Solucionario Shtgley 8Va Edicion
Páginas: 34 (8326 palabras)
Publicado: 17 de diciembre de 2012
budynas_SM_ch06.qxd
11/29/2006
17:40
FIRST PAGES
Page 147
Chapter 6
Note to the instructor: Many of the problems in this chapter are carried over from the previous
edition. The solutions have changed slightly due to some minor changes. First, the calculation
of the endurance limit of a rotating-beam specimen Se is given by Se = 0.5 Sut instead of
Se = 0.504 Sut . Second, whenthe fatigue stress calculation is made for deterministic problems,
only one approach is given, which uses the notch sensitivity factor, q, together with Eq. (6-32).
Neuber’s equation, Eq. (6-33), is simply another form of this. These changes were made to hopefully make the calculations less confusing, and diminish the idea that stress life calculations are
precise.
6-1 H B = 490
Eq. (2-17):Sut = 0.495(490) = 242.6 kpsi > 212 kpsi
Se = 100 kpsi
Eq. (6-8):
a = 1.34, b = −0.085
Table 6-2:
ka = 1.34(242.6) −0.085 = 0.840
Eq. (6-19):
1/4
0.3
−0.107
Eq. (6-20):
kb =
= 1.02
Eq. (6-18):
Se = ka kb Se = 0.840(1.02)(100) = 85.7 kpsi
Ans .
6-2
(a) Sut = 68 kpsi, Se = 0.5(68) = 34 kpsi
(b) Sut = 112 kpsi, Se = 0.5(112) = 56 kpsi
(c) 2024T3 has noendurance limit
(d) Eq. (6-8): Se = 100 kpsi
6-3
Eq. (2-11):
Eq. (6-8):
Ans .
Ans .
Ans.
Ans .
σ F = σ0 εm = 115(0.90) 0.22 = 112.4 kpsi
Se = 0.5(66.2) = 33.1 kpsi
log(112.4/33.1)
= −0.084 26
log(2 · 106 )
Eq. (6-12):
b=−
Eq. (6-10):
f=
112.4
(2 · 103 ) −0.084 26 = 0.8949
66.2
Eq. (6-14):
a=
[0.8949(66.2)]2
= 106.0 kpsi
33.1
Eq. (6-13):
S f =a N b = 106.0(12 500) −0.084 26 = 47.9 kpsi Ans.
Eq. (6-16):
N=
σa
a
1/b
=
36
106.0
−1/0.084 26
= 368 250 cycles Ans.
budynas_SM_ch06.qxd
148
11/29/2006
17:40
FIRST PAGES
Page 148
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-4 From S f = a N b
log S f = log a + b log N
Substituting (1, Sut )
logSut = log a + b log (1)
a = Sut
From which
Substituting (10 , f Sut ) and a = Sut
3
log f Sut = log Sut + b log 103
From which
b=
∴
1
log f
3
S f = Sut N (log f )/3
1 ≤ N ≤ 103
For 500 cycles as in Prob. 6-3
S f ≥ 66.2(500) (log 0.8949)/3 = 59.9 kpsi
Ans.
6-5 Read from graph: (103, 90) and (106, 50). From S = a N b
log S1 = log a + b log N1
log S2 = log a + b logN2
From which
log a =
log S1 log N2 − log S2 log N1
log N2 / N1
log 90 log 106 − log 50 log 103
log 106 /103
= 2.2095
=
a = 10log a = 102.2095 = 162.0
b=
log 50/90
= −0.085 09
3
( S f ) ax = 162−0.085 09
Check:
103 ≤ N ≤ 106 in kpsi Ans.
103 ( S f ) ax = 162(103 ) −0.085 09 = 90 kpsi
106 ( S f ) ax = 162(106 ) −0.085 09 = 50 kpsi
The end points agree.
6-6
Eq.(6-8):
Table 6-2:
Eq. (6-19):
Se = 0.5(710) = 355 MPa
a = 4.51,
b = −0.265
ka = 4.51(710) −0.265 = 0.792
budynas_SM_ch06.qxd
11/29/2006
17:40
FIRST PAGES
Page 149
149
Chapter 6
d
7.62
−0.107
=
32
7.62
−0.107
Eq. (6-20):
kb =
= 0.858
Eq. (6-18):
Se = ka kb Se = 0.792(0.858)(355) = 241 MPa Ans.
6-7 For AISI 4340 as forged steel,Se = 100 kpsi
Eq. (6-8):
a = 39.9,
Table 6-2:
b = −0.995
ka = 39.9(260) −0.995 = 0.158
Eq. (6-19):
0.75
0.30
Each of the other Marin factors is unity.
kb =
Eq. (6-20):
−0.107
= 0.907
Se = 0.158(0.907)(100) = 14.3 kpsi
For AISI 1040:
Se = 0.5(113) = 56.5 kpsi
ka = 39.9(113) −0.995 = 0.362
kb = 0.907 (same as 4340)
Each of the other Marin factors is unity.
Se =0.362(0.907)(56.5) = 18.6 kpsi
Not only is AISI 1040 steel a contender, it has a superior endurance strength. Can you see
why?
6-8
(a) For an AISI 1018 CD-machined steel, the strengths are
2.5 mm
20 mm
25 mm
Eq. (2-17):
Sut = 440 MPa
⇒
HB =
S y = 370 MPa
Ssu = 0.67(440) = 295 MPa
Fig. A-15-15:
Fig. 6-21:
Eq. (6-32):
2.5
D
25
r
=
= 0.125,
=
= 1.25,
d
20...
11/29/2006
17:40
FIRST PAGES
Page 147
Chapter 6
Note to the instructor: Many of the problems in this chapter are carried over from the previous
edition. The solutions have changed slightly due to some minor changes. First, the calculation
of the endurance limit of a rotating-beam specimen Se is given by Se = 0.5 Sut instead of
Se = 0.504 Sut . Second, whenthe fatigue stress calculation is made for deterministic problems,
only one approach is given, which uses the notch sensitivity factor, q, together with Eq. (6-32).
Neuber’s equation, Eq. (6-33), is simply another form of this. These changes were made to hopefully make the calculations less confusing, and diminish the idea that stress life calculations are
precise.
6-1 H B = 490
Eq. (2-17):Sut = 0.495(490) = 242.6 kpsi > 212 kpsi
Se = 100 kpsi
Eq. (6-8):
a = 1.34, b = −0.085
Table 6-2:
ka = 1.34(242.6) −0.085 = 0.840
Eq. (6-19):
1/4
0.3
−0.107
Eq. (6-20):
kb =
= 1.02
Eq. (6-18):
Se = ka kb Se = 0.840(1.02)(100) = 85.7 kpsi
Ans .
6-2
(a) Sut = 68 kpsi, Se = 0.5(68) = 34 kpsi
(b) Sut = 112 kpsi, Se = 0.5(112) = 56 kpsi
(c) 2024T3 has noendurance limit
(d) Eq. (6-8): Se = 100 kpsi
6-3
Eq. (2-11):
Eq. (6-8):
Ans .
Ans .
Ans.
Ans .
σ F = σ0 εm = 115(0.90) 0.22 = 112.4 kpsi
Se = 0.5(66.2) = 33.1 kpsi
log(112.4/33.1)
= −0.084 26
log(2 · 106 )
Eq. (6-12):
b=−
Eq. (6-10):
f=
112.4
(2 · 103 ) −0.084 26 = 0.8949
66.2
Eq. (6-14):
a=
[0.8949(66.2)]2
= 106.0 kpsi
33.1
Eq. (6-13):
S f =a N b = 106.0(12 500) −0.084 26 = 47.9 kpsi Ans.
Eq. (6-16):
N=
σa
a
1/b
=
36
106.0
−1/0.084 26
= 368 250 cycles Ans.
budynas_SM_ch06.qxd
148
11/29/2006
17:40
FIRST PAGES
Page 148
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-4 From S f = a N b
log S f = log a + b log N
Substituting (1, Sut )
logSut = log a + b log (1)
a = Sut
From which
Substituting (10 , f Sut ) and a = Sut
3
log f Sut = log Sut + b log 103
From which
b=
∴
1
log f
3
S f = Sut N (log f )/3
1 ≤ N ≤ 103
For 500 cycles as in Prob. 6-3
S f ≥ 66.2(500) (log 0.8949)/3 = 59.9 kpsi
Ans.
6-5 Read from graph: (103, 90) and (106, 50). From S = a N b
log S1 = log a + b log N1
log S2 = log a + b logN2
From which
log a =
log S1 log N2 − log S2 log N1
log N2 / N1
log 90 log 106 − log 50 log 103
log 106 /103
= 2.2095
=
a = 10log a = 102.2095 = 162.0
b=
log 50/90
= −0.085 09
3
( S f ) ax = 162−0.085 09
Check:
103 ≤ N ≤ 106 in kpsi Ans.
103 ( S f ) ax = 162(103 ) −0.085 09 = 90 kpsi
106 ( S f ) ax = 162(106 ) −0.085 09 = 50 kpsi
The end points agree.
6-6
Eq.(6-8):
Table 6-2:
Eq. (6-19):
Se = 0.5(710) = 355 MPa
a = 4.51,
b = −0.265
ka = 4.51(710) −0.265 = 0.792
budynas_SM_ch06.qxd
11/29/2006
17:40
FIRST PAGES
Page 149
149
Chapter 6
d
7.62
−0.107
=
32
7.62
−0.107
Eq. (6-20):
kb =
= 0.858
Eq. (6-18):
Se = ka kb Se = 0.792(0.858)(355) = 241 MPa Ans.
6-7 For AISI 4340 as forged steel,Se = 100 kpsi
Eq. (6-8):
a = 39.9,
Table 6-2:
b = −0.995
ka = 39.9(260) −0.995 = 0.158
Eq. (6-19):
0.75
0.30
Each of the other Marin factors is unity.
kb =
Eq. (6-20):
−0.107
= 0.907
Se = 0.158(0.907)(100) = 14.3 kpsi
For AISI 1040:
Se = 0.5(113) = 56.5 kpsi
ka = 39.9(113) −0.995 = 0.362
kb = 0.907 (same as 4340)
Each of the other Marin factors is unity.
Se =0.362(0.907)(56.5) = 18.6 kpsi
Not only is AISI 1040 steel a contender, it has a superior endurance strength. Can you see
why?
6-8
(a) For an AISI 1018 CD-machined steel, the strengths are
2.5 mm
20 mm
25 mm
Eq. (2-17):
Sut = 440 MPa
⇒
HB =
S y = 370 MPa
Ssu = 0.67(440) = 295 MPa
Fig. A-15-15:
Fig. 6-21:
Eq. (6-32):
2.5
D
25
r
=
= 0.125,
=
= 1.25,
d
20...
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