Shigkeys 11th Ed Ch 3 Solutions
Chapter 6
6-1 Eq. (2-21): SHut 3.4 B 3.4(300) 1020 MPa
Eq. (6-8): SSeu 0.5 t0.5(1020) 510 MPa
Table 6-2: ab1.58, 0.
Eq. (6-19):
0. 1.58(1020) 0.
b kaauSt
Eq. (6-20):
0.107 0. kdb 1.24 1.24(10) 0.
Eq. (6-18): SkkSeabe (0.877)(0.969)(510)433 MPa Ans.
6-2 (a) Table A-20: Sut = 80 kpsi
Eq. (6-8): SAe0.5(80) 40 kpsi ns.
ns
ns
(b) Table A-20: Sut = 90 kpsi
Eq. (6-8): SAe0.5(90) 45 kpsi.
(c) Aluminum has no endurance limit. Ans.
(d) Eq. (6-8): Sut > 200 kpsi, SAe100 kpsi.
__
6-3 Sut120 kpsi, rev 70 kpsi
Fig. 6-18: f 0.
Eq. (6-8): SSee0.5(120) 60 kpsi
Eq. (6-14):
2 2 ()0.82(120) 161.4 kpsi 60
ut
e
fS a S
Eq. (6-15):
1 1 0.82(120)
log log 0. 3360
ut
e
fS b S
Eq. (6-16):
1/ 1 0. rev 70 116 700 cycles.
161.
b
NA a
ns
__
6-4 Sut1600 MPa, 900 rev MPa
Fig. 6-18: Sut = 1600 MPa = 232 kpsi. Off the graph, so estimate f = 0.77.
Eq. (6-8): Sut > 1400 MPa, so Se = 700 MPa
Eq. (6-14):
2 2 ()0.77(1600) 2168.3 MPa 700
ut
e
fS a S
Eq. (6-15):
1 1 0.77(1600)
log log 0. 3 3 700
ut
e
fS b S
Eq. (6-16):
1/ 1 0. rev 900 46 400 cycles. 2168.
b
NA a
ns
__
6-5 SNut230 kpsi, 150 000 cycles
Fig. 6-18, point is off the graph, so estimate: f = 0.
Eq. (6-8): Sut > 200 kpsi, so SSee 100 kpsi
Eq. (6-14):
2 2 ()0.77(230) 313.6 kpsi 100
ut
e
fS a S
Eq. (6-15):
1 1 0.77(230)
log log 0. 3 3 100
ut
e
fS b S
Eq. (6-13):
0. 313.6(150 000) 117.0 kpsi.
b Saf N Ans
__
6-6 Sut1100 MPa= 160 kpsi
Fig. 6-18: f = 0.
Eq. (6-8): SSee0.5(1100) 550 MPa
Eq. (6-14):
2 2 ()0.79(1100) 1373 MPa 550
ut
e
fS a S
Eq. (6-15):
1 1 0.79(1100)
log log 0. 3 3 550
ut
e
fS b S
Eq. (6-13):
0. 1373(150 000) 624 MPa.
b Saf N Ans
__
6-
SS Nut150 kpsi, yt 135 kpsi, 500 cycles
Fig. 6-18: f = 0.
From Fig. 6-10, we note that below 10
3 cycles on the S-N diagram constitutes the low-
cycle region, in which Eq. (6-17) is applicable.
Check:
3
6
3 0. 10 6 0. 10
( ) 162(10 ) 90 kpsi
( ) 162(10 ) 50 kpsi
fax
fax
S
S
The end points agree.
6-10 d = 1.5 in, Sut = 110 kpsi
Eq. (6-8): Se0.5(110) 55 kpsi
Table 6-2: a = 2.70, b = 0.
Eq. (6-19):
0. 2.70(110) 0.
b kaauSt
Since the loading situation is not specified, we'll assume rotating bending or torsion so
Eq. (6-20) is applicable. This would be the worst case.
0.107 0. 0.879 0.879(1.5) 0.
Eq. (6-18): 0.777(0.842)(55) 36.0 kpsi.
b
eabe
kd
SkkS Ans
__
6-11 For AISI 4340 as-forged steel,
Eq. (6-8): Se = 100 kpsi
Table 6-2: a = 39.9, b = 0.
Eq. (6-19): ka = 39.9(260)
0. = 0.
Eq. (6-20):
0. 0. 0. 0.
kb
Each of the other modifying factors is unity.
Se = 0.158(0.907)(100) = 14.3 kpsi
For AISI 1040:
0.
0.5(113) 56.5 kpsi
39.9(113) 0.
0.907 (same as 4340)
e
a
b
S
k
k
Each of the other modifying factors is unity
Se0.362(0.907)(56.5) 18.6 kpsi
Not only is AISI 1040 steel a contender, it has a superior endurance strength.
6-12 D = 1 in, d = 0.8 in, T = 1800 lbfin, f = 0.9, and from Table A-20 for AISI 1020 CD,
Sut = 68 kpsi, and Sy = 57 kpsi.
(a)
0.1 1
Fig. A-15-15: 0.125, 1.25, 1. 0.8 0.
ts
rD K dd
Get the notch sensitivity either from Fig. 6-21, or from the curve-fit Eqs. (6-34) and
(6-35b). We'll use the equations.
35 8 23 a 0.190 2.51 10 68 1.35 10 68 2.67 10 68 0.
11
0.
0.
1 1
0.
qs a
r
Eq. (6-32): Kfs = 1 + qs (Kts 1) = 1 + 0.812(1.40 1) = 1.
For a purely reversing torque of T = 1800 lbfin,
33
16 1.32(16)(1800)
23 635 psi 23.6 kpsi (0.8)
fs afs
Tr KT K Jd
Eq. (6-8): Se0.5(68) 34 kpsi
Eq. (6-19): ka = 2.70(68)
0. = 0.
Eq. (6-20): kb = 0.879(0.8)
0. = 0.
Eq. (6-26): kc = 0.
Eq. (6-18) (labeling for shear): Sse = 0.883(0.900)(0.59)(34) = 15.9 kpsi
For purely reversing torsion, use Eq. (6-54) for the ultimate strength in shear.
Eq. (6-54): Ssu = 0.67 Sut = 0.67(68) = 45.6 kpsi
Adjusting the fatigue strength equations for shear,
Eq. (6-14):
22 0.9(45.6) 105.9 kpsi 15.
su
se
fS a S
Eq. (6-15):
1 1 0.9(45.6)
log log 0.137 27 3 3 15.
su
se
fS b S
Eq. (6-16):
1 1 23.3 0.137 27 3 61.7 10 cycles. 105.
a b NA a
ns
6 Sf 403 10 3 1. a 7200 /
n b
b = 0.0299 m Select b = 30 mm.
Since the size factor was guessed, go back and check it now.
Eq. (6-25):
1/ 2 dhe0.808 b b0.808 0.808 30 24.24 mm
Eq. (6-20):
0. 24. 0. 7.
kb
Our guess of 0.85 was slightly conservative, so we will accept the result of
b = 30 mm. Ans.
Checking yield,
6 max 3
7200
10 267 MPa 0.
max
420
1.
267
y y
S
n
__
-14 Given: w =2.5 in, t = 3/8 in, d = 0.5 in, nd = 2. From Table A-20, for AISI 1020 CD,
Eq. (6-8):
b = 1 (axial loading)
Eq. (6-18): Se = 0.88(1)(0.85)(34) = 25.4 kpsi
notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
h
6
Sut = 68 kpsi and Sy = 57 kpsi.
Se0.5(68) 34 kpsi
Table 6-2: k 88
0. 2.70(68) 0.
a
Eq. (6-21): k
Eq. (6-26): kc = 0.
Table A-15-1: /dKw0.5 / 2.5 0.2, t 2.
Get the
(6-35a). The relatively large radius is off the graph of Fig. 6-20, so we'll assume the
curves continue according to the same trend and use the equations to estimate the notc
sensitivity.
352 83 a 0.246 3.08 10 68 1.51 10 68 2.67 10 68 0.
11
0.
0.
1 1
0.
q a
r
Eq. (6-32): Kqft 1 (K1) 1 0.836(2.5 1) 2.
2.
=
(3 / 8)(2.5 0.5)
aa af
FF
KF
A
a
e life was not mentioned, we'll assume infinite life is desired, so the Since a finit
completely reversed stress must stay below the endurance limit.
25.
2
3
e f aa
S
n F
FAa4.23 kips ns.
ble A-20, for AISI 1095 HR, Sut = 120 kpsi and Sy = 66 kpsi.
6 -15 Given: max min
Dd r M 2 in, 1.8 in, 0.1 in, 25 000 lbf in, M0.
From Ta
Eq. (6-8): SSeu 0.5 t0.5 120 60 kpsi
Eq. (6-19):
0. 2.70(120) 0.
b kaauSt
Eq. (6-24): dde 0.370 0.370(1.8) 0.666 n i
Eq. (6-20):
0.107 0.10 7 kdbe0.879 0.879(0.666) 0.
Fig. A-15-14:
Eq. (6-26): 1kc
Eq. (6-18): SkkkSeabce (0.76)(0.92)(1)(60)42.0 kpsi
Dd/ 2 / 1.8 1.11, /rd 0.1 / 1.8 0.056 Kt2.
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). We'll use the equations.
352 83 a 0.246 3.08 10 120 1.51 10 120 2.67 10 120 0.
11
0.
0.
1 1
0.
q a
r
Eq. (6-32): Kqft 1 (K1) 1 0.87(2.1 1) 1.96
4 44 Id ( / 64)( / 64)(1.8) 0.5153 in
max
min
25 000(1.8 / 2)
43 664 psi 43.7 kpsi 0.
0
Mc
I
Eq. (6-8):
' SSeu 0.5 t0.5(470) 235 MPa
Eq. (6-19):
0. 4.51(470) 0.
b kaauSt
Eq. (6-24):
0.107 0. kdb 1.24 1.24(35) 0.
Eq. (6-26): kc 1
Eq. (6-18):
' SkkkSeabce(0.88)(0.85)(1)(235) 176 MPa
rev
176
1.14 Infinite life is predicted.. 1.55 99.
e f f
S
nA K
ns
__
6-17 From a free-body diagram analysis, the
bearing reaction forces are found to be RA =
2000 lbf and RB = 1500 lbf. The shear-force
and bending-moment diagrams are shown. The critical location will be at the shoulder
fillet between the 1-5/8 in and the 1-7/8 in
diameters, where the bending moment is
large, the diameter is smaller, and the stress
concentration exists.
M = 16 000 – 500 (2.5) = 14 750 lbf · in
With a rotating shaft, the bending stress will
be completely reversed.
rev 4
14 750(1.625 / 2)
35.0 kpsi ( / 64)(1.625)
Mc
I
This stress is far below the yield strength of 71 kpsi, so yielding is not predicted.
Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). We will use the equations.
35238 a 0.246 3.08 10 85 1.51 10 85 2.67 10 85 0.
11
0.
0.
1 1
0.
q a
r
.
Eq. (6-32): Kqft 1 (K1) 1 0.76(1.95 1) 1.72
Eq. (6-8): SS
' eu 0.5 t0.5(85) 42.5 kpsi
Eq. (6-19):
0. 2.70(85) 0.
b kaauSt
Eq. (6-20):
0.107 0. kdb 0.879 0.879(1.625) 0.
Eq. (6-26): kc 1
Eq. (6-18):
' SkkkSeabce(0.832)(0.835)(1)(42.5)29.5 kpsi
rev
29.
0..
1.72 35.
e f f
S
nA K
ns
Infinite life is not predicted. Use the S-N diagram to estimate the life.
Fig. 6-18: f = 0.
22 0.867(85) Eq. (6-14): 184. 29.
1 1 0.867(85)
Eq. (6-15): log log 0. 3 3 29.
ut
e
ut
e
fS a S
fS b S
11
rev (1.72)(35.0) 0. Eq. (6-16): 4611 cycles 184.
b Kf N a
N = 4600 cycles Ans.
6-18 From a free-body diagram analysis, the
bearing reaction forces are found to be RA =
1600 lbf and RB = 2000 lbf. The shear-force
and bending-moment diagrams are shown.
The critical location will be at the shoulder fillet between the 1-5/8 in and the 1-7/8 in
diameters, where the bending moment is
large, the diameter is smaller, and the stress
concentration exists.
M = 12 800 + 400 (2.5) = 13 800 lbf · in
With a rotating shaft, the bending stress will
be completely reversed.
rev 4
13 800(1.625 / 2
)
32.8 kpsi ( / 64)(1.625)
Mc
I
This stress is far below the yield strength of 71 kpsi, so yielding is not predicted.
Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.
First, we'll evaluate the stress. From a free-body diagram analysis, the reaction forces at
the bearings are R 1 = 2 kips and R 2 = 6 kips. The critical stress location is in the middle
of the span at the shoulder, where the bending moment is high, the shaft diameter is smaller, and a stress concentration factor exists. If the critical location is not obvious,
prepare a complete bending moment diagram and evaluate at any potentially critical
locations. Evaluating at the critical shoulder,
M2 kip 10 in20 kip in
rev 4333
/ 2 32 32 20 203.
kpsi /
Mc Md M
Id d d d
Now we'll get the notch sensitivity and stress concentration factor. The notch sensitivity
depends on the fillet radius, which depends on the unknown diameter. For now, we'll
estimate a value for q = 0.85 from observation of Fig. 6-20, and check it later.
Fig. A-15-9: Dd/ 1.4 /d d1.4, /rd0.1 /d d0.1, Kt 1.
Eq. (6-32): Kqft 1 (K1) 1 0.85(1.65 1) 1.55
Now we will evaluate the fatigue strength.
'
0.
0.5(120) 60 kpsi
2.70(120) 0.
e
a
S
k
Since the diameter is not yet known, assume a typical value of k b = 0.85 and check later. All other modifiers are equal to one.
Se = (0.76)(0.85)(60) = 38.8 kpsi
Determine the desired fatigue strength from the S-N diagram.
Fig. 6-18: f = 0.
22 0.82(120) Eq. (6-14): 249. 38.
1 1 0.82(120)
Eq. (6-15): log log 0. 3 3 38.
ut
e
ut
e
fS a S
fS b S
0. Eq. (6-13): 249.6(570 000) 41.9 kpsi
b Saf N
Compare strength to stress and solve for the necessary d.
3 rev
d = 2.29 in
41.
1.
1.55 203.7 /
f f f
S
n K d
Since the size factor and notch sensitivity were guessed, go back and check them now.
Eq. (6-20):
0.157 0. kdb 0.91 0.91 2.29 0.
Our guess of 0.85 was conservative. From Fig. 6-20 with r = d/10 = 0.229 in, we are off
the graph, but it appears our guess for q is low. Assuming the trend of the graph
continues, we'll choose q = 0.91 and iterate the problem with the new values of kb and q.
Intermediate results are Se = 36.5 kpsi, Sf = 39.6 kpsi, and Kf = 1.59. This gives
3 rev
39.
1.
1.59 203.
d = 2.36 in Ans.
/
f f f
S
n K d
a
A quick check of kb and q show that our estimates are still reasonable for this diameter.
6-20 SS Seyu 40 kpsi, 60 kpsi, tm a m80 kpsi, 15 kpsi, 25 kpsi, 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
1/ 2 2 1/ 2 22 2
1/ 2 2 1/ 2 22 2
3 25 3 0 25.00 kpsi
3 0 3 15 25.98 kpsi
aaa
mmm
1/ 2 22 1/ 2 22 max max max
1/ 2 22
33
25 3 15 36.06 kpsi
am am
max
60
1..
36.
y y
S
nA
ns
(a) Modified Goodman, Table 6-
1
1..
(25.00 / 40) (25.98 / 80)
nAf
ns
(b) Gerber, Table 6-
2 2 1 80 25.00 2(25.98)(40) 1 1 1.. 2 25.98 40 80(25.00)
nAf
ns
1/ 2 22 1/ 2 22 max max max
22 1/ 2
33
12 0 3 10 15 44.93 kpsi
am am
max
60
1..
44.
y y
S
nA
ns
(a) Modified Goodman, Table 6-
1
1..
(21.07 / 40) (25.98 / 80)
nAf
ns
(b) Gerber, Table 6-
2 2 1 80 21.07 2(25.98)(40) 1 1 1.. 2 25.98 40 80(21.07)
nAf
ns
(c) ASME-Elliptic, Table 6-
22
1
1..
(21.07 / 40) (25.98 / 60)
nAf
ns
a
__
6-23 SS Seyu 40 kpsi, 60 kpsi, ta m80 kpsi, 30 kpsi, a 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
1/ 2 2 1/ 2 22 2
22 1/ 2
3 0 3 30 51.96 kpsi
3 0 kpsi
aaa
mmm
1/ 2 22 1/ 2 22 max max max
2 1/ 2
33
3 30 51.96 kpsi
am am
max
60
1..
51.
y y
S
nA
ns
(a) Modified Goodman, Table 6-
1
0..
(51.96 / 40)
nAfns
(b) Gerber criterion of Table 6-7 is only valid for m > 0; therefore use Eq. (6-47).
40
10
51.
ae ff ea
S
nn S
.77Ans.
(c) ASME-Elliptic, Table 6-
2
1
0..
(51.96 / 40)
nAfns
Since infinite life is not predicted, estimate a life from the S-N diagram. Since 'm = 0,
the stress state is completely reversed and the S-N diagram is applicable for 'a.
Fig. 6-18: f = 0.
Eq. (6-14):
2 2 ()0.875(80) 122. 40
ut
e
fS a S
Eq. (6-15):
1 1 0.875(80)
log log 0. 3340
ut
e
fS b S
Eq. (6-16):
1 1/ 0. rev 51. 39 600 cycles. 122.
b
NA a
ns
a
__
6-24 SS Seyu 40 kpsi, 60 kpsi, ta mm80 kpsi, 15 kpsi, 15 kpsi, 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
1/ 2 2 1/ 2 22 2 aaa 3 0 3 15 25.98 kpsi
1/ 2 2 1/ 2 22 2 mmm 3 15 3 0 15.00 kpsi
1/ 2 22 1/ 2 22 max max max
22 1/ 2
33
15 3 15 30.00 kpsi
am am
max
60
2..
30
y y
S
nA
ns
(a) Modified Goodman, Table 6-
1
1..
(25.98 / 40) (15.00 / 80)
nAf
ns
(b) Gerber, Table 6-
2 2 1 80 25.98 2(15.00)(40) 1 1 1.. 2 15.00 40 80(25.98)
nAf
ns
Eq. (6-14):
2 2 ()0.87(590) 1263 208.
ut
e
fS a S
Eq. (6-15):
1 1 0.87(590)
log log 0. 3 3 208.
ut
e
fS b S
Eq. (6-16):
1/ 1 rev 324.2 0. 33 812 cycles 1263
b
N a
N = 34 000 cycles Ans.
6-26 SSFFut 590 MPa, y 490 MPa, max 28 kN, min 12 kN
Check for yielding
max 2 max
28 000
147.4 N/mm 147.4 MPa 10(25 6)
F
A
max
490
3..
147.
y y
S
nA
ns
Determine the fatigue factor of safety based on infinite life.
From Prob. 6-25: SKef208.6 MPa, 2.
max min 28 000 12 000 2.2 92.63 MPa 2 2(10)(25 6)
af
FF
K
A
max min 28 000 12 000 2.2 231.6 MPa 2 2(10)(25 6)
mf
FF
K
A
Modified Goodman criteria:
1 92.63 231.
208.6 590
am
nSSfeut
nAf 1.20 ns.
Gerber criteria:
22 1 2 11 2
ut a m e f me uta
SS
n SS
2 2 1 590 92.63 2(231.6)(208.6) 11 2 231.6 208.6 590(92.63)
nAf1.49 ns.
ASME-Elliptic criteria:
22 2
11
( / ) ( / ) (92.63 / 208.6) (231.6 / 490)
f ae m y
n SS
2
= 1.54 Ans.
The results are consistent with Fig. 6-27, where for a mean stress that is about half of the
yield strength, the Modified Goodman line should predict failure significantly before the other two.
6-27 SSut590 MPa, y 490 MPa
(a) FFmax28 kN, min 0 kN
Check for yielding
max 2 max
28 000
147.4 N/mm 147.4 MPa 10(25 6)
F
A
max
490
3..
147.
y y
S
nA
ns
From Prob. 6-25: SKef208.6 MPa, 2.
max min
max min
28 000 0
2.2 162.1 MPa 2 2(10)(25 6)
28 000 0
2.2 162.1 MPa 2 2(10)(25 6)
af
mf
FF
K
A
FF
K
A
1 162.1 162.
208.6 590
am
nSSfeut
nAf0.95 ns.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
rev
162.
223.5 MPa 1 ( / ) 1 (162.1 / 590)
a
mutS
Fig. 6-18: f = 0.
Eq. (6-14):
2 2 ()0.87(590) 1263 208.
ut
e
fS a S
Shigkeys 11th Ed Ch 3 Solutions
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