assignment static

8/13/2019 assignment static

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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.html

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