Engineering Mech(Presentation)

7/28/2019 Engineering Mech(Presentation)

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A REVIEW:

FEATI AERO BOARD REFRESHER

Prepared by: Contado, Clovis B.

ENGINEERING MECHANICS


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Mechanics is the branch of physics thatconsiders the action of forces on bodies or fluids

that are both at rest and in motion

Engineering Mechanics is the branch of

engineering that applies the principles of

mechanics to any design that must take into

account the effect of forces


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Outline - Statics

Chapter I. Principles of Statics

Chapter II. Resultants of Force Systems

Chapter III. Equilibrium of Force Systems

Chapter IV. Analysis of Structure Chapter V. Friction

Chapter VI. Force Systems in Space

Chapter VII. Centroids & Centers of Gravity

Chapter VIII. Moments of Inertia


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ENGINERRING

MECHANICS

Statics

Force Systems

Concurrent

Parallel

Non-

Concurrent

Application

Trusses

Centroids

Friction

Dynamics

Kinematics Kinetics

Fig 1.1 Outline of Engineering

Mechanics


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Chapter I

FUNDAMENTAL CONCEPTS &

DEFINITIONS


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Chapter I. Fundamental Concepts &

Definition

Engineering Mechanics the science which considersthe effects of forces on rigid bodies

Statics Consider the effects and distribution of

forces on rigid bodies

Dynamics consider the motion of rigid bodiescaused by the forces acting on them. It deals with

objects or structures with a non-zero acceleration.

Force that which changes , or tend to change the

state of motion of body.

Note: External effects of forces are considered in Engineering

Mechanics; Internal effects, in Strength of Material


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Definition

Characteristics of a Force: (1) has magnitude (2)position of its line of action (3) the direction orsense

Classification of Force Systems

1. Coplanar Force System the line of action of all

forces lie on one plane

2. Non-Coplanar (Space Systems of Forces) the

line of action of all forces do not lie on a same

plane

3. Concurrent forces lines of forces pass trough acommon point

4. Non-concurrent

5. Parallel Force System


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Definition

Axioms Of Mechanics

1. The Parallelogram Law the resultant of two

forces is the diagonal formed on the vectors of

these forces

2. Two forces are in equilibrium only when equal in

magnitude, opposite in direction, and collinear

in action

3. A set of forces in equilibrium may be added toany system of forces without changing the effect

if the original system

4. Action and reaction forces are equal but

oppositely directed,


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Chapter II.

RESULTANTS OF FORCE

SYSTEMS


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.Systems

The effect of a system of forces on a body isusually expressed in terms of a resultant.

1. Fx = Fcos x

2. Fy = F Sin x

3.

4.


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.Systems

Problems

1. Determine the resultant of

the concurrent forces

shown

2. The resultant of theconcurrent forces shown in

the fig is 300lb pointing up

along the Y-axis. Compute

the values of F and

required to give theresultant.

500

lb240

lb

30


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ap er . esu an s o orceSystems

3. A boat moving at12kph is crossing a river

500m wide in which a

current is flowing at

4kph. In what directionshould the boat head if

it is to reach a point on

the other side of the

river directly opposite its

starting point?


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.Systems

4. Beam AB in the fig below, supports a load whichvaries from an intensity of 50lb/ft at one end to

200lb/ft at the other. Calculate the magnitude &

position of the resultant.


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.Systems

Moment of a Force the moment of a force about a

point or axis measures the tendency of the force tocause the body to rotate around that axis or point.

M= Fd

Where: d perpendicular distance

F Resultant Force


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.Systems

A. Varignons Theorem the moment of the

resultant is equivalent to the moment sum of itscomponent


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.Systems

Problems

1. In a certain non-concurrent

concurrent force system, it is

is found that X = -80 lb, Y

Y = 160lb, M = 480 ft-lb in

a counterclockwise sense.Determine at which the

resultant intersects the x-axis.

2. Two forces P & Q passthrough a point A which is 4ft

to the right of and 3ft above

a moment center O. Force P

is 200lb directed up to the

right at 30 with thehorizontal and force Q is

100lb directed up to the left

at 60 with the horizontal.

Determine the moment of theresultant of these two forces

with respect to O.

Ans: 377ft-lb


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.Systems

3.The 16-ft wing of an

airplane is subjected toa lift which varies from

zero at the tip to 360

lb/ft at the fuselage

according to = 90x^1/2 lb/ft where x is

measured from the tip.

Compute the resultant

and its location from the

wing tip.

Ans: R = 3840 lb at 9.60 ft

4. Determine completely

the resultant of theforces acting on the

step pulley shown in the

fig. Ans: 1250 lb down

to right; h = 44.3750 lb

1250

lb 250 lb

60

1.25


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.Systems

Couples. Sometimes the resultant wil be zero inmagnitude and yet have a resultant moment sum.

The special case in which the resultant has zeromagnitude but does have a moment is said to consista couple.

C = F d


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.Systems

Ex. Transform the couple

shown in the figure into anequivalent couple whose

forces are horizontal abd

act through points C & D.

Solution:

When the forces of thecouple act through points C

& D, the moment arm of

the couple becomes 3in.

Since the moment effect isconstant, the forces acting

at C & D are found from

C = Fd C = 9x4 = F x3

F = 12lb

Therefore C & D each has

magnitude of 12lb.

9lb 9lb

A B

C

D

4

3


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Systems

Replace the system of forces acting on the frame

below by a resultant R at A and a couple actinghorizontally through B & C.

Ans: R = 50lb down; B=

110lb right: C

= 110lb left

20lb

30lb 60lb

3

4

2

1

A

B

C


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Chapter III

EQUILIBRIUM OF FORCE

SYSTEMS


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.Systems

Equilibrium is the term used to designate thecondition where the resultant of a system of forces is

zero. A body is said to be in equilibrium when the

force system acting upon it has zero resultant.

The physical meaning of equilibrium, as applied to abody, is that the body either is at rest or moving in a

straight line with constant velocity


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.Systems

Equilibrium of Concurrent Force System x = 0

y = 0

Equilibrium of Parallel Force System

F = 0

M = 0


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Chapter III. Equilibrium of Force Systems

Equilibrium of Non-Concurrent Force System

x = 0 x = 0

y = 0 or MA = 0

M = 0

MB


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.Systems

Problems1. A 300 lb box is held at rest

on a smooth plane by a

force P inclined at an

angle with the plane as

shown. If = 45,

determine the value of P

and the normal force N,

exerted by the plane.


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Systems

2. A load of 100 lb is hubg

from the middle of arope, which is stretched

between two rigid walls

30ft apart. Due to the

load, the rope sags 4ftin the middle.

Determine the tension

in the rope.

Ans: 194 lbs

3. The 300 lb force and the

400lb force shown in thefig. are to be held in

equilibrium by a third

force F acting at an

unknown angle

with thehorizontal. Determine the

values of F and .

30

F

400 lb

300 lb


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.Systems

4. Determine the load P required to hold bar AB in a

horizontal position on the smooth inclines showninf Fig. Also determine the reactions at A & B.

14 24

4560

P 400 lb

BA


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Chapter IV

ANALYSIS OF STRUCTURES


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Chapter IV. Analysis of Structure

The analysis of a structure is the process by which we

determine how the loads applied are distributed

throughout a structure.

Two types of structures will be studied;

1. Pin-connected trusses the internal force in a bar

is directed along the axis of the frames

2. Pin-connected frames the members are subjected

to bending action


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A truss is a structure composed of members

fastened together in such a way to resist change

in shape: it is a rigid structure

Trusses are so constructed that all applied loads

act at the ends of the members. Such membersheld in equilibrium by only two forces are called

two-force members.

Members which are stretched are said to be in

tension, while those that are shortened are saidto be in compression.


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C f S


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Ch IV A l i f S


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Ch t IV A l i f St t


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Analysis of Structure

1. Method of Joints analysing trusses by

applying the principles of equilibrium to the

concurrent force systems

2. Method of Sections the principles of

equilibrium of non-concurrent force systems are

applied.



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Problems.

1. Using the method of sections, determine the force

on members BD,CD, & CE of the truss shown on

below figure.



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2. For the truss shown in fig, determine the force in BF

by the method of joints and then check this resultusing the method of sections.

1200 lb

F

E

D

2400 lb1200 lb

C

A

B

9

12

9

12


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Chapter V

FRICTION


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Chapter V. Friction

Friction may be defined as the contactresistance exerted by one body upon a second

body when the second body moves or tends to

move past first body.

It is a retarding force always acting opposite tothe motion or the tendency to.

Ch t V F i ti


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Chapter V. Friction

Ch t V F i ti


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Chapter V. Friction

Chapter V Friction


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Chapter V. Friction

Chapter V Friction


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Chapter V. Friction

Chapter V Friction


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Chapter V. Friction

Problems:


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Chapter VI

FORCE SYSTEMS IN SPACE

Ch t VI F S t i S


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In the preceding chapters on coplanar systems, we

have seen how two fundamental concepts, (1) thatwhich relates a force to its components and (2) the

moment effect of a force, were applied. When we

consider force systems in space, the same basic

concepts are all that are necessary, only they must beextended to include the more general case of space

forces

Ch t VI F S t i S


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Ch t VI F S t i S


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Equations:

Ch t VI F S t i S


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