exploring engineering

30
Chapter 2 Key elements in Engineering Analysis Exploring Engineering

Upload: malia

Post on 21-Mar-2016

40 views

Category:

Documents


0 download

DESCRIPTION

Exploring Engineering. Chapter 2 Key elements in Engineering Analysis. What we are going to learn. Maybe 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 - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Exploring Engineering

Chapter 2Key elements in

Engineering Analysis

Exploring Engineering

Page 2: Exploring Engineering

What we are going to learn

Maybe 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 dimensionsNewton’s 2nd law of motion

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

Significant figures

Page 3: Exploring Engineering

Basic SI Units and PrefixesPrefix Symbol Valuepico Pnano nmicro µmili mcenti cdeci ddeca dahecto hkilo kmega Mgiga Gtera T

1210

910

610

310

210

110

1102103106109101210

Quantity SI Unitlength L, l meter mmass M, m kilogram kgtime T, t second scharge Q,q Coulomb Q

Symbol Abbreviation

Page 4: Exploring Engineering

Derived SI UnitsQuantity Symbol SI Unit

Abbreviation

electric charge Q, q coulomb C electric potential V, v volt V resistance R ohm conductance G siemens S inductance L henry H capacitance C farad F frequency f hertz Hz force F,f newton N energy, work W, w joule J power P, p watt W

magnetic flux weber Wb magnetic flux density B tesla T

Page 5: Exploring Engineering

Units and DimensionsIn 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 6: Exploring Engineering

More Conversion Examples:

These use conversion factors you can paste from Convert.exe

800 m/s to mph800 [m/s][3.28 ft/m][1/5280 miles/ft][3600

s/hr]800 x 2.236 = 1790 [mph]

2,000 hp to kW2,000 [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 7: Exploring Engineering

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 accelerationor F ma

Page 8: Exploring Engineering

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

there’s more to itNewton 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,

maamFF

amma

FF

11

1

111

and

Page 9: Exploring Engineering

Force, Weight, and MassThe 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

lbfandslbfftlbm174.32 Define 2

cc g

maFg

c 2

lbm×ft 32.174 glbf ×s

Page 10: Exploring Engineering

Example 1What 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 = maHow: 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 11: Exploring Engineering

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/gcHow: 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 12: Exploring Engineering

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 unchangedSolve: F = ma/gc = 3.25 5.37 /32.2[lbm]

[ft/s2 ][lbf s2]/[lbm ft] = 0.542 lbf

Page 13: Exploring Engineering

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 14: Exploring Engineering

Significant figuresArithmetic cannot improve the accuracy of

a result10 meters, 10. meters, 10.0 meters and

10.00 meters are not identical10 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 15: Exploring Engineering

http://www.chem.sc.edu/faculty/morgan/resources/

sigfigs/index.html

Tutorial on the Use of Significant Figures

Page 16: Exploring Engineering

Rules for deciding the number of significant figures in a measured quantity:(1) All nonzero digits are significant:

1.234 g has 4 significant figures,1.2 g has 2 significant figures.

(2) Zeroes between nonzero digits are significant:1002 kg has 4 significant figures,3.07 mL has 3 significant figures

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.html

Page 17: Exploring Engineering

Rules for deciding the number of significant figures in a measured quantity:(3) Leading zeros to the left of the first

nonzero digits are not significant; such zeroes merely indicate the position of the decimal point: 0.001 oC has only 1 significant figure,0.012 g has 2 significant figures.

(4) Trailing zeroes that are also to the right of a decimal point in a number are significant: 0.0230 mL has 3 significant figures,0.20 g has 2 significant figures.

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.html

Page 18: Exploring Engineering

Rules for deciding the number of significant figures in a measured quantity:(5) When a number ends in zeroes that are not to

the right of a decimal point, the zeroes are not necessarily significant: 190 miles may be 2 or 3 significant figures,50,600 calories may be 3, 4, or 5 significant figures.

The potential ambiguity in the last rule can be avoided by the use of standard exponential, or "scientific," notation. For example, depending on whether the number of significant figures is 3, 4, or 5, we would write 50,600 calories as:5.06 × 104 calories (3 significant figures)5.060 × 104 calories (4 significant figures), or5.0600 × 104 calories (5 significant figures).

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.html

Page 19: Exploring Engineering

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

displayed in your calculatorA 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 20: Exploring Engineering

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 21: Exploring Engineering

Rules for mathematical operations

In carrying out calculations, the general rule is that the accuracy of a calculated result is limited by the least accurate measurement involved in the calculation.

(1) In addition and subtraction, the result is rounded off to the last common digit occurring furthest to the right in all components. Another way to state this rule is as follows: in addition and subtraction, the result is rounded off so that it has the same number of decimal places as the measurement having the fewest decimal places (or digits to the right). For example, 100 (assume 3 significant figures) + 23.643 (5 significant

figures) = 123.643,which should be rounded to 124 (3 significant figures). Note,

however, that it is possible two numbers have no common digits (significant figures in the same digit column).

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.html

Page 22: Exploring Engineering

Rules for mathematical operations

(2) In multiplication and division, the result should be rounded off so as to have the same number of significant figures as in the component with the least number of significant figures. For example, 3.0 (2 significant figures ) × 12.60 (4 significant

figures) = 37.8000which should be rounded to 38 (2 significant

figures).

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.html

Page 23: Exploring Engineering

Rules for rounding off numbers1) If the digit to be dropped is greater than 5, the

last retained digit is increased by one. For example, 12.6 is rounded to 13.

(2) If the digit to be dropped is less than 5, the last remaining digit is left as it is. For example, 12.4 is rounded to 12.

(3) If the digit to be dropped is 5, and if any digit following it is not zero, the last remaining digit is increased by one. For example, 12.51 is rounded to 13.

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.html

Page 24: Exploring Engineering

Rules for rounding off numbers(4) If the digit to be dropped is 5 and is followed

only by zeroes, the last remaining digit is increased by one if it is odd, but left as it is if even. For example, 11.5 is rounded to 12, 12.5 is rounded to 12.

This rule means that if the digit to be dropped is 5 followed only by zeroes, the result is always rounded to the even digit. The rationale for this rule is to avoid bias in rounding: half of the time we round up, half the time we round down.

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.html

Page 25: Exploring Engineering

Rules for rounding off numbers When using a calculator, if you work the entirety of a long calculation

without writing down any intermediate results, you may not be able to tell if an error is made. Further, even if you realize that one has occurred, you may not be able to tell where the error is.

In a long calculation involving mixed operations, carry as many digits as possible through the entire set of calculations and then round the final result appropriately. For example,

(5.00 / 1.235) + 3.000 + (6.35 / 4.0)=4.04858... + 3.000 + 1.5875=8.630829...

The first division should result in 3 significant figures. The last division should result in 2 significant figures. The three numbers added together should result in a number that is rounded off to the last common significant digit occurring furthest to the right; in this case, the final result should be rounded with 1 digit after the decimal. Thus, the correct rounded final result should be 8.6. This final result has been limited by the accuracy in the last division.

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/sigfigs3.html

Page 26: Exploring Engineering

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 placeThus 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 27: Exploring Engineering

HomeworkBook Problems18, 19, 22, 25, 26, 27, 28

Page 28: Exploring Engineering

18

19

22

Page 29: Exploring Engineering
Page 30: Exploring Engineering

SummaryEngineering 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 EarthSignificant Figures are important in engineeringcalculations.