me a chapter and 2

8/8/2019 Me a Chapter and 2

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EMB201: MHENDSLKMEKAN STATK

DERSEGR


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Hoca Hakknda

Dersin Hocas: Dr. Cokun ZALP

Oda: Faklteler Binas B Blok, 1. Kat Email: [email protected]

Tel. No. : 2512

alma Saatleri: Sabah 8.00 Aksam17.00


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COURSE OBJECTIVESUpon successful completion of this course, studentsshould be able to:

) Determine the resultant ofcoplanar and space for cesystems.

(ii) Determine the centroid and center of mass ofplane areas and volumes.

(iii) Distinguish between concurrent, coplanar andspace force systems

(iv) Draw free body diagrams.


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COURSE OBJECTIVES CONTD.

(v) Analyze the reactions and pin forcesinduces in coplanar and space systemsusing equilibrium equations and free bodydiagrams.

(vi) Determine friction forces and theirinfluence upon the equilibrium of a system.

(vii) Apply sound analytical techniques andlogical procedures in the solution ofengineering problems.


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Course Content

(i) Introduction, Forces in a plane, Forces in space (ii) Statics of Rigid bodies (iii) Equilibrium of Rigid bodies (2 and 3

dimensions)(iv) Centroids and Centres ofgravity

(v) Moments of inertia of areas and masses (vi) Analysis of structures (Trusses, Frames and Machines) (vii) Forces in Beams (viii)Friction


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Course Textbook and Lecture Times

Vector Mechanics For Engineers By F.P.Beer and E.R. Johnston (Third MetricEdition), McGraw-Hill.

Lectures: Wednesday, 1.00 to 1.50 p.m.

Thursday , 10.10 to 11.00 a.m.

Tutorials: Monday, 1.00 to 4.00 p.m. [Once in

Two Weeks]

Attendance at Lectures and Tutorials is Compulsory


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TutorialOutlineChapter 2 STATICS OF PARTICLES

2.39*, 41, 42*, 55, 85*, 86, 93*, 95, 99*, 104, 107*, 113

Chapter 3 RIGID BODIES: EQUIVALENT SYSTEM OF FORCES

3.1*, 4, 7*, 21, 24*, 38, 37*, 47, 48*, 49, 70*, 71, 94*, 96, 148*, 155

Chapter 4 EQUILIBRIUM OF RIGID BODIES

4.4*, 5, 9*, 12, 15*, 20, 21*, 31, 61*, 65, 67*, 93, 115*

Chapters 5 and 9 CENTROIDS AND CENTRES OF GRAVITY, MOMENTS OFINERTIA

5.1*, 5, 7*, 21, 41*, 42, 43*, 45, 75*, 77 9.1*, 2, 10*, 13, 31*, 43, 44*

Chapter 6 ANALYSIS OF STRUCTURES

6.1*, 2, 6*, 9, 43*, 45, 75*, 87, 88*, 95, 122*, 152, 166*, 169

Chapters 7 and 8 FORCES IN BEAMS AND FRICTION

7.30 , 35, 36, 81, 85 8.25, 21, 65

* ForChapters 1 to 6 and 9, two groups will do the problems in asterisks; the other twogroups will do the other ones. All the groups will solve all the questions in Chapters 7

and 8.


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Time-Table For Tutorials/Labs

M O N D A Y 1 :0 0 - 4 :0 0 P .M .

WeekG r o u p

1,5,9 2,6,10 3,7,11, 4 ,8,12

K - M E 1 3 A M E 1 6 A

(3,7)

M E 1 3 A

L M E 1 3A - M E 1 3A M E 1 6A

(4,8)

M M E 1 6 A(5,9)

M E 1 3 A - M E 1 3 A

N M E 1 3A M E 1 6A(6,10)

M E 1 3 A -


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ourse ssessmen

(i) One (1) mid-semester test, 1-hourduration counting for 20% of the totalcourse.

(ii) One (1) End-of-semesterexamination, 2 hours duration countingfor 80% of the total course marks.


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ME13A: ENGINEERINGSTATICS

CHAPTER ONE:INTRODUCTION


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1.1 MECHANICS

Body of Knowledge which

Deals with the Study andPrediction of the State of Rest

or Motion of Particles andBodies under the action ofForces


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PARTS OF MECHANICS


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1.2 STATICS

Statics Deals With the Equilibrium

of Bodies, That Is Those That AreEither at Rest or Move With aConstant Velocity.

Dynamics Is Concerned With theAccelerated Motion of Bodies andWill Be Dealt in the NextSemester.


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ME13A: ENGINEERINGSTATICS

CHAPTER TWO:STATICS OFPARTICLES


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A particle has a mass but a size that canbe neglected.

When a body is idealised as a particle,the principles of mechanics reduce t o asimplified form, since the geometry of

the body will not be concerned in theanalysis of the problem.

2.1 PARTICLE


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PARTICLE CONTINUED

All the for ces acting on a

body will be assumed to beapplied at the same point,

that is the for ces areassumed concurrent.


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Force on a Particle Contd.

Note: Point P is the point of action offorce and and are directions. Tonotify that F is a vector, it is printed inbold as in the text book.

Its magnitude is denoted as |F| or

simply F.

E


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Force on a Particle Contd.

There can be many for ces acting on aparticle.

The resultant of a system of forceson a particle is the single forcewhich has the same effect as the

system of forces. The resultant oftwo forces can be found using theparalleolegram law.


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2.2.VECTOROPERATIONS

2.3.1 EQUAL VECTORS

Twov

ectors are equal if they are equalin magnitude and act in the samedirection.

pP

Q


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Equal Vectors Contd.

Forces equalin Magnitude can act inopposite Directions

S

R


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Vector Addition Contd.

Triangle Rule: Draw the first Vector. Jointhe tail of the Second to the head of theFirst and then join the head of the third tothe tail of the first force to get the resultantforce, R

Q

PR = Q + P


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Triangle Rule Contd.

Also:

P

Q

R = P + Q

Q + P = P + Q. This is the cummutative law of

vector addition


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Polygon Rule

Can be used for the addition of morethan two vectors. Two vectors are

actually summed and added to thethird.


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Polygon Rule contd.

P

QS

P

Q

S

R

R=

P+ Q +

S

(P + Q)


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Polygon Rule Contd.

P + Q = (P + Q) . Triangle Rule

i.e.P

+ Q +S

= (P

+ Q) +S

=R

The method ofdrawing the vectors is

immaterial . The following method can

be used.


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Polygon Rule contd.

P

QS

P

Q

S

R

R=

P+ Q +

S

(Q + S)


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Polygon Rule Concluded

Q + S = (Q + S) . Triangle Rule

P + Q + S = P + (Q + S) = R i.e. P + Q + S = (P + Q) + S = P + (Q + S)

This is the associative Law of Vector

Addition


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2.3.3. Vector Subtraction

P - Q = P + (- Q)

P

Q

P

-Q

P-Q

Q

P

P- Q

Parm. RuleTriangle Rule


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2.4 Resolution of Forces

It has been shown that theresultant of forces acting at thesame point (concurrent forces) canbe found.

In the same way, a given force, Fcan be resolved into components.

There are two majorcases.


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Resolution of Forces: Case 1

(a)When one of the two components, P isknown: The second component Q is

obtained using the triangle rule. Join the tipof P to the tip of F. The magnitude anddirection of Q are determined graphically orby trignometry.

F

PQ

i.e. F = P + Q


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Resolution of Forces: Case 2

(b) When the line of action of each component is known: The force, F can be

resolved into two components having lines of action along lines a and b using th

paralleogram law. From the head ofF, extend a line parallel to a until it intersects b.

Likewise, a line parallel to b is drawn from the head ofF to the point of intersection with

a. The two components P and Q are then drawn such that they extend from the tail o

F to points of intersection.

a

Q F

P b


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Trignometric Solution

Usin thecosine law:

= 9 + 6 - 9 6 cos

= 9 .6 = 9

Usin the sine law:

R

Bi e B

The angle of the resul t

sin sin. . sin

sin

.tan . .

135

600 600 135

1391

17 830 17 8 47 8

1S

S

S

S

! !

!

! !

ie. = 9

4 .o

R

900 N

600N

135o

30oB


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Solution

Sol tion: Usin Trian le r le:

o k

k o

U o

R

=

+

- cos

- cosine law= 4 .

Usin sine r le:

4013

105

20 20 105

401328 8

288 25 38

401 38

1. sin

..

. .

. . , .

N

Sin Sinand Sin

Angle R

i e R N

o

oo

o o o

o

! ! !

! !

!

U


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2.5 RECTANGULAR

COMPONENTSOFFORCE

x

F

j

iFx = Fx i

Fy = Fy j

y


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RECTANGULARCOMPONENTS

OFFORCE CONTD.

In many problems, it is desirable to resolveforce F into two perpendicularcomponents in

the x and y directions. Fx and Fy are called rectangular vector

components. In two-dimensions, the cartesian unit vectors

i andj are used to designate the directions ofx and y axes. Fx = Fx i and Fy = Fy j i.e. F = Fx i Fyj Fx and Fy are scalarcomponents ofF


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RECTANGULARCOMPONENTS

OFFORCE CONTD.

While the scalars, Fx and Fy may be positive or negative, depending on the sense ofFx

and Fy, their absolute values are respectively equal to the magnitudes of the component

forces Fx and Fy,

Scalarcomponents of F have magnitudes:

Fx = F cos U and Fy = F sin U

F is the magnitude of force F.


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Example

Determine the resultant of the threeforces below.

25o45o

350 N

800 N600 N

60o

y

x

o u on


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o u on

= cos

o

+ cos

o

- 6 cos 6

o

= . + .6 - = 9 .

y = sin o + sin o + 6 sin 6 o

= 4 .9 + + 9.6 = 4 9.

i.e. F = 9 . i + 4 9. j

es ltant,

F N! !

! !

2908 1419 3 1449

14193

290878 4

2 2

1 0

. .

tan.

..U

F

= 449 .4

o

25o45o

350 N

800 N

600 N

60o

y


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Example

A hoist trolley is subjected to the threeforces shown. Knowing that = 40o ,determine (a) the magnitude of force, P for which the resultant of the three for ces isvertical (b) the corresponding magnitude of

the resultant.

E

1000 N

P

2000 N

EE


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Solution

1000 N

P

2000 N

40o40o

(a) The resul ant bein vertical meansthatthe

hori ontalcomponentis zero.

F x = 1000 sin40o + P - 2000 cos40o = 0

P = 2000 cos40o - 1000 sin40o =

1532.1 - 642.8 = 889.3 = 889 kN

(b) Fy = - 2000 sin40o - 1000 cos40o =

- 1285.6 - 766 = - 2052 N = 2052 N


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2.6. EQUILIBRIUMOF A PARTICLE

A particle is said to be at equilibrium when the resultant of all the forces acting on it is

zero. It two forces are involved on a body in equilibrium, then the forces are equal and

opposite.

.. 150 N 150 N

If there are three forces, when resolving, the triangle of forces will close, if they are in

equilibrium.

F2 F1 F2

F3

F1

F3


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EQUILIBRIUM OF A PARTICLECONTD.

If there are more than three forces, the polygon of forces will be closed if the particle is

in equilibrium.

F3

F2 F2F3 F1 F

F1

F

The closed polygon provides a graphical expression of the equilibrium of forces.

Mathematically: For equilibrium:

R = F = 0

i.e. ( Fx i Fy j) = 0 or (Fx) i (Fy) j


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FREE BODY DIAGRAMS


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FREE BODYDIAGRAMS:

Space diagram represents the sketchof the physical problem. The free bodydiagram selects the significant particleor points and draws the for ce systemon that particle or point.

Steps:

1. Imagine the particle to be isolated or

cut free from its surroundings. Draw or sketch its outlined shape.


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Free Body Diagrams Contd.

3. Label known for ces with their magnitudes and directions. use lettersto represent magnitudes and directionsof unknown forces.

Assume direction of for ce which may

be corrected later.

l


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Example

The crate below has a weight of 50 kg. Drawa free body diagram of the crate, the cord BD

and the ring at B.

CRATE

B ring C

A

D

45o

S l ti


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Solution(a) rate

FD ( forceofcordactin oncrate)

50 kg (wt. ofcrate)

(b) ord D

FB (forceof ringactingoncord)

FD(forceofcrateactingoncord)

CRATE

C45o

B

A

D

S l ti C td


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Solution Contd.

(c) ing

FA (ForceofcordBAacting along ring)

FC(forceof cordBCacting on ring)

FB (forceofcordBDacting on ring)


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Example

S l i C d


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Solution Contd.

FF

FBCAC

o

o AC! !

sin

cos. .............( )

75

753 73 1

Fy=

0 i.e. FBC sin 75

o

- FACcos 75

o

- 1 2=

0

FF

FBCAC

AC!

! 1962 0 26

096620312 0 27 2

.

.. . ......( )

From Equations (1) and (2), 3.73 FAC = 2031.2 0.27 FAC

FAC= 587 N

From (1), FBC = 3.73 x 587 = 2190 N


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RECTANGULAR COMPONENTS

OFFORCE (REVISITED)

x

j

iFx = Fx i

Fy = Fy j

y

F = Fx + Fy

F = |Fx| .i + |Fy| .j

|F|2 = |Fx|2 + |Fy|2

F | | | | | | F Fx Fy! 2 2


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RectangularComponents of a Forcein Space

F = Fx + Fy + Fz

F = |Fx| .i + |Fy| .j + |Fz| .k

|F|2 = |Fx|2 + |Fy|2 + |Fz|2

| | | | | | | | F Fx Fy Fz! 2 2 2

| | | | cos | | | | cos | | | |cos, cos

,

Fx F Fy F Fz F

Cos Cos and Cos are called direction ines of

angles and

x y z

x y z

x y z

! ! !U U U

U U U

U U U

F i S C td


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Forces in Space Contd.

i.e. F = F ( cos Ux i cos Uy j cos Uz k) = F P

F can therefore be expressed as the product of scalar, F

and the unit vector P where: P = cos Ux i cos Uy j cos Uz k.

P is a unit vector of magnitude 1 and of the same direction as F.

P is a unit vector along the line of action of F.


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Force Defined by Magnitude and two Pointson its Line of Action Contd.

Unit vector, P along the lineofactionof F = MN/MN

MN is thedistance, d fromM toN.

P = MN/MN = 1/d( dxi + dyj + dzk)

ecall that: F = F P

F = FP = F/d (dxi + dyj + dzk)

F Fdd

F Fdd

F Fdd

d x x d y y d z z

d d d d

d

d

d

d

d

d

xx

y

y

zz

x y z

x y z

x x y y z z

! ! !

! ! !

!

! ! !

, ,

, ,

cos , cos , cos

2 1 2 1 2 1

2 2 2

U U U


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2.8.3 Addition of Concurrent Forces

in Space

The result t, of t oormore for es i space isobtai edby

summi their rectangularcomponents i.e.

F

i.e. i y j ( i yj )

( ) i ( y)j ( )k

R Fx, Ry Fy , Rz Fz

R Rx2 Ry

2 Rz2

cos Ux Rx/R cos Uy Ry/R cos Uz Rz/R


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Solution


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Solution

Solution:

Position vector ofBH = 0. m i 1.2 mj - 1.2 m k

Magnitude, BH= 0 6 12 12 182 2 2. . . . ! m

P

P

BH

BH BH BH BH

BH

x y z

BH

BHm i m j m k

T T TBH

BH

N

mm i m j m k

T N i N j N k

F N F N F N

! !

! ! !

!

! ! !

p

pp

p

p p p

| | .( . . . )

| |. | || | .

. . .

( ) (500 ) (500 )

, ,

1

180 6 12 12

75018

0 6 12 12

250

250 500 500


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me a chapter and 2

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