engg[1].mechanics study matl

7/29/2019 Engg[1].Mechanics Study Matl

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BASIC CONCEPTS:

Newtons Laws of motion

First Law of Motion: law of Inertia

An object at rest will remain at rest unless acted upon by an external and unbalanced force.An object in motion will remain in motion unless acted upon by an external and unbalancedforce.

This law is also called the law ofinertia.

The net force on an object is the vector sum of all the forces acting on the object. Newton's

first law says that if this sum is zero, the state of motion of the object does not change.

Essentially, it makes the following two points:

An object that is not moving will not move until a net force acts upon it.

An object that is in motion will not change its velocity (accelerate) until a net force acts

upon it.

Second Law of Motion: law of acceleration

The rate of change of momentum of a body is proportional to the resultant force acting on thebody and is in the same direction.

Newton's second law of motion explains how an object will change velocity if it ispushed or pulled upon.

Firstly, this law states that if you do place a force on an object, it will accelerate,i.e., change its velocity, and it will change its velocity in the direction of the force.

Secondly, this acceleration is directly proportional to the force. For example, if youare pushing on an object, causing it to accelerate, and then you push, say, threetimes harder, the acceleration will be three times greater.


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Thirdly, this acceleration is inversely proportional to the mass of the object. Forexample, if you are pushing equally on two objects, and one of the objects has fivetimes more mass than the other, it will accelerate at one fifth the acceleration of theother.

In mathematical terms, Newton's second law can be written as a differential equation:

where:

is forceis mass

is velocity

is time.

The product of the mass and velocity is the momentum of the object.

If mass of an object in question is known to be constant, this differential equation can be

rewritten (using the definition ofacceleration) as:

where:

is the acceleration.

Newton's third law: law of reciprocal actions

All forces occur in pairs, and these two forces are equal in magnitude and opposite indirection.


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This law of motion is commonly paraphrased as: "To every action force there is an equal,

but opposite, reaction force".

The concept of mass is introduced here for the first time in the words "reciprocally

proportional to the bodies" which are now traditionally added to Newton's second law as

"inversely proportional to the mass of the object

The skaters' forces on each other are equal in magnitude, and in opposite directions.

Although the forces are equal, the accelerations are not: the less massive skater will have a

greater acceleration due to Newton's second law. It is important to note that theaction/reaction pair act on different objects and do not cancel each other out. The two

forces in Newton's third law are of the same type, e.g., if the road exerts a forward

frictional force on an accelerating car's tires, then it is also a frictional force that Newton'sthird law predicts for the tires pushing backward on the road.

The third law follows mathematically from the law ofconservation of momentum.

EQUATIONS OF MOTION

The first equation of motion

An object moving with a constant acceleration a (sometimes called a uniformacceleration) starts with an initial velocity u and achieves a final velocity v in a time

oft seconds

The quantities u, v, a, and t is linked mathematically by the equation:

a=v-u/t


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Click to make v the subject of the above equation and note that

v=u+at

This is called the first equation of motion.

The second equation of motion

the fact that the acceleration is uniform between speeds u and v allows us to state

that the average speed is

The second equation of motion can be expressed as

S=ut+1/2 at2

The third equation of motion

The second equation of motion states that

S=ut+1/2 at2

and the first states that

a=v-u/t

By combining the first and second equations of motion we have been able toderive the third:

V2=u2+2as


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A. Angular Distance

Suppose the wheel mounted on an axle through its center, as shown below, is

rotated through and angle. There are three different ways in which this rotationis measured.

The two common ways to measure are with revolutions and degree units. Arevolution is defined to be on complete turn, and one complete turn is defined to be360 degrees. These two units are simply related by

1 rev = 360

Although you are probably most familiar with these two methods for measuringangular displacement, it is important that you realize that both of these units arearbitrary. There is no physical reason why there should be 360 degrees in one

revolution. We could divide a complete revolution into 10, 20, 100, or 1000 partsand define a different relationship between revolutions and degrees.

Angular Velocity and Acceleration

Definition: Angular Velocity

Angular velocity of a rigid body is the rate of change of its angular position.

Thus, if = 1at t = t1 , and = at t = t2 , then the average angular velocity of

the body over the time interval t = t2 - t1 is:

= = .

(4)

As for linear motion, the instantaneous velocity is obtained by making the timeinterval very small:


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= .

(5)

The units of angular velocity are most conveniently given in rads/sec, but can alsobe expressed in revolutions/sec or degrees/sec using the conversions givenabove.

Definition: Angular Acceleration

Angular Acceleration is the rate of change of angular velocity with time. Theaverage angular acceleration of a rigid body over a time interval t = t2 - t1 is:

+The instantaneous angular acceleration is obtained by taking a very small timeinterval:

= .

(7)

The units of angular acceleration are normally radians/sec2.

=s

r(6.1)

Where is measured in radians (rad).

Since s = 2r for one full turn of the wheel, we see that

1 rev = 360 = 2 rad

Note: s/r is simply a ratio of lengths, so, strictly speaking, it has no units. Even so,we shall say, for example, the angle is rad (radians) or 180 or rev to makeit clear how we are measuring angles.

Centrifugal and Centripetal force

Centripetal force: Real forceWhenever an object moves in a circular path we know the object is acceleratingbecause the velocity is constantly changing direction. All accelerations are causedby a net force acting on an object. In the case of an object moving in a circularpath, the net force is a special force called the centripetal force (not centrifugal!).Soa centripetal force is a center seeking force which means that the force is alwaysdirected toward the center of the circle. Without this force, an object will simplycontinue moving in straight line motion.


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Centrifugal force:Pseudo forceCentrifugal force is a virtual force. It is not really a force.

A real or "reactive" centrifugal force occurs in reaction to a centripetal accelerationacting on a mass. This centrifugal force is equal in magnitude to the centripetalforce, directed away from the center of rotation, and is exerted by the rotating

object upon the object which imposes the centripetal acceleration.

RELATION BETWEEN ANGULAR ACCELARATION ,VELOCITY WITH LINEAR

Relationship between linear and angular motion

Relationship between linear speed and angular speed

If a point P move round a circle of radius r with constant linear speed, v, (see Figure 1.7)

then the angular speed,, will be constant at

Figure 1.7: Circular Motion

Where t is the time to move from Q to P along the arc QP of the curve.

However, arc length QP is rwhen is measured in radians. Hence linear speed v is

Equation 1.9


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Substituting Equation 1.8 into Equation 1.9 leads to the this relationship for circular

motion:

Relationship between linear acceleration and angular acceleration

and

as r is constant this can be written

and as is linear accelerationa,

.

DEGREES OF FREEDOM

In mechanics, degrees of freedom (DOF) are the set of independent


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displacements that specify completely the displaced or deformed position of thebody or system. This is a fundamental concept relating to systems of movingbodies in mechanical engineering, aeronautical engineering, robotics, structuralengineering, etc.The number of DOF that a manipulator possesses is the number of independent

position variables that would have to be specified in order to locate all parts of themechanism. In other words, it refers to the number of different ways in which aassembly component can move.

WORK, ENERGY & POWER

"The WORK done on an object is the product of the average force on it and thedistance travelled in the direction of the force."

The unit of work in the modern system is thejoule J . (Very old units include thecalorie, BTU and the erg. )

GRAPHICALLY

Work has no sense of direction. We do not ascribe arrows to work or energy.

Distance is used rather than displacement in the simple definition because theforce acting may take a windy path. The total path taken which is important is thedistance rather than the displacement.

"ENERGY is the ABILITY of an object to do work for whatever reason."

PRINCIPLE OF CONSERVATION OF ENERGY; " In any closed system, the totalamount of energy remains constant regardless of any process which takes place."

GRAVITATIONAL POTENTIAL ENERGY;


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In falling through a height "h" which is in the same direction as the force, the workdone by gravity is

Work done = force. Distance = Mg.h

Thus Grav. Pot. Energy (Ep) = Mgh

This is a stored energy available to be converted into movement energy onrelease.

KINETIC ENERGY; Energy available because of the object's motion".

Consider a mass, m, which is moving with a speed, v, and does work which bringsit to rest.

The unbalanced force, F, which it exerts in doing the work is, by Newton's ThirdLaw also exerted on it , bringing it to a halt.

Funbal = ma

so, Work done = Funbal . di st = mas

( we are assuming all of this takes place in a straight line so that distance anddisplacement are essentially the same )

Using 2as = v2 - vo2

We get mas = 1/2 .mv2 = Work done

Ek = Kinetic energy = 1/2 mv2

All forms of energy can have such formulae worked out for them .

POWER

"Power is the rate of doing work or changing energy."


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P = Work Done = Energy

t t

FrictionFriction is the force that opposes the relative motion or tendency toward suchmotion of two surfaces in contact. It is not a fundamental force, as it is made up ofelectromagnetic forces between atoms. When contacting surfaces move relative toeach other, the friction between the two objects converts kinetic energy intothermal energy, orheat.

The coefficient of friction (also known as the frictional coefficient) is adimensionless scalarvalue which describes the ratio of the force of frictionbetween two bodies and the force pressing them together. The coefficient of

friction depends on the materials used -- for example, ice on steel has a lowcoefficient of friction (the two materials slide past each other easily), while rubberon pavement has a high coefficient of friction (the materials do not slide past eachother easily). Coefficients of friction range from near zero to greater than one -under good conditions.

Types of Friction:

Static friction

Static friction occurs when the two objects are not moving relative to each other(like a rock on a table). The coefficient of static friction is typically denoted as

s.

The initial force to get an object moving is often dominated by static friction. Thestatic friction is in most cases higher than the kinetic friction

Kinetic friction

Kinetic (or dynamic) friction occurs when two objects are moving relative to eachother and rub together (like a sled on the ground). The coefficient of kinetic frictionis typically denoted as k, and is usually less than the coefficient of static friction.

Rolling Friction

Rolling friction is caused primarily by the interference of small indentations formedas one surface rolls over another. This is the idea behind the frictional forcesinvolved with wheels, cylinders, and spheres. The coefficient of rolling friction istypically denoted as r, and is usually less than the coefficient of staticfriction.

Laws of Friction:


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They are 5 laws of friction. They are:

1)When the surfaces of two objects are in rough contact, and have a tendency to

move relative to each other, equal and opposite frictional forces act, one on eachof the objects, so as to oppose the potential movement.

2) Until it reaches its limiting value, the magnitude of the frictional force F is justsufficient to prevent motion.

3) When the limiting value is reached, F= uR, where R is the normal reactionbetween the surfaces and u is the coefficient of friction for those two surfaces.

4) For all rough contacts 0 < F


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The Final momentum = m1 v1 + m2 v2

By the law of conservation we have

m1u1+ m2u2= m1v1+ m2v.

TYPES OF IMPACT

An impact force is a high force orshock applied over a short time period. Such a

force can have a greater effect than a lower force applied over a proportionally

longer time period.

COEFFICIENT OF RESTITUTION:

The coefficient of restitution is the ratio of the differences in velocities before and

after the collision. In other words, the difference in the velocities of the two colliding

objects after the collision, divided by the difference in their velocities before the

collision. In symbolic language:

c = coefficient of restitution

v1 = linear velocity of the racquet mass center before impact

s1 = linear velocity of the ball before impact (will be negative according to our

convention that away from the player is positive)

v2 = linear velocity of the racquet mass center after impact

s2 = linear velocity of the ball after impact


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SIMPLY SUPPORTED BEAMS

Beam supported at two ends is called a simply supported beam.

Different types of support provided can be roller, pin etc.

For the loaded, simply supported beam shown, determine expressions for the internal shear

forces and bending moments in each section of the beam, and draw shear force and bending

moment diagrams for the beam.

Solution: We first need to determine the external support reaction by applying our standardstatic equilibrium conditions and procedure..


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PART ASTEP 1: Draw a free body diagram showing and labeling all load forces and support(reaction) forces, as well as any needed angles and dimensions.

STEP 2: Resolve any forces not already in x and y direction into their x and y components.

STEP 3: Apply the equilibrium conditions.

Sum Fx =0 (no external x-forces acting on structure.)Sum Fy=(-800 lbs/ft)(8 ft) - (1,200 lbs/ft)(8 ft) +Ay +Cy =0Sum TA =(Cy)(12 ft) - (800 lbs/ft)(8 ft)(4 ft) - (1,200 lbs/ft)(8 ft)(12 ft) =0Solving for the unknowns:Cy =11,700 lbs; Ay =4,270 lbs

Part B: Determine the Shear Force and Bending Moment expressions for each section ofthe loaded beam. For this process we will cut the beam into sections, and then use Statics

- Sum of Forces to determine the Shear Force expressions, and Integration to determine the

Bending Moment expressions in each section of the beam.

Section 1: We note that the loading of the beam (800 lb./ft) remains uniform until 8 feet,where it changes to 1200 lb./ft. As a result, for our first beam section, we cut the beam at an

arbitary position x, where 0 < x < 8 ft. Then we analyze the left hand beam section.

1. Draw a FBD of the beam section showing and labeling all forces and toque acting -including the shear force and bending moment (which act as an external force and torque at

the point where we cut the beam.) (See Diagram - Section 1) Notice we have drawn the

shear force and bending moment in their positive directions according to the defined signconvention discussed earlier, and have labeled them as V1 and M1, as this is section 1 of

the beam.


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2. Resolve all forces into x & y components (yes)

3. Apply translational equilibrium conditions (forces only) to the section 1 of the beam:Sum Fx =0 (no net external x- forces)Sum Fy =4,300 lbs - 800 lbs/ft(x)ft - V1 =0;

Solving for the shear force: V1 =[4,300 - 800x] lbs

4. We can find the bending moment from static equilibrium principles; summing torqueabout the left end of the beam. Referring to the free body diagram for beam section 1, we

can write:

Sum Torque left end = -800 lb/ft * (x) * (x/2) - V1 (x) +M1 =0To make sure we understand this equation, let's examine each term. The first term is the

torque due to the uniformly distributed load - 800 lb./ft * (x) ft (this is the load) times (x/2)

which is the perpendicular distance, since the uniform load may be considered to act in thecenter, which is x/2 from the left end. Then we have the shear force V1 times x feet to the

left end, and finally we have the bending moment M1 (which needs no distance since it is

already a torque).Next we substitute the expression for V1 (V1 = [-4,300 - 800x] lb.) from our sum of forces

result above into the torque equation to get:Sum Torque left end =-1000 lb/ft * (x) * (x/2) -[-4,300 - 800x] (x) +M1 =0 ;and solving forM1 =[-400x2+4,300x] ft-lbs for 0 < x < 8 ft.

We will now also find the bending moment expression by integration of the shear force

equation. Integration , solving M1 =-400x2 +4,300x +C1Our boundary condition to find the integration constant, C1, is at x =0, M =0 (since this is

a simply support beam end.)Applying the boundary condition: 0 =-400(0)2 +4,300(0) +C1, and solving gives us: C1 =0.Therefore the bending moment expression for section 1 of the beam is:M1 =[-400x2+4,300x] ft-lbs for 0 < x < 8 ft.

(The shear force and bending moment diagrams are shown at the bottom of this example

page.)


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Section 2: We continue in the same manner with beam section 2. We note that the loadingchanges once more at 12 ft, due to the upward support force acting at that point. So forbeam section 2, we cut the beam at location x, where 8 < x


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We next substitute the value forV2 (V2 =[7,500 - 1,200x] lb.) from above and obtain:- (800 lb./ft * 8 ft) * 4 ft. -1200 * (x - 8')*[(x-8')/2+8'] - [7,500 - 1,200x]* x +M2 =0;

and then solving for M2 we find:M2 =[-600x2 +7,500x - 12,800] ft-lbs for 8 < x < 12

Next we find the bending moment, M2, from integration of shear force expression,V2.

Integration: , and solving, M2 =-600x2 +7,500x+C2We obtain our boundary condition for beam section 2 by remembering that the bending

moment must be continuous along the beam. This means that value of the bending moment

at the end of section 1 (at x = 8 ft.) must also be the value of the bending moment at thebeginning of section 2 (at x = 8 ft.). Thus our boundary condition to find C2 is: at x =8 ftM =8,560 ft-lbs(from equation M1). Now applying the boundary condition and solvingfor the integration constant, C2, we have:

8560 ft-lbs =-600(8)2 +7500(8) +C2, and solving: C2 =-13,000 ft-lbsTherefore our bending moment expression is:M2 =[-600x2 +7,500x - 12,800] ft-lbs for 8 < x < 12(The shear force and bending moment diagrams are shown at the bottom of this example

page.)

Section 3: Finally, we continue with the last section of the beam, cutting the beam atlocation x, where 12 < x < 16 ft., and analyzing the left hand beam section from x to the

left end of the beams.

1. Draw a FBD of the beam section showing and labeling all forces and toque acting -

including the shear force and bending moment (which act as an external force and torque atthe point where we cut the beam.) We have labeled them as V3 and M3, as this is section 3

of the beam.


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2. All forces in x & y components (yes)3. Apply translational equilibrium conditions (forces only):Sum Fx =0 (no net external x- forces)Sum Fy =4,300 lbs - 800 lbs/ft(8 ft) +11,700 lbs - 1,200lbs/ft (x - 8)ft - V3 =0Solving for the shear force expressing: V3 =[-1,200x +19,200] lbs

4. We may determine the bending moment expression by applying rotational equilibriumconditions, or by integration. Once more we will do it both ways for this section.Rotational Equilibrium:Sum of Toqueleft end =- (800 lb./ft * 8 ft) * 4 ft. - 1200 lb. *(x- 8 ft.)*[(x-8')/2 +8'] +11,700 lb. * 12 ft -V3 * x +M3 =0;

Once again the distance from end A at which the effective load due to the uniform load of

1200 lb/ft [1200 lb/ft * (x-8')]may be considered to act must be determined carefully. Thatdistance [(x-8')/2 +8')] is shown at the top of the adjacent diagram.

We next substitute the value forV3 (V3 =[-1,200x +19,200] lb.) from above and obtain:- (800 lb./ft * 8 ft) * 4 ft. - 1200 lb. *(x- 8 ft.)*[(x-8')/2 +8'] +11,700 lb. * 12 ft - [-1,200x+19,200] lb* x +M3 =0; and then solving for M3 we find: M3 =[-600x2 +19,200x -153,000] ft-lbsfor 12 < x < 16.


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Determine the bending moment by Integration: ,or, M3 =-600x2 +19,200x +C3

We find our boundary condition for beam section 3, by realizing at the end of the beam (afree end) the bending moment must go to zero, so our boundary condition to find C3 is: at x=16 ft M =0 ft-lb.Appling the boundary condition, and solving for the integration constant C3, we have:BC: 0 = -600(16)

2+ 19,200(16) + C3; and then C3 =-153,000 ft-lbs

So the final expressionn for the bending moment on section 3 will be:M3 =[-600x2 +19,200x - 153,000] ft-lbsfor 12 < x < 16

PART C: Shear Force and Bending Moment Diagrams: Now using the expressions foundin Part B above, we can draw the shear force and bending moment diagrams for our loaded

beam.V1 =-800x +4,300 lb.; V2 =-1,200x +7,500 lb.; V3 =-1,200x +19,200lbM1 =-400x2 +4,300x ft-lb.; M2 =-600x2 +7,500 - 12,800 ft-lb.; M3 =-600x2 +19,200x -153,000ft-lb

CANTILEVER BEAMSA cantileveris a beam supported on only one end. The beam carries the load to the

support where it is resisted by moment and shear stress. Cantilever construction

allows for overhanging structures without external bracing. Cantilevers can also be

constructed with trusses orslabs.


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STRUCTURES

A structure is any connected system of members built to support or transfer forces and

to safely withstand the loads applied to it.

Types of structures:

1. Trusses2. Frames3. Machines

Structures are distinguished as statically determinate and statically indeterminate for

determination of forces acting on it.

Statically Determinate Structures do not have more supporting constraints than are

necessary to maintain an equilibrium configuration.

Statically Indeterminate Structures are the ones with redundant members or

supporting links which can be removed still maintain the equilibrium condition.

Lets discuss on this with an example


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Statically determinate members

In a situation where values of all external forces acting on a body canbe determined by the equations of static equilibrium alone , then the

force system is statically determinate.

Example

The system is statically determinate since there are three equations of equilibriumavailable for the system and these are sufficient to determine the three unknowns.

Statically indeterminate structures

Statically indeterminate structures are a principle topic of the theory ofstructures and the subject of this chapter. The fact that the equilibrium equations in

this case have no unique solution is annoying to the extent that a given load shouldproduce a unique response from a linear system. The problem is, of course, thattheequations of statics do not completely define a structure; to complete thedescription.it is necessary to introduce member stiffnesses into the formulation. Once that hasbeen done the expected unique system response follows directly.The role of member stiffness is obvious in the two cases shown in Figure4.1. Intuitively, stiffer members in a structure carry more of the load. Put another


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way, as the area of bar 2 goes to zero in each case bar 1 is forced to carry theentire load P; the argument is symmetric to the extent that as bar 1 goes to zerobar 2 must carry the load. There are thus simple bounds on the manner in whichthese systems can respond.What eliminates the lack of uniqueness found in the equilibrium equations

is the fact that the pieces of a deformed structure must fit together. That fact will beenforced here in two different ways and thus give rise to two different methods ofanalysis, a force method and a displacement method. For example, if the lowersupport is removed in the case of the two bars in series, the structure becomesstatically determinate and the displacement at this support can be computed by

summing the length changes of both bars

Since a force (F2) is used as the unknown here this type of analysis is called theforce method.The parallel bars of Figure 4.1 are obvious candidates for another type ofanalysis. In this case it is clear that if the displacement is known, the bar forcescan be computed asF1 = K1 and F2 = K2 which implies an applied load of(K1 + K2 ) But since the applied load must be P, it follows thatP = (K1 + K2 ) = P / (K1 + K2 )andF1 = K1 = K1 P / (K1 + K2 ) and F2 = K2 = K2 P / (K1 + K2 )Since a displacement is used as the unknown in this analysis it is called adisplacement method.


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Statically Indeterminate StructuresThis chapter is concerned with formalizing the force and displacementmethods so that they can be applied to arbitrary structures. The methods arerelated to the extent that they both begin by introducing procedures that reduce thestructure to some case that can be dealt with easily (such as a staticallydeterminate structure). This reduction will be shown to violate the definition of thegiven structure. Finally a solution is constructed to repair these violations.These introductory comments attempt to show that the logic of the analysisof statically indeterminate structures is simple. The fact that these methodssometimes appear complex in application is due to the inherent geometricalcomplexity of structures which surfaces when displacements or forces must becomputed.

4.1 THE FORCE METHODGiven the idea of the force method just presented, this section develops asystematic approach with three steps. It begins with the discussion of a second


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single degree of freedom structure which is more practical than the two-bar trussdiscussed above. It then moves on to a two-degree of freedom system whichfinally leads to a general statement of the force method.4.1.1 A Single Degree of Freedom SystemThe force method can easily be described as a sequence of steps:

Step 1. Reduce the structure to a statically determinate structure. That is done inExample 4.1 by replacing the right-hand horizontal support by a roller. This stepallows the structure to displace where it was formerly fixed. The reaction R which istemporarily set to zero can be used to define a displacement (discontinuity) anditsdirection.Step 2. Compute the value ofdue to the load. The method of virtual work is usedto compute the displacement of the structure at its right support. In this case thereduced structure is the real structure and the structure marked R = 1 is the virtualstructure. Since this support moves to the right while R and have been taken aspositive to the left, comes out to he negative.

Step 3. Compute the value ofdue to a unit value of R.Again, virtual work is usedto compute the motion at the support. In this case the structure marked R = 1

servesas both the virtual structure and the real structure.Step 4. Solve for the reaction R. R is of course the value of the reaction which isrequired in order to push the structure back into place.Step 5. Compute and plot the final stress resultant diagrams.

A Two Degree of Freedom SystemThe trouble with the single degree of freedom system is that there is an interactionbetween redundants which it does not demonstrate. Example 4.2 shows thisinteraction clearly. It is again convenient to proceed in steps:Step 1. Reduce the structure to a statically determinate structure. In the case ofExample 4.2, this requires that two cuts be made thus defining two redundants.(These redundants are taken to be positive when they place their respective barsin tension.)Step 2. Analyze the structure. It will be necessary below to have the bar forces forthree cases of load: the structure under the given load and the structure under unitvalues of the redundants R1 and R2.Step 3. Set up the equations of superposition and compute the coefficients . Twosimultaneous equations will be used to determine the bar forces R1 and R2,(4.1)This is now the heart of the force method. Physically it is required to select R1 andR2 so that the ends of the cut bars are not separated in the final solution. ('Thepieces must fit together.') The interaction mentioned above arises because achange in R1 causes the ends of bar 2 to separate.


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The General CaseStep 1. Introduce releases to make the structure statically determinate. The basicidea is to reduce a structure to something that is workable. In this case workableimplies a statically determinate structure for which it is possible to compute bothforces and displacements. In order to do so releases are introduced into thestructure. As remarked in chapter 3, a release is a mechanical device which forcesa particular stress resultant called a redundant to be zero. When a release isinserted

110 Statically Indeterminate StructuresIn a structure it creates, by definition, a discontinuity. In the final solution, theredundants are selected so that the value of each discontinuity is zero.With regard to this text, releases will be introduced on a trial and errorbasis. When introduced properly, releases can create a statically determinatestructure; when introduced improperly the resulting structure will either bestatically indeterminate or unstable. The number of releases required to make astructure statically determinate is called the degree of statical indeterminacy, k.For more comments concerning the degree of statical indeterminacy the readershould refer to Appendix 6.Step 2. Analysis. It is necessary to solve the reduced structure for k + 1 loadingconditions: k cases of unit loads corresponding to individual redundants and onecase of the reduced structure under the applied toads.Step 3. Set up and solve the superposition equations. In general, these equationshave the form.


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or more simply in matrix formR + 0 = 0where ij is the discontinuity at release i due to a unit value of redundant j , i0 isthe discontinuity at release i due to the applied 'loads' or other external effectssuch as temperature and settlement, and Ri is unknown value of the i redundant.To compute the coefficient ij again requires combining the i and jsolutions in the virtual work expression. As remarked above, each of theseequations requires the discontinuity at a specific release to be zero in the finalsolution where all effects are present For a structure which is staticallyindeterminate to the kth degree, the force method requires the solution of k

simultaneous equations.Step 4. Combine solutions and plot the results. As before the final solution is thecombination of all effects,final solution = zero solution + R1 one solution + R2 two solution +... +RK kth solution.

THE DISPLACEMENT METHODThere is a less common (manual) method of structural analysis called thedisplacement method. In this method constraints are added to the structure until itbecomes workable or falls within the realm of some known solutions. Thedisplacements associated with the constraints are then selected so that the

fictitious forces associated with the constraints are zero in the final solution. This ismost easily explained through examples.4.2.1 A structure with a single displacement degree of freedomExample 4 4 describes an application of the displacement method to a simple rigidframe. It can be argued in the following manner. In general a rigid frame joint hasthree degrees of freedom: a horizontal displacement, a vertical displacement, anda rotation. In this particular case, if: member length changes are neglected as it issometimes common to do, one kinematic degree of freedom remains, the joint114 Statically Indeterminate Structuresrotation . When is specified, it is possible to compute member forces using onlycommonly available beam solutions .

As a parallel to the force method, the rotation

can be specified to be zero(i.e. a constraint can be introduced), but to do so requires the application of afictitious external moment F10. A unit value of again requires the application ofan external fictitious moment, say F11. In the final solution the value of isselected so that this fictitious external moment is


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The Displacement MethodSolutions are again superimposed to obtain the final moment diagram.

4.2.2 A structure with two degrees of freedomExample 4.5 describes a two-degree of freedom structure analyzed by the


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displacement method. In this case, a three-span beam is to be solved. This is atwo degree of freedom system, since knowing the two beam rotations at the centersupports means that the structure


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has been reduced to well known solutions for single span beams. In order toemphasize the idea of a mechanical constraint to which moments may be applied,fictitious shafts are indicated in the figure (of Example 4.5). As in the case of theforce method, a two-degree of freedom structure requires three analyses, a zerocasewhich provides for the external load and two cases of unit rotations at the

constraints. Having these analyses it is then possible to write the two superpositionequations which require that the fictitious constraint forces go to zero in the finalsolution,Here

Here 1 , 2 are rotations at constraints 1 and 2, F11 is the constraint moment at 1due to a unit value of1 , F12 = F21 is the constraint moment at 1 (joint 2) due to aunit rotation at joint 2 (joint 1), F22 is the constraint moment at joint 2 due to a unit

rotation at joint 2, F10 is the constraint moment at joint 1 due to load, and F20 istheconstraint moment at joint 2 due to load.When the load is applied but the interior joints are not allowed to rotate, aconstraint of wL2 / 8 must be applied externally to the first joint as indicated in thefigure (of Example 4.5). It follows thatF10 = - w L2/8 and F20 = 0If a unit rotation is applied to the first constraint, external moments must


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be applied as indicated in the figure (of Example 4.5). It follows thatF11 = 7 EI/L and F21 = 2 EI/Land from symmetry thatF22 = 7 EI/L and F12 = 2 EI/LIn the case of the example under discussion the superposition equations become


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Trusses on functionality are divided into:

1. Axial Member2. Non-axial member

Axial Memberis a member which is only in simple tension or compression.The internal force in the member is constant and acts only along the axis of themember .

Non-Axial Memberis a member which is not simply in tension or compression.It may have shear forces acting perpendicular to the member and/or there may bedifferent values of tension and compression forces in different parts of the member.


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FRAMES AND MACHINES

Frame : A structure is called as frame if one of its individual member is multi-forcemember and it is designed to support applied loads and they are usually in fixedpositions.

Machine: Machines are structure which contain moving parts and are designed totransmit forces or couples from input values to output values .

RESONANCE

Natural frequency: The frequency or frequencies at which an object tends tovibrate with when hit, struck, plucked, strummed or somehow disturbed is knownas the natural frequency of the object.

All objects that can be made to vibrate have a certain frequency at which they will

vibrate most strongly (i.e.. with maximum amplitude). If a body is excited with awhole range of frequencies it will vibrate approximately equally in response to themall except those frequencies nearest to its own natural frequency. At one frequencyit will vibrate most strongly. This frequency is called the resonant frequency andthe condition is called resonance.

The simplest vibratory system can be described by a single mass connected to aspring . The mass is allowed to travel only along the spring elongation direction.Such systems are called Single Degree-of-Freedom (SDOF) systems and areshown in the following figure,

Single Degree-of-Freedom System


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Wherem is the massc is the viscous damping coefficientk is the stiffnessx is the absolute displacement of the massy is the base input displacement

VIBRATIONS

Vibrations refers to mechanical oscillations about an equilibrium point. The

oscillations may be periodic such as the motion of a pendulum orrandom such asthe movement of a tire on a gravel road.Vibration is occasionally desirable. For example the motion of a tuning fork, thereed in a woodwind instrument orharmonica, or the cone of a loudspeakerisdesirable vibration, necessary for the correct

function ing of thevarious devices.


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Types of Vibration

1.Free Vibration occurs when a mechanical system is set off with an initial input

and then allowed to vibrate freely

2.Forced Vibration is when an alternating force or motion is applied to amechanical system.

Vibration isolation is the process of isolating an object, such as a piece ofequipment, from the source ofvibrations

Housed Spring Mountings

Open Spring Mountings


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