1. 2014 POE Final Review

2. Design Process Overview
Introduction to Engineering DesignTM
Unit 1 – Lesson 1.1 – Introduction to Design Process
Example Design
Process
There are many design processes that guide professionals in developing solutions to problems. The example that you see here is the design process that we will use for this course and the rest of the Project Lead The Way, Inc. courses you will take.
Project Lead The Way, Inc.
Copyright 2007

3. Engineering & Engineering Technology
Manufacturing
Test and Evaluation
Development
Routine Design
Complex Design
Production
Operation, Service,
And Maintenance
Complex Analysis
Distribution and Sales
Research
Image courtesy of
More Mathematical
Less Mathematical

4. Mechanisms

5. Ideal Mechanical Advantage (IMA)
Theory-based calculation
Friction loss is not taken into consideration
Ratio of distance traveled by effort and resistance force
Used in efficiency and safety factor design calculations
DE = Distance traveled by effort force
DR = Distance traveled by resistance force

6. Actual Mechanical Advantage (AMA)
Inquiry-based calculation
Frictional losses are taken into consideration
Used in efficiency calculations
Ratio of force magnitudes
FR = Magnitude of resistance force
FE = Magnitude of effort force

7. Mechanical Advantage Example
A mechanical advantage of 4:1 tells us
what about a mechanism?
Magnitude of Force:
Effort force magnitude is 4 times less than the magnitude of the resistance force.
Distance Traveled by Forces:
Effort force travels 4 times greater distance than the resistance force.

8. Pulley IMA Movable Pulley - 2nd class lever with an IMA of 2
Fixed Pulley
- 1st class lever with an IMA of 1
- Changes the direction of force
10 lb
5 lb
5 lb
Movable Pulley
- 2nd class lever with an IMA of 2
- Force directions stay constant
10 lb
10 lb

9. Pulley Problem
Figure 5 represents a belt driven system. Pulley B, which has a diameter of 16 inches, is being driven by Pulley A, which has a diameter of 4 inches. If Pulley A is spinning at 60 RPMs, then Pulley B is spinning at ______ RPMs.
Spring 2007 #13 A
G: DIA of B= 16”, DIA of A=4, A RPM=60
U: B RPM
E: DIA-B/DIA- A=RPM-B/RPM-A
S: 16/4=60/B RPM
S: 15=B RPM

10. Gear Problem
The gear train in Figure 3 consists of a 40-tooth (input), 20-tooth, and 30-tooth (output) gear. If the input gear rotates 10 times, how many times will the output gear rotate? (Spring 2009 # 7 Part A Practice)
G:40 g, 20 g, 30 g, Nin 10
U: Nout
E: Nin*R=Nout*R
S: 40*10=Nout*30
S: 13.3= Nout
40/20=2/1 *10= 20
20/30=2/3 *5=20/3

11. Electrical

12. V=IR I=V/R R=V/I Ohm’s Law
Current in a resistor varies in direct proportion to the voltage applied to it and is inversely proportional to the resistor’s value
The mathematical relationship between current, voltage, and resistance
If you know 2 of the 3 quantities, you can solve for the third.
Quantities
Abbreviations
Units
Symbols
Voltage V
Volts
Current I
Amperes A
Resistance R
Ohms Ω
V=IR
I=V/R
R=V/I

13. There is only one path for the electrons to flow
A Basic Circuit
Battery
Source = Battery
Wire
Conductor = Wire
Lamp
Load = Lamp
Switch
Control = Switch
There is only one path for the electrons to flow

14. A Series Circuit Source = Battery Load = 2 Lamps Conductor = Wire
Control = Switch
Electrons can only flow along one path, and MUST go through each component before getting to the next one.
Switch

15. A Parallel Circuit Source = Battery Load = 3 Lamps 3 Lamps Battery
Conductor = Wire
Control = Switch
Switch
Wire
Two or more components are connected so that current can flow to one of them WITHOUT going through another

16. A Series-Parallel Circuit
A combination of components both in Series and in Parallel

17. To measure current with a multimeter,
the “leads” must be placed in
series.
To measure voltage with a multimeter, the “leads” must be placed in
Parallel (across the load).

18. 5a
Figure 1
Figure 2
The images in Figures 1 and 2 show the voltmeter configurations that two
different POE students used to take a voltage reading within a simple circuit.
Only one of the two students was able to measure the voltage value in the
simple circuit. Use the information given in the figures to answer the
following questions.
5a. Which of the two setups (Figure 7 or Figure 8) shows the correct way to
measure voltage? [1 point]

19. voltage value of the power supply? (accuracy = 0) [3 points]
5b. If the amount of current in the circuit is equal to 0.5A, what is the
voltage value of the power supply? (accuracy = 0) [3 points]
To solve the problem use the following equation:
V = I * R ( Voltage = Current * Resistance)
G: I = 0.5 A, R = 16.0 Ω
U: V
E: V = I * R
S: V = 0.5 A * 16.0 Ω
S: V = 8v

20. Thermodynamics

21. Heat Transfer
Conduction: the flow of thermal energy through a substance from a higher- to a lower-temperature region.
Convection: the transfer of heat by the circulation or movement of the heated parts of a liquid or gas.
Radiation: the process in which energy is emitted as particles or waves.
R-Value ability to resist the transfer of heat. The higher the R-Value, the more effective the insulation

22. Thermal Energy Transfer Equations
Introduction to Thermodynamics
Thermal Energy Transfer Equations
Principles Of EngineeringTM
Unit 1 – Lesson 1.2 Energy Forms
Project Lead The Way, Inc.
Copyright 2008

23. Thermal Energy Transfer Equations
Introduction to Thermodynamics
Thermal Energy Transfer Equations
Principles Of EngineeringTM
Unit 1 – Lesson 1.2 Energy Forms
Project Lead The Way, Inc.
Copyright 2008

24. Introduction to Thermodynamics
U-Value
Introduction to Thermodynamics
Principles Of EngineeringTM
Unit 1 – Lesson 1.2 Energy Forms
Thermal Conductivity of a Material
Overall heat coefficient The measure of a material’s ability to conduct heat
Metric system
U.S. customary system
Project Lead The Way, Inc.
Copyright 2008

25. Introduction to Thermodynamics
R-Value
Introduction to Thermodynamics
Principles Of EngineeringTM
Unit 1 – Lesson 1.2 Energy Forms
Thermal Resistance of a Material
The measure of a material’s ability to resist heat
The higher the R-value, the higher the resistance
The R-value is equal to the reciprocal of a material’s U-value.
Bulk R-value =
R-value Object 1 + R-value Object 2 + … = Total R-Value
Project Lead The Way, Inc.
Copyright 2008

26. Control Systems

27. VEX Sensors Subsystem Provide inputs to sense the environment
Bumper Switch
Limit Switch
Line Follower
Optical Encoder
Ultrasonic

28. Boolean Logic Program decisions are always based on questions
Only two possible answers
yes or no
true or false
Statements that can be only true or false are called Boolean statements
Their true-or-false value is called a truth value.

29. 2 types of Control Systems
open loop system a process provides no feedback to the device.
a closed loop system provides feedback from sensors and the program responds based on the values.

30. Open/Closed Loop Systems
“Many devices function without ever knowing whether they are doing the job that they were programmed to do. They might run for a specific amount of time or perform one function and then stop. For example if you set the clothes dryer to run for 45 minutes, your clothes might be dry or they might not be dry. A clothes dryer is an open loop system because the process provides no feedback to the device. Newer clothes dryers possess moisture sensors. The moisture sensors inform the machine when the clothes are dry, at which point the dryer can stop running. The feedback provided by the sensor makes this a closed loop system.”

31. Fluid Power

32. Pascal’s Law
Pressure exerted by a confined fluid acts undiminished equally in all directions.
Pressure: The force per unit area exerted by a fluid against a surface
Symbol
Definition
Example Unit p
Pressure
lb/in.2 F
Force lb A
Area
in.2
The area of a cylinder will be the surface area of the piston.

33. Pneumatics vs. Hydraulics
Pneumatic Systems . . .
Use a compressible gas
Possess a quicker, jumpier motion
Are not as precise
Require a lubricant
Are generally cleaner
Often operate at pressures around 100 psi
Generally produce less power

34. Boyle’s Law
The volume of a gas at constant temperature varies inversely with the pressure exerted on it.
NASA
p1 (V1) = p2 (V2)
Symbol
Definition
Example Unit V
Volume
in.3

35. Charles’ Law
Volume of gas increases or decreases as the temperature increases or decreases, provided the amount of gas and pressure remain constant.
NASA
Note: T1 and T2 refer to absolute temperature.

36. Gay-Lussac’s Law
Absolute pressure of a gas increases or decreases as the temperature increases or decreases, provided the amount of gas and the volume remain constant.
Note: T1 and T2 refer to absolute temperature.
p1 and p2 refer to absolute pressure.

37. Statics

38. Centroids
Centroid Location
Principles of EngineeringTM
Unit 4 – Lesson Statics
The centroid of a square or rectangle is located at a distance of 1/2 its height and 1/2 its base. H B

39. Free Body Diagram Procedure
Free Body Diagrams
Principles of EngineeringTM
Unit 4 – Lesson Statics
Free Body Diagram Procedure
4. Label dimensions.
For more complex free body diagrams, proper dimensioning is required, including length, height, and angles.
N=5 lbf
45°
8 ft
10 ft
38.6°
PLTW – DE book
Free body diagrams should be drawn completely isolated from all other objects. We “free” the body from its system. Instead of taking time to draw the object in detail, we substitute a simple geometric shape, such as a square or a circle. For these exercises the shape we draw is considered dimensionless.
W=5 lbf

40. Calculating Moment of Inertia – Rectangles
Moment of Inertia Principles
Moment of Inertia
Principles of EngineeringTM
Unit 4 – Lesson Statics
Why did beam B have greater deformation than beam A?
Difference in moment of inertia due to the orientation of the beam
Calculating Moment of Inertia – Rectangles

41. Truss Problem – Spring 2006 Figure 1 Figure 2
Directions: Part C is an open-notes, closed-book test. No software applications may be used to assist you during this test. To receive full credit on any problem that requires calculations, you must: 1) identify the formula you are using, 2) show substitutions, and 3) state the answer with the correct units. You have 45 minutes to complete the following questions.

42. Truss Problem – Spring 2006 1a Figure 1 Figure 2
Study the truss in Figure 1 and its free body diagram in Figure 2, and answer the following questions.
a. Draw a point free body diagram for joint C and label all of the given information for that node (assume all member forces are tension).
[2 points]
What steps do you take to draw a free body diagram?

43. Truss Problem – Spring 2006 That’s it! 1a Isolate joint C:
Draw the force of AC
Draw the force of BC
Draw the force of F1
Figure 2 C
FAC
30°
That’s it!
FBC
F1 = 100 lbs.

44. Truss Problem – Spring 2006 1b Figure 1 Figure 2
B. Calculate the length of truss member BC. (answer precision = 0.000)
[3 points]
What do we know from looking at Figure 1 & 2?

45. Truss Problem – Spring 2006 1b Figure 1 Figure 2
What do we know from looking at Figure 1 & 2?
Length of AC = 4 feet
Angle between AC and BC is 30°
What can you use to solve for the length of BC? (SOHCAHTOA)
Use Cosine θ!

46. Truss Problem – Spring 2006 1b Figure 1 Figure 2
Draw your diagram and use the GUESS Method
G: AC = 4 ft, θ= 30°
U: BC
E: Cos θ= adjacent/hypotenuse
S: = BC/4ft
S: BC = ft A B C
4 ft
30°

47. Truss Problem – Spring 2006 1c 3.464 ft. Figure 1 Figure 2
C. Using joint C, determine the magnitude and type of force (tension or compression) that is being carried by truss member BC. (answer precision = 0.0) [4 points]
What do we need to know to solve this problem?

48. Truss Problem – Spring 2006 1c 3.464 ft. Figure 1 Figure 2
First we need to isolate joint C and identify the forces that we know.
Look back at the free body diagram you drew in problem A. C
FAC
FBC
F1 = 100 lbs.
30°

49. Truss Problem – Spring 2006 1c 3.464 ft. Figure 1 Figure 2
What formula is used for calculating the force on member BC?
∑FCY = 0 = F1 + FBCY
Now begin the GUESS Method C
FAC
FBC
F1 = 100 lbs.
30°

50. Truss Problem – Spring 2006 G: F1= 100 lbs. U: FBC
C. Using joint C, determine the magnitude and type of force (tension or compression) that is being carried by truss member BC. (answer precision = 0.0) [4 points] C
FAC
FBC
F1 = 100 lbs.
30°
G: F1= 100 lbs.
U: FBC
E: ∑FCY = 0 = F1 + FBCY
S: ∑FCY = 0 = -100 lbs. + -(FBCsin30°)
∑FCY = 0 = -100 lbs. + -(FBC • 0.5)
S: FBC = -200 lbs.

51. Strengths of Materials

52. Material Testing
Materials testing is basically divided into two major groups, destructive testing, and nondestructive testing.
Destructive testing is defined as a process where a material is subjected to a load in some manner which will cause that material to fail.
When non-destructive testing is performed on a material, the part is not permanently affected by the test, and the part is usually still serviceable. The purpose of that test is to determine if the material contains discontinuities (an interruption in the normal physical structure or configuration of a part) or defects (a discontinuity whose size, shape or location adversely affects the usefulness of a part).
During testing of a material sample, the stress–strain curve is created and shows a graphical representation of the relationship between stress, derived from measuring the load applied on the sample, and strain, derived from measuring the deformation of the sample

53. Tensile Test Data
Calculate the stress in the dog bone with a 430 lb applied force.

54. Stress-Strain Curve/ Diagram
Stress at the proportional limit
proportional limit
Strain at the proportional limit
What is illustrated in a Stress-Strain Curve/ Diagram?
The proportional limit: The greatest stress at which a material is capable of sustaining the applied load without deviating from the proportionality of stress to strain
The stress at the proportional limit
The strain at the proportional limit

55. Properties of Materials Symbols
D – the change in
d – total deformation (length and diameter)
s – stress, force per unit area (psi)
e – strain (inches per inch)
E – modulus of elasticity, Young’s modulus (ratio of stress to strain for a given material or the measure of the stiffness of a material.)
P – axial forces (along the same line as an axis (coaxial) or centerline)

56. Properties of Materials Symbols
Formulae you might use are:
σ = P/A (Stress)
= /L (Strain)
= PL/A (Total Deformation)
= σ/ (modulus of elasticity)

57. 3a
What are we solving for and what does our diagram provide us?
We are solving for the force so use the equation F p. l. = A O x sp. l.
What do we know from the diagram?
sp. l. = 40,000 psi (Stress at proportional limit)
Information from the problem statement?
0.2 in2 = cross sectional area (AO)
3. A test sample, having a cross-sectional area of 0.2 in2 and a 2 inch test length, was pulled apart in a tensile test machine. Figure 4 shows the resulting Stress-Strain diagram. Use the information in the diagram to answer the following questions.
a. Calculate the force that the sample experienced at the proportional limit. (answer precision = 0.0) [3 points]

58. Now solve the problem: 3a
Calculate the force that the sample experienced at the proportional limit. (answer precision = 0.0) [3 points]
Use the GUESS Method to solve the problem
G: AO = 0.2 in2, s=40,000 psi
U: F P.I. (Force at the proportional limit)
E: F p. l. = A O x s p. l. (Area x Stress)
S: F p. l. = 0.2 in2 x 40,000 psi
S: F p. l. = 8,000 lbs.

59. 3b
3b. Starting at the origin and ending at the proportional limit, calculate the modulus of elasticity for this material. [3 points]
To solve this problem we are going to use the equation E = s/∈
Using the GUESS Method here is what we have.
G: ∈ = in/in, s =40,000 psi
U: E (modulus of elasticity)
E: E = s/∈ (Stress/Strain)
S: E = 40,000/ 0.005
S: E = 8,000,000 psi

60. Properties of Materials
Figure 10 shows a 100 lb. normal force being applied to a 12” long x 10” diameter cylinder. What is the resulting compressive stress in the cylinder? (Spring 2009 #28 Part A Practice)
G: F=100 lb., Dia.=10”
U: Compressive Stress
E: 1) A = π * r2
2) S = F/A
S: 1) A = 3.14 * 52
2) S = 100 lbs/78.5 in2
S: S = 1.27 psi

61. Kinematics

62. Trajectory Motion
There are few concepts that you need to understand to get through this problem. Let’s start by defining kinematics……
Speed (average velocity) = D/T d=distance and t=time
Kinematics is the study of motion allowing us to predict the path of an object when traveling at some angle with respect to the Earth’s surface.
It is easy to calculate if the force of Gravity remains constant and we ignore the effects of air resistance

63. Human Cannonball Training Calculating Initial Velocity
Gravity should be substituted in as a negative number. Because the formula has a negative sign in front of g it becomes a positive value for the purpose of solving the equation.

64. Human Cannonball Training Calculating Horizontal Distance

65. Human Cannonball Training Calculating Horizontal Distance

66. Trajectory Motion – Spring 2006
Take-off angle = 35°
Take-off speed = ft/sec
2. Study Figure 3 and answer the following questions.
What was the motorcyclist’s initial horizontal velocity? (answer precision = 0.00) [3 points]
Start by identifying what we know from the information provided.

67. Trajectory Motion – Spring 2006
Vertical Velocity
VIY = VI sinθ
Might be good to know, hint, hint
Take-off angle = 35°
Take-off speed = ft/sec
G: θ= 35°, Take-off speed ft/sec
U: VIX (horizontal velocity)
What equation is used to calculate horizontal velocity?
E: VIX = VI cosine θ
S: VIX = ft/sec cosine 35°
VIX = ft/sec • 0.82
S: VIX = 30.33ft/sec
35°
36.99 ft/sec
VIX

68. Trajectory Motion – Spring 20062b
Take-off angle = 35°
Take-off speed = ft/sec
b. What was the horizontal distance between the take-off and landing points? Assume that both points exist on the same horizontal plane. Use ft/sec2 for acceleration due to gravity. (answer precision = 0.00) [3 points]
35°
36.99 ft/sec
VIX = 30.33ft/sec

69. Trajectory Motion – Spring 20062b
What do we know based on the problem we just solved?
G: θ= 35°, Take-off speed ft/sec, VIX = 30.33ft/sec
U: X
E: X = VI2 sin(2*θ)/g
S: X = (36.99 ft/sec)2 sin(2*35°) / ft/sec2
X = ft2/sec2sin 70º / ft/sec2
X = ft * 0.94 / 32.15
X = ft / 32.15
S: X = ft
35°
36.99 ft/sec
VIX = 30.33ft/sec

70. Probability

71. Bernoulli Process P = Probability
Principles of EngineeringTM
Unit 4 – Lesson Statistics
Bernoulli Process
P = Probability
x = Number of times an outcome occurs within n trials
n = Number of trials
p = Probability of success on a single trial
q = Probability of failure on a single trial
*Technically a Bernoulli process happens only once (flip one coin), while a binomial process comes by adding many Bernoulli processes. The formula here is for a binomial process (combining the results of n independent Bernoulli trials).
n! (called n factorial) is the number n times each number that is smaller than n. For instance, 5! = 5*4*3*2*1 = 120
Most scientific and graphing calculators have a ! key.

72. Probability Distribution
Principles of EngineeringTM
Unit 4 – Lesson Statistics
What is the probability of tossing a coin three times and it landing heads up two times?