Design for Strength and Endurance – Chapter 2
Chapter 2
Two Dimensional Stress Analysis
Screen Titles
Definitions – Stress Components
Rectilinear Components
Rotated Stress System
Normal Stress – X’ direction
Shear Stress – Y’ direction
Normal Stress – Y’ direction
Double Angle Formulation
Average Stress (introduction)
Average Stress (continuation)
Graphical Interpretation
Principal Stresses
Mohr’s Circle Construction
Orientation – Principal Stresses
General Stress State
Principal Stresses – 3D Orientation
Principal Stress Equation – 3D
Mohr’s Circle – 3D Stress State
Review Exercise
Off line Exercise
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Two Dimensional Stresses
- 29 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
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Two Dimensional Stresses
- 30 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
1. Title page
2. Page Index
Chapter two covers the definition and analysis of
a two dimensional rectilinear state of stress
involving both normal and shear components.
The topics discussed include the equilibrium of
cross shear stresses, development of the
equations defining stress components with
respect to a rotated axis system, graphical
interpretation of rotated axis system stress
equations, principal stress components, Mohr’s
circle construction and use together with a brief
introduction to the analysis of a generalized
three dimensional state of stress.
Several
sample problems demonstrating the application
of the theory presented are also included.
Listed on this page are all the individual pages in
Chapter 2 with the exception of the sample
problems. Each title is hyperlinked to its specific
page and can be accessed by clicking on the
title. It is suggested that the reader first proceed
through all pages sequentially. Clicking on the
text button at the bottom of the page provides a
pop up window with the text for that page. The
text page is closed by clicking on the x in the top
right corner of the frame. Clicking on the index
button returns the presentation to the chapter
page index.
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Two Dimensional Stresses
- 31 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
3. Definitions – Stress Components
4. Rectilinear Stress Components
The figure illustrates some general solid body in
equilibrium under the action of a number of
external forces. If a plane is passed through the
body there must exist a distribution of force
elements delta F on incremental area elements
delta A of the cut surface to keep the bottom
portion of the body in equilibrium.
The
incremental force element delta F can be broken
down into two components, one that is normal to
the surface of delta A designated delta Fn and a
second component parallel to the surface
element designated delta Ft. The limit of the
ratio of delta Fn to delta A as delta A goes to
zero is mathematically defined as the normal
stress sigma at that point. The limit of the ratio
of delta Ft to delta A as delta A approaches zero
is defined as the shear stress at that point. Both
the normal stress and the shear stress may vary
over the surface depending on the shape of the
body and the location and direction of the
externally applied loads.
Consider a cubical element of material with
dimensions dx, dy and dz within a loaded body
with the dx and dy edges parallel to an xy axes
system as shown. The third dz edge is parallel
to a z-axis, which is normal to the page. In a
two dimensional state of stress the right face
perpendicular to the x axis is acted on by a
normal component of stress sigma x and a
positive shear stress component tau xy pointing
in the y direction. To satisfy both horizontal and
vertical equilibrium the stress components on
the left face perpendicular to the x-axis must be
as shown. In the most general two-dimensional
case the upper face normal to the y-axis will also
be acted on by a normal stress component
sigma y and a positive shear component tau yx
pointing in the x direction. To again satisfy
horizontal and vertical equilibrium the normal
and shear components on the lower face
perpendicular to the y axis must contain normal
and shear components as shown. To satisfy
moment equilibrium about the center of the
element the force couples created by tau xy and
tau yx stresses must be equal to each other.
Assuming the dz length of the cube is unity
equality of these couples results in tau xy equal
to tau yx. This is commonly referred to as
equilibrium of cross shears. Hence, in the most
general case a two dimensional stress state in
rectangular coordinates consists of two normal
stresses, sigma x and sigma y together with only
a single shear stress, tau xy.
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Two Dimensional Stresses
- 32 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
5. Rotated Stress System
6. Normal Stress – x’ direction
Given a two dimensional state of stress with
respect to a given xy axis system the question
then is what is the state of stress with respect to
some x’y’ axis system that makes an angle theta
with the original axis system as shown in the
figure. Do the values of the normal stresses
sigma x’ and sigma y’ and shear stress tau xy’
measured relative to planes perpendicular to the
x’y’ axes change and how are they related to the
given state of stress. To answer this question it
is necessary to perform an equilibrium analysis
on elements that share planes perpendicular to
the x’y’ axes and the xy axes. This is carried out
on the next two pages.
To determine the normal stress in the x’
direction the triangular element on the left is
employed. The plane normal to x’ is acted on by
the stresses sigma x’ and tau xy’ while the
original stress components sigma x, sigma y and
tau xy act on the two faces normal to the x and
y-axes. Force equilibrium is now applied in the
x’ direction. In this development the element is
assumed to be of unit depth in the z direction
and the length of the inclined edge is also
assumed to be of unit length. Thus the edge
parallel to the x-axis is sin theta and the edge
parallel to the y-axis is cos theta. The force due
to sigma x’ is simply the stress times the area it
acts over which is one by one. The force
contribution of sigma x in the x’ direction is the
stress times the area, cos theta times one, times
cos theta to give the component in the x’
direction.
In a similar fashion the force
contribution of tau xy on the x face is the stress
times the area, cos theta times 1, times sin theta
to give the component in the x’ direction. The
force contribution of sigma y in the x’ direction is
the stress times the area, sin theta times one,
times sin theta to give the component in the x’
direction. Like wise the force contribution of tau
xy on the y face is the stress times the area, sin
theta times 1, times cos theta to give the
component in the x’ direction. All contribution
due to the original stress state are in the
negative x’ direction. Combining trigonometric
terms gives the final equation at the bottom of
the page. It is seen that all components of the
original stress state contribute to the magnitude
of sigma x’.
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Two Dimensional Stresses
- 33 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
7. Shear Stress – y’ direction
8. Normal Stress – y’ direction
The triangular element from the previous page
can also be used to determine an equation for
tau xy’ in terms of the original stress state by
applying the condition for force equilibrium in the
y’ direction. The force due to the tau xy’ stress
is simply tau xy’ times the area it acts on which
is one by one. The force contribution of sigma x
will be positive and is given by the stress times
its area, cos theta times 1, times sin theta to
give the component in the y’ direction. The
contribution of tau xy on the x face which will be
negative is the stress times the area, cos theta
times 1, times cos theta for its component in the
y’ direction. On the y face the contribution of
sigma y will be the stress times the area, sin
theta times 1, times cos theta for its component
in the y’ direction. Finally the positive force
contribution of tau xy on the y face will be the
stress times the area sin theta times 1 times sin
theta to give the component in the y’ direction.
Appropriately combining terms results in the
equation for tau xy’ at the bottom of the page.
Again all components of the original stress state
contribute to the magnitude of tau xy’.
To satisfy yourself that you really understand the
developments on the previous two pages
undertake the exercise of generating the
equation for the normal stress sigma y’ in the y’
direction in terms of the original xy axis stress
state and the angle theta. This will require a
different triangular element, one that possesses
a plane perpendicular to the y’ direction as well
as planes perpendicular to the x and y axes.
The correct result for sigma y’ is given by the
equation at the bottom of the page. If needed
you can review the solution to this exercise by
clicking on the solution button, or when you
complete the development go on to the next
page.
(Solution on Page 47)
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Two Dimensional Stresses
- 34 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
9. Double Angle Formulation
10. Average Stress - Introduction
For reasons that will soon become apparent it is
convenient to write the equations developed for
sigma x’, sigma y’ and tau xy’ in terms of the
double angle of axis rotation, two theta. In
particular the identities which prove useful for
this purpose are sin theta cos theta equal to sin
two theta, sin squared theta equal to ½ - ½ cos
two theta and cos squared theta equal to ½ + ½
cos two theta. This permits the earlier equations
written in terms of theta to be expressed in
terms of half the sum of sigma x and sigma y,
half the difference between sigma x and sigma
y, tau xy and the sin and cos of two theta as
shown on the page. It is of interest to note that
these are periodic functions all of which have
maximum and minimum values for some
prescribed values of theta. To show that this is
true and to determine what these extreme
values are a further rearrangement of these
equations will be undertaken on the next two
pages.
This further development is begun by defining
sigma average as one half the sum of sigma x
plus sigma y. This term is then subtracted from
the equation for sigma x’ and the result is
squared. The gives an equation with the rather
extended right side as shown on the page. A
further manipulation of this result is continued on
the next page.
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Two Dimensional Stresses
- 35 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
11. Average Stress (continued)
12. Graphical Interpretation
The equation for tau xy’ is now squared and
added to the equation of the previous page.
This significantly simplifies the right side of the
resulting equation to give the expression shown
in the middle of the page. By defining R
squared as the sum of the quantity sigma x
minus sigma y all over two squared plus tau xy
squared the even simpler form of the equation
shown at the bottom of the page is obtained.
This should be recognized as the equation of a
circle in the sigma x’, tau xy’ axes system whose
center is on the sigma x’ axis at the location
sigma avg with a radius of R. A graphical
representation is shown on the next page.
A graphical representation of the circle
represented by the last equation on the previous
page is shown on the left plotted on a sigma x’,
tau xy’ axis system. The center is located at
sigma avg on the sigma x’ axis and the circle is
drawn with a radius R. The expressions for the
circle and the definitions of sigma avg and the
radius R in terms of the original stress
components are given to the right. Any point on
the circle represents the sigma and tau stress
components on some rotated plane from that of
the original defined stress state. It is observed
that the normal stress will have both a maximum
and minimum stress where the circle cuts the
sigma x’ axis. At this location the accompanying
value of the shear stress will be zero. These
extreme values of the normal stress are called
the principal stress of the stress state. It is also
observed that the shear stress attains its
maximum value at the highest point on the circle
and is just equal to the radius. On the plane
orientation where the shear stress is maximum
the normal stress is just equal to sigma avg or
the average of the normal stresses in the
original stress state.
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Two Dimensional Stresses
- 36 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
13. Principal Stresses
14. Sample Problem 1
Using the graphical representation from the
previous page it is now a simple matter to write
the equations for the maximum and minimum
normal stresses and maximum shear stress in
terms of the xy axis stress components. The
maximum normal stress is equal to the average
normal stress plus the radius of the stress circle.
This become the quantity sigma x plus sigma y
over two plus the square root of the quantity
sigma x minus sigma y over two squared plus
tau xy squared. The minimum normal stress
from the circle becomes the average stress
minus the radius of the circle. Hence sigma min
is made up of the two terms that make up sigma
max with the exception that the terms are
subtracted rather than added. This makes the
two equations easy to remember. Finally the
maximum shear stress is simply given by the
radius of the circle, which is the square root term
in the expressions for the two principal stresses.
This sample problem is included to help you
understand the application of the equations for
the principal stresses and maximum shear
stress for a given initial stress state. Solve this
problem before proceeding further. Click on the
solution button to check your answer and/or
review the solution process then go on to the
next page.
(Solution on Pages 47 and 48)
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Two Dimensional Stresses
- 37 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
15. Mohr circle construction
The circular representation of the general two
dimensional stress equations can also be used
to determine the stress state on any set of
rotated axes relative to the axis system of the
original stress state. This construction is well
known as Mohr’s circle, named after its
developer.
The question to be answered is:
given a set of two dimensional stresses relative
to an xy axis system as shown on the left what
will be the normal and shear stress on a plane
perpendicular to an x’ axis rotated through an
angle theta in the counter clockwise direction.
The graphical construction will take place on the
sigma’ - tau’ coordinate system shown on the
right. A point whose coordinates are sigma x
and tau xy is plotted together with a second
point whose coordinates are sigma y and minus
tau xy in this axis field. A straight line is then
drawn connecting these two points. This line is
the diameter of the stress circle. Where it
crosses the sigma ‘ axis is the center of the
circle. A circle defined in this fashion is now
drawn. Next, an angle two theta is measured
counterclockwise from the radius defined by
point sigma x, tau xy. At this angular position
another diameter is drawn on the circle. The
point of intersection of the diameter with the
circle represents the coordinates sigma x’ and
tau xy’ the two desired stress on the plane
perpendicular to the x’ direction. The point
where the far end of this second diameter
intersects the circle represents the coordinates
sigma y’ and minus tau xy’ acting on a plane
perpendicular to the y’ direction. It is further
observed that the angle two theta between the
planes on which the principal stress act and the
maximum shear stresses act is 90 degrees.
This means the physical planes on which these
two sets of stresses act are at 45 degrees with
respect to each other.
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Two Dimensional Stresses
- 38 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
16. Orientation – Principal Stresses
17. Sample problem -2
It will be helpful to graphically visualize the
orientation of the planes of principal stress to the
planes of maximum shear stress discussed on
the previous page. On the left is shown an
element on which the principal stress sigma max
and sigma min are assumed to be acting relative
to axes that are horizontal and vertical. The
planes on which the maximum shear stress act
will be perpendicular to an axis system that is
rotated 45 degrees counter clockwise as shown
on the figure on the right. Satisfy yourself that
the normal stresses on these planes of
maximum shear stress are just equal to sigma
average or sigma max plus sigma min divided
by two. Also determine what the value of the
maximum shear stress will be if sigma max is
equal to minus sigma min. This is referred to as
a state of pure shear such as occurs in the
simply torsion of a shaft.
This sample problem will help you understand
the application of Mohr’s circle. For the same
initial stress state as in Sample Problem –1
determine the direction of the principal stress
planes. That is, calculate the value and the
direction of the orientation of the axis for the
perpendicular plane on which sigma max acts.
Solve this problem before proceeding further.
Click on the solution button to check your
answer and/or review the process of solution
then go on to the next page.
(Solution on Pages 48 and 49)
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Two Dimensional Stresses
- 39 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
18. General Stress State
19. Principal Planes – 3 D Orientation
The discussions and developments so far have
dealt with two dimensional stress states. Before
leaving this topic it is worthwhile including some
introductory material on three dimensional stress
states. First consider the type and number of
components associated with a general threedimensional state of stress.
This requires
considering what stress components can exist
on a plane perpendicular to the z-axis and the
effect of these components on the two
dimension stresses relative to an xy axis
system. In general three stresses will exist on
the z-axis plane. These will consist of a normal
stress sigma z and two components of shear
stress tau xz and tau yz. The two shear
stresses require cross shears to exist on the
planes perpendicular to the xy axes. By the
principal of equal cross shears they will have the
magnitude of the shears on the z face and the
directions shown in the figure. Hence, in a
general three-dimensional stress state there
exist three normal components of stress, sigma
x, sigma y and sigma z. There are also three
shear stress components tau xy, tau xz and tau
yz
Just as in a two dimensional state of stress it is
possible to show that there exist a set of rotated
coordinates axes in three dimensions for which
there are only normal stresses on a cubical
element whose planes are perpendicular to this
axis system. This is illustrated in the graphic
where no shear stresses are shown on the
rotated planes.
Hence, the normal stresses
measured relative to the rotated axis system are
the principal stresses for the given threedimensional stress state. The analysis that
leads to the equations for the directions of these
rotated axes and the magnitudes of the
associated normal stress are beyond the scope
of this study. In addition these specific results
are seldom of use in real problems. However,
there are some special instances in which the
general knowledge of this behavior is important.
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Two Dimensional Stresses
- 40 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
20. Principal Stress Equation in 3D
21. Mohr Circles – 3D Stress State
Should an instance occur where the principal
stresses for a general three-dimensional state of
stress are required they can be determined by
solving the cubic equation listed on this page.
Note that the coefficients of this third order
equation involve a variety of combinations of all
the stress state components relative to the xyz
axis system. Satisfy yourself before proceeding
on that if all shear stresses in this equation are
set equal to zero substituting either sigma x,
sigma y or sigma z into the resulting
simplification will indeed satisfy the equation.
Each set of planes associated with the principal
axes in a three dimensional stress state will
possess a set of principal normal stress. One
such set can be expressed as sigma 1 and
sigma 2 relative to the x’y’ axis system. A
second set would be sigma 2 and sigma 3
relative to the y’z’ axes and finally sigma 1 and
sigma 3 relative to the x’z’ axes. Each of these
sets of principle stresses can be represented
graphically as a Mohr circle as shown on the
right. Note that all three circles are tangent to
one another as they pass through the points
sigma 1, sigma 2 and sigma 3 on the horizontal
axis. This is because each pair shares common
stresses with each other. Also these stress
values lie on the horizontal sigma axis since the
shear stresses on planes of principal normal
stress are all zero. The radius of each of the
circles corresponds to the maximum shear
stress associated with that particular primed axis
system. Thus the maximum shear stress for the
total state of stress is tau 3, the radius of the
circle of largest diameter.
Being able to
determine this maximum shear stress is
sometimes an important requirement and can be
missed in dealing with two dimensional stress
states. This is illustrated in Sample Problem – 3
on the next page.
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Two Dimensional Stresses
- 41 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
22. Sample Problem – 3
23. Review Exercise
In this sample problem you are asked to
determine the maximum shear stress in the axial
portion of a thin walled pressure vessel
subjected to an internal pressure of 1300 psi
with the dimensions given in the figure. The
correct value can only be calculated when the
three dimensional stress considerations of the
previous slide are properly taken into account.
Solve this problem before proceeding further.
Click on the solution button to check your
answer and/or to review the process of solution
then go on to the next page.
In this review exercise select the answer to the
question by clicking on the appropriate button in
the group of choices below the question. An
immediate feedback will be provided indicating
whether the selection was correct or what
material should be reviewed to obtain the right
result. You can also click on the hot word in the
question to pop up the relevant page from the
chapter. To remove the feedback click the
mouse and proceed to the next question. When
all questions have been answered correctly
proceed to the next page.
(Solution on Pages 48,49 and 50)
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Two Dimensional Stresses
- 42 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
24. Off Line Exercise
25. Off Line Exercise 9 (continued)
This off line exercise requires the application of
the two dimensional stress analysis theory and
equations developed in this chapter. If you can
correctly answer all four questions for the four
example stress states listed on the next page
you will have a good understanding of two
dimensional stress analysis.
This ability is
fundamental to the topics to be taken up in
subsequent chapters.
When finished with the information on this page
click on the exit button to leave the chapter or
the index button to select any specific page to
be revisited.
(Solution in Appendix)
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Two Dimensional Stresses
- 43 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
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Two Dimensional Stresses
- 44 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
Chapter 2
Two Dimensional Stress Analysis
Problem Solutions
Screen Titles
Normal Stress - Y’ direction
Maximum Normal Stress
Minimum Normal Stress / Shear Stress
Generic Principal Axis Formula
Principal Axis Orientation
Axial Stress
Hoop Stress
Stress Calculations
Complete Stress State
Maximum Shear Stress
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Two Dimensional Stresses
- 45 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
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Two Dimensional Stresses
- 46 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
1. Normal Stress - y’ direction
2. Maximum Normal Stress
The element shown on the left is chosen to
determine the normal stress in the y’ direction
since it contains planes perpendicular to the x, y
and y’ axes. Again force equilibrium is applied
in the y’ direction taking into account the
contributions of all the stress components acting
on the element. First recognize that if the length
of the inclined edge of the element is taken to be
unity then the length of the edge parallel to the x
axis is cos theta and the edge parallel to the y
axis is sin theta. It is also assumed that the
element is of unit depth in the z direction. To
apply equilibrium all stress components must be
multiplied by the area over which they act before
being multiplied by the appropriate trigonometric
function of theta to give their contibution in the y’
direction. The force contribution of sigma y’ is
simply the stress times the area one time one.
On the x face sigma x has a negative
contribution equal to the stress times the area
sin theta times one times the sin of theta. The
positive force contribution of tau xy on the x face
is the stress times the area times cos theta. On
the y face the negative force contribution of
sigma y in the y’ direction is the stress times the
area cos theta time one times cos theta. Tau xy
on the y face contributes a positive force
contribution in the y’ direction equal to the stress
times the area times sin theta. Carrying through
the multiplication, combining the tau xy terms
Begin with the equation for the maximum normal
stress, sigma 1, that is the sum of sigma
average, the quantity sigma x plus sigma y
divided by two, plus the radius of the stress
circle, the square root of the quantity sigma x
minus sigma y over 2 squared plus tau xy
squared. Next substitute the values of the given
stress state for the parameters into the equation.
These are sigma x equal to 11,000 psi, sigma y
equal to zero and tau xy equal to 6,900 psi.
Carrying out the indicated mathematical
operations gives a final answer for sigma 1, the
maximum normal stress, of 14,320 psi. tension.
and solving for sigma y’ gives the equation
at the bottom of the page.
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Two Dimensional Stresses
- 47 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
3. Min Normal Stress/ Max Shear Stress
4. Generic Principal Axis Formula
Now begin with the equation for sigma 2, which
is the difference between the value of the sigma
average and the radius of the stress circle.
Substituting the value of the stress components
sigma x, sigma y and tau xy, into this equation
gives the difference of the two numbers used to
determine sigma 1. This results in sigma 2
having a final value of - 3,320 psi compression.
The value of the maximum shear stress is just
equal to the radius of the stress circle, which is
the square root term in the sigma 1 and sigma 2
equations. The result is a maximum shear
stress of 8,820 psi.
To determine the direction of the principal axes
attention is directed to the generic Mohr circle
shown on the left. The highlighted red triangle
can be used to graphically define the tangent of
two theta that represents the rotation of the
initial stress state circle diameter defined by
sigma x and tau xy to the horizontal sigma axis
that defines the principal stress orientation.
From the geometry of the figure it is seen that
the height of the triangle is simply tau xy while
the base is equal to sigma x minus sigma
average, that is sigma x plus sigma y over 2,
giving the result sigma x minus sigma y over 2.
Thus the tangent of two theta that defines the
orientation of the principal axes becomes 2 tau
xy divided by the quantity sigma x minus sigma
y.
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Two Dimensional Stresses
- 48 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
5. Principal Axis Orientation
6. Axial Stress
The values of the given stress state are now
substituted into the formula from the previous
page for the tangent of two theta. This gives a
value of two theta equal to 51.4 degrees. Recall
that this is double the actual physical angle of
rotation. Hence, the orientation of the principal
axes is 25.7 degrees counter clockwise from the
x-axis since the physical rotation is opposite
from that on the circle. The figure on the right
illustrates graphically the orientation of the
principal axes relative to the xy axes system.
If the cylinder is cut by a plane perpendicular to
its central axis then the resulting free body must
exhibit an axial stress distribution on the cut
plane to balance the force of the pressure acting
on the interior surface of the spherical end cap.
Thus the pressure load must be equal to the
axial stress load resultant to satisfy equilibrium.
The pressure load is just the pressure multiplied
by the projected area of the surface it is acting
on. In this case the projected area is just Pi
times R squared. The axial stress resultant is
equal to the stress sigma a times the area it acts
over which is the circumference 2 Pi times R
times t. Setting the pressure load equal to the
stress reaction force and solving for sigma a
gives simply PR over 2t. As the length L of the
cylinder of the free body is shrunk to zero it is
seen that the axial stress on the spherical end
cap is also the same as the axial stress in the
cylinder.
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Two Dimensional Stresses
- 49 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
7. Hoop Stress
8. Stress Calculations
Next consider a free body portion of the cylinder
that is one unit in length along the axis of the
cylinder and cut by a horizontal plane that
contains the axis of the cylinder. In the figure
shown the radial effect of the pressure acting in
the interior of this free body must be balanced
by hoop stress reactions acting vertically down
on the cut surfaces of the thickness. Again the
effective pressure force acting upward must
equal the hoop stress reaction forces acting
down to satisfy equilibrium.
The effective
pressure load is the pressure times the effective
area which is just 2R times the unit depth of the
free body. The stress reaction force on one side
is the hoop stress times the area it acts on which
is t times the unit depth. Hence the equilibrium
condition becomes P times 2R is equal to 2
sigma h times t. Solving for sigma h gives PR
over t.
Employing the formulas for the axial stress,
sigma a, and the hoop stress, sigma h,
developed on the previous two pages the values
of these stresses can now be calculated using
the numerical parameters given in this problem.
Note that both stresses are tensile and the hoop
stress is twice the axial stress. Carrying out
these calculations gives an axial stress of
31,200 psi and a hoop stress of 62,400 psi.
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Two Dimensional Stresses
- 50 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
9. Complete Stress State
10. Maximum Shear Stress
Now consider a cubical element at the inside
surface of the cylinder with its edges oriented in
the axial, circumferential and radial directions as
shown. Acting on the axial face is the normal
tensile stress, sigma a. Acting on the face
perpendicular to the circumferential direction is
the tensile hoop stress, sigma h. Finally, acting
on the face perpendicular to the radial direction
on the inside surface of the tank is the internal
pressure P which can be represented by a
compressive normal stress, sigma p, equal to 1300 psi. Since there are no shear stresses
acting on any of the cubical element faces the
three normal stresses represent a threedimensional principal stress state.
The maximum shear stress can now be
determined by drawing Mohr circles for the
principal stresses from the previous page. Its
numeric value will be equal to the radius of the
largest circle. The graphic on the left illustrates
the three Mohr circles for this stress state. It is
easily seen that the largest circle is the one
corresponding to the axis system associated
with the circumferential and radial directions.
Thus the maximum shear stress is given by the
hoop stress minus the pressure stress divided
by two. For this problem the numerical value is
31,850 psi. It should be noted that if the
problem had been treated as a two-dimensional
stress state involving only the hoop and axial
stresses an incorrect
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Two Dimensional Stresses
- 51 C.F. Zorowski  2002
Design for Strength and Endurance – Chapter 2
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Two Dimensional Stresses
- 52 C.F. Zorowski  2002