Intro to Mech

7/31/2019 Intro to Mech

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Statics of Rigid Bodies

STATICS OF RIGIDSTATICS OF RIGIDBODIESBODIES

STATICS OF RIGIDSTATICS OF RIGIDBODIESBODIES

enterDepartment of Engineering Sciences

Chapter 1: Introduction

DEFINITIONDEFINITION

Mechanics the study of the relationship among forces and

their effects on bodies.


the science which describes and predicts theconditions for rest and motion of bodies underthe action of forces.

a physical science (for it deals with physicalphenomena)

Department of Engineering Sciences Jump to ShowStopPrev Next

MECHANICSMECHANICS


MECHANICS

FLUIDS


BODIES

STATICS DYNAMICS INCOMPRESSIBLE COMPRESSIBLE

bodies at

rest

bodies in motion


What is a FORCE?What is a FORCE?

represents the action of one body on anotherthat tends to change the state or state of motionof a body.


may be exerted by actual contact or at adistance (e.g. gravitational and magneticforces).characterized by its point of application,magnitude and direction.represented by a vector.


Effects of a FORCEEffects of a FORCE

development of other forces(reactions or internal forces)

deformation of the body


accee ra on o e o y

Applied Force

Development of other forcesDevelopment of other forces


Applied Force

Development of force or forces at


Reaction

points of contact with other

bodies (reactions).


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Development of other forcesDevelopment of other forces


Applied Force


Development of forces within the

body itself (internal forces)A A


Deformation of the bodyDeformation of the body

Applied Force



Acceleration of the bodyAcceleration of the body

A lied Force



Properties of a FORCEProperties of a FORCEA force represents the action of one body on anotherand is generally characterized by its point of application,its magnitude, and its direction. Forces acting on a givenparticle, however, have the same point of application.


10N

30o10N

30o

Magnitude 10N

Direction 30 degrees (upwards or downwards)

Point of Application line of action (LOA) dashed linehaving an angle 30 degrees


DEFINITIONDEFINITIONParticles - in the context of this course, does notindicate smallness of size, rather, it means thatthe shape and size of the object does not


under consideration.

Rigid Bodies - the problems considered in thiscourse are assumed to be non-deformable.Again, such assumption does not significantlyaffect the solution of the problems underconsideration.

BASIC QUANTITIES & UNITSBASIC QUANTITIES & UNITS


Quantity SI ENGLISH

Length m (meter) ft (feet)


Mass kg (kilogram) slugs

Time s (seconds) s (seconds)

Force kg*m/s^2 or N lbsNewton (pounds)

Acceleration due to gravity

g = 9.81 m/s^2 or g = 32.2 ft/s^2


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RESULTANT OF THERESULTANT OF THE

ORIGINAL FORCESORIGINAL FORCES

RESULTANT OF THERESULTANT OF THE

ORIGINAL FORCESORIGINAL FORCESsingle equivalent force having the same effect

enter Department of Engineering Sciences

.

P

Q

A

R

A

Moment of a Force

RESULTANT CAN BERESULTANT CAN BE

OBTAINED THRU:OBTAINED THRU:

RESULTANT CAN BERESULTANT CAN BE

OBTAINED THRU:OBTAINED THRU:Parallelo ram Law or Vector Addition


Triangle Rule


PARALLELOGRAM LAWPARALLELOGRAM LAWPARALLELOGRAM LAWPARALLELOGRAM LAWThe resultant of two forces is the diagonalof the parallelogram formed on the vectorsof these forces drawn tail-to-tail.


.

P

QA

R


TRIANGLE LAWTRIANGLE LAWTRIANGLE LAWTRIANGLE LAW

If two forces are represented by their free vectorsplaced head-to-tail, their resultant vector is the third


.

P

QA

R


VECTORSVECTORSVECTORSVECTORS Vectors are defined as mathematical expressionspossessing magnitude and direction, which add

according to the parallelogram law.


P

QA

P+Q

QQPR

BPQQPRrrr

+=+= cos2222

Law of cosines,

Law of sines,

P

C

R

B

Q

A sinsinsin==


ADDITION OF VECTORSADDITION OF VECTORSADDITION OF VECTORSADDITION OF VECTORS

The addition of two vectors is commutative.

PP

vvvv

+=+


P

QA

P+Q

Q P

QA

Q+P

P


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SUM OF VECTORSSUM OF VECTORSSUM OF VECTORSSUM OF VECTORS If the vectors are coplanar, the resultant may be

obtained by using the polygon rule for the addition ofvectors arrange the given vectors in a tip-to-tail fashionand connect the tail of the first vector with the tip of the last


one

P

QA

Q

S

R


VECTOR ADDITIONVECTOR ADDITIONVECTOR ADDITIONVECTOR ADDITION

Vector addition is associative.

R = P+Q+S = P+S+Q


P

QA

Q

S

R


VECTOR ADDITIONVECTOR ADDITIONVECTOR ADDITIONVECTOR ADDITION Vector addition is associative.

R = P+Q+S = P+S+Q


P

QA

Q

S

R


PRODUCT OF A SCALARPRODUCT OF A SCALAR& VECTOR& VECTOR

PRODUCT OF A SCALARPRODUCT OF A SCALAR& VECTOR& VECTOR

=P

P


Pnv

- have the same direction as Pwith magnitude Pnv

P 1.5P -2P


Resultant of Several Concurrent ForcesResultant of Several Concurrent Forces

Concurrent forces: set of forces whichall pass through the same point.

A set of concurrent forces applied to a

particle may be replaced by a single

resultant force which is the vector sum


of the applied forces.

Vector force components: two or moreforce vectors which, together, have the

same effect as a single force vector.


Resultant of Several Concurrent ForcesResultant of Several Concurrent ForcesSOLUTION:

Trigonometric solution - use thetriangle rule for vector addition in

conjunction with the law of cosinesand law of sines to find the resultant.


The two forces act on a bolt at

A. Determine their resultant.


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Trigonometric solution -Apply the triangle rule.

From the Law of Cosines,

( ) ( ) ( )( ) +=

+=

155cosN60N402N60N40

cos222

222BPQQPR

N73.97=R



From the Law of Sines,

QBA

RB

QA

=

=

sinsin

sinsin


A

A

+==

=

20

04.15N73.97

N60155sin

= 04.35

Resultant of Several ConcurrentForces

Resultant of Several ConcurrentForces

Example 2

Determine graphically,the magnitude anddirection of theresultant of the two

900 N

600 N

45o

28

forces using (a)Parallelogram law and(b) the triangle rule.

30o

Graphical Solution

A parallelogram with sides equal to 900 N and 600 N is drawn to scaleas shown.

From the scaled drawing ofthe forces, the resultant is

600 N

R

29

R 1400 N900 N

30o

45o 46o

Note: The triangle rule may also be

used. Join the forces in a tip to tailfashion and measure the magnitude

and direction of the resultant.

Trigonometric Solution

For the magnitude of R, using the cosinelaw:

( )( ) oR 135cos6009002600900 222 +=

NR 13916.1390 ==

600 N

135o

R

30

For angle , using sine law:

sin

600

135sin=

O

R

30o

OO8.1730 +=The angle of the resultant:

OO

8.171391

135sin600sin 1 =

=


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Example 3

Two forces are applied as shown toa hook suppor t. Knowing that themagnitude of P is 35 N, determine

50 N

25O

31

by trigonometry (a) the requiredangle i f t he resul tan t of the twoforces applied to the hook support isto be horizontal, (b) thecorresponding magnitude of R.

P

Solution (a)

50 N

25O

P

RR

O sin

35

25sin

50

sin==

32

(a) Determining &

OO

14.3735

25sin50sin

1 =

=

OO 25180 =

O86.117=

Solution (b)

50 N

25O

P

RR

O sin

35

25sin

50

sin==

33

(b) Determining R

sin

sin50=R NR 22.73=

Resolution of ForcesResolution of Forces

It has been shown that the resultant of

forces acting at the same point(concurrent forces) can be found.

34

In the same way, a given force, F can

be resolved into components.

There are two major cases.

Resolution of Forces: Case 1

(a) When one of the two components, P isknown: The second component Q isobtained using the triangle rule. Join the tail of

P to the tail of R. The magnitude and direction

35

o are e ermne grap ca y or ytrigonometry.

R

PQ

i.e. R = P + Q

Resolution of Forces: Case 2

(b) When the line of action of each component isknown: The force R can be resolved into twocomponents having lines of actions along a and busing the parallelogram law. From the head of R,extend a line parallel to a until it intersects b.Likewise, a line arallel to b is drawn from the head of

36

F to the point of intersection with a. The twocomponents P and Q are then drawn such that theyextend from the tail of F to points of intersections.

a

Q R

P b


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Solution

25o

45o350 N

800 N

600 N

60o

y

x

RX = F x = 350 cos 25O + 800 cos 70O -600 cos 60O

43

RX = 317.2 + 273.6 300 = 290.8 N

RY = F y = 350 sin 25 + 800 sin 70O +600 sin 60O

RY = 147.9 + 751 + 519.6 = 1419.3 N

F= 290.8 N i +1419.3 N j

Resultant, F NF 14493.14198.290 22 =+= O4.788.290

3.1419tan 1 ==

F = 1449 N 78.4O

Forces in SpaceForces in SpaceThe resultant of forces in 3-D s ace canbe determined using parallelogramlaw/triangle law.

Rectangular components can also beused to get resultant.

Forces in Space

The resultant ofconcurrent forcesacting on a particle ins ace will also act at

z

F1

F2

the same particle.

Only the magnitudeand direction are to bedetermined.

y

x

o

F3

Note: In this illustration, the magnitude and direction of all of the

forces are given.

Force Polygon

From the force polygon, theresultant can be drawn from the tailof the first force to the head of the

last force.

The magnitude and direction of the

z

F1

F2 F3

z

y

R

resultant can be computed usingsuccessive use of the triangle law. y

x

o

x

Note: The sine and cosine lawsare hard to implement because

usually the given angles areabsolute.

Rectangular Components of a Force

The rectangular components of a force can be determined easilydependingon thegivencharacteristics of theforce.

Given the Magnitude and Two Angles

y

z

x

o

F

z

y

y

z

x

o

F

z

xy

xyF

zF

zxyFF sin=

zz FF cos=



z

FzF

z

zF

xyzxyxyy FFF cossincos ==

y

x

o

z

xy

xyF

zxy FF sin=

y

x

o

xy

xyF

xF

yF

xyzxyxyx FFF sinsinsin ==


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z

zF

xyzxyxyx FFF sinsinsin ==

zz FF cos=

y

x

o

xy

xyF

xF

yF xyzxyxyyFFF cossincos ==

222

FzFyFxF

kFzjFyiFxF

++=

++=

In vector form,


Given the Magnitude and Three Absolute Angles

Fx = Fcos x=

z

Fz = Fcos zwhere cos x, cos yand cos z are directioncosinesF = Fxi + Fyj + Fzk

y

x

o

x

F

y

z

Example 7

A force of 500N forms angles of 600, 450 and 1200, respectively, with thex,y and z axes. Find the components Fx, Fy and Fz of the force. Find alsothe vector representation of the force.

zx = cos x = cosFx = +250NFy = F cos y = (500N) cos 450

Fy = +354NFz = F cos z = (500N) cos 1200

Fz = -250N

y

x

o

F

x

y

z

Fx

Fy

Fz

F = 250N i + 354N j 250N k

Example 7

y

z

o

x

y

z

Fx

Fy

Note: Theangle a force F forms withan axis shouldbe measured from thepositive side of the axis and willalways be between 0 and1800.

x

F

Fz

F = 250N i + 354N j 250N k

Direction Cosines

( )kzjyixFF coscoscos ++=let = cos x i + cos y j+ cos z k

= unit vectorx= cos x

y= cos yz= cos z

x2 + y2 + z2 = 1

cos2 x + cos2 y + cos2 z = 1

it follows that,

F = FThe force vector is equal to the product of themagnitude of the force and the unit vector.

Direction Cosines

cos x = Fx/ F

cos y = Fy/ F

x = cos-1 (Fx/ F)

y = cos-1 (Fy/ F)

cos z = Fz/ F

F = Fx2 + Fy2 + Fz2

z = cos-1 (Fz/ F)


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Example 8

A force has the components Fx = 20N, Fy = -30N, Fz = 60N. Determine

its magnitude F and the angles x, y, z it forms with the coordinateaxes.

z

y

x

o

Fx

Fy

Fz222 FxFyFxF ++=

NNF

NNNF

704900

)60()30()20( 222

==

++=

Example 8

x = cos-1 (Fx/ F) = cos-1 (20/70)

x = 73.40

= cos-1 (F / F ) = cos-1 (-30/70)

z

oFy

FzF

F = 70 N

y = 115.40

z = cos-1 (Fz/ F) = cos-1 (60/70)

y = 31.00

y

x

Fx


Given the Magnitude and Two Points where Force Passes

yyd

xxd

oey

oex

=

=z ( )eee zyxE ,,

F

222 dzdydxd

zzd oez

++=

=

Fd

dxFx = F

d

dyFy = F

d

dzFz =

kFzjFyiFxF ++=

y

x

o

( )ooo zyxO ,,


kFzjFyiFxF ++=

z( )eee zyxE ,,

F

FxF

Fz

y

x

o

( )eee zyxO ,,

d

dxx =cos

d

dyy =cos

d

dzz =cos

Example 9

Determine the vector representation of the given force.

z

2.4m E

x

O

1.5m

F=1.6kN

1.2m

1.5m

y

Example 9


z

2.4m E O(1.2, 1.5, 0.0)

E(0.0, 2.4, 1.5)

x

O

1.5m

F=1.6kN

1.2m

1.5m

y dx = 0.0 -1.2 = -1.2dy = 2.4 -1.5 = +0.9

dz = 1.5 - 0.0 = +1.5

( ) ( ) ( )222 5.19.02.1 ++=d

d = 2.12m


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Example 9


kNkNFx 905.0)6.1(12.2

2.1 ==

z

E

kNkNFy 679.0)6.1(12.2

.==

kNkNFz 131.1)6.1(12.2

5.1=

+=

kkNjkNikNF 131.1679.0905.0 ++=

Fx

Fz

x

O

y

Fy

Example

Sample Problem 2.10 : (2.89) A frame ABC is supported in part by

cable DBE which passes through a frictionless ring at B. Knowing

that the tension in the cable is 385 N, determine the components ofthe force exerted by the cable on the support D.

210mm

280mm

400mm510mm

480mm600mm

xz

A

B

C

DE

Example

O(0, 510, 280)

E(480, 0, 600)

dx = xE xO = 480-0=480

dy = yE yO = 0-510=-510400mm

210mm

280mm

510mm

DE

TDB = 385N

TDBY

TDBXTDBZ

O(0, 510, 280)

dz = zE zO = 600-280=320

d = 770 mm

480mm

600mm

x

z

A

B

C

E(480, 0, 600)

( ) NTd

dT

DBx

DBX 240385770

480===

( ) NTd

dT DB

y

DBY 255385770

510=

==

( ) NTd

dTDB

zDBZ

160385770320 ===( )NkjikTjTiTT DBZDBYDBXDB 160255240 +=++=

r

Solution:

Position vector of BH = 0.6 m i + 1.2 m j - 1.2 m k

Magnitude, BH = 0 6 12 12 182 2 2

. . . .+ + = m

BH

1

BH

BH BH BH BH

BH

x y z

BHm m m

T T TBH

BH

N

mm i m j m k

T N i N j N k

F N F N F N

= =

= = = +

= +

= = =

| | .. . .

| |. | || | .

. . .

( ) (500 ) (500 )

18

750

180 6 12 12

250

250 500 500

Addition of Concurrent Forcesin Space


The resultant R of two or more forces in space willbe determined by summing their rectangularcomponents. Graphical or trigonometric methods are

generally not practical in the case of forces in space.

= FR ( ) ++=++ FzFyjFxiRzkRyiRxi( ) ( ) ( )kFzjFyiFx ++=

= FxRx = FyRy = FzRz


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222

RzRyRxR ++=

R

Rxx =cos

R

Ryy =cos

R

Rzz =cos

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Intro to Mech