lec 02 (statics review)-- jan 11

Dr. Ali Keshavarz

General Principles

Equilibrium of Bodies

At rest or moving with constant

velocity

Mechanics of Materials

Physical science concerned with the state of rest or

motion of bodies that are subject to forces

Solid Mechanics - Lec. 2 1

Dr. Ali Keshavarz

Newtons Laws of Motion First Law.

A particle originally at rest or moving in a straight line with constant velocity, will remain in this state provided that the particle is not subjected to an unbalanced force.

Second Law. A particle acted upon by an unbalanced force F experiences an

acceleration, a that has the same direction as the force and a magnitude that is proportional to the force.

Third Law. The mutual forces of action and reaction between two particles

are equal, opposite, and collinear.

mF a


Dr. Ali Keshavarz

Newtons Law of Gravitational Attraction

1 22

m mF Gr

1 2

12 3 2

: Force of gravitation between two particles and : Mass of each particle

: Distance between two particles: Universal constant of gravitation

66.73 10 m / kg s

Fm mrGG


Dr. Ali Keshavarz

Significant Figures Accuracy of a number is specified by the number of significant figures it contains. A

significant figure is any digit, including a zero, provided it is not used to specify the location of the decimal point for the number, i.e., 0.5 has only one significant digit.

Example 57 098 and 44.893 (Both numbers have 5 significant digits) When numbers begin or end with zeros, it gets little confusing. Consider 400. In this kind of situations, express the number in engineering notation

(exponent is used in multiples of 3). Then, 400 = 0.4x103. Only 1 Significant digit. 2500 for example can be written as 2.5x103 with 2 significant digits or can be written

as 2.50x103 with 3 significant digits. This is done to specify more accuracy. 0.00546 can be written as 5.46x10-3 with 3 significant digits. Rounding Off Numbers

Rule: Use of 3 significant digits is usually enough in final answers. In intermediate calculations keep a higher number of significant digits.


Dr. Ali Keshavarz

Force Vectors Scalar. A quantity characterized by a positive or negative number such as mass,

volume, length. Vector. A quantity that has

A magnitude (how big is your vectors length compared to a given reference) A direction (on a line you can have 2 direction choices) A line of action

Examples: Position, velocity, force When specifying direction: Always know your reference, i.e., CW from an axis.


Dr. Ali Keshavarz

Vector Math Multiplication and division of a vector with a scalar

Changes the magnitude only (aA or A/a or -A) Vector addition (Commutative Law: A + B = B + A)

Parallelogram law or triangular construction Vector subtraction: (A B = A + (-B))

Same as addition concept Just multiply the vector being subtracted and add two


Dr. Ali Keshavarz

Vector Resolution Resolution of a vector

Resolve into 2 components on 2 known line of actions 2 known line of actions are not necessarily perpendicular WHY? We may need to resolve due to geometry


Dr. Ali Keshavarz

Trigonometric Laws and Force Notation Sine Law and Cosine

Law Trigonometric

relations Geometric relations Force Notation

Scalar and Vector500 N and F or F

F


Dr. Ali Keshavarz

Vector Addition of 3 or More Forces Application of successive parallelogram law will

lead to the result. This can get involved and can be error prone in

terms of geometry and trigonometry.


Dr. Ali Keshavarz

Adding Coplanar Forces: Cartesian Vector Notation

x yF F F i j, : unit vectors

magnitude = 1 (unity)direction: +/- sign

i j


Dr. Ali Keshavarz

Resultant of Coplanar Forces

1 1 1

2 2 2

3 3 3

yx

yx

yx

F FF F

F F

F i jF i jF i j


Dr. Ali Keshavarz

The Resultant

1 2 3R F F F FR Rx RyF F F i j

1 2 3

1 2 3

Rx x x x

Ry y y y

F F F FF F F F

Rx x

Ry y

F FF F

2 2

1tan

R Rx Ry

Ry

Rx

F F F

FF


Dr. Ali Keshavarz

Unit Vector is used to specify direction

Rectangular 3D Components of A Vector

x y z

x y zA A A

A A A AA i j k

or

0

A AAAA

Au A u


Dr. Ali Keshavarz

Revisit the Unit Vector

or

0

A AAAA

Au A u

cos cos cos

yx zA

A

AA AA A A A

Au i j k

u i j k2 2 2cos cos cos 1

cos cos cosA

x y z

AA A AA A A

A ui j k

i j k


Dr. Ali Keshavarz

Addition and Subtraction of Cartesian Vectors

x y z

x y z

A A AB B B

A i j kB i j k

( ) ( ) ( )x x y y z xA B A B A B

R A Bi j k

Concurrent Force Systems:

R x y zF F F F F i j kSolid Mechanics - Lec. 2 15

Dr. Ali Keshavarz

Position Vectors

x y z r i j kSolid Mechanics - Lec. 2 16

Dr. Ali Keshavarz

Dot Product How do we find angle

between two lines? Dot Product = Scalar

Product (result is scalar) Commutative Law:

AB = BA Multiply by a Scalar:

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

Distributive Law: A(B+D) = (AB) + (AD)

cos(0 180 )

AB

A B

10

i i j j k ki j i k k j

x y z

x y z

A A AB B B

A i j kB i j k

x x y y z zA B A B A B A BSolid Mechanics - Lec. 2 17

Dr. Ali Keshavarz

Applications of Dot Product

The angle formed between 2 vectors

Components of a vector parallel/perpendicular to a line

1cos 0 180AB

A B

||

||

coscos ( )

A AA

A u A u u


Dr. Ali Keshavarz

Force System ResultantsMoment of a Force

Moment: A measure of the tendency of the

force to cause a body to rotate about a point or

axis


Dr. Ali Keshavarz

Moment of a Force

Scalar Formulation Magnitude:

Direction: Right-Hand Rule

OM Fd


Dr. Ali Keshavarz

Cross Product The result is a vector The order of multiplication

does matter Magnitude:

Direction:

sinC AB

( sin ) CAB C AB u


Dr. Ali Keshavarz

Cross ProductCartesian Vector Formulation

i i = j j = kk = 0i j k jk i k i = j

j i k k j i ik = j

x y z

x y z

A A AB B B

A i j kB i j k

( ) ( ) ( )x y z y z z y x z z x x y y xx y z

A A A A B A B A B A B A B A BB B B

i j k

AB i j k

MINUS SIGNImportant!


Dr. Ali Keshavarz

Moment of a ForceVector Formulation The moment of force F about point O The moment of force F about the

moment axis passing through O and perpendicular to the plane containing O and F

r is the position vector from O to anypoint lying on the line of action of F

Magnitude:

Direction: Right-hand rule

O M rF

sin ( sin )OM rF F r Fd


Dr. Ali Keshavarz

Moment of a ForceCartesian Vector Formulation

x y z

x y z

r r rF F F

r i j kF i j k

( )( ) ( )

( ) ( ) ( )

( ) ( ) ( )O yO x O z

O x y z

x y z

y z z y x z z x x y y x

MM M

O x O y O z

r r rF F F

r F r F r F r F r F r F

M M M

i j kM rF

i j k

i j k


Dr. Ali Keshavarz

Resultant Moment ofA System of Forces

( )OR M rF


Dr. Ali Keshavarz

Equilibrium Conditions

O

F 0M 0

FOR A RIGID BODY


Dr. Ali Keshavarz

Springs

Linear spring (ideal)

= stiffness of the spring (N/m or lb/ft)

Undeformed length of the spring,

F ks

0l

k

if 0, Pushing Force, if 0, Pulling Force, ss

FF


Dr. Ali Keshavarz

Cables and Pulleys

Assumptions: They have NO weight They DONT stretch They ONLY can

support tension Assuming NO friction

in the pulley: For equilibrium, applied

force is balanced at the other end of the pulley.


Dr. Ali Keshavarz

Equilibrium in 2D: Free-Body Diagrams Due to symmetry of geometry and loading

conditions, many engineering problems can be simplified to 2 dimensions.

All known (given) and unknown external forcesacting on a body must be identified.

FBD: An isolated sketch of a body from its surroundings showing all the forces and couple moments that the surroundings exert on the body.


Dr. Ali Keshavarz

FBD Procedure Draw the outline shape

Make sure the body is totally isolated or cut free from its constraints and connections and sketch an outlined shape

Show all forces and couple moments Identify all external forces and couple moments that act on the

body Applied loadings Reactions occurring at the supports and contact with other bodies Weight of a body

Identify each loading and give directions Label given forces and couple moments with their magnitudes

and directions. Label with letters unknowns resolved into x, y components, i.e., Ax, Ay, etc., when necessary.


Dr. Ali Keshavarz

Weight and Center of Gravity

Forces due to gravitational field acting on a body can be considered as a system of parallel forces.

Using our previous analysis, we can find a single resultant force: Weight of the body.

Location of Weights point of application is called the center of gravity.


Dr. Ali Keshavarz

Support Reactions in 2D

There are common engineering joints/supports. Consider attachments of a simple horizontal beam.


Dr. Ali Keshavarz

Common 2D Supports


Dr. Ali Keshavarz

Examples: Support ReactionsCable

Rocker

SmoothSurface

Fixed

SmoothPin


Dr. Ali Keshavarz

Equations of Equilibrium: 2D

O

F 0M 0

000

x

y

O

FFM

MO is the sum of couple moments and moments of all forces about an axis

perpendicular to x-y plane and passing through the arbitrary point O. O can be

on the body or off the body.


Dr. Ali Keshavarz Solid Mechanics - Lec. 2 36

Dr. Ali Keshavarz

Procedure for Analysis Free-Body Diagram

Establish x-y axes in any suitable orientation Draw an outlined shape of the body Show all the forces and couple moments acting on the body Label all loadings and specify their directions with respect to x, y axes. For an unknown load (force/moment), directions can be assumed along

the known line of action Indicate dimensions necessary for calculating moments

Equations of Equilibrium Apply the moment equation of equilibrium about a point O that lies at

the intersection of the lines of actions of two unknown forces. This way, you can eliminate the moment effect of those forces

Choose the most convenient x-y axes so that the resolutions of forces along x and y would be easier

If a negative result found, it indicates the chosen direction must be reversed


Dr. Ali Keshavarz

Equilibrium in 3D: Support Reactions


Dr. Ali Keshavarz

Examples: 3D Supports

SmoothPin

ThrustBearing


Dr. Ali Keshavarz

Equations of Equilibrium: 3D

O

F 0M 0

000

x

y

z

FFF

000

x

y

z

MMM

x y z

O x y z

F F FM M M

F i j k 0M i j k 0

You can use these equations to solve for 6 unknowns!


Dr. Ali Keshavarz

Structural Analysis

Simple Trusses


Dr. Ali Keshavarz

Planar Trusses: Roof


Dr. Ali Keshavarz

Planar Trusses: Bridge


Dr. Ali Keshavarz

Method of Joints

To design a truss, we must know the force in each member.

If FBD of the entire truss is used, the forces in the members are internal forces and they cannot be determined.

Considering the FBD of a joint solves this problem.


Dr. Ali Keshavarz

Method of Joints

Analysis must start from a joint with at least one known force and at most 2 unknowns. 0

0x

y

FF


Dr. Ali Keshavarz

Determining Directions

Always assume the unknown member forces to be tension.Positive result Tension (T)Negative result Compression (C)Once an unknown force is found, use the

correct direction (T or C) in subsequent step. Determine the correct direction by

inspection when possible.


Dr. Ali Keshavarz

Method of Sections

If a body is in equilibrium, any part of that body must also be in equilibrium.

Using the same idea we can cut or section the members of an entire truss.

This can be desirable to determine force of a specific member.


Dr. Ali Keshavarz

Method of Sections

Since we have 3 equations, try to cut through a maximum of 3 members with unknown forces.

0, 0, 0x y OF F M


Dr. Ali Keshavarz

Determining Directions

Always assume the unknown member forces at the cut section to be tension.Positive result Tension (T)Negative result Compression (C)

Determine the correct direction by inspection when possible


Dr. Ali Keshavarz

APPLICATIONS

To design the structure for supporting a water tank, we will need to know the weights of the tank and water as well as the locations where the resultant forces representing these distributed loads are acting.

How can we determine these weights and their locations?


Dr. Ali Keshavarz

APPLICATIONS (continued)

One concern about a sport utility vehicle (SUVs) is that it might tip over while taking a sharp turn.

One of the important factors in determining its stability is the SUVs center of mass.

Should it be higher or lower for making a SUV more stable?

How do you determine its location?Solid Mechanics - Lec. 2 62

Dr. Ali Keshavarz

Center of Gravity and Centroid for a Body

xdW ydW zdW

x y zdW dW dW

W V dW dV Specific WeightWeight per unit

volume

x dV y dV z dV

x y zdV dV dV


Dr. Ali Keshavarz

Center of Massg

DensityMass per unit

volume

x dV y dV z dV

x y zdV dV dV

dm dV

xdm ydm zdmx y z

dm dm dm


Dr. Ali Keshavarz

CentroidCentroid

Geometric centerHomogenous

V V V

V V V

xdV ydV zdVx y z

dV dV dV

A A A

A A A

xdA ydA zdAx y z

dA dA dA

L L L

L L L

xdL ydL zdLx y z

dL dL dL


Dr. Ali Keshavarz

Symmetry


Dr. Ali Keshavarz

STEPS FOR DETERMING AREA CENTROID

1. Choose an appropriate differential element dA at a general point (x,y). Hint: Generally, if y is easily expressed in terms of x(e.g., y = x2 + 1), use a vertical rectangular element. If the converse is true, then use a horizontal rectangular element.

2. Express dA in terms of the differentiating element dx (or dy).

3. Determine coordinates of the centroid of the rectangular element in terms of the general point (x,y).

4. Express all the variables and integral limits in the formula using either x or y depending on whether the differential element is in terms of dx or dy, respectively, and integrate.

Note: Similar steps are used for determining CG, CM, etc.. These steps will become clearer by doing a few examples.

)~,~( yx


Dr. Ali Keshavarz

EXAMPLE

Solution

1. Since y is given in terms of x, choose dA as a vertical rectangular strip.

x,y

x , y~~2. dA = y dx = (9 x2) dx

3. x = x and y = y / 2~~

Given: The area as shown.

Find: The centroid location (x , y)

Plan: Follow the steps.


Dr. Ali Keshavarz

EXAMPLE (continued)

4. x = ( A x dA ) / ( A dA )~

0

0

0 x ( 9 x2) d x [ 9 (x2)/2 (x4) / 4] 30 ( 9 x2) d x [ 9 x (x3) / 3 ] 3

= ( 9 ( 9 ) / 2 81 / 4 ) / ( 9 ( 3 ) ( 27 / 3 ) )

= 1.13 ft

3= =

3

33.60 ftA y dA 0 ( 9 x2) ( 9 x2) dx

A dA 0 ( 9 x2) d x

3

=y = = ~


Dr. Ali Keshavarz

More Example

Given: The area as shown.

Find: The x of the centroid.

Plan: Follow the steps.

Solution

1. Choose dA as a horizontal rectangular strip.(x1,,y) (x2,y)

2. dA = ( x2 x1) dy

= ((2 y) y2) dy

3. x = ( x1 + x2) / 2

= 0.5 (( 2 y) + y2 )

~


Dr. Ali Keshavarz

More Example (continued)

4. x = ( A x dA ) / ( A dA )~

A dA = 0 ( 2 y y2) dy[ 2 y y2 / 2 y3 / 3] 1 = 1.167 m2

1

0

A x dA = 0 0.5 ( 2 y + y2 ) ( 2 y y2 ) dy= 0.5 0 ( 4 4 y + y2 y4 ) dy= 0.5 [ 4 y 4 y2 / 2 + y3 / 3 y5 / 5 ] 1

= 1.067 m30

1

1

~

x = 1.067 / 1.167 = 0.914 m


Dr. Ali Keshavarz

Composite Bodies

xW yW zWx y zW W W

CompositeMade of connected

simpler shapes


Dr. Ali Keshavarz

APPLICATIONS

The I-beam is commonly used in building structures.

When doing a stress analysis on an I - beam, the location of the centroid is very important.

How can we easily determine the location of the centroid for a given beam shape?


Dr. Ali Keshavarz

APPLICATIONS (continued)

Cars, trucks, bikes, etc., are assembled using many individual components.

When designing for stability on the road, it is important to know the location of the bikes center of gravity (CG).

If we know the weight and CG of individual components, how can we determine the location of the CG of the assembled unit?


Dr. Ali Keshavarz

CONCEPT OF A COMPOSITE BODY

Many industrial objects can be considered as composite bodies made up of a series of connected simpler shaped parts or holes, like a rectangle, triangle, and semicircle.

Knowing the location of the centroid, C, or center of gravity, G, of the simpler shaped parts, we can easily determine the location of the C or G for the more complex composite body.

a

be

d

ab

ed


Dr. Ali Keshavarz

This can be done by considering each part as a particle and following the procedure as described.

This is a simple, effective, and practical method of determining the location of the centroid or center of gravity.

CONCEPT OF A COMPOSITE BODY(continued)

a

be

d

ab

ed


Dr. Ali Keshavarz

STEPS FOR ANALYSIS

1. Divide the body into pieces that are known shapes. Holes are considered as pieces with negative weight or size.

2. Make a table with the first column for segment number, the second column for weight, mass, or size (depending on the problem), the next set of columns for the moment arms, and, finally, several columns for recording results of simple intermediate calculations.

3. Fix the coordinate axes, determine the coordinates of the center of gravity of centroid of each piece, and then fill-in the table.

4. Sum the columns to get x, y, and z. Use formulas like

x = ( xi Ai ) / ( Ai ) or x = ( xi Wi ) / ( Wi )This approach will become clear by doing examples!


Dr. Ali Keshavarz

EXAMPLE

Given: The part shown.

Find: The centroid of the part.

Plan: Follow the steps for analysis.

Solution:

1. This body can be divided into the following pieces: rectangle (a) + triangle (b) + quarter circular (c) semicircular area (d)

a

bc

d


Dr. Ali Keshavarz

EXAMPLE (continued)

Steps 2 & 3: Make up and fill the table using parts a, b, c, and d.

39.8376.528.0

274.59

- 2/3

5431.5 9 0

1.51

4(3) / (3 )4(1) / (3 )

37

4(3) / (3 )0

184.5

9 / 4 / 2

RectangleTriangleQ. CircleSemi-Circle

A y( in3)

A x( in3)

y(in)

x(in)

Area A(in2)

Segment

abc

d


Dr. Ali Keshavarz

x = ( x A) / ( A ) = 76.5 in3/ 28.0 in2 = 2.73 iny = ( y A) / ( A ) = 39.83 in3 / 28.0 in2 = 1.42 in

4. Now use the table data and these formulas to find the coordinates of the centroid.

EXAMPLE (continued)

C


Dr. Ali Keshavarz

More Example

Given: Two blocks of different materials are assembled as shown.

The densities of the materials are:

A = 150 lb / ft3 andB = 400 lb / ft3.Find: The center of gravity of this

assembly.

Plan: Follow the steps for analysis.Solution

1. In this problem, the blocks A and B can be considered as two segments.


Dr. Ali Keshavarz


Weight = w = (Volume in ft3)wA = 150 (0.5) (6) (6) (2) / (12)3 = 3.125 lb

wB = 400 (6) (6) (2) / (12)3 = 16.67 lb

56.2553.1229.1719.79

6.2550.00

3.12550.00

12.516.67

23

13

41

3.12516.67

AB

wz (lbin)

w y (lbin)

w x (lbin)

z (in)y (in)x (in)w (lb)Segment


Dr. Ali Keshavarz

~x = ( x w) / ( w ) = 29.17/19.79 = 1.47 iny = ( y w) / ( w ) = 53.12/ 19.79 = 2.68 inz = ( z w) / ( w ) = 56.25 / 19.79 = 2.84 in ~

~



Dr. Ali Keshavarz

Moment of Inertia for Areas

kz dF dA kzdA

2dM dFz kz dA 2M k z dA

C


Dr. Ali Keshavarz

Moment of Inertia

2

2

x A

y A

I y dA

I x dA

2o x yAJ r dA I I

UNITSm4, mm4, ft4, in4

2 2 2r x y Polar Moment of Inertia


Dr. Ali Keshavarz

Parallel Axis Theorem2

22

( )

2

x yA

y yA A A

I y d dA

y dA d y dA d dA

2

2

x x y

y y x

I I Ad

I I Ad

2O CJ J Ad

Moment of Inertia for Composite Areas:If moment of inertia of simple shapes thatmake up the final shape is known or can bedetermined about a specified axis, then themoment of inertia of the composite shape isthe algebraic sum of the moments of its parts.


lec 02 (statics review)-- jan 11

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