© Katja Hölttä-Otto 2009EGR 102 03/24/2009 SYLLABUS REVIEW.

© Katja Hölttä-Otto 2009

EGR 102 03/24/2009

SYLLABUS REVIEW


EGR 102 03/24/2009

Syllabus - objectives

During this class You will• Learn about engineering profession and

ethics governing engineers at work• Learn basic concepts of engineering• Learn open-ended problem solving skills

through work on projects in engineering• Learn to work in teams• Learn to use a computational tool: Excel• Learn good engineering communication

skills: written and oral

Debates

Lectures

Homework

In-class tasks

Teardown project

Design project


EGR 102 03/24/2009

Syllabus - policies• Course policies:

– Attendance is mandatory– You may miss 2 classes without penalty– Late work will not be accepted

• Plagiarism: – Plagiarism is a serious offense – Follow ethics guidelines described in student handbook:

http://www.umassd.edu/studenthandbook/academicregs/ethicalstandards.cfm

• Students with Disabilities– In accordance with University policy, if you have a documented

disability and require accommodations to obtain equal access in this course, please meet with the instructor at the beginning of the semester and provide the appropriate paperwork from the Disabled Students Services Office. The necessary paperwork is obtained when you bring proper documentation to the Disabled Students Services Office (DSS), which is located in Group I, Room 016, phone: 508-999-8711.


EGR 102 03/24/2009

Syllabus – schedule (cont’d)Date Topic

Tue 3/23 Engineering Design PracticeThu 3/25 Mass and Force, Ethics Debate 4Tue 3/30 ForceThu 4/1 ForceTue 4/6 Force, Energy & Power, Thu 4/8 Energy & Power, Ethics Debate 5Tue 4/13 Energy & Power Thu 4/15 Energy & Power, Ethics Debate 6Tue 4/20 Review (Area Moment of Inertia)Thu 4/22 Product Teardown, Bring your product!Tue 4/27 Final Design Project-preparation and analysisThu 4/29 Area Moment of InertiaThu 5/4 Area Moment of InertiaTue 5/6 Senior Design PresentationsTue 5/11 Final Design Project-CompetitionTue 5/18 Final Exam (11:30 AM - 2:30 PM)


EGR 102 03/24/2009

Mass and Force


EGR 102 03/24/2009

Agenda• Mass

– Mass– Density– Mass Flow– Mass Moment of Inertia– Momentum

• Force– Force basics and applications– Spring forces– Friction forces– Newton’s laws– Pressure


EGR 102 03/24/2009

Mass

• SI unit kg• Weight is a force, unit N• Mass is involved in multiple

engineering principles!

What is the mass of your calculator? How about its weight on Earth? And on Mars?


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Density• Density=mass/volume

• Important in e.g. material selection• Note: Density can change due to e.g. temperature and pressure

V

m

3m

kg

V

m

Which one of these three materials would you choose for a part (size 0.001m3) of an airplane interior decor? How about as a counter weight for an elevator?

Material Density (kg/m3)

Wood (oak) 750

Cement 1920

Aluminum 2740


EGR 102 03/24/2009

Lava Lamp Example

How does the lava lamp work?

www.google-store.com


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Mass Flow

• Mass flow = mass/time

• Remember Volume Flow?

t

mm

s

kg

t

mm

t

VV

Vt

V

t

mm

V Vmm


EGR 102 03/24/2009

Volume or mass flow?

• Depends on application• Volume flow preferred

– when filling a tank of a specific volume with liquids of different densities

– when a process can accept only a limited volume at a time

• Mass flow preferred– in chemical reactions, where the number of

reactant molecules (mass) is important– when measuring gas flow– When goods sold based on weight


EGR 102 03/24/2009

Mass Flow - Task

Estimate the mass flow of a gas pump. Density of regular gasoline is 720 kg/m3.


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Mass Moment of Inertia

• Measure of how hard it is to rotate something with respect to the center of rotation, or resistance to rate of change of rotation

• For a single mass particle:

• For a system of mass particles:

mrI zz2

i

iizz mrI 2

z-axes

r

m2

r2

m

dmrI zz2


EGR 102 03/24/2009

Mass moment of inertia - example

Which one of the following object is harder to rotate around the z-axes? Both are made of steel (=7860 kg/m3).

R=5 cm

Ø=20 cm

h 1=

30 c

m

h 1=

4 cm

z-axis


EGR 102 03/24/2009

How are these related to mass moment of inertia?

flywheel


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Momentum

• Momentum p (or L)

• Momentum is directionalVelocity (a component of momentum) is

directional

vmp

vmL


EGR 102 03/24/2009

Momentum - Example

www.aerospaceweb.org/question/investigations/columbia/foam-impact.jpg

Investigators into the Columbia accident have estimated that the dislodged foam was about 48 x 29 x 14 cm (19 x 11.5 x 5.5 in) , weighed about 0.75 kg (1.7 lb) and impacted the Shuttle at nearly 850 km/h (530 mph). For the sake of a rough comparison, this block of foam would be about the same size and weight as a large loaf of bread. (www.aerospaceweb.org/question/investigations/q0131.shtml)

p (or L) = mv = 0.75kg * 850,000m/3600s = 177 kg m/s

Same momentum as a 5 kg (11 lb) brick hitting you driving 127.5km/h (80mph) !


EGR 102 03/24/2009

Momentum - Task

Which has greater momentum?A)An Olympic 100m runner at

speed 10 m/sB)A 1000kg car pulling out of a

parking lot at 2 km/h


EGR 102 03/24/2009

Mass and Weight

• Mass - scalar (SI unit kg)• Weight – vector, it’s force (SI unit N)


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Basics

• Force is the interaction of two objects, typically one pushes or pulls the other– Direct contact: you pulling a door open– No direct contact: gravity pulling you toward the center

of the earth

• Force causes objects to move, lengthen, shorten, twist, bend, etc.

• SI Unit: Newton [F]=N– F=ma N=kg·m/s2

• U.S. Customary unit: pound force lbf

– F=ma lbf = 1slug·ft/s2 (1lbf=4.448 N)


EGR 102 03/24/2009

Applications


EGR 102 03/24/2009

Force is a vector quantity(on

whiteboard)

F1 F2


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Spring Forces

• Springs widely used in engineering– Store energy– Return to original position– Dampen vibration

• Spring types:– Linear, torsional

www.motorsportscenter.com/uploads/suspension.jpg

www.pleasanthillgrain.com/bag_clip_bag_clips_stainless.asp

www.pharma-pen.com

http://rclsgi.eng.ohio-state.edu/~gnwashin/me481/mech_sys.html


EGR 102 03/24/2009

Hooke’s law

• Applicable in the elastic range of the spring– Elastic means there is no permanent deformation after the force is removed

kxF F = applied force (N)k = spring constant (N/mm, N/m)x = deformation of the spring (mm, m)

x

F


EGR 102 03/24/2009

Whiteboard example

• A compression spring is 10 cm long when no force is applied. When a force is applied, the deformed length is 8 cm. The spring has a spring constant of 10 N/m. Calculated the applied force.


EGR 102 03/24/2009

Determining the spring constant

• In-class task: determine the spring constant of one of the scales in the back of the room

• Plot your data in Excel, explain all the steps you take

• Prepare to present in front of the class• See EF example 10.1 for help



EGR 102 03/24/2009

Homework and Teardown• Homework – due Thursday 04/01 before the class

– From the course book: 9.5 (10p), 9.6 (less than 1 page typed)(20p), 9.12 (10p), and 9.23 (10p)

• Look for (broken) products to take apart later – good product will replace one bad assignment grade– Bring products early!


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Friction Forces

• “Frictionless” systems, commonly used in physics, do not really exist

• Friction can be useful• Types of friction:

– Dry friction• Static friction• Dynamic (kinetic) friction

– Viscous friction (fluid friction)

www.garageboy.com

www.respo.net/respo_school/respo_school_006/pics/pour_oil_01.jpg

NFfriction


EGR 102 03/24/2009

Applied force and friction

Fric

tion f

orc

e (

N)

Applied force (N)

Maximum static friction force

Dynamic friction force


EGR 102 03/24/2009

Whiteboard Example

• The static coefficient of friction between an object and a horizontal surface is 0.85. The object’s mass is 0.550 kg. If the object is pushed on the surface (force horizontal) of 5N, will the object move?


EGR 102 03/24/2009

Friction In-Class Task

How would you calculate the static coefficient of friction of your calculator starting to slide on your course book?


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Free body diagrams

• Free body diagram shows all external forces acting on the body.

• Commonly used in statics, dynamics, and mechanics of materials

• Steps to draw the free body diagram– Make a simplified drawing of the body in question– Draw all force vectors acting on it

• Do not forget weight, unless gravitational forces are ignored

– Label all forces– Define fore coordinate system


EGR 102 03/24/2009

Practice (white board)

• Steps to draw the free body diagram– Make a simplified drawing of the body in question– Draw all force vectors acting on it

• Do not forget weight, unless gravitational forces are ignored

– Label all forces– Define fore coordinate system


EGR 102 03/24/2009

In-class task

• Draw a free body diagram for the two pipes in a v-shaped channel.

40 20


EGR 102 03/24/2009

Newton’s Laws

I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it

II.

III. For every action there is an equal and opposite reaction

amF


EGR 102 03/24/2009

Newton’s Laws

I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it

11m

1.5m

v0vy0

100m

vx0

This also applies to an object in rest – they will not move unless acted upon by an unbalanced force

Remember what happened to the cannon ball in both x and y-directions?


EGR 102 03/24/2009

Newton’s LawsI.

II. ,amF

m

a

F

Notice the relation between the magnitudes and directions of F and a!

dt

vdmF


EGR 102 03/24/2009

Newton’s LawsI.

II. ,amF

m

a

1F

dt

vdmF

2F

amFFF 21


EGR 102 03/24/2009

Newton’s LawsI. S

II. s

III. For every action there is an equal and opposite reaction

m

N

mg

gmN

gmN

Why are the absolute value signs on N and g, not mg or m?

Both the magnitude and direction of the two forces are equal


EGR 102 03/24/2009

Force Basics - revisited

• Force is the interaction of two objects, typically one pushes or pulls the other– Direct contact: you pulling a door (from the

handle) open the door pulling the handle so it does not come off

– No direct contact: gravity pulling you toward the center of the earth the surface of the earth (pavement?) pushing you so you do not sink in the earth


EGR 102 03/24/2009

Team Assignment – Due Thu 10/17 12 noon

• Problem/Design project:– Design a mass-spring system that can be taken to Mars

to measure the acceleration due to gravity at the surface of Mars.

• Explain the basis of your design – The governing equations & law’s of physics and how they relate to your design– Decisions on materials, components, attachment methods or working principle– Decisions that relate to the ability to take it to Mars

• Include a drawing of your design– Include rough dimensions– Include a parts list. The level of detail can be “spring, glue, screw, metal plate”, so no

need to find the actual part numbers and exact materials for the components.

• Explain how your design should be calibrated and used• No need to build the system.• Hand in a report including equations and figures. The length can be

anything from 2-4 pages typed. The length will not be graded. Only content is graded. Max 50p.


EGR 102 03/24/2009

Assignment

• Individual assignment (=homework)• Due Tue 10/22 12 noon• Problems:

– 10.11, 10.15, 10.19, 10.21, 10.22 (10p each)

• Follow format in course book EF chapter 4


EGR 102 03/24/2009

Reminders

• Look for (broken) products to take apart later – good product will replace one bad assignment

grade


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Newton’s Law - Application

• Equilibrium of forces and moments

System:

If F is 50N, what is FA+ FB, why?

i

ixF 0, i

iyF 0, i

iOM 0,

FBX

FFree Body Diagram:

FBYFA

F


EGR 102 03/24/2009

Whiteboard example

System: jiF 43

i

j


EGR 102 03/24/2009

Whiteboard example 2You are pushing a lawn mower with force of 703N. The lawn mower weighs 30 kg. The mower is not moving in the y-direction. What is the acceleration of the mower on the grass? The friction coefficient between the wheels and the grass is 0.60. At what angle should you push to maximize the acceleration? What is a reasonable angle if you were to redesign the mower?

45°

F

mg

N

F

y

x


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Pressure

• Pressure: Force acting over an area

• SI unit Pascal:

• US customary unit:

• Pressure analogous to stress

A

Fp

Pa

m

N

A

Fp

2

A1=r12 A2=r2

2A

F F=mg

F=mg

psi

in

lb

A

Fp f

2


EGR 102 03/24/2009

Pressure in Engineering

• Pressure and stress

• Hydrostatics

• Hydrodynamics

• Hydraulics/Pneumatics

• Aerodynamics

AP

www.howstuffworks.com


EGR 102 03/24/2009

Pressure - Fluids

• Pascal’s law: for fluid at rest, pressure at a point is the same in all directions.

• For a fluid at rest, pressure increases with the depth of fluid:

h

ghp

Pam

N

ms

mkgm

s

m

m

kghgp

22223

1


EGR 102 03/24/2009

Pressure - Air

• Analogous to fluids• Atmospheric pressure 101.325 kPa (14.696

psi)– Based on the weight of air in the atmosphere above the

surface of the earth divided by the area at the base of the column

hhh1

h2


EGR 102 03/24/2009

Measuring Pressure

• Absolute pressure vs gauge pressure

atmospheregaugeabsolute ppp


EGR 102 03/24/2009

Pressure Application

• Hydraulic systems

F1 F2

1

11 A

Fp

2

22 A

Fp

11

22

2

2

1

121 F

A

AF

A

F

A

Fpp

www.immersivetechnologies.com/Images/machine/cat5130.jpg


EGR 102 03/24/2009

Hydraulic system – whiteboard example

F1

F2

r1 = 1 cmr2 = 3 cmF1 = 10 NF2 = ?


EGR 102 03/24/2009

Example (whiteboard)

F1

A1

A2 = 20.0 A1

F1 = ?k=3000.0 N/mx=5.00 cm

Rigid beam

How much force (F1) do you need to apply to compress the spring by 5.00cm?How much force (F1) do you need to apply to compress the spring by 5.00cm?


EGR 102 03/24/2009

Agenda• Mass




EGR 102 03/24/2009

Hooke’s law - revisited

• Applicable in the elastic range of the spring– Elastic means there is no permanent deformation after the force is removed


x

F


EGR 102 03/24/2009

Hooke’s law cont’d

• Applicable in the elastic range of the material– Elastic means there is no permanent deformation after the force is removed

kxF F = applied forcek = a constant x = deformation

x

F

Elastic stress is not typically visible


EGR 102 03/24/2009

Stress-Strain curve

Str

ess

(Pa)

Strain (mm/mm)

Upper yield stressLower yield stress

Ultimate stress

Elastic stress

Fracture stress

L

LE

Remember: Hooke’s law valid in “elastic range”

Hooke’s law restated:

E

Modulus of Elasticity E


EGR 102 03/24/2009

Modulus of Elasticity & Shear Modulus

• Material property• Modulus of elasticity, E,

for linear pulling (or pushing) of material

• Shear modulus for shearing or twisting material

F F

F

F

FF


EGR 102 03/24/2009

Tensile Strength & Compressive Strength

• Tensile strength is the stress when a material is pulled apart– Yield strength in the elastic region

(commonly used in engineering)– Ultimate stress in the maximum

stress material can handle before failure

• Compressive strength is the stress when a material is compressed

F F

F F

F


EGR 102 03/24/2009

Modulus of elasticity of materials


EGR 102 03/24/2009

Stress-Strain curve Brittle and Ductile material

Str

ess

(Pa)

Strain (mm/mm)

Str

ess

(Pa)


EGR 102 03/24/2009

Strength of materials

• When designing a piece, it is typically designed to withstand loads to its yield stress + a factor of safety


EGR 102 03/24/2009

Example (whiteboard)

• A rod is pulled in a tensile testing machine with force 50.0 kN. The rod has a diameter of 4.0 cm. The stain is 0.00001 What material could the rod be?

L

LE

Hooke’s law restated:

E

Modulus of Elasticity E


EGR 102 03/24/2009

Replacing an engine - problem

You are replacing an engine in your car using a rope as shown in figure. How much force do you need to lift the engine? What is the tension force on the supporting cable? Choose a rope diameter to accommodate the largest stress. Use a reasonable factor of safety. The tensile strength of the rope is 42 MPa.The engine weighs 321 kg.

Engine

10°

80°

ring, not a pulley

Supporting cable

To solve, draw a free body diagram, make the sum of forces (as vectors or as components) add to 0 (equilibrium), and solve for unknown forces. From forces, calculate the needed diameter for the worst case. Mathematical answer for diameter is 1.0 cm, what is your engineering answer?

To solve, draw a free body diagram, make the sum of forces (as vectors or as components) add to 0 (equilibrium), and solve for unknown forces. From forces, calculate the needed diameter for the worst case. Mathematical answer for diameter is 1.0 cm, what is your engineering answer?


EGR 102 03/24/2009

Reminders

• Look for (broken) products to take apart later – good product will replace one bad assignment

grade


EGR 102 03/24/2009

Agenda

• Force basics and applications• Spring forces• Friction forces• Free body diagrams• Newton’s laws• Pressure• Stress and strain• Moments• Work


EGR 102 03/24/2009

Moment

• Moment is force acting at a distance

FdM NmdFM

F

dF

r


EGR 102 03/24/2009

Moment & Torque in Engineering

http://eml.ou.edu/Photo/Struct/Traffic%20light2.jpg www.craneoperator.com

www.mech.uwa.edu.au/DANotes/gears/intro/gearbox.jpeg

http://content.answers.com/main/content/wp/en/thumb/b/b4/500px-The_Little_Belt_Bridge_(1935).jpeg


EGR 102 03/24/2009

Calculating moments

1. Draw a free body diagram2. Write the force and moment

equilibrium equations3. Solve for unknowns


EGR 102 03/24/2009

Moment – Example 1

FSystem:

L

L/2

FFree Body Diagram:

FByFA

FBx




EGR 102 03/24/2009


FSystem:

L

L/2

FFree Body Diagram:

FAy

M

FAx




EGR 102 03/24/2009


• EF problem 10.25




EGR 102 03/24/2009

Moment – in class task

m1g

Calculate the forces and moments in terms on given mi , L, and g. L

L

FAyM FAx

L/2

L/4

L/8

m2g




EGR 102 03/24/2009

Agenda

• Force basics and applications• Spring forces• Friction forces• Free body diagrams• Newton’s laws• Pressure• Stress and strain• Moments• Work


EGR 102 03/24/2009

Work

• Work is the force to the direction of the movement need to move an item

FdW JNmdFW

F

d

F

Fwork Fwork

dFW


EGR 102 03/24/2009

Work in Engineering

www.otis.com

mowabb.com/aimages/images/2005/07-15-05.jpg

www.mhinfo.com/mhi_new/images/conveyor.jpg


EGR 102 03/24/2009

Work - Example

• How much work is required to push a lawn mower for 100m? How much force do you need to push the mower? F=200N. Pushing angle is 60°.

Fwork

60°

F

Fy

mg

N

F


EGR 102 03/24/2009

Work - Example

• How much work is required to push a lawn mower for 100m?

=200i+231jF

mg

N


EGR 102 03/24/2009

Work – In class task

• How much work is required to push w/ force F the crate 1m down the ramp at constant speed. Friction force is 800N. The crate weighs 100 kg.

F

mg

NF

40


EGR 102 03/24/2009

Assignment

• Individual assignment (=homework)• Due Thu 4/16 8:00am• Problems:

– 10.23, 10.24, 10.27, + the pulley problem on the next slide (10p for each problem in EF + 20p for the pulley problem)

• Follow format in course book EF chapter 4


EGR 102 03/24/2009

• A pulley is hanging on 2 cables on the ceiling. You are pulling a known mass on a rope as shown in the figure. What percentage of the effort (force) you use goes to work done to lift the mass? Which rope/cable is under most stress?

40 30

m

25

Cable A Cable B

Rope

You

are

pul

ling

here


EGR 102 03/24/2009

40 30

25

TA

G

Fpul

l

TBTAy

TAx

TBy

TBx

40 30

Fpull_x

Fpull_y

© Katja Hölttä-Otto 2009EGR 102 03/24/2009 SYLLABUS REVIEW.

Documents

Transcript of © Katja Hölttä-Otto 2009EGR 102 03/24/2009 SYLLABUS REVIEW.