ch03 GearTeam

shi20396_ch03.qxd 8/18/03 10:18 AM Page 40 Chapter 3 3-1 From Table A-20 Sut = 470 MPa (68 kpsi), S y = 390 MPa (57 ...

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Chapter 3 3-1 From Table A-20 Sut = 470 MPa (68 kpsi), S y = 390 MPa (57 kpsi)

Ans.

3-2 From Table A-20 Sut = 620 MPa (90 kpsi), S y = 340 MPa (49.5 kpsi) Ans. 3-3 Comparison of yield strengths: Sut of G10 500 HR is S yt of SAE1020 CD is

620 = 1.32 times larger than SAE1020 CD 470

Ans.

390 = 1.15 times larger than G10500 HR Ans. 340

From Table A-20, the ductilities (reduction in areas) show, SAE1020 CD is

40 = 1.14 times larger than G10500 35

Ans.

The stiffness values of these materials are identical Ans. Sut MPa (kpsi)

Sy MPa (kpsi)

Table A-20 Ductility R%

SAE1020 CD 470(68) UNS10500 HR 620(90)

390 (57) 340(495)

40 35

Table A-5 Stiffness GPa (Mpsi) 207(30) 207(30)

3-4 From Table A-21 1040 Q&T S¯ y = 593 (86) MPa (kpsi)

at 205◦C (400◦F)

Ans.

3-5 From Table A-21 1040 Q&T

R = 65%

at 650◦C (1200◦F)

Ans.

3-6 Using Table A-5, the specific strengths are: Sy 39.5(103 ) = = 1.40(105 ) in Ans. UNS G10350 HR steel: W 0.282 Sy 43(103 ) = = 4.39(105 ) in Ans. 2024 T4 aluminum: W 0.098 Ti-6Al-4V titanium:

Sy 140(103 ) = = 8.75(105 ) in W 0.16

ASTM 30 gray cast iron has no yield strength.

Ans.

Ans.

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Chapter 3

3-7 The specific moduli are: 30(106 ) E = = 1.06(108 ) in W 0.282

UNS G10350 HR steel:

Ans.

2024 T4 aluminum:

10.3(106 ) E = = 1.05(108 ) in W 0.098

Ans.

Ti-6Al-4V titanium:

16.5(106 ) E = = 1.03(108 ) in W 0.16

Ans.

Gray cast iron:

14.5(106 ) E = = 5.58(107 ) in W 0.26

Ans.

2G(1 + ν) = E

3-8

⇒ ν=

E − 2G 2G

From Table A-5 Steel: ν = Aluminum: ν = Beryllium copper: ν = Gray cast iron: ν =

30 − 2(11.5) = 0.304 2(11.5)

Ans.

10.4 − 2(3.90) = 0.333 2(3.90) 18 − 2(7) = 0.286 2(7) 14.5 − 2(6) = 0.208 2(6)

Ans.

Ans. Ans.

3-9 E U 80 70 60 Stress P兾A0 kpsi

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50 Y

40

Su ⫽ 85.5 kpsi Ans. Sy ⫽ 45.5 kpsi Ans.

30

E ⫽ 90兾0.003 ⫽ 30 000 kpsi Ans.

20

R⫽

A 0 ⫺ AF 0.1987 ⫺ 0.1077 (100) ⫽ 45.8% Ans. ⫽ 0.1987 A0

10

⫽

A0 l ⫺ l0 ⌬l l ⫽ ⫽ ⫺ 1⫽ ⫺1 A l0 l0 l0

0

0

0.002 0.1

0.004 0.2

0.006 0.3

0.008 0.4 Strain, 

0.010 0.5

0.012 0.6

0.014 0.7

0.016 (Lower curve) 0.8 (Upper curve)

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3-10

To plot σtrue vs. ε, the following equations are applied to the data. A0 = Eq. (3-4)

π(0.503) 2 = 0.1987 in2 4

ε = ln

l l0

ε = ln

A0 A

σtrue =

for 0 ≤ L ≤ 0.0028 in for L > 0.0028 in

P A

The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log σ vs log ε m = 0.2306

The curve fit gives

log σ0 = 5.1852 ⇒ σ0 = 153.2 kpsi

Ans.

For 20% cold work, Eq. (3-10) and Eq. (3-13) give, A = A0 (1 − W ) = 0.1987(1 − 0.2) = 0.1590 in2 ε = ln

0.1987 A0 = ln = 0.2231 A 0.1590

Eq. (3-14): S y = σ0 εm = 153.2(0.2231) 0.2306 = 108.4 kpsi

Ans.

Eq. (3-15), with Su = 85.5 kpsi from Prob. 3-9, Su = P 0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800

85.5 Su = = 106.9 kpsi 1−W 1 − 0.2

Ans.

L

A

ε

σtrue

log ε

log σtrue

0 0.0004 0.0006 0.0010 0.0013 0.0023 0.0028 0.0036 0.0089

0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 4 0.197 8 0.196 3 0.192 4 0.187 5 0.156 3 0.130 7 0.107 7

0 0.000 2 0.000 3 0.000 5 0.000 65 0.001 149 0.001 399 0.001 575 0.004 604 0.012 216 0.032 284 0.058 082 0.240 083 0.418 956 0.612 511

0 5032.388 10 064.78 15 097.17 20 129.55 35 226.72 42 272.06 44 354.84 46 511.63 46 357.62 68 607.07 81 066.67 108 765.2 125 478.2 137 418.8

−3.699 01 −3.522 94 −3.301 14 −3.187 23 −2.939 55 −2.854 18 −2.802 61 −2.336 85 −1.913 05 −1.491 01 −1.235 96 −0.619 64 −0.377 83 −0.212 89

3.701 774 4.002 804 4.178 895 4.303 834 4.546 872 4.626 053 4.646 941 4.667 562 4.666 121 4.836 369 4.908 842 5.036 49 5.098 568 5.138 046

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Chapter 3

160000 140000 120000 true (psi)

100000 80000 60000 40000 20000 0

0

0.1

0.2

0.3

0.4 true

0.5

0.6

0.7

5.2

y ⫽ 0.2306x ⫹ 5.1852 5

log 

5.1

4.9

⫺1.6

3-11

⫺1.4

Tangent modulus at σ = 0 is E0 = At σ = 20 kpsi

⫺1

⫺0.8 log 

⫺0.6

⫺0.4

⫺0.2

0

4.8

5000 − 0 σ . = = 25(106 ) psi ε 0.2(10−3 ) − 0

. (26 − 19)(103 ) E 20 = = 14.0(106 ) psi Ans. − 3 (1.5 − 1)(10 )

ε(10−3 )

σ (kpsi)

60

0 5 10 16 19 26 32 40 46 49 54

50

0 0.20 0.44 0.80 1.0 1.5 2.0 2.8 3.4 4.0 5.0 3-12

⫺1.2

40  (kpsi)

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(Sy)0.001 ⫽ ˙ 35 kpsi Ans.

30 20 10 0

0

1

2

3  (10⫺3)

From Prob. 2-8, for y = a1 x + a2 x 2 a1 =

yx 3 − x yx 2 xx 3 − (x 2 ) 2

a2 =

xx y − yx 2 xx 3 − (x 2 ) 2

4

5

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Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

Let x represent ε(10−3 ) and y represent σ (kpsi), x 0 0.2 0.44 0.80 1.0 1.5 2.0 2.8 3.4 4.0 5.0  = 21.14

y

x2

x3

xy

0 5 10 16 19 26 32 40 46 49 54 297

0 0.04 0.1936 0.64 1.00 2.25 4.00 7.84 11.56 16.00 25.00 68.5236

0 0.008 0.085 184 0.512 1.000 3.375 8.000 21.952 39.304 64.000 125.000 263.2362

0 1.0 4.4 12.8 19.0 39.0 64.0 112.0 156.4 196.0 270.0 874.6

Substituting, 297(263.2362) − 874.6(68.5236) = 20.993 67 21.14(263.2362) − (68.5236) 2 21.14(874.6) − 297(68.5236) a2 = = −2.142 42 21.14(263.2362) − (68.5236) 2 The tangent modulus is a1 =

dy dσ = = 20.993 67 − 2(2.142 42)x = 20.993 67 − 4.284 83x dx dε At σ = 0, E 0 = 20.99 Mpsi Ans. At σ = 20 kpsi 20 = 20.993 67x − 2.142 42x 2 ⇒ x = 1.069, 8.73 Taking the first root, ε = 1.069 and the tangent modulus is E 20 = 20.993 67 − 4.284 83(1.069) = 16.41 Mpsi Ans. Determine the equation for the 0.1 percent offset line y = 20.99x + b at y = 0, x = 1 ∴ b = −20.99 y = 20.99x − 20.99 = 20.993 67x − 2.142 42x 2 2.142 42x 2 − 20.99 = 0 ⇒ x = 3.130 (S y ) 0.001 = 20.99(3.13) − 2.142(3.13) 2 = 44.7 kpsi 3-13

Since |εo | = |εi |

       R+h      R R + N ln  = ln  = −ln   R+ N  R+ N  R  R+N R+h = R+N R ( R + N ) 2 = R( R + h)

From which,

N 2 + 2R N − Rh = 0

Ans.

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Chapter 3



   h 1/2 N = R −1 ± 1 + R

The roots are:

The + sign being significant,

   h 1/2 N=R 1+ −1 R

Ans.

Substitute for N in εo = ln  Gives

  ε0 = ln  

R+h R+N

R+h

  h 1/2 R+ R 1+ −R R

    h 1/2   = ln 1 +  R

Ans.

These constitute a useful pair of equations in cold-forming situations, allowing the surface strains to be found so that cold-working strength enhancement can be estimated. 3-14 16T 16T 10−6 = = 2.6076T MPa πd 3 π(12.5) 3 (10−3 ) 3     π π ◦ ◦ θ r (12.5) θ 180 180 = = 6.2333(10−4 )θ ◦ γ = L 350 τ=

For G, take the first 10 data points for the linear part of the curve.

T

θ (deg.) γ (10−3 )

0 0 7.7 0.38 15.3 0.80 23.0 1.24 30.7 1.64 38.3 2.01 46.0 2.40 53.7 2.85 61.4 3.25 69.0 3.80 76.7 4.50 80.0 5.10 85.0 6.48 90.0 8.01 95.0 9.58 100.0 11.18

0 0.236 865 0.498 664 0.772 929 1.022 261 1.252 893 1.495 992 1.776 491 2.025 823 2.368 654 2.804 985 3.178 983 4.039 178 4.992 873 5.971 501 6.968 829

τ (MPa) 0 20.078 52 39.896 28 59.974 8 80.053 32 99.871 08 119.949 6 140.028 1 160.106 6 179.924 4 200.002 9  = 208.608 221.646 234.684 247.722 260.76

γ (10−3 ) x

τ (MPa) y

x2

0 0.236 865 0.498 664 0.772 929 1.022 261 1.252 893 1.495 992 1.776 491 2.025 823 2.368 654 11.450 57

0 20.078 52 39.896 28 59.974 8 80.053 32 99.871 08 119.949 6 140.028 1 160.106 6 179.924 4 899.882 8

0 0.056 105 0.248 666 0.597 420 1.045 018 1.569 742 2.237 992 3.155 918 4.103 957 5.610 522 18.625 34

xy 0 4.7559 19.8948 46.3563 81.8354 125.1278 179.4436 248.7586 324.3476 426.1786 1456.6986

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300 250

 (MPa)

200 150 100 50 0

0

1

2

3

4

5

6

7

 (10⫺3)

y = mx + b,

τ = y,

γ = x where m is the shear modulus G,

m=

MPa N x y − xy = 77.3 −3 = 77.3 GPa 2 2 N x − (x) 10

b=

y − mx = 1.462 MPa N

Ans.

. From curve S ys = 200 MPa Ans. Note since τ is not uniform, the offset yield does not apply, so we are using the elastic limit as an approximation. 3-15 x 38.5 39.5 40.5 41.5 42.5 43.5 44.5 45.5 46.5 47.5 48.5 49.5  = 528.0

f

fx

f x2

2 9 30 65 101 112 90 54 25 9 2 1 500

77.0 355.5 1215.0 2697.5 4292.5 4872.0 4005.0 2457.0 1162.5 427.5 97.0 49.5 21 708.0

2 964.50 14 042.25 49 207.50 111 946.30 182 431.30 211 932.00 178 222.50 111793.50 54 056.25 20 306.25 4 704.50 2 450.25 944 057.00

x¯ = 21 708/500 = 43.416, σˆ x =

944 057 − (21 7082 /500) = 1.7808 500 − 1

C x = 1.7808/43.416 = 0.041 02, y¯ = ln 43.416 − ln(1 + 0.041 022 ) = 3.7691

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Chapter 3

ln(1 + 0.041 022 ) = 0.0410,    1 ln x − 3.7691 2 1 g(x) = √ exp − 2 0.0410 x(0.0410) 2π σˆ y =

x

f /( N w)

g(x)

x

f /( N w)

g(x)

38 38 39 39 40 40 41 41 42 42 43 43 44 44

0 0.004 0.004 0.018 0.018 0.060 0.060 0.130 0.130 0.202 0.202 0.224 0.224 0.180

0.001 488 0.001 488 0.009 057 0.009 057 0.035 793 0.035 793 0.094 704 0.094 704 0.172 538 0.172 538 0.222 074 0.222 074 0.206 748 0.206 748

45 45 46 46 47 47 48 48 49 49 50 50

0.180 0.108 0.108 0.050 0.050 0.018 0.018 0.004 0.004 0.002 0.002 0

0.142 268 0.142 268 0.073 814 0.073 814 0.029 410 0.029 410 0.009 152 0.009 152 0.002 259 0.002 259 0.000 449 0.000 449

f (x) 0.25

Histogram PDF

0.2

0.15

0.1

0.05

0 35

40

45

S y = LN(43.42, 1.781) kpsi 3-16

50

x

Ans.

From Table A-22 AISI 1212

S y = 28.0 kpsi, σ f = 106 kpsi, Sut = 61.5 kpsi σ0 = 110 kpsi,

From Eq. (3-12)

m = 0.24,

ε f = 0.85

εu = m = 0.24

Eq. (3-10)

1 1 A0 = = 1.25 =  Ai 1−W 1 − 0.2

Eq. (3-13)

εi = ln 1.25 = 0.2231 ⇒ εi < εu

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S y = σ0 εim = 110(0.2231) 0.24 = 76.7 kpsi Ans.

Eq. (3-14)

Su =

Eq. (3-15) 3-17

Page 48

61.5 Su = = 76.9 kpsi Ans. 1−W 1 − 0.2

For H B = 250, Su = 0.495 (250) = 124 kpsi = 3.41 (250) = 853 MPa

Eq. (3-17)

3-18

For the data given,  H B = 2530

Ans.



H B2 = 640 226 640 226 − (2530) 2 /10 2530 = 3.887 = 253 σˆ H B = H¯ B = 9 10

Eq. (3-17)

S¯u = 0.495(253) = 125.2 kpsi

Ans.

σ¯ su = 0.495(3.887) = 1.92 kpsi Ans. 3-19

From Prob. 3-18,

H¯ B = 253

and σˆ HB = 3.887

Eq. (3-18) S¯u = 0.23(253) − 12.5 = 45.7 kpsi σˆ su = 0.23(3.887) = 0.894 kpsi 3-20

. 45.52 = 34.5 in · lbf/in3 uR = 2(30)

(a)

Ans.

Ans.

Ans.

(b) P

L

0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800

0 0.0004 0.0006 0.0010 0.0013 0.0023 0.0028 0.0036 0.0089

A

0.1963 0.1924 0.1875 0.1563 0.1307 0.1077

A0 /A − 1

ε

0.012 291 0.032 811 0.059 802 0.271 355 0.520 373 0.845059

0 0.0002 0.0003 0.0005 0.000 65 0.001 15 0.0014 0.0018 0.004 45 0.012 291 0.032 811 0.059 802 0.271 355 0.520 373 0.845 059

σ = P/A0 0 5 032.39 10 064.78 15 097.17 20 129.55 35 226.72 42 272.06 44 285.02 46 297.97 45 794.73 66 427.53 76 492.30 85 550.60 82 531.17 74 479.35

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Chapter 3  90000 80000 70000 60000 50000 40000 30000 20000 10000 0

0

0.2

0.4 All data points

0.6

0.8



 50000 45000 40000 35000 30000 25000 20000 A1

15000

A2

10000 5000 0

0

0.001

0.002 0.003 First 9 data points

0.004

0.005



 90000 80000 70000 60000 50000 40000

A4

A5

30000 20000 A3 10000 0

0

0.2

0.4 0.6 Last 6 data points

0.8



5 1 .  uT = Ai = (43 000)(0.001 5) + 45 000(0.004 45 − 0.001 5) 2 i=1 1 + (45 000 + 76 500)(0.059 8 − 0.004 45) 2 +81 000(0.4 − 0.059 8) + 80 000(0.845 − 0.4)

. = 66.7(103 )in · lbf/in3

Ans.