Machine Design PE Technical Study Guide Errata

Machine Design PE Technical Study Guide Errata This product has been updated to incorporate all changes shown in the comments on the webpage and email comments as of August 20, 2018. If you have purchased this product prior to this date and wish for the latest version then please email Justin Kauwale at contact@engproguides.com. The following attached pages summarize the changes from October 31, 2017 till August 20, 2018. Machine Design PE Technical Study Guide Errata -1

8.0 PRACTICE PROBLEMS 8.1 PROBLEM 1 - ECONOMICS Background: A client is contemplating on purchasing a new high efficiency pump and motor, with an initial cost of $10,000. The pump has a lifetime of 15 years and is estimated to save approximately $1,000 per year. There is an additional maintenance cost of $300 per year associated with this new pump. The pump will have a salvage value of $0 at the end of its lifetime. Assume the interest rate is 4%. Problem: What is the annual value of the pump? (a) -$499 (b) -$199 (c) $199 (d) $499 8.2 PROBLEM 2 - ECONOMICS Background: A client is contemplating between two separate turbines. Turbine 1 has a life of 25 years, an initial cost of $50,000, an ongoing maintenance/electricity cost totaling $1,000 per year. Turbine 2 has a life of 25 years, an initial cost of $35,000 and an ongoing maintenance/electricity cost totaling $1,500 per year. Assume interest rate is equal to 4%. Problem: What is the present value of the two turbines? (a) Chiller 1 = -$65,622 ; Chiller 2 = -$58,433 (b) Chiller 1 = -$91,646 ; Chiller 2 = -$103,455 (c) Chiller 1 = $23,646 ; Chiller 2 = $97,469 (d) Chiller 1 = $12,646 ; Chiller 2 = $103,455 Basic Engineering Practice-33

Simple beam with a uniformly distributed load Weight of beam (W) 0 x L 0 x L Figure 14: Simple beam with uniformed load diagrams This diagram is typically used when a beam is supportedd at both ends for only the weight of the beam. The downward force is equal to the weight of thee beam. The weight is evenly distributed evenly the entire length of the beam, which is why the vectors are off equal size. Total Wei ght of Beamm W lbf = w (lbf/ft) * L (ft) There will be upward reaction forces at the support that will counteract the weight of the beam. The force at each support will be equal to one-half the entire weight. R R W 2 lbf f The next figure shows the shear force diagram. The shear force acting at any point x on the beam is governed by the below equation. The shear force is at its maximum at the supports. You can see that when 0 and L are inserted into the equation for x, the value will be of magnitude equal to ½ the weight. V W 1 2 x lb f L Engineering Science & Mechanics-15

V 0 W 1 2 0 L W 1 and V L 1/2W lbf 2 The bending moment acting at any point x on the beam is governed by the following equation. The moment is at its maximum at the center. The moment will be equal to zero when x is equal to L or 0. M Wx 2 L x in lbs Maximum Moment M / WL 8 in lbs The deflection at any point x on the beam is governed by the following equation. The deflection is at its maximum at the center. Δ Wx 24 E I l 2 l x x in where E modulus of elasticity psi and l length in Δ 5WL 384 E I in The moment can then be used to find the maximum stress in the beam. The maximum stress in the beam will help to influence the beam dimensions and material choices. The maximum stress will be discussed in the Strength of Materials section under Bending. Engineering Science & Mechanics-16

5.4 S SOLUTION 4 - BEAMS Use the following diagram to find the moment at the center of the beam. The first step is to create a free body diagram. Next, cut the beam at the center. Now look at only the forces to the right of the beam and solve for the moment at the center. M wx 2 The answer is most nearly, (b) 1,250 lb-ft 25 lbs ft 1 10ft 10 ft /2 1,250 lb ft Engineering Science & Mechanics-41

When a pipe is heated or cooled, the length of the pipe will change based on the temperature and a coefficient of thermal expansion which is dependent on the material. L L α T T where L change in length of pipe ft ; L α T T Figure 19: Increasing the temperature causes thermal expansion. Bending stress: The bending stress of a pipe due to external forces or the weight of a pipe can be found using simple free body diagrams as shown in the following section. 4.5 THERMAL CONDUCTIVITY Thermal conductivity is the ability of a material to conduct heat with a given temperature difference. Thermal conductivity is the material property that measures the rate of change in heat per unit distance per unit temperature difference. q k dt dx q heat flow per unit time per unit distance W m or Btu h ft W k thermal conductivity m K or Btu h ft F More commonly on the PE exam you will use the following equation to conduct an energy balance as energy is transferred from one material to another material. The first step is to find the heat flow rate through an area of the subject material, as shown below. Q k T Area; x distance m or ft ; T temp. diff.;q total heatbtu x h ; Material Properties-27

5.5 PROBLEM 5: STRESS-STRAIN A 6 long cylindrical rod of unknown diameter must be able to hold 100,000 lbf without elongating more than 2. The rod is made of a material with a modulus of elasticity of 15 x 10 6 psi. What is the minimum diameter of the rod to meet the maximum elongation requirement? a) 0.08 in b) 0.17 in c) 0.25 in d) 0.5 in 5.6 PROBLEM 6: STRESS-STRAIN A cylindrical rod has a diameter of 0.5 inches. This rod is stressed in tension with a force of 3,500 lbf. This causes the diameter of the rod to be reduced to 0.4998 inches. The rod is made of a material with a modulus of elasticity of 15 x 10 6 psi. What is Poisson s ratio of this material? a) 0.08 b) 0.21 c) 0.29 d) 0.55 Material Properties-33

6.5 SOLUTION 5: STRESS-STRAIN A 6 long cylindrical rod of unknown diameter must be able to hold 100,000 lbf without elongating more than 2. The rod is made of a material with a modulus of elasticity of 15 x 10 6 psi. What is the minimum diameter of the rod to meet the maximum elongation requirement? The first step is to calculate the maximum strain. 2" ε 6 ft 12 in.0278 ft Next use this value to calculate the maximum stress with the modulus of elasticity. E Stress Strain 15 x 10 Stress. 0278 σ 0.417 x 10 psi Finally, stress is equal to the force per unit area. You are given the design force and the maximum stress value. So you need to use the stress equation to solve for area and ultimately the diameter of the rod. σ F A The correct answer is most nearly, d) 0.55 in. 100,000 lbf πd 4 100,000 lbf 4 D π 0.417 x 10 psi D 0.55 in Material Properties-38

Step 2: Find radius of gyration The radius of gyration compares the moment of inertia and the cross sectional area. The larger cross sectional area will cause the radius of gyration to increase, which means the object is less slender. You will need to compute the radius of gyration for both the x and y axis and use the minimum radius of gyration for the next step. Radius of Gyration r I A Step 3: Find the slenderness ratio Once you have the minimum radius of gyration and the effective length, then you can find the slenderness ratio with the following equation. Step 4: Find column constant Slenderness Ratio R L r ; The column constant takes into account the modulus of elasticity and the yield stress of the material. A material with a larger yield stress will mean the column is much stronger and thus the column can be longer before it will be considered slender and therefore subject to buckling. Column Constant C 2π E σ E Young s Modulus Step 5: Compare slenderness ratio and column constant Finally, compare the slenderness ratio and column constant. If the slenderness ratio is less than the column constant then the object is considered to be slender and long. Thus the column will be subject to buckling R C Column is considered slender Column is subject to buckling Use Euler formula to determine critical buckling load R C Column is not slender Column might not be subject to buckling before yield stress. Use Johnson formula to determine critical buckling load 5.1 CRITICAL BUCKLING LOAD The force at which a skinny column will fail under compression (buckling) is directly related to the material s Young s Modulus and moment of inertia. The force is inversely related to the square of the length and a constant which describes the degrees of freedom of the column. The Strength of Materials-15

6.4 PARALLEL AXIS THEOREM The polar second moment of area is still the second moment of area. The polar term just refers to the fact that these equations focus on circular cross sections. Similar to the second moment of area section, the parallel axis theorem also applies. When a shaft is off-center and the axis needs to be adjusted to account for this difference, then you can use the parallel axis theorem. This theorem states that the second moment of area about an axis can be adjusted to another axis by adding the cross sectional area multiplied by the distance between the two axes squared. 6.5 TORSION FAILURE The failure of a shaft during torsion will depend on whether the material is ductile versus brittle. If the material is ductile, then the material will most likely fail due to a maximum shear stress. If the material is brittle then the material will most likely fail due to a maximum tensile stress. This concept is a little difficult to grasp, but you can try to imagine a brittle material twisting and at the same time the material is twisting, the material is elongating and creating tension in a direction that is 45 degrees from the longitudinal direction. Figure 12: Ductile materials (left) will flex in torsion, until the maximum shear stress is met and the material will fail at a near 90 degree angle. Brittle materials (right) will not flex in torsion and will fail in tension at an angle near 45 degrees to the longitudinal direction. When you are completing a torsion problem and you are finding the shear stress due to torsion, you can compare this shear stress value to either the shear or tensile strength of the material based on the type of material. On the PE exam and in practice, you will most likely not encounter a ductile material loaded in torsion. Ductile materials like copper and aluminum are rarely loaded in torsion. It is most common that all the various types of steel will be loaded in torsion, such that you can use the shear strength when comparing the design shear stress. Strength of Materials-20

9.0 PRACTICE PROBLEMS 9.1 PROBLEM 1 - BENDING A wood beam is situated as shown in the figure below. The material has strength of 900 psi. The beam shall be designed to have a safety factor of 1.0. What should be the dimension of the height of the beam? Assume the height of the beam is 2 times the width of the beam. (a) 0.89 in (b) 2.03 in (c) 2.55 in (d) 5.84 in 9.2 PROBLEM 2 - BENDING A W 8 x 10 beam has an allowable stress of 50 ksi and dimensions as shown below. What is the maximum weight per foot that the beam can support? (a) 120 lb/ft (b) 210 lb/ft (c) 289 lb/ft (d) 420 lb/ft Strength of Materials-28

9.3 PROBLEM 3 - BUCKLING A W 6 X 9 steel column with yield strength of 40 ksi and a modulus of elasticity of 29,000 ksi will be used to support an unknown load. What is the critical buckling load of this column? Assume the column is slender. (a) 9 kips (b) 19 kips (c) 26 kips (d) 52 kips 9.4 PROBLEM 4 - TORSION A handle is torqued to close a valve against the point shown in the figure below. What is the maximum shear stress developed at the outer wall of the solid stem? The length of the stem is 6 and the handle has a diameter of 4. The stem has a radius of 2. The forces are applied at the end of the handle at equal distances from the center. (a) 1.6 psi (b) 3.2 psi (c) 5.0 psi (d) 10.1 psi Strength of Materials-29

10.0 SOLUTIONS 10.1 SOLUTION 1 - BENDING A wood beam is situated as shown in the figure below. The material has strength of 900 psi. The beam shall be designed to have a safety factor of 1.0. What should be the dimension of the height of the beam? Assume the height of the beam is 2 times the width of the beam. First, use your beam diagrams from either your Machinery s Handbook or the link that was previously discussed and solve for the reactionary forces. 25 30 30 2 10 2 10 25 30 30 2 10 2 10 375 375 Next use these reaction forces to solve for the max moment of inertia. 25 10 1,250 2 Finally, use the section modulus equation to solve for the dimensions of the beam. The correct answer is most nearly, (d) 5.84 in. M σ ;where I bh 12 and c h 2 and b 1 2 h σ 900psi 1 1 12 2 1,250 ; /2 900 psi 1,250 12 12 ; 900 psi 200; 5.84 Strength of Materials-31

10.2 SOLUTION 2 - BENDING A W 8 x 10 beam has an allowable stress of 50 ksi and dimensions as shown below. What is the maximum weight per foot that the beam can support? In order to solve this problem you should use the sectional modulus equation. S M σ ; For the W 8 x 10 beam, you should refer to your Machinery s Handbook for the sectional modulus. Make sure you use the x-direction, since the load is perpendicular to the x-direction. S I 30.8 in 7.81 in h/2 7.89in/2 The maximum moment can also be found in your Machinery s Handbook. M wl 8 But you also know from the sectional modulus equation that the moment is also equal to S*allowable stress. Now solve for w (lb/ft). The correct answer is most nearly, (c) 289 lb/ft M 7.81 in 50ksi M 7.81 in 50,000 psi 390,500 lb in M 32,542 lb ft wl 32,542 lb ft 8 w 41,653 lb/ft Strength of Materials-32

10.3 SOLUTION 3 - BUCKLING A W 6 X 9 steel column with yield strength of 40 ksi and a modulus of elasticity of 29,000 ksi will be used to support an unknown load. What is the critical buckling load of this column? Assume the column is slender. First, use your Machinery s Handbook to find the necessary properties for the column. A 2.68 in ; I 16.4 in ;I 2.2 in ;S 5.56 in Next use the critical force equation below, since you can assume the column is slender. F π EI KL Since the load is pin connected on both sides, K = 1. Convert the length to inches. F The correct answer is most nearly, (a) 9 kips. π 29,000,000 lbs in 2.2 in 25 12 in 6,996 lbs or 6.996 kips Strength of Materials-33

10.4 SOLUTION 4 - TORSION A handle is torqued to close a valve against the point shown in the figure below. What is the maximum shear stress developed at the outer wall of the solid stem? The length of the stem is 6 and the handle has a diameter of 4. The stem has a radius of 2. The forces are applied at the end of the handle at equal distances from the center. First, you need to calculate the torque on the stem. So you need to balance the moments upon the center of the stem at the handle. Then use the maximum shear stress equation. 0 20 2" 15lbf*2" 10 ; 2 The correct answer is most nearly, (a) 1.6 psi. 20 2 ; 2 2 1.59 Strength of Materials-34

3.0 DAMPING Once the natural and disturbing frequencies have been determined, it is necessary to design an appropriate damping system to reduce unwanted vibrations. The previous discussions and equations assumed that the mechanical system was undamped. If no energy is lost or dissipated in friction or other resistance during oscillation, the vibration is known as undamped vibration. If any energy is lost, then it is called damped vibration. In many physical systems, the amount of damping is so small that it can be disregarded for most engineering purposes. However, consideration of damping becomes extremely important in analyzing vibratory systems near resonance. Figure 3: A spring system can be modeled similar to the undamped system, except for the addition of the damping coefficient. There are three main types of damping, (1) Coulomb Damping, (2) Hysteresis or Inherent Damping and (3) Viscous Damping. (1) Coulomb Damping: This type of damping force is constant throughout the entire displacement and also does not depend on the velocity of the system. This type of damping is also called dry friction damping. The best example is a mass moving on a surface. The frictional force acting upon the mass does not depend on the speed or location of the mass. This type of damping is not used on the PE exam. (2) Hysteresis or Inherent Damping: This type of damping uses the elastic properties of elastomers to dampen vibrations. This is typical of rubber pads or rubber bushings. (3) Viscous Damping: This type of damping is used on the PE exam. This type of damping is typical of spring vibration isolators. This type of damping changes its force in response to the velocity and deflection from equilibrium. The following discussions only apply to viscous damping. The damping value is shown in the previous figure as a damping coefficient labeled C. This value has the units as shown below. SI Units C kg sec US Units C lb sec Vibration-8

The critical damping coefficient is the value that causes the machine design system to most quickly return back to equilibrium. This value is found through the following equation. Notice that similar to the natural frequency, the critical damping coefficient is only a function of k and mass. SI Units SI Units C 2 k mass kg m s m kg kg m s kg s US Units C 2 k mass/g US Units lbf lbm ft s ft 1 lbf lbm lbm s lbm s The actual damping coefficient will be determined by the manufacturer of the vibration dampener or spring. 3.1 DAMPING RATIO Once the critical damping coefficient and actual damping coefficient are found, then the damping ratio can be computed. The damping ratio has dimensionless units and it describes how slowly or quickly a system is returned to its original conditions after a temporary change to the original conditions. The damping ratio is defined as the ratio of the actual damping to the critical damping. Damping Ratio ζ C C C 2 k mass The ratio will determine the damping classification of the machine design system into one of the four categories. 3.1.1 Undamped If there is no damping then the damping ratio will equal 0. This situation was previously discussed and is called undamped. There will be no decay in the sinusoid and the amplitude of the sinusoid will remain the same as time progresses. The undamped equation was previously given, but is shown again here in a different format, since natural frequency has been introduced. Undamped ζ 0 y t A sin ω t ψ ; Vibration-9

The typical questions asked with actuators are to calculate the output force, the required pressure to output a force, the speed of the cylinder, the flow needed for a cylinder speed and the cylinder area required. 13.3.1 Cylinder force The force that a cylinder can produce is a result of the pressure of the fluid within the cylinder and the area that the cylinder acts upon. Force lbf Pressre psi Area in In the equation above, the pressure describes the pressure of the hydraulic or pneumatic fluid. The area is the area that is conducting the mechanical power. 13.3.2 Fluid pressure If the force required and the cylinder area is given, then you can use the reverse of the previous equation to find the required pressure to achieve the mechanical work. Pressure psi Force lbf Area in 13.3.3 Cylinder speed The next type of question is determining the cylinder speed, which is based upon the volume of the cylinder and the flow rate of the pressurized fluid. As the fluid fills the cylinder, it moves the cylinder. The rate at which the cylinder is moved can be found with the below equation. Cylinder speed ft ft Volumetric flow rate min min Area ft 13.3.4 Fluid flow The next type of question is the opposite of the cylinder speed type question. If you are given the required cylinder speed and the area of the cylinder, then you must determine the required fluid flow. The below equation can be used for this type of problem. Volumetric flow rate ft min Area ft Cylinder speed ft min 13.3.5 Bulk modulus The term Bulk Modulus is a property of a fluid that describes the compressibility of the fluid. Bulk modulus, β, is defined in the equation below. This term is used to determine how much a Mechanical Components-73

15.7 PROBLEM 7 BELTS AND PULLEYS A pulley as shown below has the below tension on the tight and slack sides. The angle of wrap will be increased with another pulley from 180 degrees to 200 degrees. What is the increase in torque capacity at the large pulley? Assume the slack tension remains the same. The coefficient of friction is 0.2. T=145 lbf (a) 16% (b) 33% (c) 49% (d) 71% 15.8 PROBLEM 8 BELTS AND PULLEYS The force on the tight side of a belt and pulley drive is 100 lbf and 50 lbf on the slack side. The diameter of the small pulley is 3 and the diameter of the large pulley is 6. The large pulley is moving at a speed of 1,750 rpm. What is the power transmitted by the large pulley? (a) 0.5 HP (b) 1.9 HP (c) 3.1 HP (d) 4.2 HP Mechanical Components-87

15.17 PROBLEM 17 HYDRAULICS & PNEUMATICS A hydraulic system is comprised of a slave cylinder of diameter 12. The master cylinder is 1. If a force of 200 LBF is required at the slave cylinder, how much force is required at the master cylinder? The answer is most nearly, (a) 1.4 lbf (b) 5.8 lbf (c) 10.0 lbf (d) 16.8 lbf 15.18 PROBLEM 18 - HYDRAULICS & PNEUMATICS A piston compresses a hydraulic fluid with a bulk modulus of 3.0 x 10 5 psi. If a force of 25,000 lbs is applied to the piston, what is the change in height of the hydraulic fluid in inches? The hydraulic fluid sits in a cylinder that is 6 in diameter and 12 in height at its maximum. The answer is most nearly (a) 1.0 X 10-4 (b) 7.8 X 10-4 (c) 5.9 X 10-3 (d) 3.6 X 10-2 Mechanical Components-92

16.0 SOLUTIONS 16.1 SOLUTION 1 PRESSURE VESSELS A pipe is carrying compressed air at a pressure of 500 psi. The pipe has an internal diameter of 19.5 and the pipe has a thickness of 0.5. The ends of the 10 long pipe is capped at both ends and sealed airtight. What is the hoop stress developed in the pipe? The pipe can be treated as a thin walled vessel because the ratio of the radius to the thickness is greater than 10. 9.75" 0.5" 10 So you can use the thin walled pressure vessel equation to find the hoop stress. 500 9.75" 0.5" 9,750 The correct answer is most nearly, (d) 9,750 psi. 16.2 SOLUTION 2 PRESSURE VESSEL A cylindrical pressure vessel has an internal pressure of 1,000 psi. The pressure vessel has an internal diameter of 15.5 and a thickness of 0.25. One end of the pressure vessel will be capped with a bolt-nut system. What force should the cap be capable of withstanding? In this question you must find the pressure acting upon the capped end. This is equal to the internal pressure multiplied by the area of the capped end. 1,000 15.5 0.25 188,691 The correct answer is most nearly, (a). Mechanical Components-95

16.7 SOLUTION 7 BELTS & PULLEYS A pulley as shown below has the below tension on the tight and slack sides. The angle of wrap will be increased with another pulley from 180 degrees to 200 degrees. What is the increase in torque capacity at the large pulley? Assume the slack tension remains the same. The coefficient of friction is 0.2. T=145 lbf The ratio of the tight and slack forces is equal to the natural logarithm of the angle of wrap and the coefficient of friction. 180 180 3.14; 200 3.49; 180,.. ;,..,,,,.. ;,,..,, 1.87;,, 2.01 Next use the ratios and the fact that the new slack tension is the same as the original slack tension. 2, 1.87 2, 2.01 Finally, cancel out the slack tension value and radius value and find the percentage difference from the original to the new conditions., 2.01, 1.87 2.01 1 1.87 1 1.16 The correct answer is most nearly (a) 16% increase. Mechanical Components-100

16.8 SOLUTION 8 - BELTS & PULLEYS The force on the tight side of a belt and pulley drive is 100 lbf and 50 lbf on the slack side. The diameter of the small pulley is 3 and the diameter of the large pulley is 6. The large pulley is moving at a speed of 1,750 rpm. What is the power transmitted by the large pulley? The power transmitted is found through the below equation. 550 & ; ; But first, convert the speed from rpm to radians/minute and then to ft/sec. 2 2 0.5 1,750 2 The correct answer is most nearly, (d) 4.2 HP. 2 1 60 1 60 100 50 45.8 550 4.2 45.8 / Mechanical Components-101

F A F A F π 1" 4 200 lbf π 12" 4 F 200 lbf 1 12 F 1.4 lbf The correct answer is most nearly, (a) 1.4 lbf 16.18 SOLUTION 18 HYDRAULICS & PNEUMATICS A piston compresses a hydraulic fluid with a bulk modulus of 3.0 x 10 5 psi. If a force of 25,000 lbs is applied to the piston, what is the change in height of the hydraulic fluid in inches? The hydraulic fluid sits in a cylinder that is 6 in diameter and 12 in height at its maximum. First, use the bulk modulus equation. β P ;psi V/V From this equation you know you need the applied pressure, which is simply the force divided by the area. P F 25,000 lbf 25,000 lbf 884.2 psi A π D π 6 4 4 Next, use the bulk modulus equation to solve for change in volume. 3.0 X 10 psi 884.2 psi V V 1.0 in π 6" 4 12" Finally, the change in volume will be equal to the area multiplied by the change in height. 1.0 in π 6" 4 H 0.035 in The correct answer is most nearly, (d) 3.6 x 10-2 in Mechanical Components-107

The correct answer is most nearly, (b) 11.15 in. 6.7 S SOLUTION 7 WELD DING The figure shows a weld under torsion. The fillet welds each have a throat of 1/8. The horizontal distance from A to the external load is 7. What is the secondary shear stresss that acts upon point, A, solely in the vertical direction? Thee centroid is located at point (0.8 in, -1.8 in), where the origin is at the intersection of the vertical and horizontal welds in green. The first step is to find the centroid for the given weld pattern, through the below equation. 4 in 16 in x 2 4 in 6 0.8 in in 20 in 6 in 36 in y 2 4 in 6 1.8 in in 20 in Next, draw the centroid and the directional forces. The red force is the secondary force that opposes torsion. First, you need to calculate the total moment acting upon the centroid. This is found by multiplying the external force by the radial distance between the centroid and the external force location. r 10.2 in M 5,000 lbs 10.2 in 51,000 lb inn The shear stress due to the vertical component of thee red force will be equal to the Moment times the horizontal distance between the centroid andd point A divided by the second polar Joints and Fasteners-41

moment of area. You do not use the radial distance, because this will give you the total force as opposed to solely the vertical component. J b d 6 b d 12 b d σ M r 51,000 lb in 3.2 in 2, J 54.5 in 994 psi The correct answer is most nearly, (a) 3,890 psi 4 6 6 4 6 54.5 in 12 4 6 6.8 S SOLUTION 8 WELD DING The figure shows a weld under torsion. The fillet welds each have a throat of 1/8. The horizontal distance from A to the external load is 7. What is the primary shear stress thatt acts upon point, A? The centroid is located at point (0.88 in, -1.8 in), where the origin, (0,0) is located at the intersection of the horizontal and vertical welds in green. First, you need the throat area for the given weld pattern. Area 0.707 h b d 0.707 0.125 in 4 in 6 in 0.884 in The primary shear stress that opposes the downward external shear force can be calculated by dividing the external shear force by the total weld area. σ 5,00 00 lbs 5,658 psi 0.884 in The correct answer is most nearly, (c) 5,660 psi. 6.9 S SOLUTION 9 ADHES SIVES A beam is made by gluing two beams together as shown in the cross section below. What is the shear stress in the glue? Joints and Fasteners-42

The first number designation, 209 is the number assigned to the standard. The following number after the dash, 14 stands for the year it was issued. In this case, the standard was issued in 2014. The important standards for the exam to be aware of are ASTM A and B. Do not purchase these standards, just be aware that they exist and what these standards cover. 4.2 AWS The American Welding Society s standards covers all aspects of welding, like safety, materials, corrosion, different types of welding, different welding conditions and different welding applications. You should go through the website, to get a feel of the available standards and the type of material that is covered by AWS. https://pubs.aws.org/ 4.3 ANSI American National Standards Institute or ANSI includes information on quality management, medical devices, IT security, fall protection and a lot more topics, including topics within Machine Design. The standard that would be most applicable to Machine design would be the one shown below on threads. However, there are similar standard specifications for items like roller chains, bearings, shafts and many other machine design elements that require specifications to ensure that manufacturer s products are compatible with one another. ASME B1.1/ANSI/ASME B1.2/ ANSI/ASME B1.20.1 Unified Screw and Pipe Threads Package 4.4 UL UL is an independent safety science company. It tests equipment, materials and products to confirm if they meet the UL safety standards. You will often find the following seal on a product, which indicates that the product has been tested and certified to meet a certain standard. UL LISTING Supportive Knowledge-17

The previous diagram shows the path in blue. The path shows that the metal never crosses into the bainite or perlite region. Thus the final result will be 100% martensite. The correct answer is most nearly, (a) 100% martensite. 8.7 SOLUTION 7 MANUFACTURING PROCESSES A lathe is used to turn carbon steel at a feed of 0.01 inches per revolution at a depth of 0.15 inches. If the diameter of the spindle is 4 and the cutting speed is 500 feet per minute, then what is the lathe spindle speed? The feed is the feed per revolution. The feed is selected to achieve a certain finish. For example, a last pass in cutting a piece will move slowly across the component at a rate of 0.005 inches per revolution. In this problem, the feed is irrelevant to the question. The cutting speed is the rotational speed of the lathe. This is typically in the range of 100 to 500 RPM, but can be as high as 2000 RPM. This problem involves converting the converting speed from 500 feet per minute to revolutions per minute. 500 1 500 1 477.5 4 12 The correct answer is most nearly, (b) 477 RPM. 8.8 SOLUTION 8 COMPUTATIONAL METHODS Finite element analysis is NOT best used to analyze stresses in a component when which of the following situations occur? (a) Material properties are not available for the component. (b) The geometry is complex and unable to be simplified. (c) There are multiple loads occurring at the same time, like impact and thermal stresses. (d) The forces acting upon the component is complex and unable to be simplified. The correct answer is most nearly, (a) Material properties are not available for the component. FEA, just like any other analysis depends on having proper inputs. FEA cannot Supportive Knowledge-30

Fillet Weld Configuration Throat Area (a thickness weld, h = throat) Centroid x, y Second Polar Moment of Area J in 0. 707 a d d/2, 0 d 12 0.707 a d 0. 707 a b d b 2 b d, 2 b d b d 6b 12 b d d 2 0.707 a d d /2, b/2 d 3b d 6 2 0.707 a d 0. 707 a b d 2 d b, b 2 8d 6db b 12 b 2b d 2 0.707 a d 2 0.707 a b d /2, b/2 b d 12 1. 44π a r 2πr This secondary shear force is applied in an opposing direction of the moment. In order to find the direction of the force, you must draw the radius from the centroid to the location in question. Then the force direction will be perpendicular to that line but in the opposite direction of the moment. If the moment is in the clockwise direction, then the force will point counterclockwise. Finally, the total stress can be vectors. calculated by adding the primary τ τ τ and secondary shear stress Joints and Fasteners-10

Fillet Weld Configuration Throat Area a = side; h =throat Centroid x, y Second Polar Moment of Area Unit J in 0. 707 a d d /2, 0 d 12 0.707 a d 0. 707 a b d b 2 b d, 2 b d b d 6b 12 b d d 2 0.707 a d d/ /2, b/2 d 3b 2 d 2 6 2 0.707 a d 0. 707 a b d 2d b, b 2 8d 6db b b 12 2b d 2 0.707 a d 2 0.707 a b d/ /2, b/2 b d 12 1. 44π a r 2πr Fillet Weld Configuration Second Moment of Area Unit I in d 12 bd 6 Cheat Sheets-288