Engg Mechanics
Engg Mechanics
Engg Mechanics
com
1. Introduction
In the early stages of scientific development, ―physics‖ mainly consisted of mechanics and
astronomy. In ancient times CLAUDIUS PTOLEMAEUS of Alexandria (*87) explained the
motionsof the sun, the moon, and the five planets known at his time. He stated that the
planets and the sun orbit the Earth in the order Mercury, Venus, Sun, Mars, Jupiter, Saturn.
This purely phenomenological model could predict the positions of the planets accurately
enough for naked-eye observations. Researchers like NIKOLAUS KOPERNIKUS (1473-1543),
TYCHO BRAHE (1546-1601) and JOHANNES KEPLER (1571-1630) described the movement of
celestial bodies by mathematical expressions, which were based on observations and a
universalhypothesis (model). GALILEO GALILEI (1564-1642) formulated the laws of free fall
of bodies and other laws of motion. His ―discorsi" on the heliocentric conception of the world
encountered fierce opposition at those times.
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After the renaissance a fast development started, linked among others with the names
CHRISTIAAN HUYGENS (1629-1695), ISAAC NEWTON (1643-1727), ROBERT HOOKE (1635-
1703) and LEONHARD EULER (1707-1783). Not only the motion of material points was
investigated, but the observations were extended to bodies having a spatial dimension. With
HOOKE’s work on elastic steel springs, the first material law was formulated. A general theory
of the strength of materials and structures was developed by mathematicians like JAKOB
BERNOULLI (1654-1705) and engineers like CHARLES AUGUSTIN COULOMB (1736-1806) and
CLAUDE LOUIS MARIE HENRI NAVIER (1785-1836), who introduced new intellectual concepts
likestress and strain.
The achievements in continuum mechanics coincided with the fast development in
mathematics: differential calculus has one of its major applications in mechanics, variational
principlesare used in analytical mechanics.
These days mechanics is mostly used in engineering practice. The problems to be solved are
manifold:
• Is the car’s suspension strong enough?
• Which material can we use for the aircraft’s fuselage?
• Will the bridge carry more the 10 trucks at the same time?
• Why did the pipeline burst and who has to pay for it?
• How can we redesign the bobsleigh to win a gold medal next time?
• Shall we immediately shut down the nuclear power plant?
For the scientist or engineer, the important questions he must find answers to are:
• How shall I formulate a problem in mechanics?
• How shall I state the governing field equations and boundary conditions?
• What kind of experiments would justify, deny or improve my hypothesis?
• How exhaustive should the investigation be?
• Where might errors appear?
• How much time is required to obtain a reasonable solution?
• How much does it cost?
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One of the most important aspects is the load–deformation behaviour of a structure. This
question is strongly connected to the choice of the appropriate mathematical model, which is
used for the investigation and the chosen material. We first have to learn something
aboutdifferent models as well as the terms motion, deformation, strain, stress and load and their
mathematical representations, which are vectors and tensors.
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nomenclature used and some rules of tensor algebra and analysis as well as theorems on
properties of tensors are included in the Appendix.
Examples: Length, l, time, t, and mass, m, can be chosen as base quantities in dynamics.
The unit of a physical quantity is the value of a chosen and defined quantity out of all
quantities of equal dimension.
Example: 1 meter is the unit of all quantities having the dimension of a length (height,
width, diameter, ...)
The numerical value of a quantity G is denoted as {G}, its unit as [G].
The invariance relation is hence G = {G} [G]
Base units are the units of base quantities. The number of base units hence always equals the
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Example: Meter, second and kilogramm can be chosen as base units in dynamics
[l] = m, [t] = s, [m] = kg
For discriminating symbols (for quantities) from units, the former are printed in italics, the
latter plain.
Fractions and multiples of SI units obtained by multiplication with factors 10±3n (n = 1, 2, ...)
have specific names and characters, which are generated by putting special prefixes before the
names and the characters of the SI units.
10-15 femto f
10-12 pico p
10-9 nano n
10-6 micro
μ
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10-3 milli m
103 kilo k
M
106 Mega
G
109 Giga
T
1012 Tera
SI Units
SI units are the seven base units of the seven base quantities of physics, namely
Name Character
Length meter m
Mass kilogramm kg
Time second s
Thermodynamic.
Temperature Kelvin K
Electrical
Curent Ampere A
Amount of Substance mol mol
Luminous Intensity candela cd
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Derived units are composed by products or ratios of base units. The same holds for the unit
characters. Some derived SI units have a special name and a special unit character
Name Char.
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mathematical equations describing the fundamental physical laws for both solids and fluids
are alike, so the different characteristics of solids and fluids have to be expressed by
constitutive equations. Obviously, the number of different constitutive equations is huge
considering the large number of materials. All of this can be written using a unified
mathematical framework and common tools. In the following we concentrate on solids.
Continuum mechanics is a phenomenological field theory based on a fundamental hypothesis
called continuum hypothesis. The governing equations comprise material independent
principles, namely
MECHANICS
• Body of Knowledge which Deals with the Study and Prediction of the State of Rest or Motion of
Particles and Bodies under the action of Forces
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2.3NEWTON ‘s Laws
Law I
Each body remains in its state of rest or motion uniform in direction
until it is made to change this state by imposed forces.
Law II
The change of motion is proportional to the imposed driving force and
occurs along a straight line in which the force acts.
Law III
To every action there is always an equal reaction: or the mutual
interactions of two bodies are always equal but directed contrary.
Force as a Vector
o Magnitude
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o Direction
o Location
The simple support structure in Figure 1.1.1 can be used to illustrate the three characteristics that
make a force equivalent to a vector.
Consider the tension load that occurs in the cable A-C due to the 100 pound load at D. The
Force FAC can be calculated using statics and is equal to 250 pounds. The 250 pound force can
be compared to the tensile capacity of member A-C. In this case, we will examine the effect of
the load on support A. Most connections are examined in terms of component forces, not total
force. The load at A will be decomposed into its orthogonal components. It is helpful then to
assign a convenient axis system which coincides with the force component, Figure 1.1.2.
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The analysis of the force components at A depend on treating the forces as vectors. Figures
1.1.3 and 1.1.4 show this representation. Any major force problem is best handled by treating
the forces as vectors that can be manipulated using trigonometry.
There are two ways in which forces can be represented in written form:
The method used depends on the type of problem being solved and the easiest approach to
finding a solution.
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Scalar notation is useful when describing a force as a set of orthogonal force components. For
example: Fx = 15N, Fy = 20N, Fz = 10N.
Vector notation is useful when vector mathematics are to be applied to a problem, such as
addition or multiplication. Vector notation is somewhat simple in form:
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Addition of Forces
Multiple forces can be applied at a point. These forces are known as concurrent forces and can be
added together to form a resultant force. If the component forces are orthogonal, then the
magnitude of the resultant force can be determined by taking the Square Root of the Sum of the
Squares (SRSS). The SRSS method is an extension of the Pythagorean Theorem to three
dimensions. Figure 1.3.1 illustrates the calculation of a vector magnitude using the SRSS
method.
Force components are sometimes not perpendicular to one another. In such cases the resultant
force vector can be found by adding the scalar values of the component vectors. Figure 1.3.2
illustrates the vector addition of three scalars to determine the resultant force.
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The magnitude of the resultant force in Figure 1.3.2 can be found by using the SRSS method:
F=[(42)^2 +(23)^2+(138)^2]^0.5=146 lbs.
The concept of a force polygon is useful when dealing with two nonperpendicular forces. The
graphical representation of the vector addition can be drawn two ways. Figure 1.3.3 shows this
addition using either the Parallelogram Method or the Head to Toe Method.
Both methods result in a triangle, or a three sided force polygon. The sides and angles of this
triangle follow the Law of Sines and the Law of Cosines. These laws can be used to determine
the magnitude and orientation of the resultant. Figure 1.3.4 outlines the process for determining
the resultant force vector.
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Forces usually occur at arbitrary angles relative to the X,Y and Z coordinate system. In
considering the effects of a force it is sometimes better to deal with the orthoganal components
Fx, Fy and Fz. The transformation from the resultant force to its components is shown in Figure
1.4.1.
If the resultant force was reported in vector notation, then determining the force components
from the scalar multipliers would be obvious. In many cases though, the resultant force is
known only by its magnitude, position and direction. If we must know the force components,
then we will have to apply mathmatics, in one of several ways, to solve for Fx, Fy and Fz from F.
As an example of the transformation of forces, consider the sign support in Figure 1.4.2.
Suppose that we know that cable AB has a tension force of 500lbs. We could measure this force
by attaching a gage to the cable while it is being loaded.
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The component forces we wish to determine occur at joint A. Knowing these components
would be important in selecting the type of hardware necessary to hold the cable in place. We
can assign a coordinate system and draw what the forces will look like, as shown in Figure 1.4.3.
There are several ways the components can be determined. The basic principle to remember is
that the shape of the cable is analogous to the shape of the force vector. This means that the
geometry, angle and aspect ratio of the cable and its dimensions are the same as the resultant
force and its components. Figure 1.4.4 illustrates this relationship.
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The first thing we will consider is the aspect ratio. Since the coordinate system is orthongonal,
the Pythagorean theorem can be applied to determine the length of the cable.
With the length of cable AB known, the aspect ratio of the side over the diagonal can be applied
to determine the component forces.
Note that the force components should sum vectorially to the original resultant force.
The SRSS summation provided a good check of the calculation. Such checks should be done
when ever you work through a problem where errors can occur.
Direction cosines are another useful method for determining rectangular components. The use of
direction cosines is illustrated in Figures 1.4.5 and 1.4.6.
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Sometimes it is useful to know how much of a force is acting in a direction other than the X,Y,
or Z direction. Such a case might involve a bolted connection with a slotted hole, as shown in
Figure 1.5.2.
Suppose that the force in Figure 1.5.1 was F=40i + 10j + 30kNewtons, and that the X' axis was
oriented along the unit vector e=0.6i + 0.5j - 0.624k. The projection of F onto X' would be the
force component acting in the direction of the slot. Now consider the possibility that the
connection would slip if Fx' was greater than 35 N. Will the connection slip? The dot product
will help us answer this question.
Figure 1.5.2 illustrates the projection of one vector onto another with the corresponding
mathematics.
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Some explanation should be given as to what A dot eb actually is. Assuming alpha, beta and
gamma represent the direction angles of A, and alpha prime, beta prime and gamma prime
represent the direction angles of eb, the dot product can be written as:
Knowing that i dot j, i dot k, and j dot k equals 0, the dot product reduces to:
Once the projected length of A onto B is known, the value of the angle between A and B can be
found:
The equations above will be used to determine the proportion of the bolt force being applied in
the direction of the slotted hole.
We already know the direction coines for the X' axis. They come from the unit vector e.
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Since the applied load along the direction of the slot is 10.3N, and the bolt strength is 35N in that
direction, X'-X' slippage will not occur. There is no guaranty that the connection will not fail in
another way though. To find out, we would have to examine the Y' and Z' loads and strengths.
Problems
Use the Law of Sines and Law of Cosines to determine the resultant force vector caused by the
two forces shown. Report your answer in vector notation.
a. Fx = 20.7# Fy = 125#
b. F = 125i + 20.7j #
c. F = 107i + 2.22j #
d. Fx = 95.2# Fy = 39.7#
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Three force vectors (F1, F2, F3) are simmultaneously applied at point A. The resultant of these
three forces is F.
Determine F3 such that: F = 3.2i + 5.5j + 2.9k kips. Write F3 in vector notation.
The unit kip represents 1000lbs, or 1 kilo lb. Sometimes this unit is just abbreviated as k.
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It's tow truck time. Yes, you turned right when you should have turned left. Fortunately, it was
just a snow filled ditch that you hit.
Because there is not much room to maneuver, the tow truck has to pull out your car at an angle
(not recommended in your owners manual). Your concern is whether to have the driver pull out
your old beater, or to leave it there without the plates.
Determine the force applied to the bumper by a tow cable tensioned at 3 kips. Report your
answer in scalar notation. Also, your bumper can only take 1500 pounds of force sideways.
Should you proceed with the pull?
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A cable with a tension force of 15.1 kN is anchored at points A and B on a structure. Compute
the direction angles for the force applied at point A. The angles should be relative to the local X',
Y', and Z' axes. The local axis system at A is parallel to the global axis system.
a.
b.
c.
d.
e.
Problems
1.The resultant of the two forces, when they act at an angle of 300 is 50N. If the forces act at 600, the
resultant is 70N. Determine the magnitude of two forces.
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3. Two elephants of mass 10 ton each are standing such that distance between their center of
gravities is 10 m. Find out the force by which they pull each other. The value of gravitational constant
is .
4. Prove that if two forces P and Q act at a point and the angle between the two forces be , then
the resultant is given by,
and the angle made by the resultant with the direction of force P is expressed
as .
Add these forces and find out the angle which the resultant makes with the x-axis.
Equation of equilibrium
One can study the equilibrium of a part of the body by isolating the part for analysis. Such a body is
called a free body. We make a free body diagram and show all the forces from the surrounding that
act on the body. Such a diagram is called a free-body diagram. For example, consider a ladder
resting against a smooth wall and floor. The free body diagram of ladder is shown in the right.
When a body is in equilibrium, the resultant of all forces acting on it is zero. Thus that resultant
force R and the resultant couple M R are both zero, and we have the equilibrium equations
..............(2.1)
These requirements are necessary and sufficient conditions.
1. Collinear forces: In this system, line of action of all the forces act along the same line. For example,
consider a rope being pulled by two players. Suppose one player is pulling with a force F1 and the other is
pulling with a force F2 . Then the resultant force, in the direction of pull of first player is (F1 - F2). By the first
condition of (2.1), it should be zero. Hence, for equilibriumF1=F2 . With this, the second condition will also be
fully satisfied, because moments of the forces about any point in the space will be of equal and opposite
magnitudes. Hence, the resultant moment will be zero.
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2. Coplanar parallel force: In this system, all forces are parallel to each other and lie in a single plane.
Consider a See-Saw. Two children are sitting in the See-Saw and it is in equilibrium. If the weights of both the
boys are W1 and W2 and they are sitting at a distance of d1 and d2 from the fulcrum, then for equilibrium, the
fulcrum reaction should be
R = W1+W2 .................(2.2)
W1d1 = W2 d2...........(2.3)
Seesaw
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Solution:
BH 1
BH (0.6 m i 12
. m j 12
. m k)
| BH | 18
.
BH 750 N
TBH | TBH |. BH | TBH | 0.6 m i 12
. m j 12
. mk
| BH | 18
. m
TBH (250 N ) i (500 N ) j (500 N ) k
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Solution:
BH 1
BH (0.6 m i 12
. m j 12
. m k)
| BH | 18
.
BH 750 N
TBH | TBH |. BH | TBH | 0.6 m i 12
. m j 12
. mk
| BH | 18
. m
TBH (250 N ) i (500 N ) j (500 N ) k
R = F
i.e. Rx i + Ry j + Rz k = ( Fx i + Fy j + Fz k )
= ( Fx) i + ( Fy)j + ( Fz )k
Rx= Fx, Ry = Fy , Rz = Fz
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= MN/MN = 1/d ( dx i + dy j + dz k )
Recall that: F = F
F = F = F/d ( dx i + dy j + dz k )
Fd x Fd y Fd z
Fx , Fy , Fz
d d d
dx x2 x1 , d y y2 y1 , d z z2 z1
2 2 2
d dx dy dz
dx dy dz
cos x , cos y , cos z
d d d
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and the angle made by the resultant with the direction of force P is expressed as
tan Ɵ = Q sinα/P+Q cosα
6. If the two forces P and Q are equal and are acting at an angle ex between them,
then the resul tant
is given by
R = 2P cosα/ 2 and angle made by the resultant is expressed as e = ~
2
7. According to Lame's theorem, "If three forces acting at a point are equilibrium,
each force will be
proportional to the sine of the angle between the other two forces."
8. The relation between newton and dyne is given by One newton = 105 dyne.
9. Moment ofa force about a point = Force x perpendicular distance ofthe line of
action of the force from that point.
10. The force causes linear displacement while moment causes angular
displacement. Abody will be
in equilibrium if (i) resultant force in any direction is zero and (ii) the net moment
of the forces
about any point is zero.
11. Gravitational law of attraction is given by,
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28. If the resultant of a number of parallel forces is not zero, the system can be
reduced to a sin
force, whose magnitude is equal to the algebraic sum of all forces. The point of
application oft
single force is obtained by equating the moment of this single force about any
point to the al
braic sum of moments of all forces acting on the system about the same point.
29. If the resultant of a number of parallel forces is zero, then the system may have
a resultf
couple or may be in equilibrium. If the algebraic sum of moments of all forces
about any poilll
not zero, then system will have a resultant couple. But if the algebraic sum of
moments of
forces about any point is zero, the system will be in equilibrium.
~~~~~~~~~~~~~~~~~( EXERCISE 1 )1---------------------------------
10.
-11.
12.
What do you mean by scalar and vector quantities?
Define the law of parallelogram of forces. What is the use of this law?
State triangle law of forces and Lame's theorem.
Two forces P and Q are acting at a point in a plane. The angle between the forces
is 'a'. Prove tl
the resultant (R) of the two forces is given by R = ~ p2 + Q2 + 2PQ cos a .
Define the following terms: dyne, newton, meganewton and moment of a force.
Prove that one newton is equal to 105 dyne.
Explain the terms: clockwise moments and anti-clockwise moments.
What is the effect of force and moment on a body?
Indicate whether the following statement is true or false.
"The resultant components of the forces acting on a body along any direction is
zero but the n
moment ofthe forces about any point is not zero, the body will be in equilibrium'.
[Ans. Fah
Write the S.L units of: Force, moment and velocity.
What do you mean by resolution of a force?
A number of coplanar forces are acting at a point making different angles with x-
axis. Find;
expression for the resultant force. Find also the angle made by the resultant force
with x-axil
State and explain the pinciple of transmissibility of forces.
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X
F1 = 1000 N
19. The four coplanar forces are acting at a point as shown in
Fig. 1.93. One of the forces is unknown and its magnitude is
shown by P. The resultant is having a magnitude 500 N and is
acting along x-axis. Determine the unknown force P and its inclination
with x-axis. [Ans. P = 286.5 Nand El = 53° 15']
(iii) Coplanar Parallel Forces
20. Four forces of magnitudes 20 N, 40 N, 60 Nand 80 N are
acting respectively along the four sides of a square ABCD
as shown in Fig. 1.94. Determine the resultant moment
about point A.
Each side of square is 2 m. [Ans. 200 Nm anti-clockwise]
21. A force of 50 N is acting at a point A as shown in Fig. 1.95.
Determine the moment of this force about O.
[Ans. 100 Nm clockwise]
22. Three like parallel forces 20 N, 40 Nand 60 N are acting at
points, A, Band C respectively on a straight line ABC. The
distances are AB = 3 m and BC = 4 m. Find the resultant and
also the distance of the resultant from point A on line ABC.
[Ans. 120 N, 4.5 m]
The three like parallel forces 101 N, F and 300 N are
acting as shown in Fig. 1.96. If the resultant R = 600 N
and is acting at a distance of 45 em from A, then find the
magnitude of force F and distance of F and A.
[Ans. 200 N, 30 em]
Resultant
= 500 N
200 N
Fig. 1.93
40N
~
e12m
B
~I20 N
2m
Fig. 1.94
y
A
50~ ~~
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~~ 30°
0
~
\+--4 m-.j X
Fig. 1.95
R = 600 N
F t 300 N .I II Ie
Ie
I
100 N t!4--x---.j
~ 45 em ---<~"I""- 25 em -.j
Fig. 1.96
24. Four parallel forces of magnitudes 100 N, 200 N, 50 N
and 400 N are shown in Fig. 1.97. Determine the magnitude
of the resultant and also the distance ofthe resultant
from pointA. [Ans. R = 350 N, 3.07 m]
100 N 200 N 50 N 400 N J_L J_t ~ 1 m --.j+-1.5 m~ 1 m-.j
Fig. 1.97
25. A system of parallel forces are acting on a rigid bar as shown in Fig. 1.98.
Reduce this system to:
(i) a single force, [Ans. (i) R = 120 N at 2.83 m from A
(ii) a single force and a couple atA (ii) R = 120 Nand MA = - 340 Nm
(iii) a single force and a couple at B. (iii) R = 120 Nand Mll = 120 Nm]
20 N 100 N 40 N 80 N
t====-A 1_oI__ D ====t B
Fig. 1.98
26. Five forces are acting on a body as shown in Fig. 1.99. Determine the resultant.
[Ans. R = 0, Resultant couple = 10Nm]
20 NL
r---2.5m f.--2m
20 N 40 N 30 N 10 N J~1_D L
Fig. 1.99
27. Determine the resultant of the parallel forces shown in Fig. 1.100.
[Ans. Body is in equilibrium]
10 N 40 N 30 N 10 N 10 N
1 lJ lJ I.• 2m I 0.5m I 1.0 m 10.5 m I ~.. ~.. ~.. ~
Fig. 1.100
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UNIT-2–
EQUILIBRIUM OF RIGID BODIES
elements of stable equilibrium – Moments and Couples – Moment of a force about a point and
about an axis – Vectorial representation of moments and couples – Scalar components of a
moment – Varignon’s theorem – Equilibrium of Rigid bodies in two dimensions – Equilibrium of
Rigid bodies in three dimensions – Examples
A rigid body is one which does not suffer deformation. It can be continuous or
connected members.
F2 F3
F1
P2
P1
F3 F1 F2
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which is written:
of P and Q and sine of the angle between their tails ( 0 < < 180o).
Thus : V = P Q Sin
counterclockwise the rotation through that brings the vector P in line with vector Q.
V = P x Q = (P Q sin ) v
Where: the scalar PQ sin defines the magnitude of V and the unit vector, v
V = PxQ V
Q PQ sin =V
P P
Laws of Operation
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j x
k = (i x j) i
i x j = i . j . sin 90o = 1 x 1 x 1 = 1.
Using the right hand rule, the resultant vector points in the k direction.
at the tip of - k.
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Likewise: i x j = k i x k = -j i xi =0 j
j xk=i j x i = -k j xj =0 k i
kxi =j k x j =-i kx k =0
Rules: If a circle is constructed as shown, then the vector product of two unit vectors in
a counterclockwise fashion around the circle yields the positive third unit vector eg.
= Px Qx (i x i) + Px Qy (i x j) + Px Q z ( i x k)
+ Py Qx ( j x i) + Py Qy (j x j) + Py Q z (j x k)
+ Pz Q x (k x i) + Pz Qy (k x j) + Pz Qz (k x k)
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RECTANGULAR COMPONENTS OF
THE
MOMENT OF A FORCE
(CARTESIAN VECTOR
FORMULATION)
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Finally, denoting by the angle between the lines of action of the position vector, r and
Mo = r F sin = Fd
F
F
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An ancient alien artifact has been located in neutral space. Three starships have been dispatched
to tow the object into their respective territories: Ferengi, Cardassian, and Romulan. A
Federation ship has been sent to keep the object within neutral space.
If the three opposing vessels are applying the following force vectors to the object, using their
tractor beams:
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g.
h. The moment, M, which is causing the door to rotate, equals:
i. M=F*d
j. Because moments are derived from forces times distrances, they will have units such as
inch-pounds, foot-kips, or Newton-meters. It is vital that you pay close attention to the
units. If you expect a moment in your answer, the proper units should occur. It is
helpful to write your units down when ever possible. Writing down units will help you
avoid errors and provide some tangible meaning to the numbers that you generate.
k. Enough soap box stuff. Moments play an important roll in the function of machines and
even body movement. Consider the simple example of a wrench in Figure 2.1.2. The
torque, or moment to the nut, is what would loosen it.
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l.
m. Notice how the 50# force is parallel to the y-axis. Is the moment to the nut 50# times
12" or 600 in-#? The answer is no. We can see why by breaking the load into
components that are parallel and perpendicular to the wrench handle, see Figure 2.1.3.
n.
o. Our wrench problem can now be redrawn, Figure 2.1.4.
p.
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q. With Figure 2.1.4 you can see that the 17.1# load is pointed directly at the nut and its
eccentricity to the nut is zero. The 47.0# load, though, is perpendicular to the 12"
moment arm. As a matter of fact, if you considered F2 as a long line, the closest it would
come to the nut is 12 inches. This minimum distance is one of the properties of a
perpendicular.
r. The moment at the nut can be calculated as:
u.
v. The moment at O can be considered as the sum of three components Mx, My and Mz, just
as the force at A can be considered the sum of three components Fx, Fy and Fz. By first
considering the moment at O in three parts, we can better visualize the value of the
moment and how it is oriented.
w. First break the force at A into the orthongonal components, Figure 2.2.2.
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x.
y. Now consider the possibility of a moment about the X axis. Notice how the Fx force is
parallel to the X axis. The Fx force will not cause any twisting or rotation about the X
axis. The Fy and Fz forces are not parallelto the X axis and they also have an eccentric
distance. Remember that moment equals force times eccentric distance. Note however,
that the forces cause twisting in opposite directions about the X axis. Some cancellation
has to occur. To account for this cancellation, we use the Right Hand Rule.
z. Imagine a moment about the X axis. This moment is like the curving of the fingers
when you form a fist with your rigth hand. The thumb points in the positive direction
along the X axis. The moment can be visuallized as a double headed vector. Any
moment vector that points in the positive direction is positive. Vise versa for negitive.
Figure 2.2.3 illustrates the Right Hand Rule.
aa.
bb. With the Right Hand Rule in mind, let us now compute Mx:
cc. Mx = 4#(16") - 2#(24") = 16 in-#
dd. We can apply similar logic to My and Mz. Remember moment as a double headed
vector, with positive consistent with the axis system. Also remember that moment
comes from force times the eccentric distance. Since Fz is parallel to axis Z, it contribute
no moment to My and Mz, respectively.
ee. My = 2#(20") - 3#(16") = -8 in-#
ff. Mz = 3#(24") - 4#(20") = -8 in-#
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hh.
ii. We can also use the SRSS method to determine the magnitude of the resultant moment
Mo:
jj.
kk. Writing Mo in vector form:
ll. |Mo| = 16i - 8j - 8k in-#
mm. Finally, Figure 2.2.5 will illustrate the single force at A causing the resultant
moment at O.
nn.
oo.
pp.
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into account the sign of the moments and the fact that Fx has no effect on Mx, Fy on My
and Fz on Mz.
ss. Mathematically, the cross product can be represented by the three vectors shown in
Figure 2.3.1.
tt.
uu. The magnitude of the cross product is found from the sine equation; the direction of C is
perpendicular to the plane formed by vectors A and B.
vv. The cross product can also be written in equation form:
ww. A x B = (AyBz -Az By)i + (AzBx-AxBz)j + (AxBy-AyBx)k
xx. An interesting graphical form of the cross product is illustrated in Figure 2.3.2. The
terms of the cross product are determined by adding or subtracting (which ever is
appropriate) the diagonal multiplication products.
yy.
zz. You might ask what does the cross product have to do with the moment caused by an
eccentric skewed load. Well, if A represented the vector going from the moment
location to the force, and B represented the force vector, then C would be the moment
vector caused by the force.
aaa. Let's check this out by running a cross product calculation on the eccentric
skewed force example from the last section. The example is redrawn in figure 2.3.3,
with the relative position of the force shown as vector A.
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bbb.
ccc. The graphical approach is used in Figure 2.3.4. This approach is fairly easy to
remember, provided the proper sign convention is followed.
ddd.
eee. The dot product method for solving for a three dimensional moment is very
efficient. However, it does not provide a physical feel for how each of the force
components contributes to the moment components, as was seen in section II-ii.
The last section explained the calculation of a moment caused by a skewed excentric load in
terms of perpendicular distances times orthagonal forces. There is another method for
determining moments based on the cross product. The cross product automatically takes
into account the sign of the moments and the fact that Fx has no effect on Mx, Fy on My and Fz
on Mz.
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Mathematically, the cross product can be represented by the three vectors shown in Figure
2.3.1.
fff.
The magnitude of the cross product is found from the sine equation; the direction of C is
perpendicular to the plane formed by vectors A and B.
The cross product can also be written in equation form:
A x B = (AyBz -Az By)i + (AzBx-AxBz)j + (AxBy-AyBx)k
An interesting graphical form of the cross product is illustrated in Figure 2.3.2. The
terms of the cross product are determined by adding or subtracting (which ever is
appropriate) the diagonal multiplication products.
You might ask what does the cross product have to do with the moment caused by an
eccentric skewed load. Well, if A represented the vector going from the moment
location to the force, and B represented the force vector, then C would be the moment
vector caused by the force.
Let's check this out by running a cross product calculation on the eccentric skewed force
example from the last section. The example is redrawn in figure 2.3.3, with the relative
position of the force shown as vector A.
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The graphical approach is used in Figure 2.3.4. This approach is fairly easy to
remember, provided the proper sign convention is followed.
The dot product method for solving for a three dimensional moment is very efficient.
However, it does not provide a physical feel for how each of the force components
contributes to the moment components, as was seen in section II-ii.
A wrench is being used to tighten a nut. Because of obstructions, the wrench end can only be
pushed at a skewed angle. Determine the tightening torque, My, that is applied to the nut.
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Balancing a Fulcrum
A balance beam is used to hold a weight in position. Two forces are applied at one end of the
beam; one force is known, the other is unknown. Based on the magnitudes and dimensions given,
determine the magnitude of force FB such that the fulcrum is in equilibrium about the piviot.
Assume that the piviot can resist any twisting about an axis other than its principle axis of
rotation.
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a. FB = 2.33 kN
b. FB = 19 kN
c. FB = 11.5 kN
Compute the resultant moment at point A, caused by a force vector at point B. Report the
moment using vector notation.
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A cellar door needs to be lifted open. You grasp the handle and pull, but at a skewed angle
relative to the door and hinge. If you pull with a force of 40 pounds(#), how mush moment is
applied toward opening the door, i.e. moment about the hinge axis?
a. 32.9 ft-#
b. 45.2 ft-#
c. 473 ft-#
A cellar door needs to be lifted open. You grasp the handle and pull, but at a skewed angle
relative to the door and hinge. If you pull with a force of 40 pounds(#), how mush moment is
applied toward opening the door, i.e. moment about the hinge axis?
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a. 32.9 ft-#
b. 45.2 ft-#
c. 473 ft-#
d.
Centroid of a Truck
A truck is driving over a bridge. The response of the bridge to the truck will partially depend on
the weight of the truck and the center of gravity of this weight. Based on the axle load and
configuration shown, determine the centroid of the truck weight. Report the location in number
of feet back from the front bumper.
a. 2.90ft
b. 16.8ft
c. 6.2ft
d. 15.9ft
e. Somewhere between 10.5ft and 11.5ft
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A 1.5m length of aircraft wing has a net upward pressure distribution as shown. Determine the
magnitude of uplift on the segment of wing and the X location of the upward resultant force.
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X = 6.15", Y = 4.91"
X = 4.61", Y = 6.31"
X = 6.00", Y = 5.00"
X = 5.79", Y = 5.13"
Concept of Equilibrium
Any kind of object or component must have a net balance of zero forces and moments
applied to it in order to remain at rest. This type of balance is called static equilibrium,
where nothing is moving. Figure 4.1.1 illustrates a weight, pully and block situatuion
where all of the forces, including friction beneath the block, sum to zero.
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Upon inspection of the forces in Figure 4.1.1, you will see that they sum to zero. Hence,
there is equilibrium and no movement.
Determining the forces on a body is an important part of mechanics. It helps us answer
questions such as: Will an object move? How much force or stress is in a part? Will
something break? Do you think the stationary block in Figure 4.1.1 would move if the
hanging mass was 50kg instead of 25kg? What if the cable were to break?
There are two vital items which must be understood in order to evaluate forces. These
two items are boundry conditions and the free body diagram. Both of these items can be
used in a visual way to help see a problem such as the previous weight and pully
example.
BoundryCondictions
Boundry condition is a fancy way of saying what is happening at the edge of an object. What
happens at the edge depends on how an object is connected to, or resting on its neighbor. Take
the example of a soda can resting on a dashboard (Figure 4.2.1).
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The question you should ask is: Is the can open? Because if it is, someone might get soaked! Ok,
Obviously this is an equilibrium problem. The can could slide off the dash, tip over, or stay put.
Much depends on how the can is supported at the boundry. In this case, the boundry is the
base/dash interface. On the surface of the dash we can have friction and normal forces (Figure
4.2.2).
Remember from physics how the maximum frictional force is a function of the normal force:
Where is the normal force coming from? In this case normal force is a fraction of the weight of
the can.
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Figure 4.2.3 illustrates analysis of the sliding part of the problem. A typical can of soda is
125mm tall, 65mm in diameter, and weighs 3.90N. Let us suppose that the dash has not been
cleaned in a while and the coeffient of friction is 0.20.
Notice in Figure 4.2.3 that the tangential force (due to gravity) is about 25% greater than the
maximum possible friction force. Friction is what holds the can in place. So, in this example, the
can will move.
There are many more types of boundry conditions than just a can sitting on a dashboard. How a
boundry performs depends on its configuration. Figure 4.2.4 illustrates several types of
boundries that commonly occur. Notice how the type of reactionary force developed depends on
the type of boundry.
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When we are faced with the problem of resolving the focus in a system, we must not only think
about the applied forces such as the loads or self weight, but also the reactionary forces. There
must be an equilibrium of all the forces combined.
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The best way to think about FBDs is to draw them out. What is helpful about the
technique is that the rules apply to any object or component, regardless of the size.
Therefore, you can take a big problem and break it into smaller more manageable pieces.
Figure 4.3.1 illustrates a pair of pliers being used to grip a pin.
The pin is being pulled from a mechanism. The question that could arise here is whether
sufficient force is being used to pull out the pin, and whether the pliers might slip off
due to inadequate friction.
First, let us create a FBD of the overall system, in order to determine the reaction at the
pin. It is helpful to consider the applied load as orthogonal components:
H = 25# * Sin40° = 16.1#
The reactions at point A are H, V and M. To compute these reactions we can consider
the moments about point A. They should all add up to zero. If we assign counter
clockwise as positive and write the equation for the moment at A:
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In a similar way, we will find that V=0. We are finally left with H, which we see by
must be 2 * 19.2# = 38.4# to the left. Can the pliers pull out the pin? Well, that depends
on how well it's jammed in there.
Now suppose that the pin does not move. Will the pliers slip off? Slippage would occur
if the sliding force exceeded the maximum available frictional force. The applied sliding
force is 16.1# per side. The frictional force must be calculated from the normal force.
To determine the normal force, let us construct another FBD (Figure 4.3.3). This time,
we will use only one half of the pliers (that is where the pair comes from). The bolt that
would normally serve to hold the two parts together acts as a pin or hinge, providing
support to horizontal and vertical forces, but not movement.
Let's assume that the 16.1# force on the left end of the diagram is in line with point B.
Now we will take moments about point B to determine N.
With the normal force known, the maximum friction force can be calculated. If you look
in reference for the coefficient of static friction of metal on metal, you will find a value
of about 0.5. So, in this case, the maximum sliding force that can be applied before
slippage starts is:
0.5 * 63.0# = 31.5#
Since the applied force is only 16.1#, the pliers will hold tight. As a matter of fact, the
pulling force could be nearly doubled before slippage occurs.
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For such a brace, a major concern would be the amount of force applied at the ends. In a
real world situation, a brace could buckle under too great a load. We would therefore
want to determine the load in the member and compare it to the allowable capacity.
Recall that a pin support can resist forces in two directions, but not a moment. With this
in mind, consider what loads could be transmitted at the ends of the brace. Figure 4.4.2
illustrates these unknown forces.
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Because Figure 4.4.2 is actually a free body diagram, the vertical forces are equal and
opposite as well as the horizontal forces:
Also, because the pinned ends do not provide any moment resistance, taking moments
about C leads to:
Notice how the tangent ratio applies to the force components, just the same as if it were
to apply to the opposite and adjacent sides of the triangle. This means that the triangle
formed by V and H at B, or V and H at C is the same as the triangle formed by the
height and width of the diagonal brace. The direction of the force resultant coincides
with the direction of the member axis, or vector BC, to be exact. These new resultant
forces are shown in Figure 4.4.3. These two forces demonstrate the meaning of a two
force member.
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With the orientation of the forces at pin B, we can now procede to evaluate the
remaining structure.
Consider the free body diagram in Figure 4.4.4. For convenience, the reactions at A and
B are illustrated as orthogonal components. Notice how if H at B points to the right (it
has to if the brace is in compression), then H at A must point to the left for balance.
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We know the relationship between V at B and H at B from the geometery of the brace.
The load on the platform from the pin at B is equal and opposite to the load on the brace
from the pin. Figure 4.4.5 illustrates this conundrum:
With all the information in Figure 4.4.5, we can easily compute the force in the brace
member. We could use the SRSS method or trigonometery.
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Internal Hinges
In the two force member example of the previous section, a joint had to be analyzed.
This joint was an internal hinge which acted as a pin connection between two members.
An understanding of the load distribution in the structure hinges (excuse the pun!) on
knowing the magnitude of the forces transmitted through the connection.
Sometimes statics problems require the analysis of member forces. While other
instances ask for the magnitude of loads at a support. In any event, being able to
determine forces at a hinge is important. Another way to think of the problem is to
consider a hinge as a possible boundry condition that is equal and opposite (balanced)
for all the members at the joint.
A detailed example will be useful in demonstrating the concept of the free body
diagram, two force members, and the internal hinge.
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To analyze the mechanism, we can break it into parts. Component CBD is illustrated in
Figure 4.5.2 as a free body diagram. The 50 lbs force is shown in its orthogonal
component form to assist in the calculations.
Equilibrium Rules! Cried the rebels as they crashed through the gates of the castle. And
if that is the case, the sum of the moments about point D equals zero.
Now, recalling that positive means H at D points to the right tells that:
We still do not know what V at H and V at D equal. We can, however, put the free body
diagram of component AB to work, as is done in Figure 4.5.3.
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Now, the only unknown left is V at D, which can be found by the vertical equilibrium of
member CBD. Treating V at D as a positive upward force:
All the forces were found for the pin at B and the reactions at A and D by using
equilibrium equations. Because the forces were originally unknown, a direction had to
be assumed. If the unknown came out of the equations as a positive, the assumed
direction was correct. If it came out as a negitive, the opposite direction was correct.
The best advise for doing these types of analysis problems is to divide the components
up at the connections or boundries. The smaller problems then make it easier to solve the
multitude of unknowns one at a time.
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Another possible mode in which instability can occur is due to member rotation. If it is
possible for all the members to rotate with a common arc direction or common center of
rotation (instantaneous center), then the structure can move at the slightest push. Figure
4.6.2 illustrates a few examples of how rotational instability can occur. Notice how the
arc movement is common at the joint in the left figure, or at the instant center in the
right figure.
The two examples in Figure 4.6.2 can be modified to prevent instability. To do this, we
can reorient the members or the roller planes, so that the common arc path or instant
center no longer exists. These stable structures are shown in Figure 4.6.3
Determinacy comes into play when dealing with stability. An indeterminate structure is
usually very stable. By its very nature, indeterminacy means that one or more supports
can be taken away and the structure will remain intact. Examples of three indeterminate
structures are shown in Figure 4.6.4. Imagine taking one support away to make them
determinate.
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Determine the reation forces at supports A and C. Consider a force up and to the right as
positive.
A beam is used to support a pile of bricks. You must figure out the tension in cable AB, and the
reactions at supports B and C. Draw out your solution in a sketch
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a.
b.
c.
d.
Stability of a Component
A machine is to be built with an internal component that has three supporting bearings. Is this
component stable? If not, what would a stable design look like?
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a. Stable
b.
c.
A small structure is composed of a cantilever bar and two linkages, held together by pins. A
single load is applied at point B.
Determine the force in member CD and the reactions at joint E. Tension is positive in CD, and
the reactions at E are positive for upward force, a force to the right, and a counter clockwise
moment.
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A shaft is used to support two pulleys. One rope is pulled in a horizontal direction while the
other is pulled downward. The size of the pulleys and the magnitude of the forces are such that
the moment about the shaft is zero.
Determine the reactions at supports A and D. Report the forces in terms of the sign convention
shown in the problem sketch.
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h.
i. When setting up a FBD to solve for V and M, it is best to first compute the support
reactions. An example that uses support reactions is shown in Figure 5.1.2.
j.
k. Once a FBD is properly set-up, a calculation can be made to determine the value of the
forces on the cut face. Looking at the two previous examples, Figures 5.1.1 and 5.1.2,
you can see how the X value is important in calculating the value of the moment M,
and/or the value of the shear. These calculations are based of the equilibrium of the FBD.
The process of drawing the free body diagram of a beam leads to the unknown forces at a cut
face. It also creates an additional variable, X, that is the location of the cut face. With X as a
independent (controllable) variable, an equation can be written for V and for M. Hence, V
and M become functions of X.
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m. Two major items must be considered before writing equations for V and M. The first is
that equations may not be necessarily be constant over the length of a beam; this is
depends on the discontinuities. The second consideration is the sign convention. There
must be a consistent understanding of what is treated as a positive and negitive shear and
moment.
n. Figure 5.2.1 illustrates the sign convention for shear and moment applied to an end or a
cut face of a beam.
o.
p. When solving for an unknown shear or moment, it is good to draw the V and M forces as
positive, as shown in Figures 5.2.1 and 5.2.2. If the solution results in a positive V or M,
then you know the assumed direction is correct. If the solution is negitive, then simply
reverse the direction of either V or M, which ever is appropriate.
q. Figure 5.2.2 illustrates V and M equations derived from a beam example. In this example,
the V and M are written in terms of X. Since there is a discontinuity in load in beam
ABC, at B, two sets of equations must be written.
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s. You can see in Figure 5.2.2 that the equation for V and M can vary, depending on what
point of the beam is being considered. As a general rule, a new set of equations have to
be written for V and M when ever X extends beyond a new external load or reaction.
t.
You may know that a function of X can be plotted, especially a function that lies in a
plane. The same idea can be applied towards the equations for V and M. One way of
doing this is to compute the values of V and M for various values of X, then plot the
points on a graph. If the points were then connected by lines, the shear and bending
moment diagrams would appear. Figure 5.3.1 illustrates the construction of a V and M
diagram by connecting the dots.
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v.
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w. The plot of the V and M diagrams have several simple but important items to note. First
off, the diagrams are clearly defined with a description of what they are, what constitutes
positive and negative, the important magnitudes, and the units being written. Without
these items, ambiguity could occur later in the calculation process.
x. The second interesting thing to note is that the load is a constant 5kN/m downward along
the beam. Knowing this, consider the relationship between the following 3 equations:
y.
z. Do you see how shear is the integral of load, and moment is the integral of shear. The 10
creeped in as a boundary condition or constant of integration, V = 10, when x = 0.
aa. The generation of V and M diagrams by the connect the dot method works, but it is very
slow. It may help in the very begining when you are just starting out. The real tool to use
here, though, is the knowledge of the fact that V and M are integrals of the load, or
conversely, L and V are differentials of the moment:
bb.
cc. The change in sign occurs because a downward load is usually considered as positive.
Yet, a downward load on the end of a beam would cause a negitive shear:
dd.
ee. The calculus based relationship between L, V and M can be used in two different ways.
One method would utilize the integrals of the function to determine V and M. This
integration of equations is almost like what was done in Section 5.2. Indeed, knowing
that M is the integral of V would help in checking the solution.
ff. Another approach which benefits from calculus is the visual integration of load, shear and
moment diagrams. Visual integration is neat because an analysis can be made in seconds
rather than minutes. After several dozen practices, the method becomes almost intuituve.
Figure 5.3.4 illustrates several visual integration examples, with brief explanations
discussing the background theory.
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hh. The reason why V and M diagrams need to be drawn is because this is how many
member fail, regardless of whether these members are part of a bridge, fuel pump, or
computer power supply. One of the principal tenets of engineering is to understand why
things occur and anticipate or preclude failure. V and M diagrams provide such
information.
ii. The drawing of V and M diagrams start with the correct calculation of beams reactions.
These calculations themselves depend on proper modeling of loads and boundary
conditions. In a way, construction of V and M diagrams embody all the theory and
methods discussed in these past five chapters. It would be fitting to end this tutorial with
the statement:
jj.
Calculate the reactions and draw the shear and bending moment diagrams for the given partially
loaded span.
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a.
b.
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c.
d.
Overhanging Cantilever
Calculate the reactions and draw the shear and bending moment diagrams for the given
overhanging cantilever beam.
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a.
b.
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c.
d.
Calculate the support reactions and draw the shear and bending moment diagrams for the given
beam.
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a.
b.
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c.
d.
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Write the shear and bending moment equations for segment AB and BC. Use an x that is
measured relative to the left end of the beam.
a.
b.
c.
Determine the reactions and draw the shear and bending moment diagrams for the given beam.
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a.
b.
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c.
d.
In many statics problems, you must be able to quickly and efficiently create vectors in the
Cartesian plane. Luckily, you can accomplish your Cartesian vector creations easily with the
handy vector formulas in this list:
Force vectors and distance vectors are the most basic vectors that you deal with.
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Solving statics problems can be complicated; each problem requires a list of items to account for
and equations to create and solve. Solve statics problems with ease by using this checklist:
In addition to dimensions and angles, you must include four major categories of items on
a properly constructed free-body diagram:
Support reactions
Self weight
3. To solve for internal forces, identify the type of structure and write your equilibrium
equations.
After you identify the type of structure, you then know which technique to use to help
you write your equilibrium equations:
Trusses/axial members: Members are loaded with internal axial forces only. To solve,
you can use the method of joints or the method of sections.
Beams (bending members): Members are loaded with internal axial forces, shear forces,
and moments. To solve, cut the member at the desired location, draw a new free-body
diagram of the cut section, and write equilibrium equations.
Frames/machines: Members are loaded with internal axial forces, shear forces, and
moments. To solve, use the blow-it-all-apart approach to break the structure into smaller
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pieces. Look for internal hinges as common places to separate your structure and draw
free-body diagrams to solve for the connecting pin forces.
Cable structures: Members are constructed from axially loaded cables. Identify the type
of cable loading (concentrated, parabolic/uniform, or catenary). Compute the cable
tension at the location of known maximum sag (or vice versa).
Submerged surfaces: Members are subjected to fluid pressure. To solve, draw a free-
body diagram of the hydrostatic pressure diagram which is zero at the fluid surface and
linearly increases with depth. Include the weight of the fluid on objects with non-vertical
faces.
The centroid or center of area of a geometric region is the geometric center of an object's shape.
Centroid calculations are very common in statics, whether you're calculating the location of a
distributed load's resultant or determining an object's center of mass. To compute the center of
area of a region (or distributed load), you can compute the x-coordinate (and the other
coordinates similarly) from the following equations:
For discrete regions: You can break discrete regions into simple shapes such as
triangles, rectangles, circles, and so on.
For discrete shapes, creating a simple table such as the one that follows for each
coordinate can be useful. You can create as many rows as you need for as many regions
as you have.
xi Ai xiAi
For continuous regions: Continuous regions are usually defined by more complex
boundaries, so you must define them with mathematical equations such as the one that
follows:
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Shear and moment diagrams are a statics tool that engineers create to determine the internal
shear force and moments at all locations within an object. Start by locating the critical points and
then sketching the shear diagram.
o If the load area is positive, the change in shear is positive. If the shear area is
positive, the change in moment is positive.
In many statics problems, you must be able to quickly and efficiently create vectors in the
Cartesian plane. Luckily, you can accomplish your Cartesian vector creations easily with the
handy vector formulas in this list:
Force vectors and distance vectors are the most basic vectors that you deal with.
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In statics, moments are effects (of a force) that cause rotation. When computing equilibrium, you
must be able to calculate a moment for every force on your free-body diagram. To determine a
force's moment, you use one of two different calculations, as you can see in the following list.
Scalar calculation (for two dimensions): To calculate the moment about a Point O in
scalar calculations,you need the magnitude of the force and the perpendicular distance
from Point O to the line of action of the Force F.
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Vector calculation (for two or three dimensions): To compute the moment vector
about a Point O in vector calculations, you must determine the Force F in Cartesian
vector form and the position vector from Point O to the line of action of the Force F.
Forces
Forces acting on objects are vectors that are characterized by not only a magnitude (e.g. pounds force or
Newtons) but also a direction. A force vector F (vectors are usually noted by a boldface letter) can be
broken down into its components in the x, y and z directions in whatever coordinate system you‟ve drawn:
Where Fx, Fy and Fz are the magnitudes of the forces in the x, y and z directions and i, j and k are the unit
vectors in the x, y and z directions (i.e. vectors whose directions are aligned with the x, y and z coordinates and
whose magnitudes are exactly 1 (no units)).
Forces can also be expressed in terms of the magnitude = (Fx2 + Fy2 + Fz2)1/2 and direction relative to the
positive x-axis (= tan-1(Fy/Fx) in a 2-dimensional system). Note that the tan-1(Fy/Fx) function gives you an
angle between +90˚ and -90˚ whereas sometimes the resulting force is between +90˚ and +180˚ or between -
90˚ and -180˚; in these cases you‟ll have to examine the resulting force and add or subtract 180˚ from the
force to get the right direction.
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Moments of forces
Ropes are a special case of a structure that can exert a force along the line connecting the two ends of the
structure as in the above example, but cannot exert any force perpendicular to that line. Ends with pins,
bearings, etc. also fit into this category. But most structural elements can also exert a force perpendicular to
the line. This is called the moment of a force, which is the same thing as torque. Usually the term torque is
reserved for the forces on rotating, not stationary, shafts, but there is no real difference between a moment
and a torque.
The distinguishing feature of the moment of a force is that it depends not only on the vector force itself (Fi)
but also the distance (di) (I like to call it the moment arm) from the anchor point at which it acts. If you want
to loosen a stuck bolt, you want to apply whatever force your arm is capable of providing over the longest
possible di. The line through the force Fi is called the line of action. The moment arm is the distance (di again)
between the line of action and a line parallel to the line of action that passes through the anchor point. Then
the moment of force (Mi) is defined as
Mi = Fidi Equation 2
Where Fi is the magnitude of the vector F. Note that the units of Mo is force x length, e.g. ft-lbf or N-m.
This is the same as the unit of energy, but the two have nothing in common – is just coincidence. So one
could report a moment of force in units of Joules, but this is unacceptable practice – use N-m, not J.
If the two lines are co-linear (as in a rope) then there is no moment. Note also that we have to assign a sign
A
d
Line of action
Force Fi
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to Mi. Typically we will define a clockwise moment as positive and counterclockwise as negative.
In order to have equilibrium of an object, the sum of all the forces AND the moments of the forces must be
zero. In other words, there are two ways that a 2-dimensional object can translate (in the x and y directions)
and one way that in can rotate (with the axis of rotation perpendicular to the x-y plane.) So there are 3
equations that must be satisfied in order to have equilibrium, namely:
n n n
Fx, i 0; Fy, i 0; Mi 0 Equation 3
i 1 i 1 i 1
Note that the moment of forces must be zero regardless of the choice of the origin (i.e. not just at the center of
mass). So one can take the origin to be wherever it is convenient (e.g. make the moment of one of the forces
= 0.) Consider the very simple set of forces below:
200 lbf
0.707 ft 0.707 ft
A C
0.5 ft 0.5 ft
45ū 1 ft 45ū
B 0.5 ft
0.707 ft 0.707 ft
141.4 lbf 141.4 lbf
D
Figure 2. Force diagram showing different ways of computing moments
Because of the symmetry, it is easy to see that this set of forces constitutes an equilibrium condition. When
taking moments about point „B‟ we have:
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But how do we know to take moments about point B? We don‟t. But notice that if we take moments about
point „A‟ then the force balances remain the same and
The same applies if we take moments about point „C‟, or a point along the line ABC, or even a point NOT
along the line ABC. For example, taking moments about point „D‟,
MD = -141.4 lbf * (0.707 ft + 0.707 ft) + 200 lbf * 0.5 ft +141.4 lbf * 0.707 ft = 0
The location about which to take the moment can be chosen to make the problem as simple as possible, e.g.
to make some of the moments of forces = 0.
Example of “why didn’t the book just say that…?” The state of equilibrium merely requires that 3 constraint
equations are required. There is nothing in particular that requires there be 2 force and 1 moment constraint
equations. So one could have 1 force and 2 moment constraint equations:
n n n
Fx,i 0; M i, A 0; M j,B 0 Equation 4
i 1 i 1 j 1
where the coordinate direction x can be chosen to be in any direction, and moments are taken about 2
separate points A and B. Or one could even have 3 moment equations:
n n n
M i, A 0; M j,B 0; M j,C 0 Equation 5
i 1 j 1 j 1
Also, there is no reason to restrict the x and y coordinates to the horizontal and vertical directions. They can
be (for example) parallel and perpendicular to an inclined surface if that appears in the problem. In fact, the x
and y axes don‟t even have to be perpendicular to each other, as long as they are not parallel to each other, in
which case Fx = 0 and Fy = 0 would not be independent equations.
This is all fine and well for a two-dimensional (planar) situation, what about 1D or 3D? For 1D there is only
one direction that the object can move linearly and no way in which it can rotate. For 3D, there are three
directions it can move linearly and three axes about which it can rotate. Table 1 summarizes these situations.
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1 1 0 1
2 2 1 3
3 3 3 6
Table 1. Number of force and moment balance equations required for static equilibrium as
a function of the dimensionality of the system. (But note that, as just described, moment
equations can be substituted for force balances.)
A free body diagram is a diagram showing all the forces and moments of forces acting on an object. We
distinguish between two types of objects:
1. Particles that have no spatial extent and thus have no moment arm (d). An example of this would be
a satellite orbiting the earth because the spatial extent of the satellite is very small compared to the
distance from the earth to the satellite or the radius of the earth. Particles do not have moments of
forces and thus do not rotate in response to a force.
2. Rigid bodies that have a finite dimension and thus has a moment arm (d) associated with each applied
force.Rigid bodies have moments of forces and thus can rotate in response to a force.
There are several types of forces that act on particles or rigid bodies:
1. Rope, cable, etc.– Force (tension) must be along line of action; no moment (1 unknown force)
2. Rollers, frictionless surface – Force must be perpendicular to the surface; no moment (1 unknown
force)
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3. Frictionless pin or hinge – Force has components both parallel and perpendicular to the line of
action; no moment (2 unknown forces) (note that the coordinate system does not need to be parallel
and perpendicular to the bar)
4. Fixed support – Force has components both parallel and perpendicular to line of action plus a
moment of force (2 unknown forces, 1 unknown moment)
5. Contact friction – Force has components both parallel (F) and perpendicular (N) to surface, which
are related by F = µN, where µ is the coefficient of friction, which is usually assigned separate values for
static (no sliding) (µs) and dynamic (sliding) (µd) friction, with the latter being lower. (2 unknown
forces coupled by the relationF = µN). Most dry materials have friction coefficients between 0.3 and
0.6 but Teflon, for example, can have a coefficient as low as 0.04. Rubber (e.g. tires) in contact with
other surfaces (e.g. asphalt) can yield friction coefficients of almost 2.
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N F = µN
Special note: while ropes, rollers and pins do not exert a moment at the point of contact, you can still sum
up the moments acting on the free body at that point of contact. In other words, MA = 0 can be used
even if point A is a contact point with a rope, roller or pin joint, and all of the other moments about point A
(magnitude of force x distance from A to the line of action of that force) are still non-zero.
Of course, there is no guarantee that the number of force and moment balance equations will be equal to the
number of unknowns. For example, in a 2D problem, a beam supported by one pinned end and one roller
end has 3 unknown forces and 3 equations of static equilibrium. However, if both ends are pinned, there are
4 unknown forces but still only 3 equations of static equilbrium. Such a system is called statically indeterminate
and requires additional information beyond the equations of statics (e.g. material stresses and strains,
discussed in the next chapter) to determine the forces.
Examples
1. Draw a free body diagram – a free body must be a rigid object, i.e. one that cannot bend in response
to applied forces
2. Draw all of the forces acting on the free body. Is the number of unknown forces equal to the total
number of independent constraint equations shown in Table 1 (far right colum)? If not, statics can‟t
help you.
3. Decide on a coordinate system. If the primary direction of forces is parallel and perpendicular to an
inclined plane, usually it‟s most convenient to have the x and y coordinates parallel and perpendicular
to the plane, as in the cart and sliding block examples below.
4. Decide on a set of constraint equations. As mentioned above, this can be any combination of force
and moment balances that add up to the number of degrees of freedom of the system (Table 1).
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5. Decide on the locations about which to perform moment constraint equations. Generally you
should make this where the lines of action of two or more forces intersect because this will minimize
the number of unknowns in your resulting equation.
6. Write down the force and moment constraint equations. If you‟ve made good choices in steps 2 – 5,
the resulting equations will be “easy” to solve.
7. Solve these “easy” equations.
h.
i. We could also consider this beam as having a distributed load of 2 kilo Newtons per
meter (Figure 3.2.2).
j.
k. The magnitude of the load is 6kN downward, or 2kN/m times 3m. Notice how writing
out the units in the calculation gives a clearer indicaton as to whether we are dealing with
a concentrated load or a distributed load.
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l. Now, where is the centroid of the distributed load? Because the load is rectangular in
shape, it may be obvious to you that the centroid lies at the middle. This is correct, but let
us approach the problem from a more general perspective.
m. Consider the idea that the moment caused by the concentrated force about the left end of
the beam is the same as the moment caused by the distributed load about the left end.
Now, you may ask "I can see multiplying 6kN times 1.5m to get the first moment, but
what length do I multiply 2kN/m by to get the second moment?" Well, this is where
calculus comes in.
n. Imagine breaking the distributed load up into small differential widths of dX. Each
differential has a small force, 2kN/m * dX, and a location relative to the left end of the
beam, X. Figure 3.2.3 illustrates these details.
o.
p. The total moment effect is the summation or integration of all the differentials ranging
from X=0m to X=3m. This moment would be equivalent to the concentrated load times
its centroidal distance.
q.
r. We just spent the past few paragraphs proving that the centroid of the rectangular load
sits at the middle... big deal. But, the approach can be applied to any type of distributed
load. Especially ones where the centroid is not obvious.
s. By the way, how can we generally calculate the magnitude of the load? We do so by
integrating the load over the length or area where it is applied.
t. Let us look at a more complex load shape and determine the magnitude and location of
the equivalent concentrated load. Figure 3.2.4 illustrates a nonlinear load function applied
to a beam. Such a load might occur on a flat roof building during a snow storm. Drifting
can occur at locations where the roof steps up or down in elevation. Do you remember all
the roof collapses in the north-east during the winter of '93-'94. Those collapses were
caused by drifts such as the one we have here, plus the occasional rain that made the
snow extra heavy.
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u.
v. Our objective here is to determine the total force due to the load and the location of the
centroid.
w. First, determine the total load by integrating the differential over the length of the beam
(Figure 3.2.5).
x.
y. Now that we know the total force, the centroid can be computed using the equivalent
moment concept:
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z.
aa. With the total force and location known, the support reactions and the internal beam
forces can be calculated (see next two chapters). Figure 3.2.6 illustrates the two
equivalent systems.
bb.
cc. Be forwarned, the equivalency noted here is only in terms of force and moment at the
supports. As far as the interior of the beam is concerned, the applied load has a real
distributed shape, not a single point location. This distinction will be very important in
Chapter 5.
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The equivalent force and centroid problem has many guises. In some instances, it is not force
that is being integrated, but volume or area. For instance, suppose that we have bar with a
triangular cross section that is under tension. If the bar is long (>> 4 times the width), we can
assume that the tension is uniform throughout the cross section, mid-length along the member.
This is known as a uniform stress distribution (Figure 3.3.1).
A question might arrise as to what the magnitude of the force is and where it ia located. Such
answers would be critical in the design of structural or machine components.
The force in this case is equal to the uniform stress multiplied by the cross section area. The area
can be found by integrating the cross section. The location of the force coincides with the
centroid of the shape. Figure 3.3.2 illustrates the geometry of the cross section in terms of X and
Y. Figure 3.3.2 also shows the computation of the area and centroid by integration.
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Having computed the area, it can be said that the tension force is 3in^2 * 100 #/in^2 , or 300
pounds. The idealized location of the force is at the third point of the triangle.
Now, if you know your geometry (you are in college after all) you would have known the area
and center of gravity of the triangle from the begining. This could have eliminated the calculus in
Figure 3.3.2. The illustration of the calculus is helpful for understanding the calculations
necessary for more complex areas and functions. A table of common shapes, centroids and areas
can also be helpful. Figure 3.3.3 illustrates such a table.
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Multiple Shapes
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It is interesting to think how we started this chapter with a beam that carried a pile of bricks and
ended up with the area and centroid of wierd (vernacular term) shapes. This is because the two
problems are essentially the same. Another common form of this problem is the area and
centroid of a built up shape. That is, a shape composed of multiple elements, both positive and
negitive in nature.
Consider the shape in Figure 3.4.1. It might represent an angle iron used in bridge construction.
Just as with the triangular tension bar, the magnitude and location of the force in the angle will
depend on the stress level, area and centroid.
Since the angle is composed of two rectangular shapes, a descrete method can be used to
compute the area and centroid location.
Figure 3.4.2 illustrates the descrete computation of the area and centroid location. Note how the
tabular method is used to organize the data.
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Sometimes, a shape may contain holes, such as an extruded aluminum bar. In such an instance,
the area and centroid can be computed by treating the shape as a positive solid and the holes as
negitives. Figure 3.4.3 illustrates a bar containing two holes along the axis. The section
properties are computed using the tabular method.
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Determine the cross sectional area and centroid location of the structural shape. Note that the
centroidal location need not lie on the shape. If the tension stress was 5N/mm^2 , what would the
total force in the member be?
a.
b.
c.
d. None of the Above
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UNIT 4
DYNAMICS OF PARTICLES
Displacements, Velocity and acceleration, their relationship – Relative motion – Curvilinear
motion – Newton’s law – Work Energy Equation of particles – Impulse and Momentum –
Impact of elastic bodies.
Ropes
Two tugboats, the Monitor and the Merrimac, are pulling a Peace Barge due west up Chesapeake Bay toward
Washington DC. The Monitor‟s tow rope is at an angle of 53 degrees north of due west with a tension of
4000 lbf. The Merrimac‟s tow rope is at an angle of 34 degrees south of due west but their scale attached to
the rope is broken so the tension is unknown to the crew.
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Monitor y
4000 lb
53ū
x
34ū Barge
Merrimac ??? lb
(b) What is the tension trying to break the Peace Barge (i.e. in the north-south direction)?
(c) What is the force pulling the Peace Barge up Chesapeake Bay?
(d) Express the force on the Merrimac in polar coordinates (resultant force and direction, with 0˚ being due
east, as is customary)
Define x as positive in the easterly direction, y as positive in the northerly direction. In order for the
Barge to travel due west, the northerly pull by the Monitor and the Southerly pull by the Merrimac have
to be equal, or in other words the resultant force in the y direction, Ry, must be zero. The northerly pull
by the Monitor is 4000 sin(53˚) = 3195 lbf. In order for this to equal the southerly pull of the Merrimac,
we require FMerrimacsin(34˚) = 3195 lbf, thus FMerrimac = 5713 lbf.
This is just the north/south force just computed, 3195 lbf
The force exerted by the Monitor is 4000 cos(53˚) = 2407 lbf. The force exerted by the Merrimac is 5713
cos(34˚) = 4736 lbf. The resultant is Rx = 7143 lbf.
The magnitude of the force is 5713 lbf as just computed. The angle is -180˚ - 34˚ = -146˚.
Rollers
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A mining car lies on a steep track 25˚ from vertical. The weight of the car and its ore is 5500 lb. The center
of gravity is 30 inches above the track, and located half way between its frictionless wheels, which are 50 inches
apart. The car is pulled by a cable attached to the car at a point 24 inches above the track.
(a) What are the forces (Fy,A, Fy,B) on each set of wheels?
(b) What is the tension (T) in the cable?
25ū A
Fy,A
25Ó 6Ó
24Ó 5500 lbf
25Ó
B
Fy,B
(a) Take moments about points A and B as shown in the figure. These points have been chosen so that the
moment of two of the forces (due to the cable tension and through one set of wheels) are zero in each case.
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UNIT 5
FRICTION AND ELEMENTS OF RIGID BODY
DYNAMICS
Frictional force – Laws of Coloumb friction – simple contact friction – Rolling resistance – Belt
friction. Translation and Rotation of Rigid Bodies – Velocity and acceleration – General Plane
motion.
A 100lbf acts on a 300 lbf block placed on an inclined plane with a 3:4 slope. The coefficients of friction
between the block and the plane are µs = 0.25 and µd = 0.20.
b) If the block is not in equilibrium (i.e. it‟s sliding), find the net force on the block
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300 lbf
x
N F = µN
100 lbf
5 3
Which is more than the maximum available friction force, so the block will slide down the plane.
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so the net force acting on the block in the x direction (not zero since the block is not at equilbrium) is
what does this mean? lbf/lbm has units of force/mass, so it is an acceleration. But how to convert to
something useful like ft/sec2? Multiply by 1 in the funny form of gc = 1 = 32.174 lbm ft / lbf sec2, of
course!
acceleration = (-32 lbf/300 lbm) (32 lbmft / lbf sec2) = -3.43 ft/sec2
The negative sign indicates the acceleration is in the –x direction, i.e. down the slope of course.
A good function test is that the acceleration has to be less than 1 gearth, which is what you would get if you
dropped the block vertically in a frictionless environment. Obviously a block sliding down a slope (not
vertical) with friction and with an external force acting up the slope must have a smaller acceleration.
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A car of weight W is equipped with rubber tires with coefficient of static friction µs. Unlike the mining cart
example, there is no cable. The car is on a ramp of angle with respect to horizontal. The center of gravity
of the vehicle is a distance “c” above the ramp, a distance “a” behind the front wheels, and a distance “b” in
(a) What is the minimum µs required to keep the car from sliding down the ramp?
The unknowns are the resulting forces at the wheels (Fy,A and Fy,B) and the coefficient of friction µs.
Taking Fx = 0, Fy = 0 and MA = 0 yields, respectively,
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1) For = 0, µs = tan( ) = 0 (no friction required to keep the car from sliding on a level road)
2) As increases, the friction coefficient µs required to keep the car from sliding increases
3) For = 0, Fy,A = (b/(a+b)) and Fy,B = (a/(a+b)) (more weight on the wheels closer to the
center of gravity
4) Because of the – sign on the 2nd term in the numerator of Fy,A (-c sin( )) and the + sign in
the 2nd term in the numerator of Fy,B (+c sin( )), as increases, there is a transfer of weight
from the front wheels to the rear wheels.
In which case, the second equation could have been subtracted from the third to obtain:
which is the same as the first equation. So the three equations are not independent of each other,
and we can‟t solve the system. What‟s wrong? The coefficient of friction µs doesn’t appear in the set of
equations Fx = 0, MA and MB = 0. We need to have each of the three unknownsFy,A, Fy,B and µs in
at least one of the three equations. The set Fx = 0, MA and MB = 0 doesn‟t satisfy that criterion.
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(b) At what angle will the car tip over backwards, assuming that it doesn‟t start sliding down the ramp at a
smaller angle due to low µs?
This will occur when Fy,A = 0, i.e. when sin( )/cos( ) = tan( ) = b/c. This is reasonable because the
tip-over angle should increase when c is made larger (center of gravity closer to the ground) or b
made smaller (center of gravity shifted forward). Notice also that it doesn‟t depend on µs, that is, as
long as it doesn‟t slide due to low µs, the tip-over angle only depends on the force balance.
For what it‟s worth, also note that the tip-over angle equals the sliding angle when tan( ) = µs = b/c.
Since generally µs<< 1, Except for a very top-heavy (large c) or rear-weight-shifted (small b) vehicles,
the vehicle will slide down the ramp before it flips over backwards.
Pinned joint
12 in B
4 in 6 in
C
30ū
4 cos(30) 8 cos(30)
A 100 lbf
A straight bar of negligible mass 12 inches long is pinned at its lower end (call it point A) and has a roller
attached to its upper end (call it point B) as shown in the figure. The bar is at a 30˚ angle from
horizontal. A weight of 100 lbf is hung 4 inches from the lower end (call it point C).
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a) What are the forces in the x and y directions on the pinned end? What is the force in the x direction
on the roller end?
b) Would the forces change if the roller and pinned ends were reversed?
c) What would happen if the lower end were fixed rather than pinned (upper end having the roller
again)?
a) The pinned end can sustain forces in both the x and y directions, but no moment. The roller end can
sustain a force only in the x direction, and again no moment. Summing the forces in the y direction
In other words, in the y direction the vertical force at point A must be +100 lbf since that is the only
force available to counteract the 100 lbf weight. Next, taking moments about point A (since the lines of
FA,x+ FB,x + FC,x= 0 FA,x= -FB,x - FC,x = -(-57.7 lbf) – 0 = +57.7 lbf
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Taking moments about point C just for variety (not the easiest way, since neither FA,y nor FB,y are known,
c) In this case there are 4 unknown quantities (FA,x, FA,y, MA and FB,x) but only 3 equations ( Fx = 0, Fy
= 0, M = 0) so the system is statically indeterminate. If one takes away the roller end entirely, then
obviously FA,y = 100 lbf, FA,x = 0 and MA = +(100 lbf)(4 in)(cos(30˚)) = 346.4 in lbf.
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npnn
FACULTY OF ENGINEERING
DEPARTMENT OF MECHANICAL ENGINEERING
QUESTION GRADE
TOTAL:
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NOTES:
6) Every page of the submitted work must have its margins. (Drawing the margins is
not essential, but the marginal spaces must be there!)
7) Hand written title pages will not be accepted!
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COURSE OBJECTIVES:
1. TO PROVIDE THE STUDENT WITH A CLEAR AND THOROUGH UNDERSTANDING OF THE THEORY AND
APPLICATIONS OF ENGINEERING MECHANICS, COVERING BOTH STATICS AND DYNAMICS; AND IN
THIS CONTEXT;
2. TO PROVIDE THE STUDENT WITH A THOROUGH UNDERSTANDING OF THE CONCEPT, DRAWING, AND
THE USE OF FREE-BODY DIAGRAMS (WHICH IS ABSOLUTELY ESSENTIAL WHEN SOLVING PROBLEMS
IN MECHANICS.)
3. TO PROVIDE A MEANS FOR DEVELOPING THE STUDENT’S ABILITY TO FORMULATE EQUILIBRIUM
EQUATIONS;
4. TO PROVIDE THE STUDENT WITH GOOD PROBLEM SOLVING SKILLS;
5. TO PROVIDE THE STUDENT WITH THE FUNDAMENTAL CONCEPTS AND METHODS OF STATICS AND
DYNAMICS;
6. TO ENHANCE STUDENT’S CREATIVITY AND IMAGINATION AS WELL AS KNOWLEDGE AND
SYSTEMATIC THINKING CAPABILITIES AS REQUIRED BY THE ENGINEERING DISCIPLINE.
7. TO PROVIDE THE STUDENT WITH SIMPLE DESIGN AND SAFETY IDEAS.
COURSE OUTCOMES:
1. UNDERSTAND THE BASIC PRINCIPLES OF MECHANICS AND TO APPLY THEM TO REAL-LIFE PROBLEMS
OR TO NEW SITUATIONS.
2. DRAW FREE-BODY DIAGRAMS
3. FORMULATE AND USE THE EQUILIBRIUM EQUATIONS
4. IDENTIFY THE PRINCIPLES AND EQUATIONS THAT APPLY, AND USE THEM IN SOLVING STATIC AND
DYNAMIC PROBLEMS.
5. USE THE MATHEMATICAL TOOLS AND THE STANDARD PROCEDURES AS WELL AS PERFORMING
NUMERICAL CALCULATIONS.
6. ANALYZE ONE, TWO AND SOME THREE DIMENTIONAL FORCE SYSTEMS AND TO DETERMINE THE
RESULTANT OF FORCE SYSTEMS; INCLUDING MOMENTS/COUPLES, REACTIONS, AND INTERNAL
LOADS.
7. DEVELOP EQUATIONS OF EQUILIBRIUM AND EQUATIONS OF MOTION, AND SOLVE THESE FOR
FORCES AND/OR DIFFERENT MOTION PARAMETERS, SUCH AS DISPLACEMENT, VELOCITY, AND
ACCELERATION OF PARTICLES, RIGID BODIES, AND SIMPLE MECHANICAL SYSTEMS.
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Quizzes: 20 %,
Midterm(s): 30 %,
Final Exam: 36 %.
ATTENDANCE:
Attendance to lectures and tutorials are essential and expected. Attendance shall be taken. You must be there
to participate. If you miss more than one midterm/monthly exams and/or quizes or if your attendance record is
less than 75 %, your condition may be treated as “Non-Gradable” and shall receive an “NG” grade.
PLAGIARISM
This is intentionally failing to give credit to sources used in writing regardless of whether they are published or
unpublished. Plagiarism (which also includes any kind of cheating in exams) is a disciplinary offence and will be
dealt with accordingly.
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OTHER NOTES:
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Second Semester
GE 131 - ENGINEERING MECHANICS
1. Determine the resultant of the three forces F1 = 2.0i + 3.3j – 2.6k; F2 = - i + 5.2j – 2.9k; and F3 =
8.3i – 6.6j + 5.8k, which are concurrent at the point (2, 2, -5.). The forces are in newtons and the
distances are in metres.
2. A force F = (6N)i – (3Nj – (4N)k is acting at a point P whose position vector from the origin ‘O’ of
the coordinate axes is (8 mm)i + (6 mm)j – (4 mm)k. Find the moment of the force about the
origin.
4. State the theorems of Pappus and Guldinus to find out the surface area and the volume of a
body.
6. A belt embraces an angle of 200 over the surface of a pulley of 500 mm diameter. If the tight
side tension of the belt is 2.5 kN. Find out the slack side tension of the belt. The coefficient of
friction between the belt and the pulley can be taken as 0.3.
7. The motion of a particle in defined by the relation x = t3 – 15 t2 – 20, where ‘x’ is expressed in
metres and ‘t’ in seconds. Determine the acceleration of the particle at t = 3 seconds.
8. A mass of 50 kg. has an initial velocity of 15 m/s. horizontally on a smooth surface. Determine
the value of horizontal force that will bring the mass to rest in 4 seconds.
PART – B (5 x 16 = 80 Marks)
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11. A force F acts at the origin of a coordinate system in a direction defined by the angles θx = 69.3
and θz = 57.9 . If the component of the force F along y direction is = -174N, determine.
12. (a) A load P of 3500 N is acting on the boom, which is held by a cable BC as shown in Fig.12(a). The
weight of the boom can be neglected.
(i) Draw the free body diagram of the boom. (4 Marks)
(ii) Find out the tension in cable BC. (8 Marks)
(iii) Determine the reaction at A. (4 Marks)
(OR)
(b) Three forces +20N, -10N and +30N are acting perpendicular to x z plane as shown in Fig. Q.12(b).
The lines of action of all the forces are parallel to y-axis. The coordinates of the point of action
of these forces along x and z directions are respectively (2,3), (4,2) and (7,4), all the distances
being referred in metres. Find out.
(i) The magnitude of the resultant force. (4 Marks) (ii) The
13.(a) (i) Determine the coordinates of the centroid of the shaded area shown in Fig. Q.13 (a) if the area removed is semicircular.
(4 Marks) (ii) Find the moment of inertia of the shaded area about the centroidal axes, the axes being parallel to x and y-axes.
(6 Marks) (iii) Find the product of inertia of the shaded area about the centroidal axes. .
(6 Marks)
(OR)
(b) Find the mass moment of inertia of the rectangular block shown in Fig. Q.13(b), about the x and y
axes. A cuboid of 20 mm x 20 mm x 20 mm has been removed from the rectangular block as
shown in the figure. The mass density of the material of the block is 7850 kg/m3.
14.(a) Two masses m1 and m2 are tied together by a rope parallel to the inclined plane surface, as shown
in Fig. Q.14(a). Their masses are 22.5 kg. and 14 kg. respectively. The coefficient of friction
between m1 and the plane is 0.25, while that of mass m2 and the plane is 0.5. Determine.
(i) the value of the inclination of the plane surface, θ, for which the masses will
just start sliding. (6 Marks) (ii) the
tension in the rope. (6 Marks) (iii) what will be the
friction forces at the mass surfaces. (4 Marks)
(OR)
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(b) The driver of an automobile decreases the speed at a constant rate from 72 to 48 km./hour over
a distance of 230 m. along a curve of 460 m. radius. Determine the magnitude of the total
acceleration of the automobile, after the automobile, after the automobile has traveled 150 m.
15.(a) Two steel blocks, shown in Fig. Q. 15(a), slide without friction on a horizontal surface. The
velocities of the blocks immediately before impact are as shown. If the coefficient of restitution
between the blocks is 0.75, determine (i) the velocities of the blocks after impact
and (12 Marks) (ii) the energy loss during impact.
(4 Marks)
(OR)
*******
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