ch 1- 3

Strength of Materials - Lec. 1 - Ch. 1-3 1Dr. Ali Keshavarz

Ch. 1 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


Fundamental Concepts

Basic QuantitiesLENGTH

Position, geometryTIME

Succession of eventsMASS

Property of matterAmount of matter

FORCEPush or pull effect

Some IdealizationsParticle

Has a massSize neglected

Rigid-BodyNot deforming

Concentrated ForceActing at a point


Newton’s Laws of MotionFirst 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.

m=F a


Newton’s 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 −= × ⋅


Concept of Weight

Force of gravitation between the earth and a particle with a mass = Weight.

2emMW G

r=

24

6

: Force of gravitation: Mass of the earth ( 6 10 kg)

: Mass of the particle: Distance between the earth's center and the particle

6.39 10 m (Radius of the earth at sea level with 45 lattitude

e

WMmrr

= ×

= ×STANDARD LOCATION

2

2

)

(Gravitational acceleration)

9.81 m/s

eGMgr

g

=

=W mg=


Units of Measurement

SI UnitsLENGTH meter (m)TIME second (s)MASS kilogram (kg)FORCE Newton (N)

US Customary UnitsLENGTH foot (ft)TIME second (s)FORCE pound (lb)MASS slug (slug)

2

kg mN = s⋅

29.81 m/sg =

2lb sslug = ft⋅

232.2 ft/sg =


Conversion of Units

Quantity US Customary

SI System

FORCE 1 lb = 4.4482 N

MASS 1 slug = 14.5938 kg

LENGTH 1 ft = 0.3048 m


Useful Information on UnitsSome Conversions

1 ft = 12 in. (inch)5280 ft = 1 mi (mile)1000 lb = 1 kip (kilo-pound)2000 lb = 1 ton

Time Conversions1 min. = 60 s1 h = 3600 s1 s = 1000 ms1 s = 1000000 μs

PrefixesMULTIPLE

109 giga (G)106 mega (M)103 kilo (k)

SUBMULTIPLE10-3 milli (m)10-6 micro (μ)10-9 nano (n)

π = 3.141592…= 3.14Conversion for Angles

180° = π rad


More on Prefixes of SI UnitsGm Mm km m mm μm nm

1 Gm = 10+9

m 1 Mm = 10+6

m 1 km = 10+3

m 1 m = 1 m 1 mm = 10-3

m 1

μm = 10-6 m 1 nm = 10-9 m

… h min s ms μs ns

… 1 h = 3600 s 1 min = 60 s 1 s = 1 s 1 ms = 10-3

s 1

μs = 10-6 s 1 ns = 10-9 s

GN MN kN N mN μN nN

1 GN = 10+9

N 1 MN = 10+6

N 1 kN

= 10+3

N 1 N = 1 N 1 mN

= 10-3

N 1

μN = 10-6 N 1 nN

= 10-9 N

Gg Mg kg g mg μg ng

1 Gg

= 10+6

kg 1 Mg = 10+3

kg 1 kg = 1 kg 1 g = 10-3

kg 1 mg = 10-6 kg 1

μg = 10-9 kg 1 nm = 10-9 kg


Significant FiguresAccuracy 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.


Improper Application of Statics


Problem Solving Strategy

InterpretRead carefully and determine what is given and what is to be found/delivered. Ask, if not clear. If necessary, make assumptions and indicate them.

PlanThink about major steps (or a road map) that you will take to solve a given problem. Think of alternative/creative solutions and choose the best one.

ExecuteCarry out your steps (symbolically as much as possible). Use appropriate diagrams and equations. Estimate your answers. Avoid simple calculation mistakes. Reflect on/revise your work.


Ch 2. Force VectorsScalar. 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 vector’s length compared to a given reference)A direction (on a line you can have 2 direction choices)A line of action

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


Vector MathMultiplication 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 constructionVector subtraction: (A – B = A + (-B))

Same as addition conceptJust multiply the vector being subtracted and add two


Vector ResolutionResolution of a vector

Resolve into 2 components on 2 known line of actions2 known line of actions are not necessarily perpendicularWHY? We may need to resolve due to geometry


Resolution due to Geometry


Trigonometric Laws and Force Notation

Sine Law and Cosine LawTrigonometric relationsGeometric relationsForce Notation

Scalar and Vector500 N and F or F

→

= F


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.


Addition of A System of Coplanar Forces

Using a rectangular coordinate system is the common method such as x-y axes.Scalar Notation

We can find components of a vector (force) along specified axes. Then we can add components on the same axis algebraically (scalar).We have to be careful with signs (directions)

x y= +F F F

x y′ ′ ′= +F F F


Adding Coplanar Forces: Cartesian Vector Notation

x yF F= +F i j, : unit vectors

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

i j


Resultant of Coplanar Forces

1 1 1

2 2 2

3 3 3

yx

yx

yx

F F

F FF F

= +

= − +

= −

F i jF i jF i j


The Resultant

1 2 3R = + +F F F F

R 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

= Σ= Σ

2 2

1tan

R Rx Ry

Ry

Rx

F F F

FF

θ −

= +

=


Angle Specification for the Resultant Vector

θ

θ

θθ

θθ

R

R

R

R


Cartesian Vectors in 3DUsing Cartesian vector notation greatly simplifies solving problems in 3 dimensional space.Right-handed coordinate system

Thumb +z (Zenith direction, height, altitude)Out between fingers +xToward arm +y

In 2D space, +z in always outward and perpendicular to the page.


Unit Vector is used to specify direction

Rectangular 3D Components of A Vector

x y z

x y zA A A

= + +

= + +

A A A A

A i j k

or

0

A AAA

A

= =

≠

Au A u


Magnitude and Direction

2 2 2x y zA A A A= + +

cos

cos

cos

x

y

z

AAAA

AA

α

β

γ

=

=

=

Coordinate direction angles (direction cosines) are

measured between tail of A and + x, y, z axes


Revisit the Unit Vector

or

0

A AAA

A

= =

≠

Au A u

cos cos cos

yx zA

A

AA AA A A A

α β γ

= = + +

= + +

Au i j k

u i j k2 2 2cos cos cos 1α β γ+ + =

cos cos cosA

x y z

AA A AA A A

α β γ== + += + +

A ui j k

i j k


Addition and Subtraction of Cartesian Vectors

x y z

x y z

A A A

B B B

= + +

= + +

A i j k

B 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= Σ = Σ +Σ +ΣF F i j k


3D Force Vector Example


Position Vectors

x y z= + +r i j k


Position Vectors Generalized

A A A A

B B B B

x y zx y z

= + += + +

r i j kr i j k

A B B A+ = → = −r r r r r r

( ) ( ) ( )AB B A B A B Ax x y y z z= = − + − + −r r i j k

2 2 2( ) ( ) ( )AB B A B A B Ar x x y y z z= − + − + −

1 1 1( ) ( ) ( )cos cos cosB A B A B A

AB AB AB

x x y y z zr r r

α β γ− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − −= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

ABAB

ABr=

ru


Force Vector Directed Along A Line

F Fr

⎛ ⎞= = ⎜ ⎟⎝ ⎠

rF u

ABAB

AB

F Fr

⎛ ⎞= = ⎜ ⎟

⎝ ⎠

rF u


Real Applications


Dot ProductHow do we find angle between two lines?Dot Product = Scalar Product (result is scalar)Commutative Law:

A·B = B·AMultiply by a Scalar:

a(A·B) = (aA)·B = A·(aB) =(A·B)a

Distributive Law:A·(B+D) = (A·B) + (A·D)

cos(0 180 )

AB θ

θ

⋅ =

≤ ≤

A B

10

⋅ = ⋅ = ⋅ =⋅ = ⋅ = ⋅ =

i i j j k ki j i k k j

x y z

x y z

A A A

B B B

= + +

= + +

A i j k

B i j k

x x y y z zA B A B A B⋅ = + +A B


Applications of Dot Product

The angle formed between 2 vectors

Components of a vector parallel/perpendicular to a line

1cos 0 180AB

θ θ− ⋅⎛ ⎞= ≤ ≤⎜ ⎟⎝ ⎠

A B

||

||

coscos ( )

A AA

θ

θ

=

= = ⋅A u A u u


Real Applications


Ch 3. 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


Moment of a Force

Scalar FormulationMagnitude:

Direction:Right-Hand Rule

OM Fd=


Resultant Moment of a System of Coplanar Forces

All forces are in the same plane (x-y)Resultant Moment:

Algebraic sum of each moment created by each forceCCW = +zCW = -z

ORM Fd+ = Σ


Real Example


Cross ProductThe result is a vectorThe order of multiplication does matterMagnitude:

Direction:

sinC AB θ=

( sin ) CAB θ= =C A×B u


Cross Product Laws of Operation

Not Commutative

Multiplication by a scalar

Distributive Law:

≠= −

A×B B× AA×B B× A

( ) ( ) ( ) ( )a a a a= = =A×B A ×B A× B A×B

( ) ( ) ( )+ = +A× B D A×B A×D


Cross Product Cartesian Vector Formulation

= == − = − −

i × i = j× j = k ×k = 0i× j k j×k i k ×i = j

j× i k k × j i i×k = j

x y z

x y z

A A A

B B B

= + +

= + +

A i j k

B i j k

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

x y z

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

= = − − − + −i j k

A×B i j k

MINUS SIGNImportant!


Moment of a Force Vector Formulation

The moment of force F about point OThe 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 FMagnitude:

Direction: Right-hand rule

O =M r×F

sin ( sin )OM rF F r Fdθ θ= = =


Principle of TransmissibilityThe force F applied at point Acreates a moment about O

r can extend from O to any point on the line of action of force F. Therefore, F can be applied at A, B, or CF is a sliding vector

O A=M r ×F

O A B C= = =M r ×F r ×F r ×F


Moment of a Force Cartesian Vector Formulation

x y z

x y z

r r r

F F F

= + +

= + +

r i j k

F 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 r×F

i j k

i j k


Resultant Moment of A System of Forces

( )OR = ΣM r×F


Real Applications


Principle of MomentsVarignon’s Theorem:

The moment of a force about a point is equal to the sum of the moments of the force’s components about the point

Why is this important?Easier to find moments of components

1 2

1 2( )O = += +=

1 2F = F + FM r×F r×F

r× F Fr×F


Moment of a Force about a Specified Axis: Scalar Analysis

The moment vector and its axis is always perpendicular to the plane that contains the force and the moment armIt is sometimes important to find the component of this moment along a specified axis that passes through this point (Moment may be given or not)If the line of action of a force F is perpendicular to any specified axis aa, then the magnitude of the moment of Fabout the axis can be determined from the equation

Direction: Right-hand rule

a aM Fd=


Real Example


Moment of a Force about a Specified Axis: Vector Analysis

(0.3 0.4 ) m20 N

(0.3 0.4 ) ( 20 )8 6 Nm

( 8 6 ) 6 Nm

A

O A

y O AM

= += −== + −= − += ⋅

= − + ⋅ =

r i jF k

M r ×Fi j × k

i jM u

i j j


Moment about Specified Axis: Generalized Vector Analysis

( ) ( )a a aM = ⋅ = ⋅u r×F r×F u

( )

( )

x y z

x y z

a a a a x y z

x y z

a a a

a a x y z

x y z

M u u u r r rF F F

u u u

M r r rF F F

= + + ⋅

= ⋅ =

i j ki j k

u r×F

[ ( )]a a a a aM= = ⋅M u u r×F u

[ ( )] ( )a a aM = Σ ⋅ = ⋅Σu r×F u r×FResultant moment about

aa’

axis of series of forces


Couple DefinedA couple is defined as two parallel forces that have the same magnitude, have opposite directions, and are separated by a perpendicular distance d.Since the resultant force is zero, only effect of a couple is to produce a rotation (or tendency of rotation) in a specified direction


Moment of a Couple (Couple Moment)

Free vector (can act at any point)Scalar Formulation:

Vector Formulation:r is crossed with the force F to which it is directed

( ) ( )O A B= − +M r × F r ×F

O is any point

A =M r×F( )A B A= −

r

M r r ×F

M Fd=

=M r×F


Equivalent Couples Resultant Couple Moment

Equivalent Couples. Two couples are said to be equivalent if they produce to same moment. Forces of equal couples lie either in the same plane on in planes that are parallel to one anotherResultant Couple Moment

( )R = ΣM r×F

1 2R = +M M M


Real Examples


Equivalent System: Point O is on the Line of Action of Force

Simply force F can be moved from point A to point O. Principle of transmissibility

External effects: Remain unchanged, i.e., support forcesInternal effects: Higher intensity around A than O

We will study in Mechanics of Materials


Equivalent System: Point O is not

on the Line of Action of Force

Force F can be moved from point A to point O.However, the couple introduced creates a moment.This is a free vector, couple moment, can act at any point of the body: P or O or A.Force F is now acting at point O with a couple moment created on the body.


Equivalent System Concepts Illustrated


Resultant of a Force and Couple System

When F is moved to O, moments M1 and M2 are created.Mc can be moved to O since it is a free vector.The resultant force and moment:Generalized:

1 2

1 2O

R

R C

= += + +

F F FM M M M

O

R

R C O

= Σ= Σ +Σ

F FM M M

x

y

O

R x

R y

R C O

F F

F F

M M M

= Σ

= Σ

= Σ +Σ


Resultant Force and Couple Moment Concept Illustrated


Further Reduction of a Force and Couple System: Single Resultant Force

Special Case Only: FR and MROare perpendicular to each other


Concurrent Force Systems

Since all forces intersect at the point P, there is no resultant couple moment and only resultant force that is the sum of all forces


Coplanar Force Systems: 2D

( )O

R

R C

= Σ= Σ +Σ

F FM M r×F

OR

R

Md

F=


Parallel Force Systems

( )O

R

R C

= Σ= Σ +Σ

F FM M r×F

OR

R

Md

F=Couple

moments are perpendicular to the forces.


Parallel Force Systems Illustrated


Reduction to a Wrench

Documents

ch 1- 3