STATICS

Universiti Tunku Abdul Rahman

CHAPTER 2a: FORCE VECTORS

Prepared by KS Woon @ 2013,ohb 2014

Chapter Outline

1. Scalars and Vectors2. Vector Operations3. Vector Addition of Forces4. Addition of a System of Coplanar Forces5. Cartesian Vectors6. Addition and Subtraction of Cartesian Vectors7. Position Vectors8. Force Vector Directed along a Line9. Dot Product10. Cross Product

2.1 Scalars and Vectors

Scalar A quantity characterized by a positive or negative number. Indicated by letters in italic such as A. e.g. Mass, volume and length.

2.1 Scalars and Vectors

Vector A quantity that has magnitude and direction,

e.g. position, force and moment Represent by a letter with an arrow over it, A

r Represent by a letter with an arrow over it, Magnitude is designated as In this subject, vector is presented as A and its

magnitude (positive quantity) as A

AAr

2.2 Vector Operations

Multiplication and Division of a Vector by a Scalar- Product of vector A and scalar a = aA- Magnitude = - Law of multiplication applies e.g. A/a = ( 1/a ) A, a0

aA- Law of multiplication applies e.g. A/a = ( 1/a ) A, a0


Vector Addition- Addition of two vectors A and B gives a resultant vector R by the parallelogram law.

- Resultant of R can be found by triangle construction.


Vector Addition

- Hence, R = A + B = B + A- Special case: Vectors A and B are collinear (both have the same line of action)


Vector Subtraction- The resultant of the difference between two vectors A and B of the same type may be expressed as:

R = A B = A + ( - B )- Rules of Vector Addition Applies.

2.3 Vector Addition of Forces

2.3.1 Finding a Resultant Force Parallelogram law is carried out to find the resultant

force

Resultant, FR = ( F1 + F2 )


Procedure for Analysisa)Parallelogram Law

Make a sketch using the parallelogram law 2 components forces add to form the resultant force 2 components forces add to form the resultant force Resultant force is shown by the diagonal of the

parallelogram The components is shown by the sides of the

parallelogram


Procedure for Analysisb)Trigonometry

Redraw half portion of the parallelogram Magnitude of the resultant force can be determined Magnitude of the resultant force can be determined

by the law of cosines. Direction of the resultant force, and magnitude of

the two components can be determined by the law of sines.

Example 2.1

The screw eye is subjected to two forces, F1 and F2. Determine the magnitude and direction of the resultant force.

Solution

Parallelogram LawUnknown: magnitude of FR and angle

Solution

TrigonometryLaw of Cosines

( ) ( ) ( )( )( ) NN

NNNNFR2136.2124226.0300002250010000

115cos1501002150100 22

==+=

+= o

Law of Sines

( ) NN 2136.2124226.0300002250010000 ==+=

( )o

o

8.39

9063.06.212

150sin

115sin6.212

sin150

=

=

=

NN

NN

The device is used for surgical replacement of the knee joint. If the force acting along the leg is 360 N, determine its components along the x and y axes.

Example (add on)


Self Reading..

More examples and practices can be found in:Hibbeler, R. C. (2004) Engineering mechanics Hibbeler, R. C. (2004) Engineering mechanics statics. 13th ed. Pearson Prentice Hall, Chapter 2

2.4 Addition of a System of Coplanar Forces

Scalar Notation x and y axes are designated positive and negative Components of forces expressed as algebraic

scalars

sin and cos FFFFFFF

yx

yx

==

+=


Cartesian Vector Notation Cartesian unit vectors i and j are used to designate

the x and y directions Unit vectors i and j have dimensionless magnitude

of unity ( = 1 ) of unity ( = 1 ) Magnitude is always a positive quantity,

represented by scalars Fx and Fy

jFiFF yx +=


Coplanar Force ResultantsTo determine resultant of several coplanar forces: Resolve force into x and y components Addition of the respective components using scalar Addition of the respective components using scalar

algebra Resultant force is found using the parallelogram

law Cartesian vector notation:

jFiFFjFiFF

jFiFF

yx

yx

yx

333

222

111

=

+=

+=


Coplanar Force Resultants Vector resultant is therefore

( ) ( ) jFiFFFFFR

+=

++=

321

If scalar notation are used

( ) ( ) jFiF RyRx +=

yyyRy

xxxRx

FFFFFFFF

321

321

+=

+=


Coplanar Force Resultants In all cases we have

=

= xRx

FF

FF

* Take note of sign conventions

Magnitude of FR can be found by Pythagorean Theorem

= yRy FF

Rx

RyRyRxR F

FFFF 1-22 tan and =+=

* Take note of sign conventions

Example 2.5

Determine x and y components of F1 and F2 acting on the boom. Express each force as a Cartesian vector.

Solution

Scalar Notation:

======

NNNFNNNF

y

x

17317330cos20010010030sin200

1

1o

o

Hence, from the slope triangle, we have

=

125

tan 1

Solution

By similar triangles we have

N1005260

N24013122602

=

=

=

=x

F

F

Scalar Notation:

Cartesian Vector Notation:

N1001352602 =

=yF

===

NNFNF

y

x

100100240

2

2

{ }{ }NjiF

NjiF100240

173100

2

1

=

+=

Example 2.6

The link is subjected to two forces F1 and F2. Determine the magnitude and orientation of the resultant force.

Solution I

Scalar Notation:

=

=

=

NNNF

FF

Rx

xRx

8.23645sin40030cos600

:

oo

=+=

==

N

NNF

FFN

Ry

yRy

8.582

45cos40030sin600

:

8.236

oo

Solution I

Resultant Force

From vector addition, direction angle is

( ) ( )N

NNFR629

8.5828.236 22

=

+=

From vector addition, direction angle is

o9.678.2368.582

tan 1

=

=

NN

Solution II

Cartesian Vector Notation:F1 = { 600cos30i + 600sin30j } NF2 = { -400sin45i + 400cos45j } N

Thus, FR = F1 + F2

= (600cos30N - 400sin45N)i+ (600sin30N + 400cos45N)j

= {236.8i + 582.8j}NThe magnitude and direction of FR are determined in the same manner as before.

Calculate the magnitude of FA and its direction , so that the resultant force is directed along the positive x axis and has a magnitude of 1250 N.

Example (add on)

2.5 Cartesian Vectors

Right-Handed Coordinate System:A rectangular or Cartesian coordinate system is said to be right-handed provided: Thumb of right hand points in the direction of the

positive z axispositive z axis z-axis for the 2D problem would be perpendicular,

directed out of the page.


Rectangular Components of a Vector: A vector A may have one, two or three rectangular

components along the x, y and z axes, depending on orientation

By two successive application of the parallelogram law By two successive application of the parallelogram lawA = A + AzA = Ax + Ay

Combing the equations, A can be expressed as:

A = Ax + Ay + Az


Unit Vector: Direction of A can be specified using unit vectors, i,

j ,k to designate the directions of x, y, z axes respectively.

Unit vector has a magnitude of 1. Unit vector has a magnitude of 1. The positive Cartesian unit vectors are shown in

figure below.


Cartesian Vector Representations: 3 components of A act in the positive i, j and k

directions

A = A i + A j + A kA = Axi + Ayj + AZk

*Note the magnitude and direction of each componentsare separated, easing vector algebraic operations.


Magnitude of a Cartesian Vector: From the colored triangle,

From the shaded triangle, 22' yx AAA +=

22' zAAA +=

Combining the equations gives magnitude of A.

222zyx AAAA ++=

yx


Direction of a Cartesian Vector: Orientation of A is defined as the coordinate

direction angles , and measured between the tail of A and the positive x, y and z axes.

0 , and 180 0 , and 180 The direction cosines of A is:

AAx

=cosAAy

=cos

AAz

=cos


Direction of a Cartesian Vector: Angles , and can be determined by the

inverse cosines.Given that,

A = Axi + Ayj + Azkthen, unit vector UA in the direction of A:

uA = A /A = 1 = (Ax/A)i + (Ay/A)j + (AZ/A)k

where 222zyx AAAA ++=

A = vector quantitiesA = vector magnitude


Direction of a Cartesian Vector: uA can also be expressed as:

uA = cosi + cosj + cosk

Since and u = 1, we have222 AAAA ++= Since and uA = 1, we have

A as expressed in Cartesian vector form is:A = AuA

= Acosi + Acosj + Acosk= Axi + Ayj + AZk

222zyx AAAA ++=

1coscoscos 222 =++

2.6 Addition and Subtraction of Cartesian Vectors

Concurrent Force Systems: Force resultant is the vector sum of all the forces in

the system.

If A = Axi + Ayj + AzkB = Bxi + Byj + Bzk

Therefore, R = A + B

= (Ax + Bx)i + (Ay + By)j + (Az + Bz)k

Hence, FR = F = Fxi + Fyj + Fzk

Example 2.8

Express the force F as Cartesian vector.

Solution

Since two angles are specified, the third angle is found by

( ) ( )oo

5.0707.05.01cos

145cos60coscos1coscoscos

22

222

222

==

=++

=++

Two possibilities exit, namely

( ) o1205.0cos 1 == ( ) o605.0cos 1 ==

( ) ( ) 5.0707.05.01cos 22 ==

Solution

By inspection, = 60 since Fx is in the +x directionGiven F = 200N

F = Fcosi + Fcosj + Fcosk= (200cos60N)i + (200cos60N)j

k+ (200cos45N)k= {100.0i + 100.0j + 141.4k}N

Checking:

( ) ( ) ( ) NFFFF zyx

2004.1410.1000.100 222

222

=++=

++=

Determine the magnitude of F2 and its coordinate direction angles so that the resultant force, FR acts along the positive y axis with a magnitude of 800 N.

Example (add on)

Self Reading..

More examples and practices can be found in:Hibbeler, R. C. (2004) Engineering mechanics Hibbeler, R. C. (2004) Engineering mechanics statics. 13th ed. Pearson Prentice Hall, Chapter 2

2.7 Position Vectors

x,y,z Coordinates Right-handed coordinate system Positive z axis points upwards, measuring the height of

an object or the altitude of a point Points are measured relative Points are measured relative

to the origin, O.


Position Vector Position vector r is defined as a fixed vector which

locates a point in space relative to another point. E.g. r = xi + yj + zk


Position Vector Vector addition gives rA + r = rB Solving

r = rB rA = (xB xA)i + (yB yA)j + (zB zA)kr B A B A B A B A

or r = (xB xA)i + (yB yA)j + (zB zA)k


Length and direction of cable AB can be found by measuring A and B using the x, y, z axes

Position vector r can be established Magnitude r represent the length of cable Angles, , and represent the direction of the cable Unit vector, u = r/r

Example 2.12

An elastic rubber band is attached to points A and B. Determine its length and its direction measured from A towards B.

Solution

Position vectorr = [-2m 1m]i + [2m 0]j + [3m (-3m)]k= {-3i + 2j + 6k}m

Magnitude = length of the rubber band

Unit vector in the director of ru = r /r

= -3/7i + 2/7j + 6/7k

( ) ( ) ( ) mr 7623 222 =++=

Solution

= cos-1(-3/7) = 115 = cos-1(2/7) = 73.4 = cos-1(6/7) = 31.0

2.8 Force Vector Directed along a Line

In 3D problems, direction of F is specified by 2 points, through which its line of action lies

F can be formulated as a Cartesian vector:

F u rF = F u = F (r/r)

Note that F has units of forces (N) unlike r, with units of length (m)

2.8 Force Vector Directed along a Line

Force F acting along the chain can be presented as a Cartesian vector by- Establish x, y, z axes- Form a position vector r along length of chain

u r Unit vector, u = r/r that defines the direction of both the chain and the force

We get F = Fu

Example 2.13

The man pulls on the cord with a force of 350N. Represent this force acting on the support A, as a Cartesian vector and determine its direction.

Solution

End points of the cord are A (0m, 0m, 7.5m) and B (3m, -2m, 1.5m)r = (3m 0m)i + (-2m 0m)j + (1.5m 7.5m)k= {3i 2j 6k}m

Magnitude = length of cord AB

Unit vector, u = r /r

= 3/7i - 2/7j - 6/7k

( ) ( ) ( ) mmmmr 7623 222 =++=

Solution

Force F has a magnitude of 350N, direction specified by u.

F = Fu= 350N(3/7i - 2/7j - 6/7k)= {150i - 100j - 300k} N= {150i - 100j - 300k} N

= cos-1(3/7) = 64.6 = cos-1(-2/7) = 107 = cos-1(-6/7) = 149

Example (add on)

The tower is held in place by 3 cables. If the force of each cable acting on the tower is shown, determine the magnitude & coordinate direction angles , , of the resultant force. Take x = 20 m, y = 15 m.

2.9 Dot Product

Dot product of vectors A and B is written as AB(Read A dot B)

Define the magnitudes of A and B and the angle between their tails

AB = AB cos where 0 180AB = AB cos where 0 180 Referred to as scalar product of vectors as result is a

scalar

2.9 Dot Product

Laws of Operation1. Commutative law

AB = BA2. Multiplication by a scalar2. Multiplication by a scalar

a(AB) = (aA)B = A(aB) = (AB)a3. Distribution law

A(B + D) = (AB) + (AD)

2.9 Dot Product

Cartesian Vector Formulation- Dot product of Cartesian unit vectors

ii = (1)(1)cos0 = 1; ij = (1)(1)cos90 = 0- Similarly- Similarly

ii = 1 jj = 1 kk = 1; ij = 0 ik = 0 jk = 0

- If AB in 3D cartesian:AB = (Axi + Ayj + Azk) (Bxi + Byj + Bzk)

= AxBx (ii) + AxBy (ij) + AxBz (ik) + AyBx (ji) + AyBy (jj) + AyBz (jk) + AzBx (ki) + AzBy (kj) + AzBz (kk)

2.9 Dot Product

Cartesian Vector Formulation Dot product of 2 vectors A and B

AB = AxBx + AyBy + AzBz Applications: Applications:

The angle formed between two vectors or intersecting lines. = cos-1 [(AB)/(AB)] 0 180

The components of a vector parallel and perpendicular to a line.Aa = A cos = Au

Example 2.17

The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB.

Solution

Since

( ) ( ) ( )kji

kjir

ru

B

BB

429.0857.0286.0362

362222

++=

++

++==

rrr

rrr

r

rr

Thus

( ) ( )N

kjijuFFF

B

AB

1.257)429.0)(0()857.0)(300()286.0)(0(429.0857.0286.0300.

cos

=

++=

++==

=

rrrrrr

rr

Solution

Since result is a positive scalar, FAB has the same sense of direction as uB. Express in Cartesian form

( )( )Nkji

kjiNuFF ABABAB

}1102205.73{429.0857.0286.01.257

rrr

rrr

rrr

++=

++=

=

Perpendicular componentNkjikjijFFF

Nkji

AB }110805.73{)1102205.73(300

}1102205.73{

rrrrrrrrrr

rrr

+=++==

++=

Solution

Magnitude can be determined from F or from Pythagorean Theorem,

FFF AB22

=

rrr

( ) ( )N

NN155

1.257300 22

=

=

Example (add on)Calculate the magnitude and angle of the projected component of the force along the pipe AO.

2.10 Cross Product

Cross product of two vectors A and B yields C, which is written as

C = A X BMagnitude Magnitude of C is the product of

the magnitudes of A and B For angle , 0 180

C = AB sin

2.10 Cross Product

Direction Vector C has a direction that is perpendicular to the

plane containing A and B such that C is specified by the right hand rule

Expressing vector C when magnitude and direction are Expressing vector C when magnitude and direction are known as:

C = A X B = (AB sin)uC

Scalar for magnitude of C

Unit vector of C direction

2.10 Cross Product

Laws of Operations1. Commutative law is not valid

A X B B X ARather, Rather,

A X B = - B X A Cross product A X B yields a

vector opposite in direction to C

B X A = -C

2.10 Cross Product

Laws of Operations2. Multiplication by a Scalar

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

3. Distributive LawA X ( B + D ) = ( A X B ) + ( A X D )

Proper order of the cross product must be maintained since they are not commutative

2.10 Cross Product

Cartesian Vector Formulation Use C = AB sin on pair of Cartesian unit vectors A more compact determinant in the form as:

rrr

The application of using cross product to calculate moment of a force will be discussed in Topic 4.

zyx

zyx

BBBAAAkji

BA

rrr

rr=

QUIZ

1. Which one of the following is a scalar quantity?A) Force B) Position C) Mass D) Velocity

2. For vector addition, you have to use ______ law.A) Newtons SecondB) the arithmeticC) Pascals D) the parallelogram

QUIZ

3. Can you resolve a 2-D vector along two directions, which are not at 90 to each other?A) Yes, but not uniquely.B) No.C) Yes, uniquely.

4. Can you resolve a 2-D vector along three directions (say at 0, 60, and 120)?A) Yes, but not uniquely.B) No.C) Yes, uniquely.

QUIZ

5. Resolve F along x and y axes and write it in vector form. F = { ___________ } N

A) 80 cos (30) i 80 sin (30) jB) 80 sin (30) i + 80 cos (30) j

i j30

xy

C) 80 sin (30) i 80 cos (30) jD) 80 cos (30) i + 80 sin (30) j

6. Determine the magnitude of the resultant (F1 + F2) force in N when F1={ 10i + 20j }N and F2={ 20i + 20j } N .

A) 30 N B) 40 N C) 50 ND) 60 N E) 70 N

30

F = 80 N

QUIZ

7. Vector algebra, as we are going to use it, is based on a ___________ coordinate system.

A) Euclidean B) Left-handedC) Greek D) Right-handed E) Egyptian

8. The symbols , , and designate the __________ of a 3-D Cartesian vector.

A) Unit vectors B) Coordinate direction anglesC) Greek societies D) X, Y and Z components

QUIZ

9. What is not true about an unit vector, uA ?A) It is dimensionless.B) Its magnitude is one.C) It always points in the direction of positive X- axis.D) It always points in the direction of vector A.D) It always points in the direction of vector A.

10. If F = {10 i + 10 j + 10 k} N andG = {20 i + 20 j + 20 k } N, then F + G = { ____ } N

A) 10 i + 10 j + 10 kB) 30 i + 20 j + 30 kC) 10 i 10 j 10 kD) 30 i + 30 j + 30 k

QUIZ

11. A position vector, rPQ, is obtained byA) Coordinates of Q minus coordinates of PB) Coordinates of P minus coordinates of QC) Coordinates of Q minus coordinates of the originD) Coordinates of the origin minus coordinates of PD) Coordinates of the origin minus coordinates of P

12. A force of magnitude F, directed along a unit vector U, is given by F = ______ .A) F (U)B) U / FC) F / UD) F + UE) F U

QUIZ

13. P and Q are two points in a 3-D space. How are the position vectors rPQ and rQP related?

A) rPQ = rQP B) rPQ = - rQPC) rPQ = 1/rQP D) rPQ = 2 rQP

14. If F and r are force vector and position vectors, respectively, in SI units, what are the units of the expression (r * (F / F)) ?

A) Newton B) DimensionlessC) Meter D) Newton - MeterE) The expression is algebraically illegal.

QUIZ

15. Two points in 3 D space have coordinates of P (1, 2, 3) and Q (4, 5, 6) meters. The position vector rQP is given byA) {3 i + 3 j + 3 k} mB) { 3 i 3 j 3 k} mB) { 3 i 3 j 3 k} mC) {5 i + 7 j + 9 k} mD) { 3 i + 3 j + 3 k} mE) {4 i + 5 j + 6 k} m

16. Force vector, F, directed along a line PQ is given byA) (F/ F) rPQ B) rPQ/rPQC) F(rPQ/rPQ) D) F(rPQ/rPQ)

QUIZ

17. The dot product of two vectors P and Q is defined asA) P Q cos B) P Q sin C) P Q tan D) P Q sec

P

Q

18. The dot product of two vectors results in a _________ quantity.A) Scalar B) VectorC) Complex D) Zero

Q

QUIZ

19. If a dot product of two non-zero vectors is 0, then the two vectors must be _____________ to each other.A) Parallel (pointing in the same direction)B) Parallel (pointing in the opposite direction)C) PerpendicularD) Cannot be determined.

20. If a dot product of two non-zero vectors equals -1, then the vectors must be ________ to each other.A) Parallel (pointing in the same direction)B) Parallel (pointing in the opposite direction)C) PerpendicularD) Cannot be determined.

QUIZ

1. The dot product can be used to find all of the following except ____ .A) sum of two vectorsB) angle between two vectorsC) component of a vector parallel to another lineC) component of a vector parallel to another lineD) component of a vector perpendicular to another line

2. Find the dot product of the two vectors P and Q.P = {5 i + 2 j + 3 k} m

Q = {-2 i + 5 j + 4 k} mA) -12 m B) 12 m C) 12 m2D) -12 m2 E) 10 m2

Universiti Tunku Abdul Rahman

CHAPTER 2b: MOMENTS OF INERTIA

Prepared by KS Woon @ 2013, ohb_201405

Chapter Outline

1. Definitions of Moments of Inertia for Areas2. Parallel-Axis Theorem for an Area3. Moments of Inertia for Composite Areas

10.1 Definition of Moments of Inertia for Areas

Centroid for an area is determined by the first moment of an area about an axis.

Second moment of an area is referred as the moment of inertia.

Moment of inertia of an area originates whenever one Moment of inertia of an area originates whenever one relates the normal stress, or force per unit area


Moment of Inertia Consider area A lying in the x-y plane By definition, moments of inertia of the differential

plane area dA about the x and y axes == dAxdIdAydI 22

For entire area, moments of inertia are given by

=

=

==

Ay

Ax

yx

dAxI

dAyI

dAxdIdAydI

2

2

22


Moment of Inertia Formulate the second moment of dA about the pole O

or z axis This is known as the polar axis

dArdJ 2=where r is perpendicular from the pole (z axis) to the element dA

Polar moment of inertia for entire area,

dArdJO2

=

yxAOIIdArJ +==

2

10.2 Parallel Axis Theorem for an Area

For moment of inertia of an area known about an axis passing through its centroid, determine the moment of inertia of area about a corresponding parallel axis using the parallel axis theorem

Consider moment of inertia of the shaded area Consider moment of inertia of the shaded area A differential element dA is

located at an arbitrary distance yfrom the centroidal x axis


The fixed distance between the parallel x and x axes is defined as dy

For moment of inertia of dA about x axis ( )+= yx dAdydI 2'

For entire area

First integral represent the moment of inertia of the area about the centroidal axis

( )

++=

+=

AyAyA

A yx

dAddAyddAy

dAdyI22

2

'2'

'


Second integral = 0 since x passes through the areas centroid C

Third integral represents the total area A2

0;0' ydAydAy

+=

===

Similarly

For polar moment of inertia about an axis perpendicular to the x-y plane and passing through pole O (z axis)

2

2

2

AdJJ

AdII

AdII

CO

xyy

yxx

+=

+=

+=

Example 10.1

Determine the moment of inertia for the rectangular area with respect to (a) the centroidal x axis, (b) the axis xbpassing through the base of the rectangular, and (c) the pole or z axis perpendicular to the x-y plane and passing through the centroid C.passing through the centroid C.

Solution

Part (a)Differential element chosen, distance y from x axis.Since dA = b dy,

32/ 22/ 22 1')'('' bhdyybdyydAyI hh ====

Part (b)By applying parallel axis theorem,

32/

2

2/22

12')'('' bhdyybdyydAyI

hhAx====

32

32

31

2121 bhhbhbhAdII xxb =

+=+=

Solution

Part (c)For polar moment of inertia about point C,

121 3

'hbI y =

)(121

1222

'bhbhIIJ yxC +=+=

10.4 Moments of Inertia for Composite Areas

Composite area consist of a series of connected simpler parts or shapes

Moment of inertia of the composite area = algebraic sum of the moments of inertia of all its parts

Procedure for AnalysisComposite Parts Divide area into its composite parts and indicate the

centroid of each part to the reference axisParallel Axis Theorem Moment of inertia of each part is determined about its

centroidal axis

10.4 Moments of Inertia for Composite Areas

Procedure for AnalysisParallel Axis Theorem When centroidal axis does not coincide with the

reference axis, the parallel axis theorem is used Summation Moment of inertia of the entire area about the

reference axis is determined by summing the results of its composite parts

Example 10.4

Compute the moment of inertia of the composite area about the x axis.

Solution

Composite PartsComposite area obtained by subtracting the circle from the rectangle.Centroid of each area is located in the figure below.

Solution

Parallel Axis TheoremCircle

( ) ( ) ( ) ( )2

'

1AdII yxx +=

Rectangle

( ) ( ) ( ) ( ) 46224 104.1175252541

mm=+= pipi

( )( ) ( )( )( ) ( ) 46232

'

105.11275150100150100121

mm

AdII yxx

=+=

+=

Solution

SummationFor moment of inertia for the composite area,

( ) ( )( ) 46

66

10101105.112104.11

mm

I x=

+=

( ) 4610101 mm=

Example (add on)

Locate the centroid, of the cross-section area. Then, determine the moment of inertia of the area about the centroidal y axis.

x

QUIZ

1. The definition of the Moment of Inertia for an areainvolves an integral of the formA) x dA. B) x2 dA.C) x2 dm. D) m dA.

2. Select the SI units for the Moment of Inertia for an area.A) m3B) m4C) kgm2 D) kgm3

QUIZ

3. A pipe is subjected to a bending moment as shown. Which property of the pipe will result in lower stress (assuming a constant cross-sectional area)?A) Smaller Ix B) Smaller IyC) Larger Ix D) Larger Iy

My

M

xC) Larger Ix D) Larger Iy

4. In the figure to the right, what is the differential moment of inertia of the element with respect to the y-axis (dIy)?A) x2 ydx B) (1/12)x3dyC) y2 x dy D) (1/3)ydy

Pipe sectionx

x

y=x3

x,y

y

QUIZ

5. The parallel-axis theorem for an area is applied betweenA) An axis passing through its centroid and any corresponding parallel axis.B) Any two parallel axis.C) Two horizontal axes only.D) Two vertical axes only.

6. The moment of inertia of a composite area equals the ____ of the MoI of all of its parts.A) Vector sumB) Algebraic sum (addition or subtraction)C) Addition D) Product

QUIZ

7. For the area A, we know the centroids (C) location, area, distances between the four parallel axes, and the MoI about axis 1. We can determine the MoI about axis 2by applying the parallel axis theorem ___ .A) Directly between the axes 1 and 2.A) Directly between the axes 1 and 2.B) Between axes 1 and 3 and

then between the axes 3 and 2.C) Between axes 1 and 4 and

then axes 4 and 2.D) None of the above.

d3d2

d1

4321

A

C

Axis

QUIZ

8. For the same case, consider the MoI about each of the four axes. About which axis will the MoI be the smallest number?A) Axis 1B) Axis 2 A

AxisB) Axis 2C) Axis 3D) Axis 4E) Can not tell.

d3d2d1

4321

A

C

QUIZ

9. For the given area, the moment of inertia about axis 1 is 200 cm4 . What is the MoI about axis 3?A) 90 cm4 B) 110 cm4C) 60 cm4 D) 40 cm4

A=10 cm2

Cd2

d1

321

C

C) 60 cm4 D) 40 cm4

10. The moment of inertia of the rectangle about the x-axis equalsA) 8 cm4. B) 56 cm4 .C) 24 cm4 . D) 26 cm4 .

d1 1

d1 = d2 = 2 cm

2cm

2cm

3cm

x

STATICS

Documents

resultant vector r

vector addition of forces2

vector addition of forces4

vector operations3

force vectors

resultant of r

parallelogram magnitude

aa magnitude