Design for Strength and Endurance – Chapter 1
Chapter 1
Review of Basic Concepts
Screen Titles
Strength and Endurance Factors
Review of Basic Concepts
Normal Stress Definition
Normal Strain Definition
Shear Stress Definition
Shear Strain Definition
Hooke’s Law
Axial Elongation
Internal Beam Loading
Beam Loading Problem
Bending and Shear Characteristics
Bending Stress Distribution
Shear Stress IN Bending
Qmax for Rectangle
Maximum Shear Stress
Shaft in Torsion
Shear Stress Distribution
Angular Twist
Review Exercises
Off Line Exercises
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Design for Strength and Endurance – Chapter 1
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Design for Strength and Endurance – Chapter 1
1. Title page
2. Page Index
Chapter one provides an introductory overview
of the factors that influence the strength and
endurance of mechanical components as well as
a review of some basic concepts that underlie a
detailed study of this subject. The factors
discussed include applied loading, geometric
property characteristics of the mechanical
components, material properties and failure
criteria. The basic concepts reviewed cover the
definition of normal and shear stress and strain,
elongation and shear deformation, Hooke’s law
and the analysis of simple beams and shafts.
Listed on this page are all the individual pages in
Chapter 1. Each title is hyperlinked to its
specific page and can be accessed by clicking
on the title. It is suggested, however, that the
reader first proceed through all pages
sequentially Clicking on the text button at the
bottom of the page will provide a pop up window
with the text for that page. Clicking on the x in
the top right corner closes the text page.
Clicking on the index button returns the
presentation to the chapter page index.
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Design for Strength and Endurance – Chapter 1
3. Strength and Endurance Factors
4. Review of Basic Concepts
There are three generic factors that directly
influence the strength and endurance of
mechanical components. One factor is the
loading applied to the component.
More
specifically, three important characteristics of the
loading are its magnitude, its location and
direction and whether the load is static or
dynamic.
A second factor deals with the
mechanical component itself. In this case the
important characteristics influencing strength
and endurance are the shape and size of the
component. The third factor is the material from
which the component has been made. Here the
important characteristics are the material’s
stress and strain behavior under both static and
dynamic loading.
A fourth consideration that
also needs to be considered is the failure criteria
that define the strength and endurance
characteristics.
Failure may not simply be
designated by fracture of the component. There
may also be design considerations dealing with
excessive deformation or permanent distortion
that will dictate the definition of strength and
endurance.
The brief review that follows deals primarily with
some of the more important definitions, material
behavior assumptions and specific component
behavior related to the topic called strength of
materials. The first concept reviewed is that of
normal stress and shear stress within a loaded
body resulting from external loading. This is
followed by the definition of elongation and
shear strain that accompany these stresses.
Hooke’s law is introduced to relate stress to
strain in terms of the moduli of the material from
which the component is made. Finally these
concepts are applied to the determination of
normal and shear stress distributions in
transverse loaded beams and twisted shafts. All
these topics should indeed be a review for the
reader but in no way does this material
represent a complete review of basic strength of
materials.
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5. Normal Stress Definition
6. Normal Strain Definition
Normal stress is most easily visualized and
defined in terms of the behavior of a bar in
simple tension or compression as shown in the
figure. Consider cutting the bar in tension
transversely and treat the left portion as a free
body diagram. This portion of the bar must be
kept in equilibrium by the distribution of internal
force elements across the cut cross-section that
balances the applied external load P. These
internal force elements are given the units of
load per unit area or a kind of pressure and are
called the normal stress. Assuming these force
element distribution is uniform across the
section the axial normal stress in a bar
subjected to an axial load is the magnitude of
the load divided by the cross sectional area of
the bar. Hence, the units of normal stress are
lbs. per square in. or PSI. For a bar under
compression the normal stress is negative since
the applied load that is compressive is
considered as being a negative force.
The axial bar from the previous page will
lengthen under the action of a tensile load and
shorten when compressed. The change in
length of the bar under axial load is used to
define a generic parameter that represents this
axial deformation.
The increase in length
divided by the original length of the bar is
defined as the positive axial strain, epsilon. In
order words, epsilon is simply delta L divided by
L. Note that axial strain is dimensionless since
both delta L and L have the units of length. For
a bar shortened by compression the strain will
be negative since the shortened length delta L is
considered negative.
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Design for Strength and Endurance – Chapter 1
7. Shear Stress Definition
8. Shear Strain Definition
Consider a block element that is fixed along its
bottom edge and subjected to a horizontal force
V applied parallel to the top edge acting to the
right as shown in the figure on the left. To keep
this block in horizontal equilibrium there must
exist distributed force elements along its bottom
surface directed to the left to balance the
magnitude of the applied force V. Assuming that
these force elements are uniform over the fixed
surface of the block this leads to the concept of
shear stress acting parallel to the bottom surface
as the magnitude of the force V divided by the
surface area of the block where it is fixed. In
other words, shear stress is defined as the shear
force V divided by the area over which it acts.
Note that the dimensions of shear stress are
pressure or force per unit area expressed as PSI
similar to that of normal stress. The basic
difference between normal and shear stress is
that the first acts normal to a surface while the
second act parallel to a surface.
The application of the shear force V on the
previous page produces an angular distortion of
the block as indicated by the dotted
parallelogram on the figure. When this occurs
the right angles of the corners of the original
block are distorted.
The change in angle
gamma of the bottom left corner is defined as
the shear strain associated with the shear stress
resulting from the shear force V. Since this is a
change in angle and is measured in radians the
units of shear strain are dimensionless just like
the units of normal strain.
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Design for Strength and Endurance – Chapter 1
9. Hooke’s Law
10. Axial Elongation
Robert Hooke, an English experimental physicist
is credited with discovering the first law of
elasticity in the late 17th century. This law simply
states that stress and strain are linearly related.
That is normal stress is proportional to normal
strain and shear stress is proportional to shear
strain. The constant of proportionality for normal
stress is designated E and is called the modulus
of elasticity or Young’s modulus. The constant
of proportionality for shear strain is designated G
and is called the shear modulus of elasticity.
Both of these physical quantities must be
determined experimentally for a given specific
material. Most solid materials and in particular
metals behave in accordance with this law at
least for small values of strain. More detail on
this behavior will be discussed in a later chapter.
Using the definitions of normal stress and strain
together with Hook’s law it is possible to express
the elongation of a bar in tension in terms of the
applied load P. First begin with the relationship
expressing Hooke’s law, that is the normal
stress in the bar, sigma, is equal to E times
epsilon. Now replace sigma with its equivalent
which is the load P divided by the cross
sectional area of the bar A. Also replace the
strain epsilon with its definition which is the
change in length divided by the original bar
length L. Now solve the resulting relationship for
the bar elongation delta L. This gives the final
expression PL over AE.
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Design for Strength and Endurance – Chapter 1
11. Internal Beam Loading
12. Beam Loading Problem
Transversely loaded simple beams will now be
reviewed. Before proceeding to normal and
shear stresses in bending it is first necessary to
determine the internal bending moment and
shear force distribution due to the effect of
external transverse loadings. The internal shear
force and bending moment at any point along a
beam is determined as the force and moment
which must exist at the location of a cut cross
section of beam to satisfy equilibrium treating
the remaining beam portion either to the right or
left of that location as a free body diagram.
Applying the principle that the sum of all vertical
forces acting on the free body beam portion
must be zero will result in an equation from
which the magnitude and direction of the shear
force at the cut cross section can be determined.
Setting to zero the sum of all moments of all
external forces acting on that portion of the
beam about the cut cross section location
together with the internal bending moment will
result in an equation defining the direction and
magnitude of the internal bending moment. The
sign conventions for bending moment and shear
force are given below the figure.
The
distributions of these internal reactions along the
length of the beam are referred to as the
bending moment and shear force diagrams.
Before going on consider the application of the
concept and method of analysis described on
the previous page to the beam loaded and
supported as shown in the figure. Assume that
the internal and left end supports can only
sustain vertical reaction forces.
For the
dimensions and loading given calculate
expressions for the shear force and bending
moment distributions in the beam. Plot the shear
and bending moment diagrams along the length
of the beam and determine the maximum
bending moment. When you have completed
you solution to the problem click on the solution
button to check your results against mine.
(Solution on Pages 27 and 28)
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Design for Strength and Endurance – Chapter 1
13. Bending and Shear Characteristics
14. Bending Stress Distribution
Listed on this page are some characteristics of
shear force and bending moment diagram
behavior that are useful aids in developing these
characteristics of internal loading in beam. The
slope of the shear diagram will be equal to the
magnitude of any distributed load on the beam
at each point. That is if the loading is 200 lb/ft
then the shear will change at the rate of 200 lbs
per foot.
It also follows that between
concentrated external loads the slope of the
shear diagram will be zero. The slope of the
bending moment diagram is always equal to the
magnitude of the shear force at each point.
Thus if the shear force is constant over a length
section of the beam the slope of the bending
moment diagram over that same section will be
a constant.
Applied external forces cause
vertical jumps in the shear force diagram equal
to the magnitude of the applied load. In a similar
fashion applied external moments cause jumps
in the moment diagram but do not effect the
shear force distribution. To see how some of
these characteristics apply you may wish to
revisit the problem solution on the previous
page.
A positive bending moment applied to a beam
will cause it to assume a concave upward
deformation. This in turn means that the top
fibers of the beam will be in compression while
the bottom fibers of the beam will be put in
tension. The strain distribution is linear and
passes through zero at the centroid of the
section. This location is referred to as the
neutral axis of the beam. The accompanying
internal normal stress distribution is also linear
and given by the expression sigma equal to the
bending moment M times the distance
measured from the neutral axis y divided by the
area moment of inertia I of the cross section
through its centroid about an axis perpendicular
to the plane of the external loading. Thus the
normal bending stress is zero at the neutral axis
and is maximum at the top and bottom of the
beam cross section. For a rectangular cross
section the moment of inertia I is given as 1/12
the base b or width of the section times the
height h cubed. It is important to remember that
this stress formula is only valid for a beam
loaded in a plane of symmetry of the cross
section.
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Design for Strength and Endurance – Chapter 1
15. Shear Stress in Bending
16. Qmax for Rectangle
If a section of beam is in pure bending it will
experience no shearing stresses at an internal
cross section over this section. This is because
the internal shear force is zero since the
moment is a constant. However, if an internal
shear force exists then there will be
corresponding shear stress that accompanies
the normal bending stress. The magnitude of
the shear stress tau is given by the expression
of the product of the shear force V times the first
moment of the area of the cross section above
the point of the stress about the an axis through
the centroid perpendicular to the plane of
external loading divided by the area moment of
inertia I of the cross section and the width of the
cross section b. For a rectangular cross section
this gives rise to a parabolic shear distribution as
shown in which the shear stress is maximum at
the neutral axis and zero at the top and bottom
of the cross section.
To determine the maximum shear stress for a
rectangular cross section it is first necessary to
determine the first moment Q of the top half of
the cross section since the maximum shear
stress will be at the neutral axis through the
centroid of the cross section. This is computed
by multiplying the area of the top half of the
cross section by the square of the distance from
the centroid of the top half to the centroid of the
total cross section. Hence Q max becomes the
product of half the height h/2 times the width b
times the square of h/4. The final result for Q
max is given by 1/8 bh2.
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Design for Strength and Endurance – Chapter 1
17. Maximum Shear Stress in Bending
18. Shaft in Torsion
The expression VQ over I b is now used to
calculate the magnitude of the maximum shear
stress in a rectangular cross section.
Substituting Q equal to 1/8 bh2 and I equal to
1/12 bh3 results in a maximum shear stress of
3/2 times the shear force divided by the area of
the cross section. In other words the maximum
shear stress in bending in a rectangular cross
section is 150 % higher than what would be
expected if the it were assumed that the shear
stress was uniform over the section and just
equal to V divided by A. This multiplier for the
maximum shear stress as compared to the
average shear stress of V/A is of course a
function of the shape of the cross section and
must be determine d for other geometries.
A circular shaft in torsion will experience an
internal shear stress distribution on cross
section perpendicular to the axis of the shaft.
The formula for its calculation is similar in
appearance and form to that for normal bending
stress except that M is replaced by the torque T
acting at the section, the distance y is replaced
by the radius r measured from the center to the
point of stress and I is replaced by J which is the
polar moment of inertia of the circular cross
section. Since the torque is in the units of force
times distance or length and J is in the units of
length to the fourth power and r has the units of
length the shear stress has the units of force per
length squared or PSI as expected.
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Design for Strength and Endurance – Chapter 1
19. Shear Stress Distribution
20. Angular Twist
Since the magnitude of the shear stress in a
twisted shaft is proportional to r or the distance
measured from the center of the section then the
shear stress distribution is liner and maximum at
the outer surface of the shaft. It is important that
this analysis is only applicable to shafts of
circular cross section, however they need not be
solid. Hence the analysis is also applicable to
hollow shafts.
The angular twist, theta, of one end of the shaft
relative to the other requires the introduction of
the shear Modulus G and is given by the
expression TL over JG. In this expression T is
the Torque carried by the shaft, L is the length of
the shaft, J is the polar moment of inertia of the
circular cross section and G is the shear
modulus. Satisfy yourself that if G is given in the
units of force per unit area that the units of theta
the angle of twist are dimensionless. Thus the
magnitude of the twist that will be calculated will
be in radians.
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Design for Strength and Endurance – Chapter 1
21. Review Exercises
22. Off Line Exercises
At the end of each chapter one or more review
exercises are provided to help readers satisfy
themselves that they under stand the material
contents presented. In this exercise the items in
the list on the left are to be matched with the
mathematical relationships on the right. Place
the cursor over an item on the left and hold
down the left button. A pencil will appear that
can be dragged to one of the green dots on the
right. If the right choice is made the arrow will
remain. If the selection is incorrect the arrow will
disappear. After the exercise is completed
proceed to the next page.
Each chapter includes one or more off-line
exercises. These are to be completed by the
reader and submitted at required by the
calendar provided by the course instructor.
When you have finished with this page click on
the main menu button. This will return the
program to the chapter index page to select
another chapter or exit the module.
(Solution in Appendix)
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Design for Strength and Endurance – Chapter 1
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Design for Strength and Endurance – Chapter 1
Chapter 1
Review of Basic Concepts
Problem Solutions
Screen Titles
Support Reactions
Shear Force Distribution
Bending Moment Distribution
Shear and Moment Diagrams
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Design for Strength and Endurance – Chapter 1
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Design for Strength and Endurance – Chapter 1
1. Support Reactions
2. Shear Force Distribution
The reaction forces on the beam at the support
points A and B must be determined first using
equilibrium. The reactions are represented by
vertical forces F1 and F2 since A and B are
simple supports.
Summing moments about
point A permits F1 to be determined directly as
3750 lbs in the direction assumed since it is
positive. Summing forces vertically with F1 now
known permits F2 to be calculated as minus
1250 lbs. The negative sign indicates that it acts
down at A rather than up as initially assumed.
Assume the internal shear force V at section 1 is
positive. Equilibrium of the beam from section 1
to point C indicates that the shear force is a
constant value of 2500 lbs. from point C to B
since section 1 is a general location. Assume
again that the shear force V is positive at section
2. Equilibrium applied to the beam from section
2 to point C gives a value for V of –1250 lbs.
The negative sign indicates that the positive
shear assumption was incorrect. Since section
2 is a general location between point A and B
the shear in this section of –1250 N is again
constant.
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Design for Strength and Endurance – Chapter 1
3. Bending Moment Distribution
4. Shear and Moment Diagrams
The bending moment at section 1 is now
determined by applying moment equilibrium to
the beam section from 1 to C. It is necessary to
define the location of section 1 by the dimension
x since moment arms need to be established.
Setting clockwise moments equal to zero on the
length x to C on the beam gives the moment M
at section 1 as minus 2500x.
Hence the
moment varies linearly from C to B and is
negative.
Summing clockwise moments ay
section two for the beam from 2 to C and setting
them equal to zero gives a second linear
equation in x for the moment. Note that the
moment value at B is 12500 ft lbs whether it is
calculated from the equation for the portion C to
B or for the portion B to A. Also note that the
moment at location A from the second equation
is zero.
The shear diagram has a constant value of 2500
lbs. from point C to B as calculated previously.
At B it changes its value to –1250 lbs. due to the
effect of the reaction at B and continues constant
over the beam length from B to A. The bending
moment diagram is plotted directly below the
shear diagram to see the effects of the shear on
the bending moment. From the equations on the
previous slide the bending moment starts at zero
at point C and increases negatively in linear
fashion until it reaches the value of 12500 ft. lbs.
at point B. At this point the second equation for
the moment becomes valid which decreases its
negative value from 12500 ft. lbs. at B to zero at
point A. Placing the shear force diagram
immediately above the bending moment diagram
leads to a very useful observation. The bending
moment diagram is the integral of the shear
diagram beginning from the left end of the beam.
Although no general proof of this is given here it
is applicable to all beam bending problems.
Another interesting observation is that the
maximum bending moment occurs where the
shear force passes through zero. This is a useful
way to quickly determine where the largest
values of bending moment will occur.
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