microsoft power point - statics handouts 3

8/6/2019 Microsoft Power Point - STATICS Handouts 3

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V. Analysis of 3DForce Systems

Most of the physical quantities in mechanics canbe expressed mathematically by means ofscalars and vectors

A scalar quantity is characterized by a positiveor negative number, i.e., mass, volume andlength

A vector quantity has both magnitude anddirection, i.e., force and moment

Unit vector

A unit vector is a vector having a magnitude of 1. If Ais a vector having a magnitude A 0, then a unitvector having the same direction as A is representedby

where A represents the magnitude of and defines the direction

and sense of

Au

Au

A

A

=

=

A

uA

1

Cartesian Unit Vectors

In threedimensions, the setof Cartesian unitvectors I, j, k, isused to designatethe directions of thex, y, and z axe

respectively,z

y

xi

j

k


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Cartesian Vector Representation

x

z

y

Ayj

Axi

Azk

j

i

k

kAjAiA zyx++=

Cartesian vector form of components:

Magnitude of a Cartesian Vector

x

z

y

Ayj

Axi

Azk

j

i

k

( ) ( ) ( )222

zyx AAAA++=

Ay

Az

Ax

A

Direction of a Cartesian Vector

x

z

y

Ayj

Axi

Azk

A

Ax=cosDirection cosines of A: , ,

A

Ay=cos A

Az=cos

kAjAiA

kAjAiA

Au

kyiu

kA

Aj

A

Ai

A

A

Au

zyx

A

A

zyxA

++=

++=

=

=++

++=

++=

=

coscoscos

1coscoscos

coscoscos

222

If magnitude and coordinate direction angles of A are given,the Cartesian vector form of A is,


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Vector Operations

x

z

y

(Ax + Bx)i

kAjAiA zyx ++=

R

B

A

(Az + Bz)k

(Ay + By)j

kBjBiBzyx ++=

Given:

( ) ( ) ( )

( ) ( ) ( )kBAjBAiBAR

kBAjBAiBAR

zzyyxx

zzyyxx

++==

+++++=+=

'

Vector Addition:

Vector Subtraction:

Resultant of Force Systems

Concurrent Force System

The concept of vector addition may be generalizedand applied to a system of several concurrentforces.

The force resultant is the vector sum of all the forcesin the system

ZkYjXiFR ++==

Problem 1

Determine themagnitude andcoordinatedirection anglesof the resultantforce acting onthe ring.

x

y

z

F1 = {300i+400j}NF2 = {- 500i+1000j+1000k}N

Problem 2

Two forces act onthe hook. Specifythe coordinatedirection angles ofF2 so that theresultant force FRacts along thepositive x axis andhas a magnitude of

800N

y

z

x

F2=700N

F1=300N

600

450

1200


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Position vector

x

z

y

( )zyxP ,,

yj

xi

zk

zkyjxir ++=

The position vector r isdefined as a fixed vectorwhich locates a point inspace relative to another

point.

O

The position vector may ne denoted from pointA to point B in space

( )AAA zyxA ,,

x

y

z

r

rA

rB

( )ixx AB

( ) ( )( ) ( ) ( )kzzjyyixxr

kzjyixkzjyixrrr

rrr

ABABAB

BBABBBAB

BA

++=

++++==

=+

x

z

y

A

B

( )jyy AB

( )kzz AB

Dot product

The dot product of vectors A and B is defined asthe product of the magnitudes of A and B andthe cosine of the angle between their tails

cosAB=

A

B

Laws of Operation

Commutative law

Multiplication by a scalar

Distributive law

=

( ) ( ) ( ) ( )aaaa ===

( ) ( ) ( )DD =


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The dot product of two general vectors A and Bexpressed in Cartesian vector form

Applications

Angle formed between two vectors or intersecting lines

Components of a vector parallel and perpendicular to a line

( ) ( )

zzyyxx

zyxzyx

BABABA

kBjBiBkAjAiA

++=

++++=

Cross Product

The cross product of two vectors A and Byields the vector C,

C = A x B

The magnitude of C is defined as the productof the magnitudes of A and B and the sine ofthe angle between their tails

C = AB sin

Vector C has a direction perpendicular to

the plane containing A and B

C = A x B = (AB sin ) ucC

BA

ucC = AB sin

Laws of Operation

A x B = -B x A

Multiplication by a scalar

a(A x B) = (aA) x B = A x (aB) = (A x B)a

Distributive law

A x (B + D) = (A x B) + (A x D)


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The cross product of two general vectors A andB expressed in Cartesian vector form

The above equation may be expressed indeterminant form

( ) ( )

( ) ( ) ( )kBABAjBABAiBABA

kBjBiBkAjAiA

xyyxxzzxyzzy

zyxzyx

+=

++++=

z

z

y

y

x

x

B

A

k

B

A

j

B

A

i

=

Moment of a Force

The moment of a force F about point or, orabout the moment axis passing through O andperpendicular to the plane containing O and F,can be expressed using the vector crossproduct,

FrMO =

F

MO

OA r

d

Moment axis

The magnitude of the above cross product

( ) FdrFrFMO

=== sinsin zyx

kyxO

FFF

rrr

ijk

rxFM =


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Equilibrium in Three Dimensions

Support reactionsCable

Smooth surface

Roller

Ball and socket

Single smooth pin

Single hinge

Fixed supportFxFz

Fy

Mx

My

Mz

Fx

Fy

Fz

Conditions of Equilibrium

0

0

0

0

0

=

=

=

=++=

=++=

z

y

x

zyxO

zyx

F

F

F

kMjMiMM

kFjFiFF

0

0

0

=

=

=

z

y

x

M

M

M

Problem

Determine thetension incables BC andBD and thereactions atthe ball-andsocket joint Afor the mast

shown.z

y

x

F = 1KN

6m

6m

6m 3m

Solution program (Vector Analysis)

Draw the FBD

Represent each force on the FBD in vector form

Apply the force equation of equilibrium

Take moment about point A to eliminate theother unknowns

Evaluate the cross product and combine liketerms and equate to zero to solve for the

unknows.


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{ }

( ) ( )

( ) ( ) ( )

++

++=

=

+

+=

=

++=

=

222

22

zx

zyx

magnitude

ctorpositionve

z

zy

magnitude

ctorpositionve

BDyBDBD

BDBDBD

BD

BD

BD

BDBD

BCBCy

BCBC

BC

BC

BC

BCBC

zyx

rrr

krjrirT

r

rTT

rr

krjrT

r

rTT

kAjAiAA

NiF

( ) ( )

03

1

2

2

0

03

2

2

2

0

03

21

0

03

1

3

2

3

2

2

2

2

2

0

0

=

=

=

=

=+

=

=

+

+

++++

=+++

=

BDBCz

BDBCy

BDx

BDBDBDBCBCzyx

BDBC

TTA

Z

TTA

Y

TA

X

kTjTiTkTjTkAjAiAi

TTAF

F

( )

( ) ( )

( )

046

0

0223

0

3

1

3

2

3

2

2

2

2

26

0

0

=

=

=

=

+

+

+

=++

=

BD

z

BDBC

x

BDBDBDBCBC

BDBCAB

A

T

M

TT

M

kTjTiTkTjTij

TTFr

M

0

5.1

0

707.02

2

5.1

:

=

=

=

==

=

z

y

x

BC

BD

A

KNA

A

KNT

KNT

Answer

Problem

Rod AB issubjected tothe 200N force.Determine thereactions at theball-and-socketjoint A and thetension incables BD and

BE.

A

E

B

C

D

1.5m

1.5m

2m

2m

200N


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Solution program (Vector Analysis)

Draw the FBD

Represent each force on the FBD in vector form

Apply the force equation of equilibrium

Take moment about point A to eliminate theother unknowns

Expand the equation and equate to zero to solvefor the unknowns

{ }

iTT

kTT

kAjAiAA

NiF

BDBD

BEBE

zyx

=

=

+=

= 200

( ) ( ) ( ) ( )

0

0

200

0200

0

0

0

0200

0

0

=+

=

=

=+

=

=+

=

=++++

=+++

=

BEz

y

y

BDx

BDBEzyx

BDBE

TA

Z

NA

A

Y

TA

X

iTkTkAjAiAj

TTAF

F( )

( ) ( ) ( ) ( )

0200

0

022

0

0200

0

0222005.0

0

0

=+

=

=+

=

=

=

=++++

=++

=

BD

z

BDBE

y

BE

x

BDBE

BDBEABAC

A

T

M

TT

M

T

M

iTkTkjijkji

TTrFr

M

NA

NATNAT

Answers

y

xBD

zBE

200

200200

:

=

==

==


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Problem Set 2Problem Set 3

Instructions for Problem Set

Use short-size bondpaper as answer

sheet

Show complete solution

Deadline of Submission: Monday, August

15, 2011

Problem

The uniformconcrete slabhas a weight of25KN. Determinethe tension ineach of the threeparallelsupportingcables when theslab is held in the

horizontal plane.

8m

y

zx

TC

A

B

C

TA

TB

Problem

A vertical load P =3600N applied tothe tripod causes acompressive forceof 1200N in leg ABand a compressiveforce of 1300N inleg AC. Determinethe force in leg ADand the coordinatesxDand zDof its

lower end D.

3m

2.4m

2.4m

1.8m

xD

zDO

C

A

B

D

P


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microsoft power point - statics handouts 3

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