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2014 POE Final Review.

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Presentation on theme: "2014 POE Final Review."— Presentation transcript:

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 2006
2b 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 2006
2b 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?


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