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MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-1-1 PROBLEM 1-1 Statement: It is often said, "Build a better mousetrap and the world will beat a path to your door." Consider this problem and write a goal statement and a set of at least 12 task specifications that you would apply to its solution. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts. Solution: Goal Statement: Create a mouse-free environment. Task Specifications: 1. Cost less than $1.00 per use or application. 2. Allow disposal without human contact with mouse. 3. Be safe for other animals such as house pets. 4. Provide no threat to children or adults in normal use. 5. Be a humane method for the mouse. 6. Be environmentally friendly. 7. Have a shelf-life of at least 3 months. 8. Leave no residue. 9. Create minimum audible noise in use. 10. Create no detectable odors within 1 day of use. 11. Be biodegradable. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be

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  1. 1. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-1-1 PROBLEM 1-1 Statement: It is often said, "Build a better mousetrap and the world will beat a path to your door." Consider this problem and write a goal statement and a set of at least 12 task specifications that you would apply to its solution. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts. Solution: Goal Statement: Create a mouse-free environment. Task Specifications: 1. Cost less than $1.00 per use or application. 2. Allow disposal without human contact with mouse. 3. Be safe for other animals such as house pets. 4. Provide no threat to children or adults in normal use. 5. Be a humane method for the mouse. 6. Be environmentally friendly. 7. Have a shelf-life of at least 3 months. 8. Leave no residue. 9. Create minimum audible noise in use. 10. Create no detectable odors within 1 day of use. 11. Be biodegradable. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities. P0101.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  2. 2. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-2-1 PROBLEM 1-2 Statement: A bowling machine is desired to allow quadriplegic youths, who can only move a joystick, to engage in the sport of bowling at a conventional bowling alley. Consider the factors involved, write a goal statement, and develop a set of at least 12 task specifications that constrain this problem. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts. Solution: Goal Statement: Create a means to allow a quadriplegic to bowl. Task Specifications: 1. Cost no more than $2 000. 2. Portable by no more than two able-bodied adults. 3. Fit through a standard doorway. 4. Provide no threat of injury to user in normal use. 5. Operate from a 110 V, 60 Hz, 20 amp circuit. 6. Be visually unthreatening. 7. Be easily positioned at bowling alley. 8. Have ball-aiming ability, controllable by user. 9. Automatically reload returned balls. 10. Require no more than 1 able-bodied adult for assistance in use. 11. Ball release requires no more than a mouth stick-switch closure. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities. P0102.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  3. 3. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-3-1 PROBLEM 1-3 Statement: A quadriplegic needs an automated page turner to allow her to read books without assistance. Consider the factors involved, write a goal statement, and develop a set of at least 12 task specifications that constrain this problem. Then suggest 3 possible concepts to achieve the goal. Make annotated, freehand sketches of the concepts. Solution: Goal Statement: Create a means to allow a quadriplegic to read standard books with minimum assistance. Task Specifications: 1. Cost no more than $1 000. 2. Useable in bed or from a seated position 3. Accept standard books from 8.5 x 11 in to 4 x 6 in in planform and up to 1.5 in thick. 4. Book may be placed, and device set up, by able-bodied person. 5. Operate from a 110 V, 60 Hz, 15 amp circuit or by battery power. 6. Be visually unthreatening and safe to use. 7. Require no more than 1 able-bodied adult for assistance in use. 8. Useable in absence of assistant once set up. 9. Not damage books. 10. Timing controlled by user. 11. Page turning requires no more than a mouth stick-switch closure. 12. Be simple to use with minimal written instructions necessary. Concepts and sketches are left to the student. There are an infinity of possibilities. P0103.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  4. 4. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-4-1 PROBLEM 1-4 Statement: Convert a mass of 1 000 lbm to (a) lbf, (b) slugs, (c) blobs, (d) kg. Units: blob lbf sec 2 in := Given: Mass M 1000 lb:= Solution: See Mathcad file P0104. 1. To determine the weight of the given mass, multiply the mass value by the acceleration due to gravity, g. W M g:= W 1000lbf= 2. Convert mass units by assigning different units to the units place-holder when displaying the mass value. Slugs M 31.081 slug= Blobs M 2.59 blob= Kilograms M 453.592 kg= P0104.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  5. 5. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-5-1 PROBLEM 1-5 Statement: A 250-lbm mass is accelerated at 40 in/sec2. Find the force in lb needed for this acceleration. Given: Mass M 250 lb:= Acceleration a 40 in sec 2 := Solution: See Mathcad file P0105. 1. To determine the force required, multiply the mass value, in slugs, by the acceleration in feet per second squared. Convert mass to slugs: M 7.770 slug= Convert acceleration to feet per second squared: a 3.333s 2- ft= F M a:= F 25.9 lbf= P0105.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  6. 6. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-6-1 PROBLEM 1-6 Statement: Express a 100-kg mass in units of slugs, blobs, and lbm. How much does this mass weigh? Units: blob lbf sec 2 in Given: M 100 kg Assumptions: The mass is at sea-level and the gravitational acceleration is g 32.174 ft sec 2 or g 386.089 in sec 2 or g 9.807 m sec 2 Solution: See Mathcad file P0106. 1. Convert mass units by assigning different units to the units place-holder when displaying the mass value. The mass, in slugs, is M 6.85 slug The mass, in blobs, is M 0.571 blob The mass, in lbm, is M 220.5 lb Note: Mathcad uses lbf for pound-force, and lb for pound-mass. 2. To determine the weight of the given mass, multiply the mass value by the acceleration due to gravity, g. The weight, in lbf, is W M g W 220.5 lbf The weight, in N, is W M g W 980.7 N 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  7. 7. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-7-1 PROBLEM 1-7 Statement: Prepare an interactive computer program (using, for example, Excell, Mathcad, or TKSolver) from which the cross-sectional properties for the shapes shown in the inside front cover can be calculated. Arrange the program to deal with both ips and SI unit systems and convert the results between those systems. Solution: See the inside front cover and Mathcad file P0107. 1. Rectangle, let: b 3 in h 4 in Area A b h A 12.000 in 2 A 7742 mm 2 Moment about x-axis Ix b h 3 12 Ix 16.000 in 4 Ix 6.660 10 6 mm 4 Moment about y-axis Iy h b 3 12 Iy 9.000 in 4 Iy 3.746 10 6 mm 4 Radius of gyration about x-axis kx Ix A kx 1.155 in kx 29.329 mm Radius of gyration about y-axis ky Iy A ky 0.866 in ky 21.997 mm Polar moment of inertia Jz Ix Iy Jz 25.000 in 4 Jz 1.041 10 7 mm 4 2. Solid circle, let: D 3 in Area A D 2 4 A 7.069 in 2 A 4560 mm 2 Moment about x-axis Ix D 4 64 Ix 3.976 in 4 Ix 1.655 10 6 mm 4 Moment about y-axis Iy D 4 64 Iy 3.976 in 4 Iy 1.655 10 6 mm 4 Radius of gyration about x-axis kx Ix A kx 0.750 in kx 19.05 mm P0107.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  8. 8. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-7-2 Radius of gyration about y-axis ky Iy A ky 0.750 in ky 19.05 mm Polar moment of inertia Jz D 4 32 Jz 7.952 in 4 Jz 3.310 10 6 mm 4 3. Hollow circle, let: D 3 in d 1 in Area A 4 D 2 d 2 A 6.283 in 2 A 4054 mm 2 Moment about x-axis Ix 64 D 4 d 4 Ix 3.927 in 4 Ix 1.635 10 6 mm 4 Moment about y-axis Iy 64 D 4 d 4 Iy 3.927 in 4 Iy 1.635 10 6 mm 4 Radius of gyration about x-axis kx Ix A kx 0.791 in kx 20.08 mm Radius of gyration about y-axis ky Iy A ky 0.791 in ky 20.08 mm Polar moment of inertia Jz 32 D 4 d 4 Jz 7.854 in 4 Jz 3.269 10 6 mm 4 4. Solid semicircle, let: D 3 in R 0.5 D R 1.5 in Area A D 2 8 A 3.534 in 2 A 2280 mm 2 Moment about x-axis Ix 0.1098 R 4 Ix 0.556 in 4 Ix 2.314 10 5 mm 4 Moment about y-axis Iy R 4 8 Iy 1.988 in 4 Iy 8.275 10 5 mm 4 P0107.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  9. 9. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-7-3 Radius of gyration about x-axis kx Ix A kx 0.397 in kx 10.073 mm Radius of gyration about y-axis ky Iy A ky 0.750 in ky 19.05 mm Polar moment of inertia Jz Ix Iy Jz 2.544 in 4 Jz 1.059 10 6 mm 4 Distances to centroid a 0.4244 R a 0.637 in a 16.17 mm b 0.5756 R b 0.863 in b 21.93 mm 5. Right triangle, let: b 2 in h 1 in Area A b h 2 A 1.000 in 2 A 645 mm 2 Moment about x-axis Ix b h 3 36 Ix 0.056 in 4 Ix 2.312 10 4 mm 4 Moment about y-axis Iy h b 3 36 Iy 0.222 in 4 Iy 9.250 10 4 mm 4 Radius of gyration about x-axis kx Ix A kx 0.236 in kx 5.987 mm Radius of gyration about y-axis ky Iy A ky 0.471 in ky 11.974 mm Polar moment of inertia Jz Ix Iy Jz 0.278 in 4 Jz 1.156 10 5 mm 4 P0107.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  10. 10. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-1 PROBLEM 1-8 Statement: Prepare an interactive computer program (using, for example, Excell, Mathcad, or TKSolver) from which the mass properties for the solids shown in the page opposite the inside front cover can be calculated. Arrange the program to deal with both ips and SI unit systems and convert the results between those systems. Units: blob lbf sec 2 in Solution: See the page opposite the inside front cover and Mathcad file P0108. 1. Rectangular prism, let: a 2 in b 3 in c 4 in 0.28 lbf in 3 Volume V a b c V 24.000 in 3 V 393290 mm 3 Mass M V g M 0.017 blob M 3.048 kg Moment about x-axis Ix M a 2 b 2 12 Ix 0.019 blob in 2 Ix 2130.4 kg mm 2 Moment about y-axis Iy M a 2 c 2 12 Iy 0.029 blob in 2 Iy 3277.6 kg mm 2 Moment about z-axis Iz M b 2 c 2 12 Iz 0.036 blob in 2 Iz 4097.0 kg mm 2 Radius of gyration about x-axis kx Ix M kx 1.041 in kx 26.437 mm Radius of gyration about y-axis ky Iy M ky 1.291 in ky 32.791 mm Radius of gyration about z-axis kz Iz M kz 1.443 in kz 36.662 mm 2.Cylinder, let: r 2 in L 3 in 0.30 lbf in 3 Volume V r 2 L V 37.699 in 3 V 617778 mm 3 Mass M V g M 0.029 blob M 5.13 kg 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  11. 11. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-2 Moment about x-axis Ix M r 2 2 Ix 0.059 blob in 2 Ix 6619.4 kg mm 2 Moment about y-axis Iy M 3 r 2 L 2 12 Iy 0.051 blob in 2 Iy 5791.9 kg mm 2 Moment about z-axis Iz M 3 r 2 L 2 12 Iz 0.051 blob in 2 Iz 5791.9 kg mm 2 Radius of gyration about x-axis kx Ix M kx 1.414 in kx 35.921 mm Radius of gyration about y-axis ky Iy M ky 1.323 in ky 33.601 mm Radius of gyration about z-axis kz Iz M kz 1.323 in kz 33.601 mm 3. Hollow cylinder, let: a 2 in b 3 in L 4 in 0.28 lbf in 3 Volume V b 2 a 2 L V 62.832 in 3 V 1029630 mm 3 Mass M V g M 0.046 blob M 7.98 kg Moment about x-axis Ix M 2 a 2 b 2 Ix 0.296 blob in 2 Ix 3.3 10 4 kg mm 2 Moment about y-axis Iy M 12 3 a 2 3 b 2 L 2 Iy 0.209 blob in 2 Iy 2.4 10 4 kg mm 2 Moment about z-axis Iz M 12 3 a 2 3 b 2 L 2 Iz 0.209 blob in 2 Iz 2.4 10 4 kg mm 2 Radius of gyration about x-axis kx Ix M kx 2.550 in kx 64.758 mm Radius of gyration about y-axis ky Iy M ky 2.141 in ky 54.378 mm 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  12. 12. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-3 Radius of gyration about z-axis kz Iz M kz 2.141 in kz 54.378 mm 4. Right circular cone, let: r 2 in h 5 in 0.28 lbf in 3 Volume V r 2 h 3 V 20.944 in 3 V 343210 mm 3 Mass M V g M 0.015 blob M 2.66 kg Moment about x-axis Ix 3 10 M r 2 Ix 0.018 blob in 2 Ix 2059.4 kg mm 2 Moment about y-axis Iy M 12 r 2 3 h 2 80 Iy 0.023 blob in 2 Iy 2638.5 kg mm 2 Moment about z-axis Iz M 12 r 2 3 h 2 80 Iz 0.023 blob in 2 Iz 2638.5 kg mm 2 Radius of gyration about x-axis kx Ix M kx 1.095 in kx 27.824 mm Radius of gyration about y-axis ky Iy M ky 1.240 in ky 31.495 mm Radius of gyration about z-axis kz Iz M kz 1.240 in kz 31.495 mm 5. Sphere, let: r 3 in Volume V 4 3 r 3 V 113.097 in 3 V 1853333 mm 3 Mass M V g M 0.082 blob M 14.364 kg Moment about x-axis Ix 2 5 M r 2 Ix 0.295 blob in 2 Ix 33362 kg mm 2 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  13. 13. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-8-4 Moment about y-axis Iy 2 5 M r 2 Iy 0.295 blob in 2 Iy 33362 kg mm 2 Moment about z-axis Iz 2 5 M r 2 Iz 0.295 blob in 2 Iz 33362 kg mm 2 Radius of gyration about x-axis kx Ix M kx 1.897 in kx 48.193 mm Radius of gyration about y-axis ky Iy M ky 1.897 in ky 48.193 mm Radius of gyration about z-axis kz Iz M kz 1.897 in kz 48.193 mm 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  14. 14. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-9-1 PROBLEM 1-9 Statement: Convert the template in Problem 1-7 to have and use a set of functions or subroutines that can be called from within any program in that language to solve for the cross-sectional properties of the shapes shown on the inside front cover. Solution: See inside front cover and Mathcad file P0109. 1. Rectangle: Area A b h( ) b h Moment about x-axis Ix b h( ) b h 3 12 Moment about y-axis Iy b h( ) h b 3 12 2. Solid circle: Area A D( ) D 2 4 Moment about x-axis Ix D( ) D 4 64 Moment about y-axis Iy D( ) D 4 64 3. Hollow circle: Area A D d( ) 4 D 2 d 2 Moment about x-axis Ix D d( ) 64 D 4 d 4 Moment about y-axis Iy D d( ) 64 D 4 d 4 4. Solid semicircle: Area A D( ) D 2 8 Moment about x-axis Ix R( ) 0.1098 R 4 Moment about y-axis Iy R( ) R 4 8 5. Right triangle: Area A b h( ) b h 2 Moment about x-axis Ix b h( ) b h 3 36 Moment about y-axis Iy b h( ) h b 3 36 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  15. 15. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-10-1 PROBLEM 1-10 Statement: Convert the template in Problem 1-8 to have and use a set of functions or subroutines that can be called from within any program in that language to solve for the cross-sectional properties of the shapes shown on the page opposite the inside front cover. Solution: See the page opposite the inside front cover and Mathcad file P0110. 1 Rectangular prism: Volume V a b c( ) a b c Mass M a b c ( ) V a b c( ) g Moment about x-axis Ix a b c ( ) M a b c ( ) a 2 b 2 12 Moment about y-axis Iy a b c ( ) M a b c ( ) a 2 c 2 12 Moment about z-axis Iz a b c ( ) M a b c ( ) b 2 c 2 12 2. Cylinder: Volume V r L( ) r 2 L Mass M r L ( ) V r L( ) g Moment about x-axis Ix r L ( ) M r L ( ) r 2 2 Moment about y-axis Iy r L ( ) M r L ( ) 3 r 2 L 2 12 Moment about z-axis Iz r L ( ) M r L ( ) 3 r 2 L 2 12 3. Hollow cylinder: Volume V a b L( ) b 2 a 2 L Mass M a b L ( ) V a b L( ) g Moment about x-axis Ix a b L ( ) M a b L ( ) 2 a 2 b 2 Moment about y-axis Iy a b L ( ) M a b L ( ) 12 3 a 2 3 b 2 L 2 Moment about z-axis Iz a b L ( ) M a b L ( ) 12 3 a 2 3 b 2 L 2 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  16. 16. MACHINE DESIGN - An Integrated Approach, 4th Ed. 1-10-2 4. Right circular cone: Volume V r h( ) r 2 h 3 Mass M r h ( ) V r h( ) g Moment about x-axis Ix r h ( ) 3 10 M r h ( ) r 2 Moment about y-axis Iy r h ( ) M r h ( ) 12 r 2 3 h 2 80 Moment about z-axis Iz r h ( ) M r h ( ) 12 r 2 3 h 2 80 5. Sphere: Volume V r( ) 4 3 r 3 Mass M r ( ) V r( ) g Moment about x-axis Ix r ( ) 2 5 M r ( ) r 2 Moment about y-axis Iy r ( ) 2 5 M r ( ) r 2 Moment about z-axis Iz r ( ) 2 5 M r ( ) r 2 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  17. 17. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-1-1 PROBLEM 2-1 Statement: Figure P2-1 shows stress-strain curves for three failed tensile-test specimens. All are plotted on the same scale. (a) Characterize each material as brittle or ductile. (b) Which is the stiffest? (c) Which has the highest ultimate strength? (d) Which has the largest modulus of resilience? (e) Which has the largest modulus of toughness? Solution: See Figure P2-1 and Mathcad file P0201. 1. The material in Figure P2-1(a) has a moderate amount of strain beyond the yield point, P2-1(b) has very little, and P2-1(c) has considerably more than either of the other two. Based on this observation, the material in Figure P2-1(a) is mildly ductile, that in P2-1(b)is brittle, and that in P2-1(c) is ductile. 2. The stiffest material is the one with the grearesr slope in the elastic range. Determine this by dividing the rise by the run of the straight-line portion of each curve. The material in Figure P2-1(c) has a slope of 5 stress units per strain unit, which is the greatest of the three. Therefore, P2-1(c) is the stiffest. 3. Ultimate strength corresponds to the highest stress that is achieved by a material under test. The material in Figure P2-1(b) has a maximum stress of 10 units, which is considerably more than either of the other two. Therefore, P2-1(b) has the highest ultimate strength. 4. The modulus of resilience is the area under the elastic portion of the stress-starin curve. From observation of the three graphs, the stress and strain values at the yield points are: P2-1(a) ya 5:= ya 5:= P2-1(b) yb 9:= yb 2:= P2-1(c) yc 5:= yc 1:= Using equation (2.7), the modulus of resiliency for each material is, approximately, P21a 1 2 ya ya:= P21a 12.5= P21b 1 2 yb yb:= P21b 9= P21c 1 2 yc yc:= P21c 2.5= P2-1 (a) has the largest modulus of resilience 5. The modulus of toughness is the area under the stress-starin curve up to the point of fracture. By inspection, P2-1 (c) has the largest area under the stress-strain curve therefore, it has the largest modulus of toughness. P0201.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  18. 18. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-2-1 PROBLEM 2-2 Statement: Determine an approximate ratio between the yield strength and ultimate strength for each material shown in Figure P2-1. Solution: See Figure P2-1 and Mathcad file P0202. 1. The yield strength is the value of stress at which the stress-strain curve begins to be nonlinear. The ultimate strength is the maximum value of stress attained during the test. From the figure, we have the following values of yield strength and tensile strength: Figure P2-1(a) Sya 5:= Sua 6:= Figure P2-1(b) Syb 9:= Sub 10:= Figure P2-1(c) Syc 5:= Suc 8:= 2. The ratio of yield strength to ultimate strength for each material is: Figure P2-1(a) ratioa Sya Sua := ratioa 0.83= Figure P2-1(b) ratiob Syb Sub := ratiob 0.90= Figure P2-1(c) ratioc Syc Suc := ratioc 0.63= P0202.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  19. 19. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-3-1 PROBLEM 2-3 Statement: Which of the steel alloys shown in Figure 2-19 would you choose to obtain (a) Maximum strength (b) Maximum modulus of resilience (c) Maximum modulus of toughness (d) Maximum stiffness Given: Young's modulus for steel E 207 GPa Solution: See Figure 2-19 and Mathcad file P0203. 1. Determine from the graph: values for yield strength, ultimate strength and strain at fracture for each material. Steel Yield Strength Ultimate Strength Fracture Strain AISI 1020: Sy1020 300 MPa Sut1020 400 MPa f1020 0.365 AISI 1095: Sy1095 550 MPa Sut1095 1050 MPa f1095 0.11 AISI 4142: Sy4142 1600 MPa Sut4142 2430 MPa f4142 0.06 Note: The 0.2% offset method was used to define a yield strength for the AISI 1095 and the 4142 steels. 2. From the values of Sut above it is clear that the AISI 4142 has maximum strength. 3. Using equation (2-7) and the data above, determine the modulus of resilience. UR1020 1 2 Sy1020 2 E UR1020 0.22 MN m m 3 UR1095 1 2 Sy1095 2 E UR1095 0.73 MN m m 3 UR4142 1 2 Sy4142 2 E UR4142 6.18 MN m m 3 Even though the data is approximate, the AISI 4142 clearly has the largest modulus of resilience. 4. Using equation (2-8) and the data above, determine the modulus of toughness. UT1020 1 2 Sy1020 Sut1020 f1020 UT1020 128 MN m m 3 UT1095 1 2 Sy1095 Sut1095 f1095 UT1095 88 MN m m 3 UT4142 1 2 Sy4142 Sut4142 f4142 UT4142 121 MN m m 3 Since the data is approximate, there is no significant difference between the 1020 and 4142 steels. Because of the wide difference in shape and character of the curves, one should also determine the area under the curves by graphical means. When this is done, the area under the curve is about 62 square units for 1020 and 66 for 4142. Thus, they seem to have about equal toughness, which is about 50% greater than that for the 1095 steel. 5. All three materials are steel therefore, the stiffnesses are the same. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  20. 20. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-4-1 PROBLEM 2-4 Statement: Which of the aluminum alloys shown in Figure 2-21 would you choose to obtain (a) Maximum strength (b) Maximum modulus of resilience (c) Maximum modulus of toughness (d) Maximum stiffness Given: Young's modulus for aluminum E 71.7 GPa Solution: See Figure 2-21 and Mathcad file P0204. 1. Determine, from the graph, values for yield strength, ultimate strength and strain at fracture for each material. Alum Yield Strength Ultimate Strength Fracture Strain 1100: Sy1100 120 MPa Sut1100 130 MPa f1100 0.170 2024-T351: Sy2024 330 MPa Sut2024 480 MPa f2024 0.195 7075-T6: Sy7075 510 MPa Sut7075 560 MPa f7075 0.165 Note: The 0.2% offset method was used to define a yield strength for all of the aluminums. 2. From the values of Sut above it is clear that the 7075-T6 has maximum strength. 3. Using equation (2-7) and the data above, determine the modulus of resilience. UR1100 1 2 Sy1100 2 E UR1100 0.10 MN m m 3 UR2024 1 2 Sy2024 2 E UR2024 0.76 MN m m 3 UR7075 1 2 Sy7075 2 E UR7075 1.81 MN m m 3 Even though the data is approximate, the 7075-T6 clearly has the largest modulus of resilience. 4. Using equation (2-8) and the data above, determine the modulus of toughness. UT1100 1 2 Sy1100 Sut1100 f1100 UT1100 21 MN m m 3 UT2024 1 2 Sy2024 Sut2024 f2024 UT2024 79 MN m m 3 UT7075 1 2 Sy7075 Sut7075 f7075 UT7075 88 MN m m 3 Even though the data is approximate, the 7075-T6 has the largest modulus of toughness. 5. All three materials are aluminum therefore, the stiffnesses are the same. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  21. 21. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-5-1 PROBLEM 2-5 Statement: Which of the thermoplastic polymers shown in Figure 2-22 would you choose to obtain (a) Maximum strength (b) Maximum modulus of resilience (c) Maximum modulus of toughness (d) Maximum stiffness Solution: See Figure 2-22 and Mathcad file P0205. 1. Determine, from the graph, values for yield strength, ultimate strength, strain at fracture, and modulus of elasticity for each material. Plastic Yield Strength Ultimate Strength Fracture Strain Mod of Elasticity Nylon 101: SyNylon 63 MPa SutNylon 80 MPa fNylon 0.52 ENylon 1.1 GPa HDPE: SyHDPE 15 MPa SutHDPE 23 MPa fHDPE 3.0 EHDPE 0.7 GPa PTFE: SyPTFE 8.3 MPa SutPTFE 13 MPa fPTFE 0.51 EPTFE 0.8 GPa 2. From the values of Sut above it is clear that the Nylon 101 has maximum strength. 3. Using equation (2-7) and the data above, determine the modulus of resilience. URNylon 1 2 SyNylon 2 ENylon URNylon 1.8 MN m m 3 URHDPE 1 2 SyHDPE 2 EHDPE URHDPE 0.16 MN m m 3 URPTFE 1 2 SyPTFE 2 EPTFE URPTFE 0.04 MN m m 3 Even though the data is approximate, the Nylon 101 clearly has the largest modulus of resilience. 4. Using equation (2-8) and the data above, determine the modulus of toughness. UTNylon 1 2 SyNylon SutNylon fNylon UTNylon 37 MN m m 3 UTHDPE 1 2 SyHDPE SutHDPE fHDPE UTHDPE 57 MN m m 3 UTPTFE 1 2 SyPTFE SutPTFE fPTFE UTPTFE 5 MN m m 3 Even though the data is approximate, the HDPE has the largest modulus of toughness. 5. The Nylon 101 has the steepest slope in the (approximately) elastic range and is, therefore, the stiffest of the three materials.. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  22. 22. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-6-1 PROBLEM 2-6 Statement: A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 12.8-mm dia and has a 50-mm gage length. Can you define the type of metal based on the given data? Given: Elastic limit: Strength Sel 414 MPa Strain el 0.002 Test specimen: Diameter do 12.8 mm Length Lo 50 mm Solution: See Mathcad file P0206. 1. The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is E Sel el E 207 GPa 2. The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or U'el 1 2 Sel el U'el 414 kN m m 3 The total strain energy in the specimen is the strain energy per unit volume times the volume, Uel U'el do 2 4 Lo Uel 2.7 N m 3. Based on the modulus of elasticity and using Table C-1, the material is steel. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  23. 23. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-7-1 PROBLEM 2-7 Statement: A metal has a strength of 41.2 kpsi (284 MPa) at its elastic limit and the strain at that point is 0.004. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 0.505-in dia and has a 2-in gage length. Can you define the type of metal based on the given data? Given: Elastic limit: Strength Sel 41.2 ksi Strain el 0.004 Sel 284 MPa Test specimen: Diameter do 0.505 in Length Lo 2.00 in Solution: See Mathcad file P0207. 1. The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is E Sel el E 10.3 10 6 psi E 71 GPa 2. The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or U'el 1 2 Sel el U'el 82.4 lbf in in 3 U'el 568 kN m m 3 The total strain energy in the specimen is the strain energy per unit volume times the volume, Uel U'el do 2 4 Lo Uel 33.0 in lbf 3. Based on the modulus of elasticity and using Table C-1, the material is aluminum. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  24. 24. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-8-1 PROBLEM 2-8 Statement: A metal has a strength of 134 MPa at its elastic limit and the strain at that point is 0.006. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 12.8-mm dia and has a 50-mm gage length. Can you define the type of metal based on the given data? Given: Elastic limit: Strength Sel 134 MPa Strain el 0.003 Test specimen: Diameter do 12.8 mm Length Lo 50 mm Solution: See Mathcad file P0208. 1. The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is E Sel el E 45 GPa 2. The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or U'el 1 2 Sel el U'el 201 kN m m 3 The total strain energy in the specimen is the strain energy per unit volume times the volume, Uel U'el do 2 4 Lo Uel 1.3 N m 3. Based on the modulus of elasticity and using Table C-1, the material is magnesium. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  25. 25. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-9-1 PROBLEM 2-9 Statement: A metal has a strength of 100 kpsi (689 MPa) at its elastic limit and the strain at that point is 0.006. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 0.505-in dia and has a 2-in gage length. Can you define the type of metal based on the given data? Given: Elastic limit: Strength Sel 100 ksi Strain el 0.006 Sel 689 MPa Test specimen: Diameter do 0.505 in Length Lo 2.00 in Solution: See Mathcad file P0209. 1. The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is E Sel el E 16.7 10 6 psi E 115 GPa 2. The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or U'el 1 2 Sel el U'el 300 lbf in in 3 U'el 2 10 3 kN m m 3 The total strain energy in the specimen is the strain energy per unit volume times the volume, Uel U'el do 2 4 Lo Uel 120.18 in lbf 3. Based on the modulus of elasticity and using Table C-1, the material is titanium. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  26. 26. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-10-1 PROBLEM 2-10 Statement: A material has a yield strength of 689 MPa at an offset of 0.6% strain. What is its modulus of resilience? Units: MJ 10 6 joule Given: Yield strength Sy 689 MPa Yield strain y 0.006 Solution: See Mathcad file P0210. 1. The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately UR 1 2 Sy y UR 2.067 MJ m 3 UR 2.1 MPa 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  27. 27. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-11-1 PROBLEM 2-11 Statement: A material has a yield strength of 60 ksi (414 MPa) at an offset of 0.2% strain. What is its modulus of resilience? Units: MJ 10 6 joule Given: Yield strength Sy 60 ksi Sy 414 MPa Yield strain y 0.002 Solution: See Mathcad file P0211. 1. The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately UR 1 2 Sy y UR 60 in lbf in 3 UR 0.414 MJ m 3 UR 0.414 MPa 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  28. 28. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-12-1 PROBLEM 2-12 Statement: A steel has a yield strength of 414 MPa, an ultimate tensile strength of 689 MPa, and an elongation at fracture of 15%. What is its approximate modulus of toughness? What is the approximate modulus of resilience? Given: Sy 414 MPa Sut 689 MPa f 0.15 Solution: See Mathcad file P0212. 1. Determine the modulus of toughness using Equation (2.8). UT Sy Sut 2 f UT 82.7 MN m m 3 UT 82.7 MPa 2. Determine the modulus of resilience using Equation (2.7) and Young's modulus for steel: E 207 GPa UR 1 2 Sy 2 E UR 414 kN m m 3 UR 0.41 MPa 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  29. 29. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-13-1 PROBLEM 2-13 Statement: The Brinell hardness of a steel specimen was measured to be 250 HB. What is the material's approximate tensile strength? What is the hardness on the Vickers scale? The Rockwell scale? Given: Brinell hardness of specimen HB 250 Solution: See Mathcad file P0213. 1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3. Sut 0.5 HB ksi Sut 125 ksi Sut 862 MPa 2. From Table 2-3 (using linear interpolation) the hardness on the Vickers scale is HV HB 241 277 241 292 253( ) 253 HV 263 3. From Table 2-3 (using linear interpolation) the hardness on the Rockwell C scale is HRC HB 241 277 241 28.8 22.8( ) 22.8 HRC 24.3 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  30. 30. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-14-1 PROBLEM 2-14 Statement: The Brinell hardness of a steel specimen was measured to be 340 HB. What is the material's approximate tensile strength? What is the hardness on the Vickers scale? The Rockwell scale? Given: Brinell hardness of specimen HB 340 Solution: See Mathcad file P0214. 1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3. Sut 0.5 HB ksi Sut 170 ksi Sut 1172 MPa 2. From Table 2-3 (using linear interpolation) the hardness on the Vickers scale is HV HB 311 341 311 360 328( ) 328 HV 359 3. From Table 2-3 (using linear interpolation) the hardness on the Rockwell C scale is HRC HB 311 341 311 36.6 33.1( ) 33.1 HRC 36.5 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  31. 31. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-15-1 PROBLEM 2-15 Statement: What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it hardenable? By what techniques? Solution: See Mathcad file P0215. 1. Determine the principal alloying elements from Table 2-5 for 43xx steel.. 1.82% Nickel 0.50 or 0.80% Chromium 0.25% Molybdenum 2. From "Steel Numbering Systems" in Section 2.6, the carbon content is From the last two digits, the carbon content is 0.40%. 3. Is it hardenable? Yes, all of the alloying elements increase the hardenability. By what techniques? It can be through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4). P0215.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  32. 32. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-16-1 PROBLEM 2-16 Statement: What are the principal alloy elements of an AISI 1095 steel? How much carbon does it have? Is it hardenable? By what techniques? Solution: See Mathcad file P0216. 1. Determine the principal alloying elements from Table 2-5 for 10xx steel. Carbon only, no alloying elements 2. From "Steel Numbering Systems" in Section 2.6, the carbon content is From the last two digits, the carbon content is 0.95%. 3. Is it hardenable? Yes, as a high-carbon steel, it has sufficient carbon content for hardening. By what techniques? It can be through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4). P0216.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  33. 33. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-17-1 PROBLEM 2-17 Statement: What are the principal alloy elements of an AISI 6180 steel? How much carbon does it have? Is it hardenable? By what techniques? Solution: See Mathcad file P0217. 1. Determine the principal alloying elements from Table 2-5 for 61xx steel.. 0.15% Vanadium 0.60 to 0.95% Chromium 2. From "Steel Numbering Systems" in Section 2.6, the carbon content is From the last two digits, the carbon content is 0.80%. 3. Is it hardenable? Yes, all of the alloying elements increase the hardenability. By what techniques? It can be through hardened by heating, quenching and tempering; and it can also be case hardened (See Section 2.4). P0217.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  34. 34. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-18-1 PROBLEM 2-18 Statement: Which of the steels in Problems 2-15, 2-16, and 2-17 is the stiffest? Solution: See Mathcad file P0218. 1. None. All steel alloys have the same Young's modulus, which determines stiffness. P0218.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  35. 35. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-19-1 PROBLEM 2-19 Statement: Calculate the specific strength and specific stiffness of the following materials and pick one for use in an aircraft wing spar. Given: Material Code Ultimate Strength Young's Modulus Weight Density Steel st 0 Sut st 80 ksi E st 30 10 6 psi st 0.28 lbf in 3 Aluminum al 1 Sut al 60 ksi E al 10.4 10 6 psi al 0.10 lbf in 3 Titanium ti 2 Sut ti 90 ksi E ti 16.5 10 6 psi ti 0.16 lbf in 3 Index i 0 1 2 Solution: See Mathcad file P0219. 1. Specific strength is the ultimate tensile strength divided by the weight density and specific stiffness is the modulus of elasticity divided by the weight density. The text does not give a symbol to these quantities. Specific strength Sut i i 1 in 328610 360010 356310 Specific stiffness E i i 1 in 610710 610410 610310 Steel Aluminum Titanium 2. Based on the results above, all three materials have the same specific stiffness but the aluminum has the largest specific strength. Aluminum for the aircraft wing spar is recommended. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  36. 36. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-20-1 PROBLEM 2-20 Statement: If maximum impact resistance were desired in a part, which material properties would you look for? Solution: See Mathcad file P0220. 1. Ductility and a large modulus of toughness (see "Impact Resistance" in Section 2.1). P0220.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  37. 37. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-21-1 PROBLEM 2-21 _____ Statement: Refer to the tables of material data in Appendix A and determine the strength-to-weight ratios of the following material alloys based on their tensile yield strengths: heat-treated 2024 aluminum, SAE 1040 cold-rolled steel, Ti-75A titanium, type 302 cold-rolled stainless steel. Given: Material Yield Strength Specific Weight Mat 1 "2024 Aluminum, HT" Sy 1 290 MPa 1 0.10 lbf in 3 1 27.14 kN m 3 Mat 2 "1040 CR Steel" Sy 2 490 MPa 2 0.28 lbf in 3 2 76.01 kN m 3 Mat 3 "Ti-75A Titanium" Sy 3 517 MPa 3 0.16 lbf in 3 3 43.43 kN m 3 Mat 4 "Type 302 CR SS" Sy 4 1138 MPa 4 0.28 lbf in 3 4 76.01 kN m 3 i 1 2 4 Solution: See Mathcad file P0221. 1. Calculate the strength-to-weight ratio for each material as described in Section 2.1. SWR i Sy i i SWR i 10 4 m 1.068 0.645 1.190 1.497 Mat i "2024 Aluminum, HT" "1040 CR Steel" "Ti-75A Titanium" "Type 302 CR SS" 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  38. 38. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-22-1 PROBLEM 2-22 _____ Statement: Refer to the tables of material data in Appendix A and determine the strength-to-weight ratios of the following material alloys based on their ultimate tensile strengths: heat-treated 2024 aluminum SAE 1040 cold-rolled steel, unfilled acetal plastic, Ti-75A titanium, type 302 cold-rolled stainless steel. Given: Material Tensile Strength Specific Weight Mat 1 "2024 Aluminum, HT" Sut 1 441 MPa 1 0.10 lbf in 3 1 27.14 kN m 3 Mat 2 "1040 CR Steel" Sut 2 586 MPa 2 0.28 lbf in 3 2 76.01 kN m 3 Mat 3 "Acetal, unfilled" Sut 3 60.7 MPa 3 0.051 lbf in 3 3 13.84 kN m 3 Mat 4 "Ti-75A Titanium" Sut 4 586 MPa 4 0.16 lbf in 3 4 43.43 kN m 3 Mat 5 "Type 302 CR SS" Sut 5 1310 MPa 5 0.28 lbf in 3 5 76.01 kN m 3 i 1 2 5 Solution: See Mathcad file P0222. 1. Calculate the strength-to-weight ratio for each material as described in Section 2.1. SWR i Sut i i SWR i 10 4 m 1.625 0.771 0.438 1.349 1.724 Mat i "2024 Aluminum, HT" "1040 CR Steel" "Acetal, unfilled" "Ti-75A Titanium" "Type 302 CR SS" 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  39. 39. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-23-1 PROBLEM 2-23 _____ Statement: Refer to the tables of material data in Appendix A and calculate the specific stiffness of aluminum, titanium, gray cast iron, ductile iron, bronze, carbon steel, and stainless steel. Rank them in increasing order of this property and discuss the engineering significance of these data. Units: Mg 10 3 kg Given: Material Modulus of Elasticity Density Mat 1 "Aluminum" E 1 71.7 GPa 1 2.8 Mg m 3 Mat 2 "Titanium" E 2 113.8 GPa 2 4.4 Mg m 3 Mat 3 "Gray cast iron" E 3 103.4 GPa 3 7.2 Mg m 3 Mat 4 "Ductile iron" E 4 168.9 GPa 4 6.9 Mg m 3 Mat 5 "Bronze" E 5 110.3 GPa 5 8.6 Mg m 3 Mat 6 "Carbon steel" E 6 206.8 GPa 6 7.8 Mg m 3 Mat 7 "Stainless steel" E 7 189.6 GPa 7 7.8 Mg m 3 i 1 2 7 Solution: See Mathcad file P0223. 1. Calculate the specific stiffness for each material as described in Section 2.1. E' i E i i E' i 10 6 s 2 m 2 25.6 25.9 14.4 24.5 12.8 26.5 24.3 Mat i "Aluminum" "Titanium" "Gray cast iron" "Ductile iron" "Bronze" "Carbon steel" "Stainless steel" 2. Rank them in increasing order of specific stiffness. Mat 5 "Bronze" E' 5 10 6 s 2 m 2 12.8 Mat 3 "Gray cast iron" E' 3 10 6 s 2 m 2 14.4 Mat 7 "Stainless steel" E' 7 10 6 s 2 m 2 24.3 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  40. 40. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-23-2 Mat 4 "Ductile iron" E' 4 10 6 s 2 m 2 24.5 Mat 1 "Aluminum" E' 1 10 6 s 2 m 2 25.6 Mat 2 "Titanium" E' 2 10 6 s 2 m 2 25.9 Mat 6 "Carbon steel" E' 6 10 6 s 2 m 2 26.5 3. Bending and axial deflection are inversely proportional to the modulus of elasticity. For the same shape and dimensions, the material with the highest specific stiffness will give the smallest deflection. Or, put another way, for a given deflection, using the material with the highest specific stiffness will result in the least weight. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  41. 41. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-24-1 PROBLEM 2-24 Statement: Call your local steel and aluminum distributors (consult the Yellow Pages) and obtain current costs per pound for round stock of consistent size in low-carbon (SAE 1020) steel, SAE 4340 steel, 2024-T4 aluminum, and 6061-T6 aluminum. Calculate a strength/dollar ratio and a stiffness/dollar ratio for each alloy. Which would be your first choice on a cost-efficiency basis for an axial-tension-loaded round rod (a) If maximum strength were needed? (b) If maximum stiffness were needed? Solution: Left to the student as data will vary with time and location. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  42. 42. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-25-1 PROBLEM 2-25 Statement: Call your local plastic stock-shapes distributors (consult the Yellow Pages) and obtain current costs per pound for round rod or tubing of consistent size in plexiglass, acetal, nylon 6/6, and PVC. Calculate a strength/dollar ratio and a stiffness/dollar ratio for each alloy. Which would be your first choice on a cost-efficiency basis for an axial-tension-loaded round rod or tube of particular diameters. (a) If maximum strength were needed? (b) If maximum stiffness were needed? Solution: Left to the student as data will vary with time and location. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  43. 43. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-26-1 PROBLEM 2-26 Statement: A part has been designed and its dimensions cannot be changed. To minimize its deflections under the same loading in all directions irrespective of stress levels, which material woulod you choose among the following: aluminum, titanium, steel, or stainless steel? Solution: See Mathcad file P0226. 1. Choose the material with the highest modulus of elasticity because deflection is inversely proportional to modulus of elasticity. Thus, choose steel unless there is a corrosive atmosphere, in which case, choose stainless steel. P0226.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  44. 44. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-27-1 PROBLEM 2-27 Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the tensile yield strength for 1050 steel quenched and tempered at 400F if a reliability of 99.9% is required? Given: Mean yield strength Sy 117 ksi Sy 807 MPa Solution: See Mathcad file P0227. 1. From Table 2-2 the reliability factor for 99.9% is Re 0.753 . Applying this to the mean tensile strength gives Sy99.9 Sy Re Sy99.9 88.1 ksi Sy99.9 607 MPa 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  45. 45. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-28-1 PROBLEM 2-28 Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the ultimate tensile strength for 4340 steel quenched and tempered at 800F if a reliability of 99.99% is required? Given: Mean ultimate tensile strength Sut 213 ksi Sut 1469 MPa Solution: See Mathcad file P0228. 1. From Table 2-2 the reliability factor for 99.99% is Re 0.702 . Applying this to the mean ultimate tensile strength gives Sut99.99 Sut Re Sut99.99 150 ksi Sut99.99 1031 MPa 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  46. 46. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-29-1 PROBLEM 2-29 Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the ultimate tensile strength for 4130 steel quenched and tempered at 400F if a reliability of 90% is required? Given: Mean ultimate tensile strength Sut 236 ksi Sut 1627 MPa Solution: See Mathcad file P0229. 1. From Table 2-2 the reliability factor for 90% is Re 0.897 . Applying this to the mean ultimate tensile strength gives Sut99.99 Sut Re Sut99.99 212 ksi Sut99.99 1460 MPa 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  47. 47. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-30-1 PROBLEM 2-30 Statement: Assuming that the mechanical properties data given in Appendix Table A-9 for some carbon steels represents mean values, what is the value of the tensile yield strength for 4140 steel quenched and tempered at 800F if a reliability of 99.999% is required? Given: Mean yield strength Sy 165 ksi Sy 1138 MPa Solution: See Mathcad file P0230. 1. From Table 2-2 the reliability factor for 99.999% is Re 0.659 . Applying this to the mean tensile strength gives Sy99.9 Sy Re Sy99.9 109 ksi Sy99.9 750 MPa 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  48. 48. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-31-1 PROBLEM 2-31 Statement: A steel part is to be plated to give it better corrosion resistance. Two materials are being considered: cadmium and nickel. Considering only the problem of galvanic action, which would you chose? Why? Solution: See Mathcad file P0231. 1. From Table 2-4 we see that cadmium is closer to steel than nickel. Therefore, from the standpoint of reduced galvanic action, cadmium is the better choice. Also, since cadmium is less noble than steel it will be the material that is consumed by the galvanic action. If nickel were used the steel would be consumed by galvanic action. P0231.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  49. 49. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-32-1 PROBLEM 2-32 Statement: A steel part with many holes and sharp corners is to be plated with nickel. Two processes are being considered: electroplating and electroless plating. Which process would you chose? Why? Solution: See Mathcad file P0232. 1. Electroless plating is the better choice since it will give a uniform coating thickness in the sharp corners and in the holes. It also provides a relatively hard surface of about 43 HRC. P0232.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  50. 50. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-33-1 PROBLEM 2-33 Statement: What is the common treatment used on aluminum to prevent oxidation? What other metals can also be treated with this method? What options are available with this method? Solution: See Mathcad file P0233. 1. Aluminum is commonly treated by anodizing, which creates a thin layer of aluminum oxide on the surface. Titanium, magnesium, and zinc can also be anodized. Common options include tinting to give various colors to the surface and the use of "hard anodizing" to create a thicker, harder surface. P0233.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  51. 51. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-34-1 PROBLEM 2-34 Statement: Steel is often plated with a less nobel metal that acts as a sacrificial anode that will corrode instead of the steel. What metal is commonly used for this purpose (when the finished product will not be exposed to saltwater), what is the coating process called, and what are the common processes used to obtain the finished product? Solution: See Mathcad file P0234. 1. The most commonly used metal is zinc. The process is called "galvanizing" and it is accomplished by electroplating or hot dipping. P0234.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  52. 52. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-35-1 PROBLEM 2-35 Statement: A low-carbon steel part is to be heat-treated to increase its strength. If an ultimate tensile strength of approximately 550 MPa is required, what mean Brinell hardness should the part have after treatment? What is the equivalent hardness on the Rockwell scale? Given: Approximate tensile strength Sut 550 MPa Solution: See Mathcad file P0235. 1. Use equation (2.10), solving for the Brinell hardness, HB. Sut 3.45 HB= HB Sut 3.45 MPa HB 159 2. From Table 2-3, the equivalent hardness on the Rocwell scale is 83.9HRB. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  53. 53. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-36-1 PROBLEM 2-36 Statement: A low-carbon steel part has been tested for hardness using the Brinell method and is found to have a hardness of 220 HB. What are the approximate lower and upper limits of the ultimate tensile strength of this part in MPa? Given: Hardness HB 220 Solution: See Mathcad file P0236. 1. Use equation (2.10), solving for ultimate tensile strength. Minimum: Sutmin 3.45 HB 0.2 HB( ) MPa Sutmin 715 MPa Maximum: Sutmax 3.45 HB 0.2 HB( ) MPa Sutmax 803 MPa 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  54. 54. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-37-1 PROBLEM 2-37 Statement: Figure 2-24 shows "guide lines" for minimum weight design when failure is the criterion. The guide line, or index, for minimizing the weight of a beam in bending is f 2/3/, where f is the yield strength of a material and is its mass density. For a given cross-section shape the weight of a beam with given loading will be minimized when this index is maximized. The following materials are being considered for a beam application: 5052 aluminum, cold rolled; CA-170 beryllium copper, hard plus aged; and 4130 steel, Q & T @ 1200F. The use of which of these three materials will result in the least-weight beam? Units: Mg kg 3 Given: 5052 Aluminum Sya 255 MPa a 2.8 Mg m 3 CA-170 beryllium copper Syb 1172 MPa b 8.3 Mg m 3 4130 steel Sys 703 MPa s 7.8 Mg m 3 Solution: See Mathcad file P0237. 1. The values for the mass density are taken from Appendix Table A-1 and the values of yield strength come from from Tables A-2, A-4, and A-9 for aluminum, beryllium copper, and steel, respectively. 2. Calculate the index value for each material. Index Sy Sy 0.667 Mg m 3 MPa 0.667 Aluminum Ia Index Sya a Ia 14.4 Beryllium copper Ib Index Syb b Ib 13.4 Steel Is Index Sys s Is 10.2 The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  55. 55. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-38-1 PROBLEM 2-38 Statement: Figure 2-24 shows "guide lines" for minimum weight design when failure is the criterion. The guide line, or index, for minimizing the weight of a member in tension is f/, where f is the yield strength of a material and is its mass density. The weight of a member with given loading will be minimized when this index is maximized. For the three materials given in Problem 2-37, which will result in the least weight tension member? Units: Mg kg 3 Given: 5052 Aluminum Sya 255 MPa a 2.8 Mg m 3 CA-170 beryllium copper Syb 1172 MPa b 8.3 Mg m 3 4130 steel Sys 703 MPa s 7.8 Mg m 3 Solution: See Mathcad file P0238. 1. The values for the mass density are taken from Appendix Table A-1 and the values of yield strength come from from Tables A-2, A-4, and A-9 for aluminum, beryllium copper, and steel, respectively. 2. Calculate the index value for each material. Index Sy Sy Mg m 3 MPa Aluminum Ia Index Sya a Ia 91.1 Beryllium copper Ib Index Syb b Ib 141.2 Steel Is Index Sys s Is 90.1 The beryllium copper has the highest value of the index and would be the best choice to minimize weight. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  56. 56. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-39-1 PROBLEM 2-39 Statement: Figure 2-23 shows "guide lines" for minimum weight design when stiffness is the criterion. The guide line, or index, for minimizing the weight of a beam in bending is E1/2/, where E is the modulus of elasticity of a material and is its mass density. For a given cross-section shape the weight of a beam with given stiffness will be minimized when this index is maximized. The following materials are being considered for a beam application: 5052 aluminum, cold rolled; CA-170 beryllium copper, hard plus aged; and 4130 steel, Q & T @ 1200F. The use of which of these three materials will result in the least-weight beam? Units: Mg kg 3 Given: 5052 Aluminum Ea 71.7 GPa a 2.8 Mg m 3 CA-170 beryllium copper Eb 127.6 GPa b 8.3 Mg m 3 4130 steel Es 206.8 GPa s 7.8 Mg m 3 Solution: See Mathcad file P0239. 1. The values for the mass density and modulus are taken from Appendix Table A-1. 2. Calculate the index value for each material. Index E ( ) E 0.5 Mg m 3 GPa 0.5 Aluminum Ia Index Ea a Ia 3.0 Beryllium copper Ib Index Eb b Ib 1.4 Steel Is Index Es s Is 1.8 The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  57. 57. MACHINE DESIGN - An Integrated Approach, 4th Ed. 2-40-1 PROBLEM 2-40 Statement: Figure 2-24 shows "guide lines" for minimum weight design when stiffness is the criterion. The guide line, or index, for minimizing the weight of a member in tension is E/, where E is the modulus of elasticity of a material and is its mass density. The weight of a member with given stiffness will be minimized when this index is maximized. For the three materials given in Problem 2-39, which will result in the least weight tension member? Units: Mg kg 3 Given: 5052 Aluminum Ea 71.7 GPa a 2.8 Mg m 3 CA-170 beryllium copper Eb 127.6 GPa b 8.3 Mg m 3 4130 steel Es 206.8 GPa s 7.8 Mg m 3 Solution: See Mathcad file P0240. 1. The values for the mass density and modulus are taken from Appendix Table A-1. 2. Calculate the index value for each material. Index E ( ) E Mg m 3 GPa Aluminum Ia Index Ea a Ia 25.6 Beryllium copper Ib Index Eb b Ib 15.4 Steel Is Index Es s Is 26.5 The steel has the highest value of the index and would be the best choice to minimize weight. 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  58. 58. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-1-1 PROBLEM 3-1 Statement: Which load class from Table 3-1 best suits these systems? (a) Bicycle frame (b) Flag pole (c) Boat oar (d) Diving board (e) Pipe wrench (f) Golf club. Solution: See Mathcad file P0301. 1. Determine whether the system has stationary or moving elements, and whether the there are constant or time-varying loads. (a) Bicycle frame Class 4 (Moving element, time-varying loads) (b) Flag pole Class 2 (Stationary element, time-varying loads) (c) Boat oar Class 2 (Low acceleration element, time-varying loads) (d) Diving board Class 2 (Stationary element, time-varying loads) (e) Pipe wrench Class 2 (Low acceleration elements, time-varying loads) (f) Golf club Class 4 (Moving element, time-varying loads) P0301.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  59. 59. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-2a-1 PROBLEM 3-2a Statement: Draw free-body diagrams for the system of Problem 3-1a (bicycle frame). Assumptions: 1. A two-dimensional model is adequate. 2. The lower front-fork bearing at C takes all of the thrust load from the front forks. 3. There are no significant forces on the handle bars. Solution: See Mathcad file P0302a. 1. A typical bicycle frame is shown in Figure 3-2a. There are five points on the frame where external forces and moments are present. The rider's seat is mounted through a tube at A. This is a rigid connection, capable of transmitting two force components and a moment. The handle bars and front-wheel forks are supported by the frame through two bearings, located at B and C. These bearings are capable of transmitting radial and axial loads. The pedal-arm assembly is supported by bearings at D. These bearings are capable of transmitting radial loads. The rear wheel-sprocket assembly is supported by bearings mounted on an axle fixed to the frame at E. dyF ex E F eyF dxF D eR axF ayF a aM A R dR bR Rc C ctbr B F F crF FIGURE 3-2a Free Body Diagram for Problem 3-2a 2. The loads at B and C can be determined by analyzing a FBD of the front wheel-front forks assembly. The loads at D can be determined by analyzing a FBD of the pedal-arm and front sprocket (see Problem 3-3), and the loads at E can be determined by analyzing a FBD of the rear wheel-sprocket assembly. 3. With the loads at B, C, D, and E known, we can apply equations 3.3b to the FBD of the frame and solve for Fax, Fay, and Ma. Fx : Fax Fbr cos ( ) Fcr cos ( )+ Fct sin ( ) Fdx Fex+ 0= (1) Fy : Fay Fbr sin ( ) Fcr sin ( )+ Fct cos ( )+ Fdy Fey+ 0= (2) Mz : Ma Rbx Fby Rby Fbx( )+ Rcx Fcy Rcy Fcx( )+ Rdx Fdy Rdy Fdx( ) Rex Fey Rey Fex( )++ ... 0= (3) P0302a.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  60. 60. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-2e-1 PROBLEM 3-2e Statement: Draw free-body diagrams for the system of Problem 3-1e (pipe wrench). Assumptions: A two-dimensional model is adequate. Solution: See Mathcad file P0302e. 1. A typical pipe wrench with a pipe clamped in its jaw is shown in Figure 3-2e(a). When a force Fhand is applied on the wrench, the piping system provides an equal and opposite force and a resisting torque, Tpipe. T F pipe hand hand a F (a) FBD of pipe wrench and pipe bd Fay Fax A Fbt Fbn (b) FBD of pipe wrench only FIGURE 3-2e Free Body Diagrams for Problem 3-2e 2. The pipe reacts with the wrench at the points of contact A and B. The forces here will be directed along the common normals and tangents. The jaws are slightly tapered and, as a result, the action of Fhand tends to drive the wrench further into the taper, increasing the normal forces. This, in turn, allows for increasing tangential forces. It is the tangential forces that produce the turning torque. 3. Applying equations 3.3b to the FBD of the pipe wrench, Fx : Fax Fbn cos ( )+ Fbt sin ( ) 0= (1) Fy : Fay Fbn sin ( )+ Fbt cos ( )+ Fhand 0= (2) MA : d Fbt cos ( ) Fbn sin ( )+( ) d a+( ) Fhand 0= (3) 4. These equations can be solved for the vertical forces if we assume is small so that sin() = 0 and cos () = 1. P0302e.xmcd 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  61. 61. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-3-1 PROBLEM 3-3 Statement: Draw a free-body diagram of the pedal-arm assembly from a bicycle with the pedal arms in the horizontal position and dimensions as shown in Figure P3-1. (Consider the two arms, pedals and pivot as one piece). Assuming a rider-applied force of 1500 N at the pedal, determine the torque applied to the chain sprocket and the maximum bending moment and torque in the pedal arm. Given: a 170 mm b 60 mm Frider 1.5 kN Assumptions: The pedal-arm assembly is supprted by bearings at A and at B. Solution: See Figure 3-3 and Mathcad file P0303. 1. The free-body diagram (FBD) of the pedal-arm assembly (including the sprocket) is shown in Figure 3-3a. The rider-applied force is Frider and the force applied by the chain (not shown) is Fchain. The radial bearing reactions are Fax, Faz, Fbx, and Fbz. Thus, there are five unknowns Fchain, Fax, Faz, Fbx, and Fbz. In general, we can write six equilibrium equations for a three-dimensional force system, but in this system there are no forces in the y-direction so five equations are available to solve for the unknowns. Frider Pedal x b Arm a Fax Fbx A B Arm (sectioned) y Sprocket Faz Fbz Fchain z (a) FBD of complete pedal-arm assembly x Pedal b Arm Frider a cF Mc y z Tc (b) FBD of pedal and arm with section through the origin FIGURE 3-3 Dimensions and Free Body Diagram of the pedal-arm assembly for Problem 3-3 2. The torque available to turn the sprocket is found by summing moments about the sprocket axis. From Figure 3-3a, it is 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  62. 62. MACHINE DESIGN - An Integrated Approach, 4th Ed. 3-3-2 Ty-axis : a Frider r Fchain a Frider Tsprocket= 0= Tsprocket a Frider Tsprocket 255 N m where r is the sprocket pitch radius. 3. In order to determine the bending moment and twisting torque in the pedal arm, we will cut the arm with a section plane that goes through the origin and is parallel to the y-z plane, removing everything beyond that plane and replacing it with the internal forces and moments in the pedal arm at the section. The resulting FBD is shown in Figure 3-3b. The internal force at section C is Fc the internal bending moment is Mc, and the internal twisting moment (torque) is Tc. We can write three equilibrium equations to solve for these three unknowns: Shear force in pedal arm at section C Fz : Fc Frider 0= Fc Frider Fc 1.5 kN Bending moment in pedal arm at section C My-axis : a Frider Mc 0= Mc a Frider Mc 255 N m Twisting moment in pedal arm at section C Mx-axis : b Frider Tc 0= Tc b Frider Tc 90 N m 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
  63. 63. MACHINE DESIGN - An Integrated Approach,4th Ed. 3-4-1 PROBLEM 3-4 Statement: The trailer hitch from Figure 1-1 has loads applied as shown in Figure P3-2. The tongue weight of 100 kg acts downward and the pull force of 4905 N acts horizontally. Using the dimensions of the ball bracket in Figure 1-5 (p. 15), draw a free-body diagram of the ball bracket and find the tensile and shear loads applied to the two bolts that attach the bracket to the channel in Figure 1-1. Given: a 40 mm b 31 mm c 70 mm d 20 mm Mtongue 100 kg Fpull 4.905 kN t 19 mm Assumptions: 1. The nuts are just snug-tight (no pre-load), which is

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