IF YOU HAVE
LINKS STARTING WITH-
"The most valuable of
is that of never using two words where one will do"-Thomas
I am revising and updating our earlier Statics
and presenting the results below for our Class EGM
2511. You can use it to gain further
into the basic concepts of statics. Also it contains extra
of use to you such as what topics are to be discussed in
lectures. The book we are using is the
edition of ENGINEERING
R.C.Hibbeler. It is easy to read and understand and
errors. The knight on horseback points to the next lecture. If
have questions please contact me at firstname.lastname@example.org
Do you recognize the
above? They are of Galileo Galilei (1564-1642) and Sir Isaac
the two most important early contributors to classical
mechanics , and
the statue of the Thinker by Rodin(about 1910)
Newton's three laws of motion, plus the universal
law, SI and US systems of units.
Properties of vectors, Forces as vectors,Vector addition ,
and multiplication, Position vectors.
AND LENGTH SCALES: A few helpful conversions are: 1
1 km=0.6215mile, 1 lb=4.448N, 1 slug=14.593kg, 1 ft-lb=1.3558
hp= 550 ft-lb/sec=745.70 watts(or joules/sec), 1 mile/hr=
and 1 lb/in^2=6894.76 pascal. While we're at it-
One Light Year=186,000
x 3600 sec/hr x (24 x 365) hr/yr=5.87x10^12
The diameter of our
Way) is about 100,000 light years , the sun is 500 light
the earth, the next nearest star (Alpha Centauri) is 4.3
and we can see out a distance of between 9 and 18 billion
the best optical and radio telescopes. Radiation coming from
distance started its journey 18 billion years ago at the
universe( big bang theory). If you want some really small
1 micron=10^-6 meter for the size of a typical microchip
angstrom=0.529x10^-10 meter for the radius of the first Bohr
hydrogen atom , and 2.4x10^-15 meter for the diameter of a
web has available many convenient sites for converting
between SI and
CLIMB ANGLES AT THE
demonstrate the use of vectors in carrying out certain
we were asking in class what are the angles associated with
great pyramid of Cheops at Giza in Egypt. One can look up on
that the base of this pyramid has sides of length b=756 ft .
of this pyramid are in the form of equilateral triangles.
of a cartesian coordinate system at the center of the base,
it is easy
to show that the coordinate of the pyramid vertex is at
Thus the position vector of length b running along one of
its edges is
R=-bi/2-bj/2+bk/sqrt(2). Taking a dot product of this vector
cos(theta)=1/sqrt(2), or a climb angle of 45 deg. Note that
starting from the middle of one of the sides is much steeper
54.73 deg. , again shown by dot product manipulations. If
you ever get
a chance to go over there sometime you will be truly amazed
marvel of the ancient world. They don't let you climb the
anymore but you can go inside those of Cheops and Cephren
doing back in 1983) .
on Forces, Equlibrium of a particle in 2D , Concept of unit
a given direction.
Equilibrium Problems. Working out in detail several equilibrium
from the book.
of a Couple. Working out several more problems in static
WHAT ARE MOMENTS AND
moment M is defined as the cross product between a position
the point about which the moment is taken and a force F,
that is M=rxF.
The simplest way to evaluate this product is via a three by
in which the first row elements are the base vectors i,j and
row consists of the position vector components and the third
force components. Thus r=2i+5j-3k and F=20i-40j+30k yields
The dimensions of M are length-force and so given in ft-lb
in the US
of units and in met-newton in SI. A couple C consists of two
opposite forces F and -F separated by a position vector r.
value is simply rxF and will retain this same value no
point the moments for the two forces are taken. The couple
F=10j acting at (3,0,0) and -F=-10j acting at (-3,0,0)is
of the combined couple, moment and force problem acting on a
from the book worked out in detail.
BASIC LAWS OF STATICS: We
have now reached the point in the course where we can state
of Static Equilibrium and the rest of the course will be their
to all sorts of problems, which, as you will find, is not
The Laws are-
(1)-THE SUM OF ALL FORCES
ON A BODY IN STATIC EQUILIBRIUM MUST BE ZERO.
(2)-THE SUM OF ALL
COUPLES) ACTING ABOUT ANY POINT ON THE BODY MUST BE ZERO.
of a moment about a given line. Examples worked out in
MOMENT ABOUT A SPECIFIED
IN SPACE:-In solving some statics
problems one can often determine an unknown force(such as a
a rope attached to the body) by taking moments about a line
two points on the body at which there exist other unknown
forces . By
doing you avoid having to solve for these other forces. To
procedure requires knowledge of the moment about such a line.
how this is done. Consider a cube extending over the range
. A force F1=10i+20j-30k acts at corner(x=1,y=0,z=1)and a
F2=-10i+40k acts at corner (1,1,1). The moment produced by
about the origin at (0,0,0) is
Now lets ask what is the projection of this moment on a line
the origin (0,0,0) and (1,1,0). We have this line defined by
u=(i+j)/sqrt(2) so that dotting it with M yields 10/sqrt(2).
about the line becomes 5i+5j. Run through this calculation
a sketch and also see if you can get this result another way.
we took the original moment about a point on the line chosen.
be able to obtain a moment projection along a given line if
moment is taken about a point not on the line?
of Force and Couple Systems. Concurrent, Coplanar, and
Systems. The Wrench. Location of Resultant Forces.
EQUIVALENT FORCE SYSTEMS:
In many statics problem involving the effect of several forces
a body, it is often convenient to replace these with a single
force plus a couple, before proceding further in the
you how one can establish such an equivalent system, look at the
Body Diagrams . Forces and Moments acting at different Supports
Rollers, Rockers, Hinges, plus Ball and Socket Joints.
HOW TO HANDLE DISTRIBUTED
When dealing with continous force distributions on extended
convenient to replace these with a single resultant force R
acting at a
given point P. To see how one determines the value of the
and its location, we consider the case of a weighless
length x=L which is loaded by a distributed downward linear
w(x)=a+bx (expressed in force per length) and supported
at A and B. In this case the moment about the z axis produced
force about the left pivot point at A has magnitude
In static equilibrium this plus LBy, where By is the vertical
the beam at the right support, must be zero. Hence the
force R=Int[w(x)dx] acts at a distance x*=LBy/R. This turns
out to be
x*=L[(a/2+bL/3)/(a+bL/2)] for this linear force distribution
/3)L when a=0. Click HEREto
see a picture of the configuration.
out more Statics problems in 2D and 3D for all sorts of
forget about your First Statics Exam coming
up next week.
more statics problems in 2D and 3D from Chapter 5.
Some of you have asked me about posing an old
is one from 1997 which you can use to practice and get up to
STACKING OF BRICKS TO FORM
STATICALLY STABLE ARCH: Every
once in a
one discovers certain facinating problems even in topics as
as statics. One of these I ran across this weekend is an
the problem of the furthest distance three bricks
stacked on top
of each other can have their centers of gravity moved to the
causing the pile to tumble. The three brick problem is easy to
It is clear from a moment balance that the second brick can be
half brick length to the right of the top brick without
two bricks in the top two rows have their center of gravity at
1/4 of a
brick length so that the bottom brick can be placed at 3/4 of
to the right of the top brick. Now comes the interesting
n rows containing one brick each. Again starting with the top
next to the top row is placed at 1/2 to the right for a unit
The one after that is at 3/4. Continuing on, we find the brick
row from the top is at x[n+1]=x[n]+1/(2n)
with x=0, x=1/2, x=3/4, x=11/12, x=25/24, etc.
canned program MAPLE to evaluate this for a 20 row stack, we
result looks very much like half of a parabolic arch and becomes
arch when we also allow the same type of stacking on the other
that you wouldn't need any binder to build such an arch.
stacking by early man is how he first came up with the idea of
It is interesting to note that the ancient Greeks did not
arch in their architecture* and
not until Roman times that the circular arch was invented and
and building construction. The arch reached its high point
ages with the discovery of the gothic arch. This last structure,
in conjunction with a flying buttress, is extremely stable since
loads lie strictly along the arch and accounts for the fact that
the European cathederals have remained intact for nearly a
*-Several years ago
visited an ancient tomb in Mycenae in Greece built some 3000
The tomb has an entrance arch formed by stacking up stone
the overlap the same for all rows. In view of the problem solved
you can see why this was not the best solution and not really a
FOR THE FIRST EXAM. The exam will be closed book and
chapters 1 through 5. You can answer any three of the four
Analysis(beginning of Chapter 6). Simple Trusses.
and Tension in Members. Method of Joints.
ANALYSIS OF A TRUSS: We
have shown in class that the concatenation of triangular
by three members connected at their ends produces a stable
is the basic building block for most structures be they roof
bridges , furniture, etc. In the analysis of such structures
solves for all the external forces acting on the truss and
the internal forces by either the method of joints or method
We give HERE
a very simple example of such an analysis. The important thing
is that the internal forces produce a compression
a member when the calculated force points toward the pin at
When your calulations show the force going away from the pin
is in tension.(just
the opposite of what a cursory glance would suggest)
on Trusses. The Crane Problem. Method of Sections.
discussion on trusses in 2D by application of both the Method of
and the Method of Sections.
APPLICATION OF THE
SECTIONS: When one is
interested in just
of the internal forces of a member of a truss, especially
has many members, it is convenient to use the method
of sections. This involves
putting a cut
the truss including the member in whose internal force we
Next treating the unknown forces in the cut members as
one can rapidly find the value of the force in the member
can do so much faster than when using the method of joints.
this solution method for one particular truss HERE.
you ever wondered how many members M and joints J a truss
composed of N
triangles has? You can arrive at a relation between these
by slowly working up from N=1. Thus when N=2 you have M=5 and
for N=3 you find M=7 and J=5. Continuing on, you soon realize
M=2N+1. Thus a truss with 80
will have 161 members and 82 joints.
Lecture #15: Trusses
in 3D(Space Trusses). Introduction to Frames.
THE "A" FRAME: A
example of a frame stuctrure is the A frame used in a variety of
including tables and vacation homes. it consists of two two side
one cross bar anchored together by pins at the three joints. You
a problem of this type by looking at the three members
determining the external reactions on the frame. If you go HERE
you will see the analysis when the members are weightless and a
W is hung from the center of the cross-bar. Note how one always
the forces at a joint in going from one member to the other. If
assume the wrong direction of the force at a joint, this will
at the end of the calculations by giving you a minus sign.
More on Frames and Machines. The Crank-Piston Problem. Analysis
ANALYSIS OF A PULLEY:
We all learned back in high school physics that there are six
namely, (1)the lever, (2)the screw, (3)the wheel and axle,
(5)the inclined plane, and (6)the pulley. Of these , the pulley
the one hardest to understand. In view of our discussion on
is actually quite easy to grasp by analyzing each wheel
as shown HERE.
At wheel A we see, via a vertical force balance,(assuming
of the wheel) that the force in the rope going about the wheel
to F while the rope holding wheel A has a tension of 2F. going
B we see that the rope hung from the wheel axis must be 2F by a
Continnuing this analysis on through wheel D, one sees that it
a weight of 8F. By using n wheels in such a pulley the
gained will be 2^(n-1). That is, one could use such a pulley to
640lb car engine with a force of just F=40lb, when the pulley
of discussion on Frames and Machines, plus Introduction to
Drawing Shear and Bending
Diagrams: The basic formula for determining shear in a
dV/dx=-w(x), where w(x) is the x dependent downward loading and
shear. The shear is thus a constant where the loading is zero
vary linearly with x for constant loading. The bending moment
dM/dx=V(x), with M vanishing at the end supports(except at those
can sustain a moment such as a cantilever). Note that if V is a
of order n then the bending moment M will be one of order n+1 .
in V at a point load is just equal to the value of this load. By
you can see a typical example of a shear(orange) and bending
diagram for a cantilever beam subjected to a linearly varying
out Shear and Bending Moment Diagrams for a variety of
No classes next week due to
spring break. Enjoy the vacation!
on Shear and Bending Moment Diagrams. Also Shear and Normal
N,V, AND M FOR A CURVED
far we have looked at normal (N) and shear(V) forces
moments(M) in only 1D beams for which the simple formulas
dM/dx=V(x) apply. Let us next extend the discussion to curved
the case of a weightless semicircular beam of radius r=a
a smooth floor by a single force Fo applied at its
The analysis proceeds as usual by first determining the
These are simply Ay=By=Fo/2.
N, V, and M at some point C along the curved beam, we apply to
segment B-C the conditions that the summation of forces in the
x and y
direction are zero and that the sum of the moments taken
also be zero . A simple calculation then yields V=-(Fo/2)sin(q),
N=(Fo/2)cos(q) and M=(aFo/2)[1-cos(q)],
in the range 0<q<p/2. For p/2<q<p
have a change in sign for V and the bending moment becomes M=(aFo/2)[1+cos(q)].
Note the maximum bending moment occurs at the top of the
and vanishes where the semicircle touches the floor.
Lecture #20: Completion
of stuff on Shear and Bending Moments. How to handle a hook
a discussion on Cables and the Shape of a Suspension Bridge
Shape of a Suspension
Cable: In discussing cables during the previous lecture
the question what would be the shape of a cable of if it is
a uniform x dependent loading w(x)=Const. This is the well
suspension bridge problem which predicts, as shown HERE,
that the shape of the cable is that of a parabola
y=const x2. Note that a cable hanging under
has a different shape , namely, that of a catenary curve
Lecture #21: Introduction
to Coulomb Friction. Angle of Repose and Problem of
when resting on a rough floor.
on Dry Friction including discussion of the Wedge Problem.
Lecture #23: Belt
Friction, Tipping and Sliding.
THE WEDGE PROBLEM:
of the problems which can be well explained by using Coulomb's
law f=mN is that of the wedge.
with wedge angle a used to hold up
W as shown HERE.
The external forces acting are indicated and we are
interested in determining the ratio of F/W as a function of m
, where the coefficient of friction at
the wall is zero but has the same value of m
both wedge surfaces. Looking at the system as two bodies with
forces f1 and f2 parallel to the two wedge surfaces, one finds
of six equations governing equilibrium which can be solved to
B, f1, f2, and alpha for a given weight W and friction
We have neglected the wedge weight w in our calulations since
of discussion on dry friction by working some problems from the
When you wrap a rope about the circumference of a cylinder,
finds that there is a difference in the rope tension at the
where the rope leaves the cylinder. This tension difference is
effects and can become quite large when wrapping the
times around the cyliner. The equation for the tension change
where m is the coefficient of
the rope and the cylinder surface and q
total angle the rope is wrapped around the cylinder. Since the
term is greater than unity, it is clear that T>To. This
means that the friction force f will always be opposite
of impending motion. Go HERE
to see the derivation of the belt friction equation. Such belt
is employeed by capstans used to tie up large boats at harbour
in devices for slowly lowering heavy weights.
HOUR EXAM: Same format as
Beginning of Chapter 9 on Center of Gravity and Centroids.
and use of calculus to find CGs for various 2D and 3D
Lecture #26: Centroids
CGs for Composite Bodies.
CENTROID OF COMPOSITE
WITH HOLES: We have shown in
class how the
centroid of a 2D composite body is calculated by looking at
of the sum of the products of xbar and the sub-area divided by
area. When a body contains a hole the procedure stays the same
the sub-area representing the hole is preceded by a negative
take the circular disc x^2+y^2=4 into which is cut a square
-1/2<y<1/2. Here by symmetry ybar is zero but xbar
-0.043228. So xbar is negative as expected. Click HERE
to see a schematic of this calculation.
CENTROID FOR A COMPOSITE
To get some practice on finding the centroid of some 2D
the centroid of the block letter combination UF ( University
).What is x bar and y bar for this set up assuming both
letters U and F
are constructed of unit width rectangles and the two letters
by a unit distance? To make the calculation we break the U up
rectangles. The first of these has corners at (-4,5),
The second has corners at (-3,0), (-1,0), (-1,1) and (-3,1)
corners at (-1,0), (0,0), (0,5) and (-1,5). Applying our
formula to these three rectangles yields a centroid for the U
centroid of U is at (-2.00,2.1667). Next look at F. Breaking
it also up
into three rectangles with the largest having corners at
The other rectangle has corners at (2,4), (5,4), (5,5) and
remaining square bounded by corners (2,2), (3,2), (3,3) and
finds for the F the centroid is at (xbarF,
combining the centroids for U and F yields
and ybar=[2.2778*12+3.16667*9]/21=2.5952. Thus the UF
have zero net moment about the point (-0.16667, 2.5952). I
several years ago as one of the problems on a statics exam
and, if I
correctly , about half got it right.
Lecture #27: More
calculations for the CG and Centroid of Bodies plus the
THEOREMS OF PAPPUS
are two theorems associated with the the names of Pappus of
290-350AD) and Guldinus( 1577-1643AD). The first of these
product of the length of a curve and the distance moved by
when the curve is rotated about an axis equals the surface
As an example consider the half circle x^2+y^2=R^2, y>0.
R*Pi and its centroid is at 2*R/Pi. Thus by the first
area will be R*Pi*(2*Pi)*(2*R/Pi)=4*Pi*R^2, a well known
Use this theorem to calculate the surface area of a toroid.
theorem states that "The product of an area and the distance
by its centroid when this area is rotated about an axis
of the solid generated". As a demonstration of this consider
-a<x<a, 0<y<b rotated about the x axis. Its area
is 2ab and
its ybar is b/2. The volume generated is thus V=2a*b^2*Pi.
as expected, with the volume of a cylinder of height 2a and
Go HERE for a
the Pappus Theorems.
for Distributed Loadings. Fluid Pressure, Pascal's Law
FORCE ON SUBMERGED
The centroid concept can be applied to loading of surfaces due
forces. Click HERE
see the general development for the resultant force on a
Also consider the case of the resultant force R on a vertical
h and width w. By Pascal's law we know the fluid pressure at
be p=rgz so the force on the dam
will be R=Int[rgz
That is, the force equals the dam area w*h times the average
pressure rgh/2. Note, R is
on how much water is stored behind the dam. On the other
very large for dams such as the Hoover dam in Nevada where
the resultant force R acts at the center of pressure which
lies at the
centroid of the pressure triangle and thus at h/3 from the
bottom for a
filled , vertical walled, dam.
Lecture #29: Introduction
to Moments of Inertia(Chapter 10). Basic definition and
Using calculus to find Ixx, Iyy, and Izz.
Go HERE to see how
is used to determine these for various bodies.
AREA MOMENTS OF
A RECTANGULAR PLATE: A good
of area moments of inertia is given by looking at the
0<x<a, 0<y<b. In this case the values Ixx=Int[y^2dA]=a*b3/3
Iyy=Int[x^2dA]=b*a3/3. Also Izz=Jo=Int[r2dA]=Ixx+Iyy=[(a*b)/3]*[b2+a2].
last relation between the three area moments of inertia holds
for lamina and is often referred to as the perpendicular
axis theorem. We also have from
axis theorem that the Izz
have its minimum value at the plate centroid at x=a/2, y=b/2.
value is Izz(at lower left corner)=Izz(at
area)*(square of the distance between the two points).
Here this yields Izz(at plate centroid)=(a*b/12)*(a2+b2)
which is seen to be four times less than that about any of the
corners. Note the area moments of inertia have the
length to the 4th power, while mass moments of inertia have
of mass times length squared.
Moments of Inertia of Composite Bodies.
AREA MOMENT OF INERTIA FOR
COMPOSITE LAMINA: Moments of
so that one can readily calculate the I values of a composite
any chosen axis by knowing the values for the sub-bodies and
the parallel axis theorem. HERE
is an example of such a calculation.
Product of Inertia Ixy, Moment of Inertia about any
Moments of Inertia, Mohr Circle.
MOMENT OF INERTIA ABOUT AN
AXIS IN THE PLANE OF A LAMINA:
lamina with a superimposed x and y axis. Introduce a second
axis u and v where the u axis makes an angle of A with respect
to the x
axis. From the geometry it is easy to express u and v as
functions of x
and y , as we have done in class, and then work out the area
inertia Iuu and Ivv, plus the product of inertia Iuv. Go HEREto
see the details. Note also the Mohr Circle follows from this
Mass Moments of Inertia defined and evaluated using Calculus.
CALCULATING THE MASS MOMENT
OF INERTIA FOR A UNIFORM SPHERE:
the mass moment of inertia about the z axis of a body as
where the integral extends over the entire body with the mass
dm=rho*dxdydz. In many cases one does not need to carry out
of such triple integrals directly if one knows the value of
dIzz of the
sub-bodies making up the body of interest. We show you HEREthe
calculation for finding Izz of a uniform sphere of radius R=a
known dIzz for the discs making up this sphere.
Virtual Work(Chapter 11)and some Elementary
EXAMPLE OF A VIRTUAL WORK
have shown in class that an alternative method for solving
problems is that of Virtual Work. The technique is based on
the product of the forces acting on a body times their virtual
must add up to zero under static equlibrium. The virtual work
and applies only to those points where the force F can cause
dr. We show you
on Virtual work. Conservative Forces, and Potential Energy. Also
filling out of teacher and course evaluation forms for Dr.
SIDE DIVERSION ON PI
FOR THIRD EXAM : This exam
will emphasize Chapters 9, 10
11 on Centroids, Center of Gravity, Moments of Inertia, and
One of the problems will also contain material (such as possibly
and friction)covered in the earlier chapters.
November 27, 2010