Post on 12-Nov-2014
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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