Topic 4 (Shaft):
1. A shaft is loaded in bending and torsion such that Ma = 600 lbf·in, Ta = 400 lbf·in, Mm =
500 lbf·in, and Tm = 300 lbf·in. For the shaft, Su = 100 kpsi and Sy = 80 kpsi, and a fully
corrected endurance limit of Se = 30 kpsi is assumed. Let Kf = 2.2 and Kfs = 1.8. With a
design factor of 2.0 determine the minimum acceptable diameter of the shaft using the
(a) DE-Gerber criterion.
(b) DE-elliptic criterion.
(c) DE-Soderberg criterion.
(d) DE-Goodman criterion.
Discuss and compare the results.
Solution:
Given: Ma = 600 lbf·in, Ta = 400 lbf·in, Mm = 500 lbf·in, Tm = 300 lbf·in, Su = 100 kpsi, Sy =
80 kpsi, Se = 30 kpsi, Kf = 2.2, Kfs = 1.8, n = 2.0.
2
2
2
2
𝐴 = √4(𝐾𝑓 𝑀𝑎 ) + 3(𝐾𝑓𝑠 𝑇𝑎 ) = √4(2.2 ∙ 600)2 + 3(1.8 ∙ 400)2 = 2920 lbf ∙ in
𝐵 = √4(𝐾𝑓 𝑀𝑚 ) + 3(𝐾𝑓𝑠 𝑇𝑚 ) = √4(2.2 ∙ 500)2 + 3(1.8 ∙ 300)2 = 2391 lbf ∙ in
(a) DE-Gerber criterion:
16𝑛 𝐴
2𝐵𝑆𝑒 2
) ]
𝑑=(
∙
{1 + [1 + (
𝜋 2𝑆𝑒
𝐴𝑆𝑢𝑡
1⁄3
1⁄2
})
1⁄3
1⁄2
16 ∙ 2
2920
2 ∙ 2391 ∙ 30000 2
) ]
=(
∙
{1 + [1 + (
𝜋
2 ∙ 30000
2920 ∙ 100000
= 1.016 in
})
𝐴𝑛𝑠.
(b) DE-elliptic criterion:
1⁄3
1⁄3
𝐴2
2
𝐵2
16𝑛
𝑑=(
√ 2 + 2)
𝜋
𝑆𝑒 𝑆𝑦
= 1.012 in
2
16 ∙ 2
2920
2391
√(
) +(
) )
=(
𝜋
30000
80000
𝐴𝑛𝑠.
(c) DE-Soderberg criterion:
1⁄3
16𝑛 𝐴
𝐵
𝑑=[
( + )]
𝜋 𝑆𝑒 𝑆𝑦
= 1.090 in
16 ∙ 2 2920
2391 1⁄3
(
)]
=[
+
𝜋
30000 80000
𝐴𝑛𝑠.
1
(d) DE-Goodman criterion:
𝑑=[
16𝑛 𝐴
𝐵 1⁄3
16 ∙ 2 2920
2391 1⁄3
( +
)]
(
)]
=[
+
𝜋 𝑆𝑒 𝑆𝑢𝑡
𝜋
30000 100000
= 1.073 in
𝐴𝑛𝑠.
Criterion
DE-Gerber
DE-elliptic
DE-Soderberg
DE-Goodman
d (in)
Compared to DE-Gerber
1.016
1.012 0.4% lower less conservative
1.090 7.3% higher more conservative
1.073 5.6% higher more conservative
2. The rotating solid steel shaft is simply supported by bearings at points B and C and is driven
by a gear (not shown) which meshes with the spur gear at D, which has a 150-mm pitch
diameter. The force F from the drive gear acts at a pressure angle of 20°. The shaft transmits
a torque to point A of TA = 340 N·m. The shaft is machined from steel with Sy = 420 MPa
and Sut = 560 MPa. Using a factor of safety of 2.5, determine the minimum allowable
diameter of the 250-mm section of the shaft based on
(a) a static yield analysis using the distortion energy theory and
(b) a fatigue-failure analysis. Assume sharp fillet radii at the bearing shoulders for
estimating stress-concentration factors.
Solution:
Given: d = 0.15 m, ϕ = 20°, TA = 340 N·m, Sy = 420 MPa, Sut = 560 MPa, n = 2.5.
𝑇𝐴 = 𝐹 cos 20°
𝐹=
𝑑
2
2𝑇𝐴
2 ∙ 340
=
= 4824 N
𝑑 cos 20° 0.150 cos 20°
2
The maximum bending moment will be at point C, with MC = F·dCD = 4824(0.1) = 482.4 N·m.
Due to the rotation, the bending is completely reversed, while the torsion is constant.
Thus, Ma = 482.4 N·m, Tm = 340 N·m, Mm = Ta = 0.
For sharp fillet radii at the shoulders, from Table 7-1, Kt = 2.7, and Kts = 2.2. Examining Figs.
6-20 and 6-21 with Sut = 560 MPa, conservatively estimate q = 0.8 and qs = 0.9. These estimates
can be checked once a specific fillet radius is determined.
𝐾𝑓 = 1 + 𝑞(𝐾𝑡 − 1) = 1 + 0.8(2.7 − 1) = 2.4
𝐾𝑓𝑠 = 1 + 𝑞𝑠 (𝐾𝑡𝑠 − 1) = 1 + 0.9(2.2 − 1) = 2.1
(a) We will choose to include fatigue stress concentration factors even for the static
analysis to avoid localized yielding.
2 1⁄2
2
32𝐾𝑓 (𝑀𝑚 + 𝑀𝑎 )
16𝐾𝑓𝑠 (𝑇𝑚 + 𝑇𝑎 )
′
𝜎𝑚𝑎𝑥
= [(
)
+
3
(
) ]
𝜋𝑑3
𝜋𝑑3
𝑛=
𝑆𝑦
𝜋𝑑3 𝑆𝑦
2
2 −1⁄2
=
[4(𝐾
𝑀
)
+
3(𝐾
𝑇
)
]
𝑓
𝑎
𝑓𝑠
𝑚
′
𝜎𝑚𝑎𝑥
16
Solving for d,
1⁄3
16𝑛
2
2 1⁄2
𝑑=(
{[4(𝐾𝑓 𝑀𝑎 ) + 3(𝐾𝑓𝑠 𝑇𝑚 ) ] })
𝜋𝑆𝑦
⁄
1 3
16 ∙ 2.5
2
2 ]1⁄2
=(
{[4(2.4 ∙ 482.4) + 3(2.1 ∙ 340)
})
𝜋 ∙ 420 ∙ 106
= 0.0430 m = 43.0 mm
𝐴𝑛𝑠.
𝑏
(b) Marin surface factor: 𝑘𝑎 = 𝑎𝑆𝑢𝑡
= 4.51(560)−0.265 = 0.84
Marin size factor: Assume kb = 0.85 for now. Check later once a diameter is known.
𝑆𝑒 = 𝑘𝑎 𝑘𝑏 𝑆𝑒′ = 𝑘𝑎 𝑘𝑏 ∙ 0.5𝑆𝑢𝑡 = 0.84(0.85)(0.5)(560) = 200 MPa
Selecting the DE-ASME Elliptic criteria, with Mm = Ta = 0.
2
1⁄3
𝐾𝑓 𝑀𝑎 2
𝐾𝑓𝑠 𝑇𝑚
16𝑛
√
) + 3(
𝑑=(
4(
) )
𝜋
𝑆𝑒
𝑆𝑦
1⁄3
2
=(
2
16 ∙ 2.5
2.4 ∙ 482.4
2.1 ∙ 340
√4 (
) + 3(
) )
6
𝜋
200 ∙ 10
420 ∙ 106
= 0.0534 m = 53.4 mm
With this diameter, we can refine our estimates for kb and q.
𝑘𝑏 = 1.51𝑑 −0.157 = 1.51(53.4)−0.157 = 0.81
Assuming a sharp fillet radius, from Table 7-1, r = 0.02d = 0.02(53.4) = 1.07 mm.
3
Fig. (6-20): q = 0.72
Fig. (6-21): qs = 0.77
Iterating with these new estimates,
𝐾𝑓 = 1 + 𝑞(𝐾𝑡 − 1) = 1 + 0.72(2.7 − 1) = 2.2
𝐾𝑓𝑠 = 1 + 𝑞𝑠 (𝐾𝑡𝑠 − 1) = 1 + 0.77(2.2 − 1) = 1.9
𝑆𝑒 = 𝑘𝑎 𝑘𝑏 𝑆𝑒′ = 𝑘𝑎 𝑘𝑏 ∙ 0.5𝑆𝑢𝑡 = 0.84(0.81)(0.5)(560) = 191 MPa
2
𝑑=(
2
1⁄3
𝐾𝑀
𝐾 𝑇
16𝑛
√4 ( 𝑓 𝑎 ) + 3 ( 𝑓𝑠 𝑚 ) )
𝜋
𝑆𝑒
𝑆𝑦
= 53 mm
𝐴𝑛𝑠.
Further iteration does not change the results.
Topic 5 (Screw):
3. Find the power required to drive a 1.5-in power screw having double square threads with a
pitch of 1/4 in. The nut is to move at a velocity of 2 in/s and move a load of F = 2.2 kips.
The frictional coefficients are 0.10 for the threads and 0.15 for the collar. The frictional
diameter of the collar is 2.25 in.
Solution:
Given: d = 1.5 in, p = 0.25 in, n = 2, F = 2.2 kips, dc = 2.25 in, f = 0.10, fc = 0.15, V = 2 in/s.
dm = d - p/2 = 1.5 - 0.25/2 = 1.375 in
l = np = 2(0.25) = 0.5 in
𝑇𝑅 =
𝐹𝑑𝑚 𝑙 + 𝜋𝑓𝑑𝑚
𝐹𝑓𝑐 𝑑𝑐
(
)+
2 𝜋𝑑𝑚 − 𝑓𝑙
2
2.2 ∙ 103 ∙ 1.375 0.5 + 𝜋 ∙ 0.10 ∙ 1.375
2.2 ∙ 103 ∙ 0.15 ∙ 2.25
(
)+
=
2
𝜋 ∙ 1.375 − 0.10 ∙ 0.5
2
= 330 + 371 = 701 lbf ∙ in
Since N = V/l = 2/0.5 = 4 rev/s = 240 rev/min
so the power is
𝐻=
𝑇𝑁
701 ∙ 240
=
= 2.67 hp
63025
63025
𝐴𝑛𝑠.
4. The figure shows a cast-iron bearing block that is to be bolted to a steel ceiling joist and is
to support a gravity load of 18 kN. Bolts used are M24 ISO 8.8 with coarse threads and
with 4.6-mm-thick steel washers under the bolt head and nut. The joist flanges are 20 mm
4
in thickness, and the dimension A, shown in the figure, is 20 mm. The modulus of elasticity
of the bearing block is 135 GPa.
(a) Find the wrench torque required if the fasteners are lubricated during assembly and the
joint is to be permanent.
(b) Determine the factors of safety guarding against yielding, overload, and joint separation.
Solution:
(a) From Table 8-11, Sp = 600 MPa.
From Table 8-1, At = 353 mm2.
𝐹𝑖 = 0.9𝐴𝑡 𝑆𝑝 = 0.9 ∙ 353 ∙ 10−6 ∙ 600 ∙ 106 = 190.6 kN
Table 8-15: K = 0.18
𝑇 = 𝐾𝐹𝑖 𝑑 = 0.18 ∙ 190.6 ∙ 24 = 823 N ∙ m
𝐴𝑛𝑠.
(b) Washers: tw = 4.6 mm, d = 24 mm, D = dw = 1.5d = 1.5(24) = 36 mm, E = 207 GPa.
𝑘1 =
=
0.5774𝜋𝐸𝑑
(1.155𝑡 + 𝐷 − 𝑑)(𝐷 + 𝑑)
ln
(1.155𝑡 + 𝐷 + 𝑑)(𝐷 − 𝑑)
0.5774𝜋 ∙ 207 ∙ 24
= 31990 MN⁄m
(1.155𝑡 + 36 − 24)(36 + 24)
ln
(1.155𝑡 + 36 + 24)(36 − 24)
Cast iron: tb = 20 mm, d = 24 mm, D = dw + 2tw∙tan  = 36 + 2(4.6) tan 30° = 41.31
mm, E = 135 GPa.
𝑘2 =
0.5774𝜋𝐸𝑑
= 10785 MN⁄m
(1.155𝑡 + 𝐷 − 𝑑)(𝐷 + 𝑑)
ln
(1.155𝑡 + 𝐷 + 𝑑)(𝐷 − 𝑑)
Steel joist: tj = 20 mm, d = 24 mm, D = 41.31 mm, E = 207 GPa.
𝑘3 =
𝑘𝑚 =
0.5774𝜋𝐸𝑑
= 16537 MN⁄m
(1.155𝑡 + 𝐷 − 𝑑)(𝐷 + 𝑑)
ln
(1.155𝑡 + 𝐷 + 𝑑)(𝐷 − 𝑑)
1
=
2⁄𝑘1 + 1⁄𝑘2 + 1⁄𝑘3
1
= 4636 MN⁄m
2
1
1
31990 + 10785 + 16537
5
Bolt: l = 2tw + tb + tj = 2(4.6) + 2(20) = 49.2 mm. Nut, Table A-31, H = 21.5 mm. L > l
+ H = 49.2 + 21.5 = 70.7 mm. From Table A-17, use L = 80 mm.
LT = 2d + 6 = 2(24) + 6 = 54 mm, ld = L - LT = 80 - 54 = 26 mm, lt = l - ld = 49.2 - 26 =
23.2 mm
From Table (8-1), At = 353 mm2, Ad = πd2 / 4 = π (242) / 4 = 452.4 mm2
𝑘𝑏 =
𝐴𝑑 𝐴𝑡 𝐸
452.4 ∙ 353 ∙ 207
=
= 1680 MN⁄m
𝐴𝑑 𝑙𝑡 + 𝐴𝑡 𝑙𝑑 452.4 ∙ 23.2 + 353 ∙ 26
C = kb / (kb + km) = 1680 / (1680 + 4636) = 0.266, Sp = 600 MPa, Fi = 190.6 kN, P =
Ptotal / N = 18/4 = 4.5 kN
Yield:
𝑛𝑝 =
𝑆𝑝 𝐴𝑡
600 ∙ 353 ∙ 10−3
=
= 1.10
𝐶𝑃 + 𝐹𝑖 0.266 ∙ 4.5 + 190.6
𝐴𝑛𝑠.
Load factor:
𝑛𝐿 =
𝑆𝑝 𝐴𝑡 − 𝐹𝑖 600 ∙ 353 ∙ 10−3 − 190.6
=
= 17.7
𝐶𝑃
0.266 ∙ 4.5
𝐴𝑛𝑠.
Separation:
𝑛0 =
𝐹𝑖
190.6
=
= 57.5
𝑃(1 − 𝐶) 4.5(1 − 0.266)
𝐴𝑛𝑠.
As was stated in the text, bolts are typically preloaded such that the yielding factor of
safety is not much greater than unity which is the case for this problem. However, the
other load factors indicate that the bolts are oversized for the external load.
5. The cantilever bracket is bolted to a column with three M12 × 1.75 ISO 5.8 bolts. The
bracket is made from AISI 1020 hot-rolled steel. Find the factors of safety for the following
failure modes: shear of bolts, bearing of bolts, bearing of bracket, and bending of bracket.
6
Solution:
Given: F = 12 kN,
Bolts, from Table 8-11, Sy = 420 MPa
Bracket, from Table A-20, Sy = 210 MPa
𝐹′ =
𝐹 12
=
= 4 kN
3
3
𝑀 = 𝐹𝑑𝐹 = 12 ∙ 200 = 2400 N ∙ m
𝐹𝐴″ = 𝐹𝐵″ =
𝑀
2400
=
= 37.5 kN
𝑑𝐹 ″
64
𝐹𝐴 = 𝐹𝐵 = √(𝐹𝐴′ )2 + (𝐹𝐴″ )2 = √42 + 37.52 = 37.7 kN
𝐹𝑂 = 4 kN
Bolt shear:
The shoulder bolt shear area, As = πd2 / 4 = π (122) / 4 = 113.1 mm2
𝑆𝑠𝑦 = 0.577𝑆𝑦 = 0.577 ∙ 420 = 242.3 MPa
𝜏=
𝐹𝐴
37.7 ∙ 103
=
= 333 MPa
𝐴𝑠 113.1 ∙ 10−6
𝑛=
𝑆𝑠𝑦 242.3
=
= 0.728
𝜏
333
𝐴𝑛𝑠.
Bearing on bolts:
𝐴𝑏 = 𝑑𝑏 = 12 ∙ 8 = 96 mm2
𝐹𝐴
37.7 ∙ 103
𝜎𝑏 = −
=−
= −393 MPa
𝐴𝑏
96 ∙ 10−6
7
𝑛=
𝑆𝑦𝑐
420
=
= 1.07
|𝜎𝑏 | 393
𝐴𝑛𝑠.
Bearing on member:
𝜎𝑏 = −393 MPa
𝑛=
𝑆𝑦𝑐
210
=
= 0.534
|𝜎𝑏 | 393
𝐴𝑛𝑠.
Bending stress in plate:
𝑏ℎ3 𝑏𝑑3
𝑏𝑑 3
𝐼=
−
− 2(
+ 𝑎2 𝑏𝑑)
12
12
12
8 ∙ 1363 8 ∙ 123
8 ∙ 123
=
−
− 2(
+ 322 ∙ 8 ∙ 12)
12
12
12
= 1.48 ∙ 106 mm4
𝐴𝑛𝑠.
𝑀𝑐 2400 ∙ (32 + 36) ∙ 10−3
𝜎=
=
= 110 MPa
𝐼
1.48 ∙ 106 ∙ 10−12
𝑆𝑦 210
𝑛=
=
= 1.91 𝐴𝑛𝑠.
𝜎
110
Failure is predicted for bolt shear and bearing on member.
6. Figure 8–19 is a cross section of a grade 25 cast-iron pressure vessel. A total of N bolts are
to be used to resist a separating force of 36 kip.
(a) Determine kb, km, and C.
(b) Find the number of bolts required for a load factor of 2 where the bolts may be reused
when the joint is taken apart.
(c) With the number of bolts obtained in part (b), determine the realized load factor for
overload, the yielding factor of safety, and the load factor for joint separation.
8
Solution:
(a) The grip is l = 1.50 in. From Table A–31, the nut thickness is 35/64 in. Adding two
threads beyond the nut of 2/11 in gives a bolt length of
𝐿=
35
2
+ 1.50 +
= 2.229 in
64
11
1
From Table A–17 the next fraction size bolt is L = 2 4 in. From Eq. (8–13), the thread
length is LT = 2d + 0.25 = 2(0.625) + 0.25 = 1.50 in. Thus, the length of the unthreaded
portion in the grip is ld = L - LT = 2.25 − 1.50 = 0.75 in. The threaded length in the grip
is lt = l − ld = 0.75 in. From Table 8–2, At = 0.226 in2. The major-diameter area is Ad =
πd2/4 = π(0.625)2/4 = 0.3068 in2. The bolt stiffness is then
𝑘𝑏 =
𝐴𝑑 𝐴𝑡 𝐸
0.3068 ∙ 0.226 ∙ 30
=
𝐴𝑑 𝑙𝑡 + 𝐴𝑡 𝑙𝑑 0.3068 ∙ 0.75 + 0.226 ∙ 0.75
= 5.21 Mlbf⁄in
𝐴𝑛𝑠.
From Table A–24, for no. 25 cast iron we will use E = 14 Mpsi. The stiffness of the
members, from Eq. (8–22), is
𝑘𝑚 =
0.5774𝜋𝐸𝑑
0.5774𝜋 ∙ 14 ∙ 0.625
=
0.5774𝑙 + 0.5𝑑
0.5774 ∙ 1.5 + 0.5 ∙ 0.625
2 ln (5
) 2 ln (5
)
0.5774𝑙 + 2.5𝑑
0.5774 ∙ 1.5 + 2.5 ∙ 0.625
= 8.95 Mlbf⁄in
𝐴𝑛𝑠.
If you are using Eq. (8–23), from Table 8–8, A = 0.77871 and B = 0.61616, and
𝑘𝑚 = 𝐸𝑑𝐴 exp(𝐵𝑑 ⁄𝑙 )
= 14 ∙ 0.625 ∙ 0.77871 exp(0.61616 ∙ 0.625⁄1.5)
= 8.81 Mlbf⁄in
which is only 1.6 percent lower than the previous result.
From the first calculation for km, the stiffness constant C is
𝐶=
𝑘𝑏
5.21
=
= 0.368
𝑘𝑏 + 𝑘𝑚 5.21 + 8.95
𝐴𝑛𝑠.
(b) From Table 8–9, Sp = 85 kpsi. Then, using Eqs. (8–31) and (8–32), we find the
recommended preload to be
𝐹𝑖 = 0.75𝐴𝑡 𝑆𝑝 = 0.75 ∙ 0.226 ∙ 85 = 14.4 kip
For N bolts, Eq. (8–29) can be written
𝑛𝐿 =
𝑁=
𝑆𝑝 𝐴𝑡 − 𝐹𝑖
𝐶(𝑃total ⁄𝑁)
𝐶𝑛𝐿 𝑃total
0.368 ∙ 2 ∙ 36
=
= 5.52
𝑆𝑝 𝐴𝑡 − 𝐹𝑖 85 ∙ 0.226 − 14.4
9
Six bolts should be used to provide the specified load factor.
Ans.
(c) With six bolts, the load factor actually realized is
𝑛𝐿 =
𝑆𝑝 𝐴𝑡 − 𝐹𝑖
85 ∙ 0.226 − 14.4
=
= 2.18
𝐶(𝑃total ⁄𝑁)
0.368 ∙ 36⁄6
𝐴𝑛𝑠.
From Eq. (8–28), the yielding factor of safety is
𝑛𝑝 =
𝑆𝑝 𝐴𝑡
85 ∙ 0.226
=
= 1.16
𝐶(𝑃total ⁄𝑁) + 𝐹𝑖 0.368 ∙ 36⁄6 + 14.4
𝐴𝑛𝑠.
From Eq. (8–30), the load factor guarding against joint separation is
𝑛0 =
𝐹𝑖
14.4
=
= 3.80
(𝑃total ⁄𝑁)(1 − 𝐶) 36⁄6 ∙ (1 − 0.368)
𝐴𝑛𝑠.
Topic 6 (Spring):
7. A helical compression spring is wound using 0.105-in-diameter music wire. The spring has
an outside diameter of 1.225 in with plain ground ends, and 12 total coils.
(a) What should the free length be to ensure that when the spring is compressed solid the
torsional stress does not exceed the yield strength, that is, that it is solid-safe?
(b) What force is needed to compress this spring to closure?
(c) Estimate the spring rate.
(d) Is there a possibility that the spring might buckle in service?
Solution:
Given: Music wire, d = 0.105 in, OD = 1.225 in, plain ground ends, Nt = 12 coils.
Table 10-1: Na = Nt − 1 = 12 − 1 = 11, Ls = dNt = 0.105(12) = 1.26 in
Table 10-4: A = 201, m = 0.145
(a) Eq. (10-14):
𝑆𝑢𝑡 =
𝐴
201
=
= 278.7 kpsi
𝑚
𝑑
0.1050.145
Table 10-6: Ssy = 0.45Sut = 0.45(278.7) = 125.4 kpsi, D = OD - d = 1.225 − 0.105 =
1.120 in, C = D/d = 1.120/0.105 = 10.67
𝐾𝐵 =
4𝐶 + 2 4 ∙ 10.67 + 2
=
= 1.126
4𝐶 − 3 4 ∙ 10.67 − 3
𝜋𝑑 3 𝑆𝑠𝑦 𝜋 ∙ 0.1053 ∙ 125.4 ∙ 103
𝐹|𝑆𝑠𝑦 =
=
= 45.2 lbf
8𝐾𝐵 𝐷
8 ∙ 1.126 ∙ 1.120
𝑘=
𝑑4𝐺
0.1054 ∙ 11.75 ∙ 106
=
= 11.55 lbf⁄in
8𝐷3 𝑁𝑎
8 ∙ 1.1203 ∙ 11
10
𝐿0 =
(b) 𝐹|𝑆𝑠𝑦 = 45.2 lbf
2.63𝐷
𝛼
=
𝑘
+ 𝐿𝑆 =
45.2
+ 1.26 = 5.17 in
11.55
𝐴𝑛𝑠.
𝐴𝑛𝑠.
(c) 𝑘 = 11.55 lbf⁄in
(d) (𝐿0 )cr =
𝐹|𝑆𝑠𝑦
𝐴𝑛𝑠.
2.63∙1.120
0.5
= 5.89 in
Many designers provide (𝐿0 )cr ≥ 5 or more; therefore, plain ground ends are not often
used in machinery due to buckling uncertainty.
8. A helical compression spring is to be made of oil-tempered wire of 4-mm diameter with a
spring index of C = 10. The spring is to operate inside a hole, so buckling is not a problem
and the ends can be left plain. The free length of the spring should be 80 mm. A force of
50 N should deflect the spring 15 mm.
(a) Determine the spring rate.
(b) Determine the minimum hole diameter for the spring to operate in.
(c) Determine the total number of coils needed.
(d) Determine the solid length.
(e) Determine a static factor of safety based on the yielding of the spring if it is compressed
to its solid length.
Solution:
Given: Oil-tempered wire, d = 4 mm, C = 10, plain ends, L0 = 80 mm, and at F = 50 N, y = 15
mm.
(a) k = F/y = 50/15 = 3.333 N/mm
Ans.
(b) D = Cd = 10(4) = 40 mm, OD = D + d = 40 + 4 = 44 mm
Ans.
(c) From Table 10-5, G = 77.2 GPa
𝑁𝑎 =
𝑑4 𝐺
44 ∙ 77.2 ∙ 103
=
= 11.6 coils
8𝑘𝐷3 8 ∙ 3.333 ∙ 403
Table 10-1: Nt = Na = 11.6 coils
Ans.
(d) Table 10-1: Ls = d (Nt + 1) = 4(11.6 + 1) = 50.4 mm
Ans.
(e) Table 10-4: m = 0.187, A = 1855 MPa·mmm
𝑆𝑢𝑡 =
𝐴
1855
=
= 1431 MPa
𝑑𝑚 40.187
Table 10-6: Ssy = 0.50Sut = 0.50(1431) = 715.5 MPa
ys = L0 - Ls = 80 - 50.4 = 29.6 mm
Fs = kys = 3.333(29.6) = 98.66 N
11
4𝐶 + 2 4 ∙ 10 + 2
=
= 1.135
4𝐶 − 3 4 ∙ 10 − 3
8𝐹𝑠 𝐷
8 ∙ 98.66 ∙ 40
𝜏𝑠 = 𝐾𝐵
=
1.135
= 178.2 MPa
𝜋𝑑3
𝜋 ∙ 43
𝑆𝑠𝑦 715.5
𝑛𝑠 =
=
= 4.02 𝐴𝑛𝑠.
𝜏𝑠
178.2
𝐾𝐵 =
9. Consider the steel spring in the illustration.
(a) Find the pitch, solid height, and number of active turns.
(b) Find the spring rate. Assume the material is A227 HD steel.
(c) Find the force Fs required to close the spring solid.
(d) Find the shear stress in the spring due to the force Fs.
Solution:
Given: A227 HD steel.
From the figure: L0 = 4.75 in, OD = 2 in, and d = 0.135 in. Thus, D = OD - d = 2 - 0.135 =
1.865 in
(a) By counting, Nt = 12.5 coils. Since the ends are squared along 1/4 turn on each end,
Na = Nt - 2∙0.25 = 12.5 – 0.5 = 12 turns
p = L0 / Na = 4.75 / 12 = 0.396 in
Ans.
Ans.
The solid stack is 13 wire diameters
Ls = 13d = 13(0.135) = 1.755 in
Ans.
(b) From Table 10-5, G = 11.4 Mpsi
𝑑4𝐺
0.1354 ∙ 11.4 ∙ 106
𝑘=
=
= 6.08 lbf⁄in
8𝐷3 𝑁𝑎
8 ∙ 1.8653 ∙ 12
(c) 𝐹𝑠 = 𝑘(𝐿0 − 𝐿𝑠 ) = 6.08(4.75 − 1.755) ∙ 10−3 = 18.2 lbf
𝐴𝑛𝑠.
𝐴𝑛𝑠.
(d) C = D/d = 1.865/0.135 = 13.81
4𝐶 + 2 4 ∙ 13.81 + 2
=
= 1.096
4𝐶 − 3 4 ∙ 13.81 − 3
8𝐹𝑠 𝐷
8 ∙ 18.2 ∙ 1.865
𝜏𝑠 = 𝐾𝐵
=
1.096
= 38.5 ∙ 103 psi = 38.5 kpsi
𝜋𝑑 3
𝜋 ∙ 0.1353
𝐾𝐵 =
12
𝐴𝑛𝑠.
10. The extension spring shown in the figure has full-twisted loop ends. The material is AISI
1065 OQ&T wire. The spring has 84 coils and is close-wound with a preload of 16 lbf.
(a) Find the closed length of the spring.
(b) Find the torsional stress in the spring corresponding to the preload.
(c) Estimate the spring rate.
(d) What load would cause permanent deformation?
(e) What is the spring deflection corresponding to the load found in part (d)?
Solution:
Given: Nb = 84 coils, Fi = 16 lbf, OQ&T steel, OD = 1.5 in, d = 0.162 in. D = OD - d = 1.5 0.162 = 1.338 in
(a) Eq. (10-39):
L0 = 2(D - d) + (Nb + 1)d = 2(1.338 - 0.162) + (84 + 1)(0.162) = 16.12 in
or L0 = 2d + L0 = 2(0.162) + 16.12 = 16.45 in overall
(b) C = D/d = 1.338/0.162 = 8.26
4𝐶 + 2 4 ∙ 8.26 + 2
=
= 1.166
4𝐶 − 3 4 ∙ 8.26 − 3
8𝐹𝑖 𝐷
8 ∙ 16 ∙ 1.338
𝜏𝑖 = 𝐾𝐵
= 1.166
= 14950 psi
3
𝜋𝑑
𝜋 ∙ 0.1623
𝐾𝐵 =
𝐴𝑛𝑠.
(c) From Table 10-5 use: G = 11.4(106) psi and E = 28.5(106) psi
Na = Nb + G/E = 84 + 11.4/28.5 = 84.4 turns
𝑘=
𝑑4𝐺
0.1624 ∙ 11.4 ∙ 106
=
= 4.855 lbf⁄in
8𝐷3 𝑁𝑎
8 ∙ 1.3383 ∙ 84.4
𝐴𝑛𝑠.
(d) Table 10-4: A = 147 psi·inm, m = 0.187
𝑆𝑢𝑡 =
𝐴
147
=
= 207.1 kpsi
𝑑 𝑚 0.1620.187
Sy = 0.75Sut = 0.75(207.1) = 155.3 kpsi
Ssy = 0.50Sut = 0.50(207.1) = 103.5 kpsi
Body
𝐹=
𝜋𝑑3 𝑆𝑠𝑦 𝜋 ∙ 0.1623 ∙ 103.5 ∙ 103
=
= 110.8 lbf
8𝐾𝐵 𝐷
8 ∙ 1.166 ∙ 1.338
13
Ans.
Torsional stress on hook point B
2𝑟2 2(0.25 + 0.162⁄2)
=
= 4.086
𝑑
0.162
4𝐶2 − 1 4 ∙ 4.086 − 1
(𝐾)𝐵 =
=
= 1.243
4𝐶2 − 4 4 ∙ 4.086 − 4
𝐶2 =
𝜋𝑑3 𝑆𝑠𝑦
𝜋 ∙ 0.1623 ∙ 103.5 ∙ 103
𝐹=
=
= 103.9 lbf
8(𝐾)𝐵 𝐷
8 ∙ 1.243 ∙ 1.338
Normal stress on hook point A
𝐶1 =
(𝐾)𝐴 =
2𝑟1 1.338
=
= 8.26
𝑑
0.162
4𝐶12 − 𝐶1 − 1 4 ∙ 8.262 − 8.26 − 1
=
= 1.099
4𝐶1 (𝐶1 − 1)
4 ∙ 8.26(8.26 − 1)
𝑆𝑦𝑡 = 𝜎 = 𝐹 [
𝐹=
16(𝐾)𝐴 𝐷
4
+
]
𝜋𝑑 3
𝜋𝑑2
𝑆𝑦𝑡
155.3 ∙ 103
=
= 85.8 lbf
16 ∙ 1.099 ∙ 1.338
4
16(𝐾)𝐴 𝐷
4
+
+ 2
𝜋 ∙ 0.1623
𝜋 ∙ 0.1622
𝜋𝑑3
𝜋𝑑
𝐹 = 𝑚𝑖𝑛(110.8, 103.9, 85.8) = 85.8 lbf
𝐴𝑛𝑠.
(e) Eq. (10-48):
𝑦=
𝐹 − 𝐹𝑖 85.8 − 16
=
= 14.4 in
𝑘
4.855
𝐴𝑛𝑠.
Topic 7 (Bearing):
11. A certain application requires a ball bearing with the inner ring rotating, with a design life
of 30 000 h at a speed of 300 rev/min. The radial load is 1.898 kN and an application factor
of 1.2 is appropriate. The reliability goal is 0.90. Find the multiple of rating life required,
xD, and the catalog rating C10 with which to enter a bearing table. Choose a 02-series deepgroove ball bearing from Table 11–2, and estimate the reliability in use.
Solution:
For the deep-groove 02-series ball bearing with R = 0.90, the design life xD, in multiples of
rating life, is
𝑥𝐷 =
𝐿𝐷 60ℒ𝐷 𝑛𝐷 60 ∙ 30000 ∙ 300
=
=
= 540
𝐿𝑅
𝐿10
106
The design radial load FD is FD = 1.2(1.898) = 2.278 kN
From Eq. (11-6),
14
𝐴𝑛𝑠.
1⁄𝑎
𝑥𝐷
)
𝐶10 = 𝑎𝑓 𝐹𝐷 (
𝑥0 + (𝜃 − 𝑥0 )(ln 1⁄𝑅𝐷 )1⁄𝑏
⁄
1 3
540
)
= 2.278 (
0.02 + (4.459 − 0.02)(ln 1⁄0.9)1⁄1.483
= 18.59 kN
𝐴𝑛𝑠.
Table 11-2: Choose a 02-30 mm with C10 = 19.5 kN.
Ans.
Eq. (11-18):
𝑅 = exp −
𝑎𝑓 𝐹𝐷 𝑎
𝑥𝐷 ( 𝐶 ) − 𝑥0
10
𝜃 − 𝑥0
{ [
= 0.919
𝑏
1.483
2.278 3
540 (
) − 0.02
19.5
= exp − [
]
4.459 − 0.02
] }
{
}
𝐴𝑛𝑠.
12. A straight (cylindrical) roller bearing is subjected to a radial load of 20 kN. The life is to
be 8000 h at a speed of 950 rev/min and exhibit a reliability of 0.95. What basic load rating
should be used in selecting the bearing from a catalog of manufacturer 2?
Solution:
For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.95 and FR
= 20 kN.
𝑥𝐷 =
𝐿𝐷 60ℒ𝐷 𝑛𝐷 60 ∙ 8000 ∙ 950
=
=
= 456
𝐿𝑅
𝐿10
106
1⁄𝑎
𝑥𝐷
)
𝐶10 = 𝐹𝐷 (
𝑥0 + (𝜃 − 𝑥0 )(ln 1⁄𝑅𝐷 )1⁄𝑏
⁄
= 20 (
3 10
456
)
0.02 + (4.459 − 0.02)(ln 1⁄0.95)1⁄1.483
= 145 kN
𝐴𝑛𝑠.
13. A full journal bearing has a shaft journal with a diameter of 1.25 in and a unilateral tolerance
of −0.0006 in. The bushing bore has a diameter of 1.252 in with a unilateral tolerance of
0.0014 in. The bushing bore is 2 in in length. The bearing load is 620 lbf and the journal
rotates at 1120 rev/min. Analyze the minimum clearance assembly and find the minimum
film thickness, the coefficient of friction, and the total oil flow if the average viscosity is
8.5 μreyn.
15
Solution:
Given: dmax = 1.25 in, bmin = 1.252 in, l = 2 in, W = 620 lbf, µ = 8.5 µreyn, and N = 1120
rev/min.
cmin = (bmin - dmax)/2 = (1.252 – 1.25)/2 = 0.001 in
r = d / 2 = 1.25 / 2 = 0.625 in
r / c = 0.625 / 0.001 = 625
N = 1120 / 60 = 18.67 rev/s
𝑃=
𝑊
620
=
= 248 psi
𝑙𝑑 1.25 ∙ 2
𝑟 2 𝜇𝑁
8.5 ∙ 10−6 ∙ 18.67
2
𝑆=( )
= 625
= 0.250
𝑐 𝑃
248
l / d = 2 / 1.25 = 1.6
From Eq. (12-16), and Figs. 12-16, 12-18, and 12-19
h0/c
fr/c
Q/rcNl
l/d
1.6
1.6
1.6
y∞
0.9
4.5
3
h0 = 0.69c = 0.69(0.001) =0.000 69 in
f = 4.92/(r/c) = 4.92/625 = 0.007 87
y1
0.58
5.3
3.98
y1/2
0.36
6.5
4.97
y1/4
0.185
8
5.6
yl/d
0.69
4.92
3.59
Ans.
Ans.
Q = 3.59rcNl = 3.59(0.625) 0.001(18.67) 2 = 0.0838 in3/s
Ans.
14. A full journal bearing is 28 mm long. The shaft journal has a diameter of 56 mm with a
unilateral tolerance of −0.012 mm. The bushing bore has a diameter of 56.05 mm with a
unilateral tolerance of 0.012 mm. The load is 2.4 kN and the journal speed is 900 rev/min.
For the minimum clearance assembly find the minimum oil-film thickness, the power loss,
and the side flow if the operating temperature is 65°C and SAE 40 lubricating oil is used.
Solution:
Given: dmax = 56.00 mm, bmin = 56.05 mm, l = 28 mm, W = 2.4 kN, N = 900 rev/min, and SAE
40 at 65C.
cmin = (bmin - dmax)/2 = (56.05 – 56)/2 = 0.025 mm
r = d / 2 = 56 / 2 = 28 mm
r / c = 28 / 0.025 = 1120
l / d = 28 / 56 = 0.5
N = 900 / 60 = 15 rev/s
16
𝑃=
𝑊
2400
=
= 1.53 MPa
𝑙𝑑 28 ∙ 56
Fig. 12-13: SAE 40 at 65C, μ = 30 mPa·s
𝑟 2 𝜇𝑁
30 ∙ 10−3 ∙ 15
2
𝑆=( )
= 1120
= 0.369
𝑐 𝑃
1.53 ∙ 106
From Figures 12-16, 12-18, 12-19 and 12-20:
h0 / c = 0.44, fr / c = 8.5, Qs / Q = 0.71, Q / (rcNl) = 4.85
h0 = 0.44c = 0.44(0.025) = 0.011 mm
Ans.
f = 8.5 / (r/c) = 8.5 / 1120 = 0.00759
T = fWr = 0.00759(2.4)(28) = 0.51 N·m
H = 2πTN = 2π(0.51)(15) = 48.1 W
Ans.
Q = 4.85 (rcNl) = 4.85(28)(0.025)(15)(28) = 1426 mm3/s
Qs = 0.71Q = 0.71(1426) = 1012 mm3/s
Ans.
Topic 8 (Gear):
15. The tooth numbers for the automotive differential shown in the figure are N2 = 16, N3 = 48,
N4 = 14, N5 = N6 = 20. The drive shaft turns at 900 rev/min.
(a) What are the wheel speeds if the car is traveling in a straight line on a good road surface?
(b) Suppose the right wheel is jacked up and the left wheel resting on a good road surface.
What is the speed of the right wheel?
(c) Suppose, with a rear-wheel drive vehicle, the auto is parked with the right wheel resting
on a wet icy surface. Does the answer to part (b) give you any hint as to what would happen
if you started the car and attempted to drive on?
17
Solution:
(a) The planet gears act as keys and the wheel speeds are the same as that of the ring gear.
Thus,
𝑛𝐴 = 𝑛3 = 𝑛2
𝑁2
16
= 900 ∙
= 300 rev⁄min
𝑁3
48
𝐴𝑛𝑠.
(b) nF = n5 = 0, nL = n6
𝑁5 𝑁4
20 14
𝑒 = − ( ) ( ) = − ( ) ( ) = −1
𝑁4 𝑁6
14 20
𝑒=
n6 = 600 rev/min
𝑛𝐿 − 𝑛𝐴 𝑛6 − 300
=
= −1
𝑛𝐹 − 𝑛𝐴
0 − 300
Ans.
(c) The wheel spins freely on icy surfaces, leaving no traction for the other wheel. The car
is stalled.
Ans.
16. The tooth numbers for the gear train illustrated are N2 = 20, N3 = 16, N4 = 30, N6 = 36, and
N7 = 46. Gear 7 is fixed. If shaft a is turned through 10 revolutions, how many turns will
shaft b make?
Solution:
Let nF = n2, then nL = n7 = 0.
𝑒=−
𝑁2 𝑁3 𝑁6
20 16 36
∙
∙
=−
∙
∙
= −0.5217
𝑁3 𝑁4 𝑁7
16 30 46
𝑒=
𝑛𝐿 − 𝑛5
0 − 𝑛5
=
= −0.5217
𝑛𝐹 − 𝑛5 10 − 𝑛5
n5 = 3.428 turns, nb = n5 in same direction
18
17. The figure shows a pair of shaft-mounted spur gears having a diametral pitch of 5 teeth/in
with an 18-tooth 20° pinion driving a 45-tooth gear. The horsepower input is 32 maximum
at 1800 rev/min. Find the direction and magnitude of the maximum forces acting on
bearings A, B, C, and D.
Solution:
Given: P = 5 teeth/in, N2 = 18T, N3 = 45T, ϕn = 20, H = 32 hp, n2 = 1800 rev/min
Gear 2:
Tin = 63025·H/n2 = 63025(32)/1800 = 1120 lbf·in
dP = N2/P = 18/5 = 3.600 in
dG = N3/P = 45/5 = 9.000 in
𝑡
𝑊32
=
𝑇in
1120
=
= 622 lbf
𝑑𝑃 ⁄2 3.6⁄2
𝑡
𝑟
𝑊32
= 𝑊32
tan 20° = 622 tan 20° = 226 lbf
𝑡
𝑡
𝐹𝑎2
= 𝑊32
= 622 lbf
𝑟
𝑟
𝐹𝑎2
= 𝑊32
= 226 lbf
𝑡 )2
𝑟 )2
𝐹𝑎2 = √(𝐹𝑎2
+ (𝐹𝑎2
= √6222 + 2262 = 662 lbf
Each bearing on shaft a has the same radial load of RA = RB = 662/2 = 331 lbf.
19
Gear 3:
𝑡
𝑡
𝑊23
= 𝑊32
= 622 lbf
𝑟
𝑟
𝑊23
= 𝑊32
= 226 lbf
Fb3 = Fa2 = 662 lbf, RC = RD = 662 / 2 = 331 lbf
Each bearing on shaft b has the same radial load which is equal to the radial load of bearings A
and B. Thus, all four bearings have the same radial load of 331 lbf.
Ans.
18. A gear train is composed of four helical gears with the three shaft axes in a single plane, as
shown in the figure. The gears have a normal pressure angle of 20° and a 30° helix angle.
Shaft b is an idler and the transmitted load acting on gear 3 is 500 lbf. The gears on shaft b
both have a normal diametral pitch of 7 teeth/in and have 54 and 14 teeth, respectively.
Find the forces exerted by gears 3 and 4 on shaft b.
Solution:
20
Given: Pn = 7 teeth/in, ϕ = 20°, ψ = 30°,
Gear 3:
Pt = Pn cos ψ = 7 cos 30° = 6.062 teeth/in
tan ϕt = tan ϕ / cos ψ = tan 20°/cos 30° = 0.4203, ϕt = 22.8°
d3 = N3/Pt = 54/6.062 = 8.908 in
𝑊3𝑡 = 500 lbf
𝑊3𝑎 = 𝑊3𝑡 tan 30° = 288.7 lbf
𝑊3𝑟 = 𝑊3𝑡 tan 22.8° = 210.2 lbf
W3 = 210.2i + 288.7j-500k lbf
Ans.
Gear 4:
d4 = N4/Pt = 14/6.062 = 2.309 in
𝑊4𝑡 = 𝑊3𝑡
𝑑3
8.908
= 500 ∙
= 1929 lbf
𝑑4
2.309
𝑊4𝑎 = 𝑊4𝑡 tan 30° = 1114 lbf
𝑊4𝑟 = 𝑊4𝑡 tan 22.8° = 811 lbf
W4 = -811i+1114j-1929k lbf
Ans.
21