Spring 2021
Mechanics Notes
Statics and Mechanics of Materials
Lisa Andersen Gielfeldt
06-04-2021
Contents
Contents ................................................................................................................................................................................................ 1
L01 STATICS - Basics ..................................................................................................................................................................... 2
L02 STATICS - Structures............................................................................................................................................................. 2
L03 MECHANICS OF MATERIALS - Introduction .............................................................................................................. 2
L04 MECHANICS OF MATERIALS - Axially Loaded Members and Torsion ......................................................... 14
L05 MECHANICS OF MATERIALS - Bending and Stresses in Beams ...................................................................... 24
L06 MECHANICS OF MATERIALS - Shear Stresses in and Deflection of Beams ................................................ 36
L07 MECHANICS OF MATERIALS - Columns and Statically Indeterminate Beams ......................................... 46
L08 MECHANICS OF MATERIALS - An overview............................................................................................................. 59
L09 DYNAMICS - Kinematics of particles............................................................................................................................ 59
L10 DYNAMICS - Kinematics and kinetics of particles ................................................................................................. 67
L11 DYNAMICS - Work and energy, impulse and momentum .................................................................................. 70
L12 DYNAMICS - Kinetics of Particle and System of Particles ................................................................................... 76
L13 DYNAMICS - Mass flow and variable mass ................................................................................................................ 85
L14 MECHANICAL VIBRATIONS - Fundamentals of Vibration ................................................................................. 86
L15 MECHANICAL VIBRATIONS - Free Vibration of Particles .................................................................................. 86
L16 MECHANICAL VIBRATIONS - Forced Vibration of Particles, Vibration of Rigid Bodies ....................... 88
L17 DYNAMICS & MECHANICAL VIBRATIONS - An overview .................................................................................. 89
L01 STATICS
- Basics
L02 STATICS
- Structures
L03 MECHANICS OF MATERIALS
- Introduction
1.2 Problem Solving Approach (PSA)
1. Conceptualize
a. Hypothesize and sketch.
b. List all the relevant data and draw a sketch showing all apply forces, support/boundary
conditions, and interactions between adjacent bodies.
c. Development and refinement of the free body diagram is an essential part of this step.
2. Categorize
a. Simplify classify identify the unknowns and the problems and make any necessary
assumptions to simplify the problem and streamline the solution process.
3. Analyze
a. Evaluate select relevant equations carry out mathematical solution apply appropriate
theories set up the necessary equations for the chosen mathematical model and then
solve for the unknowns.
4. Finalize
a. Conclude, examine answer
b. Does it make sense, are the units correct, how does it compare to similar problem
solutions?
1.3 Statics Review
Equilibrium Equations
𝑅 = ∑𝐹 = 0
𝑀 = ∑ 𝑀 = ∑(𝑟 × 𝐹) = 0
2 dimensions:
∑ 𝐹𝑥 = 0
3 dimensions
∑ 𝐹𝑦 = 0
∑ 𝑀𝑧 = 0
∑ 𝐹𝑥 = 0
∑ 𝐹𝑦 = 0
∑ 𝑀𝑧 = 0
∑ 𝑀𝑥 = 0
∑ 𝑀𝑦 = 0
∑ 𝑀𝑧 = 0
Applied Forces
Concentrated forces and moments: Acting at a certain point on a structure/body.
𝐹 = [𝑁]
𝑀 = [𝑁𝑚]
Distributed forces and moments:
𝐹 = [𝑁/𝑚2 ]
Body Force
𝑤 = [𝑁/𝑚3 ]
Can be replaced by
𝑊𝑠 = 𝑊𝑝 = [𝑁]
(the force at the center of gravity)
Reactive Forces and Support Conditions
Reaction force at support:
Moment restraint at support
Internal Forces (Stress Resultant)
Stress resultants are usually taken along and normal to the member (i.e., local or member axes are
used.)
Deformation sign convention: Tension is positive & compression is negative.
1. 4 Normal Stress and Strain
Stress: Intensity of internal force
When the bar is stretched by the forces P, the stresses are tensile stresses, if the forces are reversed in
direction causing the bar to be compressed, they are compressive stresses.
Normal stress
Normal stresses act in a direction perpendicular to the cut surface. Normal stresses may be either
tensile or compressive. Shear stress act parallel to the surface.
Stress:
𝜎=
𝑃
𝑁
𝑁
] = [𝑀𝑝𝑎]
= [ 2] ⇒ [
𝐴
𝑚
𝑚𝑚2
we know the formula from
𝑃𝑡𝑟𝑦𝑘 =
𝐹
𝐴
Limitations
The axial forces are applied at the centroids of the cross sections unless specifically stated otherwise.
As a practical rule, the formula 𝜎 = 𝑃/𝐴 may be used with good accuracy at any point within a
prismatic bar that is at least as far away from the stress concentration as the largest a lateral
dimension of the bar.
It is customary when solving textbook problems to omit (leave out) the weight of the structure unless
specifically instructed to included.
Normal Strain
Strain: Percent of how much the material has stretched
Strain:
𝜀=
Length before: 𝐿
𝛿
= [−]
𝐿
Length after: 𝐿 + 𝛿
Relative length: 𝛿
Uniaxial Stress and Strain
The equations above requires that the deformation of the bar be uniform throughout its volume, which
in turn requires that the bar be prismatic, the loads act through the centroids of the cross sections, and
the material be homogeneous and isotropic (that is, the same throughout all parts of the bar ). the
resulting state of stress and strain is called uniaxial stress and strain (although lateral strain exists as
discussed later )
1.5 Mechanical Properties of Materials
Stress Strain Diagrams
Stresses and Strains for Structural Steel
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
Stress-strain diagram for steel in tension
Displacement control
Typical diagram
5 zones
O-A: Linear region, definite performance of material. Linear elasticity
A: proportional limit
A-B: none linear elasticity. Yield stress
B-C: Perfect plasticity or yielding: behaves like fluid and does not resist elongation.
C-D: increase in stress, much shallower than in O-A. Strain hardening.
D: ultimate stress. Here necking starts- deformation of material at a certain point. No longer
uniform deformation. Necking happens at the weakest link of the material.
D-E: Necking at weakest point is happening. engineering stress, using original area.
E: fracture happens
D-E’: dashed: force divided by area - the actual area.
C-E’: substantial difference when using actual area. As the rod is stretched the rod becomes
longer and thinner → cross sectional area becomes smaller and impacts 𝜎. The actual fracture
happens at the highest possible stress. Fracture will happen at D if you maintain the force
magnitude.
Temperature growth in the area of necking, due to internal friction forces, has do to with
moving and dislocation.
Ductility
o
o
o
o
Metals such as structural steel that undergo large permanent strains before failure are
classified as ducktile.
Ductility is a property that enables a bar of steel to be bent into a circular arch or drawn into a
wire without breaking.
A desirable feature of ductile materials is that visible distortions occur if the loads become too
large, thus providing an opportunity to take remedial action before an actual fracture occurs.
Also, materials exhibiting ductile behavior are capable of solving large amounts of strain
energy prior to fracture.
Aluminum Alloys
o
o
o
Although they have considerable ductility, aluminum alloys typically do not have a clearly
definable yield point, as shown by the stress strain diagram.
However, they do have an initial linear region with a recognizable proportional limit.
When a material such as aluminum does not have an obvious yield point yet undergoes large
strains after the proportional limit is exceeded and arbitrary yield stress point may be
determined by the offset method.
o A straight line is drawn on the stress strain diagram parallel to the initial linear part of
the crew but offset by some standard strain, such as 0.002 (or 0.2%).
o The intersection of the offset line and the stress strain curve defines the yield stress.
o Because this stress is determined by an arbitrary rule and is not an inherent physical
property of the material, it should be distinguished from a true yield stress by referring
to it as the offset yield stress.
Rubber
o
o
o
o
Rubber maintains a linear relationship between stress and strain up to relatively large strange
(as compared to metals).
Beyond the proportional limit, the behavior depends upon the type of rubber . Some kinds of
soft rubber stretch enormously without failure, reaching length several times their original
links .
The material eventually offers increasing resistance to the load, and the stress strain curve
turns markedly upward.
Note that although rubber exhibits very large strange, it is not a dock tile material because the
strains are not permanent . It is of course an elastic material.
Ductility and Elongation
Ductile materials can withstand large deformation of material. Example rubber. Here you do not see
the first part of the diagram as we saw in steel. A to B is very small.
Percent elongation:
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 = 𝜀𝑥 % =
𝐿1 − 𝐿0
· 100
𝐿0
Percent reduction:
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 = 𝜀𝑦 % =
𝐴1 − 𝐴0
· 100
𝐴0
Brittle Materials
o
o
o
o
o
Materials that fail in tension at relatively low values of strain are classified as brittle. Brittle
materials failed with only little elongation after the proportional limit.
Furthermore, the reduction in area is insignificant, and so the nominal fracture stress is the
same as the true ultimate stress.
Example cast iron. Right diagram. The first part is like the steel diagram. The second part is
from A-B B is fracture point - instantaneous death of material.
A: generation of crack.
Brittle materials are not capable to withstand large deformation, they just crack.
Plastics
o
o
o
o
Some plastics are brittle others are ducktile.
When designing with plaques plastics, it is important to realize that their properties are greatly
affected by both temperature changes and the passage of time.
Also, a loaded plastic may stretch gradually overtime until it is no longer serviceable.
This phenomenon is called creep.
Composites
o
o
o
A filament reinforced material consists of a base material in which high strengths filaments,
fibers, or whiskers are embedded.
The resulting composite material is much greater strength than the base material. As an
example of the use of glass fibers can more than double the strength of a plastic matrix.
Composites are widely used in aircrafts, boats, rockets, and space vehicles where higher
strength and light weight are needed.
Compression
Deformation is negative
No fracture, just maximum level of force from machine.
Important: the further in the diagram the more heat the material will gain, due to internal friction
forces, has do to with moving and dislocation.
1.6 Elasticities, Plasticity and Creep
Two types of deformation: elastic and plastic
o
o
Elastic: fully reversable
Plastic: non reversable
Ductile materials: Have both elastic and plastic
Brittle does this and then break down.
Reloading of material
(a) you can say whether this is elastic or plastic. Purely elastic
(b) when you drop the force, you get a straight line, which is parallel to the tangent of the curved
diagram (it will parallel to O-A from the steel diagram). O-C residual strain. C-D elastic recovery.
(c) shows reloading case of (b). now it follows a new path up until the point it was stretched to before
and then follows the original curve afterwards(B-F)
Creep:
Tactile materials. Not brittle materials.
Creep is dependent on stress and time.
You have a material and apply a force and instantaneously have an elongation 𝛿 and the gradually over
time (hours/days/months) the material will deform.
Creep can be very dangerous. Creep deformation is extremely sensitive to temperature. If the
temperature increases the creep increases rapidly.
O-t_0 is a fraction of a second- instantaneously. And from then on we have stress decrease and the
material will relax. Stress decrease is dependent of temperature.
Irrelevant for brittle materials.
Only plasticity
1.7 Linear Elasticity, Hooke’s Law Poisson Ratio
Only in linear region: If we are in the area of linear elasticity (steel
diagram O-A)
Hooke’ law:
𝐸=
𝜎
𝜀
⇔
𝜎=𝐸·𝜀
𝐸 = [𝑃𝑎] ⇒ [𝑀𝑃𝑎]
o
o
o
Modulus of elasticity, E
Young’s modulus: (E is very big and 𝜀 is very small)
E is the hældning in the linear region
Possions ratio:
𝜈=−
𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
𝜀 ′ 𝜀𝑦
=− =
𝑎𝑥𝑖𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
𝜀
𝜀𝑥
𝜈=−
𝜀′
𝜀
⇔
Lateral strain: decrease in cross section
𝜈=−
𝜀𝑦
𝜀𝑥
Axial strain: increase in length
Limitations
Homogeneous (First condition)
•
•
•
•
•
First, the material must be homogeneous, that is, it must have the same composition (and
hence the same elastic properties) at every point.
However, having a homogeneous material does not mean that the elastic properties at a
particular point are the same in all directions.
For instance, the modulus of elasticity could be different in the actual and lateral directions, as
in the case of a wood pole.
Therefore, a second condition for uniformity in the lateral strains is that the elastic properties
must be the same in all directions perpendicular to the longitudinal axis.
When the preceding conditions are met, as it's often the case with metals, lateral strain in a
prismatic bar subjected to uniform tension will be the same at every point in the bar and the
same in all lateral directions.
Isotropic (Second condition)
Materials having the same properties in all directions (whether actual, lateral, or any other direction)
are said to be isotropic.
Anisotropic; If the properties differ in various directions, the material is called anisotropic.
In this book all examples and problems are sold with the assumption that the materialist linearly
elastic, homogeneous, and isotropic - unless a specific statement is made to the contrary.
1.8 Shear Stress and Strain
Shear force: parallel
Normal: perpendicular
Shear deformation
L04 MECHANICS OF MATERIALS
- Axially Loaded Members and Torsion
Description:
•
•
•
•
•
Changes in Lengths of Axially Loaded Members
Statically Indeterminate Structures
Thermal Effects, Misfits and Prestrains
Torsional Deformation of a Circular Bar
Circular Bars of Linearly Elastic Materials
Literature: Mechanics of Materials, Seventh Edition J. M. Gere, B.J Goodno, Sections 2.2; 2.4; 2.5;3.2;
3.3;
Exercises: Problems 2.2-3; 2.3-4; 2.4-7; 3.2-1
2.2 changes in lengths of axially loaded members
Springs
𝑃=𝑘·𝛿
𝛿 =𝑓·𝑃
𝐿 = natural length / unstressed length / relaxed length
𝑃 = force applied / tension applied
𝛿 = 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝𝑟𝑖𝑛𝑔 (𝑤𝑖𝑙𝑙 𝑏𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑖𝑓 𝑖𝑡 ′ 𝑠 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑒𝑑)
𝑘 = 𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝𝑟𝑖𝑛𝑔 = 𝑠𝑝𝑟𝑖𝑛𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑓 = 𝑓𝑙𝑒𝑥𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝𝑟𝑖𝑛𝑔 = 𝑠𝑝𝑟𝑖𝑛𝑔 𝑐𝑜𝑚𝑝𝑙𝑖𝑎𝑛𝑐𝑒
𝑘 and 𝑓 are the reciprocal of each other:
𝑘=
1
𝑓
⇔
𝑓=
1
𝑘
𝜀=
𝛿
𝐿
Prismatic bars
REP
𝜎=
𝑃
=𝐸·𝜀
𝐴
Put formulas for spring and bar together and you get:
𝛿=
Cables
𝑃𝐿
𝐸𝐴
𝑘=
𝐸𝐴
𝐿
2.4 Statically Indeterminant Structures
∑ 𝐹𝑣𝑒𝑟𝑡 = 0
𝑅𝐴 − 𝑃 + 𝑅𝐵 = 0
𝛿𝐴𝐵 = 0
1. Equilibrium:
𝑅𝐴 − 𝑃 + 𝑅𝐵 = 0
2. Compatibility
𝛿𝐴𝐵 = 0
3. Force displacement relation (Hooke’s law):
𝛿𝐴𝐶 =
𝑅𝐴 𝑎
𝐸𝐴
𝛿𝐶𝐵 = −
𝛿𝐴𝐵 = 𝛿𝐴𝐶 + 𝛿𝐶𝑏 =
𝑅𝐵 𝑏
𝐸𝐴
𝑅𝐴 𝑎 𝑅𝐵 𝑏
−
=0
𝐸𝐴
𝐸𝐴
End result:
𝑅𝐴 =
𝑃𝑏
𝐿
𝛿𝐶 = 𝛿𝐴𝐶 =
General Comments
𝑅𝐵 =
𝑃𝑎
𝐿
𝑅𝐴 𝑎 𝑃𝑎𝑏
=
𝐸𝐴
𝐿𝐸𝐴
2.5 Thermal Effect, Misfits, And Prestrains
Thermal Effects
𝜀𝑇 = 𝛼(Δ𝑇)
𝜎𝑇 = 𝐸𝛼(ΔT)
𝛿 𝑇 = 𝜀𝑇 𝐿 = 𝛼(Δ𝑇)𝐿
𝛼 = 𝑘𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛
Misfits and Prestrains
Boats and Turnbuckles
𝛿 = 𝑛𝑝
𝛿 = 2𝑛𝑝
3.2 Torsional Deformations of a Circular Bar
Shear Strain at The Outer Surface
𝛾𝑚𝑎𝑥 =
𝑏𝑏′
𝑎𝑏
𝛾𝑚𝑎𝑥 =
𝑟𝑑𝜙
𝑑𝑥
𝜃=
𝛾𝑚𝑎𝑥 =
𝑑𝜙
𝑑𝑥
𝑟𝑑𝜙
= 𝑟𝜃
𝑑𝑥
𝛾𝑚𝑎𝑥 = 𝑟𝜃 =
𝑟𝜙
𝐿
Shear Strain Within the Bar
𝛾 = 𝜌𝜃 =
𝜌
·𝛾
𝑟 𝑚𝑎𝑥
Circular Tubes
𝛾𝑚𝑎𝑥 =
𝛾𝑚𝑖𝑛 =
𝑅2 𝜙
𝐿
𝑟1
𝑟1 𝜙
𝛾𝑚𝑎𝑥 =
𝑟2
𝐿
3.3 Circular Bars of Linearly Elastic Materials
𝜏 = 𝐺𝛾
𝜏𝑚𝑎𝑥 = 𝐺𝑟𝜃
𝜏 = 𝐺𝜌𝜃 =
𝜌
𝜏
𝑟 𝑚𝑎𝑥
The Torsion Formula
𝑑𝑀 = 𝜏𝜌𝑑𝐴 =
𝜏𝑚𝑎𝑥 2
· 𝜌 𝑑𝐴
𝑟
𝑇 = ∫ 𝑑𝑀 =
𝐴
𝜏𝑚𝑎𝑥
𝜏𝑚𝑎𝑥
· ∫ 𝜌2 𝑑𝐴 =
· 𝐼𝑝
𝑟
𝑟
𝐴
𝐼𝑝 = ∫ 𝜌2 𝑑𝐴
𝐴
𝐼𝑝 =
𝜋𝑟 4 𝜋𝑑 4
=
2
32
𝜏𝑚𝑎𝑥 =
𝜏𝑚𝑎𝑥 =
𝜏=
𝑇𝑟
𝐼𝑝
16𝑇
𝜋𝑑 3
𝜌
𝑇𝜌
𝜏𝑚𝑎𝑥 =
𝑟
𝐼𝑝
Angle of Twist
𝜃=
𝜙=
𝑘𝑇 =
𝐺𝐼𝑝
𝐿
𝑇
𝐺𝐼𝑝
𝑇𝐿
𝐺𝐼𝑝
𝑓𝑇 =
𝐿
𝐺𝐼𝑝
Circular Tubes
𝐼𝑝 =
𝐼𝑝 =
𝜋 4
𝜋 4
(𝑟2 − 𝑟14 ) =
(𝑑 − 𝑑14 )
2
32 2
𝜋𝑟𝑡
𝜋𝑑𝑡
(4𝑟 2 + 𝑡 2 ) =
= (𝑑 2 + 𝑡 2 )
2
4
𝐼𝑝 ≈
Limitations
2𝜋𝑟 2 𝑡
𝜋𝑑 3 𝑡
=
4
L05 MECHANICS OF MATERIALS
- Bending and Stresses in Beams
4.1 Introduction
4.2 Types of beans loads and reactions
Types of loads
Reactions
Internal releases
4.3 Shear forces and bending moments
∑ 𝐹𝑣𝑒𝑟𝑡 = 0
∑𝑀 = 0
𝑃−𝑉 =0
𝑀 − 𝑃𝑥 = 0
𝑜𝑟
𝑉=𝑃
𝑜𝑟 𝑀 = 𝑃𝑥
Sign convention
4.4 Relationships among loads shear forces and bending moments
Distributed loads
Shear force
∑ 𝐹𝑣𝑒𝑟𝑡 = 0
𝑉 − 𝑞𝑑𝑥 − (𝑉 + 𝑑𝑉 ) = 0
𝑑𝑉
= −𝑞
𝑑𝑥
𝑞0 𝑥
𝑞=
𝐿
𝑞0
𝑉 (𝑥) =
· (𝐿2 − 3𝑥 2 )
6𝐿
𝑑𝑉
𝑑 𝑞0
𝑞0 𝑥
=
[ · (𝐿2 − 3𝑥 2 )] = −
= −𝑞
𝑑𝑥 𝑑𝑥 6𝐿
𝐿
𝐵
𝐵
∫ 𝑑𝑉 = − ∫ 𝑞𝑑𝑥
𝐴
𝐴
𝐵
𝑉𝐵 − 𝑉𝐴 = − ∫ 𝑞𝑑𝑥 = −(𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑𝑖𝑛𝑔 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴 𝑎𝑛𝑑 𝐵)
𝐴
Bending moment
∑𝑀 = 0
− 𝑀 − 𝑞𝑑𝑥 (
𝑑𝑥
) − (𝑉 + 𝑑𝑉)𝑑 + 𝑀 + 𝑑𝑀 = 0
2
𝑑𝑀
=𝑉
𝑑𝑥
𝑞0 𝑥 2
(𝐿 − 𝑥 2 )
𝑀(𝑥) =
6𝐿
𝑑𝑀
𝑑 𝑞0 𝑥 2
𝑞
(𝐿 − 𝑥 2 )] = 0 (𝐿2 − 3𝑥 2 )
=
[
𝑑𝑥 𝑑𝑥 6𝐿
6𝐿
𝐵
𝐵
∫ 𝑑𝑀 = ∫ 𝑉𝑑𝑥
𝐴
𝐴
𝐵
𝑀𝐵 − 𝑀𝐴 = ∫ 𝑉𝑑𝑥 = (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐴 𝑎𝑛𝑑 𝐵)
𝐴
Concentrated loads
Loads in the form of couples
4.5
5.2 Pure bending and nonuniform bending
5.3 Curvature of a beam
5.4 Longitudinal strains and beans
5.5 Normal stresses in beams (linearly elastic materials )
Location of neutral axis
Moment curvature relationship
Flexure formula
Maximum stresses at a cross section
Doubly symmetric shapes
Properties of beam cross sections
Limitations
5.6 Design of beams for painting stresses
Beams of standardized shapes and sizes
Relative efficiency of various beam shapes
L06 MECHANICS OF MATERIALS
- Shear Stresses in and Deflection of Beams
4.5 Shear Force and Bending Moment Diagrams – reminder
5.8 Shear Stresses in Beams of Rectangular Cross Section
Derivation of shear formula
𝜎1 = −
𝑀𝑦
𝐼
𝜎2 = −
𝐹1 = ∫ 𝜎1 · 𝑑𝐴 = ∫
(𝑀 + 𝑑𝑀)𝑦
𝐼
𝑀𝑦
𝑑𝐴
𝐼
𝐹3 = 𝐹2 − 𝐹1 = ∫
𝐹2 = ∫ 𝜎2 · 𝑑𝐴 = ∫
(𝑀 + 𝑑𝑀)𝑦
𝑑𝐴
𝐼
(𝑀 + 𝑑𝑀)𝑦
(𝑑𝑀)𝑦
𝑀𝑦
𝑑𝐴 − ∫
𝑑𝐴 = ∫
𝑑𝐴
𝐼
𝐼
𝐼
𝐹3 = 𝜏𝑏𝑑𝑥
𝜏=
𝑑𝑀 1
( ) ∫ 𝑦𝑑𝐴
𝑑𝑥 𝐼𝑏
𝜏=
𝑉𝑄
𝐼𝑏
𝑄 = ∫ 𝑦𝑑𝐴
𝜏=
𝑉
ℎ2
· ( − 𝑦12 )
2𝐼
4
𝜏𝑚𝑎𝑥 =
𝑉ℎ2 3𝑉
=
8𝐼
2𝐴
5.8 Shear Stresses in Beams of Circular Cross Section
𝜏𝑚𝑎𝑥
2𝑟 2
𝑉
·
(
𝑉𝑄
4𝑉
4𝑉
3 )
=
=
=
=
4
2
𝜋𝑟
𝐼𝑏
3𝜋𝑟
3𝐴
( 4 ) · (2𝑟)
9.2 Differential Equations of the Deflection Curve
9.3 Deflections by Integration of the Bending-Moment Equation
L07 MECHANICS OF MATERIALS
- Columns and Statically Indeterminate Beams
L08 MECHANICS OF MATERIALS
- An overview
L09 DYNAMICS
- Kinematics of particles
2/1 Introduction to kinematics of particles
Particle Motion: When the curvature of an object is much bigger than the object. The object can be
seen as a particle.
Constrained motion: If a particle is confined to a specified path. An example is an object constrained
by a string and whirled in a circle.
Unconstrained Motion: The particle can move freely / there is no physical guides. An example is the
previous one but where the string breaks and afterwards the motion is unconstrained.
Choice of Coordinates:
•
•
•
•
Rectangular
Cylindrical
Spherical
Tangent and normal
Fixed reference axis:
Moving reference axis (relative motion analysis)
2/2 Rectilinear Motion
Velocity and Acceleration
𝑣=
𝑎=
Δ𝑠
Δ𝑡
⇔
𝑑𝑣
= 𝑣̇
𝑑𝑡
𝑣=
𝑑𝑠
= 𝑠̇
𝑑𝑡
𝑑2 𝑠
= 𝑠̈
𝑑𝑡 2
⇔
𝑎=
⇔
𝑠̇ 𝑑 𝑠̇ = 𝑠̈ 𝑑𝑠
Eliminating time variable:
𝑣𝑑𝑣 = 𝑎𝑑𝑠
Or:
𝑣 · 𝑑𝑣 = 𝑎 · 𝑑𝑠
𝑎=𝑣·
𝑑𝑣
= 𝑣 · 𝑣̇
𝑑𝑠
Or:
𝑣=
𝑎
𝑣̇
2/3 Plane Curvilinear Motion
Instantaneous velocity
𝒗=
𝑑𝒓
= 𝒓̇
𝑑𝑡
⇔
𝑣⃗ =
𝑑𝑟⃗
= 𝑟̇⃗
𝑑𝑡
Magnitude of v is a scalar:
𝑑𝑠
= 𝑠̇
𝑑𝑡
𝑣 = |𝑣⃗| =
𝑣 = |𝒗| =
𝑑𝑠
= 𝑠̇
𝑑𝑡
Distinction between:
Magnitude of the derivative:
Represents the magnitude of the velocity or speed of the particle
|
𝑑𝒓
| = |𝒓̇ | = |𝑠̇ | = |𝒗| = 𝑣
𝑑𝑡
⇔
|
𝑑𝑟⃗
| = |𝑟̇⃗| = |𝑠̇ | = |𝑣⃗| = 𝑣
𝑑𝑡
Derivative of the magnitude:
Represents the rate at which the length of the position vector r is changing.
𝑑|𝒓| 𝑑𝑟
=
= 𝑟̇
𝑑𝑡
𝑑𝑡
Instantaneous acceleration
𝒂=
𝑑𝒗
= 𝒗̇
𝑑𝑡
⇔
𝑎⃗ =
𝑑𝑣⃗
= 𝑣̇⃗
𝑑𝑡
Rules for differentiating vector terms are the same as scalars except for cross product where the order
of the terms must be preserved - see Art C/7 in appendix C for rules.
Choice of Coordinates:
•
•
•
Rectangular
Normal and Tangential
Polar
2/4 Rectangular Coordinates
Vector representation
𝑟⃗ = 𝑥 · 𝑖̂ + 𝑦 · 𝑗̂
𝑣⃗ = 𝑟⃗̇ = 𝑥̇ · 𝑖̂ + 𝑦̇ · 𝑗̂
𝑎⃗ = 𝑣⃗̇ = 𝑟⃗̈ = 𝑥̈ · 𝑖̂ + 𝑦̈ · 𝑗̂
𝑣𝑥 = 𝑥̇
𝑎𝑥 = 𝑣𝑥̇ = 𝑥̈
𝑣𝑦 = 𝑦̇
𝑎𝑦 = 𝑣𝑦̇ = 𝑦̈
𝑣 = √𝑣𝑥2 + 𝑣𝑦2
𝑎 = √𝑎𝑥2 + 𝑎𝑦2
𝜃 = tan−1 (
𝑣𝑦
)
𝑣𝑥
Projectile Motion
𝑎𝑥 = 0
𝑣𝑥 = 𝑣𝑥,0
𝑥 = 𝑥0 + 𝑣𝑥,0 · 𝑡
𝑎𝑦 = −𝑔
𝑣𝑦 = 𝑣𝑦,0 − 𝑔𝑡
1
𝑦 = 𝑦0 + 𝑣𝑦,0 · 𝑡 − 𝑔 · 𝑡 2
2
2 − 2𝑔(𝑦 − 𝑦 )
𝑣𝑦2 = 𝑣𝑦,0
0
2/5 Normal and Tangential Coordinates (n-t)
n and t coordinates move along the path of the particle.
The positive direction of n is always taken toward the center of the curvature
Velocity and Acceleration
𝑒⃗⃗⃗⃗⃗
𝑛 = 𝒆𝒏 = 𝑈𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑛 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑒⃗⃗⃗⃗𝑡 = 𝒆𝒕 = 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑡 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑑𝑠 = 𝜌 · 𝑑𝛽
𝑣=
𝑑𝑠
𝑑𝑏
=𝜌
𝑑𝑡
𝑑𝑡
𝒗 = 𝑣 · 𝒆𝒕 = 𝜌𝛽̇ · 𝒆𝒕
⇔
𝑣⃗ = 𝑣 · 𝑒⃗⃗⃗⃗𝑡 = 𝜌 · 𝛽̇ · 𝑒⃗⃗⃗⃗𝑡
𝒂=
𝑑𝒗 𝑑(𝑣 · 𝒆𝒕 )
=
= 𝑣 · 𝒆𝒕̇ + 𝑣̇ · 𝒆𝒕
𝑑𝑡
𝑑𝑡
⇔
𝑎⃗ =
𝑑𝒆𝒕
= 𝒆𝒏
𝑑𝛽
⇔
𝑑𝑒⃗⃗⃗⃗𝑡
= 𝑒⃗⃗⃗⃗⃗
𝑛
𝑑𝛽
𝑒⃗⃗⃗⃗𝑡̇ = 𝛽̇ · 𝑒⃗⃗⃗⃗⃗
𝑛
⇔
𝒆̇ = 𝛽̇ · 𝒆𝒏
𝑑𝑣⃗ 𝑑(𝑣 · 𝑒⃗⃗⃗⃗)
𝑡
=
= 𝑣 · 𝑒⃗⃗⃗⃗𝑡̇ + 𝑣̇ · 𝑒⃗⃗⃗⃗𝑡
𝑑𝑡
𝑑𝑡
𝑣2
· 𝑒⃗⃗⃗⃗⃗ + 𝑣̇ · 𝑒⃗⃗⃗⃗𝑡
𝜌 𝑛
⇔
𝑣2
𝒂=
· 𝒆𝒏 + 𝑣̇ · 𝒆𝒕
𝜌
𝑎⃗ =
Where
𝑎𝑛 =
𝑣2
= 𝜌 · 𝛽̇ 2 = 𝑣𝛽̇
𝜌
𝑎𝑡 = 𝑣̇ = 𝑠̈
𝑎 = √𝑎𝑛2 + 𝑎𝑡2
Circular Motion
𝑣 = 𝑟 · 𝜃̇
𝑣2
𝑎𝑛 =
= 𝑟 · 𝜃̇ 2 = 𝑣 · 𝜃̇
𝑟
𝑎𝑡 = 𝑣̇ = 𝑟𝜃̈
2/6 Polar Coordinates
𝒓 = 𝑟𝒆𝒓
𝑟⃗ = 𝑟 · ⃗⃗⃗⃗
𝑒𝑟
Time Derivatives of the Unit Vectors
𝑑𝑒⃗⃗⃗⃗𝑟
= 𝑒⃗⃗⃗⃗⃗
𝜃
𝑑𝜃
𝑑𝑒⃗⃗⃗⃗⃗
𝜃
= −𝑒⃗⃗⃗⃗𝑟
𝑑𝜃
⃗⃗⃗⃗
𝑒𝑟̇ = 𝜃̇ 𝑒⃗⃗⃗⃗⃗
𝜃
⃗⃗⃗⃗⃗
𝑒𝜃̇ = −𝜃̇⃗⃗⃗⃗
𝑒𝑟
Velocity
𝒗 = 𝑟̇ 𝒆𝒓 + 𝑟𝜃̇𝒆𝜽
⇔
𝑣⃗ = 𝑟̇ · ⃗⃗⃗⃗
𝑒𝑟 + 𝑟𝜃̇𝑒⃗⃗⃗⃗⃗
𝜃
𝑣𝜃 = 𝑟𝜃̇
𝑣𝑟 = 𝑟̇
𝑣 = √𝑣𝑟2 + 𝑣𝜃2
Acceleration
𝑎⃗ = (𝑟̈ − 𝑟𝜃̇ 2 )𝑒⃗⃗⃗⃗𝑟 + (𝑟𝜃̈ + 2𝑟̇ 𝜃̇)𝑒⃗⃗⃗⃗⃗
𝜃
𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2
𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇
𝑎 = √𝑎𝑟2 + 𝑎𝜃2
𝑎𝜃 =
1 𝑑 2
· (𝑟 𝜃̇)
𝑟 𝑑𝑡
Circular Motion
𝑣𝑟 = 0
𝑣𝜃 = 𝑟𝜃̇
𝑎𝑟 = −𝑟𝜃̇ 2
𝑎𝜃 = 𝑟𝜃̈
2/7 Space Curvilinear Motion
Coordinate systems
1. Rectangular (𝑥, 𝑦, 𝑧)
2. Cylindrical (𝑟, 𝜃, 𝑧)
3. Spherical (𝑟, 𝜃, 𝜙)
Not normal and tangential, cause awkward in 3D
Rectangular Coordinates (𝒙, 𝒚, 𝒛)
For 3D we use R instead of r.
𝑹 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌
𝒗 = 𝑹̇ = 𝑥̇ 𝒊 + 𝑦̇ 𝒋 + 𝑧̇ 𝒌
𝒂 = 𝒗̇ = 𝑹̈ = 𝑥̈ 𝒊 + 𝑦̈ 𝒋 + 𝑧̈ 𝒌
Cylindrical Coordinates (𝒓, 𝜽, 𝒛)
Cylindrical coordinates position
𝑹 = 𝑟𝒆𝒓 + 𝑧𝒌
Cylindrical coordinates velocity
𝒗 = 𝑟̇ 𝒆𝒓 + 𝑟𝜃̇𝒆𝜽 + 𝑧̇ 𝒌
where
𝑣𝑟 = 𝑟̇
𝑣𝜃 = 𝑟𝜃̇
𝑣𝑧 = 𝑧̇
𝑣 = √𝑣𝑟2 + 𝑣𝜃2 + 𝑣𝑧2
Cylindrical coordinates acceleration
⃗⃗
𝑎⃗ = (𝑟̈ − 𝑟𝜃̇ 2 )𝑒⃗⃗⃗⃗𝑟 + (𝑟𝜃̈ + 2𝑟̇ 𝜃̇)𝑒⃗⃗⃗⃗⃗
𝜃 + 𝑧̈ 𝑘
where
𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2
𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ =
1 𝑑 2
· (𝑟 𝜃̇)
𝑟 𝑑𝑡
𝑎𝑧 = 𝑧̈
𝑣 = √𝑎𝑟2 + 𝑎𝜃2 + 𝑎𝑧2
Spherical Coordinates (𝒓, 𝜽, 𝝓)
Spherical Coordinates velocity
𝒗 = 𝑣𝑅 𝒆𝑹 + 𝑣𝜃 𝒆𝜽 + 𝑣𝜙 𝒆𝝓
Where:
𝑣𝑅 = 𝑅̇
𝑣𝜃 = 𝑅 · 𝜃̇ cos(𝜙)
𝑣𝜙 = 𝑅𝜙̇
Spherical Coordinates acceleration
𝒂 = 𝑎𝑅 𝒆𝑹 + 𝑎𝜃 𝒆𝜽 + 𝑎𝜙 𝒆𝝓
Where:
𝑎𝑅 = 𝑅̈ − 𝑅𝜙̇ − 𝑅𝜃̇ 2 cos2 (𝜙)
𝑎𝜃 =
cos(𝜙) 𝑑
(𝑅2 𝜃̇) − 2𝑅𝜃̇𝜙̇ sin(𝜙)
𝑅 𝑑𝑡
𝑎𝜙 =
1 𝑑
(𝑅2 𝜙̇) + 𝑅𝜃 2̇ sin (𝜙)cos (𝜙)
𝑅 𝑑𝑡
L10 DYNAMICS
- Kinematics and kinetics of particles
2/8 Relative motion
Vector representation
Subscript 𝐴/𝐵 means A relative to B
𝒓𝑨 = 𝒓𝑩 + 𝒓𝑨/𝑩
𝒓̇ 𝑨 = 𝒓̇ 𝑩 + 𝒓̇ 𝑨/𝑩
𝒗𝑨 = 𝒗𝑩 + 𝒗𝑨/𝑩
𝒓̈ 𝑨 = 𝒓̈ 𝑩 + 𝒓̈ 𝑨/𝑩
𝒂𝑨 = 𝒂𝑩 + 𝒂𝑨/𝑩
Additional considerations
𝒓𝑩 = 𝒓𝑨 + 𝒓𝑩/𝑨
𝒗𝑩 = 𝒗𝑨 + 𝒗𝑩/𝑨
𝒓𝑩/𝑨 = −𝒓𝑨/𝑩
𝒗𝑩/𝑨 = −𝒗𝑨/𝑩
𝒂𝑩/𝑨 = −𝒂𝑨/𝑩
2/9 Constrained Motion of Connected Particles
No general equations
𝒂𝑩 = 𝒂𝑨 + 𝒂𝑩/𝑨
Section A - Force, Mass, and Acceleration ->
3/2 Newtons second law
𝐹1 𝐹2
𝐹
=
=⋯= =𝐶
𝑎1 𝑎2
𝑎
C = Inertia = resistance to rate of change of velocity
𝐶 = 𝑘𝑚,
𝑘 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐹 = 𝑘𝑚𝑎
𝑭 = 𝑘𝑚𝒂
𝐹⃗ = 𝑘𝑚𝑎⃗
k = unity in kinetic system and we write
𝑭 = 𝑚𝒂
We use (page 59 dynamics)
𝑚
Value of g relative to rotating earth 𝑔 = 9,81 𝑠2
𝑚
Value of g relative to non-rotation earth 𝑔 = 9,82 𝑠2
Weight of a body is given by
𝑊 = 𝑚 · 𝑔,
𝑔 = 9,81
𝑚
𝑠2
Equation of Motion and solution of Problems
∑𝑭 = 𝑚 · 𝒂
Two Types of Dynamics Problems
First type: The acceleration is known or can be calculated directly from known kinematic conditions.
Second type: The forces acting on the particle are specified, and we must determine the resulting
motion. The forces can be a function of time position or velocity. Then F=ma becomes differential
equation and must be integrated to determine the velocity and displacement.
Constrained or Unconstrained Motion
Two types of motion:
Unconstrained: Free of mechanical guides. The path of the particle is determined by its initial motion
and by the forces which are applied to it from external sources.
Constrained: path is partially or fully determined by restraining mechanical guides.
2 Degrees of freedom - 2D coordinate system
3 degrees of freedom - 3d coordinate system
3/4 Rectilinear Motion
∑ 𝐹𝑥 = 𝑚𝑎𝑥 ,
∑ 𝐹𝑦 = 𝑚𝑎𝑦 ,
∑ 𝐹𝑧 = 𝑚𝑎𝑧
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 + 𝑎𝑧 𝒌
𝑎 = |𝒂| = √𝑎𝑥2 + 𝑎𝑦2 + 𝑎𝑧2
𝒂 = 𝑎𝑥 𝒊 + 𝑎𝑦 𝒋 + 𝑎𝑧 𝒌
∑ 𝐹 = ∑ 𝐹𝑥 𝒊 + ∑ 𝐹𝑦 𝒋 + ∑ 𝐹𝑧 𝒌
2
2
|∑ 𝐹| = √(∑ 𝐹𝑥 𝒊) + (∑ 𝐹𝑦 𝒋) + (∑ 𝐹𝑧 𝒌)
3/5 Curvilinear Motion
Rectangular coordinates:
∑ 𝐹𝑥 = 𝑚𝑎𝑥
∑ 𝐹𝑦 = 𝑚𝑎𝑦 ,
Where
𝑎𝑥 = 𝑥̈ ,
𝑎𝑦 = 𝑦̈
Normal and tangential coordinates
∑ 𝐹𝑛 = 𝑚𝑎𝑛
∑ 𝐹𝑡 = 𝑚𝑎𝑡 ,
Where
𝑎𝑛 = 𝜌𝛽̇ 2 =
𝑣2
= 𝑣𝛽̇ ,
𝜌
𝑎𝑡 = 𝑣̇ ,
𝑣 = 𝜌𝛽̇
2
Polar coordinates
∑ 𝐹𝑟 = 𝑚𝑎𝑟
∑ 𝐹𝜃 = 𝑚𝑎𝜃 ,
𝑎𝑟 = 𝑟̈ − 𝑟𝜃̇ 2 ,
𝑎𝜃 = 𝑟𝜃̈ + 2𝑟̇ 𝜃̇
L11 DYNAMICS
- Work and energy, impulse and momentum
Section B - work and energy
3/6 Work and kintetic energy
Two general classes of problems where the cumulative effects of unbalanced forces are of interest:
1) Integration with respect to displacement of the particle
Leads to the equations of work and energy (discussed in section B (this section))
2) Integration with respect to the time they are applied
Leads to the equations of impulse and momentum (discussed in section C(next section))
Definition of Work
Work is:
[𝑵 · 𝒎] = [𝑱]
𝑑𝑈 = 𝑭 · 𝑑𝒓,
Magnitude of dot product
𝑑𝑈 = 𝐹 · 𝑑𝑠 · cos(𝛼)
𝛼 = 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐹 𝑎𝑛𝑑 𝑑𝑟,
𝑑𝑠 = 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑑𝑟
𝐹𝑡 = 𝐹 · cos (𝛼)
𝑑𝑈 = 𝐹𝑡 · 𝑑𝑠
Active forces: do work
Reactive forces: constraint forces and do no work
Calculation of Work
2
2
𝑈 = ∫ 𝑭 · 𝑑𝒓 = ∫ (𝐹𝑥 𝑑𝑥 + 𝐹𝑦 𝑑𝑦 + 𝐹𝑧 𝑑𝑧)
1
1
𝑠2
𝑈 = ∫ 𝐹𝑡 𝑑𝑠
𝑠1
Examples of Work
Work and curvilinear motion
2
𝑠2
𝑈1−2 = ∫ 𝑭 · 𝑑𝒓 = ∫ 𝐹𝑡 𝑑𝑠
1
𝑠1
2
2
𝑈1−2 = ∫ 𝑭 · 𝑑𝒓 = ∫ 𝑚𝒂 · 𝑑𝒓
1
𝒂 · 𝑑𝒓 = 𝑎𝑡 𝑑𝑠,
1
𝑎𝑡 = 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝑎𝑡 𝑑𝑠 = 𝑣 𝑑𝑣
2
𝑣2
1
𝑈1−2 = ∫ 𝑭 · 𝑑𝒓 = ∫ 𝑚𝑣 𝑑𝑣 = 𝑚(𝑣22 − 𝑣12 )
2
1
𝑣1
Principle of work and kinetic energy
𝑇=
1
𝑚𝑣 2
2
𝑈1−2 = 𝑇2 − 𝑇1 = Δ𝑇
𝑇1 + 𝑈1−2 = 𝑇2
Advantages of the work energy method
Power
𝑃 =𝑭·𝒗
Efficiency
𝑒𝑚 =
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑃𝑖𝑛𝑝𝑢𝑡
Total /overall efficiency:
𝑒 = 𝑒𝑚 𝑒𝑒 𝑒𝑡
𝑒𝑚 = 𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑒𝑓𝑓𝑒𝑐𝑖𝑒𝑛𝑐𝑦,
𝑒𝑒 = 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑒𝑓𝑓𝑒𝑐𝑖𝑒𝑛𝑐𝑦,
𝑒𝑡 = 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑒𝑓𝑓𝑒𝑐𝑖𝑒𝑛𝑐𝑦
3/7 potential energy
Gravitational Potential Energy
𝑉𝑔 = 𝑚𝑔ℎ
Δ𝑉𝑔 = 𝑚𝑔(ℎ2 − ℎ1 ) = 𝑚𝑔Δℎ
𝑉𝑔 = −
𝑚𝑔𝑅2
𝑟
Δ𝑉𝑔 = 𝑚𝑔𝑅2 (
1 1
− )
𝑟1 𝑟2
Elastic potential energy
𝑥
𝑉𝑒 = ∫ 𝑘𝑥 𝑑𝑥 =
0
1 2
𝑘𝑥
2
1
𝑉𝑒 = 𝑘(𝑥22 − 𝑥12 )
2
Work energy equation
𝑈 ′1−2 = Δ𝑇 + Δ𝑉
′
𝑇1 + 𝑉1 + 𝑈1−2
= 𝑇2 − 𝑉2
𝑇1 + 𝑉1 = 𝑇2 + 𝑉2
𝑜𝑟
𝐸1 = 𝐸2
Conservative force fields
𝑈 = ∫ 𝑭 · 𝑑𝒓 ∫(𝐹𝑥 𝑑𝑥 + 𝐹𝑦 𝑑𝑦 + 𝐹𝑧 𝑑𝑧)
𝑉2
𝑈2−1 = ∫ −𝑑𝑉 = −(𝑉2 − 𝑉1 )
𝑉1
𝑑𝑉 =
𝐹𝑥 =
𝛿𝑉
𝛿𝑉
𝛿𝑉
𝑑𝑥 +
𝑑𝑦 +
𝑑𝑧
𝛿𝑥
𝛿𝑦
𝛿𝑧
𝛿𝑉
,
𝛿𝑥
𝐹𝑦 =
𝛿𝑉
,
𝛿𝑦
𝐹𝑧 =
𝑭 = −𝛁𝑉
𝛁=𝒊
𝛿
𝛿
𝛿
+𝒋 +𝒌
𝛿𝑥
𝛿𝑦
𝛿𝑧
𝛿𝑉
𝛿𝑧
Section C - Impulse and Momentum ->
3/8 Impulse and momentum - introduction
3/9 linear impulse and linear momentum
∑ 𝑭 = ∑ 𝑚𝒗̇ =
𝑑
(𝑚𝒗)
𝑑𝑡
∑ 𝑭 = 𝑮̇
Where
𝐺̇ = 𝑚𝑣̇ = 𝑚𝑎
Linear momentum:
𝑘𝑔·𝑚
]
𝑠
[𝑁 · 𝑠] = [
𝐺 = 𝑚𝑣
∑ 𝐹𝑥 = 𝐺̇𝑥 , ∑ 𝐹𝑦 = 𝐺̇𝑦 , ∑ 𝐹𝑧 = 𝐺̇𝑧
The Linear Impulse-Momentum Principle
Linear impulse of a force is product of force and time and is equal to the change in impulse as the 2
equations below states:
The final linear momentum is equal to the initial linear momentum of particle plus its linear impulse
𝑡2
∫ ∑ 𝑭𝑑𝑡 = 𝑮𝟐 − 𝑮𝟏 = Δ𝑮
𝑡1
𝑡2
𝐺1 + ∫ ∑ 𝑭 𝑑𝑡 = 𝑮𝟐
𝑡2
In every direction:
𝑡2
𝑚(𝑣1 )𝑥 + ∫ ∑ 𝐹𝑥 𝑑𝑡 = 𝑚(𝑣2 )𝑥
𝑡1
𝑡2
𝑚(𝑣1 )𝑦 + ∫ ∑ 𝐹𝑦 𝑑𝑡 = 𝑚(𝑣2 )𝑦
𝑡1
𝑡2
𝑚(𝑣1 )𝑧 + ∫ ∑ 𝐹𝑧 𝑑𝑡 = 𝑚(𝑣2 )𝑧
𝑡1
Conservation of Linear Momentum
𝚫𝑮 = 𝟎
𝑜𝑟
𝐺1 = 𝐺2
3/10 Angular Impulse and Angular Momentum
The moment of the linear momentum vector about the origin is defined as the angular momentum 𝑯𝑶
of P about O and is given by:
𝑯𝒐 = 𝒓 × 𝑚𝒗
[𝑁 · 𝑚 · 𝑠] = [
𝑘𝑔 · 𝑚2
]
𝑠
The angular momentum 𝑯𝑶 is perpendicular to the
plane A
𝑯𝑶 ⊥ 𝐴
The first equation can be expanded to
𝑯𝑶 = 𝑚(𝑣𝑧 𝑦 − 𝑣𝑦 𝑧)𝒊 + 𝑚(𝑣𝑥 𝑧 − 𝑣𝑧 𝑥)𝒋 + 𝑚(𝑣𝑦 𝑥 − 𝑣𝑥 𝑦)𝒌
Also written as:
𝒊
𝑥
𝑯𝒐 = 𝑚 |
𝑣𝑥
𝒋
𝑦
𝑣𝑦
𝒌
𝑧|
𝑣𝑧
So that
𝐻𝑥 = 𝑚(𝑣𝑧 𝑦 − 𝑣𝑦 𝑧),
𝐻𝑦 = 𝑚(𝑣𝑥 𝑧 − 𝑣𝑧 𝑥),
𝐻𝑧 = 𝑚(𝑣𝑦 𝑥 − 𝑣𝑥 𝑦)
Rate of Change of Angular Momentum
Moment is the rotational force and is the derivative of the momentum
∑ 𝑴𝑶 = 𝑯̇𝑶
∑ 𝑴𝒐 = 𝒓 × ∑ 𝑭 = 𝒓 × 𝑚𝒗̇
𝑯̇𝑂 = 𝒓̇ × 𝑚𝒗 + 𝒓 × 𝑚𝒗̇
= 𝒗 × 𝑚𝒗 + 𝒓 × 𝑚𝒗̇
∑ 𝑴𝑂𝑥 = 𝑯̇𝑂𝑥 ,
∑ 𝑴𝑂 𝑦 = 𝑯̇𝑂𝑦 ,
∑ 𝑴𝑂 𝑧 = 𝑯̇𝑂𝑧
The Angular Impulse Momentum Principle
Total angular impulse on M about O equals the corresponding change in angular momentum of m
about O
𝑡2
∫ ∑ 𝑴𝑂 𝑑𝑡 = (𝑯𝑂 )2 − (𝑯𝑂 )1 = Δ𝑯𝑂
𝑡1
𝑡2
(𝑯𝑂 )1 + ∫ ∑ 𝑴𝑂 𝑑𝑡 = (𝑯𝑂 )2
𝑡1
Vector form example:
𝑡2
(𝐻𝑂𝑥 ) ∫ ∑ 𝑀𝑂𝑥 𝑑𝑡 = (𝐻𝑂𝑥 )
1
𝑡1
2
𝑡2
𝑚(𝑣𝑧 𝑦 − 𝑣𝑦 𝑧) ∫ ∑ 𝑀𝑂𝑥 𝑑𝑡 = 𝑚(𝑣𝑧 𝑦 − 𝑣𝑦 𝑧)
1
𝑡1
2
Plane-Motion Applications
When you only have an x-y plane, there is only one direction of the angular momentum
Difference between the scalar and vector forms of the angular impulse-momentum relations
𝐻𝑂1 = |𝒓𝟏 × 𝑚𝒗𝟏 | = 𝑚𝑣1 𝑑1
𝑎𝑛𝑑
𝑡2
(𝐻𝑂 )1 ∫ ∑ 𝑀𝑂 𝑑𝑡 = (𝐻𝑂 )2
𝑡1
𝐻𝑂2 = |𝒓𝟐 × 𝑚𝒗𝟐 | = 𝑚𝑣2 𝑑2
𝑡2
𝑜𝑟
𝑚𝑣1 𝑑1 ∫ ∑ 𝐹𝑟 sin(𝜃) 𝑑𝑡 = 𝑚𝑣2 𝑑2
𝑡1
Conservation of Angular Momentum
There is conservation when:
The resultant moment about a fixed point O of all forces acting on the particle is zero during an
interval of time. 𝑀𝑂 = 0 ⇒ 𝐻𝑂 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑒𝑑
Can be conserved about one axis, but change about another axis.
Principle of conservation of angular momentum:
𝚫𝑯𝑂 = 0
𝑜𝑟
(𝑯𝑂 )1 = (𝑯𝑂 )2
L12 DYNAMICS
- Kinetics of Particle and System of Particles
Section D Special Applications
3/11 Introduction
Treated in section D
1. Impact
2. Central-force motion
3. Relative motion
3/12 Impact
Impact: collision of two bodies. Relatively large contact forces acting during a short interval of time.
Direct Central Impact
Direct central impact: 𝑣1 > 𝑣2 collision occurs with the contact forces directed along the line of
centers. After impact 𝑣1 ′ < 𝑣2 ′
Because forces are equal and opposite, there is conservation of linear momentum.
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2 ′
Coefficient of Restitution
The equation has two unknowns:
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2 ′
Introducing coefficient of restitution:
For particle 1:
𝑡
𝑒=
∫𝑡 𝐹𝑟 𝑑𝑡
0
𝑡
∫0 0 𝐹𝑑 𝑑𝑡
=
𝑚1 (−𝑣1′ − (−𝑣0 )) 𝑣0 − 𝑣1 ′
=
𝑚1 (−𝑣0 − (−𝑣1 )) 𝑣1 − 𝑣0
For particle 2:
𝑡
𝑒=
∫𝑡 𝐹𝑟 𝑑𝑡
0
𝑡
∫0 0 𝐹𝑑 𝑑𝑡
=
𝑚2 (𝑣2′ − 𝑣0 ) 𝑣2′ − 𝑣0
=
𝑚2 (𝑣0 − 𝑣2 ) 𝑣0 − 𝑣2
Elimination 𝑣0 :
𝑒=
𝑣2′ − 𝑣0 |𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑒𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛|
=
|𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ|
𝑣 0 − 𝑣2
𝑒=
𝑣2′ − 𝑣0
𝑣0 − 𝑣2
e depends on collision speed and material.
Energy Loss During Impact
Depends on material:
Elastic impact: no energy loss and 𝑒 = 1
Inelastic impact / plastic impact: The particles cling together after impact. Energy loss is maximum
and 𝑒 = 0.
Generally, all impact conditions lie between these two extremes.
Oblique Central Impact
Oblique central impact: initial and final velocities are not parallel.
(𝑣1 )𝑛 = −𝑣1 sin(𝜃1 )
(𝑣1 )𝑡 = 𝑣1 cos(𝜃1 )
(𝑣2 )𝑛 = 𝑣2 sin(𝜃2 )
(𝑣2 )𝑡 = 𝑣2 cos(𝜃2 )
Momentum is conserved in the n- direction
𝑚1 (𝑣1 )𝑛 + 𝑚2 (𝑣2 )𝑛 = 𝑚1 (𝑣1 ′)𝑛 + 𝑚2 (𝑣2 ′)𝑛
The momentum of each particle is conserved in the t-direction
𝑚1 (𝑣1 ′)𝑡 = 𝑚1 (𝑣1 ′)𝑡
𝑚2 (𝑣2 )𝑡 = 𝑚2 (𝑣2 ′)𝑡
The coefficient of restitution
𝑒=
Finally, the angles are easily determined.
(𝑣2 ′)𝑛 − (𝑣1 ′)𝑛
(𝑣1 )𝑛 − (𝑣2 )𝑛
3/13 Central Force Application
Motion of Single Body
𝐹=𝐺
𝑚𝑚0
𝑟2
𝑚0 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑛𝑔 𝑏𝑜𝑑𝑦 − 𝑓𝑖𝑥𝑒𝑑
𝐺 = 𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑎𝑙 𝑔𝑟𝑎𝑣𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑟 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠𝑒𝑠
Most convenient coordinates: polar.
Equation may be applied directly to the 𝑟- and 𝜃-directions to give:
−𝐺
𝑚𝑚0
= 𝑚(𝑟̈ − 𝑟𝜃̇ 2 )
𝑟2
0 = 𝑚(𝑟𝜃̈ + 2𝑟̇ 𝜃̇)
Which can be rewritten to:
𝑟 2 𝜃̇ = ℎ,
ℎ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
…states that the angular momentum of 𝑚 about 𝑚0 is conserved. 𝐻𝑂 1 = 𝐻𝑂 2 if there is no moment acting
on the particle about a fixed point 𝑂.
Used for what??:
−𝐺𝑚0 𝑢2 = −ℎ2 𝑢2
𝑑𝑢2 1 2 4
− ℎ 𝑢
𝑑𝜃 2 𝑢
𝑑𝑢2
𝐺𝑚0
+𝑢= 2
𝑑𝜃 2
ℎ
Solution for the differential equation:
𝑢=
1
𝐺𝑚0
= 𝐶 cos(𝜃 + 𝛿) + 2
𝑟
ℎ
𝐶 = 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
𝛿 = 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
The phase angle 𝛿 may be eliminated by choosing the x-axis so that r is minimum when 𝜃 = 0, thus:
1
𝐺𝑚0
= 𝐶 cos(𝜃) + 2
𝑟
ℎ
Conic Sections
𝑒 is constant
𝑒=
𝑟
,
𝑑 − 𝑟 cos(𝜃)
𝑒 = 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑎 𝑝𝑜𝑖𝑛𝑡 ′ 𝑠 (𝑓𝑜𝑐𝑢𝑠 ′ ) 𝑑𝑖𝑠𝑡𝑎𝑛𝑑𝑒 𝑡𝑜 𝑎 𝑙𝑖𝑛𝑒 (𝑑𝑖𝑟𝑒𝑐𝑡𝑟𝑖𝑥)
Above equation may be rewritten as
1 1
1
= cos(𝜃) +
𝑟 𝑑
𝑒𝑑
Can be rewritten to:
𝑒=
ℎ2 𝐶
𝐺𝑚0
Ellipse: (𝒆 < 𝟏)
2𝑎 = 𝑟𝑚𝑖𝑛 + 𝑟𝑚𝑎𝑥 =
𝑒𝑑
𝑒𝑑
+
1+𝑒 1−𝑒
𝑜𝑟
𝑎=
𝑒𝑑
1 − 𝑒2
1 1 + 𝑒 cos(𝜃)
=
𝑟
𝑎(1 − 𝑒 2 )
𝑟𝑚𝑖𝑛 = 𝑎(1 − 𝑒),
𝑟𝑚𝑎𝑘𝑠 = 𝑎(1 + 𝑒)
𝑏 = 𝑎 √1 − 𝑒 2
𝜏=
𝐴
𝜋𝑎𝑏
=
1
̇
𝐴
2 ̇
2𝑟 𝜃
𝜏 = 2𝜋
𝑜𝑟
𝜏=
2𝜋𝑎𝑏
ℎ
𝑎 3/2
𝑅 √𝑔
Parabola: (𝒆 = 𝟏)
1 1
= (1 + cos (𝜃)
𝑟 𝑑
𝑎𝑛𝑑
ℎ2 𝐶 = 𝐺𝑚0
Hyperbola: (𝒆 > 𝟏)
1
1
1
= (cos (𝜃) − 𝜋) +
−𝑟 𝑑
𝑒𝑑
𝑎𝑛𝑑
1
1 cos(𝜃)
=−
+
𝑟
𝑒𝑑
𝑑
Energy Analysis
𝐸=
1
𝑚𝑔𝑅2
𝑚(𝑟̇ 2 + 𝑟 2 𝜃̇ 2 ) −
2
𝑟
1 1
𝑣 2 = 2𝑔𝑅2 ( − )
𝑟 2𝑎
𝑔 1+𝑒
𝑔 𝑟𝑚𝑎𝑥
𝑣𝑝 = 𝑅√ √
= 𝑅√ √
𝑎 1−𝑒
𝑎 𝑟𝑚𝑖𝑛
𝑔 1−𝑒
𝑔 𝑟𝑚𝑖𝑛
𝑣𝑎 = 𝑅√ √
= 𝑅√ √
𝑎 1+𝑒
𝑎 𝑟𝑚𝑎𝑥
Summery of Assumptions
Pertubed Two-body Problem
𝐺
𝑚𝑚0
𝒓 = 𝑚0 𝒓̈ 1
𝑟3
−𝐺
𝑎𝑛𝑑
−𝐺
𝑚𝑚0
𝒓 + 𝑷 = 𝑚𝒓̈ 2
𝑟3
(𝑚0 + 𝑚)
𝑷
𝒓 + = 𝒓̈ 2 − 𝒓̈ 1 = 𝒓̈
3
𝑟
𝑚
𝒓̈ + 𝐺
(𝑚0 + 𝑚)
𝑷
𝒓=
3
𝑟
𝑚
Restricted Two-body Problem
𝒓̈ + 𝐺
𝑚0
𝒓=0
𝑟3
(𝑟̈ − 𝑟𝜃̇ 2 )𝐞𝐫 + (𝑟𝜃̈ + 𝑟̇ 𝜃̇)𝐞𝛉 + 𝐺
𝜏 = 2𝜋
3/14 Relative Motion
𝑚0
(𝑟𝐞𝐫 ) = 𝟎
𝑟3
𝑎 3/2
√𝐺(𝑚0 + 𝑚)
Relative-Motion Equation
𝒂𝐴 = 𝒂𝐵 + 𝒂𝑟𝑒𝑙
∑ 𝑭 = 𝑚𝒂𝐴
∑ 𝑭 = 𝑚(𝒂𝐴 + 𝒂𝑟𝑒𝑙 )
D’Alembert’s Principle
∑ 𝑭 = 𝑚𝒂𝑟𝑒𝑙
Constant Velocity, Nonrotating Systems
1
2
𝑑𝑈𝑟𝑒𝑙 = 𝑚𝒂𝑟𝑒𝑙 · 𝑑𝒓𝑟𝑒𝑙 = 𝑚𝑣𝑟𝑒𝑙 𝑑𝑣𝑟𝑒𝑙 = 𝑑 ( 𝑚𝑣𝑟𝑒𝑙
)
2
𝑑𝑈𝑟𝑒𝑙 = 𝑑𝑇𝑟𝑒𝑙
𝑜𝑟
𝑈𝑟𝑒𝑙 = Δ𝑇𝑟𝑒𝑙
∑ 𝑭𝑑𝑡 = 𝑑(𝑚𝒗𝑟𝑒𝑙 )
∑ 𝑮̇𝑟𝑒𝑙
𝑎𝑛𝑑
∫ ∑𝑭𝑑𝑡 = Δ𝑮𝑟𝑒𝑙
∑𝑴𝐵 = (𝑯̇𝑩 )𝑟𝑒𝑙
4/1 Introduction
4/2 Generalized Newton’s Second Law
Definition of center of mass:
𝑚𝒓̅ = ∑ 𝑚𝑖 𝒓𝑖
Where 𝑚 = ∑ 𝑚𝑖 and is the total system mass.
𝑭𝟏 + 𝑭𝟐 + 𝑭𝟑 + ⋯ + 𝒇𝟏 + 𝒇𝟐 + 𝒇𝟑 + ⋯ = 𝑚𝑖 𝒓̈ 𝑖
∑ 𝑭 + ∑ 𝒇 = ∑ 𝑚𝑖 𝒓̈ 𝒊
The equation of motion of 𝑚:
∑ 𝑭 = 𝑚𝒓̅̈
̅𝑥 ,
𝑭𝑥 = 𝑚𝒂
𝑜𝑟
̅𝑦 ,
𝑭𝑦 = 𝑚𝒂
̅
𝑭 = 𝑚𝒂
̅𝑧
𝑭𝑧 = 𝑚𝒂
4/3 Work-Energy
The kinetic energy of 𝑚𝑖 is:
𝑇𝑖 =
1
𝑚 𝑣2
2 𝑖 𝐼
Work-Energy Relation
Sum of work energy equations: (work done = change in kinetic energy)
𝑈1−2 = Δ𝑇
or
𝑇1 + 𝑈1−2 = 𝑇2
Where:
𝑈1−2 = ∑(𝑈1−2 ) is the work done by all forces (external and internal) on all particles
Δ𝑇 is the change in total kinetic energy 𝑇 = ∑ 𝑇𝑖 of the system.
Accounting for the change in potential energy of the system and thus work done = change in
mechanical energy:
𝑈′1−2 = Δ𝐸
Writing out to kinetic + potential energy:
𝑈′1−2 = Δ𝑇 + Δ𝑉
or
𝑇1 + 𝑉1 + 𝑈′1−2 = 𝑇2 + 𝑉2
Kinetic Energy Expression for mass system
The velocity of the representative particle is
̅ + ρ̇ 𝑖
𝒗𝒊 = 𝒗
Where:
̅ is the velocity of the mass center G
𝒗
ρ̇ 𝑖 is the velocity of 𝑚𝑖 with respect to a translating reference frame moving with the mass center G.
The total kinetic energy of a system is:
1
1
𝑇 = 𝑚𝑣̅ 2 + ∑ 𝑚𝑖 |ρ̇ 𝑖 |2
2
2
4/4 Impulse-Momentum
Linear Momentum
The linear momentum of a mass system is:
̅
𝑮 = 𝑚𝒗
Where:
G is the linear momentum
̅ is the velocity of the center of mass
𝒗
̅ = 𝑚𝒗
̅̇
∑ 𝑭 = 𝑮̇ = 𝑚𝒂
Angular Momentum
Angular momentum of a mass system:
∑ 𝑴𝑂 = 𝑯̇𝑂
∑ 𝑴𝐺 = 𝑯̇𝐺
̅ × 𝑚𝒗
̅
𝑯𝑷 = 𝑯̇𝑮 + 𝝆
̅ × 𝑚𝒂
̅
∑ 𝑴𝑃 = 𝑯̇𝑮 + 𝝆
̅ × 𝑚𝒂
̅𝑃
∑ 𝑴𝑃 = (𝑯̇𝑃 )𝑟𝑒𝑙 + 𝝆
4/5 Conservation of Energy and Momentum
Conservation of Energy
Δ𝑇 + Δ𝑉 = 0
𝑇1 + 𝑉1 = 𝑇2 + 𝑉2
Conservation of Momentum
𝐺1 = 𝐺2
(𝐻𝑂 )1 = (𝐻𝑂 )2
𝑜𝑟
L13 DYNAMICS
- Mass flow and variable mass
4/6 Steady Mass Flow
Analysis of Flow Through a Rigid Container
Incremental Analysis
∑ 𝑭 = 𝑚′ Δ𝒗
(𝐻𝐺 )1 = (𝐻𝐺 )2
Angular Momentum in Steady-Flow Systems
∑ 𝑴𝑂 = 𝑚′ (𝐝𝟐 × 𝐯𝟐 − 𝐝𝟏 × 𝐯𝟏 )
4/7 Variable Mass
Equation of Motion
∑ 𝑭 = 𝑚𝑣̇ + 𝑚̇𝑢
Alternative Approach
Application of Rocket propulsion
4/8 Chapter review
L14 MECHANICAL VIBRATIONS
- Fundamentals of Vibration
8/1 Introduction
L15 MECHANICAL VIBRATIONS
- Free Vibration of Particles
8/2 Free Vibration of Particles
Equation of Underdamped Free Vibration
−𝑘𝑥 = 𝑚𝑥̈
𝑜𝑟
𝑚𝑥̈ + 𝑘𝑥 = 0
𝑥̈ + 𝜔𝑛2 𝑥 = 0
𝜔𝑛 = √
𝑘
𝑚
Solution for Underdamped Free Vibration
𝑥 = 𝐴 cos(𝜔𝑛 𝑡) + 𝐵 sin(𝜔𝑛 𝑡)
𝑥 = 𝐶 sin(𝜔𝑛 𝑡 + 𝜓)
𝑥0 = 𝐴
𝑎𝑛𝑑
𝑥0̇ = 𝐵𝜔𝑛
𝑥 = 𝑥0 cos(𝜔𝑛 𝑡) +
𝑥0 = 𝐶 sin(𝜓)
𝐶 = √𝑥𝑜2 + (
𝑥 = √𝑥02 + (
𝑥0̇
sin(𝜔𝑛 𝑡)
𝜔𝑛
𝑎𝑛𝑑
𝑥0̇ = 𝐶𝜔𝑛 cos(𝜓)
𝑥0̇ 2
)
𝜔𝑛
𝜓 = tan−1 (
𝑥0 𝜔𝑛
)
𝑥0̇
𝑥0̇ 2
𝑥0 𝜔𝑛
) · sin (𝜔𝑛 𝑡 + tan−1 (
))
𝜔𝑛
𝑥0̇
𝐶 = √𝐴2 + 𝐵2
𝐴
𝜓 = tan−1 ( )
𝐵
Graphical Representation of Motion
No equations
Equilibrium Position as Reference
−𝑘(𝛿𝑠𝑡 + 𝑥) + 𝑚𝑔 = 𝑚𝑥̈
−𝑘𝛿𝑠𝑡 + 𝑚𝑔 = 0
𝑚𝑥̈ + 𝑘𝑥 = 0
Equation of Motion for Damped Free Vibration
−𝑘𝑥 − 𝑐𝑥̇ = 𝑚𝑥̈
𝑜𝑟
𝜔𝑛 = √
𝜁=
𝑚𝑥̈ + 𝑐𝑥̇ + 𝑘𝑥 = 0
𝑘
𝑚
𝑐
2𝑚𝜔𝑛
𝑥̈ + 2𝜁𝜔𝑛 𝑥̇ + 𝜔𝑛2 𝑥 = 0
Solution for Damped Free Vibration
𝑥 = 𝐴𝑒 𝜆𝑡
𝜆2 + 2𝜁𝜔𝑛 𝜆 + 𝜔𝑛2 = 0
𝜆1,2 = 𝜔𝑛 (−𝜁 ± √𝜁 2 − 1)
𝜆1 = 𝜔𝑛 (−𝜁 + √𝜁 2 − 1)
𝜆2 = 𝜔𝑛 (−𝜁 − √𝜁 2 − 1)
General solution
𝜆 𝑡
𝑥 = 𝐴1 𝑒 𝜆1𝑡 + 𝐴22
𝑥 = 𝐴1 𝑒 (−𝜁+√𝜁
2 −1)𝜔
𝑛𝑡
+ 𝐴2 𝑒 (−𝜁−√𝜁
2−1)𝜔
𝑛𝑡
Categories of Damped Motion
1. 𝜁 > 1: Overdamped
2. 𝜁 = 1: Critically damped
3. 𝜁 < 1: Underdamped
Determination of Damping by Experiment
L16 MECHANICAL VIBRATIONS
- Forced Vibration of Particles, Vibration of Rigid Bodies
8/3 Forced Vibration of Particles
Harmonic Excitation
Base Excitation
Underdamped Forced Vibration
Damped Forced Vibration
Applications
Electric Circuit Analogy
8/4 Vibration of Rigid Bodies
Rotational Vibration of a Bar
Rotational Counterparts of Transitional Vibration
L17 DYNAMICS & MECHANICAL VIBRATIONS
- An overview
Overview