Ch. 05_Compound Stress or Stress Transformations-PPT.pdf

Department of Mechanical & Manufacturing Engineering, MIT, Manipal 1
Chapter 5
Compound Stresses
Or
Stress Transformations
MME 2154: STRENGTH OF MATERIALS
Subraya Krishna Bhat, PhD
Assistant Professor
Dept. of Mech & Mfg. Engg.
MIT Manipal
sk.bhat@manipal.edu

Introduction to Compound Stresses

• In the previous chapters, different types of forces were considered to act
independently. Also, stresses were determined on planes in the normal or
tangential directions.
• However, in most cases, direct and shear forces act simultaneously on a body
and the maximum value of the resultant stress may act in some other direction
than of the load application.
• It is therefore, necessary to find out stresses on planes other than those of load
application.

• Normal and shear stresses in beams, shafts, and bars can be calculated from
the basic formulas discussed in the preceding chapters.
• For instance, the stresses in a beam are given by the flexure formula (𝜎 =
𝑀𝑦/𝐼), and the stresses in a shaft are given by the torsion formula (𝜏 = 𝑇𝑅/𝐽).
• The stresses calculated from these formulas act on cross sections of the
members, but larger stresses may occur on inclined sections.
• Therefore, we will begin our analysis of stresses and strains by discussing
methods for finding the normal and shear stresses acting on inclined sections
cut through a member.
• To calculate such stresses, we determine the stresses acting on inclined planes
under a stress state known as plane stress (two-dimensional stress state).

Stress Elements
• In our discussions of plane stress we will use stress elements to represent the
state of stress at a point in a body.
• We will begin our analysis by considering an element on which the stresses are
known, and then we will derive the transformation equations that give the
stresses acting on the sides of an element oriented in a different direction.
3D Representation 2D Representation

Stress Elements
• The stress element at point 𝐶 is a small rectangular block (it doesn’t matter
whether it is a cube or a rectangular parallelepiped) with its right-hand face lying
in cross section 𝑚𝑛.
• Because it is more convenient, we usually draw a two-dimensional view of the
element instead of a three-dimensional view.
Stress element
3D Representation:
Uniaxial stress state
Stress element
2D Representation:
Uniaxial stress state

State of Stress at a Point
• When working with stress elements, we must always keep in
mind that only one intrinsic state of stress exists at a point
in a stressed body, regardless of the orientation of the
element being used to portray that state of stress.
• When we have two elements with different orientations at the
same point in a body, the stresses acting on the faces of the
two elements are different, but they still represent the same
state of stress, namely, the stress at the point under
consideration.
• This situation is analogous to the representation of a force
vector by its components — although the components are
different when the coordinate axes are rotated to a new
position, the force itself is the same.

State of Stress at a Point
• Furthermore, we must always keep in mind that
stresses are not vectors. This fact can sometimes
be confusing, because we customarily represent
stresses by arrows just as we represent force
vectors by arrows.
• Although the arrows used to represent stresses
have magnitude and direction, they are not vectors
because they do not combine according to the
parallelogram law of addition.
• Instead, stresses are much more complex quantities
than are vectors, and in mathematics they are called
tensors. Other tensor quantities in mechanics are
strains and moments of inertia.

Plane Stress
• The stress conditions that we encountered in earlier chapters when analyzing
bars in tension and compression, shafts in torsion, and beams in bending are
examples of a state of stress called plane stress.
• To explain plane stress, we will consider the stress element shown below.

Plane Stress: Normal Stresses
• The symbols for the stresses shown in the figure have the
following meanings.
• A normal stress 𝜎 has a subscript that identifies the face
on which the stress acts; for instance, the stress 𝜎𝑥 acts
on the x face of the element and the stress 𝜎𝑦 acts on the
y face of the element.
• Since the element is infinitesimal in size, equal normal
stresses act on the opposite faces.
• The sign convention for normal stresses is the familiar
one, namely, tension is positive and compression is
negative.

Plane Stress: Shear Stresses
• A shear stress 𝜏 has two subscripts — the first subscript
denotes the face on which the stress acts, and the second
gives the direction on that face.
• The sign convention for shear stresses is as follows. A
shear stress is positive when it acts on a positive face of
an element in the positive direction of an axis or on a
negative face in the negative direction. The opposite
tendency is deemed negative.
• Therefore, the stresses 𝜏𝑥𝑦 and 𝜏𝑦𝑥 shown on the positive
x and y faces and the stresses 𝜏𝑥𝑦 and 𝜏𝑦𝑥 shown on the
negative x and y faces of the element are positive.
• Also, for equilibrium of the element 𝜏𝑥𝑦 = 𝜏𝑦𝑥

Stresses on Inclined Sections
• We are now ready to consider the stresses acting on
inclined sections, assuming that the stresses 𝜎𝑥, 𝜎𝑦, and
𝜏𝑥𝑦 are known.
• To portray the stresses acting on an inclined section, we
consider a new stress element that is located at the same
point in the material as the original element.
• However, the new element has faces that are parallel and
perpendicular to the inclined direction.
• The previous conclusions regarding the shear stresses still
apply, so that 𝜏𝑥1𝑦1
= 𝜏𝑦1𝑥1
.
• This implies that if we determine the shear stress acting
on any one of those faces, the shear stresses on other
sides are known.

• By equilibrium, the stresses acting on the inclined 𝑥1𝑦1 element can be written in
terms of the stresses on the 𝑥𝑦 element. Select a wedge shaped stress
element.
Stresses Forces
𝐴𝑜

• By resolving the forces into orthogonal components, the equations of
equilibrium in the 𝑥1 and 𝑦1 directions can be written as
• Using the relationship 𝜏𝑥𝑦 = 𝜏𝑦𝑥 and simplifying
𝜎𝑥1
𝐴𝑜 sec 𝜃 − 𝜎𝑥𝐴𝑜 cos 𝜃 − 𝜏𝑥𝑦𝐴𝑜 sin 𝜃
− 𝜎𝑦𝐴𝑜 tan 𝜃 sin 𝜃 − 𝜏𝑦𝑥𝐴𝑜 tan 𝜃 cos 𝜃 = 0
𝜃
90 − 𝜃
𝜏𝑥1𝑦1
𝐴𝑜 sec 𝜃 + 𝜎𝑥𝐴𝑜 sin 𝜃 − 𝜏𝑥𝑦𝐴𝑜 cos 𝜃
− 𝜎𝑦𝐴𝑜 tan 𝜃 cos 𝜃 + 𝜏𝑦𝑥𝐴𝑜 tan 𝜃 sin 𝜃 = 0
𝜎𝑥1
= 𝜎𝑥 cos2 𝜃 + 𝜎𝑦 sin2 𝜃 + 2𝜏𝑥𝑦 sin 𝜃 cos 𝜃
𝜏𝑥1𝑦1
= − 𝜎𝑥 − 𝜎𝑦 sin 𝜃 cos 𝜃 + 𝜏𝑥𝑦 cos2
𝜃 − sin2
𝜃

Transformation Equations for Plane Stress
• By introducing the following trigonometric identities, the equations for the
stresses on an inclined section can be expressed in a more convenient form as
cos2 𝜃 =
1
2
1 + cos 2𝜃 ; sin2 𝜃 =
1
2
1 − cos 2𝜃 ; sin 𝜃 cos 𝜃 =
1
2
sin 2𝜃
• These equations are usually called the transformation equations for plane
stress because they transform the stress components from one set of axes to
another. However, note that the state of stress at the point of consideration
does not change.
• Since the transformation equations were derived solely from equilibrium of an
element, they are applicable to stresses in any kind of material, whether linear
or nonlinear, elastic or inelastic.
𝜎𝑥1
=
𝜎𝑥 + 𝜎𝑦
2
+
𝜎𝑥 − 𝜎𝑦
2
cos 2𝜃 + 𝜏𝑥𝑦 sin 2𝜃 𝜏𝑥1𝑦1
= −
2
sin 2𝜃 + 𝜏𝑥𝑦 cos 2𝜃

An Observation Concerning Normal Stresses
• As a preliminary matter, we note that the normal stress 𝜎𝑦1
acting on the 𝑦1
face of the inclined element can be obtained by substituting 𝜃 + 90° for 𝜃. The
result is the following equation for 𝜎𝑦
𝜎𝑦1
=
𝜎𝑥 + 𝜎𝑦
2
−
2
cos 2𝜃 − 𝜏𝑥𝑦 sin 2𝜃
• Summing the expressions for 𝜎𝑥1
and 𝜎𝑦1
𝜎𝑥1
+ 𝜎𝑦1
= 𝜎𝑥 + 𝜎𝑦
• This equation shows that the sum of the normal stresses acting on
perpendicular faces of plane-stress elements (at a given point in a stressed
body) is constant and independent of the angle 𝜃.

An Observation Concerning Normal Stresses
• The manner in which the normal and shear
stresses vary is shown in the figure below,
which is a graph of 𝜎𝑥1
and 𝜏𝑥1𝑦1
versus the
angle 𝜃.
• The graph is plotted for the particular case
of 𝜎𝑦 = 0.2𝜎𝑥 and 𝜏𝑥𝑦 = 0.8𝜎𝑥.
• We see from the plot that the stresses vary
continuously as the orientation of the
element is changed.
• A detailed investigation of these maximum
and minimum values is made subsequently.

Special Cases of Plane Stress
• The general case of plane stress reduces to simpler states of stress under
special conditions. Biaxial stress occurs in many kinds of structures, including
thin-walled pressure vessels. Pure shear in case of torsion of shafts
Uniaxial
Pure shear Biaxial
𝜎𝑥1
= 𝜎𝑥 1 + cos 2𝜃
𝜏𝑥1𝑦1
= −
𝜎𝑥
2
sin 2𝜃
𝜎𝑥1
= 𝜏𝑥𝑦 sin 2𝜃
𝜏𝑥1𝑦1
= 𝜏𝑥𝑦 cos 2𝜃
𝜎𝑥1
=
𝜎𝑥 + 𝜎𝑦
2
+
2
cos 2𝜃
𝜏𝑥1𝑦1
= −
2
sin 2𝜃

Example 1
A plane-stress condition exists at a point on the surface of a loaded structure, where the stresses
have the magnitudes and directions shown on the stress element shown below. Determine the
stresses acting on an element that is oriented at a clockwise angle of 15° with respect to the original
element.

Example 1
Solution:

Example 1
Solution:
• This stress can be verified by substituting 𝜃 = 75° into the stress
transformation equation for plane stress.
• As a further check on the results, we note that 𝜎𝑥1
+ 𝜎𝑦1
= 𝜎𝑥 + 𝜎𝑦.
• The stresses acting on the inclined element are shown in the figure
where the arrows indicate the true directions of the stresses.
• Again we note that both stress elements shown in Fig. 6-8 represent
the same state of stress.

Example 2
Determine the normal stress and shear stress acting on the inclined plane AB. Solve
the problem using the stress transformation equations. Show the result on the
sectioned element.

Example 2

Example 2
The results are indicated on the triangular element shown in the figures.

Principal Stresses and Maximum Shear Stresses
• The transformation equations for plane
stress show that the normal stresses 𝜎𝑥1
and the shear stresses 𝜏𝑥1𝑦1
vary
continuously as the axes are rotated through
the angle 𝜃.
• From the figure, we see that both the normal
and shear stresses reach maximum and
minimum values at 90° intervals.
• These maximum and minimum values are
usually needed for design purposes.
• For e.g., fatigue failures of structures such
as machines and aircraft are often
associated with the maximum stresses.

Principal Stresses
• The maximum and minimum normal stresses, called the principal stresses, can
be found from the transformation equation for the normal stress 𝜎𝑥1
.
• By taking the derivative of 𝜎𝑥1
with respect to 𝜃 and setting it equal to zero, the
maxima and minima, i.e., the maximum a minimum values of 𝜎𝑥1
can be found.
𝜕𝜎𝑥1
𝜕𝜃
= − 𝜎𝑥 − 𝜎𝑦 sin 2𝜃 + 2𝜏𝑥𝑦 cos 2𝜃 = 0
tan 2𝜃𝑝 =
2𝜏𝑥𝑦
• The subscript 𝑝 indicates that the angle 𝜃𝑝 defines the orientation of the
principal planes, that is, the planes on which the principal stresses act.

Principal Stresses
• Angle 𝜃𝑝 takes two values between 0° and 180°. These values differ by 90°, with
one value between 0 and 90° and the other between 90° and 180°. The two
values of 𝜃𝑝 are known as the principal angles.
• For one of these angles, the normal stress 𝜎𝑥1
is a maximum principal stress; for
the other, it is a minimum principal stress that the principal stresses occur on
mutually ss. Because the principal angles differ by 90°, we perpendicular
planes.
• The principal stresses can be calculated by substituting each of the two values of
𝜃𝑝 into the first stress-transformation equation and solving for 𝜎𝑥1
.
• By determining the principal stresses in this manner, we not only obtain the
values of the principal stresses but we also learn which principal stress is
associated with which principal angle.

Principal Stresses
• We can also obtain general formulas for the principal stresses. From the
expression of tan 2𝜃𝑝 the following right-angled triangle (see figure below) can
be deduced. Then
𝑅 =
2
2
+ 𝜏𝑥𝑦
2
; cos 2𝜃𝑝 =
2𝑅
; sin 2𝜃𝑝 =
𝜏𝑥𝑦
𝑅
• Substituting in the transformation equations and using 𝜎𝑥 + 𝜎𝑦 = 𝜎1 + 𝜎2
𝜎1,2 =
𝜎𝑥 + 𝜎𝑦
2
±
2
2
+ 𝜏𝑥𝑦
2
• The above gives the formula for the principal stresses.

Shear Stresses on the Principal Planes
• An important characteristic of the principal planes can be obtained from the
transformation equation for the shear stresses (recalling the equation below).
𝜏𝑥1𝑦1
= −
2
• If we set the shear stress 𝜏𝑥1𝑦1
equal to zero, we get an equation that is the
same as the condition obtained while determination of maxima and minima of
𝜎𝑥1
.
• In other words, the angles to the planes of zero shear stress are the same as the
angles to the principal planes.
• Thus, we can make the following important observation: The shear stresses
are zero on the principal planes.

Special Cases
• The principal planes for elements in uniaxial stress and biaxial stress are the
x and y planes themselves, because tan 2𝜃𝑝 = 0 and the two planes of 𝜃𝑝 are 0
and 90°.
• We also know that the x and y planes are the principal planes from the fact that
the shear stresses are zero on those planes.
• For an element in pure shear, the principal planes are oriented at 45° to the x
axis, because tan 2𝜃𝑝 is infinite and the two values of 𝜃𝑝 are 45° and 135°.
Uniaxial
Pure shear
Biaxial

The Third Principal Stress
• By making a more complete three-dimensional
analysis, it can be shown that the three principal planes
for a plane-stress element are the two principal planes
already described plus the z face of the element.
• By definition, 𝜎1 is algebraically larger than 𝜎2, but 𝜎3
may be algebraically larger than, between, or smaller
than 𝜎1 and 𝜎2. Of course, it is also possible for some
or all of the principal stresses to be equal. Note again
that there are no shear stresses on any of the principal
planes.
The determination of principal stresses is an example of a type of mathematical analysis
known as eigenvalue analysis, which is described in books on matrix algebra. The stress
transformation equations and the concept of principal stresses are due to the French
mathematicians A. L. Cauchy (1789–1857) and Barré de Saint-Venant (1797–1886) and
to the Scottish scientist and engineer W. J. M. Rankine (1820–1872).

Maximum Shear Stresses
• Having found the principal stresses and their directions for an element in plane
stress, we now consider the determination of the maximum shear stresses and
the planes on which they act. Taking the derivative of 𝜏𝑥1𝑦1
with respect to 𝜃 and
equating it to zero
𝜕𝜏𝑥1𝑦1
𝜕𝜃
= − 𝜎𝑥 − 𝜎𝑦 cos 2𝜃 − 2𝜏𝑥𝑦 sin 2𝜃 = 0
tan 2𝜃𝑠 = −
2𝜏𝑥𝑦
• Similar to the case of principal stresses, the maximum shear stresses occur on
perpendicular planes. Because shear stresses on perpendicular planes are
equal in absolute value, the maximum positive and negative shear stresses
differ only in sign.
• Also by comparing with expression for 𝜃𝑝 we can get 𝜃𝑠 = 𝜃𝑝 ± 45°

• The angle 𝜃𝑠1
is related to the angle 𝜃𝑝1
as follows, and
𝜃𝑝1
= 𝜃𝑠1
− 45°; cos 2𝜃𝑠1
=
𝜏𝑥𝑦
𝑅
; sin 2𝜃𝑠1
= −
2𝑅
• The corresponding maximum shear stress is obtained by substituting the
expressions for cos 2𝜃𝑠1
and sin 2𝜃𝑠1
into the second transformation equation,
yielding
𝜏𝑚𝑎𝑥 =
2
2
+ 𝜏𝑥𝑦
2
• The maximum negative shear stress 𝜏𝑚𝑖𝑛 has the same magnitude but opposite
sign. Another expression for the maximum shear stress can be obtained in terms
of the principal stresses 𝜎1 and 𝜎2 by comparing with their relationships as
𝜏𝑚𝑎𝑥 =
𝜎1 − 𝜎2
2

Example 3
The state of stress at a point is shown on the element. Determine (a) the principal
stress and (b) the maximum in-plane shear stress and average normal stress at the
point. Specify the orientation of the element in each case. Show the results on each
element.

Example 3

Example 4
The square steel plate has a thickness of 10 mm and is subjected to the edge
loading shown. Determine the maximum in-plane shear stress and the average
normal stress developed in the steel.

Example 4

Example 5
A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges
together as shown. Determine (a) the shear stress acting along the seam and (b)
the normal stress acting perpendicular to the seam, which is at 30° from the vertical,
when the tube is subjected to an axial force of 10 N. The paper is 1 mm thick and
the tube has an outer diameter of 30 mm.

Example 5
(a) Shear stress acting along the seam

Example 5
(a) Normal stress acting perpendicular to the seam

Mohr’s Circle For Plane Stress
• The transformation equations for plane stress can be represented in graphical
form by a plot known as Mohr’s circle*.
• This graphical representation is extremely useful because it enables you to
visualize the relationships between the normal and shear stresses acting on
various inclined planes at a point in a stressed body.
• It also provides a means for calculating principal stresses, maximum shear
stresses, and stresses on inclined planes.
• Furthermore, Mohr’s circle is valid not only for stresses but also for other
quantities of a similar mathematical nature, including strains and moments of
inertia.
* Mohr’s circle is named after the famous German civil engineer Otto Christian
Mohr (1835–1918), who developed the circle in 1882.

Equations of Mohr’s Circle
• The equations of Mohr’s circle can be derived from the transformation equations
for plane stress. The two equations are repeated here, but with a slight
rearrangement of the first equation:
𝜎𝑥1
−
𝜎𝑥 + 𝜎𝑦
2
=
2
cos 2𝜃 + 𝜏𝑥𝑦 sin 2𝜃
𝜏𝑥1𝑦1
= −
2
• From analytic geometry, we might recognize that that these two equations are
the equations of a circle in parametric form. Squaring and adding the two eqns.
𝜎𝑥1
−
𝜎𝑥 + 𝜎𝑦
2
2
+ 𝜏𝑥1𝑦1
2 = 𝑅2

Equations of Mohr’s Circle
• This equation can be written in simpler form by using the following notation in
Slide No. 28:
𝜎𝑎𝑣𝑔 =
𝜎𝑥 + 𝜎𝑦
2
; 𝑅 =
2
2
+ 𝜏𝑥𝑦
2
• Then the equation becomes
𝜎𝑥1
− 𝜎𝑎𝑣𝑔
2
+ 𝜏𝑥1𝑦1
2 = 𝑅2
• Which is the equation of a circle in standard algebraic form. The coordinates are
𝜎𝑥1
and 𝜏𝑥1𝑦1
, the radius 𝑅, and the center of the circle has coordinates (𝜎𝑎𝑣𝑔, 0).

Form of Mohr’s Circle
• Mohr’s circle can be plotted from the derived equations in two forms. We will stick
to the form where we plot the normal stress 𝜎𝑥1
positive to the right and the shear
stress 𝜏𝑥1𝑦1
positive downward as shown in the figure.

Construction of Mohr’s Circle
• Mohr’s circle can be constructed in a
variety of ways, depending upon which
stresses are known and which are to be
found.
• To show the basic properties of the
circle, let us assume that we know the
stresses 𝜎𝑥, 𝜎𝑦, and 𝜏𝑥𝑦 acting on the 𝑥
and 𝑦 planes of an element in plane
stress (Fig. a).
• By drawing the circle using these
information, we can determine stresses
𝜎𝑥1
, 𝜎𝑦1
, and 𝜏𝑥1
𝑦1
acting on an inclined
element (Fig. b).
(a)
(b)

With 𝜎𝑥 , 𝜎𝑦 , and 𝜏𝑥𝑦 known, procedure for
constructing Mohr’s circle is as follows:
1. Draw a set of coordinate axes with 𝜎𝑥1
as abscissa
(positive to the right) and 𝜏𝑥1𝑦1
as ordinate (positive
downward).
2. Locate the center C of the circle at the point having
coordinates (𝜎𝑎𝑣𝑔, 0).
3. Locate point A, representing the stress conditions
on the 𝑥 face of the stress element by plotting its
coordinates 𝜎𝑥1
= 𝜎𝑥 and 𝜏𝑥1𝑦1
= 𝜏𝑥𝑦 . Note that
point A on the circle corresponds to 𝜃 = 0°.
4. Locate point B, representing the stress conditions
on the 𝑦 face of the stress element by plotting its
coordinates 𝜎𝑥1
= 𝜎𝑦 and 𝜏𝑥1𝑦1
= −𝜏𝑥𝑦. Note that
point A on the circle corresponds to 𝜃 = 90°.

With 𝜎𝑥 , 𝜎𝑦 , and 𝜏𝑥𝑦 known, procedure for
constructing Mohr’s circle is as follows:
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. Points A and B, representing the stresses on
planes at 90° to each other, are at opposite ends
of the diameter (and therefore are 180° apart on
the circle).
7. Using point C as the center, draw Mohr’s circle
through points A and B. The circle drawn in this
manner has radius R.

• Now that we have drawn the circle, we can verify
by geometry that lines CA and CB are radii and
have lengths equal to R.
• We note that the abscissas of points C and A are
𝜎𝑥 + 𝜎𝑦 /2 and 𝜎𝑥, respectively.
• The difference in these abscissas is 𝜎𝑥 − 𝜎𝑦 /2,
as dimensioned in the figure.
• Also, the ordinate to point A is 𝜏𝑥𝑦. Therefore, line
CA is the hypotenuse of a right triangle having one
side of length 𝜎𝑥 − 𝜎𝑦 /2 and the other side of
length of length 𝜏𝑥𝑦.
• Taking the square root of the sum of the squares of
these two sides gives the radius R.

Stresses on an Inclined Element
• On the circle, we measure an angle 2𝜃
counterclockwise from radius CA, because point A
corresponds to 𝜃 = 0° and is the reference point
from which we measure angles.
• The angle 2𝜃 locates point D on the circle, which
has coordinates 𝜎𝑥1
and 𝜏𝑥1
𝑦1
.
• Therefore, point D represents the stresses on the
𝑥1 face of the inclined element. Consequently, this
face of the element is labeled “D”.
• Note that an angle 2𝜃 on Mohr’s circle corresponds
to an angle 𝜃 on a stress element.
• Similarly, points A and B are 180° apart on the
circle, but the corresponding faces of the element
are 90° apart.

• To show that the coordinates 𝜎𝑥1
and 𝜏𝑥1
𝑦1
of point
D on the circle are indeed given by the stress-
transformation equations.
• Let 𝛽 be the angle between the radial line CD and
the 𝜎𝑥1
axis. Then,
𝜎𝑥1
=
𝜎𝑥 + 𝜎𝑦
2
+ 𝑅 cos 𝛽 ; 𝜏𝑥1𝑦1
= 𝑅 sin 𝛽
• Since the angle between the radius CA and the
horizontal axis is 2𝜃 + 𝛽,
cos 2𝜃 + 𝛽 =
2𝑅
; sin 2𝜃 + 𝛽 =
𝜏𝑥𝑦
𝑅
• Expanding the cosine and sine functions
cos 2𝜃 cos 𝛽 − sin 2𝜃 sin 𝛽 =
2𝑅
sin 2𝜃 cos 𝛽 − cos 2𝜃 sin 𝛽 =
𝜏𝑥𝑦
𝑅

• Multiplying the first of these equations by cos 2𝜃
and the second by sin 2𝜃 and then adding,
cos 𝛽 =
1
𝑅
2
cos 2𝜃 + 𝜏𝑥𝑦 sin 2𝜃
• Also, multiplying the first of these equations by
sin 2𝜃 and the second by cos 2𝜃 and then adding,
sin 𝛽 =
1
𝑅
−
2
• Substituting these expressions of cos 𝛽 and sin 𝛽
into 𝜎𝑥1
and 𝜏𝑥1𝑦1
the stress-transformation
equations can be obtained.
• Thus, the point D on Mohr’s circle, defined by the
angle 2𝜃, represents the stress conditions 𝜎𝑥1
and
𝜏𝑥1𝑦1
on the 𝑥1 face of the stress element at an
angle 𝜃.
• Similarly, the point D’ represents 𝜎𝑦1
and −𝜏𝑥1𝑦1
.

Principal Stresses
• The determination of principal stresses is probably
the most important application of Mohr’s circle.
• As we move around Mohr’s circle, we encounter
points P1 and P2 where the normal stress reaches
its algebraically largest and smallest values,
respectively, and the shear stresses are zero.
Hence, represents the principal stresses and
principal planes.
• Algebraically they can be obtained as
𝜎1,2 =
𝜎𝑥 + 𝜎𝑦
2
± R
• The principal planes can be obtained by
cos 𝜃𝑝1
=
2𝑅
; sin 𝜃𝑝1
=
𝜏𝑥𝑦
𝑅
𝜃𝑝2
= 𝜃𝑝1
+ 90°

• Points S1 and S2, representing the planes of
maximum positive and maximum negative shear
stresses, respectively, are located at the bottom
and top of Mohr’s circle.
• These points are at angles 2𝜃 = 90° from points P1
and P2, which agrees with the fact that the planes
of maximum shear stress are oriented at 45° to the
principal planes.
• The maximum shear stresses are numerically
equal to the radius R of the circle.
• Also, the normal stresses on the planes of
maximum shear stress are equal to the abscissa of
point C, which is the average normal stress 𝜎𝑎𝑣𝑔.

A Note on Sign Convention for Shear Stresses
• Instead of thinking of the vertical axis as
having negative shear stresses plotted
upward and positive shear stresses
plotted downward (which is a bit
awkward), we can think of the vertical axis
as having clockwise shear stresses
plotted upward and counterclockwise
shear stresses plotted downward.

General Comments About the Mohr’s Circle
• Besides using Mohr’s circle to obtain the stresses on inclined planes when the stresses
on the 𝑥 and 𝑦 planes are known, we can also use the circle in the opposite manner.
• If desired, we can construct Mohr’s circle to scale and measure values of stress from
the drawing. However, it is usually preferable to obtain the stresses by numerical
calculations by using trigonometry and the geometry of the circle.
• Mohr’s circle makes it possible to visualize the relationships between stresses acting
on planes at various angles, and it also serves as a simple memory device for
calculating stresses. Although many graphical techniques are no longer used in
engineering work, Mohr’s circle remains valuable because it provides a simple and
clear picture of an otherwise complicated analysis.
• Mohr’s circle is also applicable to the transformations for plane strain and moments of
inertia of plane areas, because these quantities follow the same transformation laws as
do stresses.

Example 6
At a point on the surface of a pressurized
cylinder, the material is subjected to biaxial
stresses 𝜎𝑥 = 90 𝑀𝑃𝑎 and 𝜎𝑦 = 20 𝑀𝑃𝑎 , as
shown on the stress element shown in the
figure.
Using Mohr’s circle, determine the stresses
acting on an element inclined at an angle
𝜃 =30°. (Consider only the in-plane stresses,
and show the results on a sketch of a properly
oriented element.)

Example 6
Solution:
Construction of Mohr’s circle:
• We begin by setting up the axes for the normal and shear stresses, with 𝜎𝑥1
positive to
the right and 𝜏𝑥1𝑦1
positive downward.
• Then we place the center C of the circle on the 𝜎𝑥1
axis at the point where the stress
equals the average normal stress
𝜎𝑎𝑣𝑔 =
𝜎𝑥 + 𝜎𝑦
2
𝜎𝑎𝑣𝑔 =
90 + 20
2
= 55 𝑀𝑃𝑎
• Point A, representing the stresses on the 𝑥 face of the element (𝜃 = 0°), has coordinates
𝜎𝑥1
= 90 𝑀𝑃𝑎; 𝜏𝑥1𝑦1
= 0

Example 6
Solution:
• Similarly, the coordinates of point B, representing
the stresses on the 𝑦 face (𝜃 = 90°), are
𝜎𝑥1
= 0
• Now we draw the circle through points A and B
with center at C and radius R equal to
𝑅 =
2
2
+ 𝜏𝑥𝑦
2
𝑅 =
90 − 20
2
2
+ 0
𝑅 = 35 𝑀𝑃𝑎
𝜏𝑥1𝑦1
𝜎𝑥1
20
90
A
B
C
(𝜃 = 90°) (𝜃 = 0°)

Example 6
Solution:
Stresses on an element inclined at 𝜃 = 30°:
The stresses acting on a plane oriented at an angle
𝜃 = 30°are given by the coordinates of point D,
which is at an angle 2𝜃 = 60° from point A.
Similarly, we can find the stresses represented by
point D’, at an angle 𝜃 = 120° (or 2𝜃 = 240°).
𝜎𝑥1
= 𝜎𝑎𝑣𝑔 + 𝑅 cos 60°
𝜎𝑥1
= 55 + 35 cos 60°
𝜎𝑥1
= 72.5 𝑀𝑃𝑎
𝜏𝑥1𝑦1
= −𝑅 sin 60°
𝜏𝑥1𝑦1
= −35 sin 60°
𝜏𝑥1𝑦1
= −30.3 𝑀𝑃𝑎
𝜎𝑥1
= 𝜎𝑎𝑣𝑔 − 𝑅 cos 240°
𝜎𝑥1
= 55 − 35 cos 240°
𝜎𝑥1
= 37.5𝑀𝑃𝑎
𝜏𝑥1𝑦1
= 𝑅 sin 240°
𝜏𝑥1𝑦1
= 35 sin 240°
𝜏𝑥1𝑦1
= −30.3 𝑀𝑃𝑎
The results are shown in a
stress element oriented at 30°,
with all stresses shown in
their true directions. Note
that the sum of the normal
stresses on the inclined
element is equal to 𝜎𝑥 + 𝜎𝑦 or
110 MPa.

Example 7
At a point on the surface of a generator shaft
the stresses are 𝜎𝑥 = −50 𝑀𝑃𝑎, 𝜎𝑦 = 10 𝑀𝑃𝑎,
and 𝜏𝑥𝑦 = −40 𝑀𝑃𝑎, as shown on the stress
element shown in the figure.
Using Mohr’s circle, determine the following
quantities: (a) the stresses acting on an element
inclined at an angle 𝜃=45°, (b) the principal
stresses, and (c) the maximum shear stresses.
(Consider only the in-plane stresses, and show
the results on a sketch of a properly oriented
element.)

Example 7
Construction of Mohr’s circle:
• We begin by setting up the axes for the normal and shear stresses, with 𝜎𝑥1
positive to
the right and 𝜏𝑥1𝑦1
positive downward.
• Then center C of the circle is located on the 𝜎𝑥1
axis at the point where the stress equals
the average normal stress
𝜎𝑎𝑣𝑔 =
𝜎𝑥 + 𝜎𝑦
2
𝜎𝑎𝑣𝑔 =
−50 + 10
2
∴ 𝜎𝑎𝑣𝑔 = −20 𝑀𝑃𝑎
• Point A, representing the stresses on the 𝑥 face of the element (𝜃 = 0°), has coordinates
𝜎𝑥1
= −50 𝑀𝑃𝑎; 𝜏𝑥1𝑦1
= −40 𝑀𝑃𝑎

Example 7
• Similarly, the coordinates of point B, representing
the stresses on the 𝑦 face (𝜃 = 90°), are
𝜎𝑥1
= 40 𝑀𝑃𝑎
• The circle is now drawn through points A and B
with center at C and radius R equal to
𝑅 =
2
2
+ 𝜏𝑥𝑦
2
𝑅 =
−50 − 10
2
2
+ −402
𝑅 = 50 𝑀𝑃𝑎 𝜏𝑥1𝑦1
𝜎𝑥1
50
10
A
B
C
(𝜃 = 90°)
(𝜃 = 0°)
40
40
20
D
D’
90°
P1
P2
S1
S2

Example 7
(a) Stresses on an element inclined at 𝜃 = 45°:
The stresses acting on a plane oriented at an angle 𝜃 = 45° are
given by the coordinates of point D, which is at an angle 2𝜃 = 90°
from point A.
To evaluate these coordinates, we need to know the angle
between line CD and the negative 𝜎𝑥1
axis, i.e., angle DCP2,
which requires that the angle between line CA and negative 𝜎𝑥1
axis, i.e., angle ACP2 is known. From trigonometry,
tan 𝐴𝐶𝑃2 =
40
30
⇒ 𝐴𝐶𝑃2 = 53.13°
𝐷𝐶𝑃2 = 90° − 𝐴𝐶𝑃2 = 36.87°
Knowing these angles, coordinates of point D and D’ are
𝜎𝑥1
= −20 − 50 cos 36.87° = −60 𝑀𝑃𝑎
𝜏𝑥1𝑦1
= 50 sin 36.87° = 30 𝑀𝑃𝑎
𝜎𝑥1
= −20 + 50 cos 36.87° = 20 𝑀𝑃𝑎
𝜏𝑥1𝑦1
= −50 sin 36.87° = −30 𝑀𝑃𝑎

Example 7
(b) Principal stresses:
The principal stresses are represented by points P1 and P2 on
Mohr’s circle. As seen by the inspection of the circle, the principal
stresses are,
𝜎1 = −20 + 50 = 30 𝑀𝑃𝑎
𝜎2 = −20 − 50 = −70 𝑀𝑃𝑎
The principal planes are at angles 2𝜃𝑝1
and 2𝜃𝑝2
measured
counterclockwise as
𝐴𝐶𝑃1 = 2𝜃𝑝1
= 53.13 + 180 = 233.13° ⇒ 𝜃𝑝1
= 116.6°
𝐴𝐶𝑃2 = 2𝜃𝑝2
= 53.13° ⇒ 𝜃𝑝2
= 26.6°
The principal stresses and principal planes are shown in the
figure to the right, and again we note that the sum of the normal
stresses is equal to 𝜎𝑥 + 𝜎𝑦, or −40 𝑀𝑃𝑎.

Example 7
(c) Maximum shear stresses:
The maximum positive and negative shear stresses are represented
by points S1 and S2 on Mohr’s circle. Their magnitudes, equal to the
radius of the circle, are,
𝜏𝑚𝑎𝑥 = 50 𝑀𝑃𝑎
The angle ACS1 is the angle of the point S1, which is,
2𝜃𝑠1
= 90° + 53.13° = 143.13°
⇒ 𝜃𝑠1
= 71.6°
This is shown in the figure to the right. The maximum negative shear
stress (point S2) has the same numerical value as the positive stress.
The normal stresses acting on the planes of maximum shear stress
are equal to 𝜎𝑎𝑣𝑔, which is the coordinate of the center C of the circle
(−20 𝑀𝑃𝑎). Note that the planes of maximum shear stress are
oriented at 45° to the principal planes.

Example 8
The bent rod has a diameter of 20 mm and is subjected to the force of 400 N.
Determine the principal stress and the maximum in-plane shear stress that is
developed at point A. Show the results on a properly oriented element located at this
point.

Example 8

Example 8
Negative due to
compression at
the bottom side

Example 8

Example 9
The solid propeller shaft on a ship extends outward from the hull. During operation it
turns at 𝜔 = 15 𝑟𝑎𝑑/𝑠 when the engine develops 900 kW of power. This causes a
thrust of 𝐹 = 1.23 𝑀𝑁 on the shaft. If the shaft has an outer diameter of 250 mm,
determine the principal stresses and the maximum in-plane shear stress at any point
located on the surface of the shaft.

Example 9
1. Power Transmission:
2. Internal Torque and Force: As shown in the FBD.

Example 9
3. Section Properties:
4. Normal stress:
5. Shear stress: Applying the torsion formula

Example 9
6. In-Plane Principal Stresses: 𝜎𝑥 = −25.06 𝑀𝑃𝑎, 𝜎𝑦 = 0, and 𝜏𝑥𝑦 = 19.56 𝑀𝑃𝑎 for
any point on the shaft’s surface. Therefore,
7. Maximum In-Plane Principal Shear Stress: 𝜎𝑥 = −25.06 𝑀𝑃𝑎, 𝜎𝑦 = 0, and 𝜏𝑥𝑦 =
19.56 𝑀𝑃𝑎 for any point on the shaft’s surface. Therefore,

Example 10
The propeller shaft of the tugboat is subjected to the compressive force and torque
shown. If the shaft has an inner diameter of 100 mm and an outer diameter of 150
mm, determine the principal stress at a point A located on the outer surface.

Example 10

Ch. 05_Compound Stress or Stress Transformations-PPT.pdf

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