Machine Design Problems

(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,

(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.

P0102.xmcd

(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.

P0103.xmcd

(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 =1000 lbf

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

(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

sec2

 :=

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=3.333 s-2ft

F:= M a F =25.9 lbf

P0105.xmcd

(6)

Statement: Express a 100-kg mass in units of slugs, blobs, and lbm. How much does this mass weigh?

2



in



Given: M 100 kg

Assumptions: The mass is at sea-level and the gravitational acceleration is

g 32.174 ft sec2   or g 386.089 in sec2   or g 9.807 m sec2  

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

(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  mm4 

Moment about y-axis Iy

h b 3 12  Iy 9.000 in 4   Iy 3.746 10 6  mm4 

Radius of gyration about x-axis kx

I_x A

 kx 1.155 in

kx 29.329 mm

Radius of gyration about y-axis ky

I_y 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  mm4  2. Solid circle, let:

D 3 in Area A πD 2  4  A 7.069 in 2 A 4560 mm 2

πD4 64  Ix 3.976 in 4   Ix 1.655 10 6  mm4 

πD4 64  Iy 3.976 in 4   Iy 1.655 10 6  mm4 

I_x A

 kx 0.750 in

kx 19.05 mm

P0107.xmcd

(8)

y

A _ky

y 19.05 mm Polar moment of inertia J_z πD

4

 32

 J_z 7.952 in 4

J_z 3.310106mm4

3. Hollow circle, let:

D 3 in d 1 in Area A π 4 D 2 d2 





  A 6.283 in 2 A 4054 mm 2

Moment about x-axis I_x π

64 D 4 d4 





  I_x3.927 in 4 I_x1.635 106mm4

Moment about y-axis I_y π

64 D 4 d4 





  I_y3.927 in 4 I_y1.635 106mm4

Radius of gyration about x-axis k_x Ix A

 k_x 0.791 in

k_x 20.08 mm

Radius of gyration about y-axis k_y Iy A

 k_y 0.791 in

k_y 20.08 mm

Polar moment of inertia J_z π

32 D 4 d4 





  J_z 7.854 in 4 J_z 3.269106mm4

4. Solid semicircle, let:

D 3 in R 0.5 D R1.5 in Area A πD 2  8  A 3.534 in 2 A 2280 mm 2

Moment about x-axis I_x 0.1098 R 4 I_x0.556 in 4

I_x2.314 105mm4

Moment about y-axis I_y πR

4  8  I_y1.988 in 4 I_y8.275 105mm4 P0107.xmcd

(9)

A

kx 10.073 mm

I_y 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  mm4  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

b h 3 36  Ix 0.056 in 4   Ix 2.312 10 4  mm4 

h b 3 36  Iy 0.222 in 4   Iy 9.250 10 4  mm4 

I_x A

 kx 0.236 in

kx 5.987 mm

I_y 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  mm4  P0107.xmcd

(10)

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.

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 γ 0.28 lbf in3

Volume V a b c V 24.000 in 3 V 393290 mm 3 Mass M Vγ g  M  0.017 blob M  3.048 kg

M a



2b2



12  Ix 0.019 blob in 2    Ix 2130.4 kg mm 2   

M a



2c2



12  Iy 0.029 blob in 2    Iy 3277.6 kg mm 2   

Moment about z-axis Iz

M b



2c2



12  Iz 0.036 blob in 2    Iz 4097.0 kg mm 2   

I_x M

 kx 1.041 in

kx 26.437 mm

I_y M

 ky 1.291 in

ky 32.791 mm

Radius of gyration about z-axis kz

I_z M  kz 1.443 in kz 36.662 mm 2.Cylinder, let: r 2 in L 3 in γ 0.30 lbf in3 Volume V π Lr2 V 37.699 in 3 V 617778 mm 3 Mass M Vγ g  M  0.029 blob M  5.13 kg

(11)

Moment about x-axis I_x

2

 I_x0.059 blob in 

I_x6619.4 kg mm  2

Moment about y-axis I_y M 3 r

2  L2





 12  I_y0.051 blob in  2 I_y5791.9 kg mm  2

Moment about z-axis I_z M 3 r

2  L2





 12  I_z0.051 blob in  2 I_z5791.9 kg mm  2

Radius of gyration about x-axis k_x Ix M

 k_x 1.414 in

k_x 35.921 mm

Radius of gyration about y-axis k_y Iy M

 k_y 1.323 in

k_y 33.601 mm

Radius of gyration about z-axis k_z Iz M

 k_z 1.323 in

k_z 33.601 mm 3. Hollow cylinder, let:

a 2 in b 3 in L 4 in γ 0.28 lbf in3 Volume V π



b2 a2



L V 62.832 in 3 V 1029630 mm 3 Mass M Vγ g  M  0.046 blob M  7.98 kg

Moment about x-axis I_x M

2 a 2 b2 





  I_x0.296 blob in  2 I_x3.3104kg mm 2

Moment about y-axis I_y M

12 3 a 2   3 b 2 L2





  I_y0.209 blob in  2 I_y2.4104kg mm 2

Moment about z-axis I_z M

12 3 a 2   3 b 2 L2





  I_z0.209 blob in  2 I_z2.4104kg mm 2

 k_x 2.550 in

k_x 64.758 mm

 k_y 2.141 in

k_y 54.378 mm

(12)

M

kz 54.378 mm 4. Right circular cone, let:

r 2 in h 5 in γ 0.28 lbf in3 Volume V πr 2  h 3  V 20.944 in 3 V 343210 mm 3 Mass M Vγ g  M  0.015 blob M  2.66 kg

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

I_x M

 kx 1.095 in

kx 27.824 mm

I_y M

 ky 1.240 in

ky 31.495 mm

Radius of gyration about z-axis kz

I_z M  kz 1.240 in kz 31.495 mm 5. Sphere, let: r 3 in Volume V 4 3 rπ 3   V 113.097 in 3 V 1853333 mm 3 Mass M Vγ g  M  0.082 blob M  14.364 kg

2 5Mr 2   Ix 0.295 blob in 2    Ix 33362 kg mm 2   

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5

I_y33362 kg mm  2

Moment about z-axis I_z 2

5Mr

2



 I_z0.295 blob in  2

I_z33362 kg mm  2

 k_x 1.897 in

k_x 48.193 mm

 k_y 1.897 in

k_y 48.193 mm

Radius of gyration about z-axis k_z Iz M

 k_z 1.897 in

k_z 48.193 mm

(14)

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( ) πD

2

 4 

Moment about x-axis Ix( )D

πD4

64 

Moment about y-axis Iy( )D

πD4

64 

3. Hollow circle: Area A D d(  ) π

4 D 2 d2 





 

Moment about x-axis Ix(D d ) π 64 D 4 d4 





 

Moment about y-axis Iy(D d ) π 64 D 4 d4 





  4. Solid semicircle: Area A D( ) πD 2  8 

Moment about x-axis Ix( )R 0.1098 R 4

 

Moment about y-axis Iy( )R

πR4 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 

(15)

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 γ) V a b(  c)γ

g



Moment about x-axis Ix(a b c γ)

M a b(  c γ) a



2b2



12 

Moment about y-axis Iy(a b c γ)

M a b(  c γ) a



2c2



12 

Moment about z-axis Iz(a b c γ)

M a b(  c γ) b



2c2



12  2. Cylinder: Volume V r L(  ) π Lr2 Mass M r L(  γ) V r L(  )γ g 

Moment about x-axis Ix(r L γ)

M r L(  γ) r 2 2 

Moment about y-axis Iy(r L γ)

M r L(  γ) 3 r



 2 L2



12 

Moment about z-axis Iz(r L γ)

M r L(  γ) 3 r



 2 L2



12  3. Hollow cylinder: Volume V a b(  L ) π



b2a2



L Mass M a b(  L γ) V a b(  L )γ g 

Moment about x-axis Ix(a b L γ)

M a b(  L γ) 2 a 2 b2 





 

Moment about y-axis Iy(a b L γ)

M a b(  L γ) 12 3 a 2   3 b 2L2





 

Moment about z-axis Iz(a b L γ)

M a b(  L γ) 12 3 a 2   3 b 2L2





 

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Volume V r h(  ) πr 2  h 3  Mass M r h(  γ) V r h(  )γ g 

Moment about x-axis I_x(r h γ) 3

10M r h(  γ)r

2

 

Moment about y-axis I_y(r h γ) M r h(  γ) 12 r

2   3 h 2





80  

Moment about z-axis I_z(r h γ) M r h(  γ) 12 r

2   3 h 2





80   5. Sphere: Volume V r( ) 4 3 rπ 3   Mass M r( γ) V r( )γ g 

Moment about x-axis I_x(rγ) 2

5M r( γ)r

2

 

Moment about y-axis I_y(rγ) 2

5M r( γ)r

2

 

Moment about z-axis I_z(rγ) 2

5M r( γ)r

2

 

(17)

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) σ_ya:=5 ε_ya:=5

P2-1(b) σ_yb:=9 ε_yb:=2

P2-1(c) σ_yc:=5 ε_yc:=1

Using equation (2.7), the modulus of resiliency for each material is, approximately,

P21a 1 2 σya ⋅ ⋅ε_ya := P21a=12.5 P21b 1 2 σyb ⋅ ⋅ε_yb := P21b=9 P21c 1 2 σyc ⋅ ⋅ε_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

(18)

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 S_ua:= 6

Figure P2-1(b) Syb:= 9 S_ub:= 10

Figure P2-1(c) Syc:= 5 S_uc:= 8

2. The ratio of yield strength to ultimate strength for each material is:

Figure P2-1(a) ratioa

S_ya S_ua := ratio_a=0.83 Figure P2-1(b) ratiob S_yb S_ub := ratio_b=0.90 Figure P2-1(c) ratioc S_yc S_uc := ratio_c=0.63 P0202.xmcd

(19)

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 εf1020 0.365 AISI 1095: Sy1095 550 MPa Sut1095 1050 MPa εf1095 0.11 AISI 4142: Sy4142 1600 MPa Sut4142 2430 MPa εf4142 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 S_ut 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 Sy₁₀₂₀2 E   UR1020 0.22 MN m m3   UR1095 1 2 Sy₁₀₉₅2 E   UR1095 0.73 MN m m3   UR4142 1 2 Sy₄₁₄₂2 E   UR4142 6.18 MN m m3  

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



εf1020  UT1020 128 MN m m3   UT1095 1 2



Sy1095 Sut1095



εf1095  UT1095 88 MN m m3   UT4142 1 2



Sy4142 Sut4142



εf4142  UT4142 121 MN m m3  

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.

(20)

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 εf1100 0.170 2024-T351: Sy2024 330 MPa Sut2024 480 MPa εf2024 0.195 7075-T6: Sy7075 510 MPa Sut7075 560 MPa εf7075 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 Sy₁₁₀₀2 E   UR1100 0.10 MN m m3   UR2024 1 2 Sy₂₀₂₄2 E   UR2024 0.76 MN m m3   UR7075 1 2 Sy₇₀₇₅2 E   UR7075 1.81 MN m m3  

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



εf1100  UT1100 21 MN m m3   UT2024 1 2



Sy2024 Sut2024



εf2024  UT2024 79 MN m m3   UT7075 1 2



Sy7075 Sut7075



εf7075  UT7075 88 MN m m3  

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.

(21)

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 εfNylon 0.52 ENylon 1.1 GPa HDPE: SyHDPE 15 MPa SutHDPE 23 MPa εfHDPE 3.0 EHDPE 0.7 GPa PTFE: SyPTFE 8.3 MPa SutPTFE 13 MPa εfPTFE 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 Sy_Nylon2 E_Nylon   URNylon 1.8 MN m m3   URHDPE 1 2 Sy_HDPE2 E_HDPE   URHDPE 0.16 MN m m3   URPTFE 1 2 Sy_PTFE2 E_PTFE   URPTFE 0.04 MN m m3  

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



εfNylon

 UTNylon 37 MN m m3   UTHDPE 1

2



SyHDPESutHDPE



εfHDPE

 UTHDPE 57 MN m m3   UTPTFE 1

2



SyPTFESutPTFE



εfPTFE

 UTPTFE 5

MN m m3

 

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..

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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 εel 0.002 Test specimen: Diameter d_o 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 S_el

ε_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εel  U'el 414 kN m m3  

The total strain energy in the specimen is the strain energy per unit volume times the volume,

Uel U'el πd_o2

4  Lo

 Uel2.7 N m 

3. Based on the modulus of elasticity and using Table C-1, the material is steel.

(23)

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 εel 0.004 Sel 284 MPa Test specimen: Diameter d_o 0.505 in Length Lo 2.00 in

E S_el

ε_el

 E10.3 10 6psi E71 GPa

U'el 1 2Selεel  U'el 82.4 lbf in in3   U'el 568 kN m m3  

4  Lo

 Uel33.0 in lbf 

3. Based on the modulus of elasticity and using Table C-1, the material is aluminum.

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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 εel 0.003 Test specimen: Diameter d_o 12.8 mm Length Lo 50 mm Solution: See Mathcad file P0208.

E S_el

ε_el

 E45 GPa

U'el 1 2Selεel  U'el 201 kN m m3  

4  Lo

 Uel1.3 N m 

3. Based on the modulus of elasticity and using Table C-1, the material is magnesium.

(25)

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 εel 0.006

S_el 689 MPa

Test specimen: Diameter do 0.505 in Length Lo 2.00 in Solution: See Mathcad file P0209.

E S_el

ε_el

 E16.7 10 6psi E115 GPa

U'el 1 2Selεel  U'el 300 lbf in in3   U'el 2 10 3  kN m m3  

4  Lo

 Uel120.18 in lbf 

3. Based on the modulus of elasticity and using Table C-1, the material is titanium.

(26)

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 106joule

Given: Yield strength Sy 689 MPa Yield strain ε_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εy  UR 2.067 MJ m3   UR2.1 MPa

(27)

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 106joule

Given: Yield strength Sy 60 ksi Sy 414 MPa Yield strain ε_y 0.002

1. The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately

UR 1 2Syεy  UR 60 in lbf in3   UR 0.414 MJ m3   UR0.414 MPa

(28)

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 εf  0.15

1. Determine the modulus of toughness using Equation (2.8).

UT S_yS_ut 2













εf  UT 82.7 MN m m3   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 S_y2 E   UR 414 kN m m3   UR0.41 MPa

(29)

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

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

(30)

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

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

(31)

Statement: What are the principal alloy elements of an AISI 4340 steel? How much carbon does it have? Is it hardenable? By what techniques?

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

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1. Determine the principal alloying elements from Table 2-5 for 10xx steel. Carbon only, no alloying elements

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

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1. Determine the principal alloying elements from Table 2-5 for 61xx steel.. 0.15% Vanadium

0.60 to 0.95% Chromium

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

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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

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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 Est 30 10 6  psi  γ st 0.28 lbf in3   Aluminum al 1 Sut al 60 ksi Eal 10.4 10 6  psi  γ al 0.10 lbf in3   Titanium ti 2 Sut ti 90 ksi Eti 16.5 10 6  psi  γ ti 0.16 lbf in3   Index i 0 1 2

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 γ i 1 in  3 286·10 3 600·10 3 563·10  Specific stiffness E i γ i 1 in  6 107·10 6 104·10 6 103·10  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.

(36)

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

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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"Sy1 290 MPa γ1 0.10 lbf in 3    γ 1 27.14 kN m3   Mat 2 "1040 CR Steel" Sy2 490 MPa γ2 0.28 lbf in 3    γ 2 76.01 kN m3   Mat

3 "Ti-75A Titanium" Sy3 517 MPa γ3 0.16 lbf in 3    γ 3 43.43 kN m3   Mat 4 "Type 302 CR SS" Sy4 1138 MPa γ4 0.28 lbf in 3    γ 4 76.01 kN m3   i 1 2 4

1. Calculate the strength-to-weight ratio for each material as described in Section 2.1.

SWR i Sy i γ i  SWRi 104m 1.068 0.645 1.190 1.497  Mat i "2024 Aluminum, HT" "1040 CR Steel" "Ti-75A Titanium" "Type 302 CR SS"















(38)

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"Sut1 441 MPa γ1 0.10 lbf in 3    γ 1 27.14 kN m 3     Mat

2 "1040 CR Steel" Sut2 586 MPa γ2 0.28 lbf in 3    γ 2 76.01 kN m 3     Mat

3 "Acetal, unfilled" Sut3 60.7 MPa γ3 0.051 lbf in 3    γ 3 13.84 kN m 3     Mat

4 "Ti-75A Titanium" Sut4 586 MPa γ4 0.16 lbf in 3    γ 4 43.43 kN m 3     Mat

5 "Type 302 CR SS" Sut5 1310 MPa γ5 0.28 lbf in 3    γ 5 76.01 kN m 3     i 1 2 5

1. Calculate the strength-to-weight ratio for each material as described in Section 2.1.

SWR i Sut i γ i  SWRi 104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"