strain Objectives Stress To know