Lecture #1
Introduction, Method of Sections
Reading: 1:1-2
Mechanics of Materials is the study of the relationship between external, applied forces and
internal effects (stress & deformation).
An understanding of statics is essential. At least one half of every mechanics problem is a
statics problem. Free body diagrams, equations of equilibrium, centroids & area moments of
inertia, structural analysis (trusses & frames) and shear and bending moment diagrams are
topics that are used extensively.
To extend statics to the determination of internal forces we use the method of sections. If a
body is in equilibrium, any part of the body must also be in equilibrium. This concept means
that we can cut a body in two in order to expose a cross section on which we want to determine
internal forces, draw a free body diagram of one of the parts (putting appropriate forces on the
cut face, i.e. internal reactions), and apply equations of equilibrium.
In general, for a three dimensional problem, there are six reactions on any cross section (3
forces & 3 moments). For two dimensional problems, this can be reduced to three reactions (2
forces & 1 moment). Note that the normal force is perpendicular to the cross section while
shear forces are in the plane of the cross section. A torsional moment vector is perpendicular to
the cross section and the bending moment vectors are in the plane of the cross section.
Lecture #2
Introduction to Normal and Shear Stress
Reading: 1:3-5
Stress is the intensity with which a force affects a point in a body.
Stress is broken into components in the same way that we break forces into components.
There is a normal stress that we represent by
Normal stress is the part of the normal force
carried by an infinitesimal area at a point on a cross section divided by the differential area.
There is also a shear stress that
we represent by This is the
part of the shear force carried by
an infinitesimal area at a point
on a cross section in a particular
coordinate direction divided by
the differential area. Recognize
that there are two components of
shear stress at a point on a cross
section.
From this we see the units on stress are force per unit area.
In SI units, we would use a Pascal (Pa) which is a N/m2, or a Mega Pascal (MPa) which is 106
Pa or a N/mm2. You will notice that when I calculate stress in SI units, I will normally use N
and mm. This way the stress always has units of MPa.
In US units, we use psi which is a lb/in2, or a ksi which is a kip/in2 or 103 psi. You will notice
that when I calculate a stress in US units, I will normally use kips and inches. We never
calculate a stress in terms of feet in mechanics (always use inches).
The state of stress of a point is a picture showing all stresses that act on an infinitesimal element
at the point. The general three dimensional state of stress is shown in the figure below. There
are equal and opposite stresses on the back faces which are not shown for clarity.
The subscripts represent the direction of
the normal stress or the plane of the shear
stress. Note that there are six components
of stress for the general three dimensional
state of stress.
If we consider the general two dimensional
state of stress, we see three components of
stress. There are two normal stresses and
one shear stress.
A normal stress may be tensile or compressive. It is important to note the difference by either
writing ( T ) or ( C ) after the stress magnitude or putting a sign on the magnitude with tension
being positive and compression being negative. Do not use both conventions together since a
negative sign negates the word. In other words, -10 ksi ( C ) is that same as +10 ksi ( T ).
Later in the semester, we will be concerned about the direction of a shear stress, but for now we
will only calculate a magnitude without a sign. If the author puts a sign on shear stress in an
example or solution at this point, I want you to ignore it.
Let‘s consider a simple example of each type of stress.
The Axially Loaded Bar
If we pass a cutting plane through the bar perpendicular to its axis, we see that the cross section
must carry a normal force equal to P. The force is not actually concentrated at a point, but
rather is distributed over the area. The average normal stress can be calculated by
σAvg = P/A
If the bar is homogeneous and isotropic, and if the resultant force acting on the cross section
passes through the centroid of the cross section, then the actual stress is uniformly distributed
and equal to the average
σ = P/A
A truss is an example of a structure that is composed of axially loaded bars.
Direct Shear Stress
An example of direct shear is the lap joint.
Here, we have two plates held
together by a bolt or a rivet. If
there is an axial force applied to the
plates and friction between the
plates is negligible, then the bolt or
rivet must transfer the force from
one plate to the other.
Cutting the bolt or rivet between the
plates and looking at a free body
diagram of one plate, we see that
the force is transferred between the
plates via a shear force on the bolt
or rivet cross section.
The average shear stress on the cross section can be calculated
Avg
=V/A
The shear stress is never uniformly distributed, but we often use this equation when dealing
with small areas such as the area of a bolt or rivet. The lap joint described above is a single lap
joint. One cross section of the bolt or rivet carries the force.
A double lap joint creates a situation called double shear.
You will note that the force is
transferred between the plates via
two cross sections of the bolt or
rivet.
If we cut the bolt or rivet between
the plates and consider a free body
diagram of the center plate, we see
that each cross section of the bolt
or rivet carries half of the force.
When calculating the average shear stress, we can consider one area with half of the force or
two areas with the entire force. Either way, the stress is halved.
Avg
= V / A = (P/2) / A = P / (2A)
Lecture #3
Introduction to Design & Factor of Safety
Reading: 1:6-7
Here we introduce the idea of design or the sizing of a member based on a comparison of the
largest stress experienced and the allowable stress that the material can safely carry.
We must also introduce here the concept of factor of safety. The factor of safety is defined as
the ratio of the maximum stress a material can carry before some form of failure occurs and the
allowable stress, i.e., the stress that we are willing to accept in the material.
F.S. =
fail
/
or
allow
F.S. =
fail
/
allow
The failure stress is defined by the type of failure which might occur. Most often, we will
define the failure stress to be the yield stress, i.e., that stress that causes the material to yield.
Fail
=
or
yield
fail
=
yield
We always allow a stress that is below the failure stress, i.e., the factor of safety is always
greater that 1.
The factor of safety is sometimes called the factor of ignorance because we introduce the
concept to deal with uncertainty. We don‘t allow stresses right up to the failure stress because
of uncertainties that exist. The failure stress is a statistical quantity (an average with some
standard deviation). We are not certain of the exact value for any specimen of the material.
The actual stress is based on applied forces which may be approximations (not known with
certainty).
Anyway, a comparison between the maximum stress experienced and the allowable stress for
the material can be used to determine the minimum required size of the cross section.
max
=P/A=
Areq = P /
allow
allow
max
=V/A=
Areq = V /
allow
allow
Another stress introduced in these sections is called a bearing stress. This is a compressive
stress on the surfaces of two contacting bodies. Often this stress has a complex distribution so a
nominal value is calculated.
For example, consider the bearing stress created by the contact between the bolt and the inside
of the hole in the plate of a lap joint. Cutting the bolt between the plates and then cutting one of
the plates through the hole, we see in the figure on the next page that the force P acts on half of
the inside of the hole.
The nominal bearing stress is calculated
from a projected area - the diameter of the
hole times the thickness of the plate.
bearing
=P/(td)
Lecture #4
Introduction to Normal & Shear Strain
Reading: 2:1-2
Applied forces create deformation, i.e., changes in size and shape.
Changes in size come from normal strain. We represent the normal strain using and define it
as a change in length per unit length. The assumption here is that if the material is
homogeneous, then the change in length will be equally distributed along the length.
= L / Lo
Changes in shape come from shear strain. We represent shear strain using and define it as the
change in a right angle.
Note that strain is unitless. Therefore, must be in radians.
There are 6 components of strain. There are three normal strains - one associated with each
coordinate direction ( x , y , z ). There are also three shear strains - one associated with each
plane ( xy , xz , yz ).
We will generally assume that strains are small, i.e.,
<< 1.
Lecture #5
Material Behavior & Properties
Reading: 3:1-4,6-7
Material properties and strengths are vitally important in mechanics. To learn about a material,
we run standardized tests. One of these tests is a tension test. In a tension test, we pull on a
specimen and measure the force and the change in a prescribed gauge length.
If we plot P vs L, we don‘t actually learn anything about the
material because a specimen that has a different cross section area
or a different gauge length will behave differently. To eliminate
the dependence on the cross section, we plot stress instead of force.
To eliminate the dependence on length, we plot strain instead of the
change in length. A plot of vs will then be substantially the
same for multiple test specimens of the same material but different
sizes and lengths.
Most material vs diagrams share some common characteristics. To discuss these
characteristics, lets consider the diagram for a mild steel.
Generally, there is an initial region which is
linear. This is called the proportional region.
In this region, stress and strain are related to
one another by the slope of the curve which is a
property called the modulus of elasticity or E.
So we can write
= E . This equation is known as Hooke‘s Law.
At the top of the proportional region is the proportional limit.
Most materials also exhibit an elastic region. When a material is loaded with a stress within the
elastic region and then unloaded, the material will return to the original size and shape, i.e.,
there is no permanent deformation.
The elastic region is generally slightly larger than the proportional region, but for our purposes
can be considered to be the same. The top of the elastic region, called the elastic limit, is then
basically the same as the proportional limit.
We also define a yield point. This is the value of stress at which we start to see large increases
in strain for no increase in stress. This point is generally very near the elastic limit and
proportional limit. Mild steel exhibits a well defined yield point, y . Not all materials do. For
our purposes, the yield point, elastic limit and proportional limit will be taken to be the same.
If we stress a material beyond the yield point and unload, there will be some permanent
deformation. The unloading path is nearly linear and parallel to the proportional region.
Therefore, there is an elastic strain which is recovered when the material is unloaded, E =
Subtracting the elastic strain from the total strain gives the permanent or plastic strain, P .
/ E.
If we reload the material, the behavior will
follow the unloading path back to the original
curve. So we have essentially increased the
size of the proportional region and raised the
yield point. This phenomenon is called work
or strain hardening.
One way we classify materials is as either being ductile or brittle. A ductile material generally
has a large plastic region, i.e., it can experience considerable strain beyond the yield point. A
brittle material generally has a small plastic region. In other words, a ductile material gives a
significant warning that failure is impending, but a brittle material will give very little warning
of failure.
When a material is strain hardened, its yield strength is increased, but the material becomes
more brittle since the plastic region is reduced.
The maximum stress a material can withstand is called the ultimate strength, u .
Once the material reaches this stress, a local region of the member will begin to
experience necking. Necking is a narrowing of the cross section at the place of
ultimate failure.
Because of the reduced cross section, the applied load can be decreased and the member
continues to stretch and the neck narrow until ultimate rupture.
Earlier, it was mentioned that not all materials exhibit
a well defined yield point. In this event, it is
customary to use the 0.2% offset stress as the yield
strength. Here, we start from a strain value of 0.002
and draw a line parallel to the proportional region
until it intersects the stress vs strain curve.
When we pull on a specimen, we also note that it gets smaller in the directions perpendicular to
the load. This is called the Poisson effect.
Another material property is the Poisson Ratio,
= - (transverse strain)/(axial strain) = -
t
/
a
The negative sign is included to make a positive
quantity since t will be negative when a is positive.
We have only discussed normal stress and strain. There is also a standard test for determining a
shear stress vs shear strain diagram. The diagram for shear will be different, but the diagram
will have analogous attributes to the normal stress vs normal strain diagram. There is a
proportional region in which we can write the shear form of Hooke‘s Law, = G . G is the
shear modulus or slope of the proportional region.
There is a yield strength for shear, y , an ultimate strength, u , and other similar behaviors.
The shear modulus is a third material property, but it is not independent of E and . It has been
shown that
G = E / (2(1+ ))
There is more that we could discuss about material behavior, but this discussion is sufficient for
our purposes.
The back of the textbook has a table of properties and strength values for some typical
engineering materials.
Lecture #6
Axially Loaded Members
Read: 4:1-2
Axial Loads
We have already determined the stress due to an axial load.
=P/A
We often encounter axially loaded members that have either a cross section or an axial load that
varies over the length of the member. If we have a distributed axial load, we need to use the
method of sections to cut the member at an arbitrary location. Drawing a free body diagram of
part of the member and using equilibrium, we can determine the internal axial load as a function
of length. If there are multiple concentrated, applied axial forces, the internal axial load will be
constant between forces, but we still need to use the method of sections to cut the member at
some arbitrary location between each set of forces. Drawing a free body diagram of one part of
the member, we can use equilibrium to determine the internal axial load. Once we know how
the internal axial load changes over the length, we can calculate the stress at any point in the
member. The maximum stress will occur where P / A is maximum. This may be caused by
the largest force, the smallest area, or neither. We may need to compare stress values at a
number of locations to know for sure.
We can also use the internal load variation and cross section area variation to determine the
change in length of a member. Consider a small piece of an axially loaded member.
The small piece will elongate a small amount d under
the action of the axial load P. We know that = P / A
and = d / dx . If the material is elastic and in the
proportional region of its behavior, we know that
= E . So,
= P / A = E = E ( d / dx )
From this we find
d = ( P / (AE)) dx
If we want to determine the change in length of the entire member, we need to integrate over
the length.
= ∫ d = ∫ ( P / (AE)) dx
If P and/or A is a function of x, we will need to evaluate an integral..
If P, A and E are all constant, then
= ( P / (AE)) ∫ dx = (PL) / (AE)
If P, A and E are all constant over sections of the length, then
=
( (PL) / (AE) )i
Lecture #7
Statically Indeterminant Axially Loaded Members
Reading: 4:4
A member is statically indeterminant if the number of unknown reactions is greater than the
number of equilibrium equations.
In an axially loaded member, any reactions will also be axial forces. First, we draw a free body
diagram of the member and then apply our one equilibrium equation, Fx = 0 . In a structure
that contains axially loaded members we may have additional equilibrium equations.
Next, we look to write constraint equations which involve displacements and are based on how
the problem is supported. If a bar is supported at each end and not allowed to change length,
then our constraint equation would be
δ=0
We will work other problems where displacements of two members are the same or related to
one another in some way.
Lecture #8
Temperature Effects on Axially Loaded Members
Reading: 4:6
Now, we would like to add temperature effects.
If we have an unconstrained bar, and we increase the temperature, the bar will change length by
d
Where
T dx
is the coefficient of thermal expansion.
Within a reasonable temperature range, may be considered
a constant. The material property table in the back of the
textbook gives values for
If
T is a function of length, then
∫
If
T dx
T is constant, then
TL
If we have a member that is exposed to an axial force and a temperature change, then
( ( PL ) / ( AE) ) +
TL
For a member with multiple sections with constant P, A, E, and T, then
PL ) / ( AE) ) +
T L ]i
The change in temperature does not affect the stress in the member unless the member is
statically indeterminant and then the effect can be drastic. When a member is constrained and
the temperature is increased, significant reaction forces may be developed in order to keep the
material from expanding due to the temperature change.
Lecture #9
Torsion of Cylindrical Members
Reading: 5:1-3
Torsion of members having a circular cross section
Consider a bar with a circular cross section. First, we make a grid on the member of lines along
the length and around the circumference. Now, we place a torsional moment on the member
and watch how the grid deforms.
We note a couple of things from the deformation.
First, the vertical lines which represent a cross
section in the two dimensional view remain
vertical. This means that the cross section remains
planar. In other words, a cross section rotates as a
rigid disk.
Secondly, if we consider the shape of a single
element of the grid, we see the deformation gives
the shape that we saw earlier caused by shear
stress.
.
Now, consider a slice of the torsion member. We
will consider the left side of the slice as being
fixed and the right side as having a torque applied
The radial line CA will rotate to a new position
CA‘. The angle d is the angle of twist of the
cross section and the deformation angle of the line
PA is the max shear strain, max . The arc length
AA‘ can be written two ways.
AA‘
dx
max
=Rd
If we consider a point B a distance r from the center, it moves to position B‘. Because the angle
d if the same, the shear strain of an element on this smaller cylinder will be smaller. We will
call this angle
The arc length BB‘ can be written as
BB‘
dx
rd
From these equations, we see two expressions for d /dx .
d /dx =
So,
max
max
/R=
/r
(r/R)
From this we see that shear strain varies linearly with the distance r measured from the center of
the cylindrical member. Now, assume that the material remains within the linear, elastic range
of its behavior. We can apply Hooke‘s Law.
=G
G
max
(r/R)=
max
(r/R)
We see then that the shear stress varies linearly from the center of the cross section. Now, lets
remember that the stress distribution on the cross section must be statically equivalent to the net
torque, T.
Considering an arbitrary element of area, dA, on the
cross section a distance r from the center which has a
stress
dA = a net force dF
dF r = a net torque around the center dT
dT = r dF = r dA = r
dT =
max
Note that
max
/ R ) r2 dA = (
max
( r / R ) dA
/ R ) r2 dA
If we integrate over the cross section area, we get an equation for
T = ∫ dT = ∫ (
max
max
.
/ R ) ∫ r2 dA
∫ r2 dA = J ( Polar Moment of Inertia).
J=(
/ 2 ) R4
J=(
/ 2 ) ( Ro4 - Ri4 )
for a solid circular cross section
for a circular tube
Now, we have
max
J/R
or
max
=TR/J
Substituting back into the equation for , we get a general equation for the shear stress.
=
max
r/R=(TR/J)(r/R)=Tr/J
A torsion member may carry a torque which varies along its length. It may be a function of
length, or constant in sections. We recognize that if the applied torques are concentrated, the
internal torque will be constant between the applied torques. The internal torques are
determined using the method of sections.
An arbitrary cut must be made in each section of the member, a free body diagram must be
drawn, and an equilibrium equation written to determine the internal torque distribution.
The maximum shear stress in the member can be found by calculating the maximum shear
stress in each section of the member and comparing. If the cross section is constant, we
recognize that the maximum will come from the maximum T. But if the cross section
changes, the maximum will come from where T r / J is maximum which could be any section.
Torsion members are often used to transfer power from a power source to a machine which will
do work. Power is the rate of doing work and is related to the torque carried and the angular
velocity of the shaft.
P=T
Units are important when using this equation. In SI units, P is in Watts which is a ( N-m / s ) so
T will be in ( N-m ) and in ( rad / s ). In English units P is typically given in horsepower
(hp). One hp is equal to 550 ft-lb / s . After this conversion, T will be in ( lb– ft ) and in
( rad / s ). Make sure you use the correct units in this equation.
We are also able to design a shaft at this point. If we know the allowable shear stress,
then
allow
allow
,
= ( T R / J )max
So, from this
Jreq / Rreq = T /
allow
For a solid cross section,
R4 / R = T /
or
allow
R3 = T /
allow
From which
R=(T/(
allow
2
Keep in mind that this is a minimum required size.
Sign convention note:
When finding the internal torque, we generally assume
positive as a vector pointing away from the cross section.
However, the sign of the torque is usually ignored in a stress
calculation. The material does not distinguish between a
shear stress in one direction or the other.
Lecture #10
Angle of Twist for Cylindrical Torsion Members
Reading: 5:4
From the previous lecture, we learned that
d dx = / r
where
is the angle of twist.
Assuming the material to be linearly elastic, we can say
= /G
and
=Tr/J
So
d /dx = T r / ( G J r ) = T / ( G J )
For a general torsion member, we can find the angle of twist by integrating this equation.
d = ( T / ( G J )) dx
= ∫ ( T / ( G J )) dx
If T, G or J are functions of length, we need to integrate. If these quantities are constant over
the length, then
= ( T / ( G J )) ∫ dx = T L / ( G J )
If T, G and J are constant in sections, then we can add together the twist from each section.
( T L / ( G J ))i
When calculating the angle of twist, the sign convention for T is important. We must carry the
sign into the equation for because the angle of twist for two sections could be in opposite
directions, canceling one another to some degree over the entire length.
Lecture #11
Statically Indeterminant Torsion Members
Reading: 5:5
As in the axial load problem, if we have more reactions than equations of equilibrium, the
problem is statically indeterminant. First, we draw a free body diagram of the member. In this
case, our one equation of equilibrium is a moment equation.
T=0
Then, we write constraint equations as needed involving the angle of twist based on how the
member is constrained. If for example, a member is held at each end, we would say that
φ = 0.
We will solve other problems where the angle of twist of two members are the same or related
to one another.
Lecture #12
Torsion of Noncylindrical Members
Reading: 5:6-7
When the cross section was circular, we noted that cross sections remained planar after a torque
was applied. This does not happen when the cross section is noncircular. The cross section will
warp out of plane.
For solid noncircular cross sections, the theory is beyond the scope of this course. The textbook
presents some equations for 3 common solid cross sections. See the table in the textbook for
these equations. Although the theory is beyond this course, there is no reason why we cannot
use the results of that theory to solve problems.
Another type of noncircular cross section which we can analyze is the thin walled tube.
Although the wall must be thin (generally an order of magnitude smaller than the major inplane dimensions), the thickness of the wall does not have to be constant.
Since the wall is thin, we will assume
that the shear stress is constant
through the thickness and tangent to
the wall centerline. Consider a piece
of the tube wall
Assume a shear stress 1 at the corner
of the thickness t1 and a shear stress
of 2 at the corner of thickness t2 .
Because shear stress always occurs in
a set of four stresses, we know the
stress in the wall along the length of
our piece.
Where the thickness is t1 , the shear
stress is 1, and where the thickness
is t2 , the shear stress is 2 .
Writing
Fx = 0,
1 t1
L-
2 t2
L=0
or
1 t1
=
2 t2
We made no assumptions about the size of the piece so L and s are arbitrary. This means that
t = constant
We define this product as a new quantity called shear flow, q . Note this is a force per unit
length.
Looking back at the cross section, the shear flow around the circumference of the wall must be
statically equivalent to the applied torque.
Over a small piece of the circumference, q creates a net
force
dF = q ds
which in turn creates a torque around the axis of the shaft
dT = q ds h
The total torque is then found by integrating around the circumference.
T = ∫ dT = ∫ q h ds
But q is constant, so
T = q ∫ h ds
The integral is geometry dependent only and is equal to twice the area enclosed by the wall
centerline.
T = q 2 Am
where Am = area enclosed by wall centerline
Now, using t = q , we get
T = t 2 Am
or
= T / ( 2 Am t )
The maximum stress obviously occurs where the thickness is smallest.
The angle of twist can be derived as well to find
= (( T L ) / ( 4 Am2 G ) ∫ ( ds / t )
if T is constant over the entire length.
∫ ( ds / t ) is evaluated knowing how t varies with the circumference variable s.
If t is constant, ∫ ( ds / t ) = s / t
If t is constant in sections of the wall, ∫ ( ds / t ) =
( s / t )i
Lecture #13
Shear & Bending Moment Diagrams
Reading: 6:1
A beam is a structure that carries transverse loads. This means that a cross section will carry a
shear force and a bending moment. We want to be able to plot shear and bending moment over
the length of the beam in order to determine the maximum value of these quantities. So that we
all have the same diagrams, we introduce a sign convention.
Positive shear rotates the beam clockwise and positive bending moment points to the top of the
cross section. A positive bending moment creates compression at the top of the beam and
tension at the bottem of the beam.
One way we can plot these quantities is to use the method of sections to find V and M as
functions of length.
First, we must recognize that we may not be able to represent V and M as a single functions
over the entire length. The beam must first be divided into sections where V and M can be
represented by single functions. This occurs between points of concentrated force or moment,
supports, or a change in the functional representation of a distributed load.
Once we know how many sections we have, we can draw a free body diagram of the whole
beam and use equilibrium to calculate the reaction forces. Then, we use the method of sections
to make a cut in the interior of each section at an arbitrary distance x measured from the left
hand end of the beam. At this point, we draw a free body diagram of one part of the beam
assuming V and M in their positive directions and apply equilibrium equations to determine
functions for V and M in terms of x. Then we simply plot the functions.
Lecture #14
Shear & Bending Moment Diagrams
Reading: 6:2
If we consider a small piece of beam with a distributed load
and draw a free body diagram with positive internal forces, we
get
The shear and moment on the right side will be different from
the left because of the distributed load, but not much different
because dx is small. Let V2 = V1 + dV and M2 = M1 + dM,
then apply equilibrium equations.
Fy = 0 :
V1 + w dx - ( V1 + dV ) = 0
w dx - dV = 0
dV = w dx
M=0:
or
dV/dx = w
( M1 + dM ) - M1 - V1 dx - w dx ( dx / 2 ) = 0
dM - V1 dx - w dx2 / 2 = 0
Since dM and dx are small, dx2 is very small and can be neglected. We can also drop the
subscript on V at this time.
dM = V dx
or
dM/dx = V
These equations are the basis of what I call the summation method.
dV/dx = w says that the slope of the shear diagram is equal to the distributed load.
Integrating dV = w dx , we get
∫ dV = ∫ w dx
V2 - V1 = ∫ w dx
or
V2 = V1 + ∫ w dx
This says that the shear at some point x2 equals the shear at the starting point x1 plus the area
under the distributed load curve between the two points.
dM/dx = V says that the slope of the moment diagram is equal to the shear force.
Also, since the maximum moment occurs where dM/dx = 0 , we know that the maximum
moment will be where the shear force is zero.
Integrating dM = V dx , we get
∫ dM = ∫ V dx
M2 - M1 = ∫ V dx
or
M2 = M1 + ∫ V dx
So, the moment at a point x2 equals the moment at the starting point x1 plus the area under the
shear diagram between the two points.
These equations are valid between concentrated forces and moments. We also know that at
points of concentrated forces we get a discontinuity or a jump in the shear diagram in the
direction of and equal in magnitude of the force. At points of concentrated moments, we get a
discontinuity or a jump iin the moment diagram equal in magnitude to the concentrated
moment. The direction of the jump is up when the moment is clockwise and down when
counterclockwise.
Let‘s state a procedure for the summation method.
1. Draw a free body diagram of the beam and use equilibrium to determine the reaction forces.
2. Draw the shear diagram using the following rules.
a. Start at the left end at zero
b. When you encounter a concentrated force, follow the force. That is jump in the
direction of the force the magnitude of the force.
c. Between concentrated forces, apply
V2 = V1 + ∫ w dx
to find the shear at the end of the section and dV/dx = w
to draw the line over the section.
3. Draw the moment diagram using the following rules.
a. Start at the left end at zero.
b. When you encounter a concentrated moment, jump up if the moment is CW or down
if the moment is CCW a magnitude equal to the magnitude of the moment.
c. Between concentrated moments, apply
M2 = M1 + ∫ V dx
to find the moment at the end of a section and dM/dx = V
to draw the line over the section.
Notes:
1. Both diagrams must return to zero at the right end of the beam. If one of them does not,
then a mistake has been made. An area may have been calculated incorrectly or the
reactions may be wrong.
2. The maximum moment will occur at an end or where dM/dx = 0 . Therefore, you must find
M at all places where V = 0 .
3. This method breaks down if the distributed load is higher order than constant. For these
sections, we go back to the method of sections to find the functions of x.
Lecture #15
Centroids and Area Moments of Inertia
Reading: A:1-3
You were introduced to these topics in statics so we will only review quickly. However,
centroids and moments of inertia are very important concepts in mechanics of materials.
In general, the centroid of an area is
found by
xc = ∫ x dA / A
yc = ∫ y dA / A
where ( xc , yc ) are the coordinates of the
centroid and A = ∫ dA .
In practice, we do not often need to integrate because most areas we encounter can be treated as
a composite area - an area made up of simple shapes. Given that we have tables that locate the
centroid of simple shapes for us, we can simplify the equations to
xc =
Ai xci / A
yc =
Ai yci / A
Where ( xci , yci ) are the coordinates of the centroid of the simple shape Ai and A =
We know that if there is an axis of symmetry, the centroid must be located on that axis.
In general, area moments of inertia are found from
Ix = ∫ y2 dA
Iy = ∫ x2 dA
Ixy = ∫ x y dA J = ∫ r2 dA
where Ixy is the product of inertia and J is the polar moment of inertia.
Since r2 = x2 + y2 , we can show that J = Ix + Iy .
We know that Ix , Iy , and J are always positive, but Ixy can be positive or negative.
i
.
Again, in practice, we don‘t often need to integrate because most areas we encounter are
composite areas. Therefore, we can simplify the equations to
Ix =
Ixi
Iy =
Iyi
Ixy =
Ixyi
J=
Ji
Where the subscript i represents that the moment of inertia is for the ith simple shape.
Although we have tables that give information about moments of inertia for simple shapes, the
information given is generally for specific axes through the simple shape. Most often, the given
equations are with respect to axes that pass through the centroid. The parallel axis theorem
allows us to determine the moments of inertia for any set of axes parallel to the axes that pass
through the centroid.
Ix = Icx + A dy2
Iy = Icy + A dx2
Ixy = Icxy + A dx dy
J = Jc + A d2
Where ( dx , dy ) are the coordinates of the centroid, the bar above the moment of inertia
designates that the moment of inertia is with respect to a centroidal axis and d2 = dx2 + dy2 .
Often, information about the product of inertia is not given in the tables provided. If there is an
axis of symmetry, Ixy = 0 . Therefore, we only need equations for unsymmetric shapes. For
these shapes, the sign will depend on the orientation. For example, consider the right triangle.
Using the parallel axis theorem, we can write
Ix =
Ixy =
( Icxi + Ai dyi2 )
( Icxyi + Ai dxi dyi )
Iy =
( Icyi + Ai dxi2 )
J=
( Jci + Ai di2 )
In statics, you probably tabulated the information when finding a centroid or moment of inertia.
I will no longer do that, but feel free to make a table if that helps you.
Lecture #16
Principal Axes and Principal Moments of Inertia
Reading: A:4
Consider an area in a coordinate system x,y which is rotated an angle
to x‘,y‘ .
If we want to find the moments
of inertia with respect to the
x‘,y‘ axes, we could integrate.
Ix‘ = ∫ ( y‘ )2 dA
Iy‘ = ∫ ( x‘ )2 dA
Ix‘y‘ = ∫ ( x‘ )( y‘ ) dA
These integrals can be written in terms of the x,y coordinate system if we relate x‘ and y‘ to x
and y. Note that
x‘ = x cos
+ y sin
y‘ = y cos
- x sin
Now, we can substitute these equations into our integrals. Consider Ix‘ .
Ix‘ = ∫ ( y‘ )2 dA = ∫ ( y cos
Ix‘ = ∫ ( y2 cos2
+ x2 sin2
- x sin
)2 dA
x y sin
cos
) dA
Now, we can separate the integral into three integrals. Recognizing that
respect to the integration, we get
Ix‘ = ∫ y2 dA cos2
+ ∫ x2 dA sin2
∫ x y dA sin
Recognizing the definitions of Ix , Iy and Ixy ,
Ix‘ = Ix cos2
+ Iy sin2
- Ixy 2 sin
cos
cos
is a constant with
So we found Ix‘ in terms of Ix , Iy and Ixy . Assuming we can find the moments of inertia for one
orientation, we can now find them for any other orientation.
This equation is typically used in a different form. Knowing the trig identities
cos2
= 1/2 ( 1 + cos 2 )
sin2
cos 2 )
2 sin
cos
and
sin 2
we can substitute to find
Ix‘ = [ ( Ix + Iy )/2 ] + [ ( Ix - Iy )/2 ] cos 2 - Ixy sin 2
Following the same procedure, we also find
Iy‘ = [ ( Ix + Iy )/2 ] - [ ( Ix - Iy )/2 ] cos 2 + Ixy sin 2
and
Ixy‘ = [ ( Ix - Iy )/2 ] sin 2 + Ixy cos 2
The importance of these equations is that we can determine the maximum or minimum values
of the moments of inertia. Consider finding the extreme values of Ix‘ by taking a derivative and
setting to zero.
dIx‘/d = 0 = - [ ( Ix - Iy )/2 ] 2 sin 2 - Ixy 2 cos 2
[ ( Ix - Iy )/2 ] sin 2 = Ixy cos 2
tan 2
- 2 Ixy / ( Ix - Iy )
This locates the extreme values of Ix‘ . There are two solutions for 2 which are different by
180o or . Your calculator always gives the value between +90o and - 90o or between +
and
o
- /2. Solving for , our two roots differ by 90 . This means that we only need one of the roots,
because the second root is the opposite axis. One of the roots gives a maximum value while the
other gives a minimum value. Therefore, the maximum and minimum values of moment of
inertia occur for the same coordinate system - one being Ix‘ and the other Iy‘.
If you compare the equations for dIx‘/d and Ixy‘ , you will notice that they are the same except
for a factor or –2. Therefore, when we have the max and min values for Ix‘ and Iy‘ , the product
of inertia is zero ( Ix‘y‘ = 0 ). This coordinate system is called the principal axes and the
moments of inertia are called the principal moments of inertia. This will become important
very soon.
The maximum and minimum values can also be determined from
Imax,min = [ ( Ix + Iy ) / 2 ] ± { [ ( Ix - Iy )/2 ]2 + Ixy2 }1/2
The problem with this equation is we don‘t know which is Ix‘ and which is Iy‘ so we still need to
substitute back into one of the equations so that we know which is which. In many
circumstances, we can tell which is which by looking at the area in relation to the axes x‘, y‘. If
you can see that the area is concentrated closer to one axis and spread farther away from the
other, then the axis which has the area farther from it has the maximum moment of inertia.
Lecture #17
Stress from Pure Bending
Reading: 6:3-4
Consider the behavior of a beam which has a cross section with a vertical axis of symmetry
under the action of pure bending. If a grid of lines is marked on the beam along the length and
vertically before the moment is applied, the response of the beam to the moment can be noted.
After the moment is applied, the beam takes a curved shape. We note that the lines along the
length become arcs and the vertical lines become radial, intersecting at a common center of
curvature. We note that the top of the beam is in compression while the bottom is in tension.
Somewhere in between, there must be a point that has no stress or strain. If there were a line
along the length there, it would not have changed length. Because vertical lines remain straight,
we know that the cross section remains planar. Now, consider an infinitesimal length of the
beam.
Let the length be dx. The line that does not change
length represents a plane through the thickness called the
neutral plane or neutral surface. The intersection of the
neutral surface with the cross section is called the neutral
axis and is represented by a point in the diagram. The
radius of curvature, , is defined as the distance from the
center of curvature to the neutral surface. The angle
made by the radial lines will be called d . We know
then that dx = d . We define a distance, y, as the
distance measured from the neutral surface upward. The
change in the length of an arbitrary fiber, y above the
neutral surface, can be seen to be
du = - y d
The strain along this fiber would be
du/dx = (-y d
d
-y /
We see that the strain varies linearly from the neutral surface. If we assume that the material
stays within the linear, elastic region of its behavior, then
E
Ey
So stress varies linearly. The maximum stress or strain
occurs at the point farthest from the neutral surface. Imagine
this distance is to the top and call it ―c‖. Now looking at a
plot of versus y,
max
=-Ec/
So we could write
max
y/c
This stress distribution must be statically equivalent to the
forces on the beam cross section. Therefore,
P= ∫
Since
max
dA = 0
or
∫(
max
/ c) y dA = 0
and c are constants
∫ y dA = 0
We remember that this integral is used to locate the centroid. Therefore,
∫ y dA = A y = 0
This tells us that the neutral surface must pass through the centroid of the cross section. So the
neutral axis is a centroidal axis.
Also, the stress distribution must be equivalent to the moment on the cross section.
M=- ∫
y dA
The negative sign is included because M goes in the opposite direction caused by a tensile
stress. Substituting the equation for stress,
M=- ∫(
max
/ c) y2 dA = - (
max
/ c) ∫ y2 dA
Note that ∫ y2 dA = I, the moment of inertia of the cross section about the neutral axis.
So now,
M=-
max
I/c
or
Substituting back into the equation for
max
max
=-Mc/I
, we get
y/c=-My/I
This equation assumes that y is measured upward from the neutral axis and that M is pointed
toward the top of the cross section. It will be to our advantage later to eliminate the need for a
sign on each of these terms. Instead, from this point on the equation will be written as
=My/I
where M and y are always taken as positive (magnitudes without signs) and then a sign is
placed on the final stress value based on whether the stress is tensile(+) or compressive(-).
This can be determined from the point where you want to calculate the stress and the direction
of the bending moment. Remember that the moment always points to the side of the cross
section that is in compression.
Lecture #18
Unsymmetric Bending
Reading: 6:5
When we derived the equation for bending stress, we assumed that the cross section was
symmetric, i.e. Ixy = 0. Actually, = - M y / I can be applied to any cross section with a
bending moment as long as the moment is around a principal axis and I is a principal moment
of inertia, i.e. Ixy = 0.
We could consider two different situations.
1. What if the cross section is symmetric, so we know
where the principal axes are, but the moment vector is at
an angle to the principal coordinate system?
Consider a rectangular cross section with a moment
vector at an angle from the principal x axis in the plane
of the cross section.
If we break the moment vector into components about
the x and y axes, we can apply the bending stress
equation to each moment and use superposition to add
the stresses together.
In the equation = M y / I , we must recognize that each
letter is a placeholder or a dummy variable. The
moment must be about a principal axis, the moment of
inertia must be with respect to the same axis as the
moment, and y represents a distance measured
perpendicular to the axis the moment is around.
Consider the stress due to Mx. The equation used will be
= Mx y / Ix . There must be a sign placed on the stress
based on whether the stress is tensile or compressive. If
we want, we can write a general equation where we
allow y to be positive or negative based on the
coordinate of the point where we want to calculate the
stress.
If you consider a point where y is positive, note that the moment Mx causes tension based on the
given direction of the moment. So in general,
= + Mx y / Ix .
Now repeat this process for My to find
= - M y x / Iy .
Using superposition, we write
= + Mx y / Ix - My x / Iy .
Now, we have an equation which is good for any point in the cross section. We only need to
substitute the appropriate coordinate values of the point where we want to find the stress.
This general equation allows for the determination of the neutral axis, the line of zero stress in
the cross section. By setting our general stress equation equal to zero, we get the equation of a
line.
0 = + Mx y / Ix - My x / Iy
or
y = (( My Ix )/( Mx Iy )) x
We can plot the line or determine the angle it makes with the coordinate axes.
We can also see where the maximum stress occurs from
the general equation by finding the point farthest away
from the neutral axis.
In this case, the maximum occurs at points A and B
where the stress at A is tensile and the stress at B is
compressive.
If we are only interested in finding the stress at one
point, we do not need to develop a general expression.
We can instead take all placeholders in the equation
= M y / I to be positive and place a sign on the stress
depending on whether it is tensile or compressive at the
point. Again, the sign of the stress can be determined
from the direction of the moment and the location of the
point relative to the axis the moment goes around.
For example, the stress at B could be written as
= - Mx (h/2) / Ix - My (b/2) / Iy
The difference is that we did not write a general equation and take the x and y values to the be
coordinates of the point of interest.
The other type of problem we could be faced
with is one where we have a general cross
section which does not have an axis of
symmetry with a moment applied.
Consider, for example, a cross section as
shown with a moment around the x axis.
Let‘s assume that we can calculate the
moments of inertia Ix , Iy , Ixy and the product
of inertia is not zero.
We could use equations developed in an
earlier lesson to find the principal axes and
the principal moments of inertia. Imagine
the principal axes to be located by the angle
p and that we can determine Ix‘ and Iy‘ .
Now, the moment can be broken into
components in the x‘ and y‘ directions and
we can write a general equation for stress in
terms of x‘ and y‘ or determine the stress at a
specific point as before.
The difficult part of this second problem type is determining the coordinate values to be
substituted into the equation. Some geometry may be required.
The point where the maximum stress occurs may not be obvious. The easiest way to find the
maximum stress for a general cross section is to locate the neutral axis as before, then determine
which point in the cross section is furthest away from that axis. Then we can perform the
necessary geometry and trigonometric manipulations required to find the coordinates.
Lecture #19
Shear Stress from Shear Forces
Reading: 7:1-3
Consider a piece of a beam with constant shear force. Again, let‘s assume there is a vertical
axis of symmetry. From equilibrium,
(M+dM) - M - Vdx = 0
dM = V dx
V = dM/dx
So the moments on the two sides are not the same. If we consider the stress caused by the
bending moments (note this assumes that the material is linearly elastic), the stresses must be
larger on the right hand side.
Now, let‘s measure an arbitrary distance y from
the neutral axis and slice the piece horizontally.
If we look at the piece we have cut off, we see
that because the net horizontal force on each side
is different, a shear force develops on the cut
horizontal face.
F1 = ∫
dA = ∫ ( M y / I ) dA
F1 = ( M / I ) ∫ y dA
In the same way,
F2 = ( (M+dM) / I ) ∫ y dA
We recognize that ∫ y dA = A yc
Let‘s define this quantity as
Q = ∫ y dA = A yc =
yc
Now, from equilibrium,
Fx = 0 :
dF + F1 = F2
dF = F2 - F1 = ((M + dM) Q / I) - (M Q / I) = dM Q / I = V dx Q / I
dF/dx = VQ/I
This is a force per unit length of the beam. If we divide by the thickness of the bottom cut
surface, we get stress.
= (dF/dx)/t = VQ/(I t)
Note that by looking at an infinitesimal element at the
corners of the bottom surface, we find the stress on the
cross section is equal to that found on the bottom surface.
Also note that the shear stress on the cross section is in the
same direction as the shear force on the cross section.
Let‘s consider the shear stress variation on a rectangular
cross section. To calculate the stress at an arbitrary
location y above the neutral axis, first draw a line
perpendicular to V through the cross section at the point
of interest. Q is found from the area above this line. The
thickness t is the length of the line that cuts through
material.
Q = A y = b (h/2—y) (h/2 + y)/2 = b/2 ((h/2)2 - y2)
t=b
= VQ/(I t) = [ V (b/2)((h/2)2 - y2)]/[(bh3/12) b]
We see immediately that the stress varies parabolically or quadradically with y. If y = ± h/2,
we see that = 0. The maximum value of shear stress occurs when y = 0.
max
= [V b/2 (h/2)2]/[(bh3/12) b] = [12 V b h2]/[8 b2h3] = 3V/[2 bh] = 3V/(2A)
We have just shown that the maximum shear stress for a rectangular cross section is 50% larger
than the average shear stress.
In general, the maximum shear stress will occur where Q/t is maximum since V and I are
constants. We should note that Q is always maximum at the centroid. The thickness t of the
material along the line drawn through the cross section, however, can change the location away
from the centroid if it is small enough elsewhere.
Keep in mind that t represents the amount of material along
the line drawn through the cross section perpendicular to the
shear force at the point of interest. So a cross section with a
cutout would lower the value of t.
This equation for shear stress is a troubled equation. There are times when it does not give an
accurate stress value. In other words, it has limitations.
Consider, for example, a circular cross section. If we
consider a point a distance y above the neutral axis, we
could calculate the shear stress from
= VQ/(I t)
Which acts parallel to V along the line through the cross
section at y. Now, consider an infinitesimal volume
element at the edge of the cross section. The boundary of
the cylindrical member is called a free surface if a stress is
not applied to it. The free surface has no stress or is free
from stress. This means that on the cross section, there can
be no stress perpendicular to the boundary. The calculated
stress which is in the direction of V and was calculated to
be non zero has a component in the direction perpendicular
to the boundary. So there is a contradiction.
The actual stress will be zero at the boundary and there will be a varaiation in stress along the
line with = VQ/(I t) being the average stress along the line.
The shear stress equation will give the actual stress where the boundary is vertical since the free
surface then is parallel to the shear force and the shear stress from the equation has no
component perpendicular to the free surface.
So, we can find the stress at the centroid of the circular cross section which also turns out to be
the maximum stress.
= VQ/(I t)
Q=Ay=
r2 / 2 [ 4 r / (3
r3 / 3
= V [2 r3 / 3] / [( r4 / 4)(2 r)]
= 4 V / [3
r2 ] = 4V / (3A)
The maximum shear stress for a circular cross section is 33% above the average value.
Another limitation occurs with thin wall open sections like
wide flange beams.
The top and bottom of the flange are free surfaces. This
says that the stress at A is zero. However, we can
calculate a nonzero stress value at A using our equation.
For these types of cross sections, we apply the equation to
the web only which works out fine since the max shear
stress is at the centroid which is in the web.
Special note: When finding Q, we can take the
area above or below the cut. Remember that
Q=A y for area above the cut where we want to
calculate the stress. However, A y = A y = 0 for
the entire cross section since we are measuring y
from the centroid already. Therefore,
A yabove = - A ybelow . Taking the area below the
cut gives the same Q except with the opposite sign.
If we take the absolute value, we get the correct
value. This can work to our advantage when the
area below the cut is simpler than the area above
as shown.
Lecture #20
Beams with Built Up Cross Sections
Reading: 7:4-6
In the last lecture, we found a force per unit length of beam dF/dx. Let‘s call this force per unit
length shear flow, q. So,
q = VQ / I
If we have a built up member, that is a beam that is made of pieces nailed, bolted or glued
together, the connectors must be able to carry the shear along the beams length.
In the case of a glued joint, we can calculate the stress in the glue using
compare to the glue‘s strength.
= VQ/(I t) and
In the case of a nail or bolt, the connectors must be spaced along the length of the beam
appropriately so as to carry the shear flow.
Imagine two pieces nailed together with nails spaced s
apart. Each nail must carry the shear flow over the
length of beam which extends half way to the nail on
either side. Therefore, the nail must carry the shear
flow over the length s. The net force, q s , must be less
than the net force the nail can carry Fn . Since we
could also have multiple nails side be side attaching
two pieces, let‘s include a factor n for the number of
connectors. Now,
n Fn = q s
When finding Q, we must cut through the cross section at the place where the connectors hold
the pieces together. We have one additional concern. One assumption we used when deriving
q = VQ / I was that the cross section have a vertical axis of symmetry. Not only must this be
true here, but the area used to calculate Q must also have a vertical axis of symmetry. When we
have a closed cross section, one with an inside and an outside, we will need to separate the
connected pieces, but also another place so that we get an area with the same axis of symmetry
as the cross section. Our cuts might be horizontal or vertical.
Consider a beam built up as shown here. To find the spacing
of nails 1, we separate the pieces at A, but then we must cut at
B also. Note that there are no nails at B. We calculate Q from
the area separated, then after finding the shear flow, q, we
divide by 2 to get the shear flow over one of the cuts since
only one cut has connectors.
For the nails at 2, we would cut at A and B as shown to
separate an area to get Q and again find q and divide by 2.
The author of the text, in section 5, also discusses the calculation of shear stress in the flanges
of thin walled members. You should look at this, but it is not of tremendous importance to us
in this course. In more advanced courses, the stress in the flange is needed to find the shear
center as discussed in section 6 when the cross section does not have a vertical axis of
symmetry. We will not be covering that topic in this course.
Lecture #21
Thin Wall Pressure Vessels
Reading: 8.1
Another situation that we are able to analyze to determine equations for stress is that of a thin
walled spherical or cylindrical pressure vessel.
Consider a thin wall sphere with a pressure inside.
There will be normal stresses in the wall due to the
internal pressure. To expose the normal stress, we can
cut the sphere in half. Consider passing a horizontal
plane through the center of the sphere.
The internal pressure acting inside the half sphere creates
forces in each coordinate direction. If you consider the
force acting on an infinitesimal area dA, you can see the
components of the force.
Considering the x or y direction, we recognize that there
is another element located 180o around the z axis which
has components in the x and y direction that are equal in
magnitude but opposite in direction. Therefore,
equilibrium is satisfied identically. However, the
components in the z direction are equal and in the same
direction so they add together. The z component of the
force on dA is given by
dFz = pi sin
where dA = (ri cos
d
dA
ri d
The total force in the z direction can be found by integration.
Fz =
pi sin
0
cos
ri d
pi (ri)2
sin
0
0
Fz = pi (ri)2
sin
cos
d d
0
Fz =
(ri d
2 sin
0
cos
sin
cos
2 d
0
0
pi (ri)2
pi (ri)2
cos
0
d
(ri)2 pi sin2
0
=
(ri)2 pi
d d
Considering that the half sphere must be in equilibrium, the net force in the z direction can only
be balanced by the normal stress in the wall s.
Fz = 0 :
s
2
ravg t =
(ri)2 pi
Because the wall is thin, we can make an approximation and replace ravg by ri. Then
s
= (pi ri)/(2 t).
Note that we can cut the sphere in half with a vertical plane and we have the exact same
situation that we have just considered except rotated 90o. Therefore, there must be a normal
stress in the horizontal direction equal to s as well. In fact, we recognize that we can cut the
sphere in half in any orientation and the normal stress is the same.
The spherical pressure vessel then has a state of stress in the wall
which can be represented as
And the element could be oriented in any way on the surface of the
sphere.
Now, lets consider the thin wall cylindrical pressure vessel.
Again, because of the internal pressure, there
will be normal stresses in the wall of the
cylinder. To determine the stresses, lets first
pass a cutting plane through the cylinder
perpendicular to its length.
Again, there are forces created by the
internal pressure on the inside wall.
Considering an element dA on the side of
the cylinder, we see that the force only has
components in the x and y directions.
Since there is an element 180o around the z
axis that has components that are equal in
magnitude and opposite in direction,
equilibrium is identically satisfied in the x
and y directions. In the z direction, there is
a net force from the pressure acting against
the end wall of the cylinder.
Fz =
This force must be balanced by the normal stress,
length. Considering equilibrium
Fz = 0 :
L
2
ravg t =
L
pi (ri)2
, acting in the wall along the cylinder‘s
(ri)2 pi
Making the assumption that ravg can be replaced by ri we have
L
= (pi ri)/(2 t).
In addition to the normal stress acting along the length of the cylinder, there is also a normal
stress that acts around the circumference of the cylinder. This stress is called the hoop stress,
H.
To expose this stress, we need to cut a length
L out of the cylinder and then cut that part of
the cylinder in half by passing a cutting plane
through the cylinder‘s axis.
Since the normal stress L occurs in the wall
on both sides of the piece, equilibrium is
satisfied in the y direction. These stresses are
not shown on the figure for clarity.
Considering a strip of0 area dA, there is a force,
dF, with components in the x and z direction.
The x force component will be balanced by an
equal and opposite component acting on an
element dA on the other side of the piece.
But, the z force components add together.
dFz = pi dA sin
where dA = L ri d
Now we can find the net force in the z direction by integration.
Fz =
pi L ri sin
d = pi L ri
0
sin
d = - pi L ri cos
0
Considering equilibrium in the z direction, the normal stress
Fz = 0:
H
pi L ri
H
H
must balance the force Fz.
L t 2 = pi L ri
= (pi ri)/ t
Therefore, the state of stress for an element on the surface of the cylinder is
If the cylinder is open on the ends, then
and only the hoop stress exists.
L
will be zero
Lecture #22
Combined States of Stress (2D)
Reading: 8:2
We want to be able to find the state of stress for any point in a two dimensional problem. Since
all forces are in a plane, the state of stress will also be two dimensional. The first step in this
type of problem is to use the method of sections to cut the member to isolate the forces acting
on a cross section containing the point of interest. Then we draw a free body diagram of part of
the body and use equilibrium to determine the internal forces acting at the centroid of the cross
section. Finally, we use our stress equations to find the stress that each force creates at the
point of interest and superimpose them on an infinitesimal element representing the point.
We recognize that there are only 3 internal reaction forces in two dimensions—a normal force,
a shear force and a bending moment. Therefore, the stress equations of interest are
=P/A
=My/I
and
= (V Q)/(I t)
In general, the stress that comes from moments are
larger than those that come from forces. When
looking at the maximum normal stress, we want the
cross section with the largest moment. Also, the
maximum normal stress will often occur where the
two normal stresses ( P/A and My/I ) add together
(have the same sign) unless the distances from the
centroid to the extreme fibers in each direction are
different. Then the maximum stress could occur
where the two normal stresses oppose one another if
y is larger for that point.
The stresses act on the cross section
so the state of stress for an arbitrary
point A will consist of at most two
stresses.
If looking for the maximum shear stress, we use the same approach as for finding the maximum
shear stress from the shear force in a beam as that is the only shear stress that occurs.
Lecture #23
Combined States of Stress (3D)
Reading: 8:2
Again, we want to be able to determine the state of stress at a point, but now for a three
dimensional problem. The procedure is basically the same as for the two dimensional problem
in that we must first use the method of sections to isolate a cross section that contains the point
of interest. Then we draw a free body diagram of part of the member. However, now we have
six possible reactions acting at the centroid of the cross section ( a normal force, two shear
forces, a torque and two bending moments). We use equilibrium to determine the reactions
then apply the stress equations to find the stresses at the point of interest.
Our stress equations are
=P/A
=My/I
T r / J and
= (V Q)/(I t)
T / ( 2 Am t )
If we have a pressure vessel, we will also need
H
A
= (pi ri)/ t
and
L
= (pi ri)/(2 t)
Then we draw the stresses on an infinitesimal
element representing the point.
For a general point on the cross section, the state of stress is three dimensional and will include
three stresses as shown below for point A.
We are really most interested in the point
that has the most critical state of stress.
Remember that the largest stresses come
from the moments. Therefore, it makes
sense that the most critical point will be on
the boundary of the member since the
maximum stress from all moments are as far
as possible from the centroid.
If our point of interest is on the boundary of the cross section, the state of stress will be two
dimensional. The most difficult part of drawing the two dimensional state of stress is making
sure your element orientation is understood. You will note that I always look at the element
from the outside of the member and letter the element both on the member and the state of
stress.
When considering a two dimensional state of stress for a point in a three dimensional structure,
if the point of interest is on one of the principal axes of the cross section, you will note that one
of the shear forces and one of the bending moments will give a zero stress at the point. The
force and moment associated with the axis that the point is on will give zero stress and can be
ignored.
The the cross section to the left, the shear stress
from Vx at A is zero because the point is at the
top/bottom of the cross section with respect to the
direction of the force so Q will be zero in
(VQ)/(It).
The normal stress from Mx is zero at A because
the point A is on the centroidal axis that the
moment goes around so the perpendicular
distance y will be zero in (M y)/I.
The state of stress for point A can be seen below
with the equations used to calculate each.
Again, if this structure were a pressure vessel,
then the hoop and longitudinal stresses from the
internal stresses must be calculated and
superimposed over the state of stress from the
externally applied forces. In this case, there will
be normal stresses in both directions on the
element. Otherwise, we only have one normal
stress.
Lecture #24
Plane stress transformation equations
Reading: 9:1-3
When we calculate a state of stress, it is always referenced to some orientation of the point in
the body. This leads to the idea that if the element were oriented differently, the state of stress
would be different.
We would like to be able to determine whether a material might fail under a general two
dimensional state of stress. A large number of failure criteria have been theorized over the
years for different types of materials. Some of these failure theories depend on the maximum
normal stress or the maximum shear stress experienced by the material.
So, given a two dimensional state of stress, we would like to find equivalent states of stress for
the same point in other orientations. In particular, we are interested in determining the
maximum values of normal and shear stresses.
Lets begin with a given state of stress and determine the state of stress after a rotation of the
element by an angle
We will assume the positive directions as given in the figure for normal stress, shear stress and
angle of rotation. To relate the two orientations, we can cut the original element so as to create
a wedge with the cut face pointing in the x‘ direction.
Then, we can draw a picture of the wedge
with appropriate stresses on each face and
use equilibrium equations to relate the
stresses.
To write equilibrium equations, we must
multiply the stresses by the area on which
they act to get forces. Assuming the area of
the cut face is A, the horizontal face of the
wedge will have an area of A sin and the
vertical face will have an area of A cos
Now, considering equilibrium in the x‘
direction
Fx‘ = 0 :
xy Acos ) sin
sin
y Asin
x‘A
( x Acos ) cos
cos
xy Asin
Rearranging and dividing out the common
A, we get
x‘
x cos
2
xy
sin cos
Using the following trig identities, we can rewrite this equation
cos2
x‘
sin2
cos2
(
y)/2
x
+ ((
x
y)/2)
cos2
cos2
xy
sin cos
sin2
sin2
If we consider the y‘ direction,
Fy‘ = 0
x‘y‘
A
(
x Acos
) sin
xy Acos
) cos
Rearranging and dividing out A, we get
xy
Asin
sin
y Asin
cos
y sin
2
=
x‘y‘
(
y)
x
sin cos
xy (cos
2
sin2
Applying our trig identities,
=
x‘y‘
(
y)/2)
x
sin2
To find an equation for
y‘,
xy cos2
we can add 90o to
y‘
(
x
y)/2
+ ((
x
y)/2)
cos(2
y‘
(
x
y)/2
((
x
y)/2)
cos2
in the equation for
xy
xy
sin(2
sin2
We can find the maximum normal stress by taking a derivative of
zero.
d
x‘/d
(
x
tan 2
(
y)/2)
2
x
sin2
xy
/(
y)/2)
2 sin2
xy 2
x‘.
x‘
and setting it equal to
cos2
xy cos2
y)
x
When we take the inverse tangent to determine 2 , we get two solutions that differ by 180o. So
there are two solutions for which differ by 90o. One of these solutions corresponds to a
maximum value of and the other a minimum value. We recognize that we only need one of
the solutions since the second solution corresponds to the other coordinate direction. We use
(1/2) tan-1 [ 2
xy
/(
y)
x
]
to get one solution from our calculator, then, we substitute the angle into the equations for x‘
and y‘. One will correspond to a maximum value and the other will be a minimum value. The
maximum and minimum can also be calculated from
max,min
=(
x
y)/2
± [ (
x
2
y)/2)
+
2 1/2
xy ]
but this equations does not tell us which stress goes with which axis. Therefore, we still need to
substitute the angle into either the equation for x‘ and y‘ to determine which is which.
If you compare d x‘/d with x‘y‘ you will see that that they are the same except for a factor of
two. Since the derivative is set to zero to find , the shear stress must be identically zero for the
orientation that has the maximum and minimum normal stresses.
These normal stresses are called principal stresses and the angle is the orientation of the
principal element.
In order to find the maximum shear stress, take the derivative of
equal to zero.
d
x‘y‘/d
x
x
y)/2)
tan 2
s
y)/2)
2 cos2
x
y)/(2
tan-1[
x
xy
2 cos2
xy
x‘y‘
with respect to
and set
2 sin2
2 sin2
xy)
y)/(2
xy)]
This angle can be substituted back into the equation for x‘y‘ to get the maximum value of shear
stress. There are again two solutions s that are different by 90o. Either angle will give the
same magnitude, but with opposite signs. So one gives a maximum and the other a minimum,
but note that this corresponds to the directions of the shear stresses acting on planes 90o apart.
Also, we note that the argument inside the inverse tangent for finding the maximum normal
stress is the negative inverse of the argument for finding the maximum shear stress. From trig,
we know that this means that the inverse tangent gives angles 2 that are different by 90o. So
the angles s and p will be different by 45o.
When we have maximum shear stress, there are normal stresses. It can be shown that the
normal stresses are
x‘
y‘
=(
x
y)/2
Lecture #25 & #26
Mohr’s Circle
Reading: 9:4
It can be shown that the transformation equations for plane stress are equivalent to the equation
of a circle. This is advantageous in that we are able to do everything that we could do with the
transformation equations without actually knowing any of the equations.
Consider the equations for the stresses on the x‘ face of the rotated element.
(
x‘
x‘y‘
y)/2
x
=
(
x
+ ((
y)/2)
y)/2)
x
sin2
cos2
xy
sin2
xy cos2
Bring the constant term in the first equation to the left hand side.
(
x‘
x‘y‘
y)/2
x
=
(
x
= ((
y)/2)
y)/2)
x
sin2
cos2
xy
sin2
xy cos2
Now square both equations and add them together.
[
(
x‘
x
x
+ [ x‘y‘]2 = [( x
[( x
y)/2] xy sin2 cos2
y)/2] xy sin2 cos2
x
y)/2]
2
cos22
2
2
y)/2] sin 2
y)/2]
2
sin22
2
2
xy] cos 2
xy]
Now, combine terms and apply the trig identity cos22
[
x‘
(
x
y)/2]
2
+[
x‘y‘]
2
= [(
x
y)/2]
2
xy]
2
sin22
2
The general form of the equation for a circle is
(x - a)2 + (y - b)2 = r2
where the center of the circle is at (x,y) = (a,b) and the radius of the circle is r.
We can see that if we let the x axis represent
circle with its center at [(
x
y)/2,
and the y axis represent , the equation is a
0] and a radius of { [(
x
y)/2]
2
xy]
2
}1/2.
From this picture, we see that the principal
normal stresses occur at the farthest right
and left points of the circle.
max
=C+R
min
=C-R
We also see that these points are on the
axis, that is, the shear stress is zero at both
points.
Also, the maximum shear stress occurs at the top and bottom of the circle. The maximum shear
stress is equal to the radius, max R. We also note that at the top and bottom of the circle, there
is a normal stress equal to the coordinate of the center of the circle.
Lets set a procedure for constructing and using Mohr‘s circle using an example. Consider a
point which has the state of stress shown.
1. First, we set up our coordinate system with on the x
axis and on the y axis. Our convention is to measure
normal stress positive if tensile and negative if
compressive. For shear stress, we use a positive sign if
the stress rotates the element counterclockwise and
negative if it rotates the element clockwise. For
reasons to be explained later, we need to plot positive
down and negative up. So as not to get confused, I like
to label the axes using rotation direction with up being
CW and down being CCW.
2. Next, we will plot two known
points. The x face of the given
element has stresses of
ksi
and
3 ksi CCW
So we plot a point X = (16, 3 CCW)
The y face of the element has
stresses of
ksi and
3 ksi CW
So we plot a point Y = (8, 3 CW).
3. We recognize that these to points are 90o apart on the element, but on the circle, they will be
180o apart. This is due to the fact that the equations used to obtain the circle contained the sine
and cosine of 2 not
So if X and Y are 180o apart on the circle, they must be at opposite ends
of a diameter. Draw a line from X to Y. This line represents the diameter of the circle and the
center of the circle will be where the line intersects the axis. Note that this occurs exactly
half way between the points X and Y in the direction. So our center is at
= (16 + 8)/2 = 12 ksi.
4. Using the center point and the point X, we can construct a right triangle with sides parallel to
the and axes. The sides of the triangle can easily be determined. The horizontal side is
X - C = 16 - 12 = 4 ksi and the vertical side is 3 ksi. We can now calculate the radius from the
Pythagorean theorem. R = { 32 + 42 }1/2 = 5 ksi. Now, the circle can be drawn.
In order to use the circle, we need only use some geometry and trig.
What if we want to draw the element that experiences the principal normal stresses? Go back to
the right triangle that used the center and point X.
On the circle, we want to rotate from the point X
to the nearest principal stress which is max in this
case. The angle of rotation on the circle can easily
be found.
tan 2 = 3/4
2 = 36.9o
Note that the angle the element rotates is half of this angle = 18.45o. The direction that we
rotated around the circle to go from X to max was CCW. The rotation of the element will be in
the same direction, CCW. This is why when we set up the coordinate system, we put positive
down. Had we put positive upward, the angle of rotation of the element and around the circle
would be in opposite directions.
When we draw the stresses on the element, the stress in the new x direction is the stress that we
rotated to from the point X. This is max in our example. So the new y direction is 180o away,
or min.
= C + R = 12 + 5 = 17 ksi
Min = C - R = 12 - 5 = 7 ksi
max
Remember that there is no shear stress on
the element when we have principal normal
stress.
What if we want to draw the element that experiences the maximum shear stress? Again, go to
the triangle that used the center and point X.
Now, we want to rotate on the circle from X to
the nearest point of max shear stress, that is, the
top or bottom point of the circle. In our example,
we rotate CW to the bottom point. Since we
have already found the angle to the nearest
principal normal stress, we can easily find our
new angle by subtracting the previous angle from
90o.
2
90 - 36.9 = 53.1o
So the element must be rotated half this angle,
= 26.55o in the CW direction.
For this new orientation, we know that = C =
12 ksi and is the same on all faces of the element.
To get the direction for the shear stress, we use
the point we rotated to from point X. This point
gives the shear stress on the face pointing in the
new x direction. Our point is the bottom of the
circle in the example. We remember that a point
below the axis must rotate the element CCW.
So we can now draw the complete element.
We can also use Mohr‘s circle to perform any arbitrary rotation of the original element.
Consider that we want to rotate our original element 30o CCW. This would be a rotation of 60o
CCW around the circle.
Assuming that we have already
used the triangle to find the
angle from point X to the
horizontal and vertical axes, we
can rotate CCW 60o from X to
find our new X. We also know
that the new Y is at the other
end of the diameter, 180o away.
The angle above the horizontal axis can be found. 2 = 60 - 36.9 = 23.1o. This angle is used to
construct a new triangle using the center of the circle and the new X point.
Using trig, we can find the horizontal and vertical
sides of the triangle.
H = 5 cos 23.1 = 4.60 ksi
V = 5 sin 23.1 = 1.96 ksi
The new coordinates of the new X are (C+H, V) = (16.60, 1.96 CW)
The new Y point will have coordinates of (C-H, -V) = (7.40, 1.96 CCW)
We can now draw the element.
Using Mohr‘s circle, you can see, we can perform all of the same operations as with the
transformation equations for plane stress derived earlier, but we can do so without knowing the
equations. We simply use some geometry and trig.
Lesson #27
Mohr’s Circle for 3D States of Stress
Reading: 9:7
When we have a general 3D state of stress, Mohr‘s circle becomes three connected circles
instead of just one. If we know the three principal normal stresses for our state of stress, we can
draw a circle for each set of two principal stresses.
Assume principal normal stresses
. We have a Mohr‘s circle for the 1-2 plane
using stresses and
a Mohr‘s circle for the 2-3 plane using stresses and
and a
Mohr‘s circle for the 1-3 plane using stresses and
Note that two of the circles
are contained within the third.
In this course, our states of
stress are two dimensional,
meaning that the stresses on
the third directional face are
zero. Since the shear stresses
are zero, the normal stress of
zero is our third principal
stress.
There are also situations where our state of stress is 3D, but on there is only a shear stress in one
plane. Therefore, on the element face perpendicular to the plane of shear stress there is only a
normal stress. Since there are no shear stresses on this face, the normal stress is a principal
stress.
In problems of this type, we can neglect the third direction and draw Mohr‘s circle for the plane
that contains a shear stress just as we did previously. Then we can plot the known third
principal stress on the axis and draw two more circles using the third value and each of the
two principal normal stresses determined from the two dimensional circle.
The usefulness of these additional circles is that there is an absolute maximum shear stress in
3D which is associated with the radius of the largest circle. We will not try to draw a picture of
the element as this can be quite confusing, but the absolute maximum shear stress is indeed
useful when we want to compare to the shear strength of a material. The absolute maximum
shear stress may or may not be the same aas the in-plane maximum shear stress.
Lesson #28
Beam Design
Reading: 11:1-2
When a beam is to be designed to carry a particular load, there are two stresses to consider - the
maximum values of normal stress (tensile & compressive) and the maximum shear stress. We
have already discussed the fact that in a vast majority of beam problems, the normal stress from
bending will be much larger that the shear stress. Therefore, our design must begin with the
normal stress.
To find the maximum normal stress, we must be able to draw the shear and moment diagrams
to find Mmax (positive and negative) and find the centroid and moment of inertia of the cross
section.
Since the cross section size has not been determined yet, we define a new quantity called the
section modulus.
S = I / ymax
Now, we can write the stress as
max
Mmax / S
If the material is known so that we have an allowable stress that can be experienced, we can set
max
allow and solve for a required minimum section modulus.
Sreq = Mmax /
allow
This is not enough information to determine what the cross section dimensions are unless we
are given information about the shape of the cross section.
For example, if the cross section is to be circular, we know
S = I / ymax = ( r4 / 4)/ r =
r3 / 4
If the cross section is to be square, we can say
S = I / ymax = (b4/12)/ (b/2) = b3 / 6
If the cross section is to be rectangular, we know
S = I / ymax = (b h3/12)/ (h/2) = b h2 / 6
In this case, we still are not able to determine the dimensions unless additional information is
given. In the case of a rectangle, we might be told that the base of the height has a specific
value, or perhaps that the height is some multiple of the base. Then we can complete the
design.
Recognize that these determined dimensions are minimums, that is, something larger is okay
but not smaller. So if we are asked to round to a particular fractional size, we always round up.
If we want to design the beam using a standard manufactured section such as a Wide Flange
cross section, then we select the appropriate size from a table of standard sizes. Appendix B of
the textbook contains sample tables. You will note that one column of the tables contains the
section modulus.
When choosing the appropriate cross section, we will note that several sections may have S
larger than our Sreq. First of all, we should only consider the section modulus for the x axis of
the cross section which is larger and assumes the cross section is oriented as an I as opposed to
the H orientation. Next, we want to choose the most economical section. Since steel is
typically sold by weight, the best section will be the lightest one with S > Sreq.
Once we have determined the cross section dimensions, we must make sure that the shear stress
is acceptable. Using the selected cross section, we can calculate max and compare to allow for
the material. Most of the time, it will be okay. If not, we will need to choose a larger section
based on shear.
Generally, we also ignore the beam‘s weight when designing a beam. The reason for this is that
the beam‘s weight is normally quite small compared to the applied loads. If we want to include
the weight in the design, we have a problem that is iterative. First, we design the beam ignoring
the weight, then we can calculate the weight and include as a distributed load. Next, we
redesign the beam including the weight, but this can change the weight meaning that we have to
repeat the process.
This iterative process can be cut short by choosing a cross section that is slightly larger than
required or by rounding up. Another approach is to increase the maximum moment by 5-10%
at the beginning of the process to allow for the weight‘s contribution.
Lesson #29
Beam Deflections
Reading: 12:1-2
Deflections in beams can come from bending and shear effects, but the effect of shear forces in
small compared to that of bending and can be ignored.
Earlier in the semester, we looked at a beam in pure
bending and derived an equation for strain in the cross
section of the beam.
dx =
d
or
Du = -y d
and
d /dx = 1/
du/dx = -y d /dx = -y/
Assuming a linear elastic material so that
considering equilibrium, we derived
E and
-M y / I
Applying these equations, we find
-y/
E = -M y / (E I)
So now,
1/
M / (E I)
The curvature, one over the radius of curvature, can be found from calculus as
1/
( d2v/dx2) / [ 1 + (dv/dx)2]3/2
where v is the vertical deflection in the beam. If we make a further assumption that
displacements are small, we can say
v << 1
and
dv/dx << 1
Therefore, the denominator is essentially 1 and
1/
d2v/dx2
So now we can write
d2v/dx2 = M / (E I)
If we allow the use of ‗ to represent d/dx, then
E I v‘‘ = M
We also know from our work with shear and bending moment diagrams that
dM/dx = V
and
dV/dx = w
So we can further say that
d/dx (E I v‘‘) = V
and
d2/dx2 (E I v‘‘) = w
If we assume that E and I are constant over the length of the beam, then
E I v‘‘‘ = V
and
E I v‘‘‘‘ = w
So we have a choice between using a second, third or fourth order differential equation to
determine the deflection in a beam. Generally, we choose to use either the second or fourth
order equation. These equations require that we be able to express either the distributed load,
w, or the bending moment, M, in the beam as a function of x. We know from our work with
shear and bending moment diagrams that for an arbitrary problem, neither w nor M can be
expressed as a single function over the entire length. Rather, the beam has sections of its length
where M or w is written as a function of x. Let‘s ignore this fact for now and assume that we
have a statically determinant beam composed of one section and w and M can be expressed as a
function of x.
If we use the second order differential equation, we must first draw a free body diagram of the
beam and determine the reactions using equilibrium. Then, we use the method of sections to
cut the beam at an arbitrary distance x from the left hand end, draw a free body diagram of one
part of the beam and use equilibrium to solve for M. At this point, we substitute the function
for M into the differential equation and integrate twice to find a function for deflection.
E I v‘‘ = M
E I v‘ = ∫ M dx + A
E I v = ∫ ∫ M dx dx + A x + B
Because our solution contains two constants of integration, we need to use boundary conditions
and algebra to determine the constants and complete the solution. Boundary conditions come
from the support conditions at the two ends of the beam and must be of a lower order than the
differential equation. Since we are using a second order equation, the boundary conditions
must be on either v or v‘, that is, the deflection or slope. The typical types of boundary
conditions are shown below.
We need two boundary conditions.
Depending on the support conditions, we
might get two conditions from one end or
one condition from each end, but applying
the boundary conditions allows us to solve
for the constants of integration. Then we can
write a final function for the deflection in the
beam.
If we use the fourth order equation, we save some work on the front end of the problem because
we do not need to determine the reaction forces or the bending moment as a function of x.
Instead, we write the distributed load as a function of x. Then we substitute that function into
our differential equation and integrate.
E I v‘‘‘‘ = w
E I v‘‘‘ = ∫ w dx + A
E I v‘‘ = ∫ ∫ w dx dx + A x + B
E I v‘ =
∫ ∫ ∫ w dx dx dx + A x2 / 2 + B x + C
E I v = ∫ ∫ ∫ ∫ w dx dx dx dx + A x3 / 6 + B x2 / 2 + C x + D
So now, we see the tradeoff. We didn‘t have the front end work, but at the back end of the
problem we have to apply boundary conditions to solve for four constants of integration.
Because the starting equation was fourth order, our boundary conditions can be on deflection, v,
slope, v‘, bending moment, M = E I v‘‘, or shear force, V = E I v‘‘‘. The typical types of
boundary conditions are shown on the next page.
When using the fourth order differential equation, we always get two boundary conditions from
each end.
Both the second order and the fourth order equations can be used to find deflections in a
statically indeterminant beam as well, but when using the second order equation you can not
determine the reactions from equilibrium so the bending moment function must be found in
terms of the redundant reactions and extra boundary conditions are used to solve for the
unknown reactions as if they were constants of integration.
Lesson #30
Deflections in Beams
Reading: 12:1-2
If the bending moment or distributed load cannot be written as a single function over the entire
length of the beam, the beam must be broken into sections in the same way we would for a
shear and moment diagram. Then a differential equation could be written for each section. If
we use the second order equation, we have two constants of integration for each section. If the
fourth order equation is used, there are four constants of integration for each section. Now, the
boundary conditions alone are not sufficient to determine all the constants. We must use
matching conditions.
Matching conditions are conditions based on beam continuity and equilibrium at the point
where two sections meet. Using the second order equation, our matching conditions are v1 = v2
and v1‘ = v2’. These are called continuity conditions because the beam would be broken if the
deflection and slope from the two sections did not match at the meeting point. If there is a pin
or roller at the matching point, we get three matching conditions instead of two: v1 = 0 , v2 = 0
and v1‘ = v2’. Using the fourth order equation, we also consider equilibrium of a small piece of
beam at the matching point. Assuming that there might be a force and moment at the point,
Fy = 0:
V1 = V2 + Po
or
E I v1‘‘‘ = E I v2‘‘‘ + Po
M = 0:
M1 + Mo = M2
or
E I v1‘‘ + Mo = E I v2‘‘
If there is a roller or pin support at the matching point, we would have a reaction force on the
free body diagram. Since the reaction is not known, we can ignore the force equation and just
use the moment equation. We will still have sufficient conditions because of the extra
continuity condition. There will always be four matching conditions at each matching point
when using the fourth order differential equation.
When a beam has multiple sections, the number of constants of integration and the number of
boundary and matching conditions makes for a large system of equations that needs to be
solved algebraically. This can take considerable time and is not particularly instructive.
Therefore, when we have a beam with multiple sections, we will just set up the problem, that is,
write the differential equation for each section and then list the necessary boundary and
matching conditions. We will assume the problem can be solved given sufficient time.
Lesson #31
Deflection using the Method of Superposition
Reading: 12:5
When the deflections are small, we saw that they could be found from a linear differential
equation. The fact that the problem is a linear one allows us to apply the method of
superposition. That is, if we have a beam with a complex loading, we can break the problem
down and consider the beam with each load applied singly to the beam. Then the deflection for
the complex loading is found by adding together the simpler cases. For Example,
Over the years, the deflection problem for these beams with single loads have been solved and
tabulated so that we have tables, such as those found in Appendix C of the text, available for
use.
Appendix C is not exhaustive so for some problems, we may need to use some ingenuity and
imagination to get a solution. Also, when using the equations to get a deflection value at a
particular point in a beam, we must be very careful with units, signs and how x is measured.
Lesson #32 & #33
Statically Indeterminant Beams
Reading: 12:9
We would like to be able to solve for the reaction forces acting on a statically indeterminant
beam. There are a couple of ways to do that. We will use what is called the force method
which uses superposition and the deflection tables to accomplish the task.
The process begins by drawing a free body diagram of the beam and writing equilibrium
equations. The number of unknowns above the number of equations of equilibrium are called
redundant supports or redundant reactions. We need to write constraint equations (deflection
equations) equal in number to the number of redundant forces in order to solve for all the
reactions.
To accomplish this (and to be able to use the deflection tables), we must eliminate redundant
supports in order to create a beam that looks like one of the beams in our tables. So we must
eliminate constraints until we have a beam that is simply supported or cantilevered.
After eliminating the supports, we place the reaction forces coming from these supports on the
beam as unknown applied forces.
Next, we get our constraint equations by reimposing the released constraints or supports.
Consider the example below.
Fy = 0: A + B = 0
MA = 0: MA + BL - Mo = 0
We have 3 reaction forces, but only
two equations of equilibrium. We can
turn this beam into a cantilever if we
eliminate the roller support, but we
must include the reaction B as an
applied force.
Now, we reimpose the constraint from the roller support, that is, vB = 0. Using the tables and
superposition.
vB = vB1 + vB2 = 0 = - ( Mo L2 ) / ( 2 E I ) + ( B L3 ) / ( 3 E I )
This equation allows us to solve for B which we take back to the equilibrium equations to find
the remaining reactions.
Lesson #34
Column Buckling
Reading: 13: 1-3
When a beam or column experiences a compressive load, it can become unstable and buckle or
bow, resulting in large displacements. This could result in the member collapsing or at least
failing to perform its intended function. We would like to be able to predict these types of
failures.
Let‘s consider a simply supported beam with a compressive load and look at the deflection
problem.
As P is increased, we will eventually
reach the point of instability. At this
point, with any disturbance or
imperfection, large displacements
could result.
These displacements result in bending
moments in the beam.
Make a cut in the beam at an arbitrary
location x and draw a free body
diagram of one part of the beam.
M = 0: M + P v = 0
M=-Pv
Now apply the second order differential equation.
E I v‘‘ = M = - P v
This can be rewritten as
E I v‘‘ + P v = 0
or by dividing out the E I, we get
v‘‘ + [ P / (E I)] v = 0
So the buckling problem is governed by a second order differential equation. The solution to
this equation can be written in the following form.
v = A cos( k x ) + B sin( k x )
where k2 = P / (E I)
We then apply our boundary conditions to try to find the constants of integration.
v(x = 0) = 0 = A (1) + B (0) = A
v(x = L) = 0 = B sin( k L)
This second equation can be solved in two ways. Either B = 0 or sin( k L ) = 0. If B = 0, the
entire solution is zero and there is no buckling. This is called the trivial solution. If sin( k L ) =
0, then ( k L ) = n where n is an integer. We want to find the first value of P which creates
this situation so we can set n = 1. Solving for P, we find
Pcr = (
2
E I ) / L2
This is called Euler‘s buckling formula. Note that this equation assumes that the material is
linear and elastic.
If we have a beam with different
boundary conditions, the critical load will
be different.
Let‘s consider the case of a cantilever.
M = 0:
-M + P ( v(L) - v ) = 0
M = P v(L) - P v
Now, applying the second order differential equation
E I v‘‘ = M = P v(L) - P v
E I v‘‘ + P v = P v(L)
v‘‘ + [ P / (E I) ] v = [ P / (E I) ] v(L)
This is a nonhomogeneous differential equation. The solution is composed of the homogeneous
part which is obtained by setting the right hand side to zero (we know this solution from the
simply supported case) plus a particular part. We can use v(L) for the particular part of the
solution since when it is plugged into the differential equation it gives us the right hand side.
So now,
v = A cos( k x ) + B sin( k x ) + v(L)
There are actually three unknown constants here since v(L) is unknown. Applying boundary
conditions
v(x = 0) = 0 = A (1) + B (0) + v(L)
A = - v(L)
v‘(x) = -A k sin( k x ) + B k cos( k x )
v‘(x = 0) = 0 = -A k (0) + B k (1)
B=0
Now, v(x) = v(L) [ 1 - cos( k x ) ] and if we use as our third condition that v(x = L) = v(L), we
get
v(x = L) = v(L) = v(L) [ 1 - cos( k L )]
or
cos( k L ) = 0.
This occurs when
kL=n
/2
where n is an odd integer.
Again, setting n = 1, we find the first critical load to be
Pcr =
2
E I / ( 4 L2 )
We can write one equation to cover all types of boundary conditions by writing
Pcr =
2
E I / ( K L )2
where K is a factor dependent on boundary conditions.
Note that K corresponds to the
part of the beams length that takes
the half sine wave shape taken by
the simply supported beam.
The buckling formula is limited in application to long slender beams.
Consider the stress. Up to Pcr , the beam is stable and the displacements are small so
cr
=
2
= P / A.
E I / [ ( K L )2 A]
Now substitute I = A r2 where r is the radius of gyration.
cr
=
2
E A r2 / [ ( K L )2 A] =
2
E / ( K L / r )2
( K L / r ) is called the slenderness ratio. If we plot
figure below.
cr
versus the slenderness ratio we get the
The bucking formula is only good for
a linear, elastic material so it is not
valid above the yield strength, y.
By setting cr = y we can solve for
the slenderness ratio. Columns with
slenderness ratio below this value are
called short columns. They fail by
yielding. Experimental data shows
that the buckling formula gives good
results for cr ≤ y / 2. Columns with
slenderness ratios above the value
associated with cr = y / 2 are called
long columns and they fail by
buckling.
Beams that have critical stresses between y / 2 and y are called intermediate columns and they
fail by a combination of buckling and yield. There are a number of methods used to predict
failure in this range which we will not cover. For our work, we will assume that all columns are
long columns.