engineering mechanics(statics & dynamics) - engr.purigay (pdf format)

FEATI UNIVERSITY – AERONAUTICAL ENGINEERING REVIEW CENTER NMP


Engineering Mechanics

Mechanics of Rigid Bodies

Statics Dynamics

KinematicsKinetics

Mechanics of Deformable Bodies

1. Strength of Materials

2. Theory of Elasticity

3. Theory of Plasticity

Mechanics of Fluids

1. Ideal Fluid

2. Viscous Fluid

3. Incompressible Fluid


Mechanics of Rigid Bodies

Statics

Force Systems

Concurrent

Parallel

Non-Concurrent

Applications

Trusses

Centroids

Friction

Dynamics

Kinematics

Translation

Rotation

Plane Motion

Kinetics

Translation

Rotation

Plane Motion

Fundamental Concepts and Definitions

• Engineering Mechanics – The science which considers the effects of forces on rigid bodies.

• Statics – considers the effects and distribution of forces on rigid bodies which are and remain at rest

• Dynamics – considers the motion of rigid bodies caused by the forces acting upon them

• Kinematics – deals with pure motion of rigid bodies

• Kinetics – relates the motion to applied forces



Basic Quantities

Length – used to locate the position of a point in space and thereby describe the size of a physical system

Time – is conceived as a succession of events

Mass – is a measure of the quantity of matter that is used to compare the action of one body with that of another.

Force – a “push or pull” exerted by one body to another• External Force - changes, or tends to change, the state

of motion of a body. (independent on point of application)

• Internal Force – produces stress and deformation in the body. (dependent on point of application)

* Principle of Transmissibility – a force may be moved anywhere along its line of action without changing its external effect on a rigid body.



Idealizations

Particle – has a mass, but a size that can beneglected.

Rigid Body – can be considered as a largenumber of particles in which all the particlesremain at a fixed distance from one another,both before and after applying a load.

Concentrated Force - represents the effect ofa loading which is assumed to act at a point ona body. We can represent a load by aconcentrated force, provided the area overwhich the load is applied is very smallcompared to the overall size of the body.



Newton’s Three Laws of Motion

First Law (Law of Inertia). A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this state provided the particle is not subjected to an unbalanced force.

Second Law (Law of Acceleration). A particle acted upon by an unbalanced force experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force.

Third Law. (Law of Action-Reaction). The mutual forces of action and reaction between two particles are equal, opposite, and collinear



Newton’s Law of Gravitational Attraction

𝐹 = 𝐺𝑚1𝑚2

𝑟2


where

F = force of gravitation between the two particlesG = universal constant of gravitation; according to

experimental evidence, G = 66.73(10-12) m3/(kg · s2)m1, m2 = mass of each of the two particlesr = distance between the two particles

Weight – force on an object due to gravity

W = mg

W= 𝐺𝑚𝑚𝐸

𝑟2


Units of Measurement

F = ma ; W=mg

CGS: dyne = (g)(cm/s2) MKS: N = (kg)(m/s2) US: lbf = (slug)(ft/s2)kgf = (kgm)(9.8 m/s2) lbf = (lbm)(32.174 ft/s2)kgf = 9.8 N slug = 32.174 lbm

Under Standard Condition: g = 9.8 m/s2

kgf = kgm lbf = lbm

For Non-Standard Condition

kgf = kgm(𝑔𝑙𝑜𝑐

𝑔) lbf = lbm (

𝑔𝑙𝑜𝑐

𝑔)



Force Systems

Force System – any arrangement when two or more forces act on a body or on a group of related bodies.

• Coplanar – the lines of action of all the forces lie in one plane

• Concurrent – the lines of action pass through a common point

• Parallel – the lines of actions are parallel

• Non-Concurrent – the lines of action are neither parallel nor intersect at a common point



Axioms of Mechanics

1. The Parallelogram Law: The resultantof two forces is the diagonal of theparallelogram formed on the vectorsthese forces.

2. The forces are in equilibrium only whenequal in magnitude, opposite indirection, and collinear in action.

3. A set of forces in equilibrium may beadded to any system of forces withoutchanging the effect of the originalsystem

4. Action and reaction forces are equalbut oppositely directed.




Parallelogram Law Triangle Law

Polygon Law


Scalar and Vector Quantities

Scalars – quantities which possesmagnitude only and can be addedarithmetically.

Vectors – quantities which possesmagnitude and direction and can becombined only by geometric (vector)addition.

• Multiplication or division of a vector bya scalar will change the magnitude ofthe vector. The sense of the vector willchange if the scalar is negative.

• As a special case, if the vectors arecollinear, the resultant is formed by analgebraic or scalar addition.


Components of a Force 𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃𝑥

𝐹𝑦 = 𝐹𝑐𝑜𝑠𝜃𝑦

Resultant

𝐹 = 𝐹𝑥2 + 𝐹𝑦

2

Resultant of Three or More Concurrent Forces

𝑅 = (Σ𝐹𝑥)2 + (Σ𝐹𝑦)2

Position of Resultant

𝑡𝑎𝑛θ𝑥 =Σ𝐹𝑦

Σ𝐹𝑥


Resultant of Force Systems

Resultant – simplest system that can replace the original system without changing the effect on a rigid body

Resultant of Concurrent Forces


Resultant of Non-Perpendicular Forces

𝑅 = 𝑃12 + 𝑃2

2 + 2𝑃1𝑃2𝑐𝑜𝑠θ

Position of Resultant

𝑡𝑎𝑛ϕ =𝑃1𝑠𝑖𝑛θ

𝑃2+𝑃1𝑐𝑜𝑠θ


Determine the magnitude and direction ofthe resultant of the three forces shown.Determine also the horizontal and verticalcomponent of the resultant.

Solution:𝑅𝑥= Σ𝐹𝑥

= 50𝑐𝑜𝑠45 + 75𝑐𝑜𝑠75 − 80𝑐𝑜𝑠60𝑹𝒙 = 𝟏𝟒. 𝟕𝟕 𝐍

𝑅𝑦 = Σ𝐹𝑦

= 50𝑠𝑖𝑛45 + 75𝑠𝑖𝑛75 + 80𝑠𝑖𝑛60𝑹𝒚 = 𝟏𝟕𝟕. 𝟎𝟖 𝐍

𝑅 = (Σ𝐹𝑥)2 + (Σ𝐹𝑦)2

= 14.772 + 177.082

𝑹 = 𝟏𝟕𝟕. 𝟕𝟎 𝐍

θ = 𝑡𝑎𝑛−1 Σ𝐹𝑦

Σ𝐹𝑥= 𝑡𝑎𝑛−1 177.08

14.77

𝜽 = 𝟖𝟓. 𝟐𝟑°



Determine the magnitude and direction of Rif P1 and P2 are 100 lb and 150 lb respectively.P2 lies horizontally while P1 makes 120° withthe horizontal.

Solution 1:

𝑅 = (Σ𝐹𝑥)2 + (Σ𝐹𝑦)2

= (150 − 100𝑐𝑜𝑠60)2 + 100𝑠𝑖𝑛602

𝑹 = 𝟏𝟑𝟐. 𝟐𝟗 𝐥𝐛

ϕ = 𝑡𝑎𝑛−1 Σ𝐹𝑦

Σ𝐹𝑥= 𝑡𝑎𝑛−1 86.60

100

𝝓 = 𝟒𝟎. 𝟖𝟗°

Solution 2:

𝑅 = 𝑃12 + 𝑃2

2 + 2𝑃1𝑃2𝑐𝑜𝑠θ

= 1002 + 1502 + 2(100)(150)𝑐𝑜𝑠120𝑹 = 𝟏𝟑𝟐. 𝟐𝟗 𝐥𝐛

ϕ = 𝑡𝑎𝑛−1(𝑃1𝑠𝑖𝑛θ

𝑃2+𝑃1𝑐𝑜𝑠θ)

𝝓 = 𝟒𝟎. 𝟖𝟗°



Determine the magnitude of R if P1 and P2are 100 lb and 150 lb respectively. ϕ = 41°

Solution:

Let α be the angle opposite R and β be the angle opposite P2:

By Sine Law𝑅

𝑠𝑖𝑛α=

𝑃1

𝑠𝑖𝑛ϕ=

𝑃2

𝑠𝑖𝑛β100

𝑠𝑖𝑛41=

150

𝑠𝑖𝑛ββ = 80°α = 180 − 80 + 41 = 59100

𝑠𝑖𝑛41=

𝑅

𝑠𝑖𝑛59𝑹 = 𝟏𝟑𝟎. 𝟔𝟓 𝒍𝒃

By Cosine Law:

𝑅 = 𝑃12 + 𝑃2

2 − 2𝑃1𝑃2𝑐𝑜𝑠α

= 1002 + 1502 − 2(100)(150)𝑐𝑜𝑠59𝑹 = 𝟏𝟑𝟎. 𝟓𝟕 𝒍𝒃



A boat moving at 12kph is crossing a river 500 m widein which a current is flowing at 4 kph. In whatdirection should the boat head if it is to reach a pointon the other side of the river directly opposite itsstarting point?

𝑠𝑖𝑛θ =4

12

𝜽 = 𝟏𝟗. 𝟒𝟕°, 𝒖𝒑𝒔𝒕𝒓𝒆𝒂𝒎


Solution:


Resultant of Non-Concurrent Forces

Moment of a Force



Moment – is the measure of the ability of a force to produce turning or twisting about an axis.

𝑀𝑂 = 𝐹𝑑where d is the moment arm (perpendicular distance from the axis at point O to the line of action of the force.

The Principle of Moments (Varignon’s Theorem)

The moment of a force is equal to the sum of the moments of its components.𝑀𝑅= Σ(𝐹𝑑)𝑀𝑅= Σ𝑀 = 𝑅𝑑 𝑀𝑅 = 𝐹1𝑑1 − 𝐹2𝑑2 + 𝐹3𝑑3



Determine the resultant moment ofthe four forces acting on the rodshown below about point O.

Solution:

𝑀𝑅𝑂= Σ𝐹𝑑

= −50𝑁 2𝑚 + 60𝑁 0𝑚+ 20𝑁 3𝑠𝑖𝑛30𝑚 − (40𝑁)(4𝑚+ 3𝑐𝑜𝑠30𝑚)𝑴𝑹𝑶

= −𝟑𝟑𝟒 𝐍 · 𝒎 = 𝟑𝟑𝟒 𝑵 ↻



Couple

Couple – Two parallel, non-collinear forces that are equal in magnitude and opposite in direction𝑅 = 0 ; Σ𝑀 ≠ 0

𝐶 = 𝐹𝑑



Equivalent Couples

𝐶 = 100 𝑙𝑏 2 𝑓𝑡 = 200 𝑙𝑏 1 𝑓𝑡 = 200 𝑙𝑏 · 𝑓𝑡



Coplanar Force System



The force system shown consists ofthe couple C and four forces. If theresultant of this system is a 500-lb·in.counterclockwise couple, determineP, Q, and C.

Solution:

𝑅𝑥= Σ𝐹𝑥

= −12

13𝑄 +

4

5𝑃 + 80 = 0 (1)

𝑅𝑦 = Σ𝐹𝑦

= −5

13𝑄 +

3

5𝑃 − 20 = 0 (2)

Solving Eqns (1) and (2) simultaneously gives𝑷 = 𝟐𝟎𝟎𝒍𝒃 𝒂𝒏𝒅 𝑸 = 𝟐𝟔𝟎𝒍𝒃

𝐶𝑅 = 𝑀𝑅 = Σ𝑀𝐴

500 = −20 3 − 𝐶 + 80 4 +3

5𝑃 6 +

4

5𝑃(6)

𝑪 = 𝟏𝟒𝟒𝟎 𝒍𝒃 · 𝒊𝒏



Parallel Force System



Replace the force and couple momentsystem acting on the beam in the figure byan equivalent resultant force, and findwhere its line of action intersects the beam,measured from point O.

Solution:

𝑅𝑥= Σ𝐹𝑥

= 8 𝑘𝑁3

5= 4.8 𝑘𝑁

𝑅𝑦= Σ𝐹𝑦

= −4 𝑘𝑁 + 8 𝑘𝑁4

5= 2.4 𝑘𝑁

𝑅 = (4. 8𝑘𝑁)2 + (2.4𝑘𝑁)2

𝑹 = 𝟓. 𝟑𝟕 𝒌𝑵

θ = 𝑡𝑎𝑛−1 2.4

4.8= 𝟐𝟔. 𝟔°



Solution:

𝑀𝑅= Σ𝑀𝑂

2.4𝑘𝑁 𝑑 = −4𝑘𝑁 1.5𝑚 − 15𝑘𝑁 · 𝑚 −

8𝑘𝑁3

50.5𝑚 + 8𝑘𝑁

4

5(4.5𝑚)

𝒅 = 𝟐. 𝟐𝟓𝒎


Equilibrium of a Rigid Body

Equilibrium – A body is said to be in equilibrium if theresultant of the force system that acts on the bodyvanishes. Equilibrium means that both the resultant forceand the resultant couple are zero.

Free Body Diagram (FBD) - is a sketch of the bodyshowing all forces that act on it. The term free impliesthat all supports have been remove and replaced by theforces (reactions) that they exert on the body.

Forces that Act on a Body1. Reactive Forces (Reactions) - forces that are exerted

on a body by the supports to which it is attached.2. Applied Forces - forces acting on abody that are not

provided by the supports.



Conditions of Equilibrium

1. Graphical Condition: Under thiscondition, the forces or vectorsare transformed into a forcepolygon. For equilibrium, theforce polygon must close.



2. Directional Condition: If threeor more non-parallel forces orvectors are in equilibrium, thenthey must be concurrent. For atwo-force member, the forcesmust be equal and opposite.

3. Analytical Condition: If forcesor vectors are in equilibrium, thenit must satisfy the three staticequations:



Support Reactions



Three forces 20 N, 30 N, and 40 N are in equilibrium. Find the largest angle they make with each other.

Solution:

402 = 302 + 202 − 2 30 20 𝑐𝑜𝑠θ𝜽 = 𝟏𝟎𝟒. 𝟒𝟖°



A load of 100 lb is hung from the middleof a rope, which is stretch between tworigid walls 30 ft apart. Due to the load,the rope sags 4 ft in the middle.Determine the tension in the rope.

Solution:

𝑡𝑎𝑛θ =15

4θ = 75.068°

Σ𝐹𝑦 = 0 :

2𝑇𝑐𝑜𝑠75.068 = 100𝑻 = 𝟏𝟗𝟒 𝒍𝒃



A simply supported beam is 5m in length. Itcarries a uniformly distributed load includingits own weight of 300N/m and a concentratedload of 100 N 2m from the left end. Find thereactions if reaction A is at the left end andreaction B is at the right end.

Solution:

Σ𝑀𝐴 = 0 :100𝑁 2𝑚 + 1500𝑁 2.5𝑚 = 𝑅𝐵 5𝑚

𝑹𝑩 = 𝟕𝟗𝟎 𝑵

Σ𝐹𝑦 = 0 :

𝑅𝐴 + 𝑅𝐵 = 100𝑁 + 1500𝑁𝑹𝑨 = 𝟖𝟏𝟎 𝑵



The homogeneous 60-kg disk supported bythe rope AB rests against a rough vertical wall.Using the given FBD, determine the force inthe rope and the reaction at the wall.

Solution:

Σ𝑀𝐵 = 0 :𝑭𝑪 = 𝟎

Σ𝐹𝑦 = 0 :4

5𝑇 = 588.6𝑁

𝑻 = 𝟕𝟑𝟓. 𝟖 𝑵

Σ𝐹𝑥 = 0 :3

5𝑇 = 𝑁𝐶

𝑵𝑪 = 𝟒𝟒𝟏. 𝟓 𝑵



The homogeneous, 120-kg wooden beam issuspended from ropes at A and B. A powerwrench applies the 500-N · m clockwisecouple to tighten a bolt at C. Use the givenFBD to determine the tensions in the ropes.

Solution:

Σ𝑀𝐴 = 0 :𝑇𝐵(4𝑚) = 1177.2𝑁 3𝑚 + 500𝑁𝑚

𝑻𝑩 = 𝟏𝟎𝟎𝟕. 𝟗 𝑵

Σ𝐹𝑦 = 0 :

𝑇𝐴 + 𝑇𝐵 = 1177.2𝑻𝑨 = 𝟏𝟔𝟗. 𝟑 𝑵



The structure in Fig. (a) isloaded by the 240-lb · in.counterclockwise coupleapplied to member AB.Neglecting the weights of themembers, determine allforces acting on memberBCD.



Solution:

For Fig. (b) Σ𝑀𝐴 = 0 :𝑇𝐶𝑐𝑜𝑠30 8 + 240 = 𝑁𝐷(12)𝑁𝐷 = 0.577𝑇𝐶 + 20 (𝟏)

For Fig. (c) Σ𝑀𝐵 = 0 :𝑇𝐶𝑐𝑜𝑠30 4 + 𝑇𝐶𝑠𝑖𝑛30(3) = 𝑁𝐷(8)

𝑁𝐷 = 0.620𝑇𝐶 𝟐

Solving Eqs (1) and (2) simultaneously gives:

𝑵𝑫 = 𝟐𝟖𝟖 𝒍𝒃 ; 𝑻𝑪 = 𝟒𝟔𝟓 𝒍𝒃

Σ𝐹𝑋 = 0 :𝑁𝐷 + 𝐵𝑋 = 𝑇𝐶𝑐𝑜𝑠30

𝑩𝑿 = 𝟏𝟏𝟒 𝒍𝒃

Σ𝐹𝑦 = 0 :

𝐵𝑦 = 𝑇𝐶𝑠𝑖𝑛30

𝑩𝒚 = 𝟐𝟑𝟐 𝒍𝒃


Analysis of Structures

Truss – is a structure composed of slender membersjoined together at their end joints.

Planar Trusses - lie in a single plane and are often usedto support roofs and bridges

Simple Trusses - constructed by expanding the basictriangular truss

Simple Trusses



Truss Wing Spar

Truss-Type Fuselage. A Warren Truss uses mostly diagonal bracing.



Assumptions for Design

1. The weights of the members are negligible.2. The members are joined together by smooth pins.3. The applied forces act at the joints.

Each member of a truss is a two-force member.



Method of Joints

When using the method of joints to calculate the forces inthe members of a truss, the equilibrium equations areapplied to individual joints (or pins) of the truss.



Zero-Force Member – member that does not carry a load– contributes to the stability of the structure– can carry loads in the event that variations are introduced in

the normal external loading configuration

Σ𝐹𝑦 = 0 :

𝐺𝐶 = 0



Stable Structuresfor 𝑚 = 2𝑗 − 3 → stable

𝑚 < 2𝑗 − 3 → unstable

Determinate Structuresfor r ≤ 3𝑚 → determinate

r > 3𝑚 → indeterminate

where:m = no. of membersj = no. of jointsr = no. of reactions



Using the method of joints, determine the force in each member ofthe truss shown in the figure. Indicate whether the members are intension or compression. (One of the supports is usually designed tobe equivalent to a roller, in order to permit the elongation andcontraction of the truss with temperature changes).

Solution:

Σ𝑀𝐶 = 0 :𝑁𝐴 6 + 10 6 = 60(3)

𝑁𝐴 = 20 𝑘𝑁

Σ𝐹𝑋 = 0 :𝐶𝑋 = 10 𝑘𝑁

Σ𝐹𝑦 = 0 :

𝐶𝑦 + 𝑁𝐴 = 60

𝐶𝑦 = 40 𝑘𝑁



at pin A:

Σ𝐹𝑦 = 0 :

𝑁𝐴 + 𝐴𝐵1

2= 0

𝐴𝐵 = −28.28 𝑘𝑁𝑨𝑩 = 𝟐𝟖. 𝟐𝟖 𝒌𝑵 𝑪

Σ𝐹𝑥 = 0 :

𝐴𝐵1

2+ 𝐴𝐷 = 0

𝑨𝑫 = 𝟐𝟎 𝒌𝑵 𝑻

at pin D:

Σ𝐹𝑦 = 0 :

𝐵𝐷2

5= 60

𝑩𝑫 = 𝟔𝟕. 𝟎𝟖 𝒌𝑵 𝑻

Σ𝐹𝑥 = 0 :

𝐵𝐷1

5+ 𝐶𝐷 = 𝐴𝐷

𝐶𝐷 = −10 𝑘𝑁𝑪𝑫 = 𝟏𝟎 𝒌𝑵 𝑪

Σ𝐹𝑥 = 0 :𝐵𝐶 + 40 = 0

𝐵𝐶 = −40 𝑘𝑁𝑩𝑪 = 𝟒𝟎 𝒌𝑵 𝑪

at pin C:



Method of Sections

• Analyzing the free-body diagram of a part of a trussthat contains two or more joints is called the method ofsections.

• Principle: If the truss is in equilibrium then anysegment of the truss is also in equilibrium.

• It permits us to directly determine the force in almostany member instead of proceeding to that member byjoint-to-joint analysis.

• The cutting plane must not cut more than threemembers whose internal forces are unknown.



Using the method of sections, determine the forces in members BC, HC, HG, and DF.

Solution:

Σ𝑀𝐴 = 0 :8000 6 + 3000 12 = 𝑁𝐸(24)

𝑁𝐸 = 3500 𝑙𝑏

Σ𝐹𝑋 = 0 :𝐴𝑥 = 0

Σ𝐹𝑦 = 0 :

𝐴𝑦 + 𝑁𝐸 = 8000 + 3000

𝐴𝑦 = 7500 𝑙𝑏



Solution:

Σ𝐹𝑦 = 0 :

7500 + 𝐻𝐶4

5= 8000

𝑯𝑪 = 𝟔𝟐𝟓 𝒍𝒃 𝑻

Σ𝑀𝐻 = 0 :7500 6 + 𝐵𝐶 8 = 0

𝐵𝐶 = −5625 𝑙𝑏𝑩𝑪 = 𝟓𝟔𝟐𝟓 𝒍𝒃 𝑪

Σ𝑀𝐶 = 0 :7500 12 = 8000 6 + 𝐻𝐺(8)

𝑯𝑮 = 𝟓𝟐𝟓𝟎 𝒍𝒃 𝑻

Σ𝐹𝑦 = 0 :

𝑫𝑭 = 𝟑𝟓𝟎𝟎 𝒍𝒃 𝑻


Friction

Friction – force that resists the movement of two contacting surfaces that slide relative toone another.

1. Dry Friction - friction force that exists between two unlubricated solid surfaces.2. Fluid Friction - acts between moving surfaces that are separated by a layer of fluid.

𝑭 = 𝝁𝑵

where:F = frictional forceμ = coefficient of frictionN = normal forceφ = angle of friction

𝒕𝒂𝒏𝝓 = 𝝁


Friction

𝑭𝒎𝒂𝒙 = 𝝁𝒔𝑵𝐹𝑚𝑎𝑥 always opposes impending sliding

𝑭𝒌 = 𝝁𝒌𝑵𝐹𝑘 always opposes sliding

𝝁𝒔 > 𝝁𝒌 ; 𝑭𝒔 > 𝑭𝒌

For very low velocity:𝝁𝒔 ≈ 𝝁𝒌 ; 𝑭𝒔 ≈ 𝑭𝒌

where:𝐹𝑚𝑎𝑥 = maximum static friction𝐹𝑘 = kinetic friction𝐹𝑠 = static frictionμ𝑠 = coefficient of static frictionμ𝑘 = coefficient of kinetic friction


Friction

The 100-lb block in the figurebelow is at rest on a roughhorizontal plane before the forceP is applied. Determine themagnitude of P that would causeimpending sliding to the right

Solution:

𝑃 = 𝐹 = 𝐹𝑚𝑎𝑥 = 𝜇𝑠𝑁 = 0.5 100𝑙𝑏𝑷 = 𝟓𝟎 𝒍𝒃


Friction

A 600N block rests in a surface inclined at30°. Determine the horizontal force Prequired to prevent the block from slidingdown. Angle of friction between the blockand the inclined plane is 15°.

Solution:

Σ𝐹𝑋 = 0 :𝑃𝑐𝑜𝑠θ + 𝐹 = 𝑊𝑠𝑖𝑛θ𝑃𝑐𝑜𝑠θ + μ𝑁 = 𝑊𝑠𝑖𝑛θ

𝑃𝑐𝑜𝑠θ + 𝑡𝑎𝑛ϕ𝑁 = 𝑊𝑠𝑖𝑛θ𝑃𝑐𝑜𝑠30 + 𝑡𝑎𝑛15𝑁 = 600𝑠𝑖𝑛30 𝒂

Σ𝐹𝑦 = 0 :

−𝑃𝑠𝑖𝑛θ + 𝑁 = 𝑊𝑐𝑜𝑠θ−𝑃𝑠𝑖𝑛30 + 𝑁 = 600𝑐𝑜𝑠30 𝒃

Solving Eqns. (a) and (b) simultaneously gives:

𝑷 = 𝟏𝟔𝟎. 𝟕𝟕 𝑵𝑁 = 600 𝑁


Friction

The uniform 100-lb plank in the figure below isresting on friction surfaces at A and B. Thecoefficients of static friction are shown in the figure.If a 200-lb man starts walking from A toward B,determine the distance x when the plank will start toslide.

Solution:

𝐹𝐴 = 0.2𝑁𝐴

𝐹𝐵 = 0.5𝑁𝐵

Σ𝐹𝑋 = 0 :𝐹𝐴 + 𝐹𝐵𝑐𝑜𝑠40 − 𝑁𝐵𝑐𝑜𝑠50 = 0 (𝒂)

Σ𝐹𝑦 = 0 :

𝑁𝐴 + 𝑁𝐵𝑠𝑖𝑛50 + 𝐹𝐵𝑠𝑖𝑛40 = 300 (𝒃)Σ𝑀𝐴 = 0 :

𝑁𝐵𝑠𝑖𝑛50 10 + 𝐹𝐵𝑠𝑖𝑛40 10 − 200𝑥 = 100 5 (𝒄)

Substituting 𝐹𝐴 and 𝐹𝐵 to eqns. (a), (b), and (c) and solving simultaneously gives:𝑁𝐴 = 163.3 𝑙𝑏 ; 𝑁𝐵 = 125.7 𝑙𝑏

𝒙 = 𝟒. 𝟑𝟒 𝒇𝒕


Force Systems in Space

Six equilibrium equations in three dimensions:



Assume the three force vectors intersect at a singlepoint.

𝐹1 = 4𝑖 + 2𝑗 + 5𝑘𝐹2 = −2𝑖 + 7𝑗 − 3𝑘𝐹3 = 2𝑖 − 𝑗 + 6𝑘

What is the magnitude of the resultant force vector, R?

Solution:

𝑅 = 𝐹1 + 𝐹2 + 𝐹3

= 4𝑖 + 2𝑗 + 5𝑘 + −2𝑖 + 7𝑗 − 3𝑘 + 2𝑖 − 𝑗 + 6𝑘𝑅 = 4𝑖 + 8𝑗 + 8𝑘

𝑅 = (𝐴𝑖)2 + (𝐴𝑗)

2 + (𝐴𝑘)2

= 42 + 82 + 82

𝑹 = 𝟏𝟐 𝒖𝒏𝒊𝒕𝒔



Find the reactions for the equipment shelf shown in thesketch. The three applied loads act at the center of thevolume shown. Supports A and B cannot take reactionsin the y direction and support C cannot take a reactionin the x direction.



Solution:

Σ𝑀𝑦 = 0 :

300 5 − 400 12 + 24𝑅3 = 0𝑹𝟑 = 𝟏𝟑𝟕. 𝟓 𝒍𝒃

Σ𝐹𝑦 = 0 :

𝑹𝟒 = 𝟐𝟎𝟎 𝒍𝒃

Σ𝑀𝑥 = 0 :8𝑅3 + 16𝑅2 + 200 5 − 400(8) = 0

𝑹𝟐 = 𝟔𝟖. 𝟕𝟓 𝒍𝒃

Σ𝐹𝑧 = 0 :𝑅1 + 𝑅2 + 𝑅3 − 400 = 0

𝑹𝟏 = 𝟏𝟗𝟑. 𝟕𝟓 𝒍𝒃

Σ𝑀𝑧 = 0 :24𝑅4 − 16𝑅6 − 200 12 + 300 8 = 0

𝑹𝟔 = 𝟑𝟎𝟎 𝒍𝒃

Σ𝐹𝑥 = 0 :𝑅5 + 𝑅6 − 300 = 0

𝑹𝟓 = 𝟎


Centroid and Center of Gravity

CG or Center of Weight• It is the point at which the resultant of the gravitational forces (weight) act on a

body.• It is a property of the distribution of weight within the body.

Center of Mass• It is the point through which the resultant inertia force acts on a body.• It is a property of the distribution of mass within the body.

CG and CM



CG by Tabular Summation

CM by Tabular Summation



• It is the point at which area (or volume or line) can be concentrated• It is the point at which the static moment is zero.• The centroid represents the geometric center of a body. This point

coincides with the center of mass or the center of gravity only if thematerial composing the body is uniform or homogeneous.

• Formulas used to locate the center of gravity or the centroid simplyrepresent a balance between the sum of moments of all the parts ofthe system and the moment of the “resultant” for the system.

• In some cases the centroid is located at a point that is not on theobject, as in the case of a ring, where the centroid is at its center.Also, this point will lie on any axis of symmetry for the body.

Centroid



Centroid by Intergration

Centroid of a Volume

Centroid of an Area Centroid of a Line



Find the location of the centroidal axisthat is parallel to the base of the trianglein sketch.

Solution:

𝐴 𝑦 = 𝑦𝑑𝑎 = 𝑦𝑢𝑑𝑦

By similar triangles:𝑢

𝑏=

ℎ − 𝑦

ℎ→ 𝑢 =

𝑏

ℎℎ − 𝑦

Then

𝐴 𝑦 =𝑏ℎ

2 𝑦 =

0

ℎ 𝑏

ℎℎ − 𝑦 𝑦𝑑𝑦

=𝑏

ℎ 0

ℎ(ℎ𝑦𝑑𝑦 − 𝑦2 𝑑𝑦)

=𝑏

ℎ

ℎ𝑦2

2−

𝑦3

3

ℎ

0

=𝑏

ℎ

ℎ3

2−

ℎ3

3=

𝑏ℎ2

6

𝑦 =𝑏ℎ2

6

2

𝑏ℎ

𝒚 =𝒉

𝟑



The 16-ft wing of an airplane is subjectedto a lift which varies from zero at the tip to360 lb/ft at the fuselage according to 𝑤 =

90𝑥1

2 lb/ft where x is measured from thetip. Compute the resultant and its locationfrom the wingtip.

Solution:

𝑅 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒

𝑅 = 0

16

90 𝑥𝑑𝑥

𝑹 = 𝟑𝟖𝟒𝟎 𝒍𝒃

𝑥 = 𝑥𝑑𝑎

𝐴=

0

16𝑥 90 𝑥𝑑𝑥

3480 𝒙 = 𝟗. 𝟔𝟎 𝐟𝐭



Centroid of Common Geometric Shapes

Area and Centroid

𝐴 = 𝑏𝑑

𝑥 =1

2𝑏

𝑦 =1

2𝑑

𝐴 =1

2𝑏ℎ

𝑦 =1

3ℎ

Area and Centroid

𝐴 = π𝑟2

𝑥 = 0 𝑦 = 0

𝐴 =1

2π𝑟2

𝑥 = 0

𝑦 =4𝑟

3π



Centroid of Common Geometric Shapes

Area and Centroid

𝐴 =1

4π𝑟2

𝑥 =4𝑟

3π

𝑦 =4𝑟

3π

𝐴 = π𝑎𝑏 𝑥 = 0

𝑦 = 0

Area and Centroid

𝐴 =1

2π𝑎𝑏

𝑥 = 0

𝑦 =4𝑏

3π

𝐴 =1

4π𝑎𝑏

𝑥 =4𝑎

3π

𝑦 =4𝑏

3π


Centroid by Tabular Summation

Centroid of a Composite Area

Centroid of a Composite Curve



Find the centroidal axes of the section in the sketch.

Solution:

Element A y x Ay Ax

1 .60 2.85 1.0 1.71 .60

2 .60 .15 1.0 .09 .60

3 .48 1.50 .10 .72 .048

ΣA=1.68 ΣAy=2.52 ΣAx=1.248

𝑥 =Σ(𝐴𝑥)

Σ𝐴=

1.248

1.68=. 𝟕𝟒𝟑 𝒊𝒏

𝑦 =Σ(𝐴𝑦)

Σ𝐴=

2.52

1.68= 𝟏. 𝟓𝟎 𝒊𝒏



Moment of Inertia

Moment of Inertia• Also called the second moment of area• For structural cross-sections, the moment of inertia of interest are

those about the centroidal axes.• Used in determining the stiffness and bending stresses in beams

and the buckling loads of columns• For beams, the moment of area of interest is the one about the

bending axis, for columns, it is the minimum moment of inertia

First Moment of Area (Static Moment of Area)

𝑄 = 𝑦𝑑𝐴 = Σ𝑎𝑦

• Used to find the shear stress distribution over a cross-section of ashear carrying member

Moment of Inertia

Moment of Inertia by Integration Polar Moment of Inertia

Parallel –Axis Theorem (Transfer Formula)

where: 𝐼𝑎 = moment of inertia about an arbitrary axis 𝐼𝑎 = moment of inertia about the parallel axis that passes through the centroid𝐴 = area𝑑 = distance between the axes (transfer distance)

Moment of Inertia by Tabular Summation

𝐼𝑥 = Σ𝐼 𝑥 + ΣA𝑦2

𝑰 𝒙 = 𝑰𝒙 − 𝐀 𝒚𝟐

𝐼𝑦 = Σ𝐼 𝑦 + ΣA𝑥2

𝑰 𝒚 = 𝑰𝒚 − 𝐀 𝒙𝟐


Moment of Inertia

Centroidal Moment of Inertia (with respect to an axis passing through the centroid):


Moment of Inertia


Moment of Inertia

Moment of Inertia with respect to an axis passing through the base:


Moment of Inertia

A rectangle has a base of 3 cm and a heightof 6 cm. What is its second moment of area(in cm4) about an axis through the center ofgravity and parallel to the base.

Solution:

𝐼𝑥0= 𝐼 𝑥 =

𝑏ℎ3

12=

3(63)

12𝑰𝒙𝟎

= 𝟓𝟒 𝒄𝒎𝟒

𝐼𝑥 =𝑏ℎ3

3=

3(63)

3𝑰𝒙 = 𝟐𝟏𝟔 𝒄𝒎𝟒

Or by Transfer Formula:𝐼𝑥 = 𝐼 𝑥 + A𝑑2 = 54 + 3𝑥6 32

𝑰𝒙 = 𝟐𝟏𝟔 𝒄𝒎𝟒


Moment of Inertia

Find the moment of inertia of the sectionabout the centroidal axes parallel to axes xand y

Solution:From previous example 𝑥 = .743 𝑎𝑛𝑑 𝑦 = 1.50

Element A y x Ay2 Ax2 Ix Iy

1 .60 2.85 1.00 4.8735 .60 .0045 .200

2 .60 .15 1.00 .0135 .60 .0045 .200

3 .48 1.50 .10 1.0800 .0048 .2304 .002

1.68 5.967 1.2048 .2394 .402

𝐼𝑥 = Σ𝐼 𝑥 + ΣA𝑦2 = .2394 + 5.967 = 6.2064

𝐼 𝑥 = 𝐼𝑥 − 𝐴 𝑦2 = 6.2064 − 1.68 1.502

𝑰 𝒙 = 𝟐. 𝟒𝟐𝟔 𝒊𝒏𝟒

𝐼𝑦 = Σ𝐼 𝑦 + ΣA𝑥2 = .402 + 1.2048 = 1.6068

𝐼 𝑦 = 𝐼𝑦 − 𝐴 𝑥2 = 1.6068 − 1.68 .7432

𝑰 𝒚 =. 𝟔𝟕𝟗𝟓 𝒊𝒏𝟒



• Motion – Change of position of an object with respect to time and reference point.

• Translation – Motion involving change in displacement over a period of time.

• Rotation – Motion involving change in angle over a period of time.

• Rectilinear Motion – Straight line motion

• Curvilinear Motion – Motion along a curved path


Fundamental Concepts and Definition

• Displacement, s – Shortest distance between initial and final position of a particle.

• Velocity, V – Instantaneous rate of change of displacement with respect to time.

𝑉 =𝑑𝑠

𝑑𝑡

• Speed – Refers to the magnitude of velocity.

• Acceleration, a - Instantaneous rate of change of velocity with respect to time.

𝑎 =𝑑𝑉

𝑑𝑡


Kinematics of a Particle


Equations of Motion with Constant Acceleration:

• 𝑠 = 𝑉0𝑡 +1

2𝑎𝑡2

• 𝑉 = 𝑉0 + 𝑎𝑡

• 𝑉2 = 𝑉02

+ 2𝑎𝑠For a free falling body, 𝑉0 = 0

• 𝑦 =1

2𝑔𝑡2

Rectilinear Translation


A ball is thrown vertically upward at a speed of 20 m/s.

a. How high is it after 3s?

𝑦 = 𝑉0𝑡 +1

2𝑔𝑡2

𝑦 = 20𝑚

𝑠3𝑠 + .5 −9.8

𝑚

𝑠2 (3𝑠)2

𝒚 = 𝟏𝟓. 𝟗 𝒎

b. How high does it rise?

𝑉2 = 𝑉02

+ 2𝑔𝑦

0 = 20𝑚

𝑠

2

+ 2 −9.8𝑚

𝑠2𝑦𝑚𝑎𝑥

𝒚𝒎𝒂𝒙 = 𝟐𝟎. 𝟒𝟏 𝒎

c. How long does it take to reach the highest point?𝑉 = 𝑉0 + 𝑔𝑡

0 = 20𝑚

𝑠+ −9.8

𝑚

𝑠2 𝑡

𝒕 = 𝟐. 𝟎𝟒 𝒔

d. How long does it take for the ball to reach the ground?𝒕𝒖𝒑 = 𝒕𝒅𝒐𝒘𝒏 = 𝟐. 𝟎𝟒 𝒔

e. What is its velocity when it returns to the level from which it started?𝑽𝟎 = 𝑽𝒇 = 𝟐𝟎 𝒎/𝒔


If a particle’s position is given by the expression 𝑥 𝑡 = 3.4𝑡3 − 5.4𝑡 meters.

a. What is its velocity after t = 3 seconds?

𝑉 =𝑑𝑥

𝑑𝑡=

𝑑

𝑑𝑡(3.4𝑡3 − 5.4𝑡)

𝑡 = 3𝑽 = 𝟖𝟔. 𝟒 𝒎/𝒔

b. What is the acceleration of the particle after t = 5 seconds?

𝑉 =𝑑𝑥

𝑑𝑡=

𝑑

𝑑𝑡3.4𝑡3 − 5.4𝑡 = 10.2𝑡2 − 5.4

𝑎 =𝑑𝑉

𝑑𝑡=

𝑑

𝑑𝑡(10.2𝑡2 − 5.4)

𝑡 = 5

𝒂 = 𝟏𝟎𝟐 𝒎/𝒔𝟐




General Equation of Projectile:

𝑦 = 𝑥𝑡𝑎𝑛θ −𝑔𝑥2

2𝑉02𝑐𝑜𝑠2θ

• 𝑉𝑦 deceases as it goes up, zero at maximum height, and increases as

it goes down.• 𝑉𝑥 is constant.

𝑉0𝑥= 𝑉𝑜𝑐𝑜𝑠θ

𝑉0𝑦= 𝑉𝑜𝑠𝑖𝑛θ

Curvilinear Translation

A shot is fired at an angle of 45° with the horizontal and a velocity of 300 fps. Calculate the range of the projectile.

Solution:

𝑅 =𝑉0

2𝑠𝑖𝑛2θ

𝑔

=(300 𝑓𝑡/𝑠)2sin(2𝑥45)

32.2 𝑓𝑡/𝑠2

𝑹 = 𝟐𝟕𝟗𝟓 𝒇𝒕



A projectile leaves with a velocity of 50 m/s at anangle of 30° with the horizontal. Find the maximumheight that it could reach.

Solution:

𝐻 =𝑉0

2𝑠𝑖𝑛2θ

2𝑔

=(50 𝑚/𝑠)2(sin30)2

2 9.81𝑚𝑠2

𝑯 = 𝟑𝟏. 𝟖𝟔 𝒎




• 𝑆 = 𝑟θ• 𝑉 = 𝑟ω• 𝑎 = 𝑟αWhere:r = radiusθ, ω, and α are angular displacement, angular velocity, and angular acceleration, respectively.S, V, and a are linear dimensions.

• Linear velocity acts tangent to the point.• Linear acceleration has tangential and

normal components.

• 𝑎 = 𝑎𝑛2 + 𝑎𝑡

2

• 𝑎𝑛 =𝑉2

𝑟

• 𝑎𝑡 =𝑑𝑉

𝑑𝑡

Also,

• θ = ω0𝑡 +1

2α𝑡2

• ω = ω0 + α𝑡

• ω2 = ω02

+ 2α𝑠

Rotation

A turbine started from rest to 180 rpm in 6 minutesat a constant acceleration. Find the number ofrevolutions that it makes within the elapsed time.

Solution:

ω = ω0 + α𝑡180 𝑟𝑝𝑚 = 0 + α 6 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

α = 30𝑟𝑒𝑣

𝑚𝑖𝑛2

ω2 = ω02

+ 2α𝑠

(180 𝑟𝑝𝑚)2= 0 + 2 30𝑟𝑒𝑣

𝑚𝑖𝑛2(𝑠)

𝒔 = 𝟓𝟒𝟎 𝒓𝒆𝒗



A flywheel is 15 cm in diameter acceleratesuniformly from rest to 500 rpm in 20 seconds.What is its angular acceleration?

Solution:

500𝑟𝑒𝑣

𝑚𝑖𝑛𝑥

2π𝑟𝑎𝑑

𝑟𝑒𝑣𝑥

𝑚𝑖𝑛

60 𝑠= 52.36 𝑟𝑎𝑑/𝑠

ω = ω0 + α𝑡

52.36𝑟𝑎𝑑

𝑠= 0 + α 20 𝑠

𝜶 = 𝟐. 𝟔𝟐 𝒓𝒂𝒅/𝒔𝟐



Kinetics of a Particle


When a body is subjected to an acceleration, there exists a forceopposite the direction of motion and equal to the product ofmass and acceleration . This force is known as the reverseeffective force, REF or inertial force, ma.

𝜮𝑭 = 𝒎𝒂 =𝒘

𝒈𝒂

D’ Alembert’s Principle


What force is necessary to accelerate a 30,000 lb railwayelectric car at the rate of 1.25 ft/sec2, if the forcerequired to overcome frictional resistance is 400 lb?

Solution:

Σ𝐹ℎ = 𝑚𝑎

𝑃 − 𝐹 =𝑊

𝑔𝑎

𝑃 − 400 =30,000 𝑙𝑏

32.2 𝑓𝑝𝑠2 1.25𝑓𝑝𝑠2

𝑷 = 𝟏𝟓𝟔𝟒. 𝟔 𝒍𝒃



An elevator weighing 2,000 lb attains an upward velocityof 16 fps in 4 seconds with uniform acceleration. What isthe tension in the supporting cables?

Solution:

𝑉 = 𝑉0 + 𝑎𝑡

16𝑓𝑡

𝑠= 0 + 𝑎 4𝑠

𝑎 = 4 𝑓𝑡/𝑠2

Σ𝐹𝑣 = 𝑚𝑎

𝑇 − 𝑊 =𝑊

𝑔𝑎

𝑇 − 2000 =2000 𝑙𝑏

32.2 𝑓𝑝𝑠2 4 𝑓𝑝𝑠2

𝑷 = 𝟐𝟐𝟒𝟖. 𝟒𝟓 𝒍𝒃



Centripetal and Centrifugal Force

𝐶𝐹 = 𝑚𝑎𝑛 = 𝑚𝑉2

𝑟



A cyclist on a circular track of radius800 ft is traveling at 27 fps. Hisspeed in the tangential directionincreases at the rate of 3 fps2. Whatis the cyclist’s total acceleration?

Solution:

𝑎𝑛 =𝑉2

𝑟=

(27 𝑓𝑝𝑠)2

800 𝑓𝑡= 0.91 𝑟𝑎𝑑/𝑠2

𝑎 = 𝑎𝑡2 + 𝑎𝑛

2

𝑎 = 32 + 0.912

𝒂 = 𝟑. 𝟏𝟒 𝒇𝒑𝒔𝟐


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