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Exploring Engineering Chapter 2 Key elements in Engineering Analysis
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Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Jan 05, 2016

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Sherman Dixon
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Page 1: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Exploring Engineering

Chapter 2Key elements in

Engineering Analysis

Page 2: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

What we are going to learnMaybe the most important single lecture

in this course (which you should have already read ahead). Engineering is about units as well as

numbers. How to deal with units and dimensions Newton’s 2nd law of motion

SI and Engineering English units “gc” and “g”

Significant figures

Page 3: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Units and dimensions

All engineers will have to understand this material irrespective of sub discipline Let’s start with my favorite superhero…it’s

Superman! Why is this relevant?

Page 4: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

• Superman represents the many of the qualities that engineers must master: Consider his qualities…

1) Faster than a speeding bullet

2) More powerful than a mighty locomotive

3) Can leap tall buildings with a single bound

4) Keeps falling into kryptonite traps

Superman – Engineering Hero

Page 5: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

• Superman embodies many engineering concepts!

Faster … speeding bullet velocity/speed*

More powerful … mighty locomotive power

Can leap …tall buildings force and energy

…kryptonite traps Information (or lack of it!)

* Yes! There is a difference. We will see later!

Superman – Engineering Hero

Page 6: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Superman – Engineering Hero

Suppose I asked you what is ? Pretty good answer is 3.14, or 3.142, or 3.141593

Suppose now I ask what is Superman’s speed? Is 800 an answer? No! Not unless we add something - i.e., 800 m/s The units, meters/second, really adds some new

information..had we said 800 inches/hr Superman would be called “Supermolasses”!

Page 7: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Variable Units Number

Velocity/speed

m/s (miles per hour - mph, furlongs per fortnight)

800 (1789, 4.81 x 106)

Power hp (kW) 2,000 (1491)

Energy N-m (ft lbf) 9.81 x 104 (72,300)

Informa-tion

Bits Need enough to dodge kryptonite!

Page 8: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Units and DimensionsDid you see that we converted from one set

of units to another as in m/s converted to furlongs/fortnight?

There is a “fail-safe” method of converting*. Example: What’s the volume of a 1 ft cube in m3

if 1 m = 3.28 ft (or 3.28 [ft/m])? V = 1 ft3, V = 1/3.283 [ft3][m/ft]3 = 0.028 m3

* In simple cases the free web program Convert.exe is pretty good too!

Page 9: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Units and Dimensions

What’s the acceleration of a rocket in mph/s if you know it in SI units, a = 55 m/s2? 1 mile = 1609 m [i.e., 1609 m/mile] & 1 hour

= 3,600 s [i.e 3600 s/hr]. a = 55 3600/1609[m/s2] [s/hr][mile/m] =

120 mph/s (to 2 significant* figures)

* … of which more later

Page 10: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Units and Dimensions In this course we will require the units to be

manipulated in square brackets […] in each problem. While easy to get the previous solutions without this

method, many engineering problems are much harder than this & need this apparently clumsy methodology.

Computerized unit conversions are available in free software on the Internet (for example at:

http://joshmadison.com/software/convert-for-windows)

Page 11: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

More Conversion Examples:These use conversion factors you can

paste from Convert.exe800 m/s to mph

800 [m/s][3.28 ft/m][1/5280 miles/ft][3600 s/hr] 800 x 2.236 = 1790 [mph]

2,000 hp to kW 2,000 [hp][0.7457 kW/hp] = 1492 kW

9.81 x 104 N m to ft lbf 9.81 x 104 [N m][1/4.448 lbf/N][3.28 ft/m] 9.81 x 104 x 0.737 = 7.23 x 104 ft lbf

Page 12: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Newton’s 2nd Law and Units

What Newton discovered was not “may the force be with you”, nor “may the mass acceleration be with you” but that force is proportional to the acceleration that it produces on a given mass.

F mass acceleration

or F ma

Page 13: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Force, Weight, and Mass In high school you learned F = ma but

there’s more to it Newton said that force was proportional to mass

x acceleration (not equal to it) because the equation also defines force

So an undefined force is given by F ma and in some also undefined unit system F1 m1a1 (e.g., Force in units of wiggles, mass in carats and acceleration in furlongs/fortnight2)

Eliminate the proportionality,

maam

FF

am

ma

F

F

11

1

111

and

Page 14: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Force, Weight, and Mass The ratio (F1/m1a1) is arbitrary. Picking it

defines the unit of force. SI system: F1 1 Newton when m1 = 1 kg

and a1 = 1 m/s2

Then you can use F = ma English system: F1 1 lb force when m1 =

1 lb mass and a1 = 32.174 ft/s2

c 2

lbm×ft 32.174 g

lbf ×s

lbfandslbf

ftlbm174.32 Define

2c

c g

maFg

Page 15: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Example 1

What would the SI force on a body if its mass were 856 grams? Need: Force on a body of mass 856

g (= 0.856 kg) accelerated at 9.81 m/s2

Know: Newton’s Law of Motion, F = ma How: F in N, m in kg and a in m/s2. Solve: F = ma = 0.856 9.81 [kg] [m/s2 ]

= 8.397 = 8.40 N

Page 16: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Example 2What would the lbf force on a body if its

mass were 3.25 lb mass? Need: lbf on a body of 3.25 lbm accelerated

at 32.2 ft/s2

Know: Newton’s Law of Motion, F = ma/gc

How: gc = 32.2 lbm ft/lbf s2

Solve: F = ma/gc = 3.25 32.2 /32.2[lbm] [ft/s2 ][lbf s2]/[lbm ft] = 3.25 lbf

Weight is W = mg/gc – a special familiar force.

Page 17: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Example 3What would the lbf force on a body

located on the moon (g = 5.37 ft/s2) if its mass were 3.25 lbm? Know: Newton’s Law of Motion, F = ma/gc

How: gc = 32.2 lbm ft/lbf s2 unchanged

Solve: F = ma/gc = 3.25 5.37 /32.2[lbm] [ft/s2 ][lbf s2]/[lbm ft] = 0.542 lbf

Page 18: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Newton’s 2nd Law and Units It bears repeating: SI system is far

superior and simpler:2 provided m in kg and a in m/ sF ma

Example: How many N to accelerate 3.51 kg by 2.25 m/s2?

• Ans: F = 3.51 x 2.25 [kg][m/s2] = 7.88 N

Page 19: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Significant figuresArithmetic cannot improve the accuracy

of a result 10 meters, 10. meters, 10.0 meters and

10.00 meters are not identical 10 meters implies you have used a 10

meters scale; 10. meters implies you have used a 1 meter scale; 10.0 meters implies you have used a 0.1 meter scale and 10.00 meters implies you have used a 0.01 meter scale

Page 20: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Significant figuresThus 10/6 = 2 and not 1.66666667 etc.

as displayed in your calculator A significant figure is any one of the

digits 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0. Note that zero is a significant figure except when it is used simply to fix the decimal point or to fill the places of unknown or discarded digits.

Page 21: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Significant figures1.23 has 3 sig. figs.4567 has 4 sig. figs.0.0123 has three sig. figs.12,300 has three sig.figs. (The trailing

zeroes are place holders only)1.23 x 103, 1.230 x 103, 1.2300 x 103

have 3, 4, and 5 sig. figs. respectively

Page 22: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Significant figures – example

Round off 123.456 − 123.0 123.456 has 6 sig. figs. 123.0 has 4 sig. figs. But 123.0 is the least precise of these

numbers with just 1 figure to right of decimal place

Thus 123.456 − 123.0 = 0.456 = 0.46 = 0.5 The moral: In this course you will be

graded on significant figures – read your text for all the relevant rules of round-off!

Page 23: Exploring Engineering Chapter 2 Key elements in Engineering Analysis.

Summary

Engineering problems need precise mathematics

But not more precise than can be justified (see text, Chapter 1)

Units must be consistent […] method is very helpful in maintaining correct units

Newton’s 2nd law defines force and gives rise to different sets of units

In SI, force = ma and wt = mg In English units, force = ma/gc and wt = mg/gc

gc is a universal constant that defines force in lbf and g is merely the acceleration due to gravity on Earth

Significant Figures are important in engineeringcalculations.