Download - 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
Strength of Materials - Lec. 1 - Ch. 1-3 2Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 3Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 4Dr. Ali Keshavarz
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 −= × ⋅
Strength of Materials - Lec. 1 - Ch. 1-3 5Dr. Ali Keshavarz
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=
Strength of Materials - Lec. 1 - Ch. 1-3 6Dr. Ali Keshavarz
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 =
Strength of Materials - Lec. 1 - Ch. 1-3 7Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 8Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 9Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 10Dr. Ali Keshavarz
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.
Strength of Materials - Lec. 1 - Ch. 1-3 11Dr. Ali Keshavarz
Improper Application of Statics
Strength of Materials - Lec. 1 - Ch. 1-3 12Dr. Ali Keshavarz
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.
Strength of Materials - Lec. 1 - Ch. 1-3 13Dr. Ali Keshavarz
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.
Strength of Materials - Lec. 1 - Ch. 1-3 14Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 15Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 16Dr. Ali Keshavarz
Resolution due to Geometry
Strength of Materials - Lec. 1 - Ch. 1-3 17Dr. Ali Keshavarz
Trigonometric Laws and Force Notation
Sine Law and Cosine LawTrigonometric relationsGeometric relationsForce Notation
Scalar and Vector500 N and F or F
→
= F
Strength of Materials - Lec. 1 - Ch. 1-3 18Dr. 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.
Strength of Materials - Lec. 1 - Ch. 1-3 19Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 20Dr. Ali Keshavarz
Adding Coplanar Forces: Cartesian Vector Notation
x yF F= +F i j, : unit vectors
magnitude = 1 (unity)direction: +/- sign
i j
Strength of Materials - Lec. 1 - Ch. 1-3 21Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 22Dr. Ali Keshavarz
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 −
= +
=
Strength of Materials - Lec. 1 - Ch. 1-3 23Dr. Ali Keshavarz
Angle Specification for the Resultant Vector
θ
θ
θθ
θθ
R
R
R
R
Strength of Materials - Lec. 1 - Ch. 1-3 24Dr. Ali Keshavarz
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.
Strength of Materials - Lec. 1 - Ch. 1-3 25Dr. 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 A
A i j k
or
0
A AAA
A
= =
≠
Au A u
Strength of Materials - Lec. 1 - Ch. 1-3 26Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 27Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 28Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 29Dr. Ali Keshavarz
3D Force Vector Example
Strength of Materials - Lec. 1 - Ch. 1-3 30Dr. Ali Keshavarz
Position Vectors
x y z= + +r i j k
Strength of Materials - Lec. 1 - Ch. 1-3 31Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 32Dr. Ali Keshavarz
Force Vector Directed Along A Line
F Fr
⎛ ⎞= = ⎜ ⎟⎝ ⎠
rF u
ABAB
AB
F Fr
⎛ ⎞= = ⎜ ⎟
⎝ ⎠
rF u
Strength of Materials - Lec. 1 - Ch. 1-3 33Dr. Ali Keshavarz
Real Applications
Strength of Materials - Lec. 1 - Ch. 1-3 34Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 35Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 36Dr. Ali Keshavarz
Real Applications
Strength of Materials - Lec. 1 - Ch. 1-3 37Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 38Dr. Ali Keshavarz
Moment of a Force
Scalar FormulationMagnitude:
Direction:Right-Hand Rule
OM Fd=
Strength of Materials - Lec. 1 - Ch. 1-3 39Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 40Dr. Ali Keshavarz
Real Example
Strength of Materials - Lec. 1 - Ch. 1-3 41Dr. Ali Keshavarz
Cross ProductThe result is a vectorThe order of multiplication does matterMagnitude:
Direction:
sinC AB θ=
( sin ) CAB θ= =C A×B u
Strength of Materials - Lec. 1 - Ch. 1-3 42Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 43Dr. Ali Keshavarz
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!
Strength of Materials - Lec. 1 - Ch. 1-3 44Dr. Ali Keshavarz
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= = =
Strength of Materials - Lec. 1 - Ch. 1-3 45Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 46Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 47Dr. Ali Keshavarz
Resultant Moment of A System of Forces
( )OR = ΣM r×F
Strength of Materials - Lec. 1 - Ch. 1-3 48Dr. Ali Keshavarz
Real Applications
Strength of Materials - Lec. 1 - Ch. 1-3 49Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 50Dr. Ali Keshavarz
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=
Strength of Materials - Lec. 1 - Ch. 1-3 51Dr. Ali Keshavarz
Real Example
Strength of Materials - Lec. 1 - Ch. 1-3 52Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 53Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 54Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 55Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 56Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 57Dr. Ali Keshavarz
Real Examples
Strength of Materials - Lec. 1 - Ch. 1-3 58Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 59Dr. Ali Keshavarz
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.
Strength of Materials - Lec. 1 - Ch. 1-3 60Dr. Ali Keshavarz
Equivalent System Concepts Illustrated
Strength of Materials - Lec. 1 - Ch. 1-3 61Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 62Dr. Ali Keshavarz
Resultant Force and Couple Moment Concept Illustrated
Strength of Materials - Lec. 1 - Ch. 1-3 63Dr. Ali Keshavarz
Further Reduction of a Force and Couple System: Single Resultant Force
Special Case Only: FR and MROare perpendicular to each other
Strength of Materials - Lec. 1 - Ch. 1-3 64Dr. Ali Keshavarz
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
Strength of Materials - Lec. 1 - Ch. 1-3 65Dr. Ali Keshavarz
Coplanar Force Systems: 2D
( )O
R
R C
= Σ= Σ +Σ
F FM M r×F
OR
R
Md
F=
Strength of Materials - Lec. 1 - Ch. 1-3 66Dr. Ali Keshavarz
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.
Strength of Materials - Lec. 1 - Ch. 1-3 67Dr. Ali Keshavarz
Parallel Force Systems Illustrated
Strength of Materials - Lec. 1 - Ch. 1-3 68Dr. Ali Keshavarz
Reduction to a Wrench