# The Key to Understanding Engineering Dynamics

## What are the best steps for understanding engineering dynamics?

Dynamics is often difficult for engineering students (or even engineers).  But there is a very important key to understanding engineering dynamics.  The key is to focus on a systematic approach to solving dynamics problems. This blog post will explain how to use the systematic approach.

## Understanding Engineering Dynamics Categories

You may have noticed that there can be many different ways to solve any given dynamics problem.  I have taught dynamics several times in my academic career, and I have noticed that students have a hard time finding the best method to use for a given problem.

Any dynamics problem will fall into one of four categories:

1. Particle kinematics
2. Particle kinetics
3. Rigid body kinematics
4. Rigid body kinetics

The key for fully understanding engineering dynamics and solving dynamics problems is to identify the proper category and understand the solution procedure for that category.

## Particles and Rigid Bodies

The first step to understanding engineering dynamics problems is to know the difference between particles and rigid bodies. It is common to hear the word particle and automatically assume something small.  However, that is not necessarily the case. In dynamics problems it is not uncommon to have something like a car or train be a particle. The defining difference between particles and rigid bodies is rotation. If the object is not rotating about its own centroid it can be treated as a particle because the object’s rotational properties do not matter. If the object is rotating about its own centroid, the rotational properties are important and it is considered a rigid body.

## Kinematics and Kinetics

One very important area in understanding dynamics is the distinction between kinematics and kinetics. The determining factor here is force. Kinematics problems concentrate on motion without any regard to the force required for that motion. In other words, kinematics focuses on the geometry of motion.  Kinetics problems involve forces.

## The Systematic Approach to Understanding Engineering Dynamics

So, now we can put some ideas together and define the systematic approach. When you read a dynamics problem, the very first step is to ask yourself ‘is this problem dealing with particles or rigid bodies?’. Remember, if the object is not rotating about its own center of mass it is a particle. Also, rigid body problems will give additional information about the object such as moment of inertia.
Once you determine if the problem relates to particles or rigid bodies, the next step is to ask ‘is this a kinematics or kinetics problem?’. Force is the key here.

### Let’s look at some examples:

Example 1

The 5 inch diameter disk shown rolls on a surface without slipping. The velocity of point O is 1.5 ft/s to the right. For the position shown calculate the absolute velocity of point P.

So, the first thing we need to determine is if this is a particle or rigid body problem. Because the disk is rotating about its own center of mass it is a rigid body. The next step is to determine if the problem is in the category of kinematics or kinetics. The problem only asks about the motion (position and velocity for this problem) and does not ask or include information about forces. Therefore, the problem is a kinematics problem. Now we know the problem is a rigid body kinematics problem.

Example 2

A man completes a long jump of a distance of 21 feet. He has a horizontal velocity of 28 ft/sec at the time of takeoff. What is the vertical velocity at takeoff?
This problem does not include any rotation about the center of mass for the man. Also, notice the problem did not provide any information about the man (mass, center of gravity location, etc.). Therefore, the man can be considered a particle. The problem only includes information about motion (position and velocity) with no information about forces. So, the problem is a kinematics problem. Now we know the problem category is particle kinematics.

Example 3

A 70 kg box slides down a ramp that has an angle of 20 degrees from the horizontal and has a coefficient of kinetic friction of 0.30. The box is given an initial velocity of 3 m/s. Calculate the velocity of the box after it travels 8 meters down the ramp.
The box in this problem is not rotating about its center of gravity, so it can be treated as a particle. The question only asks for velocity, but there is a force involved in the question. The force of friction will be slowing down the box. Because a force is involved in the solution, the question is a kinetics problem. So, the problem is in the category of particle kinetics.

Example 4

The connecting rod AB for the piston assembly shown weighs 1.4 lb. The center of gravity is located 7 inches from point B, and the radius of gyration about the center of gravity is 2.5 inches. The piston head weighs 2.4 lbs. The engine is running at a constant speed of 2400 rpm. Calculate the force on the piston pin at B for the position shown.

The problem involves a piston connecting rod that is rotating about its own center of mass (it is in general plane motion). The problem also gives information about the center of mass location and the radius of gyration (which will give moment of inertia). Therefore, the problem is a rigid body problem. The question specifically asks for force, which is a kinetics problem. So, the problem is a rigid body kinetics problem.

## What’s Next?

Understanding engineering dynamics first requires the ability to determine the type of problem to be solved. This blog was all about understanding the four categories. You should be able to read a dynamics problem and identify the proper category. Future blogs will go through solution methods for each category.

Get more tips like this in the study guide

# Thermodynamic Phase Changes

This blog post will introduce the concepts of Thermodynamic phase changes of pure substances.  Before proceeding, we need to define a pure substance as a substance with fixed chemical composition throughout.  Note that a pure substance does not need to be a single chemical element.  Water, for example, is a pure substance but contains more than one chemical element.

# Phases and Thermodynamic Phase Changes

It is commonly known that substances can occur in different phases (solids, liquids, and gases). When substances are heated or cooled they can change from one phase to another. A phase change from a solid to a liquid is known as melting, and a phase change from a liquid to a gas is known as boiling (or vaporization). In the reverse direction, a phase change from a gas to a liquid is known as condensation and from a liquid to a solid is known as freezing.

The figure below illustrates phase changes in water plotted as temperature vs. heat addition.  From state a to state b the water is a solid.  Once the temperature reaches zero degrees Celsius (the melting point for water), adding more heat will result in a phase change.  The phase change from solid to liquid is represented between points b and c on the diagram.  The region between points c and d represent the liquid phase.  More heat is added until the temperature reaches 100 degrees Celsius (the boiling point for water).  As more heat is added the phase change will occur between a liquid and a gas, which is illustrated between points d and e.  The region between points e and f is the gas phase.  It is important to note that temperature remains constant during phase changes.

Thermodynamic Phase Changes

Consider now the phase diagram shown below, which is a plot of temperature (T) versus specific volume (v).   The process from points 1 to 2 occur in a liquid phase, the phase change is shown between points 2 and 3, and from points 3 to 4 the substance is a gas.  Notice again the transformation from one state to another, as shown in Figure 3 from state 2 to state 3, takes place with no change in temperature (horizontal line on the T-v diagram).  The diagram will be used to define several important terms associated with phase change processes in pure substances.

When a substance occurs in a liquid state it is called a compressed liquid (or subcooled liquid). As heat is added to a liquid, it will reach a point where any additional heat added would result in some of the liquid vaporizing. A liquid at that point where it is about to vaporize is called a saturated liquid. State 2 represents the saturated liquid state on the figure.

The line between points 2 and 3 represent the substance in the phase change between a liquid and a vapor. That region is defined as a saturated liquid-vapor mixture, because both the liquid and vapor phase exists together in equilibrium.

A vapor that is not about to condense is known as a superheated vapor. As heat is removed from the vapor it will reach the point, known as a saturated vapor, where the removal of any additional heat will cause some of the vapor to condense. At a given pressure, the substance will begin to boil at a temperature known as the saturation temperature. Or, if temperature is held constant, the substance will start to boil at the saturation pressure.

The process line 1-2-3-4 occurs at a constant pressure.  If the pressure is raised or lowered a similar process curve will exist, but the locations of the saturated liquid point and the saturated vapor point would change.  The points for different pressures would form a curve defined by the saturated liquid line and the saturated vapor line.

## Hope this helps!

I hope this short introduction to Thermodynamic Phase Changes helped!  Leave comments for ideas of other blog posts you would like to see.  Learn more like this in my FREE study guide available here!!

# Specific Heat Capacity

## Get the Specifics of Specific Heat Capacity

Today’s concept discussion is on specific heat capacity.  Basically, different fluids will ‘heat up’ differently (more specifically they store heat differently).  In Thermodynamics you need a way to take the fluid’s properties for storing heat into account.  That’s where specific heat comes in.

Before we discuss specific heat capacity we need to define heat capacity.  Heat capacity, as shown in the formula below, is the heat added divided by the change in temperature.  In other words, it is a measure of how the temperature changes for a given amount of heat added.  Obviously, the temperature change of the fluid will also depend on the amount of fluid.  Therefore, the specific heat (lower case c) is the heat capacity per unit mass.

### Specific Heat Capacity of Gases

For gases we need to define two separate specific heat values.  The temperature change will depend on whether the volume is held constant or if gas is able to expand.  We use a subscript v to define specific heat capacity at a constant volume.  A subscript p is used to define the specific heat capacity at a constant pressure.  The specific heat ratio is defined as the ratio of the constant pressure specific heat to the constant volume specific heat.

### How do we use Specific Heat Capacity in Thermodynamics?

There are several types of problems in Thermodynamics that use specific heats.  It can be used to determine final equilibrium temperatures.  It is used to determine the change in internal energy of a gas when heat is added or subtracted.  For the PE exam you want to have a table of specific heat values for common gases.

## What do you think?

I hope this short blog helped define specific heat capacity.  It is commonly used in Thermodynamics, so it is important to understand its use.  Let me know if you have thoughts or questions about this topic.  Also, head over to www.MechanicalPEAcademy.com/Facebook to like my Facebook page.  It has updates and additional content.

# Nomograms for the PE exam

## Nomograms: Antiquated technology or useful tool?

Many of today’s engineers would not consider using nomograms for the PE exam.  It would not surprise me if many of you reading this blog post have no idea what a nomogram (also called a nomograph) is.  This blog will quickly introduce them to you to help answer the question… Nomograms: Antiquated technology or useful tool?  Though they may be antiquated, you may find that they allow for very quick solutions to some types of problems.  Therefore, this blog is an introduction to the idea of using them for the PE exam to help you solve problems faster.

## What is a nomogram?

A nomogram is a graphical tool used to perform calculations, and they exist for numerous types of calculations.  They are essentially a diagram that gives a quick, approximate solution to a mathematical function.  A nomogram will have three or more lines (or curves), each for a different variable in the function.  The lines are placed on the nomogram to give the appropriate numerical relationships for the function.  If you know two or more of the variable values, you can use a straightedge to line up the values to graphically determine the unknown variable.  There will be some error, but it can be a very quick method for doing the calculations!

You can easily search the internet to get more detail on using (or making) nomograms.  There are also books teaching the lost art of nomography.  You can find books that discuss the history of nomography as well as numerous areas of application.  Omer Blodgett has a couple of great books, Design of Welded Structures and Design of Weldments, which have several great nomograms for topics in mechanics of materials.

## Here is a quick example

The figure below shows a nomogram used to solve the quadratic equation.  Though I did construct this particular nomogram I did not develop the idea (it has been around for a long time).  I am also not going to give details on how it was developed, but I will give it only as a representation of how they can be used.  The horizontal axis is the value of ‘a’ in the equation, the vertical axis is the value of ‘b’ in the equation, and the circle gives the two roots that solve the equation (values of ‘x’).  The dashed line shows an example with a = 1.5 and b = -2.  A line is drawn connecting those two points (the dashed line).  The roots are determined by the values on the circle where that line crosses.  The solutions for that particular example are x = 0.85 and x = -2.35, which are read from the values on the circle.

## These nomograms sound obsolete… are they really useful?

OK, I would agree that computers have made nomograms essentially obsolete.  However, you cannot use a computer during the PE exam.  I would also agree that the power of calculators make nomograms obsolete, but remember that you have a limited selection of acceptable calculators for the PE exam.  I am also not suggesting you use nomograms for everyday engineering design (though you could).  The real question is whether or not they can save valuable time during the PE exam.

You may not have a need for a nomogram to solve the quadratic equation, but they exist for far more complex problems.  Using them for the PE exam could allow for very rapid calculations for some common types of problems.  Machine Design Fundamentals by Hindhede, for example, provides a nomogram relating power, torsional stress, and shaft diameter that rapidly determines required shaft diameter.  Design of Welded Structures by Blodgett has rather complex nomograms that can be used to determine the required section modulus of beams (or required moment of inertia), which includes information about beam end conditions, beam length, beam loading, and allowable stress.  The same text also includes nomograms for deflection of curved beams, fatigue, torsional resistance, column effective lengths, and design aids for plate girders.  My textbook, Machine Analysis with Computer Applications (the same figure is in Mechanisms and Dynamics of Machinery by Mabie and Reinholtz), has a nomogram to determine maximum pressure angle in disk cams with roller followers.  I have also seen nomograms for fluid flow calculations to aid in pipe sizing.  All of these nomograms can save valuable time on the PE exam.  You are not limited by nomograms available in books… you can always try making some of your own for different calculations.

## Do you think nomograms are useful for the PE exam?

I hope you found this information useful.  The PE exam is all about working problems quickly and efficiently.  Nomograms can provide quick estimates of the answer (which is usually accurate enough for a multiple choice exam like the PE exam).  What are your thoughts?  Do you ever use nomograms for design?  Do you have recommendations of good nomograms to use for the PE exam (or good books that contain useful nomograms)?  Leave a comment letting everyone know and please share the information by tagging the share buttons.  Also, take a minute to go over to the Mechanical PE Academy Facebook page to get updates and exclusive content.

# Four bar Mechanism Kinematics Shortcut

## Four bar Mechanism Kinematics Shortcut to save time on the PE Exam

For mechanical engineers the PE exam will definitely contain some dynamics problems.  If you choose to take the Mechanical Systems and Materials depth session (PM portion of the exam) you will also have problems related to machine analysis.  Both dynamics and machine analysis commonly contain problems related to the kinematics of four-bar mechanisms.  This blog will give some details on a cool four bar mechanism kinematics shortcut for calculating velocities for four bar mechanisms.  It is not a method that is frequently taught in dynamics or machine analysis courses, but it can definitely save some time on the PE exam.

## When can I use this shortcut?

A common type of problem has a four-bar mechanism with a given input angular velocity of the driving link.  The problem requires you to determine the angular velocity of the output link.  Even though this may seem fairly basic, the problem can be time consuming… and you don’t have a lot of time to waste during the PE exam.  This shortcut method is great for this type of problem!  This method also allows for a very quick estimation of the answer… which may be all you need on a multiple choice exam like the PE exam.

## What is the method?

OK… let’s look at this method.  The method is not new and has actually been around for a long time.  However, it is not commonly taught.  The method is based on the fact that the velocity vector of both coupler endpoints must have the same component along the long axis of the coupler.  Confusing?  Probably…  Let’s look at this in more detail with some illustrations.

## Terminology Review

Before I get into the method let me review some quick terminology for four-bar mechanisms.  The figure below shows a typical four-bar mechanism.  Link 1 is the ground link, which does not move.  We will call link 2 the driving link (or input link).  Link 3 is the coupler, and link 4 is the output link.

## Velocity Calculations

Now that we have reviewed the names of the links, let’s discuss how to calculate velocities using this method.  Say we know the input rotational velocity of the driving link (link 2).  The velocity of point A (part a of the figure below) can be calculated.  Link 2 is in pure rotation, so the velocity of point A is the product of the length of link 2 and the rotational velocity in radians per second.  Look now at part b of the figure.  Take the component of the velocity vector at A along the length of the coupler (parallel component).  That must also be the length of the component at B (equal parallel components).  The velocity at point B can easily be found as the vector perpendicular to link 4 with that parallel component.  The rotational velocity of the output link can be determined from that velocity if needed.

## Final Thoughts

The nice thing about this method for the PE exam is that it is a quick graphical method.  Drawing a simple sketch (roughly to scale) can allow you to estimate the velocity with minimal work.  Because the PE exam is a multiple choice exam, that quick estimate may be all that is required to get the answer.

I hope this quick introduction to this method helped you work mechanism problems more efficiently.  More details on this method, along with examples, are provided in my textbook Machine Analysis with Computer Applications (you can get it on Amazon here).  As a general note, the figures used in this post are from that textbook.

# Integration by Parts – The Fast and Easy Way!

Work smarter, not harder

~Scrooge McDuck

# Integration by Parts

In this blog you will learn how to do integration by parts the fast and easy way!  First of all… the majority of problems on the PE exam will not require calculus, but it is possible.  Because it may be required… and because success on the PE exam is all about working problems quickly… I wanted to share this VERY SIMPLE and VERY FAST method for doing integration by parts!  In fact, it is so simple that you will wonder why this method is not commonly taught in calculus courses (I don’t know the reason either).  So even if you don’t need this method on the PE exam it is a great tool to have… and you can use it to impress your friends!

Why would I care… I don’t solve calculus problems by hand!

So you may be thinking… if any integration appeared on the PE exam I would look up the solution or use my calculator to solve the integral.  Though I do recommend you have a good math reference book for the exam (I highly recommend this one as a good overall math reference), I honestly believe that you could solve some basic integration by parts problems faster with the method I am about to show you because it takes a while to find the solution in the book.  As far as using the calculator, you must remember that the NCEES has a limited list of approved calculators.  So the calculator may not have all the functions for calculus.

What is integration by parts?

Before I get into the details of this VERY SIMPLE and VERY FAST method for doing integration by parts, I want to briefly remind you about the basic idea of integration by parts.  Integration by parts is based on the concept of the product rule of differentiation.  Integration by parts is a method commonly used when the function to be integrated is a product of an algebraic function and a trigonometric or exponential function.

OK… so let’s just get to the point!  What is this VERY SIMPLE and VERY FAST method?  The method presented here is one that I call the tabular method, because a simple table is developed to quickly develop the solution to the integral.  One nice advantage is that the solution is developed in one step regardless of the problem (no more repeated integration by parts in the same problem… which typically happens).

I am going to explain the process with an example.

I will illustrate the process by solving the integral shown.  The function is a product of an algebraic function and an exponential function, so integration by parts applies.

Step 1: Make the table

The first step is to construct a very simple table.  The first column is the term in the integral labeled u and the second column is the term labeled dv.  Differentiate the ‘u’ term in the first column and continue until the derivate becomes zero.  Integrate the ‘dv’ term and continue to the end of the first column.

Step 2: Cross-multiply and sum

The second step, which is the last step, is to cross-multiply and sum the results.  Before you do that you need to alternate positive and negative signs (+ and – as shown)… those will be used in the summation step.  Now simply multiply the terms as shown with the red arrows and complete the sum of all the terms (don’t forget the + and – signs).  Add the integration constant C and you are done!  Very fast, very simple, and all completed in one step!

Action Steps

What should you do now?  Go grab a calculus book and look for some example problems on integration by parts.  Solve those problems using this method to prove to yourself how much faster this process is compared to the standard method.

I hope this helped you improve your efficiency in integration!  I know it may be a little confusing from this blog post alone, so if you have questions or comments please let me know.  I can always add a couple more examples to help clarify if needed.  Also, please take a moment to share this post with others!  You can click any of the ‘share the knowledge’ tabs.  That helps me a lot!

# Mechanical PE Exam Topic Concepts Introduction

To acquire knowledge, one must study; but to acquire wisdom, one must observe.”

~ Marilyn vos Savant

Welcome to a new blog series on Mechanical PE Academy!  This blog is the overall introduction to a blog series to help you build your understanding of Mechanical PE Exam Topic Concepts… which is critical to passing the exam.

What is the overall intent of this blog series?

The main focus area for the Mechanical PE Exam Topic Concepts blog series is just as the name implies… it will focus on understanding the concepts of the different topics covered on the Mechanical PE exam.  Too many people preparing for the PE exam work problem after problem after problem and wonder why they do not pass the exam.  I believe the answer is generally that you focused on working problems and did not focus on understanding the concepts needed for solving the problems!  You cannot predict what problems you will need to solve on the PE exam.  Spending all of your study time on working problems only significantly helps you if the PE exam is filled with problems very similar (which will not happen).  You really need to understand the concepts!  You need to change your studying philosophy and ‘think outside of the box’ to focus more intently on understanding concepts.

So am I telling you to not work problems during your review?

No!  You need to work problems… lots of problems!  You should work all different types of problems.  What I am telling you is that you should not focus exclusively on working problems for the sake of working problems.  Don’t think that you have enough time during the PE exam to look through several books to find an example problem that will guide you through the solution for each problem.  Having a good grasp on the concepts will help you solve the problems quickly.

Here is a basic example…

Let’s say you are taking the PE exam and you need to solve a thermodynamics question asking about work done during an isobaric process.  If you are mostly unfamiliar with the concepts of thermodynamic processes you would need to grab your thermodynamics reference book and search for a similar style of problem to guide you through the process.  Understanding the concept, however, you would realize that an isobaric process is simply a process that has constant pressure.  Plotting that process on a p-v diagram and finding the area under the curve (which in this case is the VERY SIMPLE process of finding the area of a rectangle) will give you work done during the process.  Understanding the concepts can drastically expedite the solution process!!  You may think… that example is too easy… there would not be questions on the PE exam that only require me to solve for the area of a rectangle.  Honestly, you would be surprised how easy some questions can be if you have a good understanding of the concepts!  After all, the PE exam is testing you to see if you do understand the concepts!!

So here is the plan…

I plan to post blogs to this series frequently!  How frequently?  I am not sure yet… but frequently.  My goal is to help you understand the concepts so you can work problems faster!  I will try to give good actions for you to take after the blogs to help you reinforce your understanding of the concepts.  Check back often to get new topics.