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ADAMA SCIENCE AND TECHNOLOGY UNIVERSITY
SCHOOL OF MECHANICAL, CHEMICALAND MATERIAL
ENGINEERING
DEPARTMENT OF MECHANICAL ENGINEERING
STRENGTH OF MATERIALS II; MEng3207
CHAPTER- 2
Transformations of Stress and Strain
Prepared by: Mr. Kemal Temam (MSc)
22/05/2024
November, 2023

Contents
?Introduction
?Transformation of plane stress
?Mohr’s circle for plane stress
?General state of stress and three-dimensional analysis of stress
?Theories of failure
?Stresses in thin walled pressure vessels
?Transformation of plane strain
?Measurements of strain; strain rosette

Objectives
? Apply stress transformation equations to plane stress situations to
determine any stress component at a point.
? Apply the alternative Mohr's circle approach to perform plane stress
transformations.
?Use transformation techniques to identify key components of stress,
such as principal stresses.
?Extend Mohr's circle analysis to examine three-dimensional states of
stress.
?Examine theories of failure for ductile and brittle materials.
?Analyze plane stress states in thin-walled pressure vessels.
? Extend Mohr's circle analysis to examine the transformation of strain.

Introduction
? In general, the three dimensional state of
stress at a point in a body can be
represented by nine components:
? ?xx?yy and ?zz: Normal stresses
? ?xy ?yx ?xz ?zx ?yz and ?zy : Shear
stresses
? By equilibrium, we can show that there
are only six independent components of
the stress ?xx ,?yy
,?zz ,?xy, ?xz, and ?yz

Plane Stress
? In much of engineering stress
analysis, the condition of plane
stress applies.
? Plane Stress: one of the three
normal stresses, usually ?z vanishes
and the other two normal stresses
?x and ?y, and the shear stress ?xy
are known.

Plane Stress Transformation:
Finding Stresses on Various Planes
? General Problem:
? * Given two coordinate systems, x- y
and x' - y', and a stress state defined
relative to the first coordinate system
xyz :?x ?y, ?xy
? * Find the stress components relative to
the second coordinate system x’y’z’ :

? Consider a triangular block of
uniform thickness, t:
? For equilibrium:
Plane Stress Transformation

? Simplifying:
? Using :
Plane Stress Transformation

Transformation Equations for Plane Stress

Special Cases of Plane Stress
1. Uniaxial Stress State: ?y= ?xy=0
2. Biaxial Stress State: ?xy= 0

? 3. Pure Shear:?x = ?y= 0
Special Cases of Plane Stress

Principal Stress
? ?'x varies as a function of the angle ?
? The maximum and minimum values of ?'x are called the principal
stresses. To find the max and min values:
? Where ?p defines the orientation of the principle planes on which the
principle stress act.
? 0
d??x
d?

? Two values of 2?p in the range of 0 to 360
satisfy this equation.
? These two values differ by 180° so that ?p
has two values that differ by 90°, one
between 0 and 90° and the other between
90° and 180°.
? For one of the angles ?p , the stress is a
maximum principal stress (?1) and for the
other it is a minimum (?2).
? Because the two values of ?p are 90° apart,
==> the principal stress occur on mutually
perpendicular planes.
Principal Stress

? To calculate ?p consider the triangle
Principal Stress
? Subbing back in yields the
principal stresses:
? OR
where

The shear stress corresponding to the principal stress direction is given by:
The shear stress is identically zero in the
principal stress directions! (biaxial stress
state)
Shear Corresponding to Principal Stress

Maximum Shear Stress
? To find maximum shear:
? Where ?s defines the angle of the planes of maximum
shear stress.
? 0
d??
d?
xy

? From trigonometry
? The planes of maximum shear stress
occur at 45° to the principal planes.
? Subbing back in yields max shear:

? The normal stress corresponding to the max shear stress
direction is given by:

Example 1
For the state of plane stress shown in Fig., determine (a) the principal
planes, (b) the principal stresses, (c) the maximum shearing stress and
the corresponding normal stress.

Mohr's Circle for Plane Stress
? Recall the plane stress transformation equations:
? Rearrange to get
? The above equation is for a circle of radius R and Center ?avg

Mohr's Circle for Plane Stress
? Mohr’s circle equation
? Equation of a circle in the ?'x-?'xy
plane centered at |(?avg , 0) and radius R
* Every plane becomes a point on the circle.
* The intersection with the?’x axis defines the
principal stresses.
* The bottom and top center positions correspond to:
? R and ?min? ?R
? max

Procedures to Construct Mohr’s Circle
With ?x, ?y and ?xy known, the procedure for constructing Mohr's circle is
as follows;
1) Draw a set of coordinate axes with ? as abscissa (positive to the right) and
? as ordinate (positive upward)
2) Locate the center C of the circle at the point having coordinates(? aver, 0)
3) Locate point A, representing the stress conditions on the face A (?x, - ?xy)
4) Locate point B, representing the stress conditions on the face B (?y, ?xy)
5) Draw a line from point A to point B. This line is a diameter of the circle and passes
through the center C
6) Using point C as the center, draw Mohr's circle through points A and B.
7) On the circle, we measure an angle 2? clockwise from radius CA. The angle 2? locates
point D.
? Point D on the circle represents the stresses on the face D of the element.
? Note that an angle 2? on Mohr's circle corresponds to an angle ? on a stress element.

Procedures to Construct Mohr’s Circle

Example
For the state of plane stress shown below
(a) construct Mohr’s circle, (b) determine the principal stresses,
(c) determine the maximum shearing stress and the corresponding
normal stress.

GENERAL STATE OF STRESS
consider the general state of stress represented in figure a, and the
transformation of stress associated with the rotation of axes shown in Fig. b,
However, our analysis will be limited to the determination of the normal
stress on a plane arbitrary orientation.

GENERAL STATE OF STRESS
Three of the faces in the tetrahedron shown in Fig. are parallel to the
coordinate planes, while the fourth face, ABC, is perpendicular to the line QN.

THREE-DIMENSIONAL ANALYSIS OF STRESS
? A general state of stress is characterized by six stress components, where
the normal stress on a plane of arbitrary orientation can be expressed as a
quadratic form of the direction cosines of the normal to that plane.
? This proves the existence of three principal axes of stress and three
principal stresses at any given point.
? Rotating a small cubic element about each of the three principal axes was
used to draw the corresponding Mohr’s circles that yield the values of
σmax, σmin, and ?max In the case of plane stress when the x and y axes
are selected in the plane of stress, point C coincides with the origin O.
? If A and B are located on opposite sides of O, the maximum shearing stress
is equal to the maximum in-plane shearing stress,

THREE-DIMENSIONAL ANALYSIS OF STRESS

THEORIES OF FAILURE
Failure of a member is defined as one of two conditions.
1.Fracture of the material of which the member is made. This type of failure is the
characteristic of brittle materials.
2.Initiation of inelastic (Plastic) behavior in the material. This type of failure is the one
generally exhibited by ductile materials.
When an engineer is faced with the problem of design using a specific material, it
becomes important to place an upper limit on the state of stress that defines the
material's failure. If the material is ductile, failure is usually specified by the
initiation of yielding, whereas if the material is brittle it is specified by fracture.

THEORIES OF FAILURE
Maximum shear stress theory (Tresca) for Ductile Materials
The maximum-shear-stress theory, sometimes called the Tresca, Coulomb, or Guest
yield criterion, states that yielding will occur when the maximum shear stress reaches a
critical value equal to the shearing yield stress in the uniaxial tension test.
The maximum shear stress is given by:
Where σ1 is algebraically the largest one and σ3 is the smallest principal stress.
For a uniaxial tension test, σ1 = σo, σ2 = σ3 = 0 and σo is the yield stress in
tension test.

THEORIES OF FAILURE
Therefore, shearing yield stress for yielding simple tension is equal to one-half of the
tensile yield stress.
By substituting the above eqn. into previous eqn.
Where σ1’ and σ2’ are the principal stress deviators and ‘k’ is the yield stress for pure
shear i.e. the stress at which yielding occurs in pure shear torsion σ1 = - σ3.

THEORIES OF FAILURE
EXAMPLE
A thin-wall tube with closed ends is subjected to a maximum internal
pressure of 35 MPa in service. The mean radius of the tube is 30 cm.
a) If the tensile yield strength is 700 MPa, what minimum thickness must
be specified to prevent yielding?
b) If the material has a yield strength in shear of = 280 MPa, what
minimum thickness must be specified to prevent yielding?

THEORIES OF FAILURE
2. Von Mises, or Distortion-energy, Theory
The von Mises criterion postulates that yielding will occur when the value of the root-mean-square shear
stress reaches a critical value.

THEORIES OF FAILURE
? The Tresca and von Mises yield loci are plotted together in Figure for the same values of Y.
Note that the greatest differences occur for α = σ3/σ1 = ?1, 12 , and 2.

Plane Strain
? Plane strain is defined by the strain
state (?x ?y ?xy ) ; it is the limiting
condition in the center plane of a very
thick specimen.
? Consider a rectangular element of
material, OABC, in the xy-plane shown
in Figure; it is required to find the
normal and shearing strains in the
direction of the diagonal OB, when the
normal and shearing strains in the
directions Ox, Oy are given.
?

Strain Transformation
? Assume that strain transformation is
desired from an xy coordinate system to
an xy? set of axes, where the latter is
rotated counterclockwise (+?) from the
xy system.
? The transformation equations for plane
strain are

Principal Strains
? For isotropic materials only, principal
strains (with no shear strain) occur along
the principal axes for stress.
? In plane strain the principal strains ?1
and ?2 are expressed as
? The angular position ?p of the principal
axes (measured positive
counterclockwise) with respect to the
given xy system is determined from

Maximum Shear Strain
? Like in the case of stress, the
maximum in-plane shear strain is:
which occurs along axes at 4?? from
the principal axes, determined
from ?s
? The corresponding average
normal strain is

Mohr’s Circle of Strain
? The direct and shearing strains in
an inclined direction are given by
relations which are similar to the
Equations for the direct and
shearing stresses on an inclined
plane.
? This suggests that the strains in
any direction can be represented
graphically in a similar way to the
stress system.

? As in the case of stress, there is a
graphical overview by Mohr’s circle
of the directional dependence of the
normal and shear strain components
at a point in a material.
? This circle has a center C at ?ave =
??x+ ?y )/2 which is always on the ?
axis, but is shifting left and right in a
dynamic loading situation. The
radius R of the circle is

? For given values of ?x, ?y and ?xy it is
constructed in the following way:
? Two mutually perpendicular axes,
? and ?/2, are set up
? The points (?x, ?xy/2) and (?y, - ?xy
/2) are located; the line joining these
points is a diameter of the circle of
strain.
? The values of ? and ? /2 in an
inclined direction making an angle
? with Ox are given by the points on
the circle at the ends of a diameter
making an angle 2? with PQ; the angle
2? is measured clockwise.

? We note that the maximum
and minimum values of ?,
given by ?1 and ?2 occur when
?/2 is zero; ?1, ?2 are
called principal strains, and
occur for directions in which
there is no shearing strain.

The given state of plane stress is known to exist on the surface of a machine
component. Knowing that E = 200 GPa and G = 77.2 GPa, determine the
direction and magnitude of the three principal strains (a) by determining the
corresponding state of strain and then using Mohr’s circle for strain, (b) by
using Mohr’s circle for stress to determine the principal planes and principal
stresses and then determining the corresponding strains.

Strain Rosette
? Define the terms ?xx ?yy ?xy as the strains of
an element of size (dx*dy) at an angle
? with respect to the horizontal axis.
? Then the equations which defines these
strains are:
? If the strain at any angle could be
measured,the equation above can then be
used to determine the direct and shear
strains in the structure about the x & y
axes.
? These measurements are done using a
Strain Gauge Rosette.

Material- Property Relationships

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