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1.0INTRODUCTION
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1.1 Introduction To Static
Statics is the branch ofmechanics that is concerned with the analysis of loads
(force andtorque, or "moment")onphysical systems in static equilibrium, that is, in a
state where the relative positions of subsystems do not vary over time, or where
components and structures are at a constant velocity. When in static equilibrium, the
system is either at rest, or itscenter of mass moves at constant velocity.
ByNewton's first law, this situation implies that the net force and net torque (also
known as moment of force) on every part of the system is zero. From this constraint,
such quantities asstress orpressure can be derived. The net forces equaling zero isknown as the first condition for equilibrium, and the net torque equaling zero is
known as the second condition for equilibrium.
1.2 Objectives
Understand and define moment, determine moments of a force in 2-D and 3-D
cases.
Draw a free body diagram (FBD), apply equations of equilibrium to solve a 2-
D problem.
Define a simple truss, the forces in members of a simple truss and identify
zero-force members.
1.3 Aplication
Equilibrium O f A P article In 2-D
The particle is a model of a real body. The word "particle" does not imply that the
particle is a small body. Modelling a body as particle is equivalent to the
assumption that all forces applied on body act at the same point. This assumption
is acceptable in many practical engineering applications. The free particle and
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the constrained particle should be distinquished. The free particle (such as a planet
or a bullet) are rarely encountered in a static equilibrium problems. Most particles
are constrained. The first step when solving the equilibrium is "to free" the
particle and to sketch so called free- body diagram. To free a particle means to
isolate it from other bodies which the particle is originally joined or in touch with.
All these other bodies must be replaced by forces which they act on the particle in
question. After "freeing" the particle we have concurrent system of forces and we
solve the problem of equilibrium of this system of forces according to rules
described in section3.2.5. We usually use two component equations of
equilibrium in planar (2D) case and three component equations of equlibrium in
spatial (3D) case.
Moment
The Moment of a force is a measure of its tendency to cause a body to rotate
about a specific point or axis. This is different from the tendency for a body to
move, or translate, in the direction of the force. In order for a moment to develop,
the force must act upon the body in such a manner that the body would begin to
twist. This occurs every time a force is applied so that it does not pass through the
centroid of the body. A moment is due to a force not having an equal and
opposite force directly along it's line of action.
Imagine two people pushing on a door at the doorknob from opposite sides. If
both of them are pushing with an equal force then there is a state of equilibrium.
If one of them would suddenly jump back from the door, the push of the other
person would no longer have any opposition and the door would swing away. The
person who was still pushing on the door created a moment.
Trusses
In this chapter we focus on determination on forces internal on the structure, that
is, forces of of action and reaction between the connected members. An
engineering structures is any connected system of members built to support or
transfer forces and to safely withstand the loads applied to it. To determine the
forces internal to an engineering structure, we must dismember the structure and
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analyze separate free-body diagrams (FBD) of individual members or
combinations of members. This analysis requires careful application of Newtons
third law, which states that each action is accompanied by an equal and opposite
reaction. A framework composed of members joined at their ends to form a rigid
structure is called truss. Bridges, roof supports and others are common example
of trusses. Structural members commonly used are specially shapes which are
joined together at their ends or pins. The basic element of a plane truss is the
triangle. Three bars joined by pins at their ends, constitute a rigid frame.
Structures built from a basic triangle in the manner described are known as
simple trusses. Other than that, method of joints also one of the part of trusses.
This method for finding forces in the members of a truss consists of satisfying the
conditions of equilibrium for the forces acting on the connecting pin of each joint.
The method therefore deals with the equilibrium of concurrent forces, and only
two independent equilibrium equations are involved. Besides, the method of
sections has the basic advantage that the force in almost any desired member may
be found directly from an analysis of a section which is has cut that member.
Thus, it is not necessary to proceed with the calculation from joint to joint until
the member in question has been reached. In choosing a section of a truss, we
note that, in general, not more than three members whose forces are unknown
should be cut, since there are only three available independent equilibrium
relations.
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2.0RESEACH OFLITERATURE
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2.1 EQUILIBRIUM OF FORCE IN TWO-D
The below analysis of the forces acting upon an object in equilibrium is commonly used
to analyze situations involving objects at static equilibrium. The most common
application involves the analysis of the forces acting upon a sign that is at rest. For
example, consider the picture at the right that hangs on a wall. The picture is in a state of
equilibrium, and thus all the forces acting upon the picture must be balanced. That is, all
horizontal components must add to 0 Newton and all vertical components must add to 0
Newton. The leftward pull of cable A must balance the rightward pull of cable B and the
sum of the upward pull of cable A and cable B must balance the weight of the sign.
2.2 MOMENT
The static equilibrium of a particle is an important concept in statics. A particle is in
equilibrium only if the resultant of all forces acting on the particle is equal to zero. In a
rectangular coordinate system the equilibrium equations can be represented by three
scalar equations, where the sums of forces in all three directions are equal to zero. An
engineering application of this concept is determining the tensions of up to three cables
under load, for example the forces exerted on each cable of a hoist lifting an object or of
guy wires restraining a hot air balloon to the ground.
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2.3TRUSSESTrusses are used in a variety of applications where a lightweight yet strong structure is required.
Trusses are used extensively in bridges; buildings, particularly roofing and flooring, radio and
television towers, and space-based constructions. Space frames are extensively used in lighting
support structures. In some bridge structures, a linear truss is combined with an arch truss to
span a larger distance.
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Roof Trusses can be designed to fit our needs. Cathedral and tray ceilings, planter ledges,
and attic storage areas are just some of the features that can be incorporated into our truss
design. Roofs with a pitch or flat chord roof systems can easily be designed to fit our
needs. Each truss is individually engineered to ensure a strong, yet cost effective, roof
system.
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3.0METHODOLOGY
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3.1 Application of Equilibrium of Force in 2-D
Static equilibrium defines the state in which the sum of the forces, and torque, on each
particle of the system is zero. A particle in mechanical equilibrium is undergoing neither
linear nor rotational acceleration; however it could be translating or rotating at a constant
velocity.
Irrotational Equilibrium
The sum of the forces is equal to the mass times the acceleration. The 2nd Law tells us
that if the object or system is motionless, the acceleration is equal to zero. Therefore the
sum of the vector forces must be equal to zero.
Example:
Consider a table with four legs and a 200 kg object resting motionless in the center of the
table. What force acts upon the bottom of each table leg ?
The force upward on the legs is found by balancing these forces in the equation given by
Newton's Law. Gravity is acting downward on the 200 kg object and the 100 kg table.
Therefore, we may substitute the acceleration due to gravity on Earth which
is , for .
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Substitute in known values...
And simplifying...
The unit is equivalent to the unit of force called a Newton N. Thus when
multiplying through, the kilograms cancel out, and we may solve for ...
This means that each leg exerts a downward force of 735.75 Newtons on the floor, and
the floor simultaneously exerts the same force upward on each table leg.
Notice that it is critical that a consistent sign convention be followed throughout the entire
analysis effort. The sign convention is typically chosen in more complex problems to
ease the total amount of algebra necessary to analyze the equations in the , and
axis. We chose positive values to mean upward forces and negative values to mean
downward forces, but we could have also used the opposite, provided we wereconsistent.
Rotational Equilibrium
Lesson: In Statics the Sum of the Torques (Moments) is Equal to Zero.
The sum of all rotational forces, or torques, denoted by the capital Greek
letter tau ( ), is also zero. Commonly used units for torque areFootpounds
(ftlb)andNewtonmeters (Nm).
Newton's Second Law (applied to torques):
The sum of the torques is equal to the rotational mass or moment of inertia
(I)times the angular acceleration, denoted by the lower case Greek letter omega (
). The 2nd Law tells us that if the object or system is motionless, the angular
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acceleration is equal to zero. Therefore the sum of the vector torques must be
equal to zero.
Torques may also be calculated as forces times distances:
Example:
Consider a massless lever with two weights attached and a single massless
support:
Weight of Object 1, = 10 lb
Distance of Object 1 from fulcrum = 10 ft
Distance of support from fulcrum = 7 ft
Weight of Object 2, = 80 lb
Distance of Object 2 from fulcrum = 4 ft
What forces act upon the massless lever?
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First let's write the static torque equation for the system. Forces 1 and 2 are
actually the weights of the two objects...
Substitute in known values...
Simplify...
And solve for ...
and
therefore
Thus, a force of 60 pounds acts upward on the lever from the support in order to
balance the torques on the lever. Since the total weight on the lever is 90 pounds
and 60 of those pounds are countered by the support, the remaining 30 pounds
must act upward on the lever at the fulcrum in the lower, left corner.
Note that we didn't need to use the fulcrum as the point from which all distances
are measured, we could have chosen any point along the lever. However, as with
forces, a consistent sign convention must be used. In this case, positive may beused for clockwise (CW) torques measured from one point of view, with negative
torques used for counterclockwise (CCW) torques measured from the same POV.
The opposite system could also be used, so long as we were consistent.
Also note that the table example used previously stated that the object was in the
center of the table. If not, then balancing the torques (in two directions) would
result in more force supported by some table legs and less by others. If the object
was directly over one of the legs, for example, then it's entire weight would be
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supported by that leg, in addition to one-fourth of the table's weight. The other
legs would then only support one-fourth of the table's weight each.
3.2 Moment
Elements of a Moment
The magnitude of the moment of a force
acting about a point or axis is directly
proportinoal to the distance of the force
from the point or axis. It is defined as the
product of the force (F) and the moment arm
(d). The moment armor lever armis the
perpendicular distance between the line of
action of the force and the center of
moments.
Moment = Force x Distanceor M = (F)(d)
The Center of Momentsmay be the actual point about which the force causes
rotation. It may also be a reference point or axis about which the force may be
considered as causing rotation. It does not matter as long as a specific point is
always taken as the reference point. The latter case is much more common
situation in structural design problems.
A moment is expressed in units of foot-pounds, kip-feet, newton-meters, or
kilonewton-meters. A moment also has a sense; A clockwise rotation about the
center of moments will be considered a positive moment; while a counter-
clockwise rotation about the center of moments will be considered negative. The
most common way to express a moment is
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The example shows a wrench being applied to a nut. A 100 pound force is applied
to it at point C, the center of the nut. The force is applied at an x- distance of 12
inches from the nut. The center of moments could be point C, but could also be
points A or B or D.
Moment about C
The moment arm for calculating the moment around point C is 12 inches. The
magnitude of the moment about point C is 12 inches multiplied by the force of
100 lbs to give a total moment of 1200 inch-lbs (or 100 ft-lbs).
Moment Arm (d) = 12 inches
Magnitude (F) = 100 lbs
Moment = M = 100 lbs x 12 in. = 1200 in-lbs
Similarly, we can find the moments about any point in space.
Moment @ A B D
Moment Arm 8 inches 2 inches 0 inches
Magnitude of F 100 pounds 100 pounds 100 pounds
Total Moment 800 in- pounds 200 in- pounds 0 in- pounds
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A moment causes a rotation about a point or axis. If the moment is to be taken
about a point due to a force F, then in order for a moment to develop, the line of
action cannot pass through that point. If the line of action does go through that
point, the moment is zero because the magnitude of the moment arm is zero. Such
was the case for point D in the previous wrench poblem. The total moment was
zero because the moment arm was zero as well.
As another example, let us assume that 200 pound force is applied to the wrench
as indicated. The moment of the 200 pound force applied at C is zero because:
M = F x d = 200 lbs x 0 in = 0 in-lbs
In other words, there is no tendency for the 200 pound force to cause the wrench
to rotate the nut. One could increase the magnitude of the force until the bolt
finally broke off (shear failure).
The moment about points X, Y, and Z would also be zero because they also lie onthe line of action.
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A moment can also be considered to be the result of forces detouring from a
direct line drawn between the point of loading of a system and its supports. In this
case, the blue force is an eccentric force. In order for it to reach the base of the
column, it must make a detour through the beam. The greater the detour, the
greater the moment. The most efficient structural systems have the least amount
of detours possible. This will be discussed in more detail inLecture 37 and later
courses.
There are cases in which it is easier to calculate the moments of the componenets
of a force around a certain point than it is to calculate the moment of the force
itself. It could be that the determination of the perpendicular distance of the force
is more difficult than determining the perpendicular distance of components of
the force. The moment of several forces about a point is simply the algebraic sum
of their component moments about the same point. When adding the moments of
componenets, one must take great care to be consistant with the sense of each
moment. It is often prudent to note the sense next to the moment when
undertaking such problems.
3.3 Trusses
METHOD OF JOINTS
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We begin the analysis with any joint where at least one known load exists and
where not more than two unknown forces are present. The solution may be started
with the pin at the left hand. Its free-body diagram is shown above. The proper
directions of the forces should be evident by inspection for this sample case. The
free-body diagrams of portions of members AF and AB are also shown clearly
indicate the mechanism of the action and reaction. The member AB actually
makes contact on the left side of the pin, although the force AB is drawn from the
right side and is shown acting away from the pin. Thus, if we draw the force
arrows on the same side of the pin as the member, then tension (such as AB) will
always be indicated by an arrow away from the pin, and compression (such as
AF)will always indicated by an arrow toward the pin. The magnitude of the AF is
obtained from the equation and AB is then found from .
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Join F may be analyzed next, since it now contains only two unknowns, EF and
BF. Proceeding to next joint having no more than two unknowns, we
subsequently analyze joints B, C, E, and D in that order. Figure above shows the
free-body diagramof each joint and its corresponding force polygons, which
represents graphically the two equilibriumconditions and . The
numbers indicate the order in which the joints are analyzed. When joint D finally
reached, the reaction R2 must equilibrium with the forces in members CD and
ED, which were determined previously from the two neighbouring joints. This
requirements provide a check on the correctness of our work. Joints show the
force CE is zero when is applied.
METHOD OF SECTION
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Determine the force in the member BE, for example. An imaginary section,
indicated by the dashed line, is passed through the truss, cutting it into two parts
(figure b). This section has cut three members whose forces are initially
unknown. We can usually draw the forces with their proper senses by a visual
approximation of the equilibrium requirements. Thus, in balancing the moments
about the point B for the left-hand section, the force EF is clearly to the left,
which makes to the left, which makes it compressive, because it acts toward the
cut sectionof member EF. The load L is greater than the reaction R1, so the force
BEmust be up and to the right to supply the needed upward component for
vertical equilibrium. Force BE therefore tensile, since it acts away from the cut
section.
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4.0RESULT
OF
DISCUSSION
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5.0CONCLUSION
ADN REFRENCES
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5.1 CONCLUSION
Students will be able to determine the moment of a force about an axis using
A scalar analysis, and vector analysis. Student also can draw a free body diagram
(FBD), and, apply equations of equilibrium to solve a 2-D problem. Lastly, student
can determine the forces in members of a simple truss.
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5.2 REFRENCES
(1) http://fsinet.fsid.cvut.cz/en/u2052/node47.html
(2) http://web.mst.edu/~ide50-3/schedule/index_CH3_3Dparticle_2.html
(3) Ghazali, Mohd. Imran, 2002. Mekanik Kejuruteraan : Statik Teori, Contoh
Penyelesaian dan Masalah, Jilid 2, Unit Penerbitan Akademik, UTM.
http://fsinet.fsid.cvut.cz/en/u2052/node47.htmlhttp://web.mst.edu/~ide50-3/schedule/index_CH3_3Dparticle_2.htmlhttp://web.mst.edu/~ide50-3/schedule/index_CH3_3Dparticle_2.htmlhttp://fsinet.fsid.cvut.cz/en/u2052/node47.htmlTop Related