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Basics of Mechanical Engineering

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 BASICS OF MECHANICAL

Chapter 1. What is Mechanical Engineering?

Definition of Mechanical Engineering

My favorite definition of Mechanical Engineering is If it needs engineering but it doesn’t involve electrons, chemical reactions, arrangement of molecules, life forms, isn’t a structure (building/bridge/dam) and doesn’t fly, a mechanical engineer will take care of it… but if it does involve electrons, chemical reactions, arrangement of molecules, life forms, is a structure or does fly, mechanical engineers may handle it anyway. Although every engineering faculty member in every engineering department will claim that his/her field is the broadest engineering discipline, iin the case of Mechanical Engineering that’s actually true because the core material permeates all engineering systems (fluid mechanics, solid mechanics, heat transfer, control systems, etc. In the past 20 years, the field of Mechanical Engineering has undergone a rather remarkable transformation as a result of a number of new technological developments including

Computer Aided Design (CAD). The average non-technical person probably thinks that mechanical engineers sit in front of a drafting table drawing blueprints for devices having nuts, bolts, shafts, gears, bearings, levers, etc. While that image was somewhat true 100 years ago, today the drafting board has long since been replaced by CAD software, which enables a part to be constructed and tested virtually before any physical object is manufactured.

Simulation. CAD allows not only sizing and checking for fit and interferences, but the resulting virtual parts are tested structurally, thermally, electrically, aerodynamically, etc. and modified as necessary before committing to manufacturing.

Sensor and actuators. Nowadays even common consumer products such as automobiles have dozens of sensors to measure temperatures, pressures, flow rates, linear and rotational speeds, etc. These sensors are used not only to monitor the health and performance of the device, but also as inputs to a microcontroller. The microcontroller in turn commands actuators that adjust flow rates (e.g. of fuel into an engine), timings (e.g. of spark ignition), positions (e.g. of valves), etc.

3D printing. Traditional “subtractive manufacturing” consisted of starting with a block or casting of material and removing material by drilling, milling, grinding, etc. The shapes that can be created in this way is limited compared to modern “additive manufacturing” or “3D printing” in which a structure is built in layers.

Collaboration with other fields. Historically, a nuts-and-bolts device such as automobiles were designed almost exclusively by mechanical engineers. Modern vehicles have vast electrical and electronic systems, safety systems (e.g. air bags, seat restraints), specialized batteries (in the case of hybrids), etc., which require design contributions from electrical, bio mechanical and chemical engineers, respectively. It is essential that a modern mechanical engineer be able to understand and accommodate the requirements imposed on the system by non-mechanical considerations. These radical changes in what mechanical engineers do compared to a relatively short time ago makes the field both challenging and exciting.

Mechanical Engineering curriculum

In almost any accredited Mechanical Engineering program, the following courses are required:

• Basic sciences - math, chemistry, physics

• Breadth or distribution (called “General Education” at USC)

• Computer graphics and computer aided design

• Experimental engineering & instrumentation

• Mechanical design - nuts, bolts, gears, welds

• Computational methods - converting continuous mathematical equations into discrete equations for example

• Core “engineering science”

o Dynamics – essentially F = ma applied to many types of systems

o Strength and properties of materials

o Fluid mechanics

o Thermodynamics

o Heat transfer

o Control systems

• Senior “capstone” design project

Additionally you may participate in non-credit “enrichment” activities such as undergraduate research, undergraduate student paper competitions in ASME (American Society of Mechanical Engineers, the primary professional society for mechanical engineers, the SAE Formula race car project, etc.Many industries employ mechanical engineers; a few industries and the type of systems MEs design are listed below.

Chapter 2: Units

All engineered systems require measurements for specifying the size, weight, speed, etc. of objects as well as characterizing their performance. Understanding the application of these units is the single most important objective of this textbook because it applies to all forms of engineering and everything that one does as an engineer. Understanding units is far more than simply being able to convert from feet to meters or vice versa; combining and converting units from different sources is a challenging topic. For example, if building insulation is specified in units of BTU inches per hour per square foot per degree Fahrenheit, how can that be converted to thermal conductivity in units of Watts per meter per degree C? Or can it be converted? Are the two units measuring the same thing or not? (For example, in a new engine laboratory facility that was being built for me, the natural gas flow was insufficient… so I told the contractor I needed a system capable of supplying a minimum of 50 cubic feet per minute (cfm) of natural gas at 5 pounds per square inch (psi). His response was “what’s the conversion between cfm and psi?” Of course the answer is that there is no conversion; cfm is a measure of flow rate and psi a measure of pressure.) Engineers have to struggle with these misconceptions every day. Engineers in the United States are burdened with two systems of units and measurements:

(1) the English or USCS (US Customary System)  and (2) the metric or SI (Système International d’Unités) . Either system has a set of base unis , that is, units which are defined based on a standard measure such as a certain number of wavelengths of a particular light source. These base units include:

• Length (feet, meters); 1 meter = 100 cm = 3.281 ft = 39.37 inches

• Mass (lbm, slugs, kg); (1 kg = 2.205 lbm) (lbm = “pounds mass”)

• Time (seconds)

• Electric current (really electric charge is the base unit, and derived unit is current = charge/time) (1 coulomb = charge on 6.241506 x 1018 electrons) (1 amp = 1 coulomb/second)

• Moles – NA = Avogadro’s number 6.0221415 x 1023 (units particles/mole)

Temperature is frequently interpreted as a base unit but it is not, it is a derived unit, that is, one created from combinations of base units. Temperature is essentially a unit of energy divided by Boltzman’s constant. Specifically, the average kinetic energy of an ideal gas molecule in a 3- dimensional box is 1.5kT, where k is Boltzman’s constant = 1.380622 x 10-23 J/K (really (Joules/molecule)/K). Thus, 1 Kelvin is the  an ideal gas? By definition, ideal gas particles have only kinetic energy and negligible potential energy due to inter-molecular attraction; if there is potential energy, then we need to consider the total internal energy of the material (E, units of Joules) which is the sum of the microscopic kinetic and potential energies, in which case the temperature for any material (ideal gas or not) is defined as T ≡ ∂US # $ % & ' (V=const. (Equation 1) where S is the entropy of the material (units J/K) and V is the volume. This intimidating-looking definition of temperature, while critical to understanding thermodynamics, will not be needed in this course. Derived units are units created from combinations of base units; there are an infinite number of possible derived units. Some of the more important/common/useful ones are:

 

• Area = length2; 640 acres = 1 mile2, or 1 acre = 43,560 ft2

• Volume = length3; 1 ft3 = 7.481 gallons = 28,317 cm3; also 1 liter = 1000 cm3 = 61.02 in3

• Velocity = length/time

• Acceleration = velocity/time = length/time2 (standard gravitational acceleration on earth =g = 32.174 ft/s2 = 9.806 m/s2)

• Force = mass * acceleration = mass*length/time2

By far the biggest problem with USCS units is with mass and force. The problem is that pounds is both a unit of mass AND force. These are distinguished by lbm for pounds (mass) and lbf for pounds (force). We all know that W = mg where W = weight, m = mass, g = acceleration of gravity. So 1 lbf = 1 lbm x g = 32.174 lbm ft/sec2 (Equation 2) Sounds ok, huh? But wait, now we have an extra factor of 32.174 floating around. Is it also true That 1 lbf = 1 lbm ft/sec2 (Equation 3) which is analogous to the SI unit statement that 1 Newton = 1 kg m/sec2 (Equation 4) No, 1 lbf cannot equal 1 lbm ft/sec2 because 1 lbf equals 32.174 lbm ft/sec2. So what unit of mass satisfies the relation 1 lbf = 1 (mass unit)-ft/sec2? (Equation 5) This mass unit is called a “slug” believe it or not. By comparison of Equations (2) and (5), 1 slug = 32.174 lbm = 14.59 kg (Equation 6) Often when doing USCS conversions, one uses a conversion factor called gc:gc = 32.174 lbm ft lbf sec2 =1 One can multiply and divide any equation by gc = 1 as many times as necessary to get the units correct (an example of “why didn’t somebody just say that?”) If this seems confusing, it is to me too. That’s why I recommend that even for problems in which the givens are in USCS units and where the answer is needed in USCS units, first convert everything to SI units, do the problem, then convert back to USCS units. I disagree with some authors who say an engineer should be fluent in both systems. Still, here’s an example of how to use gc:

Note that it’s easy to write down all the formulas and conversions. The tricky part is to check to see if you’ve actually gotten all the units right. In this case I converted everything to the SI system first, then converted back to USCS units at the very end – which is a pretty good strategy for most problems.

Example 2

A 3000 pound (3000 lbm) car is moving at a velocity of 88 ft/sec. What is its kinetic energy (KE) in ft lbf? What is its kinetic energy in Joules?

Note that if you used 3000 lbf rather than 3000 lbm in the expression for KE, you’d have the wrong units – ft lbf2/lbm, which is NOT a unit of energy (or anything else that I know of…) Also note that since gc = 1, we COULD multiply by gc rather than divide by gc; the resulting units (lbm2 ft3 /lbf sec4) is still a unit of energy, but not a very useful one! Many difficulties also arise with units of temperature. There are four temperature scales in “common” use: Fahrenheit, Rankine, Celsius (or Centigrade) and Kelvin. Note that one speaks of “degrees Fahrenheit” and “degrees Celsius” but just “Rankines” or “Kelvins” (without the “degrees”).  

Special note (another example of “that’s so easy, why didn’t somebody just say that?”): when using units involving temperature (such as heat capacity, units J/kg°C, or thermal conductivity, units Watts/m°C), one can convert the temperature in these quantities these to/from USCS units (e.g. heat capacity in BTU/lbm°F or thermal conductivity in BTU/hr ft °F) simply by multiplying or dividing by 1.8. You don’t need to add or subtract 32. Why? Because these quantities are really derivatives with respect to temperature (heat capacity is the derivative of internal energy with respect to temperature) or refer to a temperature gradient (thermal conductivity is the rate of heat transfer per unit area by conduction divided by the temperature gradient, dT/dx). When one takes the derivative of the constant 32, you get zero. For example, if the temperature changes from 84°C to 17°C over a distance of 0.5 meter, the temperature gradient is (84-17)/0.5 = 134°C/m. In Fahrenheit, the gradient is [(1.8*84 +32) – (1.8*17 + 32)]/0.5 = 241.2°F/m or 241.2/3.281 = 73.5°F/ft. The important point is that the 32 cancels out when taking the difference. So for the purpose of converting between °F and °C in units like heat capacity and thermal conductivity, one can use 1°C = 1.8°F. That doesn’t mean that one can just skip the + or – 32 whenever one is lazy. Also, one often sees thermal conductivity in units of W/m°C or W/mK. How does one convert between the two? Do you have to add or subtract 273? And how do you add or subtract 273 when the units of thermal conductivity are not degrees? Again, thermal conductivity is heat transfer per unit temperature gradient. This gradient could be expressed in the above example as (84°C- 17°C)/0.5 m = 134°C/m, or in Kelvin units, [(84 + 273)K – (17 + 273)K]/0.5 m = 134K/m and thus the 273 cancels out. So one can say that 1 W/m°C = 1 W/mK, or 1 J/kg°C = 1 J/kgK. And again, that doesn’t mean that one can just skip the + or – 273 (or 460, in USCS units) whenever one is lazy.

Example 3

The thermal conductivity of a particular brand of ceramic insulating material is 0.5 BTU inch ft2 hour °F (I’m not kidding, these are the units commonly reported in commercial products!) What is the thermal conductivity in units of Watts meter °C?

0.5 BTU inch ft2 hour °F ×1055 J BTU × ft 12 inch × 3.281 ftm × hour 3600 sec ×1 Watt 1 J/sec ×1.8°F °C = 0.0721Watt m°C

Note that the thermal conductivity of air at room temperature is 0.026 Watt/m°C, i.e. about 3 times smaller than the insulation. So why don’t we use air as an insulator? We’ll discuss that in Chapter 8.

Chapter 3: “Engineering scrutiny”

“Be your own worst critic, unless you prefer that someone else be your worst critic.” - I dunno, I just made it up. But, it doesn’t sound very original.

Scrutinizing analytical formulas and results

I often see analyses that I can tell within 5 seconds must be wrong. I have three tests, which should be done in the order listed, for checking and verifying results. These tests will weed out 95% of all mistakes. I call these the “smoke test,” “function test,” and “performance test,” by analogy with building electronic devices.

1. Smoke test. In electronics, this corresponds to turning the power switch on and seeing if the device smokes or not. If it smokes, you know the device can’t possibly be working right (unless you intended for it to smoke.) In analytical engineering terms, this corresponds to checking the units. You have no idea how many results people report that can’t be correct because the units are wrong (i.e. the result was 6 kilograms, but they were trying to calculate the speed of something.) You will catch 90% of your mistakes if you just check the units. For example, if I just derived the ideal gas law for the first time and predicted PV = n/T you can quickly see that the units on the right-hand side of the equation are different from those on the left-hand side. There are several additional rules that must be followed: • Anything inside a square root, cube root, etc. must have units that are a perfect square (e.g. m2/sec2), cube, etc.) This does not mean that that every term inside the square root must be a perfect square, only that the combination of all terms must be a perfect square. For example, the speed (v) of a frictionless freely falling object in a gravitational field isv = 2gh , where g = acceleration of gravity (units length/time2) and h is the height from which the object was dropped (units length). Neither g nor h have units that are a perfect square, but when multiplied together the units are (length/time2)(length) = length2/time2, which is a perfect square, and when you take the square root, the units are v = length2 time2 = length time as required. 

• Anything inside a log, exponent, trigonometric function, etc., must be dimensionless (I don’t know how to take the log of 6 kilograms). Again, the individual terms inside the function need not all be dimensionless, but the combination must be dimensionless.

• Any two quantities that are added together must have the same units (I can’t add 6 kilograms and 19 meters/second. Also, I can add 6 miles per hour and 19 meters per second, but I have to convert 6 miles per hour into meters per second, or convert 19 meters per second into miles per hour, before adding the terms together.)

2. Function test. In electronics, this corresponds to checking to see if the device does what I designed it to do, e.g. that the red light blinks when I flip switch on, the meter reading increases when I turn the knob to the right, the bell rings when I push the button, etc. – assuming that was what I intended that it do. In analytical terms this corresponds to determining if the result gives sensible predictions. Again, there are several rules that must be followed:

• Determine if the sign (+ or -) of the result is reasonable. For example, if your prediction of the absolute temperature of something is –72 Kelvin, you should check your analysis again.

• Determine whether what happens to y as x goes up or down is reasonable or not. For example, in the ideal gas law, PV = nT:

o At fixed volume (V) and number of moles of gas (n), as T increases then P increases– reasonable

o At fixed temperature (T) and n, as V increases then P decreases – reasonable

o Etc.

• Determine what happens in the limit where x goes to special values, e.g. zero, one or infinity as appropriate. For example, consider the equation for the temperature as a function of time   T(t) of an object starting at temperature Ti at time t = 0 having surface area A (units m2),   volume V (units m3), density ρ (units kg/m3) and specific heat CP (units J/kg°C) that is suddenly dunked into a fluid at temperature T∞ with heat transfer coefficient h (units Watts/m2°C). It can be shown that in this case T(t) is given by€ T(t) = T∞ + (Ti T∞)exp – hA ρVCP t % &'()* (Equation 8)hA/ρVCP has units of (Watts/m2°C)(m2)/(kg/m3)(m3)(J/kg°C) = 1/sec, so (hA/ρVCP)t is dimensionless, thus the formula easily passes the smoke test. But does it make sense? At t =0, Ti = 0 as expected. What happens if you charge for a long time? The temperature can reach T∞ but not overshoot it. In the limit t → ∞, the term exp(-(hA/ρVCP)t) goes to zero, thus T T∞ as expected. Other scrutiny checks: if h or A increases, heat can be transferred to the object more quickly, thus the time to approach T∞ decreases. Also, if ρ, V or CP

 

increases, the “thermal inertia” (resistance to change in temperature) increases, so the time

 

required to approach T∞ increases. So the formula makes sense.

 

• If your formula contains a difference of terms, determine what happens if those 2 terms are

 

equal. For example, in the above formula, if Ti = T∞, then the formula becomes simply T(t)

 

= T∞ for all time. This makes sense because if the bar temperature and fluid temperature are

 

the same, then there is no heat transfer to or from the bar and thus its temperature never

 

changes.

 

3. Performance test. In electronics, this corresponds to determining how fast, how accurate, etc. the device is. In analytical terms this corresponds to determining how accurate the result is. This

 

means of course you have to compare it to something else, i.e. an experiment, a more

 

sophisticated analysis, someone else’s published result (of course there is no guarantee that their

 

result is correct just because it got published, but you need to check it anyway.) For example, if

 

I derived the ideal gas law and predicted PV = 7nRT, it passes the smoke and function tests with

 

no problem, but it fails the performance test miserably (by a factor of 7).

 

Scrutinizing computer solutions

 

(This part is beyond what I expect you to know for AME 101 but I include it for completeness).

 

Similar to analyses, I often see computational results that I can tell within 5 seconds must be wrong.

 

It is notoriously easy to be lulled into a sense of confidence in computed results, because

 

the computer always gives you some result, and that result always looks good when plotted

 

in a 3D shaded color orthographic projection. The corresponding “smoke test,” “function test,”and “performance test,” are as follows:

 

1. Smoke test. Start the computer program running, and see if it crashes or not. If it doesn’t crash,

 

you’ve passed the smoke test, part (a). Part (b) of the smoke test is to determine if the computed

 

result passes the global conservation test. The goal of any program is to satisfy mass, momentum, energy and atom conservation at every point in the computational domain subject to certain constituitive

 

relations (e.g., Newton’s law of viscosity τx = μ∂ux/∂y), Hooke’s Law σ = Eε) and equations of state (e.g., the ideal gas law.) This is a hard problem, and it is even hard to verify that the solution is correct once it is obtained. But it is easy to determine whether or not global conservation is satisfied, that is,

 

• Is mass conserved, that is, does the sum of all the mass fluxes at the inlets, minus the mass

 

fluxes at the outlets, equal to the rate of change of mass of the system (=0 for steady

 

problems)?

 

• Is momentum conserved in each coordinate direction?

 

• Is energy conserved?

 

• Is each type of atom conserved?

 

If not, you are 100% certain that your calculation is wrong. You would be amazed at how many

 

results are never “sanity checked” in this way, and in fact fail the sanity check when, after months or years of effort and somehow the results never look right, someone finally gets around to checking these things, the calculations fail the test and you realize all that time and effort was wasted.

 

2. Performance test. Comes before the function test in this case. For computational studies, a critical performance test is to compare your result to a known analytical result under simplified conditions. For example, if you’re computing flow in a pipe at high Reynolds numbers (where the flow is turbulent), with chemical reaction, temperature-dependent transport properties, variable density, etc., first check your result against the textbook solution that assumes constant density, constant transport properties, etc., by making all of the simplifying assumptions (in your model) that the analytical solution employs. If you don’t do this, you really have no way of knowing if your model is valid or not. You can also use previous computations by yourself or others for testing, but of course there is no absolute guarantee that those computations were correct.

 

3. Function test. Similar to function test for analyses.

 

By the way, even if you’re just doing a quick calculation, I recommend not using a calculator. Enter the data into an Excel spreadsheet so that you can add/change/scrutinize/save calculations as needed. Sometimes I see an obviously invalid result and when I ask, “How did you get that result?

 

What numbers did you use?” the answer is “I put the numbers into the calculator and this was the

 

result I got.” But how do you know you entered the numbers and formulas correctly? What if you need to re-do the calculation for a slightly different set of numbers?

 

Examples of the use of units

 

These examples, particularly the first one, also introduce the concept of “back of the envelope”

 

estimates, a powerful engineering tool.

 

Example 1. Drag force and power requirements for an automobile

 

A car with good aerodynamics has a drag coefficient (CD) of 0.2. The drag coefficient is defined as the ratio of the drag force (FD) to the dynamic pressure of the flow = . ρv2 (where ρ is the fluid density and v the fluid velocity far from the object) multiplied by the cross-section area (A) of the object, i.e. FD = 1

 

2 CDρv2A (Equation 9)

 

The density of air at standard conditions is 1.18 kg/m3.

 

(a) Estimate the power required to overcome the aerodynamic drag of such a car at 60 miles per

 

hour. Power = Force x velocity v = 60 miles/hour x (5280 ft/mile) x (m/3.28 ft) x (hour/60 min) x (min/60 sec) = 26.8 m/s

 

Estimate cross-section area of car as 2 m x 3 m = 6 m2

 

FD = 0.5 x 0.2 x 1.18 kg/m3 x (26.8 m/s)2 x 6 m2 = 510 kg m/s2 = 510 Newton

 

Power = FD x v = 510 kg m/s2 x 26.8 m/s = 1.37 x 104 kg m2/s3 = 1.37 x 104 W = 18.3

 

horsepower, which is reasonable.

 

(b) Estimate the gas mileage of such a car. The heating value of gasoline is 4.4 x 107 J/kg and its

 

density is 750 kg/m3.

 

Fuel mass flow required = power (Joules/sec) / heating value (Joules/kg)

 

= 1.37 x 104 kg m2/s3

 

/ 4.4 x 107 J/kg = 3.10 x 10-4 kg/s

 

Fuel volume flow required = mass flow / density

 

= 3.10 x 10-4 kg/s / 750 kg/m3 = 4.14 x 10-7 m3/s x (3.28 ft/m)3 x 7.48 gal/ft3

 

= 1.09 x 10-4 gal/sec

 

Gas mileage = speed / fuel volume flow rate =

 

[(60 miles/hour)/(1.09 x 10-4 gal/sec)] x (hour / 3600 sec) = 152.564627113 miles/gallon

 

Why is this value of miles/gallon so high?

 

o The main problem is that conversion of fuel energy to engine output shaft work is about

 

25% efficient at highway cruise conditions, thus the gas mileage would be 152.564627113 x

 

0.25 = 38.1411567782 mpg

 

o Also, besides air drag, there are other losses in the transmission, driveline, tires – at best the

 

drivetrain is 80% efficient – so now we’re down to 30.51292542 mpg

 

o Also – other loads on engine – air conditioning, generator, …

 

What else is wrong? There are too many significant figures; at most 2 or 3 are acceptable. When we state 30.51292542 mpg, that means we think that the miles per gallon is closer to 30.51292542 mpg than 30.51292541 mpg or 30.51292543 mpg. Of course we can’t measure the miles per gallon to anywhere near this level of accuracy. 31 is probably ok, 30.5 is questionable and 30.51 is ridiculous. You will want to carry a few extra digits of precision through the calculations to avoid round-off errors, but then at the end, round off your calculation to a reasonable number of significant figures based on the uncertainty of the most uncertain parameter. That is, if I know the drag coefficient only to the first digit, i.e. I know that it’s closer to 0.2 than 0.1 or 0.3, but not more precisely than that, there is no point in reporting the result to 3 significant figures.

 

Example 2. Scrutiny of a new formula

 

I calculated for the first time ever the rate of heat transfer (q) (in watts) as a function of time t from an aluminum bar of radius r, length L, thermal conductivity k (units Watts/m°C), thermal diffusivity α (units m2/s), heat transfer coefficient h (units Watts/m2°C) and initial temperature Tbar conducting and radiating to surroundings at temperature T∞ as

 

q = k(Tbar T∞)eαt/r2 − hr2 (Tbar T∞ −1) (Equation 10)

 

Using “engineering scrutiny,” what “obvious” mistakes can you find with this formula? What is the likely “correct” formula?

 

1. The units are wrong in the first term (Watts/m, not Watts)

 

2. The units are wrong in the second term inside the parenthesis (can’t add 1 and something

 

with units of temperature)

 

3. The first term on the right side of the equation goes to infinity as the time (t) goes to

 

infinity – probably there should be a negative sign in the exponent so that the whole term

 

goes to zero as time goes to infinity.

 

4. The length of the bar (L) doesn’t appear anywhere

 

5. The signs on (Tbar – T∞) are different in the two terms – but heat must ALWAYS be

 

transferred from hot to cold, never the reverse, so the two terms cannot have different signs.

 

One can, with equal validity, define heat transfer as being positive either to or from the bar,

 

but with either definition, you can’t have heat transfer being positive in one term and

 

negative in another term.

 

6. Only the first term on the right side of the equation is multiplied by the

 

e(−αt / r2 ) factor,

 

and thus will go to zero as t → ∞. So the other term would still be non-zero even when t →

 

∞, which doesn’t make sense since the amount of heat transfer (q) has to go to zero as t →

 

∞. So probably both terms should be multiplied by the

 

e(−αt / r2 ) factor.

 

Based on these considerations, the probable correct formula, which would pass all of the

 

smoke and function tests is q = kL(Tbar T∞)+ hr2 (Tbar T∞ ) #$ %& e−αt/r2

 

Example 3. Thermoelectric generator

 

The thermal efficiency (η) = (electrical power out) / (thermal power in) of a thermoelectric power generation device (used in outer planetary spacecraft (Figure 2), powered by heat generated from radioisotope decay, typically plutonium-238) is given by

 

€η = 1− TL TH $ % &' ()

 

1+ ZTa −1

 

1+ ZTa +TL/TH; Ta TL + TH

 

2 (Equation 11) where T is the temperature, the subscripts L, H and a refer to cold-side (low temperature), hot-side (high temperature) and average respectively, and Z is the “thermoelectric figure of merit”:

 

Z = S2

 

ρk (Equation 12)

 

where S is the Seebeck coefficient of material (units Volts/K, indicates how many volts are produced for each degree of temperature change across the material), ρ is the electrical resistivity (units ohm m) (not to be confused with density!) and k is the material’s thermal conductivity (W/mK).

 

(a) show that the units are valid (passes smoke test)

 

Everything is obviously dimensionless except for ZTa, which must itself be dimensionless so that I can add it to 1.

 

(b) show that the equation makes physical sense (passes function test)

 

o If the material Z = 0, it produces no electrical power thus the efficiency should be zero.

 

o If TL = TH, then there is no temperature difference across the thermoelectric material, and

 

thus no power can be generated. In this case

 

€η = (1−1) 1+ ZTa −1

 

1+ ZTa +1

 

= (0) 1+ ZTa −1

 

1+ ZTa +1

 

= 0 OK

 

o Even the best possible material (ZTa → ∞) cannot produce an efficiency greater than the

 

theoretically best possible efficiency (called the Carnot cycle efficiency, see page 83) = 1 –

 

TL/TH, for the same temperature range.

 

Side note #1: a good thermoelectric material such as Bi2Te3 has ZTa ≈ 1 and works up to about

 

200°C before it starts to melts.

 

By comparison, your car engine has an efficiency of about 25%. So practical thermoelectric

 

materials are, in general, not very good sources of electrical power, but are extremely useful in some niche applications, particularly when either (1) it is essential to have a device with no moving parts or (2) a “free” source of thermal energy at relatively low temperature is available, e.g. the exhaust of an internal combustion engine.

 

Side note #2: a good thermoelectric material has a high S, so produces a large voltage for a small

 

temperature change, a low ρ so that the resistance of the material to the flow of electric current is

 

low, and a low k so that the temperature across the material ΔT is high. The heat transfer rate (in

 

Watts) q = kAΔT/Δx (see Chapter 8) where A is the cross-sectional area of the material and Δx is its thickness. So for a given ΔT, a smaller k means less q is transferred across the material. One might think that less q is worse, but no. Consider this:

 

The electrical power = IV = (V/R)V = V2/R = (SΔT) 2/(ρΔx/A) = S2ΔT2A/ρΔx.

 

The thermal power = kAΔT/Δx The ratio of electrical to thermal power is [S2ΔT2A/ρΔx]/[kAΔT/Δx] = (S2/ρk)ΔT = ZΔT,

 

which is why Z is the “figure of merit” for thermoelectric generators.)

 

Figure 2. Radioisotope thermoelectric generator used for deep space missions. Note that

 

the plutonium-238 radioisotope is called simply, “General Purpose Heat Source.”

 

Example 4. Density of matter

 

Estimate the density of a neutron. Does the result make sense? The density of a white dwarf star is about 2 x 1012 kg/m3 – is this reasonable?

 

The mass of a neutron is about one atomic mass unit (AMU), where a carbon-12 atom has a mass of 12 AMU and a mole of carbon-12 atoms has a mass of 12 grams.

 

A neutron has a radius (r) of about 0.8 femtometer = 0.8 x 10-15 meter. Treating the neutron as a

 

sphere, the volume is 4πr3/3, and the density (ρ) is the mass divided by the volume, thus

 

 

ρ = mass

 

volume = 1.66 ×10-27kg

 

4π 3 (0.8 ×10−15m)3 = 7.75 ×1017 kg m3

 

By comparison, water has a density of 103 kg/m3, so the density of a neutron is far higher (by a

 

factor of 1014) that that of atoms including their electrons. This is expected since the nucleus of an atom occupies only a small portion of the total space occupied by an atom – most of the atom is empty space where the electrons reside. Also, even the density of the white dwarf star is far less than the neutrons (by a factor of 105), which shows that the electron structure is squashed by the mass of the star, but not nearly down to the neutron scale (protons have a mass and size similar to neutrons, so the same point applies to protons too.)

 

Chapter 4. Statistics

 

“There are three kinds of lies: lies, damn lies, and statistics…”

 

- Origin unknown, popularized by Mark Twain.

 

Mean and standard deviation

 

When confronted with multiple measurements y1, y2, y3, … of the same experiment (e.g. students’ scores on an exam), one typically reports at least two properties of the ensemble of scores, namely the mean value and the standard deviation:

 

Average or mean value = (sum of values of all samples) / number of samples

 

Standard deviation = square root of sum of squares of difference between each sample and the

 

mean value, also called root-mean-square deviation, often denoted by the Greek letter lower case

 

Warning: in some cases a factor of n, not (n-1), is used in the denominator of the definition of

 

standard deviation. I actually prefer n, since it passes the function test better:

 

o With n - 1 in the denominator, then when n = 1, again y1 = €

 

y , but now σ = 0/0 and thus standard deviation is undefined

 

But the definition using n – 1 connects better with other forms of statistical analysis that we won’t discuss here, so it is by far the more common definition.

 

Example:

 

On one of Prof. Ronney’s exams, the students’ scores were 50, 33, 67 and 90. What is the

 

mean and standard deviation of this data set?

 

Mean = 50 +33+ 67+ 90

 

4 = 60 (a bit lower than the average I prefer)

 

Standard deviation = (50 − 60)2 +(33− 60)2 +(67− 60)2 +(90 − 60)2

 

4 −1 = 31.12

 

Note also that (standard deviation)/mean is 31.12/60 = 0.519, which is a large spread. More

 

typically this number for my exams is 0.3 or so. In a recent class of mine, the grade

 

distribution was as follows:

 

Grade # of standard deviations above/below mean

 

A+ > 1.17 σ above mean (1.90, 1.81)

 

A 0.84 to 1.17 above mean

 

A- 0.60 to 0.67 above mean

 

B+ 0.60 above mean to 0.10 below mean

 

B 0.32 to 0.29 below mean

 

B- 0.85 to 0.68 below mean

 

C+ 1.20 to 1.07 below mean

 

C 1.67 to 1.63 below mean

 

C- > 1.67 below mean (2.04)

 

Stability of statistics

 

If I want to know the mean or standard deviation of a property, how many samples do we need?

 

For example, if I flip a coin only once, can I decide if the coin is “fair” or not, that is, does it come up heads 50% of the time? Obviously not. So obviously I need more than 1 sample. Is 2 enough, 1 time to come up heads, and another tails? Obviously not, since the coin might wind up heads or tails 2 times in a row. Below are the plots of two realizations of the coin-flipping experiment, done electronically using Excel. If you have the Word version of this file, you can double-click the plot to see the spreadsheet itself (assuming you have Excel on your computer.) Note that the first time time the first coin toss wound up tails, so the plot started with 0% heads and the second time the first coin was heads, so the plot started with 100% heads. Eventually the data smooths out to about 50% heads, but the approach is slow. For a truly random process, one can show that the uncertainty decreases as 1/√n, where n is the number of samples. So to have half as much uncertainty as 10

 

samples, you need 40 samples!

 

Figure 3. Results of two coin-toss experiments.

 

0

 

0.1

 

0.2

 

0.3

 

0.4

 

0.5

 

0.6

 

0.7

 

0.8

 

0.9

 

1

 

1.1

 

1 10 100 1000

 

Fraction heads

 

Number of samples

 

0

 

0.1

 

0.2

 

0.3

 

0.4

 

0.5

 

0.6

 

0.7

 

0.8

 

0.9

 

1

 

1.1

 

1 10 100 1000

 

Fraction heads

 

Number of samples

 

Side note: if a “fair” coin lands heads 100 times in a row, what are the chances of it landing heads on the 101st flip? 50% of course, since each flip of a fair coin is independent of the previous one.

 

Least-squares fit to a set of data

 

Suppose you have some experimental data in the form of (x1, y1), (xx, y2), (x3, y3), … (xn, yn) and you think that the data should fit a linear relationship, i.e. y = mx + b, but in plotting the data you see that the data points do not quite fit a straight line. How do you decide what is the “best fit” of the experimental data to a single value of the slope m and y-intercept b? In practice this is usually done by finding the minimum of the sum of the squares of the deviation of each of the data points (x1, y1), (xx, y2), (x3, y3), … (xn, yn) from the points on the straight line (x1, mx1+b), (x2, mx2+b), (x3, mx3+b), … ((xn, mxn+b). In other words, the goal is to find the values of m and b that minimize the sum

 

S = (y1-(mx1+b))2 + (y2-(mx2+b))2, (y3-(mx3+b))2 + … + (yn-(mxn+b))2.

 

So we take the partial derivative of S with respect to m and b and set each equal to zero to find the minimum. Note: this is the ONLY place in the lecture notes where substantial use of

 

calculus is made, so if you have trouble with this concept, don’t worry, you won’t use it

 

again in this course. A partial derivative (which is denoted by a curly “∂” compared to the straight “d” of a total derivative) is a derivative of a function of two or more variables, treating all but one of the variables as constants. For example if S(x, y, z) = x2y3 – z4, then ∂S/∂x = 2xy3, ∂S/∂y = 3x2y2 and ∂S/∂z = -4z3. So taking the partial derivatives of S with respect to m and b separately and setting both equal to zero we have:

 

These are two simultaneous linear equations for the unknowns m and b. Note that all the sums are known since you know all the xi and yi. These equations can be written in a simpler form:

 

These two linear equations can be solved in the usual way to find m and b:

 

Example

 

What is the best linear fit to the relationship between the height (x) of the group of students shown below and their final exam scores (y)? Assuming this trend was valid outside the range of these students, how tall or short would a student have to be to obtain a test score of 100? At what height would the student’s test score be zero? What test score would an amoeba (height ≈ 0) obtain?

 

Student name Height (x) (inches) Test score (y) (out of 100)

 

Juanita Hernandez 68 80

 

Julie Jones 70 77

 

Ashish Kumar 74 56

 

Fei Wong 78 47

 

Sitting Bear 63 91

 

A = 68+70+74+78+63 = 353

 

B = 80 + 77 + 56 + 47 + 91 = 351

 

C = 682 + 702 + 742 + 782 + 632 = 25053

 

D = 68*80 + 70*77 + 74*56 + 78*47 + 63*91 = 24373

 

From which we can calculate m = -3.107, b = 289.5, i.e.

 

Test score = -3.107*Height +289.5

 

For a score of 100, 100 = -3.107*Height + 289.5 or Height = 61.01 inches = 5 feet 1.01 inches

 

For a score of zero, 0 = -3.107*Height + 289.5 or Height = 93.20 inches = 7 feet 9.2 inches

 

For a height of 0, score = 289.5

 

 

 

Figure 4. Least-squares fit to data on test score vs. height for a hypothetical class

 

How does one determine how well or poorly the least-square fit actually fits the data? That is, how closely are the data points to the best-fit line? The standard measure is the so-called R2-value defined as one minus the sum of the squares of the deviations from the fit just determined (i.e. the sum of (yi-(mxi+b))2 divided by the sum of the squares of the difference between yi and the average value.

 

For a perfect fit yi = mxi + b for all i, so the sum in the numerator is zero, thus R2 = 1 is a perfect fit.

 

The example shown above is pretty good,

 

 

R2 =1−

 

(80 − (−3.107* 68 + 298.5))2

 

+ (77 − (−3.107 * 70 + 298.5))2

 

+ (56 − (−3.107* 74 + 298.5))2

 

+ (47 − (−3.107 * 78 + 298.5))2

 

+ (91− (−3.107* 63+ 298.5))2

 

(80 − 70.2)2

 

+ (77 − 70.2)2

 

+ (56 − 70.2)2

 

+ (47 − 70.2)2

 

+ (91− 70.2)2

 

= 0.9631

 

and even fairly crummy fits (i.e. as seen visually on a plot, with many of the data points far removed from the line) can have R2 > 0.9. So R2 has to be pretty close to 1 before it’s really a good-looking fit.

 

y = -3.1067x + 289.53

 

R2 = 0.9631

 

40

 

50

 

60

 

70

 

80

 

90

 

100

 

60 65 70 75 80

 

Height (inches)

 

Chapter 5. Forces in structures

 

“The Force can have a strong influence on the weak-minded”

 

- Ben Obi-wan Kenobi, explaining to Luke Skywalker how he made the famous “these aren’t the

 

Droids you’re looking for” trick work.

 

Main course in AME curriculum on this topic: AME 201 (Statics).

 

Forces

 

Forces acting on objects are vectors that are characterized by not only a magnitude (e.g. pounds force or Newtons) but also a direction. A force vector F (vectors are usually noted by a boldface letter) can be broken down into its components in the x, y and z directions in whatever coordinate system you’ve drawn:

 

F = Fxi + Fyj + Fzk Equation 20 Where Fx, Fy and Fz are the magnitudes of the forces in the x, y and z directions and i, j and k are the unit vectors in the x, y and z directions (i.e. vectors whose directions are aligned with the x, y and z coordinates and whose magnitudes are exactly 1 (no units)). Forces can also be expressed in terms of the magnitude = (Fx2 + Fy2 + Fz2)1/2 and direction relative to the positive x-axis (= tan-1(Fy/Fx) in a 2-dimensional system). Note that the tan-1(Fy/Fx) function gives you an angle between +90° and -90° whereas sometimes the resulting force is between +90° and +180° or between -90° and -180°; in these cases you’ll have to examine the resulting force and add or subtract 180° from the force to get the right direction.

 

Moments of forces

 

Some types of structures can only exert forces along the line connecting the two ends of the

 

A

 

Force Fi

 

Line of action

 

d

 

Figure 5. Force, line of action and moment (= Fd) about a point A

 

structure, but cannot exert any force perpendicular to that line. These types of structures include

 

ropes, ends with pins, and bearings. Other structural elements can also exert a force perpendicular to the line (Figure 5). This is called the moment of a force, which is the same thing as torque. Usuallythe term torque is reserved for the forces on rotating, not stationary, shafts, but  there is no real difference between a moment and a torque.

 

The distinguishing feature of the moment of a force is that it depends not only on the vector force

 

itself (Fi) but also the distance (di) from that line of force to a reference point A. (I like to call this distance the moment arm) from the anchor point at which it acts. If you want to loosen a stuck bolt, you want to apply whatever force your arm is capable of providing over the longest possible di. The line through the force Fi is called the line of action. The moment arm is the distance (di again) between the line of action and a line parallel to the line of action that passes through the anchor point. Then the moment of force (Mi) is defined as Mi = Fidi Equation 21

 

where Fi is the magnitude of the vector F. Note that the units of Mo is force x length, e.g. ft lbf or N m. This is the same as the unit of energy, but the two have nothing in common – it’s just

 

coincidence. So one could report a moment of force in units of Joules, but this is unacceptable

 

practice – use N m, not J. Note that it is necessary to assign a sign to Mi. Typically we will define a clockwise moment as positive and counterclockwise as negative, but one is free to choose the opposite definition – as long as you’re consistent within an analysis.

 

In order to have equilibrium of an object, the sum of all the forces AND the moments of the forces must be zero. In other words, there are two ways that a 2-dimensional object can translate (in the x and y directions) and one way that in can rotate (with the axis of rotation perpendicular to the x-y plane.) So there are 3 equations that must be satisfied in order to have equilibrium, namely:

 

 

Fx,i = 0;

 

i=1

 

nΣ

 

Fy,i = 0; Mi = 0

 

i=1

 

nΣ

 

i=1

 

nΣ

 

Equation 22

 

Note that the moment of forces must be zero regardless of the choice of the origin (i.e. not just at the center of mass). So one can take the origin to be wherever it is convenient (e.g. make the moment of one of the forces = 0.) Consider the very simple set of forces below:

 

24

 

141.4 lbf

 

200 lbf

 

A

 

B

 

C

 

1 ft

 

0.5 ft

 

D

 

0.5 ft

 

0.5 ft

 

45° 45°

 

141.4 lbf

 

0.707 ft 0.707 ft

 

0.707 ft 0.707 ft

 

Figure 6. Force diagram showing different ways of computing moments

 

Because of the symmetry, it is easy to see that this set of forces constitutes an equilibrium condition.

 

When taking moments about point ‘B’ we have:

 

ΣFx = +141.4 cos(45°) lbf + 0 - 141.4 cos(45°) lbf = 0

 

ΣFy = +141.4 sin(45°) lbf - 200 lbf + 141.4 sin(45°) = 0

 

ΣMB = -141.4 lbf * 0.707 ft - 200 lbf * 0 ft +141.4 lbf * 0.707 ft = 0.

 

But how do we know to take moments about point B? We don’t. But notice that if we take

 

moments about point ‘A’ then the force balances remain the same and

 

ΣMA = -141.4 lbf * 0 ft - 200 lbf * 1 ft + 141.4 lbf * 1.414 ft = 0.

 

The same applies if we take moments about point ‘C’, or a point along the line ABC, or even a point NOT along the line ABC. For example, taking moments about point ‘D’,

 

ΣMD = -141.4 lbf * (0.707 ft + 0.707 ft) + 200 lbf * 0.5 ft +141.4 lbf * 0.707 ft = 0

 

The location about which to take the moment can be chosen to make the problem as simple as

 

possible, e.g. to make some of the moments of forces = 0.

 

Example of “why didn’t the book just say that…?” The state of equilibrium merely requires that 3 constraint equations are required. There is nothing in particular that requires there be 2 force and 1 moment constraint equations. So one could have 1 force and 2 moment constraint equations:

 

where the coordinate direction x can be chosen to be in any direction, and moments are taken about 2 separate points A and B. Or one could even have 3 moment equations:

 

Also, there is no reason to restrict the x and y coordinates to the horizontal and vertical directions.

 

They can be (for example) parallel and perpendicular to an inclined surface if that appears in the

 

problem. In fact, the x and y axes don’t even have to be perpendicular to each other, as long as they are not parallel to each other, in which case ΣFx = 0 and ΣFy = 0 would not be independent

 

equations.

 

This is all fine and well for a two-dimensional (planar) situation, what about 1D or 3D? For 1D

 

there is only one direction that the object can move linearly and no way in which it can rotate. For 3D, there are three directions it can move linearly and three axes about which it can rotate. Table 1

 

summarizes these situations.

 

# of spatial

 

dimensions

 

Maximum # of

 

force balances

 

Minimum # of

 

moment balances

 

Total # of unknown

 

forces & moments

 

1 1 0 1

 

2 2 1 3

 

3 3 3 6

 

Table 1. Number of force and moment balance equations required for static equilibrium as

 

a function of the dimensionality of the system. (But note that, as just described, moment

 

equations can be substituted for force balances.)

 

Types of forces and moments

 

A free body diagram is a diagram showing all the forces and moments of forces acting on an object.

 

We distinguish between two types of objects:

 

1. Particles that have no spatial extent and thus have no moment arm (d). An example of this

 

would be a satellite orbiting the earth because the spatial extent of the satellite is very small

 

compared to the distance from the earth to the satellite or the radius of the earth. Particles

 

do not have moments of forces and thus do not rotate in response to a force.

 

2. Rigid bodies that have a finite dimension and thus has a moment arm (d) associated with each

 

applied force. Rigid bodies have moments of forces and thus can rotate in response to a

 

force.

 

There are several types of forces that act on particles or rigid bodies:

 

1. Rope, cable, etc. – Force (tension) must be along line of action; no moment (1 unknown

 

force)

 

T

 

2. Rollers, frictionless surface – Force must be perpendicular to the surface; no moment (1

 

unknown force). There cannot be a force parallel to the surface because the roller would

 

start rolling! Also the force must be away from the surface towards the roller (in other

 

words the roller must exert a force on the surface), otherwise the roller would lift off of the

 

surface.

 

F

 

3. Frictionless pin or hinge – Force has components both parallel and perpendicular to the

 

line of action; no moment (2 unknown forces) (note that the coordinate system does not

 

need to be parallel and perpendicular to the bar)

 

F

 

F||

 

4. Fixed support Force has components both parallel and perpendicular to line of action

 

plus a moment of force (2 unknown forces, 1 unknown moment)

 

F

 

F||

 

M

 

5. Contact friction Force has components both parallel (F) and perpendicular (N) to surface,

 

which are related by F = μN, where μ is the coefficient of friction, which is usually assigned

 

separate values for static (no sliding) (μs) and dynamic (sliding) (μd) friction, with the latter

 

being lower. (2 unknown forces coupled by the relation F = μN). Most dry materials have

 

friction coefficients between 0.3 and 0.6 but Teflon, for example, can have a coefficient as

 

low as 0.04. Rubber (e.g. tires) in contact with other surfaces (e.g. asphalt) can yield friction

 

coefficients of almost 2.

 

N F = μN

 

F = μN

 

N

 

Actually the statement F = μsN for static friction is not correct at all, although that’s how it’s almost always written. Consider the figure on the right, above. If there is no applied force in the horizontal direction, there is no need for friction to counter that force and keep the block from sliding, so F =

 

0. (If F ≠ 0, then the object would start moving even though there is no applied force!) Of course,

 

if a force were applied (e.g. from right to left, in the –x direction) then the friction force at the

 

interface between the block and the surface would counter the applied force with a force in the +x direction so that ΣFx = 0. On the other hand, if a force were applied from left to right, in the +x direction) then the friction force at the interface between the block and the surface would counter the applied force with a force in the -x direction so that ΣFx = 0. The expression F = μsN only applies to the maximum magnitude of the static friction force. In other words, a proper

 

statement quantifying the friction force would be |F| ≤ μsN, not F = μsN. If any larger force is

 

applied then the block would start moving and then the dynamic friction force F = μdN is the

 

applicable one – but even then this force must always be in the direction opposite the motion – so

 

|F| = μdN is an appropriate statement. Another, more precise way of writing this would be

 

F

 

dN

 

v

 

v , where v is the velocity of the block and v is the magnitude of this velocity, thus

 

v

 

v is a

 

unit vector in the direction of motion.

 

Special note: while ropes, rollers and pins do not exert a moment at the point of contact, you can still sum up the moments acting on the free body at that point of contact. In other words, ΣMA =

 

0 can be used even if point A is a contact point with a rope, roller or pin joint, and all of the other

 

moments about point A (magnitude of force x distance from A to the line of action of that force)

 

are still non-zero. Keep in mind that A can be any point, within or outside of the free body. It does not need to be a point where a force is applied, although it is often convenient to use one of those points as shown in the examples below.

 

Statically indeterminate system

 

Of course, there is no guarantee that the number of force and moment balance equations will be

 

equal to the number of unknowns. For example, in a 2D problem, a beam supported by one pinned end and one roller end has 3 unknown forces and 3 equations of static equilibrium. However, if both ends are pinned, there are 4 unknown forces but still only 3 equations of static equilibrium.

 

Such a system is called statically indeterminate and requires additional information beyond the equations of statics (e.g. material stresses and strains, discussed in the next chapter) to determine the forces.

 

 

 

Analysis of statics problems

 

A useful methodology for analyzing statics problems is as follows:

 

1. Draw a free body diagram – a free body must be a rigid object, i.e. one that cannot bend in

 

response to applied forces

 

2. Draw all of the forces acting on the free body. Is the number of unknown forces equal to

 

the total number of independent constraint equations shown in Table 1 (far right column)?

 

If not, statics can’t help you.

 

3. Decide on a coordinate system. If the primary direction of forces is parallel and

 

perpendicular to an inclined plane, usually it’s most convenient to have the x and y

 

coordinates parallel and perpendicular to the plane, as in the cart and sliding block examples

 

below.

 

4. Decide on a set of constraint equations. As mentioned above, this can be any combination

 

of force and moment balances that add up to the number of degrees of freedom of the

 

system (Table 1).

 

5. Decide on the locations about which to perform moment constraint equations. Generally

 

you should make this where the lines of action of two or more forces intersect because this

 

will minimize the number of unknowns in your resulting equation.

 

6. Write down the force and moment constraint equations. If you’ve made good choices in

 

steps 2 – 5, the resulting equations will be “easy” to solve.

 

7. Solve these “easy” equations.

 

Example 1. Ropes

 

Two tugboats, the Monitor and the Merrimac, are pulling a Peace Barge due west up Chesapeake

 

Bay toward Washington DC. The Monitor’s tow rope is at an angle of 53 degrees north of due west with a tension of 4000 lbf. The Merrimac’s tow rope is at an angle of 34 degrees south of due west but their scale attached to the rope is broken so the tension is unknown to the crew.

 

Monitor Merrimac

 

53°

 

y

 

x

 

34° Barge

 

4000 lbf

 

??? lbf

 

29

 

Figure 7. Free body diagram of Monitor-Merrimac system

 

a) What is the tension in the Merrimac’s tow rope?

 

Define x as positive in the easterly direction, y as positive in the northerly direction. In order for

 

the Barge to travel due west, the northerly pull by the Monitor and the Southerly pull by the

 

Merrimac have to be equal, or in other words the resultant force in the y direction, Ry, must be

 

zero. The northerly pull by the Monitor is 4000 sin(53°) = 3195 lbf. In order for this to equal

 

the southerly pull of the Merrimac, we require FMerrimacsin(34°) = 3195 lbf, thus FMerrimac = 5713

 

lbf.

 

b) What is the tension trying to break the Peace Barge (i.e. in the north-south direction)?

 

This is just the north/south force just computed, 3195 lbf

 

c) What is the force pulling the Peace Barge up Chesapeake Bay?

 

The force exerted by the Monitor is 4000 cos(53°) = 2407 lbf. The force exerted by the

 

Merrimac is 5713 cos(34°) = 4736 lbf. The resultant is Rx = 7143 lbf.

 

d) Express the force on the Merrimac in polar coordinates (resultant force and direction, with 0°

 

being due east, as is customary)

 

The magnitude of the force is 5713 lbf as just computed. The angle is -180° - 34° = -146°.

 

Example 2. Rollers

 

A car of weight W is being held by a cable with tension T on a ramp of angle θ with respect to

 

horizontal. The wheels are free to rotate, so there is no force exerted by the wheels in the direction parallel to the ramp surface. The center of gravity of the vehicle is a distance “c” above the ramp, a distance “a” behind the front wheels, and a distance “b” in front of the rear wheels. The cable is attached to the car a distance “d” above the ramp surface and is parallel to the ramp.

 

Figure 8. Free body diagram for car-on-ramp with cable example

 

(a) What is the tension in the cable?

 

Define x as the direction parallel to the ramp surface and y perpendicular to the surface as

 

shown. The forces in the x direction acting on the car are the cable tension T and

 

component of the vehicle weight in the x direction = Wsinθ, thus ΣFx = 0 yields

 

Wsinθ - T = 0 T = Wsinθ

 

(b) What are the forces where the wheels contact the ramp (Fy,A and Fy,B)?

 

The forces in the y direction acting on the car are Fy,A, Fy,B and component of the vehicle

 

weight in the y direction = Wcosθ. Taking moments about point A, that is ΣMA = 0 (so that

 

the moment equation does not contain Fy,A which makes the algebra simpler), and defining

 

moments as positive clockwise yields

 

(Wsinθ)(c) + (Wcosθ)(a) – Fy,B(a+b) – T(d) = 0

 

Since we already know from part (a) that T = Wsinθ, substitution yields

 

(Wsinθ)(c) + (Wcosθ)(a) – Fy,B(a+b) – (Wsinθ)(d) = 0

 

Since this equation contains only one unknown force, namely Fy,B, it can be solved directly to

 

Obtain Fy,B =W acos(θ )+(c d)sin(θ )

 

a + b

 

31

 

Finally taking ΣFy = 0 yields Fy,A + Fy,B - Wcosθ = 0

 

Which we can substitute into the previous equation to find Fy,A:

 

Fy,A =W bcos(θ )−(c d)sin(θ )

 

a + b

 

Note the function tests:

 

1) For θ = 0, T = 0 (no tension required to keep the car from rolling on a level road)

 

2) As θ increases, the tension T required to keep the car from rolling increases

 

3) For θ = 90°, T = W (all of the vehicle weight is on the cable) but note that Fy,A and

 

Fy,B are non-zero (equal magnitudes, opposite signs) unless c = d, that is, the line of

 

action of the cable tension goes through the car’s center of gravity.

 

4) For θ = 0, Fy,A = (b/(a+b)) and Fy,B = (a/(a+b)) (more weight on the wheels closer

 

to the center of gravity.)

 

5) Because of the – sign on the 2nd term in the numerator of Fy,A (-(c-d)sin(θ)) and the +

 

sign in the 2nd term in the numerator of Fy,B (+(c-d)sin(θ)), as θ increases, there is a

 

transfer of weight from the front wheels to the rear wheels.

 

Note also that Fy,A < 0 for b/(c-d) < tan(θ), at which point the front (upper) wheels lift off

 

the ground, and that Fy,A < 0 for a/(d-c) > tan(θ), at which point the back (lower) wheels lift

 

off the ground. In either case, the analysis is invalid. (Be aware that c could be larger or

 

smaller than d, so c-d could be a positive or negative quantity.)

 

Example 3. Friction

 

A 100 lbf acts on a 300 lbf block placed on an inclined plane with a 3:4 slope. The coefficients of friction between the block and the plane are μs = 0.25 and μd = 0.20.

 

a) Determine whether the block is in equilibrium

 

b) If the block is not in equilibrium (i.e. it’s sliding), find the net force on the block

 

c) If the block is not in equilibrium, find the acceleration of the block

 

32

 

5 3

 

4

 

100 lbf

 

300 lbf

 

x

 

y

 

N F = μN

 

Figure 9. Free body diagram for sliding-block example

 

(a) To maintain equilibrium, we require that ΣFx = 0 and ΣFy = 0. Choosing the x direction parallel to the surface and y perpendicular to it,

 

ΣFy = N – (4/5)(300 lbf) = 0 N = 240 lbf so the maximum possible friction force is F friction, max = μsN = 0.25 * 240 lbf = 60 lbf.

 

The force needed to prevent the block from sliding is

 

ΣFx = 100 lbf – (3/5)(300 lbf) + Fneeded = 0

 

Fneeded = -100 lbf + (3/5)(300 lbf) = 80 lbf

 

Which is more than the maximum available friction force, so the block will slide down the plane.

 

(b) The sliding friction is given by

 

Ffriction, max = μdN = 0.20 * 240 lbf = 48 lbf

 

so the net force acting on the block in the x direction (not zero since the block is not at equilbrium) is

 

ΣFx = 100 lbf – (3/5)(300 lbf) + 48 lbf = -32 lbf

 

(c) F = ma -32 lbf = 300 lbm * acceleration

 

acceleration = (-32 lbf/300 lbm) ???

 

what does this mean? lbf/lbm has units of force/mass, so it is an acceleration. But how to convert

 

to something useful like ft/sec2? Multiply by 1 in the funny form of gc = 1 =

 

32.174 lbm ft / lbf sec2, of course!

 

33

 

acceleration = (-32 lbf/300 lbm) (32.174 lbm ft / lbf sec2) = -3.43 ft/sec2

 

or, since gearth = 32.174 ft/sec2,

 

acceleration = (-3.43 ft/sec2)/(32.174 ft/sec2gearth) = -0.107 gearth.

 

The negative sign indicates the acceleration is in the –x direction, i.e. down the slope of course.

 

A good function test is that the acceleration has to be less than 1 gearth, which is what you would get if you dropped the block vertically in a frictionless environment. Obviously a block sliding down a slope (not vertical) with friction and with an external force acting up the slope must have a smaller acceleration.

 

Example 4. Rollers and friction

 

A car of weight W is equipped with rubber tires with coefficient of static friction μs. Unlike the

 

earlier example, there is no cable but the wheels are locked and thus the tires exert a friction force parallel to and in the plane of the ramp surface. As with the previous example, the car is on a ramp of angle θ with respect to horizontal. The center of gravity of the vehicle is a distance “c” above the ramp, a distance “a” behind the front wheels, and a distance “b” in front of the rear wheels.

 

example

 

(a) What is the minimum μs required to keep the car from sliding down the ramp?

 

The unknowns are the resulting forces at the wheels (Fy,A and Fy,B) and the coefficient of

 

friction μs. Taking ΣFx = 0, ΣFy = 0 and ΣMA = 0 yields, respectively,

 

34

 

-μsFy,A - μsFy,B + Wsinθ = 0

 

Fy,A + Fy,B - Wcosθ = 0

 

(Wsinθ)(c) + (Wcosθ)(a) – Fy,B(a+b) = 0

 

Which may be solved to obtain

 

 

Fy,A =W bcos(θ) − c sin(θ)

 

a + b ;Fy,B =W acos(θ) + c sin(θ)

 

a + b s = sin(θ)

 

cos(θ) = tan(θ)

 

Note the function tests

 

1) For θ = 0, μs = tan(θ) = 0 (no friction required to keep the car from sliding on a

 

level road)

 

2) As θ increases, the friction coefficient μs required to keep the car from sliding

 

increases

 

3) For θ = 0, Fy,A = (b/(a+b)) and Fy,B = (a/(a+b)) (more weight on the wheels closer

 

to the center of gravity

 

4) Because of the – sign on the 2nd term in the numerator of Fy,A (-c sin(θ)) and the +

 

sign in the 2nd term in the numerator of Fy,B (+c sin(θ)), as θ increases, there is a

 

transfer of weight from the front wheels to the rear wheels.

 

Note also that we could have also tried ΣFy = 0, ΣMA and ΣMB = 0:

 

Fy,A + Fy,B - Wcosθ = 0

 

(Wsinθ)(c) + (Wcosθ)(a) – Fy,B(a+b) = 0

 

(Wsinθ)(c) - (Wcosθ)(b) + Fy,A(a+b) = 0

 

In which case, the second equation could have been subtracted from the third to obtain:

 

Fy,A + Fy,B - Wcosθ = 0

 

which is the same as the first equation. So the three equations are not independent of each

 

other, and we can’t solve the system. What’s wrong? The coefficient of friction μs doesn’t appear in the set of equations ΣFx = 0, ΣMA and ΣMB = 0. We need to have each of the three unknownsFy,A, Fy,B and μs in at least one of the three equations. The set ΣFx = 0, ΣMA and ΣMB = 0doesn’t satisfy that criterion.

 

(b) At what angle will the car tip over backwards, assuming that it doesn’t start sliding down the

 

ramp at a smaller angle due to low μs?

 

This will occur when Fy,A = 0, i.e. when sin(θ)/cos(θ) = tan(θ) = b/c. This is reasonable

 

because the tip-over angle should increase when c is made larger (center of gravity closer to

 

the ground) or b made smaller (center of gravity shifted forward). Notice also that it doesn’t

 

depend on μs, that is, as long as it doesn’t slide due to low μs, the tip-over angle only depends

 

on the force balance. For what it’s worth, also note that the tip-over angle equals the sliding angle when tan(θ) =

 

μs = b/c. Since generally μs << 1, Except for a very top-heavy (large c) or rear-weightshifted

 

(small b) vehicles, the vehicle will slide down the ramp before it flips over backwards.

 

Example 5. Pinned joint

 

100 lbf

 

4 in

 

12 in

 

30°

 

A

 

B

 

6 in

 

C

 

4 cos(30) 8 cos(30)

 

Figure 11. Free body diagram for pinned joint example

 

A straight bar of negligible mass 12 inches long is pinned at its lower end (call it point A) and has a roller attached to its upper end (call it point B) as shown in the figure. The bar is at a 30° angle from horizontal. A weight of 100 lbf is hung 4 inches from the lower end (call it point C).

 

a) What are the forces in the x and y directions on the pinned end? What is the force in the x

 

direction on the roller end?

 

The pinned end can sustain forces in both the x and y directions, but no moment. The roller

 

end can sustain a force only in the x direction, and again no moment. Summing the forces in the

 

y direction Fy,A + Fy,B + Fy,C = 0 Fy,A + 0 - 100 lbf = 0 Fy,A = +100 lbf.

 

In other words, in the y direction the vertical force at point A must be +100 lbf since that is the

 

only force available to counteract the 100 lbf weight. Next, taking moments about point A

 

(since the lines of action of two of the unknown forces intersect at point A),

 

ΣMA = 0 -(4 in)(cos(30°))(-100 lbf) + (6 in)Fx,B = 0 Fx,B = -57.7 lbf.

 

Finally, for force balance in the x direction,

 

Fx,A + Fx,B + Fx,C = 0 Fx,A = -Fx,B - Fx,C = -(-57.7 lbf) – 0 = +57.7 lbf

 

36

 

b) Would the forces change if the roller and pinned ends were reversed?

 

In this case summing the forces in the x direction:

 

Fx,A + Fx,B + Fx,C = 0 0 + Fx,B + 0 = 0 Fx,B = 0.

 

For force balance in the y direction,

 

Fy,A + Fy,B + Fy,C = 0 Fy,A + Fy,B = 100 lbf

 

Taking moments about point C just for variety (not the easiest way, since neither Fy,A nor Fy,B are known, we just know that Fy,A + Fy,B = 100 lbf),

 

ΣMC = 0 (4 in)(cos(30°))Fy,A - (8 in)(cos(30°))Fy,B + (8 in)(sin(30°))Fx,B = 0

 

(4 in)(cos(30°))(100 lbf – Fy,B) - (8 in)(cos(30°))Fy,B + 0 = 0

 

Fy,B = +33.3 lbf Fy,A = +66.7 lbf

 

which is quite different from case (a).

 

c) What would happen if the lower end were fixed rather than pinned (upper end having the roller again)?

 

In this case there are 4 unknown quantities (Fx,A, Fy,A, MA and Fx,B) but only 3 equations (ΣFx = 0, ΣFy = 0, ΣM = 0) so the system is statically indeterminate. If one takes away the roller end entirely, then obviously Fy,A = 100 lbf, Fx,A = 0 and MA = +(100 lbf)(4 in)(cos(30°)) = 346.4 in lbf.

 

37

 

Chapter 6. Stresses, strains and material properties

 

“It is not stress that kills us, it is our reaction to it”

 

- Hans Selye, endocrinologist.

 

Main course in AME curriculum on this topic: AME 204 (Strength of Materials).

 

Stresses and strains

 

As a follow-on to the discussion of statics we need to consider if a material is subject to a given

 

tensile, compressive or side load, how much stress is imparted into the material, and will this stress cause it to break? As with many subjects in the class, you will learn about this in a lot more detail in future classes; here you’ll just get a taste of it.

 

The normal stress (σ) in a material is defined as σ ≡ F/A Equation 25

 

where F is the force (either tension or compression) acting perpendicular to an imaginary plane

 

surface passing through a piece of material and A is the cross section area. It is called “normal” not in the sense of being “typical” or “standard” but in the sense of being perpendicular or orthogonal to the cross-section of the material. Stress is defined as positive if the material is in tension (i.e. the material is being pulled apart) and negative if the material is in compression. Stress has units force/area, i.e. the same as pressure. The units are typically N/m2 or lbf/in2. Often the unit of “kips” (kilopounds per square inch = 1000 lbf/in2) is used to report stress.

 

The strain (ε) is the fractional amount of elongation or contraction in a material caused by a stress.

 

For example, if under a given amount of tensile stress, a steel bar stretches from a length (L) of 1.00

 

inch to 1.01 inch (a change in length, ΔL, of 0.01 inch) the strain = (1.01 – 1.00)/1.00 = 0.01. In

 

other words,

 

ε ≡ ΔL/L Equation 26

 

For most materials (other than gooey ones, i.e. Silly Putty™, Play-Doh™, …) the amount of strain before failure is relatively small (i.e. less that 0.1, meaning that the material deforms less than 10%

 

before failing.)

 

An elastic material has a linear relationship between stress and strain, i.e.

 

σ = εE Equation 27

 

where E is called the elastic modulus, i.e. the slope of the plot of σ vs. ε in the elastic region shown in Figure 12. Note that since ε is dimensionless, E also has units of pressure.

 

The strength of a material is generally reported in terms of the stress it withstands. For a sufficiently small stress, materials return to their original length or shape after the stress is removed. The smallest stress for which the material does not return to its original length or shape after the stress is removed is called the yield stress (σyield). Beyond this stress, generally the slope of the σ vs. ε plot becomes smaller. There is often an increase in slope as ε is increased still further, up to a maximum σ called the “ultimate stress”, beyond which σ actually decreases as ε increases, leading finally to fracture of the material.

 

Note that we can write Equation 30 in the form F/A = (ΔL/L)E or F = (EA/L)(ΔL), which looks

 

just like the force on a linear spring, F = kx, with k = EA/L. (One might wonder what happened to the – sign, that is, isn’t F = -kx? Stress is usually defined as positive in tension where as for the spring F is defined as positive in compression.) So E and the material dimensions A and L

 

determine its “spring constant.” Figure 12 is typical of a ductile material such as steel that deforms significantly before failure. This is by no means the only shape that σ vs. ε curves may have. A brittle material such as a ceramic or concrete will fail without significant yielding, that is, the σ vs. ε curve is nearly linear up to the failure point. This doesn’t mean that ceramics are necessary weak, in fact they may have higher E than ductile materials, but they are unforgiving to over-stressing (really, over-straining.)

 

Table 2. This table shows materials that are more or less isotropic, i.e. their properties are similar no matter what direction stress is applied relative to the material. A lot of engineering materials are anisotropic, i.e. they are not isotropic. A typical example of such materials is graphite-epoxy

 

composites composed of fibers of graphitic carbon (which have very high tensile strength in the

 

plane of the graphite sheets, and low strength in the direction perpendicular to this plane) that are

 

bonded to an epoxy polymer, which has poor tensile strength but good compressive and shear

 

strength. The result is a material that has very good strength for its weight. (The Boeing 787 uses

 

composites for most of the structure; this has the advantage of high strength to weight ratio, no

 

possibility of corrosion, and ease of forming into any desired shape.)

 

Of course, in any design one must employ a material with a yield strength greater than the actual

 

Figure 12. Typical stress-strain relationships: left: ductile material; right: comparison of various types of materials. Sources: http://dolbow.cee.duke.edu/TENSILE/tutorial/node4.html,

 

http://www.cyberphysics.co.uk/topics/forces/young_modulus.htm

 

E and yield or ultimate strength have the same units (Pa or lbf/in2) but there is no particular

 

relationship between E and strength. Materials can be hard (high E) but break easily (low strength) or vice versa. Some examples of material properties are shown in

 

Slope = E stress in the material; the ratio of the yield stress of the material to the actual predicted stress in the material is call the factor of safety.

 

Some factoids about materials

 

How strong are these materials? How does this compare with the strength of the attractive forces

 

between the atoms (Fatoms)? That is, can we estimate the strength of the material σ ≈ Fatoms/Aatoms, where Aatoms is the cross-section area of the atoms? How could we estimate Fatoms and Aatoms based on macroscopically measurable properties? Let’s start with the size of one atom. Let’s consider a typical material like aluminum. Its molecular weight is 27 g/mole, and its density is (by coincidence)

 

2.7 g/cm3, and 1 mole = 6.02 x 1023 atoms. So the volume occupied by each atom is

 

 

cm3

 

2.7g

 

27g

 

mole

 

mole

 

6.02 ×1023 atom

 

=1.66 ×10−23 cm3

 

atom

 

Thus each atom occupies a roughly cubic space of (1.66 x 10-23)1/3 = 2.55 x 10-8 cm = 2.55 x 10-10 m, or a cross-section area of (2.55 x 10-10 m)2 = 6.51 x 10-20 m2. What is the attractive force between the atoms? The heat of formation of Al(gas) from Al(s) is 330 kJ/mole (see http://webbook.nist.gov).

 

This is the energy needed to separate Al atoms in the solid phase from each other to make a gas.

 

On a per-atom basis this is (330,000 J/mole)(mole/6.02 x 1023 atoms) = 5.48 x 10-19 J. Then, since Energy = force x distance, we can roughly estimate the attractive force as energy/distance or

 

Fatoms = (5.48 x 10-19 J)/(2.55 x 10-10 m) = 2.15 x 10-9 N.

 

Then finally, the force per unit area is

 

Fatoms = (2.15 x 10-9 N)/(6.51 x 10-20 m2) = 3.30 x 1010 Pa = 33 GPa

 

Note that this is comparable to the elastic modulus (E), not the tensile or shear strength, which is about 1000 times smaller. Why is the strength so much smaller than the elastic modulus? For a perfect crystal with nodefects, the above estimate would be appropriate. But real materials have defects in their crystallineMaterial E (109 Pa) ν Yield strength (in tensionunless otherwise noted)

 

(106 Pa)

 

Ultimate

 

strength (106

 

Pa)

 

Aluminum, 6061-T6 68.9 0.32 276 310

 

Steel, 4340-HR 200 0.30 910 1041

 

Iron, pure 200 0.29 30 540

 

Diamond 700 – 1200 0.10 – 0.29 8680 – 16530 (compressive)

 

High-density

 

polyethylene

 

0.18 – 1.6 2.4 – 31.7 10 – 50

 

Alumina, Al2O3 370 0.22 3000 (compressive) 300

 

Solder (60% tin,

 

40% lead)

 

30 0.4 53

 

Silica aerogel 0.001 – 0.01 0.2 0.016

 

Table 2. Properties of some common materials (from http://www.matweb.com)

 

structure. The strength of materials is determined mostly by the microstructural properties like the number of defects, the size of the “grains” (individual crystals), and the response of the defects to strain. This is why small amounts of additives (like adding carbon to iron to make steel) to a material and the details of how the material is processed (e.g. heat treating, rolling, etc.) affect its strength so much, but do not significantly affect other properties such as E, ν (see below), density, etc. You’ll learn a lot more about this in MASC 310.

 

As materials deform under tension, they become longer of course, but they also become narrower. The ratio between the change in diameter (d) of a cylindrical sample and change in length (L) is called Poisson’s ratio (ν), i.e.

 

ν ≡ -(Δd/d)/(ΔL/L) Equation 28

 

The minus sign is there because under tension ΔL > 0 (sample lengthening) but Δd < 0 (sample is narrowing). Note that the volume (V) of the cylindrical sample is L*πd2/4 before applying the stress, and (L+ΔL)*π(d+Δd)2/4 after applying the stress. So (V+ΔV)/V = 1 + (L+ΔL)(d+Δd)2/ Ld2 or, keeping only terms with one Δd or ΔL (not (Δd)2, (Δd)(ΔL), (ΔL)2 etc.)

 

ΔV/V = 2(Δd/d) + (ΔL/L) = -2(ΔL/L)[-(Δd/d)/(ΔL/L)] + (ΔL/L)

 

= (ΔL/L)(1 - 2ν) Equation 29

 

Function test. From the above equation, it is apparent that for a material to have no change in volume under stress, one would need 1 - 2ν = 0 or ν = 0.5. In reality most materials have ν ≈ 0.3, which means that their volume increases under tensile load (ΔL/L > 0). Certainly one would not expect ν > 0.5, for this would imply the volume decreases under tensile load, and increases under

 

compressive load (ΔL/L < 0) – not very likely!

 

Example

 

A . inch diameter steel bar increases in length from 10 cm to 10.4 cm under an applied force of

 

10,000 lbf.

 

(a) What is the stress in the bar?

 

Stress = force/area = 10,000 lbf / (π(0.5 in)2/4) = 50920 lbf/in2.

 

(b) What is the strain in the bar?

 

Strain = ΔL/L = (10.4 cm – 10 cm)/(10 cm) = 0.04

 

(c) What is the change in diameter of the bar?

 

Note that the volume of the bar isn’t constant; to answer this question you’ll have to use

 

Poisson’s ratio.)

 

ν ≡ -(Δd/d)/(ΔL/L); if ν ≈ 0.3 for steel as in Table 2, then

 

Δd = -νd(ΔL/L) = -(0.3)(0.5 inch)((0.4 cm)/(10 cm)) = 0.006 inch

 

Shear forces

 

Tension and compression are forces that act in the direction perpendicular to a particular imaginary plane cut through the material. The force that acts parallel to a particular imaginary plane cut through the material is called the shear force (V) (why V? I dunno…). The shear stress (τ) is the shear force per unit area, i.e.

 

τ = V/A. Equation 30

 

A two-dimensional object in the x-y plane has two components of tension or compression (call

 

them σx and σy) and two shear stresses (one each in the x and y directions; call them τxy-x and τxy-y; usually these are just called τyx and τxy). As shown in Figure 13, a three-dimensional object will have three components of tension or compression (one each in the x, y and z directions) and six components of shear (in the x-y plane, in the x and y directions; in the y-z plane, in the y and z directions; and in the x-z plane, in the x and z directions). So the stress is actually not a single value but a 3 x 3 matrix called the Cauchy stress tensor:

 

Equation 31

 

Keep in mind that for each force shown in Figure 13, there is an equal and oppositely-directed force on the opposing side of the imaginary cube. This looks fairly complicated, and perhaps it is, but one saving grace is that, in order for the moments about a very small cube of material to sum to zero, one must have τxy-x = τyx-y, τzx-z = τxz-x and τzy-z = τyz-y, in other word the matrix is symmetric. So there are only 6, not 9, independent stresses.

 

Principal stresses

 

Note that normal stress is defined as the stress in the direction perpendicular to an imaginary plane and shear stress is defined as the stress in the direction parallel to that same imaginary plane – but how should that plane be chosen? For a simple shape like a cylinder it seems natural to define a plane parallel to the ends of the cylinder, but what about oddly shaped objects? Will the stresses be different depending on how one chooses the coordinate system? Will an object fail or not fail under stress depending on how one chooses the coordinate system? What is important to learn from this sub-section is that the magnitudes of both normal stress and shear stress are entirely dependent on the choice of coordinate system.

 

It can be shown that the coordinate system (x, y, z) of Figure 13 can be rotated such that all of the off-diagonal terms (i.e. all the shear stresses τ) are zero; these coordinates are called the principal directions and the corresponding stresses the principal stresses. Now we’re down to 3 independent stresses in this coordinate system. Furthermore, in the principal directions two of the three coordinates yield the maximum and minimum stress attainable from any rotation of the coordinates. But proving this or using these results is beyond the scope of this course; wait for AME 204. In this course we will consider only the simpler two-dimensional case (Figure 14). If I know in some coordinate system (x,y) the normal stresses σx and σy and the shear stress τxy, then by rotating the coordinate system by an angle θP, the principal stresses (call them σ1 and σ2, corresponding to the maximum (most positive or least negative) and minimum (most negative or least positive) stresses in the material) are obtained; their values are given by

 

Figure 14. Diagram of normal (σ) and shear (τ) stresses in a 2-dimensional system and

 

transformation to the principal stresses. From http://www.efunda.com

 

Note that σ1 and σ2 could be both positive, both negative, or one of each depending on the values of σx,, σy and τxy. If both are positive then the larger one is the only one you need to worry about in terms of material failure, and failure could occur in tension only. If both σ1 and σ2 are negative,

 

then the more negative one is the only one you need to worry about in terms of material failure, and failure could occur in compression only. If σ1 is positive and σ2 is negative or vice versa, then you need to worry about both in terms of material failure, the positive one in tension and the negative one in compression.

 

Also note that there are some good function tests you can perform on the formula for principal

 

stresses:

 

• If τxy =0, then σ1 = σx and σ2 = σy, that is, the normal stresses are the principal stresses since

 

there is no shear in this coordinate systems.

 

• If σy and τxy are both zero, that is, if there is only one normal stress and no shear stress, then

 

σ1 = σx and σ2 =0.

 

• If σy = 0 but τxy ≠ 0, then the principal stresses are

 

σ1 =

 

which makes sense because adding the shear increases σ1 to a value larger than its value if

 

there were no shear (σx), and decreases σ2 to a value smaller than its value if there were no

 

shear (0). In other words, the addition of shear increases both the minimum and maximum

 

normal stresses.

 

Also, by rotating the coordinate system by a different angle θS, the maximum shear stress (τmax) is obtained.

 

In this coordinate system, the normal stresses are the same and equal to (σx + σy)/2, i.e., the average of σx and σy.

 

Note that if in the above equations the stress in the x direction is non-zero (σx ≠ 0) but the stress in the y direction is zero (σy = 0) and the shear in the x-y plane is zero (τxy = 0), then the principal stresses are σ1 = σx, σ2 = 0 and τmax = σx/2. Note also that just because I’m only pulling on the material in one direction (say, in tension) that doesn’t mean that there is no shear stress in the material; it’s all a matter of my choice of coordinates. The important conclusion is that a material under any type of stress has both normal and shear stresses; to determine the conditions for failure, it is not sufficient just to calculate the stresses in one particular coordinate system.

 

One must determine the maximum normal and shear stresses in the material based on the above

 

equations for σ1, σ2 and τmax and choose an appropriate dimensions and materials that can withstand such stresses. An analogy of sorts is with Alfred Hitchcock movies – typically the main character is an ordinary person doing some ordinary task, then something happens to him/her that causes him/her to become involved in some terrifying event. The message of his movies is typically, “you think you’re not involved… but you ARE.” The same thing applies to stresses: “you calculate normal stress and you think you’re not involved with shear stress… but you ARE.” Note that according to Eq. 33, the only situation where the material has no shear stress at all (τmax = 0) is when

 

σx = σy and τxy = 0.c

 

Example

 

A horizontal steel bar . inch in diameter is pulled with a tension of 10,000 lbf, then a load is

 

hung on it in such a way that the shear force is 5,000 lbf.

 

(a) What is the maximum normal stress in the bar?

 

Normal stress along length of bar = σx = Force/Area

 

= (10,000 lbf)/(π(0.5 in)2/4) = 50,930 lbf/in2

 

Shear stress = τxy = (Shear force)/Area = (5,000 lbf)/(π(0.5 in)2/4) = 25,465 lbf/in2

 

From the above equation for principal stresses we find that σ1, σ2 are

 

So the maximum normal stress is +61,477 lbf/in2 (+ sign indicating tension, - sign would

 

indicate compression). Also note that the transformation from the coordinate system x, y to

 

the principal directions requires an angle of rotation given by

 

 

θ P =12

 

tan−1 2τ xy

 

σ x −σ y

 

&

 

'

 

((

 

)

 

*

 

Figure 15. Diagram of pressure vessel showing hoop stress (σh, left) and longitudinal stress

 

(σl, right). Figures from http://www.efunda.com/.

 

Pressure vessels

 

In a cylindrical vessel containing a pressure P, there are 2 stresses to be considered: the hoop stress trying to pull the cylinder apart radially, and the longitudinal stress or axial stress trying to pull it apartaxially. Referring to Figure 15, note that the total force trying to pull the cylinder apart radially is PA,where the area A = 2rL, where r is the cylinder radius and L its length. The total wall cross sectionarea resisting this force is 2τL, where τ is the wall thickness (not to be confused with τ the shearstress used above; wall thickness has units of length, shear stress units of pressure). Thus,Hoop stress (σh) = (total force)/(area of wall resisting force)

 

= (P * 2rL) / (2τL) = Pr/τ Equation 34

 

Similarly for the longitudinal stress, the total force trying to pull the cylinder apart axially is PA = P(πr2) and the total wall cross section area resisting this force is 2π * r * τ, thus

 

Longitudinal stress (σl) = (total force)/(area of wall resisting force)

 

= P(πr2)/(2πrτ) = Pr/2τ Equation 35

 

Note that the hoop stress is twice the longitudinal stress. This is why an overcooked hot dog usually cracks along the longitudinal direction first (i.e., its skin fails from hoop stress, generated by internal steam pressure). Also note that both hoop and longitudinal stress are both positive, i.e. in tension if the pressure inside the vessel is higher than that outside the vessel. Of course, if the pressure outside is higher (i.e. a vacuum chamber or submarine) then both will be hoop and longitudinal stress are both negative, i.e. in compression.

 

So for the pressure vessel we have hoop stress (call it σx, where x is the radial direction) = Pr/τ and longitudinal stress (call it σy, where y is the axial direction) = Pr/2τ. In this coordinate system there is no shear stress. Thus the principal stresses are

 

 

So for this case the principal stresses σ1 and σ2 are just the calculated stresses in the x and y

 

directions; in fact this will happen any time τxy = 0. On the other hand, the maximum shear stress for the pressure vessel is €

 

Thus, unless the yield strength in shear was less than . the yield strength in tension or compression, the material would fail in tension or compression before it failed in shear.

 

Example (to be continued below…)

 

An iron pipe 1 foot in diameter and 50 feet long has a wall thickness of 1/2 inch. The material

 

properties are: elastic modulus (E) = 30 x 106 lbf/in2, yield stress (σyield) = 30 x 103 lbf/in2 in

 

tension, yield stress = -30 x 104 lbf/in2 in compression and yield stress = 10 x 103 lbf/in2 in

 

shear. If the ends of the iron pipe are sealed and the pipe is used as a cylindrical pressure vessel,

 

with the high pressure inside:

 

(a) At what pressure (in units of lbf/in2) will the iron yield?

 

Maximum stress = hoop stress = Hoop stress (σh) = +Pr/τ (tension)

 

Pressure at yield = σyτ/r = (30 x 103 lbf/in2)(0.5 in)/(6 in) = 2,500 lbf/in2

 

(b) For what yield stress in shear (in units of lbf/in2) will the iron yield in shear rather than in

 

tension for this pressure?

 

Since the actual yield stress in shear is 10,000 lbf/in2, the pipe will not yield in shear at this

 

pressure and thus it will yield in tension instead as calculated in part (a).

 

Bending of beams

 

One of the most common problems in structural mechanics is the compute the stresses in a beam

 

subject to a load, perpendicular to the axis of the beam, distributed over the length of the beam.

 

The load is typically reported as a force per unit length along the beam (w), with units N/m or

 

(more likely) lbf/ft. As shown in

 

Figure 16 (left), this load causes a shear force in the beam (V) = wL, where L is the distance from

 

the end of the beam. Or, if w is not constant, we can say that dV = w dx and

 

 

V = w dx 0

 

L ∫ . Then

 

the moment about one end of the beam is given by (defining counterclockwise moments are

 

positive, which is standard in structural mechanics) dM = -Vdx, thus

 

dx2 = −w Equation 37

 

Knowing w(x) for the beam and the boundary conditions at the ends of the beam, one can

 

determine the moment M at any location along the beam. For example, with a constant load w per unit length and which is pinned at one end (able to withstand a force in both x and y directions, but unable to cause a moment) and has a roller at the other end (so able with withstand a force in the y direction only, so there are 3 unknown forces and 3 degrees of freedom, i.e. a statically determinate system) we have

 

d2M/dx2 = -w = constant

 

with the boundary conditions

 

M = 0 at x = 0 and x = L

 

for which the solution is

 

dM/dx = -wx + c1;

 

M = -wx2/2 + c1x + c2;

 

M = 0 at x = 0 c2 = 0;

 

M = 0 at x = L -wL2/2 + c1(L) + 0 = 0 c1 = wL/2

 

M(x) = (wx/2)(L – x) (uniformly loaded beam, pinned ends) Equation 38

 

Note that the maximum of M is at x = L/2 with a value of wL2/8.

 

The above relation applies to a uniform loading w (units force/length) along the whole beam, with pinned ends. For a point force P (units force, not force/length) at the midpoint (x = L/2) between the two ends of the beam, the above equations can be integrated to obtain

 

M(x) = Px/2 (for x ≤ L/2); M(x) = P(L-x)/2 (for x ≥ L/2) Equation 39 w

 

Figure 16. Left: Force balance on a differential element of beam of length dx, showing the

 

applied load per unit length w, shear force V and bending moment M. From

 

http://www.efunda.com/. Right: Schematic of compressive and tensile stresses in beam

 

caused by bending moment. From http://strengthandstiffness.com/6_beams/page_6b.htm.

 

Compression Tension

 

(beam with point load in the middle, pinned ends)

 

Note that for this case maximum of M is at x = L/2 with a value of PL/4. How does this compare

 

with uniform loading? Note that for uniform loading, Mmax = wL2/8 = (wL)(L/8) where (wL) is the total force exerted by the loading on the beam, which is . as much as the moment P(L/4) when the same total force is concentrated at the midpoint of the beam rather than distributed along its length (see Figure 17). Function test: note that far from the point load (i.e. away from x/L = 0.5, near x/L = 0 or x/L = 1), the bending moment is the same for uniform or point load. This makes sense because the for the same total load P = wL, far from the location of the point load one would expect the resulting bending moments to be the same for the two types of loading. Who cares about these moments? Well, what we do care about is the stress in the beam. The beam must resist this moment by the stresses in the material. In order to do that, the beam has to be (for downward loading, i.e. weight on the beam) in tension on the bottom and compression on top. In other words

 

Figure 17. Distribution of bending moments along a beam, pinned at both ends, for

 

uniform or point loading

 

Uniformly distributed load

 

Point load

 

Forces (loads) on beam Bending moments in beam Stresses in beam material

 

It’s important to understand that the bending moments cause far more stress in the beam

 

than the stresses caused by the direct application of the force w(x). It’s beyond the scope of

 

this course to derive the relationship between bending moment and stresses (you’ll learn about this in AME 204), but for a slender beam (one for which its length L is much larger than its height in the y direction) the stress σ resulting from the bending moment in the beam is given by

 

σ (x, y) = −M(x)y

 

I Equation 40

 

where y is the vertical distance from the “neutral axis” of the beam (where σ = 0) (the “neutral axis” is half-way through the beam for a symmetrical cross-section), M(x) is the moment just computed, and I is the moment of inertia of the beam cross-section (units are length4). The moment of inertia about an axis A-A’ is defined in Figure 18.

 

Note that for a given total cross-section area (thus total weight of beam) one can have large I (thus lower stress σ) by having more material at larger distances from the axis A-A’. This is the reason for using I-shaped beam sections. Formulas for I for common shapes include:

 

• Circular cross-section of diameter d: I = πd4/64

 

• Thin-wall hollow tube of diameter d and wall thickness τ: I = πd3τ/8

 

• Rectangular cross-section I = ab3/12 (a = width of beam; b = height of beam)

 

• I beam of width a, height b, thickness of central section τw and thickness of top and bottom

 

sections τh: I =

 

I = ab3 12 − (a −τ w )(b − 2τ h )3

 

Note that the I-beam formula satisfies the function tests: when τw = a or τh = b/2, the I-beam is

 

just a “filled” rectangle with I = ab3/12, and if τw = a and τh = b/2, the beam has no material thus I= 0.

 

Now let’s use this for a simple case of a uniformly-loaded or point-loaded rectangular cross-section beam (Figure 19) of height b and thickness a. For this case, the moment of inertia I = ab3/12, thus σmax (at the center of the beam (x = L/2), at the top or bottom, where y = +b/2 at the top of the beam and –b/2 at the bottom of the beam) is given by

 

€ σ max = − Mmax ymax

 

I = (wL2 /8)(±b/2)

 

ab3 /12 = 0.75 wL2

 

ab2 (uniform load)

 

σ max = − Mmax ymax

 

I = (PL/4)(±b/2)

 

ab3 /12 = 1.5 PL

 

ab2 (point load)

 

Equation 41

 

where the – sign refers to the compression at the top of the beam and the + sign refers to the

 

tension at the bottom of the beam (recall that the sign convention for stresses is that compression is negative and tension is positive.) Does this result make sense?

 

• Smoke test: The units (for uniform load) are wL2/ab2 = (Force/Length)(Length)2/(Length3)

 

= Force/Length2, which is stress – OK.

 

• Function test #1: a longer (larger L) or more heavily loaded (larger w or P) beam should have

 

more stress - OK

 

• Function test #2: a thicker (larger a) or taller (larger b) should have less stress – OK

 

Note that as mentioned below Eq. 33, for a situation in which there is normal stress σx in only one direction (σy = 0) and no shear stress in that x-y coordinate system (τxy = 0), there is still a shear stress τmax = σx/2 according to Eq. 33. So the beam would fail in shear if the yield strength in shear is less than half of the smaller of the yield strength in tension or compression.

 

Also note that for a given total load (in units of force) = wL (uniform load) or P (point load) and

 

beam length L, the stress is proportional to 1/ab2, whereas the weight is proportional to volume = abL. Thus to minimize the stress for a given weight of beam, one wants to minimize the ratio

 

(abL/ab2) = L/b, meaning that (since L is already fixed) we want to maximize b (and thus minimize a) – in other words, a tall skinny beam cross-section works better than a short fat one. That’s another reason for using the I-beam shape. Another way of thinking of this is that since σmax is proportional to 1/ab2, one gets more benefit from increasing b than increasing a. Increasing a results in a proportional decrease in stress and a proportional increase in weight, so the strength-toweight ratio doesn’t change. However, increasing b resulting in a more-than-proportional (1/b2) decrease in stress, thus the strength-to-weight ratio increases.

 

Figure 19. Schematic of example beam-loading problem

 

By examination of Fig. 19, one might notice that there is a compressive stress acting on the beam by virtue of the loading. In Fig. 19, this “direct” compressive stress would be Force/Area = wL/aL = w/a. Is this a lot or a little? By comparison the compressive stress caused by the bending moment

 

is (Eq. 41) 0.75wL2/ab2. The ratio of the bending-induced to direct compression is then

 

(0.75wL2/ab2)/(w/a) = 0.75(L/b)2. Since we have already assumed L >> b (a long, slender beam), the bending-induced compression far exceeds the “direct” compression. Similarly, for the shear force V caused by the “direct” loading, from Eq. 37 dV/dx = -w, thus V = -wx = -wL/2 at the center of the beam. This results in a shear stress = force/area = -(wL/2)/ab. The ratio of the shear stress caused by the bending moment = σx/2 = (0.75wL2/ab2) = 0.375wL2/ab2 to that caused by the “direct” loading is then 0.75(L/b). So again, since L >> b, the stress caused by the bending moment is much more than that due to direct loading, which explains why we can usually ignore the direct loading when determining the point at which a beam will fail.

 

Another property of some interest is the maximum deflection Δ (i.e. the sag in the middle of the

 

beam) due to the applied load. Its value is given by Δ = 5wL4

 

384EI (uniform load);Δ = PL3

 

48EI (point load) Equation 42

 

Note that for the same total applied load (wL = P), the maximum deflection is (1/48)/(5/384) = 1.6 times larger for the point load than the uniform load. Another common stress analysis problem is a circular disk (e.g., the top or bottom of a cylindrical pressure vessel) of radius r and thickness τ with pressure P on one side. The maximum stress

 

(including the transformation to principal stresses) is given by

 

σ max = ±c Pr2

 

τ 2 Equation 43

 

where c = 1.24 if the edges of the disk are free to pivot (e.g. like a drum head, which is not very

 

realistic) or c = 0.696 if the edges are rigidly clamped and unable to pivot (which would be the case if the disk were welded or bolted on to the end of the cylindrical part of the pressure vessel.)

 

Example

 

(a) If the iron pipe from the pressure vessel example above has no pressure inside but instead is

 

used as a beam with one pinned end and one roller end instead, what is the maximum point

 

load (P, units of force, not to be confused with pressure P) that could be applied at the

 

middle of the beam before the iron yields?

 

σ max = −Mmaxymax

 

I = − (PL / 4)(−d / 2)

 

π d3τ /8 = PL

 

π d

 

P = π d2τσ yield

 

L = π (12in)2 (0.5in)(30 ×103lbf / in2 )

 

(50 ft ×12in / ft) =11,304lbf

 

(Note: the material is weaker in tension, so the failure will occur when σy = +30 x 103

 

lbf/in2 in tension at the bottom of the beam (y = -d/2) rather than σy = -30 x 104 lbf/in2

 

in compression at the top of the beam (y = +d/2)).

 

However… we need to check for failure due to shear stress also. At the top or bottom

 

of the beam, for a given value of the normal stress in the x direction (along the length of

 

the beam) of σx, assuming no normal stress in the y direction (σy = 0) and no shear

 

stress (τxy = 0), there is a shear stress given as (Eq. 33):

 

52

 

As discussed below Eq. 33, in this particular case the maximum shear stress is half as

 

much as the normal stress σx. Thus

 

τ yield = − 1

 

So the beam yields at a lower point load P in shear than in tension (7,536 lbf vs. 11,304 lbf.)

 

(b) What is the maximum deflection of the beam?

 

 

Δ = PL3

 

48EI =

 

(11,300lbf )[(50 ft)(12in / ft)]3

 

48(30 ×106 lbf /in2 )(π(12in)3 (0.5in)/8) = 5.00in

 

Notice that the 50 foot long beam deflects/bends a maximum (i.e. at the failure load) of 5

 

inches, i.e. only 5 / (50 x 12) = 0.0083 = 0.83%.

 

(c) How much compressive force (pre-stressing) should be applied to the pipe to maximize the

 

point load that could be applied? What would this maximum point load be?

 

At the maximum load condition there is σy = -30 x 103 lbf/in2 (compression) at the top

 

of the beam and σy = +30 x 103 lbf/in2 (tension) at the top of the beam. This is sort of

 

a waste, since the top of the beam could take a lot more compressive stress before it

 

failed. So by pre-compressing the beam, one could even things out. So to have both the

 

top and bottom of the beam at their maximum stress, we would have a stress of -30 x

 

104 = PC – S at the top of the beam (- sign indicating compression, PC is the precompression,

 

S is the stress due to the applied load – on top, + on the bottom) and +30

 

x 103 = PC + S at the bottom of the beam. So combining this two equations, we have

 

-30 x 104 = PC – S

 

30 x 103 = PC + S

 

-27 x 104 = 2 PC PC = -13.5 x 104; S = 16.5 x 104

 

We already showed that without pre-compressing, a point load of 11,300 lbf produces a

 

stress of ±30 x 103 lbf/in2 in the beam. With pre-compressing, we can withstand ±16.5

 

x 104 of stress in the beam due to the loading, that is, 5.5 times more stress. Since the

 

relationship between the applied load P and the stress is linear, we can conclude that P =

 

5.5 (11,300) = 62,200 lbf.

 

(d) If the pipe has welded disk end caps of the same material and thickness as the pipe, at what

 

pressure P (again this is pressure P, not to be confused with point load P) would the end cap

 

fail?

 

σ max = ±c Pr2

 

τ 2 P = ±

 

σ maxτ 2

 

cr2 = + (30 ×103lbf / in2 )(0.5in)2

 

0.696(6in)2 = 299lbf / in2

 

where again the tensile (not compressive) strength is chosen because it the smaller value.

 

Buckling of columns

 

Another way in which a structural element under compression can fail is by buckling. This is not

 

strictly a failure of the material, but effectively eliminates the load-carrying capability of the structure.

 

The compressive force (F) at which buckling occurs in a column of length L is given by

 

Fbuckling = nπ2EI/L2 Equation 44

 

where E is the elastic modulus discussed above, I is the moment of inertia of the cross-section of

 

the column in the plane perpendicular to the direction of the applied force (which is parallel

 

to the long direction of the column) and n is a constant that depends on the way in which the

 

column ends are or are not held:

 

• Both ends pinned, i.e., free to pivot: n = 1

 

• Both ends clamped, i.e., unable to pivot: n = 4

 

• One end pinned, one end clamped: n = 2

 

It should be noted that this buckling formula is valid only for a “slender” column (where the length L is much greater than the width of the column cross section) and it assumes that the column cross-section does not change (in other words, it would not account for the crumpling of an aluminum beverage can, where the buckling occurs due to a change in the cross-section of the can. When computing I for a rectangular cross-section of a buckling column, which is the “a” dimension and which is the “b” dimension? Since the column can buckle either way, you have to use the lesser I, i.e. where a is the smaller dimension, which says that to avoid buckling, you don’t want tall skinny beam cross-sections, you want round or square ones. Note that this conflicts with the desired crosssection to minimize stress due to bending moments, i.e. it was just mentioned above that for best strength to weight ratio, you want tall skinny I-beams. Thus the optimal I-beam cross-section will be a compromise between the two shapes.

 

Example

 

What is the buckling load of a polyethylene plastic drinking straw (E ≈ 109 Pa), .” in diameter and 1/32” wall thickness, 6” long, with both ends free to pivot?

 

Fbuckling = nπ2EI/L2;

 

n = 1;

 

E ≈ 109 Pa;

 

I = πd3τ/8 = π(0.25)3(1/32)/8 = 1.92 x 10-4 in4 = 7.99 x 10-11 m4

 

L = 6 in = 0.152 m

 

Fbuckling = 1π2(109 Pa)( 7.99 x 10-11 m4)/( 0.152 m)2 = 34 N = 7.7 lbf.

 

In practice the buckling load would be less because this analysis assumes the load is exactly along the axis of the column, whereas in reality there would be some sideways (shear) load.

 

Chapter 7. Fluid mechanics

 

Main course in AME curriculum on this topic: AME 309 (Dynamics of Fluids).

 

Fluid statics

 

Fluid mechanics is just ΣF = d(mv/dt) (Newton’s 2nd Law, the sum of the forces is equal to the rate of change of momentum) applied to a fluid. What distinguishes a fluid from a solid is that a solid deforms only a finite amount due to an applied shear stress (unless it breaks), whereas the fluid continues to deform as long as the shear stress is applied. This makes fluid mechanics a lot more complicated (at least to me) than solid mechanics.

 

Hydrostatic pressure

 

Let’s look first at fluid statics, i.e. when ΣF = 0. If a fluid is not moving at all, as in a glass of water, then the fluid is static and has only a hydrostatic pressure. Imagine a column of water of height z, crosssection area A and density ρ (units M/L3, i.e. mass of fluid per unit volume.

 

Table 3 gives the density of several common liquids and gases. The weight of the water is the mass x g = density x volume x g = ρzAg. This weight is distributed over an area A, so the force per unit area (the hydrostatic pressure) is ρzAg/A = ρgz. This is added to whatever pressure P(0) exists at z = 0. So the hydrostatic pressure P(z) is P(z) = P(0) - ρgz Equation

 

where z is defined as positive upward, i.e. decreasing depth. This result assumes that the density (ρ) is constant. This is reasonable for water and practically all liquids, even at pressures of thousands of atm. It’s also ok for gases if z is not too large, i.e. such that ρgz << Po.

 

Fluid Density

 

(ρ, kg/m3)

 

Dynamic viscosity

 

(μ, kg/m sec)

 

Kinematic viscosity

 

(ν = μ/ρ, m2/sec)

 

Water 997.1 8.94 x 10-4 8.97 x 10-7

 

Air 1.18 1.77 x 10-5 1.50 x 10-5

 

Motor oil 917 0.260 2.84 x 10-4

 

Mercury 13500 1.53 x 10-3 1.13 x 10-7

 

Table 3. Properties of some common fluids at ambient temperature and pressure.

 

Buoyancy

 

According to Archimedes’ principle, an object of volume V placed in a liquid of density ρ will exert a buoyant force equal the weight of the fluid displaced = ρfgV. The net force on the object is the difference between this Archimedean (buoyant) force and the weight of the object = ρogV, where ρo is the average density of the object (just total mass/total volume). Thus the net force F acting on the object is F = (ρf - ρo)gV Equation 46, where the sign convention is such that the force is positive (directed upward) when the object density is less than the fluid density (i.e., the object floats upward). Function test: if the density of the object and the fluid are the same, the object is “neutrally buoyant,” and there is no net force on the object (F = 0).

 

Example

 

a) The deepest part of the ocean is a spot called “Challenger Deep” in the Marianas Trench in

 

the western Pacific Ocean. The depth is 35,838 feet. The density of seawater is 1026 kg/m3.

 

What is the hydrostatic pressure (in atmospheres) at this depth? Remember, at sea level, the

 

pressure is 1 atm and increases as the depth increases.

 

Ocean depth: z = -35838 ft = -10923 m; seawater density ρ = 1026 kg/m3

 

 

P = P(0) − ρgz =1atm − (1026kg/m3 )(9.81m/s2 )(−10923m) 1atm

 

101325N /m2 =1086atm

 

b) The density of air at sea level is 1.18 kg/m3. If the air density were constant (not a function

 

of elevation), at what elevation would the pressure be zero?

 

 

P = P(0) − ρairgz = 0

 

 

P(0) = ρairgz thus

 

 

z = − P(0)

 

ρairg

 

€ = − 101325N /m2

 

1.18kg/m3 × 9.81m/s2 = 8732m

 

c) Until 2012, the only vessel ever to carry people to Challenger Deep was the bathyscaphe

 

Trieste in 1960. It used gasoline (ρ = 739 kg/m3) for flotation since no air tank could be

 

made light enough to sustain an 1100 atm pressure difference and still provide positive

 

buoyancy (since the gasoline is essentially incompressible it could be contained in a thinwalled

 

tank that did not need to sustain a pressure difference between the gasoline and the

 

surrounding seawater.) The Trieste used 22,500 gallons of gasoline for flotation. How much

 

buoyant force could this much gasoline produce?

 

In this case the surrounding fluid is seawater and the “object” is the gasoline itself.

 

Bernoulli’s equation

 

One of the most common problems in fluid flows is to determine the relationship between velocity, pressure and elevation of a flowing fluid in a pipe or other duct. To do this, we enforce conservation of energy on the fluid, i.e. the energy contained by the fluid at one point in the flow is the same as any other, but the energy may be transformed from one form to another. Moreover, it is more convenient to work with power (rate of change of energy) rather than energy itself. If the flow is steady so that no energy is accumulating or dissipating within the pipe then the power (sum of all forms) must be constant. This course in general is not intended to provide derivations of formulas you will study in much greater detail in later courses, but it is worthwhile to do so for Bernoulli’s equation just as an example of the value and power of units, and the concept of conservation (e.g., of energy) applied to a fixed volume (often, a fixed mass is analyzed rather than a fixed volume).

 

There are 3 types of power that must be considered, and their sum conserved. In words, the

 

conservation of energy can be stated as:

 

(Power needed to push fluid into the tube inlet and power extracted at the tube outlet)

 

+ (power associated with the rate of change of kinetic energy of the fluid as it passes through the tube)

 

+ (power associated with gravitational potential of fluid)

 

= constant

 

Let’s compute the individual terms then add them up.

 

1. Power needed to push fluid into the tube inlet and power extracted at the tube outlet:

 

m˙ is the mass flow rate (units kg/sec), discussed in more detail in the next sub-section.

 

2. Power associated with the rate of change of kinetic energy of the fluid as it passes through the

 

tube:

 

Combine: sum of the powers at inlet (call it station 1) = sum of powers at outlet (call it station 2).

 

Assume mass flow rates are equal at inlet and outlet (if they’re not, the flow can’t be steady because mass will be accumulating or being lost from the pipe)

 

 

 

This is Bernoulli’s equation which is merely a statement of conservation of energy for an

 

incompressible (ρ = constant), inviscid (no viscosity), steady, one-dimensional flow between

 

locations 1 and 2. Recall that the term ρv2/2 is called the dynamic pressure, i.e. the increase in pressure that would occur if the fluid were decelerated (at constant z) from velocity U to a velocity of zero.

 

Function test: it was shown that for a static fluid (v1 = v2 = 0), P(z) = P(0) - ρgz, or P(z) + ρgz = P(0), which is the same as Bernoulli’s equation for z2 = 0.

 

If there are more than one inlets or outlets, we still have to conserve energy, thus the sum of the

 

Bernoulli terms must be the same at the inlet and outlet. For example, if there are two inlets (say 1a and 1b) and two outlets (say 2a and 2b) then

 

Note that Bernoulli’s equation assumes that the density (ρ) is constant. At first glance this might

 

suggest that it cannot be used for air or other gases, which are compressible. Actually, Bernoulli’s equation can be used for gases if the Mach number (ratio of velocity to sound speed) is significantly less than 1. This applies to most of our common flow situations, e.g. the air flowing over a car can for all practical purposes be considered to have constant density, as discussed later. Besides the assumption of steady 1D flow, constant density, what other significant limitations does Bernoulli’s equation have? The most important is that friction losses (viscosity) are not considered. Can we just add another term to Bernoulli’s equation to account for viscosity? No, because viscosity is dissipative and causes a loss in the total power (sum of the three terms). Where does the power go? Into thermal energy of the fluid (i.e. it gets hotter.) Moreover, the amount of power lost is path dependent, i.e. a longer or narrower tube will have more loss, whereas the above three terms don’t depend on the length of diameter of the tube. Of course, viscosity can be incorporated into fluid flow analysis, but it’s much more difficult and can’t be done with a simple equation like Bernoulli’s in which conservation of energy does not involve viscous dissipation of kinetic energy and its path dependence.

 

Conservation of mass

 

How does one determine the velocity v? For steady flow in a pipe, channel or duct, the mass flow rate € m˙ (in kg/s) has to be the same everywhere in the flow system. This mass flow rate is the product of the fluid density (ρ), velocity (v) and the cross-section area (A) of the pipe or duct

 

through which the fluid flows, i.e.,

 

59

 

m= ρ1v1A1 = ρ2v2A2 Equation 51.

 

In the case of Bernoulli’s equation, we have already assumed that the density is constant, so for this case (i.e. liquids as well as gases at low Mach number) we can simplify this to

 

v1A1 = v2A2 note that UA has units of (length/time)(length)2 = length3/time = volume/time, i.e. the volumetric flow rate, usually given the symbol Q. Thus for an incompressible fluid Q1 = Q2. If there are more than 1 inlets or outlets then the sum of the mass flows at the inlets must equal those at the outlets,

 

i.e.

 

ρ1av1aA1a +ρ1bv1bA1b = ρ2av2aA2a +ρ2bv2bA2b

 

or for incompressible flow

 

v1aA1a + v1bA1b = v2aA2a + v2bA2b or Q1a +Q1b =Q2a +Q2b

 

Example

 

Water flows from a faucet at elevation z = 0 with supply pressure (P1) of 30 lbf/in2 =

 

207,000 Pa (above atmospheric), area 5 cm2, to the roof of a house with z = 5 m and

 

through a nozzle with area 1 cm2, into ambient air with a pressure (P2) of 0 lbf/in2 above

 

atmospheric. What is the velocity of the water leaving the nozzle? Assume that viscous

 

effects are negligible.

 

Neither v1 nor v2 are known, but P1, P2, z1 and z2 are all known, so we have 2 equations

 

(Bernoulli and mass conservation) for the two unknowns.

 

First apply mass conservation: v1 = v2A2/A1 = v2(1 cm2)/(5 cm2) = 0.2 v2.

 

Viscous effects

 

Definition of viscosity

 

Bernoulli’s equation pertains only when there is no viscosity (i.e. the flow is inviscid). Fluids resist

 

motion, or more specifically resist a velocity gradient, through viscosity (μ), defined by the relation

 

(called Newton’s Law of Viscosity)

 

τ xz

 

vx

 

y Equation 52

 

where τxz is the shear stress in the x-z plane, ux is the component of velocity in the x direction, and ∂v/∂y is the velocity gradient in the y direction. This type of viscosity is called the dynamic viscosity.

 

Since τxz has units of force/area = (ML/T2)/L2 , u has units of L/T and y has units of L, the

 

viscosity μ has units of M/LT, e.g. (kg/m s). This unit has no particular name, but 1 g/cm s = 0.1

 

kg/m s = 1 Poise. The unit centipoise = 0.01 Poise = 0.001 kg/m s is frequently used because the

 

dynamic viscosity of water at ambient temperature is almost exactly 1 centipoise.

 

Another type of viscosity is the kinematic viscosity, which is just the dynamic viscosity divided by density: €ν = μρ

 

which has units of (M/LT)/(M/L3) = L2/T, e.g. m2/s. Again, this unit has no particular name but 1 cm2/s = 10-4 m2/s = 1 Stoke, and again 1 centistoke = 0.01 Stoke, which is very nearly the kinematic viscosity of water at ambient temperature. The units of kinematic viscosity, L2/T, are the same as that of diffusion coefficients, e.g. the property that describes how fast a drop of ink will spread out in a beaker of water, thus my favorite interpretation of ν is that is it the momentum diffusivity. This describes how quickly or slowly the momentum of the fluid is exchanged with the solid object passing through the fluid (or fluid passing through the solid object, as in the case of flow in a pipe.)

 

No-slip boundary condition

 

At the boundary between a fluid and a solid object, the velocity of the fluid and the solid must be

 

the same. This is called the no-slip condition. This means that any time a solid is moving through a fluid (e.g. an airplane flying through the air) or a fluid is moving through a solid (e.g. flow through a pipe) there will be a velocity gradient (i.e. ∂u/∂y in the above equation) because there is a difference between the fluid velocity far from the boundary and the fluid velocity at the boundary, and this velocity difference occurs over some finite distance. This velocity gradient is the source of the viscous drag – the velocity gradient creates a shear stress on the fluid (see definition of viscosity above) that resists the motion of the fluid. It should be noted, however, that while the action of viscosity always causes drag, not all drag is due to viscosity.

 

Reynolds number

 

How important is viscosity in a given flow? That depends on the dimensionless quantity called the Reynolds number (Re): Re ≡ ρvL μ = vL ν

 

where L is a characteristic length scale of the flow which has to be specified. (Note that the symbols “v” for velocity and “ν” (Greek letter nu) for kinematic viscosity are similar, be careful!) In the case of a wing, L is usually chosen to be the length of the wing in the streamwise direction (which is called the cord of the wing.) For flow in a pipe, L would be the pipe inside diameter. For flow around a cylinder or sphere, L would be the diameter of the cylinder or sphere. Also, the fluid velocity v changes as the fluid approaches the object, so v is chosen to be the value far away from the object. For flow inside pipes, v is the average velocity of the fluid, i.e. the volume flow rate (gallons per minute, m3/sec) divided by the cross-section area of the tube. The standard catechism of fluid mechanics states that “Reynolds number is the ratio of inertial forces to viscous forces” but this is nonsense. First of all there is no such thing as “inertial forces” in mechanics. Second, ρvL is not a unit of force, nor is μ. Here’s my interpretation of Re. Re-write

 

The first interpretation notes that ρv2/2 is the dynamic pressure noted above in the context of

 

Bernoulli’s equation and the velocity gradient ∂v/∂y is proportional to v/L, thus the shear stress τ ~

 

μv/L. The second interpretation (which is my personal favorite) notes that the time scale for any

 

type of diffusion process is L2/D where D is the diffusion coefficient for that process (the viscous diffusivity ν in this case), and the time scale for the fluid to move a distance L is simply L/v. So the second interpretation states that the Reynolds number is the ratio of the time for the momentum (or lack of momentum, as in a stationary wall with a fluid moving past it) to diffuse across a distance L to the time for the fluid to move a distance L.

 

Why is Reynolds number useful? Besides determining whether viscosity is important or not, it

 

allows one to employ scaling. For example, suppose you want to determine the drag coefficient CD (another dimensionless number, to be discussed shortly) on a 5 meter long car at 30 m/sec (about 67 mi/hr) and you don’t have a wind tunnel large enough to put a real car it in, but you have a water channel that is big enough for a 1/5 scale (1 meter long) model of a car. The kinematic viscosity of air at ambient temperature and pressure is about 1.5 x 10-5 m2/s and that of water is 1.0 x 10-6 m2/s.

 

Then by choosing the velocity of water in the water channel to get the same Reynolds number, you can obtain a valid measurement of CD:

 

Re = vairLair

 

ν air

 

= vwaterLwater

 

ν water (30 m / s)(5 m)1.5×10−5 m2 / s =

 

(vwater )(1 m)

 

1.0 ×10−6 m2 / s vwater =10 m / s

 

 

 

So a 5 meter long model in air moving at 30 m/s will have the same behavior as a 1 meter long

 

model in water moving at 10 m/s. There may be other advantages to using water, e.g. the use of

 

fluorescent dye molecules that make it easier to visualize the flow using a sheet of laser light.

 

Navier-Stokes equations As previously stated, fluid mechanics is just F = dp/dt = d(mv)/dt applied to a fluid (here p = mv ismass x velocity, i.e. the momentum of the fluid). Note that the force F, momentum p and velocity v are all vectors, hence the boldface notation. The set of equations that describe F = d(mv)/dtapplied to a fluid, including viscosity effects, is called the Navier-Stokes equations, shown here in 2dimensions, for an incompressible fluid (ρ = constant) that follows Newton’s law of viscosity (τ =μ∂v/∂y as described above):

 

Here vx and vy are the components of fluid velocity vector v in the x and y directions, respectively.

 

Of course, at every point (x, y) in the flow, the velocity components vx and vy may be different. The left-hand side of the first two equations is basically the d(mv)/dt = m(dv/dt) + v(dm/dt) terms, i.e. the rate of change of momentum of the fluid. In particular, the ρ(∂vx/∂t) terms are just m(dv/dt) = mass x acceleration and the ρv(∂vx/∂x) terms are just v(dm/dt), i.e. the increase or decrease in momentum within an infinitesimal volume due to an increase or decrease of mv within the volume. The right-hand side of the first two equations is the forces acting on the fluid due to pressure, gravity, and viscosity. The third equation is required to conserve the mass of fluid, i.e. it basically says that for a fluid of constant density, the rate of volume flow into an infinitesimal volume must equal rate of volume flow out of that volume.

 

These equations are very difficult to solve for all but the simplest situations. Certainly we will not try to do it in this course. For the purposes of this course, the key points to note about the Navier- Stokes equations are:

 

1. The first two equations are just expressions of F = d(mv)/dt applied to a fluid

 

2. There are two equations because momentum is a vector and thus there are x and y

 

components; for a three-dimensional system another equation for the z component of

 

momentum would be required

 

3. The terms on the right hand side of the first two equations are just the forces F broken

 

down into their components in the x and y directions

 

4. The terms on the left hand side of the first two equations are just the rate of change of

 

momentum d(mv)/dt broken down into their components in the x and y directions.

 

5. There are three equations for the three unknowns vx, vy and pressure P. (Unlike velocity, P

 

is a scalar so it doesn’t have x and y components.)

 

6. The equations are linear except for the vx(∂vx/∂x), vy(∂vx/∂y), vx(∂vy/∂x) and vy(∂vy/∂y)

 

terms. This nonlinearity is very significant because

 

a. It makes fluid mechanics difficult – the nonlinear terms are responsible for very

 

complicated phenomena such as flow instabilities, turbulence and shock waves.

 

b. It makes fluid flow fundamentally different than linear systems. For example, if I

 

have one solution to the Navier-Stokes equation, call it v1(x,y), P1(x,y) and a second

 

solution v2(x,y), P2(x,y), it is generally NOT the case that v1(x,y) + v2(x,y), P1(x,y) +

 

P2(x,y) is also a solution. As an example of a linear system, consider traveling waves

 

on a string. A rightward-traveling wave and a leftward-traveling wave can pass

 

through each other without any change in the waves after the passage. However, a

 

rightward-traveling flow structure (say, a spinning vortex) and a leftward-traveling

 

flow structure will interact with each other in such a way that each will be

 

permanently changed by the interaction.

 

Laminar and turbulent flow

 

When Re is low, which means that viscous effects are relatively important, the flow will be steady and smooth, which is called “laminar flow.” At higher Re, viscosity is not strong enough to suppress the instabilities (due to the nonlinear terms in the Navier-Stokes equations) and the flow becomes turbulent. While you have an intuitive feel of what turbulence is, a precise definition of what is or is not turbulent is not a simple matter.

 

The Reynolds number at the transition from laminar to turbulent flow depends on the type of flow, for example:

 

• Flow in circular pipes: Re = vd/ν ≈ 2,200 (v = mean velocity of flow in the pipe; d = inside

 

diameter of pipe)

 

• Flow along a flat plate: Re = vL/ν ≈ 500,000 (v = velocity of flow far from the plate; L =

 

distance from the “leading edge” of the plate.)

 

Lift, drag and fluid resistance

 

Lift and drag coefficients

 

Any object moving through a fluid will experience a force (FD) in the direction opposing the motion.

 

This force is called drag. Recall the definition of drag coefficient (page 12):

 

FD = 1

 

2 CDρv2A

 

where CD is the drag coefficient, ρ is the fluid density, v the fluid velocity far from the object and A is the cross-section area of the object.

 

An object moving through a fluid may also experience a force in the direction perpendicular to the direction of fluid motion. This force is called lift and is defined in a way similar to drag:

 

FL = 1

 

2 CLρv2A (Equation 56) where CL is the lift coefficient. While all objects moving through a fluid experience drag, only some

 

will experience lift. The main goal of aircraft wing design is to maximize the lift to drag ratio, i.e. CL/CD. A glider may have a lift to drag ratio of 50, whereas commercial passenger aircraft wings are in the range 15 – 20 (which is about the same as an albatross).

 

Flow around spheres and cylinders

 

In the case of laminar flow at very low Re, and only in this case, the drag coefficient CD on a

 

sphere is equal to 24/Re (for laminar flow at low Re only – got it???). Combining this result with the definition of drag coefficient and definition of Re, we obtain

 

Fdrag = 3πμvd (laminar flow around spheres) (Equation 57)

 

If the sphere is moving due to gravity alone, the buoyant force Fbuoyant = (ρfluid-ρsphere)gV =

 

(ρfluid-ρsphere)g(4π/3)r3 = (ρfluid-ρsphere)g(π/6)d3. Note that the buoyant force does not depend on v,

 

but the drag force does. Thus a dropped sphere will initially accelerate until its velocity is just that required for the drag force to equal the buoyant force, at which point there is no acceleration, and the velocity has reached a constant value called the terminal velocity. For the case of the sphere, equating Fdrag and Fbuoyant we obtain:

 

vterminal = gd2(ρfluid - ρsphere)/18μ (laminar flow around spheres) (Equation 58)

 

where the + sign is consistent with the fact that if ρfluid > ρsphere, the sphere moves upward (positive v).

 

The above terminal velocity is only valid for laminar flow. For turbulent flow, there is no simple analytical relationship between Re and CD, so one must resort to experiments or detailed (and difficult) computer simulations of the Navier-Stokes equations. Figure 20 shows a comparison of the actual CD vs. Re with that predicted by the low-Re laminar-flow model. It can be seen that the relation for laminar flow is reasonable up to about Re ≈ 3 but at higher Re, the flow is NOT laminar and thus the laminar flow result CD = 24/Re does not apply. As one would expect, CD is higher with turbulent flow. Note also that at Re ≈ 3 x 105 there is a sudden decrease in CD.

 

To provide a physical explanation of the drag coefficient plot, note that, apart from the small dip

 

near Re ≈ 3 x 105, at high Re, CD is close to 1 and doesn’t change much with Re. This is because at high Re, the high momentum (relative to viscous effects) of fluid causes the flow behind the sphere separates and there is a region behind the sphere with v ≈ 0. Thus, most of the dynamic pressure (=.ρv2) of the flow is lost, thus according to Bernoulli’s equation (which does not strictly apply because the flow is neither steady nor inviscid, but is still useful for estimation purposes) the pressure on the downstream side of the sphere is higher than that on the upstream side by .ρv2. Hence, the net force on the sphere due to this separation-induced drag is FD = .ρv2A, and thus the drag coefficient = FD/(.ρv2A) = (.ρv2A)/(.ρv2A) = 1. Thus, for any blunt object at high Re, CD is usually close to 1 (another example of “that’s easy to understand, why didn’t somebody just state that?”) At low Re (less than about 10 for the sphere), the drag is predominantly viscous and thus CD is higher.

 

Figure 20. Drag coefficients as a function of Reynolds number for spheres and cylinders.

 

A similar CD vs. Re plot for cylinders in cross-flow (i.e. with the flow in the direction perpendicular to the axis of the cylinder) is also shown in Figure 20, but for cylinders there is no simple analytical relationship analogous to the CD = 24/Re result for laminar flow over spheres. Note again the sudden decrease in CD at almost the same Re as for spheres. Note that a 3 cm golf ball hit at 70 m/s in air has a Reynolds number of (70 m/s)(0.03 m)/(1.5 x 10-5 m2/s) = 1.4 x 105. Dimpling the golf ball decreases the transition Re somewhat, and thus enables a lower CD (actually the dimpling also increases the lift due to the backspin on the ball, but that’s beyond our scope.)

 

 (b) Samantha has an unusual skydiving style. She free-falls lying perfectly straight horizontally,

 

and her shape can be treated as roughly that of a cylinder 5 ft long and 2 ft in diameter. She

 

weighs 125 lbf with all her gear. Assuming that her drag coefficient can be modeled as that

 

of a circular cylinder, what is her terminal velocity? To do this problem you will have to

 

1) Guess a terminal velocity

 

2) Compute her Reynolds number

 

3) Look up her drag coefficient in Figure 20.

 

4) Compute her drag force

 

5) Does her drag force equal her weight? If not, adjust your guess of terminal velocity

 

and go back to step 2.

 

1) “Guess” v = 113 mph = 50.51 m/s

 

2) Reynolds # Re = vd

 

Ν = (50.51m / s)(2 ft)(m / 3.281ft)

 

1.5×10−5m2 / s = 2.053×106

 

3) From Figure 20, CD ≈ 0.4

 

4) Compute drag force:

 

FD = 1

 

2 CDρv2A = 1

 

2 (0.4)(1.18kg / m3 )(50.51m / s)2 (2 ft)(5 ft)(m / 3.281ft)2

 

= 559.4N =125lbf

 

5) Does her drag force equal her weight?

 

Yes, 125 lbf = 125 lbf so mission accomplished.

 

Flow through pipes

 

For flow through pipes, drag coefficient is not used because what we’re really interested in is not the drag force but rather the pressure drop (ΔP), and thus a slightly different quantity called the friction factor (f) is used to quantify the effect of viscosity on the flow in the pipe:

 

f ≡ ΔP ρv2 2 Ld where v is the average velocity of the fluid flowing thought the pipe, ρ the fluid density, L is the length of the pipe and d its diameter. For laminar flow only in pipes, f = 64/Red where Red = ρvd/μ is the Reynolds number based on pipe diameter d, not pipe length L!, thus ΔP = (64/Red)(ρv2/2)(L/d) = (64μ/ρvd)(ρv2/2)(L/d) = 32μvL/d2 (Equation 60).

 

Sometimes it’s more convenient to deal with volume flow rate (Q) rather than velocity (v). Q is the velocity multiplied by the cross-section area of the pipe, thus Q = vπd2/4. Thus we can write one last relation:

 

ΔP = (128/π)μQL/d4 (laminar flow only!) (Equation 61).

 

Note the significance of this result: if you double the flow rate Q or the length of the pipe L, the

 

pressure drop doubles (makes sense.) Also, for a given flow rate Q, if you double the diameter of

 

the tube, the pressure drop decreases by a factor of 16! So use a bit bigger pipe in your plumbing

 

design!

 

The results leading to the last 2 equations assumed f = 64/Red and thus are valid only for laminar flow. For turbulent flow, the friction factor depends not only on Red but also the roughness of thepipe wall, which is characterized by a roughness factor = ε/d, where ε is a measure of the roughness (i.e.height of the bumps on the wall) and d is (as always) the pipe diameter. The combined effects ofroughness and Red are presented in terms of the Moody chart (Figure 21).Figure 21. “Moody Chart” showing the effect of Red (that is, Reynolds number based on pipe diameter not length) and surface roughness ε/d on the friction factor (f).

 

0.001

 

0.01

 

0.1

 

1

 

1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08

 

Friction factor (f)

 

Reynolds number (Red)

 

Laminar Turbulent

 

ε/d = 0.1

 

ε/d = 0.01

 

ε/d = 0.001

 

ε/d = 0.0001

 

ε/d = 0.00001

 

Smooth pipe

 

Laminar

 

flow

 

Smooth pipe

 

Note that laminar flow prevails up to Red = 2,200 (this value is essentially independent of the pipe roughness factor), then for higher Red, CD increases suddenly but in a way that depends on the pipe roughness – as one would expect, rougher pipes have higher CD. It’s remarkable (to me, anyway) that at high Red a tiny amount of roughness has a huge effect on f. For example, at Red = 108, f increases by a factor of 3 as one changes from a perfectly smooth pipe (ε/d = 0) to ε/d = 0.001. In other words, a roughness of one part in 1000 increases the pressure drop by a factor of 3. Size does matter!

 

Alternatively, if you don’t like using the Moody diagram, the following empirical formula for

 

turbulent flow can be used (for laminar flow, use f = 64/Red as mentioned above):

 

f = −2log ε / d

 

3.7 + 2.51

 

Red f

 

" # $$ % & '' (turbulent flow) (Equation 62)

 

but note that this formula has f on both sides of the equation, and you can’t simplify it any further, so for a given Red and ε/d, you have to guess a value of f and see if the right and left hand sides of the equations are equal, and adjust your guess of f until the two sides are equal (this is called a transcendental equation, one that cannot be solved in closed form).

 

Example: A 50 foot long garden hose has an inside diameter of 5/8” and a roughness (ε) of 1/32”. Water flows through the pipe at a velocity of 3 ft/s.

 

a) What is the flow rate in gallons per minute?

 

Flow rate = velocity x cross-sectional area

 

= 3 ft/sec x (π/4) (0.625 inch)2 x (ft / 12 inch)2 x (7.48 gallon / ft3) x (60 sec / min)

 

= 2.87 gallon / min

 

Do you think I did these calculations by hand? No way! I used an Excel sheet (double click to

 

open). The cells shaded in blue are the things you change, and the other cells are calculated values, except for the “Friction factor (guess)” cell, which you have to adjust until the left-hand side (LHS) and the right-hand side (RHS) of the equation for the friction factor are equal, and thus the “fraction error” goes to zero. (You can also use Excel’s “goal seek” feature to do this adjustment automatically.)

 

Pipe dia (in) Pipe len (ft)

 

Roughness

 

(in)

 

Viscosity

 

(m^2/s)

 

rho

 

(kg/m^3)

 

0.625 50 0.03125 1.00E-06 1000

 

Pipe dia (m) Pipe len (m) epsilon

 

0.01587423 15.23925632 0.05

 

Velocity

 

(ft/s) Velocity (m/s) Re

 

3 0.914355379 14515

 

Friction

 

factor

 

(guess) LHS RHS

 

fraction

 

error

 

0.07309956 3.698645 3.698296 9.4274E-05

 

Delta P

 

(N/m^2)

 

Pressure drop

 

(lbf/in^2)

 

29335.00 4.26

 

Compressible flow

 

All of the above discussion of fluid mechanics relates to cases with constant density (ρ), which is

 

certainly reasonable for liquids (e.g. water) under most conditions and even air if the velocity (v) is “small enough”. How small is small enough? We have to compare U to something else that also has units of velocity. That “something else” turns out to be the speed of sound (c). The ratio of these is the Mach number (M), i.e.

 

M = v/c Equation 63.

 

For an ideal gas, the sound speed c is given by the formula

 

c = (γRT)1/2 where

 

• γ is the specific heat ratio of the gas (≈1.4 for air at ambient temperature, but may be as low as

 

1 for gas molecule with many atoms, and as high as 5/3 for a monatomic gas like helium)

 

• R is the gas constant for the specific gas of interest = /M, where is the universal gas

 

constant = 8.314 J/mole K and M is the molecular mass of the gas (in kg/mole, = 0.02897

 

kg/mole for air, thus R = 287 J/kgK for air.)

 

• T is the gas temperature (in K of course)

 

How does Mach number affect density (ρ), temperature (T) and/or pressure (P)? That depends on

 

the process the gas experiences as it accelerates or decelerates. A detailed discussion of compressible gas dynamics is way beyond the scope of this course, but I’ll give you the results for the simplest case of one-dimensional steady flow of an ideal gas in a duct of changing area A with constant specific heats (CP and CV) between locations 1 and 2 assuming no heat transfer, no friction and no shock waves (we call this special case “isentropic flow,” meaning no change in the entropy of the gas) as well as no potential energy (elevation) change:

 

These equations are plotted in Figure 22 below.

 

While a lot of simplifying assumptions were made (note all the underlined words above), this

 

“isentropic flow” is still useful as the simplest model of flow in nozzles of jet and rocket engines, as well as intakes in jet engines. Note that as Mach number increases (for example, during expansion in a nozzle), pressure, density and temperature all decrease. However, to obtain transition from subsonic (M < 1) to supersonic (M > 1) flow, the area must pass through a minimum, i.e. a throat, which occurs at M = 1. Thus, rocket nozzles must have an hourglass shape in order to accelerate the exhaust to high Mach numbers and therefore produce the maximum possible thrust. Often the areas are referenced to the minimum area at the throat (A*) where M = 1, in which case

 

Isentropic flow

 

0.01

 

0.1

 

1

 

10

 

100

 

0 1 2 Mach number 3 4 5

 

Figure 22. Plot of pressure, density, temperature and area as a function of Mach

 

number for one-dimensional, isentropic flow of an idea gas with γ= 1.4. The “t”

 

subscript indicates the pressure, density or temperature when M = 0. (Double-click

 

to open Excel spreadsheet).

 

One important but often overlooked point about the above equation: the Mach number M is the

 

local (at station 1 or 2) flow velocity U divided by the local sound speed (which depends on the local temperature.) So you can’t divide the local velocity by the sound speed at ambient temperature to get the Mach number! That is, M1 = v1/(γRT1)1/2 and M2 = v2/(γRT2)1/2but you can’t say M2 = v2/(γRT1)1/2 !

 

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