HW 1 - Ch 1&2 (Concatenated)

pdf

School

Iowa State University *

*We aren’t endorsed by this school

Course

325

Subject

Mechanical Engineering

Date

Apr 3, 2024

Type

pdf

Pages

26

Uploaded by SuperRhinoceros4226

Report
1. Award: 10 out of 10.00 points Score: 10/10 Points 100 % A cylindrical part of diameter d is loaded by an axial force P . This causes a stress of P / A , where A = π d 2 /4. If the load is known with an uncertainty of ±13 percent, the diameter is known within ±2 percent (tolerances), and the stress that causes failure (strength) is known within ±17 percent, determine the minimum design factor that will guarantee that the part will not fail. The design factor that guarantee the part will not fail is 1.42 . References Numeric Response Difficulty: Medium A cylindrical part of diameter d is loaded by an axial force P . This causes a stress of P / A , where A = π d 2 /4. If the load is known with an uncertainty of ±13 percent, the diameter is known within ±2 percent (tolerances), and the stress that causes failure (strength) is known within ±17 percent, determine the minimum design factor that will guarantee that the part will not fail. The design factor that guarantee the part will not fail is 1.42 ± 5% . Explanation: Uncertainty for load, P = ±13 percent Maximum load = 1.13 P Uncertainty for diameter, D = ±2 percent Minimum area = (0.98) 2 A Uncertainty for stress = ±17 percent Minimum strength = 0.83 S The appropriate design factor is calculated as follows:
HW 1.009 Solutions With Enhancements by Scott Merkle ME325 Text: Shigley 11e Rev 2023-AUG-24b
P ??? Under this blue block is one way to visualize a real life application of this situation… Problem 01.009 - Determine design factor from uncertainties 8 percent 5 percent 17 percent
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Problem 01.009 - Determine design factor from uncertainties https://glaser.es/EN/Service-Downloads/Practical-Information/Cylinder-head-Bolts-and-Cylinder-Head-Installation.aspx P P d ??? 8 percent 5 percent 17 percent
Hints: Problem 01.009 - Determine design factor from uncertainties Note:
Problem 01.009 - Determine design factor from uncertainties ??? 8 percent 5 percent 17 percent 5 percent 17 percent 8 percent Restating The Above:
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Problem 01.009 - Determine design factor from uncertainties
Or in other words… Problem 01.009 - Determine design factor from uncertainties (newtons per meter square, for example) (newtons per meter square, for example)
Or in other words… Problem 01.009 - Determine design factor from uncertainties The stress level it can withstand before failure The stress level it will be subjected to in application (newtons per meter square, for example) (newtons per meter square, for example) Or in other words…
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Or in other words… Problem 01.009 - Determine design factor from uncertainties The stress level it can withstand before failure The stress level it will be subjected to in application (newtons per meter square, for example) (newtons per meter square, for example) Loss of function stress must be increased by 1/(1 – 17%) and Area must be increased by 1/(1 - 5%) 2 Max allowable load most decrease by 1/(1 + 8%) So, to account for these given uncertainties… Or in other words…
Or in other words… Problem 01.009 - Determine design factor from uncertainties The stress level it can withstand before failure The stress level it will be subjected to in application (newtons per meter square, for example) (newtons per meter square, for example) Loss of function stress must be increased by 1/(1 – 17%) and Area must be increased by 1/(1 - 5%) 2 Max allowable load most decrease by 1/(1 + 8%) So, to account for these given uncertainties… Or in other words… 1/(1 – 17%) * 1/(1 - 5%) 2 1/(1 + 8%)
Or in other words… Problem 01.009 - Determine design factor from uncertainties The stress level it can withstand before failure The stress level it will be subjected to in application (newtons per meter square, for example) (newtons per meter square, for example) Loss of function stress must be increased by 1/(1 – 17%) and Area must be increased by 1/(1 - 5%) 2 Max allowable load most decrease by 1/(1 + 8%) So, to account for these given uncertainties… Or in other words… 1/(1 – 17%) * 1/(1 - 5%) 2 1/(1 + 8%) (1 – 17%) * (1 - 5%) 2 (1 + 8%) (0.83) * (0.95) 2 (1.08)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Or in other words… Problem 01.009 - Determine design factor from uncertainties The stress level it can withstand before failure The stress level it will be subjected to in application (newtons per meter square, for example) (newtons per meter square, for example) Loss of function stress must be increased by 1/(1 – 17%) and Area must be increased by 1/(1 - 5%) 2 Max allowable load most decrease by 1/(1 + 8%) So, to account for these given uncertainties… Or in other words… 1/(1 – 17%) * 1/(1 - 5%) 2 1/(1 + 8%) (1 – 17%) * (1 - 5%) 2 (1 + 8%) (0.83) * (0.95) 2 (1.08) 1.44 So given the magnitude of the uncertainties, the design factor n d needs to be at least 1.44
1. Award: 10 out of 10.00 points Score: 10/10 Points 100 % A solid circular rod of diameter d undergoes a bending moment M = 1000 lbf·in inducing a stress σ = 32 M /(π d 3 ). Using a material strength of 25 kpsi and a design factor of 2.5, determine the minimum diameter of the rod. Using Table A–17 select a preferred fractional diameter and determine the resulting factor of safety. Minimum diameter of the rod is 1.006 in. Preferred fractional diameter is 1 1/4 in. The resulting factor of safety is 4.8 . References Numeric Response Difficulty: Medium A solid circular rod of diameter d undergoes a bending moment M = 1000 lbf·in inducing a stress σ = 32 M /(π d 3 ). Using a material strength of 25 kpsi and a design factor of 2.5, determine the minimum diameter of the rod. Using Table A–17 select a preferred fractional diameter and determine the resulting factor of safety. Minimum diameter of the rod is 1.006 ± 5% in. Preferred fractional diameter is 1 1/4 in. The resulting factor of safety is 4.794 ± 5% . Explanation: From Eq. (1-3), Rearranging the above equation, we get On solving for d, From Table A-17 (Preferred Sizes and Renard Numbers), The factor of safety is
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Score: 20/20 Points 100 % Three blocks A , B , and C and a grooved block D have dimensions a , b , c , and d as follows: a = 1.500 ± 0.001 in b = 2.000 ± 0.0031 in c = 3.000 ± 0.0044 in d = 6.520 ± 0.01 in NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. References Section Break Difficulty: Easy 1. Award: 10 out of 10.00 points Determine the mean gap and its tolerance. The mean gap is 0.02 ± 0.0185 in. References Numeric Response Difficulty: Easy Determine the mean gap and its tolerance. The mean gap is 0.02 ± 5% ± 0.0185 ± 5% in. Explanation: The mean width is given by, The maximum allowable tolerance is given by,
Score: 20/20 Points 100 % Three blocks A , B , and C and a grooved block D have dimensions a , b , c , and d as follows: a = 1.500 ± 0.001 in b = 2.000 ± 0.0031 in c = 3.000 ± 0.0044 in d = 6.520 ± 0.01 in NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. References Section Break Difficulty: Easy
2. Award: 10 out of 10.00 points Determine the mean size of d that will ensure that w ≥ 0.010 in. The mean size of d is 6.51 in. References Numeric Response Difficulty: Easy Determine the mean size of d that will ensure that w ≥ 0.010 in. The mean size of d is 6.5285 ± 5% in. Explanation: From part (a), w min = 0.0015 in. Thus, we must add (0.01 – w min ) = 0.0085 to d . Therefore,
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
1. Award: 10 out of 10.00 points Score: 10/10 Points 100 % A pivot in a linkage includes the pin in the figure whose dimension a ± t a is to be established. The thickness of the link clevis is 1.47 ± 0.004 in. The designer has concluded that a gap of between 0.003 and 0.07 in will satisfactorily sustain the function of the linkage pivot. Determine the dimension a and its tolerance. The dimension of a is 1.512 in. The tolerance t a is 0.028 in. References Numeric Response Difficulty: Medium A pivot in a linkage includes the pin in the figure whose dimension a ± t a is to be established. The thickness of the link clevis is 1.47 ± 0.004 in. The designer has concluded that a gap of between 0.003 and 0.07 in will satisfactorily sustain the function of the linkage pivot. Determine the dimension a and its tolerance. The dimension of a is 1.548 ± 5% in. The tolerance t a is 0.028 ± 5% in. Explanation: Start by drawing a sketch showing the three items that contribute to the pin length, the clevis link thickness, the snap ring thickness, and the desired gap. If we know the bilateral tolerance for each component in the pin length, we will be able to combine them together. So, first, determine the mean and tolerance on the gap w. and Since the gap is the result of the other dimensions, we should focus on its mean and tolerance. or The tolerances of the parts will always add. or
1. Award: 10 out of 10.00 points Score: 20/20 Points 100 % Assume that you were specifying an AISI 1050 steel for an application. References Section Break Difficulty: Medium Using Table A-21, how would you specify it if you desired to maximize the yield strength? To maximize the yield strength of AISI 1050 steel, it is quenched and tempered at a temperature of 205 ºC. References Multi-Answer Difficulty: Medium Using Table A-21, how would you specify it if you desired to maximize the yield strength? To maximize the yield strength of AISI 1050 steel, it is quenched and tempered at a temperature of 205 ± 5% ºC. Explanation: From table A-21, to maximize the yield strength of AISI 1050 steel, it is quenched and tempered at a temperature of 205ºC.
2. Award: 10 out of 10.00 points Score: 20/20 Points 100 % Assume that you were specifying an AISI 1050 steel for an application. References Section Break Difficulty: Medium Using Table A-21, how would you specify it if you desired to maximize the ductility? To maximize the ductility of AISI 1050 steel, it is quenched and tempered at a temperature of 650 ºC. References Multi-Answer Difficulty: Medium Using Table A-21, how would you specify it if you desired to maximize the ductility? To maximize the ductility of AISI 1050 steel, it is quenched and tempered at a temperature of 650 ± 5% ºC. Explanation: From table A-21, to maximize the ductility of AISI 1050 steel, it is quenched and tempered at a temperature 650ºC.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
1. Award: 10 out of 10.00 points Score: 20/20 Points 100 % A part made from hot-rolled AISI 1212 steel undergoes a 19 percent cold-work operation. References Section Break Difficulty: Medium Obtain the yield strength and ultimate strength before and after the cold-work operation. Determine the percent increase in each strength. The yield strength before the cold-work operation is 28 kpsi. The yield strength after cold-work operation is 75.7 kpsi. The ultimate strength before cold-work operation is 61.5 kpsi. The ultimate strength after cold-work operation is 75.93 kpsi. The percent increase in yield strength is 170.35 percent. The percent increase in ultimate strength is 23.46 percent. References Numeric Response Difficulty: Medium Obtain the yield strength and ultimate strength before and after the cold-work operation. Determine the percent increase in each strength. The yield strength before the cold-work operation is 28 ± 5% kpsi. The yield strength after cold-work operation is 75.7 ± 5% kpsi. The ultimate strength before cold-work operation is 61.5 ± 5% kpsi. The ultimate strength after cold-work operation is 75.93 ± 5% kpsi. The percent increase in yield strength is 170.35 ± 5% percent. The percent increase in ultimate strength is 23.46 ± 5% percent. Explanation: For cold work operation, W = 19 percent. Before cold working, the properties of hot-rolled AISI 1212 steel from Table A-22 are S y = 28 kpsi, S u = 61.5 kpsi, σ 0 = 110.0 kpsi, m = 0.24, and ε f = 0.85 After cold working, Eq. (2-23) ε u = m = 0.24 From Eq. (2-28), From Eq. (2-30), From Eq. (2-32), The percentage increase in yield strength is The percentage increase in ultimate strength is
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
2. Award: 10 out of 10.00 points Score: 20/20 Points 100 % A part made from hot-rolled AISI 1212 steel undergoes a 19 percent cold-work operation. References Section Break Difficulty: Medium Determine the ratios of ultimate strength to yield strength before and after the cold-work operation.What does the result indicate about the change of ductility of the part? The ratio of ultimate strength to yield strength before cold-work operation is 2.2 . The ratio of ultimate strength to yield strength after cold-work operation is 1 . After the cold-work operation, the ductility of the part is reduced . References Multi-Answer Difficulty: Medium Determine the ratios of ultimate strength to yield strength before and after the cold-work operation.What does the result indicate about the change of ductility of the part? The ratio of ultimate strength to yield strength before cold-work operation is 2.2 ± 5% . The ratio of ultimate strength to yield strength after cold-work operation is 1 ± 5% . After the cold-work operation, the ductility of the part is reduced . Explanation: The ratio of ultimate strength to yield strength before the cold-work operation is The ratio of ultimate strength to yield strength after the cold-work operation is
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
1. Award: 10 out of 10.00 points Score: 10/10 Points 100 % A steel member has a Brinell of H B = 300. Estimate the ultimate strength of the steel in MPa. The ultimate strength of the given steel is 1020 MPa. References Numeric Response Difficulty: Easy A steel member has a Brinell of H B = 300. Estimate the ultimate strength of the steel in MPa. The ultimate strength of the given steel is 1020 ± 5% MPa. Explanation: Given that, H B = 300, from Eq. (2-36), ultimate strength is given as follows:
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
1. Award: 10 out of 10.00 points Score: 10/10 Points 100 % A gray cast iron part has a Brinell hardness number of H B = 145. Estimate the ultimate strength of the part in kpsi. Make a reasonable assessment of the likely grade of cast iron by comparing both hardness and strength to material options in Table A–24. The ultimate strength of the part is 20.85 kpsi. This is probably ASTM No. 20 Gray cast iron. References Numeric Response Difficulty: Easy A gray cast iron part has a Brinell hardness number of H B = 145. Estimate the ultimate strength of the part in kpsi. Make a reasonable assessment of the likely grade of cast iron by comparing both hardness and strength to material options in Table A–24. The ultimate strength of the part is 20.85 ± 5% kpsi. This is probably ASTM No. 20 ± 5% Gray cast iron. Explanation: Given that, for gray cast iron, H B = 145, from Eqn (2-37), the minimum strength, as defined by the ASTM is as follows: From Table A-24, this is probably ASTM No. 20 Gray cast iron.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
1. Award: 10 out of 10.00 points Score: 10/10 Points 100 % Brinell hardness tests were made on a random sample of 10 steel parts during processing. The results were H B values of 260,262(2), 264, 265(3), 266(2), and 269. Estimate the mean and standard deviation of the ultimate strength in kpsi. The mean of ultimate strength is 132.2 kpsi. The standard deviation of the ultimate strength is 1.27 kpsi. References Numeric Response Difficulty: Medium Brinell hardness tests were made on a random sample of 10 steel parts during processing. The results were H B values of 260,262(2), 264, 265(3), 266(2), and 269. Estimate the mean and standard deviation of the ultimate strength in kpsi. The mean of ultimate strength is 132.2 ± 5% kpsi. The standard deviation of the ultimate strength is 1.27 ± 5% kpsi. Explanation: From Eq. (2-36), For the data given, converting H B to S u using Eq. (2-36), H B S u (kpsi) S u 2 (kpsi) 260 130 16900 262 131 17161 262 131 17161 264 132 17424 265 132.5 17556.25 265 132.5 17556.25 265 132.5 17556.25 266 133 17689 266 133 17689 269 134.5 18090.25 From Eqn. (1-6), the discrete mean is From Eqn. (1-7), the standard deviation is
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help

Related Documents

Physics Lab 3.pdf
Homework 1 Solutions.docx
CONS6003 Week09 Quiz.pdf
lab_3_pre_lab_2.pdf
johnsmith_module1worksheet_10082023.docx
Dent 134 Case Study Assignment updated Dec 2 2020.docx
Homework 4 solution.pdf
Homework 3 solution.pdf
Homework 1 solution.pdf
Homework 5 solution.pdf
Homework 10 solution.pdf
Prelab-Shear Intro(1).docx
Related Questions
A routine quality control test is made on a structural steel bar that is 1 in. square and 16 in. long. The data developed during the test show that the bar elongated 0.0111 in. when subjected to a tensile force of 20.5 kips. Compute the modulus of elasticity of the steel. ΔL = PL/AE f=P/A ? = ΔL/L
Do not provide handwritten solution,Maintain accuracy and quality in your answer,Take care of plagiarism,Answer completely,You will get up vote for sure.
Please all steps
A specification for a stainless steel bar is given as follows: length: 10.40 cm + 0.02 cm  Find the upper limit, the lower limit, and the tolerance interval
QUESTION 14 The maximum shear stress in a solid round bar is Tmay - 16T/n(d). The bar is subjected to a torque of 1250 N-m with a standard deviation of 125 N-m. The shear yield strength = 320 MPa with a standard deviation of 32 MPa. Assuming normal distributions for the load and strength and a reliability factor of 0.90 against yielding, determine the design factor (na). QUESTION 15 The maximum shear stress in a solid round bar is Tmax- 16T/n(d). The bar is subjected to a torque of 1250 N-m with a standard deviation of 125 N-m. The shear yield strength = 320 MPa with a standard deviation of 32 MPa. Assuming normal distributions for the load and strength and a reliability factor of 0.90 against yielding, choose the "Preferred" size from Table A-17 based on the minimum calculated diameter of the bar.
mathlab code: %Press-Fit Example Problem 117 %Steel, psi ES=30*10^6 MuS=0.292; %Cast Iron EC=14.5*10^6; MuC=0.211; di=0; d0=2.002; Di=2.000; D0=3.000; % tolerance tol=d0-D0; % Pressure Eq. (3-56) R=Di/2; r0=D0/2; ri=di/2; Hub=(1/ES)*((r0^2+R^2)/(r0^2-R^2)+MuS);     Inner=(1/EC)*((R^2+ri^2)/(R^2-ri^2)-MuC);    p=tol/(R*(Hub+Inner))
2-15 Brinell hardness tests were made on a random sample of 10 steel parts during process- ing. The results were Ha values of 230, 232(2), 234, 235(3), 236(2), and 239. Estimate the mean and standard deviation of the ultimate strength in kpsi. Ans: 50.111 kpsi Som- / -1 bpci
When design for a cyclic load, if in one case, the designed reliability is 0.99, what is the approximate reliability factor which should be used according to the textbook?     1.0     0.9     0.81     0.75
A pivot in a linkage has a pin in the figure whose dimension a + ta is to be established. The thickness of the link clevis is 37.5 ± 0.125 mm. The designer has concluded that a gap of between 0.1 and 1.25 mm will satisfactorily sustain the function of the linkage pivot. Determine the dimension a and its tolerance. Clevis Pin Snap ring a±ta 1.05+0.05 37.5±0.125
For the following statements, answer True or False and explain the answer briefly. The tolerance of standard mass used must be smaller than the uncertainty of analytical balance to be calibrated.
SOLVE CAREFULLY!! Please Write Clearly and Box the final Answer
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
SEE MORE TEXTBOOKS
Related Questions
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
SEE MORE TEXTBOOKS