SECTION 1– DESIGN FOR SIMPLE STRESSES
TENSION, COMPRESSION, SHEAR
DESIGN PROBLEMS
1.
The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load
of 8000 lb. Let h = 1.5b . If the load is repeated but not reversed, determine the
dimensions of the section with the design based on (a) ultimate strength, (b) yield
strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005
in., what should be the dimensions of the cross section?
Problems 1 – 3.
Solution:
For AISI C1045 steel, as rolled (Table AT 7)
su = 96 ksi
s y = 59 ksi
E = 30× 106 psi
F
A
where
F = 8000 lb
A = bh
but
h = 1.5b
therefore A = 1.5b 2
sd =
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
s
F
sd = u =
N A
96,000 8000
=
6
1.5b 2
5
b = 0.577 in say in .
8
2
SECTION 1– DESIGN FOR SIMPLE STRESSES
h = 1.5b =
15
in
16
(b) Based on yield strength
N = factor of safety = 3 for repeated but not reversed load (Table 1.1)
s
F
sd = u =
N A
59,000 8000
=
3
1.5b 2
9
in .
b = 0.521 in say
16
27
h = 1.5b =
in
32
(c) Elongation = δ =
FL
AE
where,
δ = 0.005 in
F = 8000 lb
E = 30×106 psi
L = 15 in
A = 1.5b 2
then,
FL
δ=
AE
(8000)(15)
0.005 =
(1.5b 2 )(30 ×106 )
3
b = 0.730 in say in .
4
1
h = 1.5b = 1 in
8
2.
The same as 1 except that the material is malleable iron, ASTM A47-52, grade 35
018.
Solution:
For malleable iron, ASTM A47-52, grade 35 018(Table AT 6)
su = 55 ksi
s y = 36.5 ksi
E = 25×10 6 psi
3
SECTION 1– DESIGN FOR SIMPLE STRESSES
F
A
where
F = 8000 lb
A = bh
but
h = 1.5b
therefore A = 1.5b 2
sd =
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
s
F
sd = u =
N A
55,000 8000
=
6
1.5b 2
7
b = 0.763 in say in .
8
5
h = 1.5b = 1 in
16
(b) Based on yield strength
N = factor of safety = 3 for repeated but not reversed load (Table 1.1)
s
F
sd = u =
N A
36,500 8000
=
3
1.5b 2
11
b = 0.622 in say
in .
16
1
h = 1.5b = 1 in
32
(c) Elongation = δ =
FL
AE
where,
δ = 0.005 in
F = 8000 lb
E = 25×10 6 psi
L = 15 in
A = 1.5b 2
then,
4
SECTION 1– DESIGN FOR SIMPLE STRESSES
δ=
FL
AE
0.005 =
(8000)(15)
(1.5b )(25 ×10 )
2
b = 0.8 in say
h = 1.5b = 1
3.
6
7
in .
8
5
in
16
The same as 1 except that the material is gray iron, ASTM 30.
Solution:
For ASTM 30 (Table AT 6)
su = 30 ksi , no s y
E = 14.5 ×106 psi
Note: since there is no s y for brittle materials. Solve only for (a) and (c)
F
A
where
F = 8000 lb
A = bh
but
h = 1.5b
therefore A = 1.5b 2
sd =
(a) Based on ultimate strength
N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1)
s
F
sd = u =
N A
30,000 8000
=
7. 5
1.5b 2
3
b = 1.1547 in say 1 in .
16
25
h = 1.5b = 1 in
32
FL
(c) Elongation = δ =
AE
where,
δ = 0.005 in
F = 8000 lb
E = 14.5 ×106 psi
5
SECTION 1– DESIGN FOR SIMPLE STRESSES
L = 15 in
A = 1.5b 2
then,
FL
δ=
AE
0.005 =
(8000)(15)
(1.5b )(14.5 ×10 )
2
6
b = 1.050 in say 1
h = 1.5b = 1
4.
1
in .
16
19
in
32
A piston rod, made of AISI 3140 steel, OQT 1000 F (Fig. AF 2), is subjected to a
repeated, reversed load. The rod is for a 20-in. air compressor, where the
maximum pressure is 125 psig. Compute the diameter of the rod using a design
factor based on (a) ultimate strength, (b) yield strength.
Solution:
From Fig. AF 2 for AISI 3140, OQT 1000 F
su = 152.5 ksi
s y = 132.5 ksi
F = force =
π
(20)2 (125) = 39,270 lb = 39.27 kips
4
From Table 1.1, page 20
Nu = 8
Ny = 4
(a) Based on ultimate strength
N F
A= u
su
π 2 (8)(39.27 )
d =
4
152.5
5
d = 1.62 in say 1 in
8
(b) Based on yield strength
NyF
A=
sy
π
4
d2 =
(4)(39.27 )
132.5
6
SECTION 1– DESIGN FOR SIMPLE STRESSES
1
d = 1.23 in say 1 in
4
5.
A hollow, short compression member, of normalized cast steel (ASTM A27-58,
65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the
ultimate strength. Determine the outside and inside diameters if Do = 2 Di .
Solution:
su = 65 ksi
Nu = 8
F = 1500 kips
A=
π
(D
4
2
o
)
− Di2 =
π
(4D
4
2
i
)
− Di2 =
3πDi2
4
3πDi2 N u F (8)(1500 )
A=
=
=
4
su
65
7
Di = 8.85 in say 8 in
8
3
 7
Do = 2 Di = 2 8  = 17 in
4
 8
6.
A short compression member with Do = 2 Di is to support a dead load of 25 tons.
The material is to be 4130 steel, WQT 1100 F. Calculate the outside and inside
diameters on the basis of (a) yield strength, (b) ultimate strength.
Solution:
From Table AT 7 for 4130, WQT 1100 F
su = 127 ksi
s y = 114 ksi
From Table 1.1 page 20, for dead load
N u = 3 ~ 4 , say 4
N y = 1.5 ~ 2 , say 2
Area, A =
π
(D
4
2
o
)
− Di2 =
π
(4D
4
2
i
)
− Di2 =
3πDi2
4
F = 25 tons = 50 kips
(a) Based on yield strength
3πDi2 N y F (2 )(50)
A=
=
=
4
sy
114
7
SECTION 1– DESIGN FOR SIMPLE STRESSES
5
in
8
5 1
Do = 2 Di = 2  = 1 in
4
8
(b) Based on ultimate strength
3πDi2 N u F (4 )(50)
A=
=
=
4
su
127
7
Di = 0.82 in say in
8
3
7
Do = 2 Di = 2  = 1 in
4
8
Di = 0.61 in say
7.
A round, steel tension member, 55 in. long, is subjected to a maximum load of
7000 lb. (a) What should be its diameter if the total elongation is not to exceed
0.030 in? (b) Choose a steel that would be suitable on the basis of yield strength if
the load is gradually applied and repeated (not reversed).
Solution:
(a) δ =
FL
FL
or A =
AE
δE
where,
F = 7000 lb
L = 55 in
δ = 0.030 in
E = 30× 10 6 psi
A=
π
4
d2 =
(7000)(55)
(0.030)(30 ×106 )
3
in
4
(b) For gradually applied and repeated (not reversed) load
Ny = 3
d = 0.74 in say
sy =
N yF
A
=
(3)(7000) = 47,534 psi
π
(0.75)2
4
s y ≈ 48 ksi
say C1015 normalized condition ( s y = 48 ksi )
8.
A centrifuge has a small bucket, weighing 0.332 lb. with contents, suspended on a
manganese bronze pin (B138-A, ½ hard) at the end of a horizontal arm. If the pin
is in double shear under the action of the centrifugal force, determine the diameter
8
SECTION 1– DESIGN FOR SIMPLE STRESSES
needed for 10,000 rpm of the arm. The center of gravity of the bucket is 12 in.
from the axis of rotation.
Solution:
From Table AT 3, for B138-A, ½ hard
sus = 48 ksi
W
F = ω 2r
g
where
W = 0.332 lb
g = 32.2 fps 2
2π n 2π (10,000 )
ω=
=
= 1047 rad sec
60
60
r = 12 in
W
0.332
F = ω 2r =
(1047)2 (1) = 11,300 lb = 11.3 kips
g
32.2
From Table 1.1, page 20
N = 3 ~ 4 , say 4
N F
A= u
su
 π  (4)(11.3)
2 d 2  =
for double shear
48
4 
25
d = 0.774 in say
in
32
CHECK PROBLEMS
9.
The link shown is made of AISIC1020 annealed steel, with b =
3
in and
4
1
h = 1 in . (a) What force will cause breakage? (b) For a design factor of 4 based
2
on the ultimate strength, what is the maximum allowable load? (c) If N = 2.5
based on the yield strength, what is the allowable load?
Problem 9.
9
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
For AISI C1020 annealed steel, from Table AT 7
su = 57 ksi
s y = 42 ksi
(a) F = su A
 3  1 
A = bh =  1  = 1.125 in 2
 4  2 
F = (57 )(1.125) = 64 kips
s A
(b) F = u
Nu
Nu = 4
 3  1 
A = bh =  1  = 1.125 in 2
 4  2 
(57 )(1.125) = 16 kips
F=
4
(c) F =
sy A
Ny
N y = 2. 5
 3  1 
A = bh =  1  = 1.125 in 2
 4  2 
(42 )(1.125) = 18.9 kips
F=
2
10.
A ¾-in.bolt, made of cold-finished B1113, has an effective stress area of 0.334 sq.
in. and an effective grip length of 5 in. The bolt is to be loaded by tightening until
the tensile stress is 80 % of the yield strength, as determined by measuring the
total elongation. What should be the total elongation?
Solution:
sL
δ=
E
from Table AT 7 for cold-finished B1113
s y = 72 ksi
then, s = 0.80 s y = 0.8(72 ) = 57.6 ksi
E = 30 ×106 psi = 30,000 ksi
sL (57.6)(5)
δ=
=
= 0.0096 in
E
30,000
10
SECTION 1– DESIGN FOR SIMPLE STRESSES
11.
A 4-lb. weight is attached by a 3/8-in. bolt to a rotating arm 14-in. from the center
of rotation. The axis of the bolts is normal to the plane in which the centrifugal
force acts and the bolt is in double shear. At what speed will the bolt shear in two
if it is made of AISI B1113, cold finish?
Solution:
From Table AT 7, sus = 62 ksi = 62,000 psi
2
 1 3
A = 2 (π )  = 0.2209 in 2
 4 8
W
F = ω 2 r = sus A
g
4
ω 2 (14) = (62,000)(0.2209)
32.2
ω = 88.74 rad sec
2π n
ω=
= 88.74
60
n = 847 rpm
12.
How many ¾-in. holes could be punched in one stroke in annealed steel plate of
AISI C1040, 3/16-in. thick, by a force of 60 tons?
Solution:
For AISI C1040, from Figure AF 1
su = 80 ksi
sus = 0.75su = 0.75(80) ksi = 60 ksi
 3  3 
A = π d t = π    = 0.4418 in2
 4  16 
F = 60 tons = 120 kips
n = number of holes
n=
13.
F
120
=
= 5 holes
Asus (0.4415)(60 )
What is the length of a bearing for a 4-in. shaft if the load on the bearing is 6400
lb. and the allowable bearing pressure is 200 psi of the projected area?
Solution:
pDL = W
where
p = 200 psi
D = 4 in
11
SECTION 1– DESIGN FOR SIMPLE STRESSES
W = 6400 lb
(200)(4)L = 6400
L = 8 in
BENDING STRESSES
DESIGN PROBLEMS
14.
A lever keyed to a shaft is L = 15 in long and has a rectangular cross section of
h = 3t . A 2000-lb load is gradually applied and reversed at the end as shown; the
material is AISI C1020, as rolled. Design for both ultimate and yield strengths. (a)
What should be the dimensions of a section at a = 13 in ? (b) at b = 4 in ? (c) What
should be the size where the load is applied?
Problem 14.
Solution:
For AISI C1020, as rolled, Table AT 7
su = 65 ksi
s y = 49 ksi
Design factors for gradually applied and reversed load
Nu = 8
Ny = 4
th 3
, moment of inertial
12
but h = 3t
h4
I=
36
I=
Moment Diagram (Load Upward)
12
SECTION 1– DESIGN FOR SIMPLE STRESSES
Based on ultimate strength
s
s= u
Nu
(a) s =
Mc Fac
=
I
I
h
2
F = 2000 lbs = 2 kips
(2)(13) h 
65
 2
s=
=
4
8
h 
 
 36 
h = 3.86 in
h 3.86
t= =
= 1.29 in
3
3
say
1
h = 4.5 in = 4 in
2
1
t = 1.5 in = 1 in
2
c=
(b) s =
Mc Fbc
=
I
I
h
2
F = 2000 lbs = 2 kips
(2)(4) h 
65
 2
s=
=
4
8
h 
 
 36 
h = 2.61 in
h 2.61
t= =
= 0.87 in
3
3
say
h = 3 in
t = 1 in
c=
(c)
13
SECTION 1– DESIGN FOR SIMPLE STRESSES
3 − h 4. 5 − 3
=
4
13 − 4
h = 2.33 in
1 − t 1.5 − 1
=
4
13 − 4
t = 0.78 in
say
5
h = 2.625 in or h = 2 in
8
15.
A simple beam 54 in. long with a load of 4 kips at the center is made of cast steel,
SAE 080. The cross section is rectangular (let h ≈ 3b ). (a) Determine the
dimensions for N = 3 based on the yield strength. (b) Compute the maximum
deflection for these dimensions. (c) What size may be used if the maximum
deflection is not to exceed 0.03 in.?
Solution:
For cast steel, SAE 080 (Table AT 6)
s y = 40 ksi
E = 30× 106 psi
14
SECTION 1– DESIGN FOR SIMPLE STRESSES
From Table AT 2
FL (4)(54 )
Max. M =
=
= 54 kips − in
4
4
bh 3
I=
12
but h = 3b
h4
I=
36
(a) s =
c=
sy
Ny
=
Mc
I
h
2
(54) h 
40
 2
=
3
 h4 
 
 36 
h = 4.18 in
h 4.18
b= =
= 1.39 in
3
3
h 4. 5
1
1
= 1.5 in = 1 in
say h = 4 in , b = =
2
3
3
2
FL3
(b) δ =
=
48 EI
(c) δ =
(4000)(54)3
= 0.0384 in
3


6 (1.5 )(4.5 )
48(30 ×10 )


12


FL3
 h4 
48E  
 36 
3
(
4000)(54 ) (36 )
0.03 =
48(30 ×106 )(h 4 )
h = 4.79 in
h 4.79
b= =
= 1.60 in
3
3
1
h 5.25
3
say h = 5.25 in = 5 in , b = =
= 1.75 in = 1 in
4
3
3
4
15
SECTION 1– DESIGN FOR SIMPLE STRESSES
16.
The same as 15, except that the beam is to have a circular cross section.
Solution:
s
Mc
(a) s = y =
Ny
I
I=
πd4
64
d
c=
2
d 
M 
32 M
2
s=  4 =
πd  πd3


 64 
40 32(54 )
=
3
πd3
d = 3.46 in
1
say d = 3 in
2
(b) δ =
I=
FL3
48 EI
πd4
64
3
64 FL3
64(4000 )(54)
δ=
=
= 0.0594 in
48 E (π d 4 ) 48(30 × 106 )(π )(3.5)4
(c) δ =
64 FL3
48 E (π d 4 )
64(4000)(54 )
0.03 =
48(30 ×106 )(π )d 4
d = 4.15 in
1
say d = 4 in
4
3
17.
A simple beam, 48 in. long, with a static load of 6000 lb. at the center, is made of
C1020 structural steel. (a) Basing your calculations on the ultimate strength,
determine the dimensions of the rectangular cross section for h = 2b . (b)
Determine the dimensions based on yield strength. (c) Determine the dimensions
using the principle of “limit design.”
16
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
From Table AT 7 and Table 1.1
su = 65 ksi
s y = 48 ksi
N u = 3 ~ 4 , say 4
N y = 1.5 ~ 2 , say 2
FL (6)(48)
=
= 72 in − kips
4
4
Mc
s=
I
h
c=
2
bh 3
I=
12
h
but b =
2
4
h
I=
24
h
M 
12M
2
s = 4  = 3
h
h
24
M=
(a) Based on ultimate strength
s
12 M
s= u = 3
Nu
h
65 12(72)
=
4
h3
h = 3.76 in
17
SECTION 1– DESIGN FOR SIMPLE STRESSES
b=
h 3.76
=
= 1.88 in
2
2
3
h 3.75
7
say h = 3.75 in = 3 in , b = =
= 1.875 in = 1 in
4
2
2
8
(b) Based on yield strength
s y 12 M
s=
= 3
Ny
h
48 12(72 )
=
2
h3
h = 3.30 in
h 3.30
b= =
= 1.65 in
2
2
h 3. 5
1
3
say h = 3.5 in = 3 in , b = =
= 1.75 in = 1 in
2
2
2
4
(c) Limit design (Eq. 1.6)
bh 2
4
h 2
 h
2
72 = (48)  
4
h = 2.29 in
h 2.29
b= =
= 1.145 in
2
2
1
h 2. 5
1
say h = 2.5 in = 2 in , b = =
= 1.25 in = 1 in
2
2
2
4
M = sy
18.
The bar shown is subjected to two vertical loads, F1 and F2 , of 3000 lb. each, that
are L = 10 in apart and 3 in. ( a , d ) from the ends of the bar. The design factor is 4
based on the ultimate strength; h = 3b . Determine the dimensions h and b if the
bar is made of (a) gray cast iron, SAE 111; (b) malleable cast iron, ASTM A4752, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and
moment diagrams approximately to scale.
18
SECTION 1– DESIGN FOR SIMPLE STRESSES
Problems18, 19.
Solution:
F1 = F2 = R1 = R2 = 3000 lb
Moment Diagram
M = R1a = (3000)(3) = 9000 lbs − in = 9 kips − in
N = factor of safety = 4 based on su
bh 3
12
h
c=
2
h 3
 h
h4
3

I=
=
12
36
I=
(a) For gray cast iron, SAE 111
su = 30 ksi , Table AT 6
 h
M 
s
Mc
18M
2
s= u =
= 4  = 3
N
I
h
h 
 
 36 
30 18(9 )
s=
= 3
4
h
h = 2.78 in
h 2.78
b= =
= 0.93 in
3
3
say h = 3.5 in , b = 1 in
(b) For malleable cast iron, ASTM A47-52, grade 35 018
19
SECTION 1– DESIGN FOR SIMPLE STRESSES
su = 55 ksi , Table AT 6
 h
M 
s
Mc
18M
2
s= u =
= 4  = 3
N
I
h
h 
 
 36 
55 18(9 )
s=
= 3
4
h
h = 2.28 in
h 2.28
b= =
= 0.76 in
3
3
1
3
say h = 2 in , b = in
4
4
(c) For AISI C1040, as rolled
su = 90 ksi , Fig. AF 1
 h
M 
s
Mc
18M
2
s= u =
= 4  = 3
N
I
h
h 
 
 36 
90 18(9 )
s=
= 3
4
h
h = 1.93 in
h 1.93
b= =
= 0.64 in
3
3
7
5
say h = 1 in , b = in
8
8
19.
The same as 18, except that F1 acts up ( F2 acts down).
Solution:
[∑ M
A
=0
]
R1 = R2 = 1875 lb
20
SECTION 1– DESIGN FOR SIMPLE STRESSES
Shear Diagram
Moment Diagram
M = maximum moment = 5625 lb-in = 5.625 kips-in
(a) For gray cast iron
su 18M
= 3
N
h
30 18(5.625)
=
4
h3
h = 2.38 in
h 2.38
b= =
= 0.79 in
3
3
1
3
say h = 2 in , b = in
4
4
(b) For malleable cast iron
s=
su 18M
= 3
N
h
55 18(5.625)
=
4
h3
h = 1.95 in
h 1.95
b= =
= 0.65 in
3
3
7
5
say h = 1 in , b = in
8
8
s=
21
SECTION 1– DESIGN FOR SIMPLE STRESSES
(c) For AISI C1040, as rolled
su 18M
= 3
N
h
90 18(5.625)
=
4
h3
h = 1.65 in
h 1.65
b= =
= 0.55 in
3
3
1
1
say h = 1 in , b = in
2
2
s=
20.
The bar shown, supported at A and B , is subjected to a static load F of 2500 lb.
at θ = 0 . Let d = 3 in , L = 10 in and h = 3b . Determine the dimensions of the
section if the bar is made of (a) gray iron, SAE 110; (b) malleable cast iron,
ASTM A47-52, grade 32 510; (c) AISI C1035 steel, as rolled. (d) For economic
reasons, the pins at A, B, and C are to be the same size. What should be their
diameter if the material is AISI C1035, as rolled, and the mounting is such that
each is in double shear? Use the basic dimensions from (c) as needed. (e) What
sectional dimensions would be used for the C1035 steel if the principle of “limit
design” governs in (c)?
Problems 20, 21.
Solution:
22
SECTION 1– DESIGN FOR SIMPLE STRESSES
[∑ M
[∑ M
A
B
=0
=0
]
3RB = 13(2500 )
]
RB = 10,833 lb
3RA = 10(2500)
RA = 8333 lb
Shear Diagram
Moment Diagram
M = (2500 )(10) = 25,000 lb − in = 25 kips − in
h = 3b
bh 3
I=
12
h4
I=
36
h
c=
2
 h
M 
Mc
18M
2
s=
= 4  = 3
I
h
h 
 
 36 
(a) For gray cast iron, SAE 110
su = 20 ksi , Table AT 6
N = 5 ~ 6 , say 6 for cast iron, dead load
s 18M
s= u = 3
N
h
20 18(25)
=
6
h3
23
SECTION 1– DESIGN FOR SIMPLE STRESSES
h = 5.13 in
h
b = = 1.71 in
3
1
3
say h = 5 in , b = 1 in
4
4
(b) For malleable cast iron, ASTM A47-32 grade 32510
su = 52 ksi , s y = 34 ksi
N = 3 ~ 4 , say 4 for ductile, dead load
s 18M
s= u = 3
N
h
52 18(25)
=
4
h3
h = 3.26 in
h
b = = 1.09 in
3
3
1
say h = 3 in , b = 1 in
4
4
(c) For AISI C1035, as rolled
su = 85 ksi , s y = 55 ksi
N = 4 , based on ultimate strength
s 18M
s= u = 3
N
h
85 18(25)
=
4
h3
h = 2.77 in
h
b = = 0.92 in
3
say h = 3 in , b = 1 in
(d) For AISI C1035, as rolled
s su = 64 ksi
N = 4 , RB = 10.833 kips
s
R
s s = su = B
N
A
π

 π
A = 2 D 2  = D 2
4
 2
64 10.833
=
ss =
π 2
4
D
2
D = 0.657 in
24
SECTION 1– DESIGN FOR SIMPLE STRESSES
11
in
16
(e) Limit Design
bh 2
M = sy
4
For AISI C1035 steel, s y = 55 ksi
say D =
b=
h
3
h 2
 h
3
M = 25 = (55)  
4
h = 1.76 in
h
b = = 0.59 in
3
7
5
say h = 1.875 in = 1 in , b = in
8
8
The same as 20, except that θ = 30o . Pin B takes all the horizontal thrust.
21.
Solution:
FV = F cos θ
[∑ M
[∑ M
A
B
=0
=0
]
3RB = 13FV
]
3RB = 13(2500 ) cos 30
RB = 9382 lb
3RA = 10 FV
3RA = 10(2500) cos 30
RA = 7217 lb
Shear Diagram
25
SECTION 1– DESIGN FOR SIMPLE STRESSES
Moment Diagram
M = (2165)(10 ) = 21,650 lb − in = 21.65 kips − in
18M
s= 3
h
(a) For gray cast iron, SAE 110
su = 20 ksi , Table AT 6
N = 5 ~ 6 , say 6 for cast iron, dead load
s 18M
s= u = 3
N
h
20 18(21.65)
=
6
h3
h = 4.89 in
h
b = = 1.63 in
3
1
3
say h = 5 in , b = 1 in
4
4
(b) For malleable cast iron, ASTM A47-32 grade 32510
su = 52 ksi , s y = 34 ksi
N = 3 ~ 4 , say 4 for ductile, dead load
s 18M
s= u = 3
N
h
(
52 18 21.65)
=
4
h3
h = 3.11 in
h
b = = 1.04 in
3
say h = 3 in , b = 1 in
(c) For AISI C1035, as rolled
su = 85 ksi , s y = 55 ksi
N = 4 , based on ultimate strength
26
SECTION 1– DESIGN FOR SIMPLE STRESSES
su 18M
= 3
N
h
85 18(21.65)
=
4
h3
h = 2.64 in
h
b = = 0.88 in
3
5
7
say h = 2 in , b = in
8
8
s=
(d) For AISI C1035, as rolled
s su = 64 ksi
N = 4 , RBV = 9382 lb
RBH = FH = F sin θ = 2500 sin 30 = 1250 lb
2
2
RB2 = RBV
+ RBH
= (9382) + (1250 )
2
2
RB = 9465 lb
s
R
s s = su = B
N
A
π
 π
A = 2 D 2  = D 2
4
 2
64 9.465
ss =
=
4 π D2
2
D = 0.614 in
5
say D = in
8
(e) Limit Design
bh 2
M = sy
4
For AISI C1035 steel, s y = 55 ksi
b=
h
3
h 2
 h
3
M = 21.65 = (55)  
4
h = 1.68 in
h
b = = 0.56 in
3
7
5
say h = 1.875 in = 1 in , b = in
8
8
27
SECTION 1– DESIGN FOR SIMPLE STRESSES
22.
A cast-iron beam, ASTM 50, as shown, is 30 in. long and supports two gradually
applied, repeated loads (in phase), one of 2000 lb. at e = 10 in from the free end,
and one of 1000 lb at the free end. (a) Determine the dimensions of the cross
section if b = c ≈ 3a . (b) The same as (a) except that the top of the tee is below.
Problem 22.
Solution:
For cast iron, ASTM 50
su = 50 ksi , suc = 164 ksi
For gradually applied, repeated load
N = 7 ~ 8 , say 8
M = F1d + F2 (d + e )
where:
F1 = 2000 lb
F2 = 1000 lb
d = 30 − 10 = 20 in
d + e = 30 in
M = (2000 )(20 ) + (1000)(30 ) = 70,000 lb − in = 70 kips − in
Mc
I
Solving for I , moment of inertia
s=
(3a )(a ) a  + (3a )(a ) 5a  = [(3a )(a ) + (3a )(a )]y
2
y=
 2 
3a
2
28
SECTION 1– DESIGN FOR SIMPLE STRESSES
I=
(3a )(a )3 + (3a )(a )(a 2 ) + (a )(3a )3 + (3a )(a )(a 2 ) = 17a 4
12
12
2
(a)
3a
2
5a
cc =
2
Based on tension
s
Mct
st = u =
N
I
(70) 3a 
50
 2
=
8
 17a 4 


 2 
a = 1.255 in
Based on compression
s
Mcc
sc = uc =
N
I
(70) 5a 
164
 2 
=
8
 17 a 4 


 2 
a = 1.001 in
Therefore a = 1.255 in
1
Or say a = 1 in
4
And b = c = 3a = 3(1.25) = 3.75 in
ct =
29
SECTION 1– DESIGN FOR SIMPLE STRESSES
3
Or b = c = 3 in
4
(b) If the top of the tee is below
5a
2
3a
cc =
2
17a 4
I=
2
M = 70 kips − in
ct =
Based on tension
s
Mct
st = u =
N
I
(70) 5a 
50
 2 
=
8
 17 a 4 


 2 
a = 1.488 in
Based on compression
s
Mcc
sc = uc =
N
I
(70) 3a 
164
 2
=
8
 17a 4 


 2 
a = 0.845 in
Therefore a = 1.488 in
1
Or say a = 1 in
2
1
And b = c = 3a = 4 in
2
CHECK PROBLEMS
30
SECTION 1– DESIGN FOR SIMPLE STRESSES
23.
An I-beam is made of structural steel, AISI C1020, as rolled. It has a depth of 3
in. and is subjected to two loads; F1 and F2 = 2F1 ; F1 is 5 in. from one end and
F2 is 5 in. from the other ends. The beam is 25 in. long; flange width is
b = 2.509 in ; I x = 2.9 in 4 . Determine (a) the approximate values of the load to
cause elastic failure, (b) the safe loads for a factor of safety of 3 based on the yield
strength, (c) the safe load allowing for flange buckling (i1.24), (f) the maximum
deflection caused by the safe loads.
Problems 23 – 25.
Solution:
[∑ M
[∑ F
V
A
=0
=0
]
]
5 F1 + 20(2 F1 ) = 25RB
RB = 1.8 F1
F1 + 2 F1 = RA + RB
RA = 3F1 − 1.8F1 = 1.2 F1
Shear Diagram
Moment Diagram
31
SECTION 1– DESIGN FOR SIMPLE STRESSES
M = 9F1 = maximum moment
For AISI C1020, as rolled
s y = 48 ksi
Mc
I
d 3
where c = = = 1.5 in
2 2
(9 F1 )(1.5)
s y = 48 =
2.9
F1 = 10.31 kips
F2 = 2 F1 = 20.62 kips
(a) s y =
sy
Mc
N
I
48 (9 F1 )(1.5)
s=
=
3
2.9
F1 = 3.44 kips
F2 = 2 F1 = 6.88 kips
(b) s =
(c)
=
L
25
=
= 9.96 < 15 (page 34)
b 2.509
sc = 20 ksi ( page 34, i1.24)
Mc
I
(9 F1 )(1.5)
20 =
2.9
F1 = 4.30 kips
F2 = 2 F1 = 8.60 kips
sc =
(d) For maximum deflection,
by method of superposition, Table AT 2
3
y max
Fb′  a( L + b′ )  2
=
 , a > b′
3EIL 
3
or
3
y max
Fa  b′(L + a )  2
=
 , b′ > a
3EIL 
3
32
SECTION 1– DESIGN FOR SIMPLE STRESSES
y max caused by F1
3
F a  b′(L + a1 )  2
y max1 = 1 1  1
 , b1′ > a1
3EIL 
3
where E = 30,000 ksi
a1 = 5 in
b1′ = 20 in
L = 25 in
I = 2.9 in 4
3
y max1
F1 (5)
 20(25 + 5)  2
=
 = 0.0022 F1
3(30,000 )(2.9 )(25) 
3
y max caused by F2
3
F b′  a (L + b2′ )  2
y max 2 = 2 2  2
 , a2 > b2′
3EIL 
3
where b2′ = 5 in
a2 = 20 in
3
y max 2
2 F1 (5)
 20(25 + 5)  2
=
 = 0.0043F1
3(30,000 )(2.9 )(25) 
3
Total deflection = δ
δ = ymax1 + ymax 2 = 0.022 F1 + 0.0043F1 = 0.0065 F1
Deflection caused by the safe loads in (a)
δ a = 0.0065(10.31) = 0.067 in
Deflection caused by the safe loads in (b)
δ b = 0.0065(3.44) = 0.022 in
Deflection caused by the safe loads in (c)
δ c = 0.0065(4.30 ) = 0.028 in
24.
The same as 23, except that the material is aluminum alloy, 2024-T4, heat treated.
Solution:
For aluminum alloy, 2024-T4, heat treated
s y = 47 ksi
(a) s y =
Mc
I
33
SECTION 1– DESIGN FOR SIMPLE STRESSES
s y = 47 =
(9 F1 )(1.5)
2. 9
F1 = 10.10 kips
F2 = 2 F1 = 20.20 kips
sy
Mc
N
I
47 (9 F1 )(1.5)
s=
=
3
2.9
F1 = 3.36 kips
F2 = 2 F1 = 6.72 kips
(b) s =
(c)
=
L
25
=
= 9.96 < 15 (page 34)
b 2.509
sc = 20 ksi ( page 34, i1.24)
Mc
I
(
9 F1 )(1.5)
20 =
2.9
F1 = 4.30 kips
F2 = 2 F1 = 8.60 kips
sc =
(d) Total deflection = δ
δ = ymax1 + ymax 2 = 0.022 F1 + 0.0043F1 = 0.0065 F1
Deflection caused by the safe loads in (a)
δ a = 0.0065(10.10 ) = 0.066 in
Deflection caused by the safe loads in (b)
δ b = 0.0065(3.36) = 0.022 in
Deflection caused by the safe loads in (c)
δ c = 0.0065(4.30 ) = 0.028 in
25.
A light I-beam is 80 in. long, simply supported, and carries a static load at the
midpoint. The cross section has a depth of d = 4 in , a flange width of b = 2.66 in ,
and I x = 6.0 in 4 (see figure). (a) What load will the beam support if it is made of
C1020, as-rolled steel, and flange buckling (i1.24) is considered? (b) Consider the
stress owing to the weight of the beam, which is 7.7 lb/ft, and decide whether or
not the safe load should be less.
34
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
(a) For C1020, as rolled, su = 65 ksi
Consider flange buckling
L
80
=
= 30
b 2.66
L
since 15 < < 40
b
22.5
22.5
sc =
=
= 15 ksi
2
2
(
30 )
L
1 +   1800 1 +
1800
b
Mc
s=
I
d 4
c = = = 2 in
2 2
From Table AT 2
FL F (80)
M=
=
= 20 F
4
4
Mc
s = sc =
I
(
20 F )(2)
15 =
6
F = 2.25 kips , safe load
(b) Considering stress owing to the weight of the beam
wL2
(Table AT 2)
8
where w = 7.7 lb ft
add’l M =
35
SECTION 1– DESIGN FOR SIMPLE STRESSES
wL2  7.7  (80 )
=
= 513 lb − in = 0.513 kips − in

8
 12  8
M = 20 F + 0.513 = total moment
Mc
s = sc =
I
(20 F + 0.513)(2 )
15 =
6
F = 2.224 kips
Therefore, the safe load should be less.
2
add’l M =
26.
What is the stress in a band-saw blade due to being bent around a 13 ¾-in. pulley?
The blade thickness is 0.0265 in. (Additional stresses arise from the initial tension
and forces of sawing.)
Solution:
t
= 0.0265 = 0.01325 in
2
r = 13.75 + 0.01325 = 13.76325 in
Using Eq. (1.4) page 11 (Text)
Ec
s=
r
where E = 30× 106 psi
c=
(30 ×10 )(0.01325) = 28,881 psi
s=
6
13.76325
27.
A cantilever beam of rectangular cross section is tapered so that the depth varies
uniformly from 4 in. at the fixed end to 1 in. at the free end. The width is 2 in. and
the length 30 in. What safe load, acting repeated with minor shock, may be
applied to the free end? The material is AISI C1020, as rolled.
Solution:
For AISI C1020, as rolled
su = 65 ksi (Table AT 7)
Designing based on ultimate strength,
N = 6 , for repeated, minor shock load
36
SECTION 1– DESIGN FOR SIMPLE STRESSES
su 65
=
= 10.8 ksi
N
6
Loading Diagram
s=
4 −1 h − 1
=
30
x
h = 0.10 x + 1
wh 3
I=
12
h
c=
2
M = Fx
(Fx ) h 
Mc
3Fx
 2  = 6 Fx = 3Fx =
=
2
2
3
I
2h
h
 wh 
(0.10 x + 1)2


 12 
Differentiating with respect to x then equate to zero to solve for x giving maximum
stress.
 (0.10 x + 1)2 (1) − 2( x )(0.10 x + 1)(0.10 ) 
ds
= 3F 
=0
dx
(0.10 x + 1)4


0.10 x + 1 − 2(0.10 x ) = 0
x = 10 in
h = 0.10(10 ) + 1 = 2 in
s
3Fx
s= u = 2
N
h
3F (10 )
10.8 =
(2)2
F = 1.44 kips
s=
TORSIONAL STRESSES
DESIGN PROBLEMS
37
SECTION 1– DESIGN FOR SIMPLE STRESSES
28.
A centrifugal pump is to be driven by a 15-hp electric motor at 1750 rpm. What
should be the diameter of the pump shaft if it is made of AISI C1045 as rolled?
Consider the load as gradually repeated.
Solution:
For C1045 as rolled,
s y = 59 ksi
sus = 72 ksi
Designing based on ultimate strength
s
s = us , N = 6 (Table 1.1)
N
72
s=
= 12 ksi
6
33,000hp 33,000(15)
Torque, T =
=
= 45 ft − lb = 540 in − lb = 0.540 in − kips
2π n
2π (1750)
For diameter,
16T
s=
π d3
16(0.540)
12 =
πd3
d = 0.612 in
5
say d = in
8
29.
A shaft in torsion only is to transmit 2500 hp at 570 rpm with medium shocks. Its
material is AISI 1137 steel, annealed. (a) What should be the diameter of a solid
shaft? (b) If the shaft is hollow, Do = 2 Di , what size is required? (c) What is the
weight per foot of length of each of these shafts? Which is the lighter? By what
percentage? (d) Which shaft is the more rigid? Compute the torsional deflection
of each for a length of 10 ft.
Solution:
33,000hp 33,000(2500 )
T=
=
= 23,036 ft − lb = 276 in − kips
2π n
2π (570 )
For AISI 1137, annealed
s y = 50 ksi (Table AT 8)
s ys = 0.6s y = 30 ksi
Designing based on yield strength
N = 3 for medium shock, one direction
38
SECTION 1– DESIGN FOR SIMPLE STRESSES
Design stress
s
30
= 10 ksi
s = ys =
N
3
(a) Let D = shaft diameter
Tc
J
π D4
J=
32
D
c=
2
16T
s=
π D3
16(276)
10 =
π D3
D = 5.20 in
1
say D = 5 in
4
s=
(b) J =
[
π (Do4 − Di4 ) π (2 Di )4 − Di4
=
32
Do 2 Di
c=
=
= Di
2
2
TDi
32T
s=
=
4
 15π Di  15π Di3


 32 
32(276 )
10 =
15π Di3
Di = 2.66 in
32
] = 15π D
4
i
32
Do = 2 Di = 5.32 in
say
5
Di = 2 in
8
1
Do = 5 in
4
(c) Density, ρ = 0.284 lb in 3 (Table AT 7)
39
SECTION 1– DESIGN FOR SIMPLE STRESSES
For solid shaft
w = weight per foot of length
π

2
w = 12 ρ  D 2  = 3πρD 2 = 3π (0.284)(5.25) = 73.8 lb ft
4

For hollow shaft
π 
2
2
w = 12 ρ   Do2 − Di2 = 3πρ Do2 − Di2 = 3π (0.284)(5.25) − (2.625) = 55.3 lb ft
4
Therefore hollow shaft is lighter
73.8 − 55.3
Percentage lightness =
(100% ) = 33.5%
55.3
(
)
(
[
)
]
(d) Torsional Deflection
TL
JG
where
L = 10 ft = 120 in
θ=
G = 11.5 ×103 ksi
For solid shaft, J =
θ=
π D4
32
(276)(120 )
 180 
o
= 0.039 rad = (0.039)
 = 2.2
π
π 
4
3


 (5.25) (11.5 × 10 )
 32 
For hollow shaft, J =
θ=
π (Do4 − Di4 )
32
(276 )(120)
 180 
o
= 0.041 rad = (0.041)
 = 2.4
π 
4
4
3
 π 
 [(5.25) − (2.625) ](11.5 × 10 )
 32 
Therefore, solid shaft is more rigid, 2.2o < 2.4o
30.
The same as 29, except that the material is AISI 4340, OQT 1200 F.
Solution:
33,000hp 33,000(2500 )
T=
=
= 23,036 ft − lb = 276 in − kips
2π n
2π (570 )
For AISI 4340, OQT 1200 F
s y = 130 ksi
s ys = 0.6s y = 0.6(130 ) = 78 ksi
Designing based on yield strength
40
SECTION 1– DESIGN FOR SIMPLE STRESSES
N = 3 for mild shock
Design stress
s
78
= 26 ksi
s = ys =
N
3
(a) Let D = shaft diameter
Tc
J
π D4
J=
32
D
c=
2
16T
s=
π D3
16(276)
26 =
π D3
D = 3.78 in
3
say D = 3 in
4
s=
(b) J =
[
π (Do4 − Di4 ) π (2 Di )4 − Di4
=
32
Do 2 Di
c=
=
= Di
2
2
TDi
32T
s=
=
4
 15π Di  15π Di3


 32 
32(276)
26 =
15π Di3
Di = 1.93 in
32
] = 15π D
4
i
32
Do = 2 Di = 3.86 in
say
Di = 2 in
Do = 4 in
(c) Density, ρ = 0.284 lb in 3 (Table AT 7)
41
SECTION 1– DESIGN FOR SIMPLE STRESSES
For solid shaft
w = weight per foot of length
π

2
w = 12 ρ  D 2  = 3πρD 2 = 3π (0.284)(3.75) = 37.6 lb ft
4

For hollow shaft
π 
2
2
w = 12 ρ   Do2 − Di2 = 3πρ Do2 − Di2 = 3π (0.284)(4) − (2) = 32.1 lb ft
4
Therefore hollow shaft is lighter
37.6 − 32.1
Percentage lightness =
(100% ) = 17.1%
32.1
(
)
(
[
)
]
(d) Torsional Deflection
TL
JG
where
L = 10 ft = 120 in
θ=
G = 11.5 ×103 ksi
For solid shaft, J =
π D4
32
(276)(120)
 180 
o
θ=
= 0.148 rad = (0.148)
 = 8.48
π
 
4
3
 π 
 (3.75) (11.5 × 10 )
 32 
For hollow shaft, J =
θ=
π (Do4 − Di4 )
32
(276)(120 )
 180 
o
= 0.122 rad = (0.122 )
 = 6.99
π
π  4
4
3


 [(4) − (2 ) ](11.5 × 10 )
 32 
Therefore, hollow shaft is more rigid, 6.99o < 8.48o .
31.
A steel shaft is transmitting 40 hp at 500 rpm with minor shock. (a) What should
be its diameter if the deflection is not to exceed 1o in 20 D ? (b) If deflection is
primary what kind of steel would be satisfactory?
Solution:
33,000hp 33,000(40 )
(a) T =
=
= 420 ft − lb = 5.04 in − kips
2π n
2π (500 )
G = 11.5 ×103 ksi
L = 20 D
42
SECTION 1– DESIGN FOR SIMPLE STRESSES
θ = 1o =
θ=
π
π
180
TL
JG
rad
(5.04)(20 D )
=
 π D4 

 11.5 ×103
32


D = 1.72 in
3
say D = 1 in
4
180
(
(b) s =
)
16T 16(5.04 )
=
= 4.8 ksi
π D 3 π (1.75)3
Based on yield strength
N =3
s ys = Ns = (3)(4.8) = 14.4 ksi
s ys
14.4
= 24 ksi
0.6 0.6
Use C1117 normalized steel s y = 35 ksi
sy =
32.
=
A square shaft of cold-finish AISI 1118 transmits a torsional moment of 1200 inlb. For medium shock, what should be its size?
Solution:
For AISI 1118 cold-finish
s y = 75 ksi
s ys = 0.6s y = 45 ksi
N = 3 for medium shock
s
T
s = ys =
N Z′
where, h = b
2b 2 h 2b 3
Z′ =
=
(Table AT 1)
9
9
T = 1200 in − lb = 1.2 in − kips
45 1.2(9)
s=
=
3
2b 3
b = h = 0.71 in
3
say b = h = in
4
43
SECTION 1– DESIGN FOR SIMPLE STRESSES
CHECK PROBLEMS
33.
A punch press is designed to exert a force sufficient to shear a 15/16-in. hole in a
½-in. steel plate, AISI C1020, as rolled. This force is exerted on the shaft at a
radius of ¾-in. (a) Compute the torsional stress in the 3.5-in. shaft (bending
neglected). (b) What will be the corresponding design factor if the shaft is made
of cold-rolled AISI 1035 steel (Table AT 10)? Considering the shock loading that
is characteristics of this machine, do you thick the design is safe enough?
Solution:
For AISI C1020, as rolled
sus = 49 ksi
F = sus (π Dt )
15
where D = in
16
1
t = in
2
 15  1 
F = 49(π )   = 72.2 kips
 16  2 
T = Fr
3
where r = in
4
3
T = (72.2 )  = 54.2 in − kips
4
16T
π d3
where d = 3.5 in
16(54.2)
s=
= 6.44 ksi
3
π (3.5)
(a) s =
(b) For AISI 1035 steel, s us = 64 ksi
for shock loading, traditional factor of safety, N = 10 ~ 15
Design factor , N =
34.
sus
64
=
= 9.94 , the design is safe ( N ≈ 10 )
s 6.44
The same as 33, except that the shaft diameter is 2 ¾ in.
Solution:
44
SECTION 1– DESIGN FOR SIMPLE STRESSES
d = 2.75 in
16T
π d3
16(54.2 )
s=
= 13.3 ksi
3
π (2.75)
(a) s =
(b) For AISI 1035 steel, s us = 64 ksi
for shock loading, traditional factor of safety, N = 10 ~ 15
Design factor , N =
sus
64
=
= 4.8 , the design is not safe ( N < 10 )
s 13.3
A hollow annealed Monel propeller shaft has an external diameter of 13 ½ in. and
an internal diameter of 6 ½ in.; it transmits 10,000 hp at 200 rpm. (a) Compute the
torsional stress in the shaft (stress from bending and propeller thrust are not
considered). (b) Compute the factor of safety. Does it look risky?
35.
Solution:
For Monel shaft,
s us = 98 ksi (Table AT 3)
N = 3 ~ 4 , for dead load, based on ultimate strength
Tc
J
π Do4 − Di4 π (13.5)4 − (6.5)4
J=
=
= 3086 in 4
32
32
Do 13.5
c=
=
= 6.75 in
2
2
33,000hp 33,000(10,000)
T=
=
= 262,606 ft − lb = 3152 in − kips
2π n
2π (200)
(3152)(6.75) = 6.9 ksi
s=
3086
(b) Factor of safety,
(a) s =
(
N=
)
[
]
sus 98
=
= 14.2 , not risky
s 6. 9
45
SECTION 1– DESIGN FOR SIMPLE STRESSES
STRESS ANALYSIS
DESIGN PROBLEMS
36.
A hook is attached to a plate as shown and supports a static load of 12,000 lb. The
material is to be AISI C1020, as rolled. (a) Set up strength equations for
dimensions d , D , h , and t . Assume that the bending in the plate is negligible.
(b) Determine the minimum permissible value of these dimensions. In estimating
the strength of the nut, let D1 = 1.2d . (c) Choose standard fractional dimensions
which you think would be satisfactory.
Problems 36 – 38.
Solution:
s = axial stress
s s = shear stress
(a)
s=
F
4F
=
2
1
πd2 πd
4
Equation (1) d =
4F
πs
46
SECTION 1– DESIGN FOR SIMPLE STRESSES
s=
F
1
π D 2 − D12
4
(
)
Equation (2) D =
ss =
4F
4F
4F
=
=
2
2
2
2
2
π D − D1
π D − 1.44d 2
π D − (1.2d )
(
)
[
] (
4F
+ 1.44d 2
πs
F
F
=
π D1h 1.2π dh
Equation (3) h =
ss =
=
F
1.2π ds s
F
π Dt
Equation (4) t =
F
π Dss
(b) Designing based on ultimate strength,
Table AT 7, AISI C1020, as rolled
su = 65 ksi
sus = 49 ksi
N = 3 ~ 4 say 4, design factor for static load
s
65
= 16 ksi
s= u =
N
4
s
49
s s = us =
= 12 ksi
N
4
F = 12,000 lb = 12 kips
From Equation (1)
4F
4(12 )
d=
=
= 0.98 in
πs
π (16 )
From Equation (2)
4F
4(12 )
2
D=
+ 1.44d 2 =
+ 1.44(0.98) = 1.53 in
πs
π (16)
From Equation (3)
F
12
h=
=
= 0.27 in
1.2π ds s 1.2π (0.98)(12 )
From Equation (4)
F
12
t=
=
= 0.21 in
π Dss π (1.53)(12 )
47
)
SECTION 1– DESIGN FOR SIMPLE STRESSES
(c) Standard fractional dimensions
d = 1 in
1
D = 1 in
2
1
h = in
4
1
t = in
4
37.
The same as 36, except that a shock load of 4000 lb. is repeatedly applied.
Solution:
(a) Same as 36.
(b) N = 10 ~ 15 for shock load, based on ultimate strength
say N = 15 , others the same.
s
65
s= u =
= 4 ksi
N 15
s
49
s s = us =
= 3 ksi
N 15
F = 4000 lb = 4 kips
From Equation (1)
4F
4(4 )
d=
=
= 1.13 in
πs
π (4)
From Equation (2)
4F
4(4)
2
D=
+ 1.44d 2 =
+ 1.44(1.13) = 1.76 in
πs
π (4 )
From Equation (3)
F
4
h=
=
= 0.31 in
1.2π ds s 1.2π (1.13)(3)
From Equation (4)
F
4
t=
=
= 0.24 in
π Dss π (1.76)(3)
48
SECTION 1– DESIGN FOR SIMPLE STRESSES
(c) Standard fractional dimensions
1
d = 1 in
8
3
D = 1 in
4
3
h = in
8
1
t = in
4
38.
The connection between the plate and hook, as shown, is to support a load F .
Determine the value of dimensions D , h , and t in terms of d if the connection
is to be as strong as the rod of diameter d . Assume that D1 = 1.2d , sus = 0.75su ,
and that bending in the plate is negligible.
Solution:
s=
F
1
πd2
4
1
F = π d 2s
4
1
s 
(1) F = π d 2  u 
4
N
49
SECTION 1– DESIGN FOR SIMPLE STRESSES
s=
F
F
=
1
1
π D 2 − D12
π D 2 − 1.44d 2
4
4
1
F = π (D 2 − 1.44d 2 )s
4
1
s 
(2) F = π D 2 − 1.44d 2  u 
4
N
F
F
ss =
=
π D1h 1.2π dh
F = 1.2π dhss
(
)
(
(
)
)
s 
 0.75su 
F = 1.2π dh us  = 1.2π dh

N
 N 
 5s 
(3) F = 0.9π dh u 
 N 
F
ss =
π Dt
F = π Dtss
s 
 0.75su 
F = π Dt  us  = π Dt 

N
 N 
s 
(4) F = 0.75π Dt  u 
N
Equate (2) and (1)
1
s  1
s 
F = π D 2 − 1.44d 2  u  = π d 2  u 
4
N 4
N
D 2 = 2.44d 2
D = 1.562d
Equate (3) and (1)
s  1
s 
F = 0.9π dh u  = π d 2  u 
N 4
N
d
h=
= 0.278d
4(0.9)
Equate (4) and (1)
s  1
s 
F = 0.75π Dt  u  = π d 2  u 
N 4
N
s  1
s 
F = 0.75π (1.562d )(t ) u  = π d 2  u 
N 4
N
d
t=
= 0.214d
4(0.75)(1.562 )
(
)
50
SECTION 1– DESIGN FOR SIMPLE STRESSES
39.
(a) For the connection shown, set up strength equations representing the various
methods by which it might fail. Neglect bending effects. (b) Design this
connection for a load of 2500 lb. Both plates and rivets are of AISI C1020, as
rolled. The load is repeated and reversed with mild shock. Make the connection
equally strong on the basis of yield strengths in tension, shear, and compression.
Problems 39, 40
Solution:
(a) s s =
F
1

5 π D 2 
4

4F
5π s s
Equation (1) D =
s=
F
t (b − 2 D )
Equation (2) b =
s=
F
+ 2D
ts
F
5 Dt
Equation (3) t =
F
5Ds
(b) For AISI C1020, as rolled
s y = 48 ksi (Table AT 7)
s ys = 0.6s y = 28 ksi
N = 4 for repeated and reversed load (mild shock) based on yield strength
48
s=
= 12 ksi
4
28
ss =
= 7 ksi
4
From Equation (1)
51
SECTION 1– DESIGN FOR SIMPLE STRESSES
D=
4F
5π s s
where
F = 2500 lb = 2.5 kips
D=
4F
4(2.5)
5
in
=
= 0.30 in say
5π s s
5π (7 )
16
From Equation (3)
F
2.5
5
t=
=
= 0.13 in say
in
5Ds
32
5
5 (12 )
 16 
From Equation (2)
F
2. 5
5
b = + 2D =
+ 2  = 1.96 in say 2 in
ts
 5 
 16 
 (12)
 32 
40.
The same as 39, except that the material is 2024-T4, aluminum alloy.
Solution:
(a) Same as 39.
(b) ) For 2024-T4, aluminum alloy
s y = 47 ksi (Table AT 3)
s ys = 0.55s y = 25 ksi
N = 4 for repeated and reversed load (mild shock) based on yield strength
47
= 12 ksi
s=
4
25
ss =
= 6 ksi
4
From Equation (1)
4F
D=
5π s s
where
F = 2500 lb = 2.5 kips
D=
4F
4(2.5)
3
=
= 0.33 in say in
5π s s
5π (6)
8
From Equation (3)
52
SECTION 1– DESIGN FOR SIMPLE STRESSES
t=
F
2.5
1
=
= 0.11 in say in
8
5Ds
3
5 (12)
8
From Equation (2)
F
2. 5
1
 3
b = + 2D =
+ 2  = 2.42 in say 2 in
ts
2
1
8
 (12)
8
41.
(a) For the connection shown, set up strength equations representing the various
methods by which it might fail. (b) Design this connection for a load of 8000 lb.
Use AISI C1015, as rolled, for the rivets, and AISI C1020, as rolled, for the
plates. Let the load be repeatedly applied with minor shock in one direction and
make the connection equally strong on the basis of ultimate strengths in tension,
shear, and compression.
Problem 41.
Solution:
(a)
F
sP =
t (b − D )
3
F
sP = 4
t (b − 2 D )
or
s sR =
F
1

4 πD 2 (2 )
4


53
Equation (1)
Equation (2)
SECTION 1– DESIGN FOR SIMPLE STRESSES
sR =
F
4 Dt
Equation (3)
(b) For AISI C1015, as rolled
suR = 61 ksi , susR = 0.75suR = 45 ksi
For AISI C1020, as rolled
suP = 65 ksi
N = 6 , based on ultimate strength
s
65
s P = uP =
= 10.8 ksi
N
6
s
61
s R = uR =
= 10.1 ksi
N
6
s
45
= 7.5 ksi
s sR = usR =
N
6
F = 8000 lb = 8 kips
Solving for D
F
s sR =
2π D 2
7
F
8
in
=
= 0.412 in say
16
2π s sR
2π (7.5)
Solving for t
F
sR =
4 Dt
F
8
1
t=
=
= 0.453 in say in
4 Ds R
2
7
4 (10.1)
 16 
Solving for b
F
Using s P =
t (b − D )
F
8
7
b=
+D=
+ = 1.92 in say 2 in
ts P
16
1
 (10.8)
2
3
F
4
Using s P =
t (b − 2 D )
D=
54
SECTION 1– DESIGN FOR SIMPLE STRESSES
3F
3(8)
7
+ 2D =
+ 2  = 1.99 in say 2 in
4ts P
16
1
4 (10.8)  
2
Therefore
b = 2 in
7
D = in
16
1
t = in
2
b=
42.
Give the strength equations for the connection shown, including that for the shear
of the plate by the cotter.
Problems 42 – 44.
Solution:
Axial Stresses
s=
F
1
π D12
4
s=
55
=
4F
π D12
F
(L − D2 )e
Equation (1)
Equation (2)
SECTION 1– DESIGN FOR SIMPLE STRESSES
s=
s=
s=
F
D2 e
Equation (3)
F
1
π a 2 − D22
4
(
F
1
π D22 − D2e
4
)
=
=
4F
Equation (4)
π a 2 − D22
(
4F
Equation (5)
π D − 4 D2e
2
2
Shear Stresses
ss =
F
2eb
ss =
F
2(L − D2 + e )t
56
)
Equation (6)
Equation (7)
SECTION 1– DESIGN FOR SIMPLE STRESSES
43.
ss =
F
π at
Equation (8)
ss =
F
π D1m
Equation (9)
ss =
F
2 D2 h
Equation (10)
A steel rod, as-rolled AISI C1035, is fastened to a 7/8-in., as-rolled C1020 plate
by means of a cotter that is made of as-rolled C1020, in the manner shown. (a)
Determine all dimensions of this joint if it is to withstand a reversed shock load
F = 10 kips , basing the design on yield strengths. (b) If all fits are free-running
fits, decide upon tolerances and allowances.
Solution: (See figure of Prob. 42)
7
t = in = 0.875 in , s sy = 0.6s y
8
For steel rod, AISI C1035, as rolled
s y1 = 55 ksi
s sy1 = 33 ksi
For plate and cotter, AISI C1020, as rolled
s y2 = 48 ksi
s sy2 = 28 ksi
N = 5 ~ 7 based on yield strength
say N = 7
From Equation (1) (Prob. 42)
sy
4F
s= 1 =
N π D12
55 4(10)
=
7 π D12
D1 = 1.27 in
1
say D1 = 1 in
4
57
SECTION 1– DESIGN FOR SIMPLE STRESSES
From Equation (9)
ssy
F
ss = 1 =
N π D1m
33
10
=
7
 1
π 1 m
 4
m = 0.54 in
9
say m = in
16
From Equation (3)
sy
F
s= 1 =
N D2e
55 10
s=
=
7 D2e
D2 e = 1.273
From Equation (5)
sy
4F
s= 1 =
2
N π D2 − 4 D2e
55
4(10 )
=
2
7 π D2 − 4(1.273)
D2 = 1.80 in
3
say D2 = 1 in
4
and D2 e = 1.273
 3
1 e = 1.273
 4
e = 0.73 in
3
say e = in
4
By further adjustment
5
Say D2 = 2 in , e = in
8
From Equation (8)
s sy
F
ss = 2 =
N π at
28
10
=
7 π a (0.875)
a = 0.91 in
say a = 1 in
58
SECTION 1– DESIGN FOR SIMPLE STRESSES
From Equation (4)
sy
4F
s= 2 =
N π a 2 − D22
48
4(10 )
=
7 π a 2 − 22
a = 2.42 in
1
say a = 2 in
2
1
use a = 2 in
2
From Equation (7)
ssy
F
ss = 2 =
N
2( L − D2 + e )t
28
10
=
5
7

2 L − 2 + (0.875)
8

L = 2.80 in
say L = 3 in
From Equation (6)
s sy
F
ss = 2 =
N
2eb
28
10
=
7
5
2 b
8
b = 2 in
From Equation (10)
s sy
F
ss = 2 =
N
2 D2 h
28
10
=
7 2(2 )h
5
h = 0.625 in = in
8
Summary of Dimensions
L = 3 in
5
h = in
8
b = 2 in
7
t = in
8
(
(
)
)
59
SECTION 1– DESIGN FOR SIMPLE STRESSES
9
in
16
1
a = 2 in
2
1
D1 = 1 in
4
D2 = 2 in
5
e = in
8
m=
(b) Tolerances and allowances, No fit, tolerance = ± 0.010 in
L = 3 ± 0.010 in
h = 0.625 ± 0.010 in
t = 0.875 ± 0.010 in
m = 0.5625 ± 0.010 in
a = 2.500 ± 0.010 in
D1 = 1.25 ± 0.010 in
For Free Running Fits (RC 7) Table 3.1
Female
Male
+ 0.0030
− 0.0040
b = 2.0
in
b = 2.0
in
− 0.0000
− 0.0058
allowance = 0.0040 in
+ 0.0030
− 0.0040
D2 = 2.0
in
D2 = 2.0
in
− 0.0000
− 0.0058
allowance = 0.0040 in
+ 0.0016
− 0.0020
e = 0.625
in
e = 0.625
in
− 0.0000
− 0.0030
allowance = 0.0020 in
44.
A 1-in. ( D1 ) steel rod (as-rolled AISI C1035) is to be anchored to a 1-in. steel
plate (as-rolled C1020) by means of a cotter (as rolled C1035) as shown. (a)
Determine all the dimensions for this connection so that all parts have the same
ultimate strength as the rod. The load F reverses direction. (b) Decide upon
tolerances and allowances for loose-running fits.
Solution: (Refer to Prob. 42)
(a) For AISI C1035, as rolled
su1 = 85 ksi
sus1 = 64 ksi
For AISI C1020, as rolled
60
SECTION 1– DESIGN FOR SIMPLE STRESSES
su2 = 65 ksi
sus2 = 48 ksi
Ultimate strength
Use Equation (1)
1

1  2
Fu = su1  π D12  = (85) π (1) = 66.8 kips
4

4 
Equation (9)
Fu = sus1 π D1m
66.8 = (64 )(π )(1)m
m = 0.33 in
3
say m = in
8
From Equation (3)
Fu = su1 D2e
66.8 = (85)D2e
D2 e = 0.7859
From Equation (5)
1

Fu = su1  π D22 − D2e 
4

1

66.8 = (85) π D22 − 0.7859 
4

D2 = 1.42 in
3
say D2 = 1 in
8
3
 
D2 e = 1 e = 0.7859
 8
e = 0.57 in
9
say e = in
16
From Equation (4)
1

Fu = su2  π a 2 − D22 
4

2
 1 
 3 
66.8 = (65) π  a 2 −1  
 4  
 8  
a = 1.79 in
3
say a = 1 in
4
From Equation (8)
(
)
61
SECTION 1– DESIGN FOR SIMPLE STRESSES
Fu = sus2 π at
66.8 = (48)(π )(a )(1)
a = 0.44 in
1
say a = in
2
3
use a = 1 in
4
From Equation (2)
Fu = su2 ( L − D2 )e
3  9 

66.8 = (65) L − 1  
8  16 

L = 3.20 in
1
say L = 3 in
4
From Equation (7)
Fu = 2 sus2 (L − D2 − e )t
3 9

66.8 = 2(48) L − 1 − (1)
8 16 

L = 1.51 in
1
say L = 1 in
2
1
use L = 3 in
4
From Equation (6)
Fu = 2 sus1 eb
 9
66.8 = 2(64 ) b
 16 
b = 0.93 in
say b = 1 in
From Equation (10)
Fu = 2 sus1 D2 h
 3
66.8 = 2(64 ) 1 h
 8
h = 0.38 in
3
say h = in
8
Dimensions
1
L = 3 in
4
62
SECTION 1– DESIGN FOR SIMPLE STRESSES
3
in
8
b = 1 in
t = 1 in
3
m = in
8
3
a = 1 in
4
D1 = 1 in
3
D2 = 1 in
8
9
e = in
16
h=
(b) Tolerances and allowances, No fit, tolerance = ± 0.010 in
L = 3.25 ± 0.010 in
h = 0.375 ± 0.010 in
t = 1.000 ± 0.010 in
m = 0.375 ± 0.010 in
a = 1.75 ± 0.010 in
D1 = 1.000 ± 0.010 in
For Loose Running Fits (RC 8) Table 3.1
Female
Male
+ 0.0035
− 0.0045
b = 1.0
in
b = 1.0
in
− 0.0000
− 0.0065
allowance = 0.0045 in
+ 0.0040
− 0.0050
D2 = 1.375
in
D2 = 1.375
in
− 0.0000
− 0.0075
allowance = 0.0050 in
+ 0.0028
− 0.0035
e = 0.5625
in
e = 0.5625
in
− 0.0000
− 0.0051
allowance = 0.0035 in
45.
Give all the simple strength equations for the connection shown. (b) Determine
the ratio of the dimensions a , b , c , d , m , and n to the dimension D so that the
connection will be equally strong in tension, shear, and compression. Base the
calculations on ultimate strengths and assume sus = 0.75su .
63
SECTION 1– DESIGN FOR SIMPLE STRESSES
Problems 45 – 47.
Solution:
(a) Neglecting bending
1

Equation (1): F = s π D 2 
4

 1

Equation (2): F = ss 2 π c 2 

 4
Equation (3): F = s (2bc )
Equation (4): F = s (ac )
Equation (5): F = s[2(d − c )b]
Equation (6): F = ss (4mb )
Equation (7): F = ss (2nb )
Equation (8): F = s (d − c )a
su
s
and s s = us
N
N
Therefore
s s = 0.75s
Equate (2) and (1)
 1
  1

F = ss 2 π c 2  = s π D 2 
  4

 4
(b) s =
1  1 
0.75s c 2  = s D 2 
2  4 
c = 0.8165 D
Equate (3) and (1)
1

F = s (2bc ) = s  π D 2 
4

1
2b(0.8165D ) = π D 2
4
b = 0.4810 D
64
SECTION 1– DESIGN FOR SIMPLE STRESSES
Equate (4) and (1)
1

F = sac = s π D 2 
4

1
a(0.8165 D ) = π D 2
4
a = 0.9619 D
Equate (5) and (1)
1

F = s[2(d − c )b] = s  π D 2 
4

1
2(d − 0.8165 D )(0.4810 ) = π D 2
4
d = 1.6329 D
Equate (6) and (1)
1

F = s s (4mb ) = s π D 2 
4

1
0.75(4m )(0.4810 D ) = π D 2
4
m = 0.5443D
Equate (7) and (1)
1

F = ss (2nb ) = s π D 2 
4

1
0.75(2n )(0.4810 D ) = π D 2
4
n = 1.0886 D
Equate (8) and (1)
1

F = s (d − c )a = s π D 2 
4

(1.6329 − D − 0.8165D )a = 1 π D 2
4
a = 0.9620 D
Summary
a = 0.9620 D
b = 0.4810 D
c = 0.8165 D
d = 1.6329 D
m = 0.5443D
n = 1.0886 D
65
SECTION 1– DESIGN FOR SIMPLE STRESSES
46.
The same as 45, except that the calculations are to be based on yield strengths. Let
s sy = 0.6s y .
Solution: (Refer to Prob. 45)
(a) Neglecting bending
1

Equation (1): F = s π D 2 
4

 1

Equation (2): F = ss 2 π c 2 

 4
Equation (3): F = s (2bc )
Equation (4): F = s (ac )
Equation (5): F = s[2(d − c )b]
Equation (6): F = ss (4mb )
Equation (7): F = ss (2nb )
Equation (8): F = s (d − c )a
(b) s =
sy
and s s =
s sy
N
N
Therefore
s s = 0.6 s
Equate (2) and (1)
 1
  1

F = ss 2 π c 2  = s π D 2 
  4

 4
1  1 
0.6 s c 2  = s D 2 
2  4 
c = 0.9129 D
Equate (3) and (1)
1

F = s (2bc ) = s  π D 2 
4

1
2b(0.9129 D ) = π D 2
4
b = 0.4302 D
Equate (4) and (1)
1

F = sac = s π D 2 
4

1
a(0.9129 D ) = π D 2
4
a = 0.8603D
66
SECTION 1– DESIGN FOR SIMPLE STRESSES
Equate (5) and (1)
1

F = s[2(d − c )b] = s  π D 2 
4

1
2(d − 0.9129 D )(0.4302) = π D 2
4
d = 1.8257 D
Equate (6) and (1)
1

F = s s (4mb ) = s π D 2 
4

1
0.6(4m )(0.4302 D ) = π D 2
4
m = 0.7607 D
Equate (7) and (1)
1

F = ss (2nb ) = s π D 2 
4

1
0.6(2n )(0.4302 D ) = π D 2
4
n = 1.5214 D
Equate (8) and (1)
1

F = s (d − c )a = s π D 2 
4

(1.8257 − D − 0.9129 D )a = 1 π D 2
4
a = 0.8604 D
Summary
a = 0.8604 D
b = 0.4302 D
c = 0.9129 D
d = 1.8257 D
m = 0.7607 D
n = 1.5214 D
47.
Design a connection similar to the one shown for a gradually applied and reversed
load of 12 kips. Base design stresses on yield strengths and let the material be
AISI C1040 steel, annealed. Examine the computed dimensions for proportion,
making changes that you deem advisable.
Solution: (See figure in Prob. 45 and refer to Prob. 46)
67
SECTION 1– DESIGN FOR SIMPLE STRESSES
N = 4 based on yield strength for gradually applied and reversed load.
For AISI C1040, annealed
s y = 47 ksi (Fig. AF 7)
s sy = 0.6s y = 28 ksi
sy
47
= 11.75 ksi
N
4
1

F = s π D 2 
4

1

12 = 11.75 π D 2 
4

D = 1.14 in
1
say D = 1 in
8
 1
a = 0.8604 D = 0.86041  = 0.97 in
 8
but a > D
1
say a = 1 in
4
 1
b = 0.43021  = 0.48 in
 8
1
say b = in
2
 1
c = 0.91291  = 1.030 in
 8
say c = 1 in
 1
d = 1.82571  = 2.05 in
 8
say d = 2 in
 1
m = 0.76071  = 0.86 in
 8
7
say m = in
8
 1
n = 1.52141  = 1.71 in
 8
3
say n = 1 in
4
s=
=
Dimension:
1
a = 1 in
4
68
SECTION 1– DESIGN FOR SIMPLE STRESSES
1
in
2
c = 1 in
d = 2 in
7
m = in
8
3
n = 1 in
4
1
D = 1 in
8
b=
48.
Give all the strength equations for the union of rods shown.
Problems 48 – 68.
Solution:
1

F = s π d 2 
4

Equation (1)
F = s s (π ad )
Equation (2)
F = ss (2tc )
Equation (3)
69
SECTION 1– DESIGN FOR SIMPLE STRESSES
F = ss [2(D − e )b]
Equation (4)
Equation (5)
F = set
F = s (D − e )t
Equation (6)
1

F = s  π k 2 − e2 
4

(
)
Equation (7)
1

F = s  π m 2 − e 2 − (m − e )t 
4

(
)
F = s s (2ef )
Equation (9)
70
Equation (8)
SECTION 1– DESIGN FOR SIMPLE STRESSES
1

F = s π e 2 − et 
4

Equation (10)
49-68. Design a union-of-rods joint similar to that shown for a reversing load and
material given in the accompanying table. The taper of cotter is to be ½ in. in 12
in. (see 172). (a) Using design stresses based on yield strengths determine all
dimensions to satisfy the necessary strength equations. (b) Modify dimensions as
necessary for good proportions, being careful not to weaken the joint. (c) Decide
upon tolerances and allowances for loose fits. (d) Sketch to scale each part of the
joint showing all dimensions needed for manufacture, with tolerances and
allowances.
Prob. No.
Load, lb.
AISI No., As Rolled
49
50
51
52
3000
3500
4000
4500
1020
1030
1117
1020
52
54
55
56
5000
5500
6000
6500
1015
1035
1040
1020
57
58
59
60
7000
7500
8000
8500
1015
1118
1022
1035
61
62
63
64
9000
9500
10,000
10,500
1040
1117
1035
1022
65
66
67
68
11,000
11,500
12,000
12,500
1137
1035
1045
1030
71
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution: (For Prob. 49 only)
(a) For AISI 1020, as rolled
s y = 48 ksi
s ys = 0.6s y = 0.6(48) = 28.8 ksi
For reversing load, N = 4 based on yield strength
s
48
= 12 ksi
s= y =
N
4
s
28.8
= 7.2 ksi
s s = ys =
N
4
F = 3000 lb = 3 kips
Equation (1)
1

F = s π d 2 
4

1

3 = 12 π d 2 
4

d = 0.5642 in
9
say d = in
16
Equation (2)
F = s s (π ad )
9
3 = 7.2(π a ) 
 16 
a = 0.236 in
1
say a = in
4
Equation (5)
F = set
3 = 12et
et = 0.25
Equation (10)
1

F = s π e 2 − et 
4

1

3 = 12 π e 2 − 0.25 
4

e = 0.798 in
13
say e = in
16
et = 0.25
72
SECTION 1– DESIGN FOR SIMPLE STRESSES
 13 
 t = 0.25
 16 
t = 0..308 in
5
say t = in
16
Equation (6)
F = s (D − e )t
13  5 

3 = 12 D −  
16  16 

D = 1.6125 in
5
say D = 1 in
8
Equation (4)
F = ss [2(D − e )b]
  5 13  
3 = 7.2 21 − b
  8 16  
b = 0.256 in
1
say b = in
4
Equation (7)
1

F = s  π k 2 − e2 
4

2
 1 
 13  
3 = 12 π  k 2 −   
 4  
 16  
k = 0.989 in
say k = 1 in
Equation (9)
F = s s (2ef )
(
)
 13 
3 = 7.2(2)  f
 16 
f = 0.256 in
1
say f = in
4
Equation (8)
1

F = s  π m 2 − e 2 − (m − e )t 
4

(
)
2
 1 
13  5 
 13   
3 = 12 π m 2 −    −  m −  
16  16 
 16   
 4 
73
SECTION 1– DESIGN FOR SIMPLE STRESSES
0.25 = 0.7854m 2 − 0.5185 − 0.3125m + 0.2539
0.7854m 2 − 0.3125m − 0.5146 = 0
m 2 − 0.3979m − 0.6552 = 0
m = 1.032 in
say m = 1 in
Equation (3)
F = ss (2tc )
5
3 = 7.2(2) c
 16 
c = 0.667 in
11
say c = in
16
DIMENSIONS:
9
d = in
16
1
a = in
4
1
b = in
4
11
c = in
16
1
f = in
4
13
e = in
16
5
t = in
16
k = 1 in
5
D = 1 in
8
m = 1 in
(b)
Modified dimensions
9
in
16
1
a = in
4
3
b = in
4
11
c = in
16
d=
74
SECTION 1– DESIGN FOR SIMPLE STRESSES
1
in
2
13
e = in
16
5
t = in
16
k = 1 in
5
D = 1 in
8
1
m = 1 in
4
f =
(c) Tolerances and allowances
No fit, ± 0.010 in
d = 0.5625 ± 0.010 in
a = 0.250 ± 0.010 in
f = 0.500 ± 0.010 in
D = 1.625 ± 0.010 in
k = 1.000 ± 0.010 in
m = 1.250 ± 0.010 in
Fits, Table 3.1, loose-running fits, say RC 8
Female
+ 0.0035
in
− 0.0000
allowance = 0.0045 in
+ 0.0028
c = 0.6875
in
− 0.0000
allowance = 0.0035 in
+ 0.0035
e = 0.8125
in
− 0.0000
allowance = 0.0045
+ 0.0022
t = 0.3125
in
− 0.0000
allowance = 0.0030 in
b = 0.750
Male
b = 0.750
− 0.0045
in
− 0.0065
c = 0.6875
− 0.0035
in
− 0.0051
e = 0.8125
− 0.0045
in
− 0.0065
t = 0.3125
75
− 0.0030
in
− 0.0040
SECTION 1– DESIGN FOR SIMPLE STRESSES
(d)
ROD
COTTER
76
SECTION 1– DESIGN FOR SIMPLE STRESSES
SOCKET
CHECK PROBLEMS
69.
1
1
The connection shown has the following dimensions: d = 1 in , D = 2 in ,
4
2
1
5
1
D1 = 1 in , h = in , t = in ; it supports a load of 15 kips. Compute the tensile,
2
8
2
compressive, and shear stresses induced in the connection. What is the
corresponding design factor based on the yield strength if the rod and nut are
made of AISI C1045, as rolled, and the plate is structural steel (1020)?
77
SECTION 1– DESIGN FOR SIMPLE STRESSES
Problem 69.
Solution:
Tensile Stresses
F
15
=
= 12.22 ksi
(1) s1 =
2
1
2
1
1


πd
π 1 
4
4  4
F
15
(2) s 2 =
=
= 8.4 ksi
2
1
2
1
1


π D1
π 1 
4
4  2
Compressive Stress
F
15
(3) s3 =
=
= 4.78 ksi
2
2
1
2
2


1
1
1
   
π (D − D1 )
π  2  − 1  
4
4  2   2  
Shear Stresses
F
15
(4) s s4 =
=
= 3.82 ksi
π Dt
 1  1 
π  2  
 2  2 
F
15
(5) s s5 =
=
= 5.09 ksi
π D1h
 1  5 
π 1  
 2  8 
For AISI C1045, as rolled (rod and nut)
s y1 = 59 ksi
s ys1 = 0.6 s y = 0.6(59 ) = 35.4 ksi
For structural steel plate (1020)
s y2 = 48 ksi
s ys1 = 0.6 s y = 0.6(48) = 28.8 ksi
Solving for design factor
78
SECTION 1– DESIGN FOR SIMPLE STRESSES
(1) N1 =
(2) N 2 =
(3) N 3 =
(4) N 4 =
(5) N 5 =
s y1
s1
s y1
s2
s y2
s3
s ys2
s s4
s ys1
s s5
=
59
= 4.83
12.22
=
59
= 6.95
8.49
=
48
= 10.04
4.78
=
28.8
= 7.54
3.82
=
35.4
= 6.96
5.09
The corresponding design factor is N = 4.83
70.
3
7
3
in , t = in , b = 3 in , and let the load, which is applied
4
16
4
centrally so that it tends to pull the plates apart, be 15 kips. (a) Compute the
stresses in the various parts of the connection. (b) If the material is AISI C1020,
as rolled, what is the design factor of the connection based on yield strengths?
In the figure, let D =
Problem 70.
Solution:
(a) Tensile stresses
F
15
s1 =
=
= 11.43 ksi
t (b − D )  7  3 3 
  3 − 
 16  4 4 
 3
3
 (15)
F
 4
4
s2 =
=
= 11.43 ksi
t (b − 2 D )  7   3  3 
  3 − 2 
 16   4  4 
Compressive bearing stress
F
15
s3 =
=
= 11.43 ksi
4 Dt
 3  7 
4  
 4  16 
79
SECTION 1– DESIGN FOR SIMPLE STRESSES
Shearing stress
F
15
ss 4 =
=
= 4.24 ksi
2
1
2
3


4 π D (2) π   (2 )
4

4
(b) For AISI C1020, as rolled
s y = 48 ksi
s ys = 0.6s y = 28.8 ksi
N=
sy
s
or N =
Using N =
s ys
ss
sy
s
s
48
= 4.2
N= y =
s 11.43
s
Using N = ys
ss
s ys 28.8
N=
=
= 6.8
s s 4.24
Therefore the design factor is N = 4.2
71.
For the connection shown, let a =
15
9
3
1
in , b = in , c = in , d = 1 in ,
16
16
4
2
3
15
in , m = n = in . The material is AISI C1040, annealed (see Fig. AF 1).
4
16
(a) For a load of 7500 lb., compute the various tensile, compressive, and shear
stresses. Determine the factor of safety based on (b) ultimate strength, (c) yield
strengths.
D=
Problem 71.
Solution:
(a) Tensile stresses
80
SECTION 1– DESIGN FOR SIMPLE STRESSES
s1 =
F
1
π D2
4
=
7.5
1 3
π 
4 4
2
= 16.98 ksi
F
7. 5
=
= 8.89 ksi
2b(d − c )
 9  1 3 
2 1 − 
 16  2 4 
F
7. 5
s3 =
=
= 10.67 ksi
a(d − c )  15  1 3 
 1 − 
 16  2 4 
Compressive Stresses (Bearing)
F
7.5
s4 =
=
= 8.89 ksi
2bc
 9  3 
2  
 16  4 
F
7. 5
s5 =
=
= 10.67 ksi
ac  15  3 
  
 16  4 
Shearing Stresses
F
7.5
s s6 =
=
= 3.56 ksi
4mb
 15  9 
4  
 16  16 
F
7.5
s s7 =
=
= 7.11 ksi
2nb
 15  9 
2  
 16  16 
For AISI C1040, annealed,Fig. AF 1
s y = 47 ksi
s2 =
su = 79 ksi
s ys = 0.6s y = 28 ksi
sus = 0.6su = 47.4 ksi
(b) Based on ultimate strength
s
79
N= u =
= 4.65
s1 16.98
(c) Based on yield strength
sy
47
N= =
= 2.77
s1 16.98
72.
The upper head of a 60,000-lb. tensile-testing machine is supported by two steel
rods, one of which A is shown. These rods A are attached to the head B by split
rings C. The test specimen is attached to the upper head B so that the tensile force
81
SECTION 1– DESIGN FOR SIMPLE STRESSES
in the specimen pulls down on the head and exerts a compressive force on the
rods A. When the machine is exerting the full load, compute (a) the compressive
stress in the rods, (b) the bearing stress between the rods and the rings, (c) the
shearing stress in the rings,
Problem 72.
Solution:
F = 60,000 lbs
(a) sc =
(b) sb =
(c) sc =
(60,000 2)
2

1  1 
π  3  − (3)2 
4  2 

(60,000 2 )
2
1  2  1 
π (4) −  3  
4 
 2  
= 11,753 psi = 11.75 ksi
= 10,186 psi = 10.19 ksi
(60,000 2) = 3,183 psi = 3.18 ksi
π (3)(1)
DEFORMATIONS
73.
A load of 22,000 lb. is gradually applied to a 2-in. round rod, 10 ft. long. The total
elongation is observed to be 0.03 in. If the stretching is entirely elastic, (a) what is
the modulus of elasticity, and (b) what material would you judge it to be, wrought
iron or stainless steel (from information available in the tables)? (c) How much
energy is absorbed by the rod? (d) Suppose that the material is aluminum alloy
3003-H14; compute its elongation for the same load. Is this within elastic action?
Solution:
F = 22,000 lbs
D = 2 in
L = 10 ft = 120 in
δ = 0.03 in
82
SECTION 1– DESIGN FOR SIMPLE STRESSES
(a) δ =
E=
FL
EA
FL
4 FL
4(22,000 )(120)
=
=
= 28 × 10 6 psi
2
2
δA δπ D
(0.03)(π )(2)
(b) Use both stainless steel, Table AT 4, E = 28× 10 6 psi and wrought iron , Table AT 7,
E = 28× 10 6 psi .
1
1
(c) Energy absorbed = Fδ = (22,000 )(0.03) = 330 lb − in
2
2
(d) For Aluminum alloy, 3003-H14
E = 10× 106 psi
s y = 21 ksi
FL
4 FL
4(22,000 )(120)
=
=
= 0.084 in
2
2
EA Eπ D
10 × 10 6 (π )(2 )
4F
4(22,000 )
s=
=
= 7003 psi = 7.0 ksi < s y , within the elastic limit.
2
πD
(π )(2)2
δ=
74.
(
)
The same as 73, except that F = 88 kips and total δ = 0.112 in . Is the
computation for part (d) valid? Explain.
Solution:
(a) F = 88 kips
δ = 0.112 in
FL
4 FL
4(88,000 )(120)
E=
=
=
= 30 ×10 6 psi
2
2
δA δπ D
(0.112)(π )(2)
(b) Use wrought steel, Table AT 4, E = 30× 106 psi
1
1
(c) Energy absorbed = Fδ = (88,000 )(0.112) = 4928 lb − in
2
2
(d) For Aluminum alloy, 3003-H14
E = 10× 106 psi
s y = 21 ksi
δ=
FL
4 FL
4(88,000)(120 )
=
=
= 0.336 in
2
2
EA Eπ D
10 × 10 6 (π )(2 )
(
)
83
SECTION 1– DESIGN FOR SIMPLE STRESSES
4F
4(88,000 )
=
= 28,011 psi = 28.0 ksi > s y , not within the elastic limit, therefore
2
πD
(π )(2)2
not valid.
s=
75.
(a) A square bar of SAE 1020, as rolled, is to carry a tensile load of 40 kips. The
bar is to be 4 ft. long. A design factor of 5 based on the ultimate stress is desired.
Moreover, the total deformation should not exceed 0.024 in. What should be the
dimensions of the section? (b) Using SAE 1045, as rolled, but with the other data
the same, find the dimensions. (c) Using SAE 4640, OQT 1000 F, but with other
data the same as in (a), find the dimensions. Is there a change in dimensions as
compared with part (b)? Explain the difference or the lack of difference in the
answers.
Solution:
L = 4 ft = 48 in
(a) For SAE 1020, as rolled
su = 65 ksi , E = 30,000 ksi
s
F
s= u =
N A
65 40
=
5 x2
x = 1.754 in
FL
δ=
EA
(40)(48)
0.024 =
(30,000)x 2
x = 1.633 in
3
Therefore say x = 1 in
4
(b) For SAE 1045, as rolled
su = 96 ksi , E = 30,000 ksi
s
F
s= u =
N A
96 40
=
5 x2
x = 1.443 in
FL
δ=
EA
(40)(48)
0.024 =
(30,000)x 2
x = 1.633 in
84
SECTION 1– DESIGN FOR SIMPLE STRESSES
5
Therefore say x = 1 in
8
(c) For SAE 4640, as rolled
su = 152 ksi , E = 30,000 ksi
s
F
s= u =
N A
152 40
= 2
5
x
x = 1.15 in
FL
δ=
EA
(40)(48)
0.024 =
(30,000)x 2
x = 1.633 in
5
Therefore say x = 1 in
8
There is lack of difference in the answers due to same dimensions required to satisfy the
required elongation.
76.
The steel rails on a railroad track are laid when the temperature is 40 F. The rails
are welded together and held in place by the ties so that no expansion is possible
due to temperature changes. What will be the stress in the rails when heated by
the sun to 120 F (i1.29)?
Solution:
s δ α L∆t
= =
E L
L
For steel α = 0.000007 in in − F
E = 30× 106 psi
s = α ∆tE = (0.000007 )(120 − 40 ) 30 × 10 6
s = 16,800 psi
(
77.
)
Two steel rivets are inserted in a riveted connection. One rivet connects plates that
have a total thickness of 2 in., while the other connects plates with a total
thickness of 3 in. If it is assumed that, after heading, the rivets cool from 600 F
and that the coefficient of expansion as given in the Text applies, compute the
stresses in each rivet after it has cooled to a temperature of 70 F, (no external
load). See i1.29. Also assume that the plates are not deformed under load. Is such
a stress likely? Why is the actual stress smaller?
85
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
s = α ∆tE
For steel α = 0.000007 in in − F
E = 30× 106 psi
s = (0.000007 )( 600 − 70 )(30,000 ) = 111.30 ksi
The stress is unlikely because it is near the ultimate strength of steel.
Actual stress must be smaller to allow for safety.
78.
Three flat plates are assembled as shown; the center one B of chromium steel,
AISI 5140 OQT 1000 F, and the outer two A and C of aluminum alloy 3003-H14,
are fastened together so that they will stretch equal amounts. The steel plate is 2 x
½ in., the aluminum plates are each 2 x 1/8 in., L = 30 in ., and the load is 24,000
lb. Determine (a) the stress in each plate, (b) the total elongation, (c) the energy
absorbed by the steel plate if the load is gradually applied, (d) the energy
absorbed by the aluminum plate. (e) What will be the stress in each plate if in
addition to the load of 24,000 lb. the temperature of the assembly is increased by
100 F?
Problem 78, 79.
Solution:
For chromium steel, AISI 5140 OQT 1000F (Table AT 7)
α1 = 0.000007 in in − F
E1 = 30 × 10 6 psi = 30,000 ksi
For aluminum alloy, 3003-H14 (Table AT 3)
α 2 = 0.0000129 in in − F
E2 = 10 × 106 psi = 10,000 ksi
(a) PA = PC
PA + PB + PC = F
(1) 2 PA + PB = F = 24 kips
1
A2 = (2)  = 0.25 in 2
8
1
A1 = (2 )  = 1 in 2
2
86
SECTION 1– DESIGN FOR SIMPLE STRESSES
δ A = δB
PA L
PL
= B
A2 E2 A1 E1
PA
PB
=
(0.25)(10,000) (1)(30,000)
(2) PB = 12 PA
(1) 2 PA + 12 PA = 24 kips
PA = 1.714 kips
PB = 12(1.714) = 20.568 kips
Stresses:
Aluminum plate
P 1.714
s A = sC = A =
= 6.856 ksi
A2 0.25
Chromium steel plate
P
20.568
sB = B =
= 20.568 ksi
A1
1
(1.714)(30) = 0.021 in
PA L
=
A2 E2 (0.25)(10,000 )
1
1
(c) Energy absorbed by steel plate = PBδ = (20.568)(0.021) = 0.216 kips − in
2
2
1
1
(d) Energy absorbed by aluminum plate = PAδ = (1.714 )(0.021) = 0.018 kips − in
2
2
(e) 2 PA + PB = F = 24 kips
δ TA + δ A = δ TB + δ B
(b) δ =
δT
δT
δT
δT
A
= α 2 L∆ t
B
= α1 L∆t
A
= (0.0000129 )(30)(100 ) = 0.0387 in
B
= (0.000007 )(30)(100 ) = 0.021 in
PA L
PA (30)
=
= 0.012 PA
A2 E2 (0.25)(10,000)
PL
PB (30)
δB = B =
= 0.001PB
A1E1 (1)(30,000 )
Then
0.0387 + 0.012 PA = 0.021 + 0.001PB
0.0177 + 0.012 PA = 0.001(24 − 2 PA )
0.0177 + 0.012 PA = 0.024 − 0.002 PA
δA =
87
SECTION 1– DESIGN FOR SIMPLE STRESSES
PA = 0.45 kips
PB = 24 − 2(0.45) = 23.1 kips
Stresses:
Aluminum plate
P
0.45
sA = A =
= 1.8 ksi
A2 0.25
Chromium steel plate
P
23.1
sB = B =
= 23.1 ksi
A1
1
79.
The same as 78, except that the outer plates are aluminum bronze, B150-1,
annealed.
Solution:
For aluminum bronze, B150-1, annealed (Table AT 3)
E2 = 15,000 ksi
α 2 = 0.0000092 in in − F
(a)
(1) 2 PA + PB = F = 24 kips
δ A = δB
PA L
PL
= B
A2 E2 A1 E1
PA
PB
=
(0.25)(15,000) (1)(30,000)
(2) PB = 8 PA
2 PA + 8PA = 24 kips
PA = 2.4 kips
PB = 8(2.4) = 19.2 kips
Stresses:
Aluminum plate
P
2.4
s A = sC = A =
= 9.6 ksi
A2 0.25
Chromium steel plate
P 19.2
sB = B =
= 19.2 ksi
A1
1
88
SECTION 1– DESIGN FOR SIMPLE STRESSES
PA L
(2.4)(30) = 0.019 in
=
A2 E2 (0.25)(15,000 )
1
1
(c) Energy absorbed by steel plate = PBδ = (19.2 )(0.019 ) = 0.182 kips − in
2
2
1
1
(d) Energy absorbed by aluminum plate = PAδ = (2.4)(0.019 ) = 0.023 kips − in
2
2
(e) 2 PA + PB = F = 24 kips
δ TA + δ A = δ TB + δ B
(b) δ =
δ T = α 2 L∆ t
δ T = α1 L∆t
δ T = (0.0000092 )(30)(100 ) = 0.0276 in
A
B
A
δ T = (0.000007 )(30)(100 ) = 0.021 in
B
PA L
PA (30 )
=
= 0.008 PA
A2 E2 (0.25)(15,000)
PL
PB (30)
δB = B =
= 0.001PB
A1E1 (1)(30,000 )
Then
0.0276 + 0.008PA = 0.021 + 0.001PB
0.0066 + 0.008PA = 0.001(24 − 2 PA )
0.0066 + 0.008PA = 0.024 − 0.002 PA
PA = 1.74 kips
PB = 24 − 2(1.74 ) = 20.52 kips
Stresses:
Aluminum plate
P 1.74
sA = A =
= 6.96 ksi
A2 0.25
Chromium steel plate
P
20.52
sB = B =
= 20.52 ksi
A1
1
δA =
80.
A machine part shown is made of AISI C1040, annealed steel; L1 = 15 in .,
3
1
L2 = 6 in ., D1 = in ., and D2 = in . Determine (a) the elongation due to a force
4
2
F = 6000 lb ., (b) the energy absorbed by each section of the part if the load is
gradually applied.
89
SECTION 1– DESIGN FOR SIMPLE STRESSES
Problems 80, 81
Solution:
For AISI C1040, annealed steel
E = 30× 106 psi
(a) δ = δ1 + δ 2
FL
(6000)(15) = 0.0068 in
δ1 = 1 =
A1 E  π  3  2
6
   30 × 10
 4  4 
FL
(6000)(6)
= 0.0061 in
δ2 = 2 =
A2 E  π  1  2
6
   30 × 10
 4  2 
δ = δ1 + δ 2 = 0.0068 + 0.0061 = 0.0129 in
(
)
(
)
(b) Energy absorbed
1
1
U1 = Fδ1 = (6000 )(0.0068) = 20.4 lb = in
2
2
1
1
U 2 = Fδ 2 = (6000 )(0.0061) = 18.3 lb = in
2
2
81.
A rod as shown is made of AISI 2340 steel, OQT 1000 F, and has the following
7
3
dimensions: L1 = 20 in ., L2 = 12 in ., D1 = in ., and D2 = in . The unit strain at
8
4
point A is measured with a strain gage and found to be 0.0025 in./in. Determine
(a) the total elongation, and (b) the force on the rod.
Solution:
For steel E = 30000 ksi
δ
F
(a) 2 = ε =
L2
A2 E
90
SECTION 1– DESIGN FOR SIMPLE STRESSES
F = ε A2 E
2
A 
D 
δ1 =
= ε  2  L1 = ε  2  L1
A1E
 A1 
 D1 
 D  2

 3
δ T = δ1 + δ 2 = ε  2  L1 + L2  = 0.0025
 D1 

 7
ε A2 EL1
2

4
 (20) + 12 = 0.067 in
8

2
 π  3 
(b) F = ε A2 E = 0.0025   (30,000) = 33.13 kips
 4  4 
82.
A rigid bar H is supported as shown in a horizontal position by the two rods
(aluminum 2024 T4, and steel AISI 1045, as rolled), whose ends were both in
contact with H before loading was applied. The ground and block B are also to be
considered rigid. What must be the cross-sectional area of the steel rod if, for the
assembly, N = 2 based on the yield strengths?
Problem 82.
Solution:
For aluminum 2024-T4 (Table AT 3)
s y1 = 47 ksi , E1 = 10,600 ksi
For steel AISI 1045, as rolled (Table AT 7)
s y2 = 59 ksi , E2 = 30,000 ksi
[∑ M
G
=0
]
R1 (24 ) + R2 (12) = 24(20 )
2 R1 + R2 = 40 Equation (1)
91
SECTION 1– DESIGN FOR SIMPLE STRESSES
δ1
=
δ2
24 12
δ1 = 2δ 2
RL
δ1 = 1 1
E1 A1
RL
δ2 = 2 2
E2 A2
L1 = 8 ft = 96 in
L2 = 12 ft = 144 in
A1 = 0.5 in 2
δ1 = 2δ 2
R1 L1 2 R2 L2
=
E1 A1 E2 A2
R1 (96)
2R2 (144)
=
(10,600)(0.5) (30,000)A2
0.53R2
R1 =
A2
But s 2 =
R2 s y2
=
A2 N
R2 59
=
= 29.5
A2
2
R1 = 0.53(29.5) = 15.64 kips
R1 s y1
=
A1 N
R1 47
=
0. 5 2
R1 = 11.75 kips
use R1 = 11.75 kips
R2 = 40 − 2(11.75) = 16.5 kips
R
16.5
A2 = 2 =
= 0.56 in 2
29.5 29.5
s1 =
92
SECTION 1– DESIGN FOR SIMPLE STRESSES
The bar shown supports a static load F = 2.5 kips with θ = 0 ; d = 3 in .,
3
L = 10 in ., h = 2 in . b = 1 in . It is made of AISI 1035, as rolled. (a) How far
4
does point C move upon gradual application of the load if the movement of A and
B is negligible? (b) How much energy is absorbed?
83.
Problem 83.
Solution:
[∑ M
[∑ M
A
B
=0
=0
]
dRB = (d + L )F
]
3RB = (3 + 10 )(2.5)
RB = 10.83 kips
dRA = LF
3RA = 10(2.5)
RA = 8.33 kips
93
SECTION 1– DESIGN FOR SIMPLE STRESSES
M = RA x − RB x − 3
d2y
M = EI 2 = 8.33 x − 10.83 x − 3
dy
dy
2
EI
= 4.165 x 2 − 5.415 x − 3 + C1
dy
3
EIy = 1.388 x 3 − 1.805 x − 3 + C1 x + C2
When x = 0 , y = 0
3
EI (0 ) = 1.388(0) − 1.805 0 + C1 (0 ) + C2
3
C2 = 0
When x = 3 , y = 0
3
EI (0 ) = 1.388(3) − 1.805 0 + C1 (3) + 0
3
C1 = −12.492
3
EIy = 1.388 x 3 − 1.805 x − 3 − 12.492 x
When x = d + L = 13 in
3
EIy = 1.388(13) − 1.805 10 − 12.492(13) = 1082
For AISI 1035, as rolled , E = 30,000 ksi
3
bh 3 (1)(2.75)
=
= 1.7331 in 4
12
12
EIy = 1082
(30,000)(1.7331)y = 1082
y = 0.021 in , upward.
3
I=
94
SECTION 1– DESIGN FOR SIMPLE STRESSES
PRESSURE VESSELS
84.
A storage tank for air, 36 in. in diameter, is to withstand an internal pressure of
200 psi with a design factor of 4 based on su . The steel has the strength
equivalent of C1020 annealed and the welded joints should have a relative
strength (efficiency) of 90 %. Determine a suitable plate thickness. Compute the
stress on a diametral section and compare it with the longitudinal stress.
Solution:
For C1020 annealed
su = 57 ksi
su 57
=
= 14.25 ksi
N
4
Solving for plate thickness
pD
s=
2 tη
p = 200 psi = 0.2 ksi
D = 36 in
(5.2)(36)
s = 14.25 =
2 t (0.9)
t = 0.281 in
5
say t = in
16
Stress on diametral section
(0.2 )(36) = 6.40 ksi
pD
s=
=
4 tη
5
4 (0.9)
 16 
Stress on longitudinal section
(0.2 )(36) = 12.80 ksi
pD
s=
=
2 tη
5
2 (0.9)
 16 
Stress on diametral section < stress on longitudinal section
s=
85.
A spherical air tank stores air at 3000 psig. The tank is to have an inside diameter
of 7 in. (a) What should be the wall thickness and weight of the tank if it is made
of 301, ¼-hard, stainless steel, with a design factor of 1.5 based on the yield
strength and a joint efficiency of 90 %. (b) Compute the wall thickness and
weight if annealed titanium (B265, gr. 5) is used? (c) What is the additional
saving in weight if the titanium is hardened? Can you think of circumstances for
which the higher cost of titanium would be justified?
95
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
(a) For 301, ¼ hard, stainless steel
s y = 75 ksi (Table AT 4)
sy
75
= 50 ksi
N 1.5
p = 3000 psi = 3 ksi
pD
s=
4 tη
(3)(7)
50 =
4t (0.90)
t = 0.117 in
s=
=
γ = 0.286 lb in 3
2
W = 4π r 2t γ = π D 2t γ = π (7 ) (0.117 )(0.286 ) = 5.2 lb
(b) For annealed titanium B265, gr. 5
s y = 130 ksi (Table AT 3)
sy
130
= 86.67 ksi
N 1.5
p = 3000 psi = 3 ksi
pD
s=
4 tη
(3)(7)
86.67 =
4t (0.90)
t = 0.061 in
s=
=
γ = 0.160 lb in 3
2
W = 4π r 2t γ = π D 2t γ = π (7 ) (0.061)(0.160) = 1.5 lb
(c) For hardened titanium
s y = 158 ksi (Table AT 3)
sy
158
= 105 ksi
N 1.5
p = 3000 psi = 3 ksi
pD
s=
4 tη
(3)(7 )
105 =
4t (0.90 )
s=
=
96
SECTION 1– DESIGN FOR SIMPLE STRESSES
t = 0.056 in
γ = 0.160 lb in 3
2
W = 4π r 2t γ = π D 2t γ = π (7 ) (0.056)(0.160 ) = 1.38 lb
1.50 − 1.38
(100%) = 8%
1.50
Circumstances: less in weight and small thickness.
Savings in weight =
86.
Decide upon a material and estimate a safe wall thickness of a cylindrical vessel
to contain helium at –300 F and 2750 psi. The welded joint should have a relative
strength ≥ 87 %, and the initial computations are to be for a 12-in.-diameter, 30ft.-long tank. (Note: Mechanical properties of metals at this low temperature are
not available in the Text. Refer to INCO Nickel Topics, vol. 16, no. 7, 1963, or
elsewhere.)
Solution:
From Kent’s Handbook, Table 8
Material – Hot Rolled Nickel
At – 300 F, su = 100 ksi , N = 4 (Table 1.1)
s 100
s= u =
= 25 ksi
N
4
pD
s=
2 tη
p = 2750 psi = 2.75 ksi
D = 12 in
η = 87%
(2.75)(12)
s = 25 =
2 t (0.87 )
t = 0.759 in
3
say t = in
4
CONTACT STRESSES
87.
(a) A 0.75-in. diameter roller is in contact with a plate-cam surface whose width is
0.5-in. The maximum load is 2.5 kips where the radius of curvature of the cam
surface is 3.333 in. Compute the Hertz compressive stress. (b) The same as (a)
except that the follower has a plane flat face. (c) The same as (a) except that the
roller runs in a grooved face and contacts the concave surface. (d) What is the
maximum shear stress for part (a) and how far below the surface does it exist?
Solution:
97
SECTION 1– DESIGN FOR SIMPLE STRESSES
(a) 2r1 = 0.75 in , r1 = 0.375 in
r2 = 3.333 in
F = 2.5 kips
b = 0.5 in
1

 1 1 2
 +  
0
.
35
F

 r1 r2  
sc max = 
  1
1  
 b +  
  E1 E2  
E = 30,000 ksi
1
sc max

1 2
 1
+

 0.35(2.5)
0.375 3.333  


=
= 279 ksi

 2 

0.5



 30,000 


1
(b) sc max

1 2
 1
(
)
0
.
35
2
.
5
+



3
.
333
3
.
333

  = 126 ksi
=

 2 

0.5



 30,000 


(c) sc max

1 2
 1
(
)
0
.
35
2
.
5
−



0.375 3.333  


=
= 249 ksi

 2 

0.5



 30,000 


1
(d) Maximum shear stress
s s max = 0.3sc max = 0.3(279 ) = 84 ksi
Location:
 1
1 
 2 
4 sc max 1 − µ 2  +  4(279 ) 1 − 0.32 

E1 E2 
30,000 


w=
=
= 0.023 in
1 
1 1
 1
+
 + 


 0.375 3.333 
 r1 r2 
(
88.
)
(
)
Two 20o involute teeth are in contact along a “line” where the radii of curvature
of the profiles are respectively 1.03 and 3.42 in. The face width of the gears is 3
in. If the maximum permissible contact stress for carburized teeth is 200 ksi, what
normal load may these teeth support?
98
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
r1 = 1.03 in
r2 = 3.42 in
b = 3 in
sc max = 200 ksi
1

 1 1 2
 +  
0
.
35
F

 r1 r2  
sc max = 
  1
1  
 b +  
  E1 E2  
E = 30,000 ksi
1

1 2
 1
+

 0.35 F 
 1.03 3.42  
sc max = 200 = 

 2 

3



 30,000 


F = 18 kips
TOLERANCES AND ALLOWANCES
89.
The pin for a yoke connection has a diameter of D of ¾ in., a total length of 2 ½
in., with a head that is 1 ¼ in. in diameter and 3/8 in. thick. The tolerance on D
(both pin and hole) is 0.003 in., with an allowance of 0.001 in., basic-hole system.
Sketch the pin showing all dimensions with appropriate tolerances.
Solution:
D = 0.75 in
For pin
+ 0.000
D = 0.749
in
− 0.003
For hole
+ 0.003
D = 0.750
in
− 0.000
Sketch
99
SECTION 1– DESIGN FOR SIMPLE STRESSES
90.
A shaft with a nominal diameter of 8 in. is to fit in a hole. Specify the allowance,
tolerances, and the limit diameters of the shaft and hole on a sketch for: (a) a close
sliding fit, (b) a precision-running fit, (c) medium-running fit, (d) a loose-running
fit.
Solution: D = 8 in
(a) For close-sliding fit, RC 1
Hole, in
+0.0008
-0.0000
Shaft, in
- 0.0006
-0.0012
Allowance = 0.0006 in
With tolerances,
+ 0.0008
Hole D = 8.0000
in
− 0.0000
+ 0.0000
Shaft D = 7.9994
in
− 0.0006
Limit dimension,
Hole D = 8.0000 to 8.0008 in
Shaft D = 7.9994 to 7.9988 in
Sketch
100
SECTION 1– DESIGN FOR SIMPLE STRESSES
(b) For a precision-running fit, RC 3
Hole, in
Shaft, in
+0.0012
-0.0020
-0.0000
-0.0032
Allowance = 0.0020 in
With tolerances,
+ 0.0012
Hole D = 8.0000
in
− 0.0000
+ 0.0000
Shaft D = 7.9980
in
− 0.0012
Limit dimension,
Hole D = 8.0000 to 8.0012 in
Shaft D = 7.9980 to 7.9968 in
Sketch
(c) For medium-running fit, RC5, RC 6. Say RC 5
Hole, in
+0.0018
-0.0000
Shaft, in
-0.0040
-0.0058
Allowance = 0.0040 in
With tolerances,
+ 0.0018
Hole D = 8.0000
in
− 0.0000
+ 0.0000
Shaft D = 7.9960
in
− 0.0018
Limit dimension,
Hole D = 8.0000 to 8.0018 in
Shaft D = 7.9960 to 7.9942 in
Sketch
101
SECTION 1– DESIGN FOR SIMPLE STRESSES
(d) For loose-running fit, RC 8, RC 9. Say RC 8
Hole, in
+0.0070
-0.0000
Shaft, in
-0.0100
-0.0145
Allowance = 0.010 in
With tolerances,
+ 0.0070
Hole D = 8.0000
in
− 0.0000
+ 0.0000
Shaft D = 7.9900
in
− 0.0045
Limit dimension,
Hole D = 8.0000 to 8.0070 in
Shaft D = 7.9900 to 7.9855 in
Sketch
91.
The same as 90, except that the nominal diameter is 4 in.
Solution:
D = 4 in
(a) For close-sliding fit, RC 1
102
SECTION 1– DESIGN FOR SIMPLE STRESSES
Hole, in
+0.0006
-0.0000
Shaft, in
-0.0005
-0.0009
Allowance = 0.0005 in
With tolerances,
+ 0.0006
Hole D = 4.0000
in
− 0.0000
+ 0.0000
Shaft D = 3.9995
in
− 0.0004
Limit dimension,
Hole D = 4.0000 to 4.0006 in
Shaft D = 3.9995 to 3.9991 in
Sketch
(b) For a precision-running fit, RC 3
Hole, in
Shaft, in
+0.0009
-0.0014
-0.0000
-0.0023
Allowance = 0.0014 in
With tolerances,
+ 0.0009
Hole D = 4.0000
in
− 0.0000
+ 0.0000
Shaft D = 3.9986
in
− 0.0009
Limit dimension,
Hole D = 4.0000 to 4.0009 in
Shaft D = 3.9986 to 3.9977 in
Sketch
103
SECTION 1– DESIGN FOR SIMPLE STRESSES
(c) For medium-running fit, RC5, RC 6. Say RC 6
Hole, in
+0.0022
-0.0000
Shaft, in
-0.0030
-0.0052
Allowance = 0.0030 in
With tolerances,
+ 0.0022
Hole D = 4.0000
in
− 0.0000
+ 0.0000
Shaft D = 3.9970
in
− 0.0022
Limit dimension,
Hole D = 4.0000 to 4.0022 in
Shaft D = 3.9970 to 7.9948 in
Sketch
(d) For loose-running fit, RC 8, RC 9. Say RC 9
Hole, in
+0.0090
-0.0000
Shaft, in
-0.0100
-0.0150
Allowance = 0.0100 in
104
SECTION 1– DESIGN FOR SIMPLE STRESSES
With tolerances,
+ 0.0090
in
− 0.0000
+ 0.0000
Shaft D = 3.9900
in
− 0.0050
Limit dimension,
Hole D = 4.0000 to 4.0090 in
Shaft D = 3.9900 to 3.9850 in
Sketch
Hole D = 4.0000
92.
A cast-iron gear is to be shrunk onto a 3-in, steel shaft. (a) Determine the
tolerance and the maximum, minimum, and average interferences of metal for
class FN 1 fit. (b) Sketch and dimension the shaft and hole with proper tolerances.
(c) Compute the stresses by the method given in the Text (i3.8) for the maximum
and minimum interferences of metal.
Solution: D = 3 in
(a) For class FN 1 fit, Table 3.2
Tolerances
Hole, in
Shaft, in
+0.0007
+0.0019
-0.0000
+0.0014
Max. interference = 0.0019 in
Min. interference = 0.0014 – 0.0007 = 0.0007 in
Ave. interference = 0.5(0.0019 + 0.0007) = 0.0013 in
(b)
105
SECTION 1– DESIGN FOR SIMPLE STRESSES
(c) For maximum interference
Ei
D
E = 23,000 ksi
(23,000 )(0.0019 ) = 14.6 ksi
s=
3
s=
For minimum interference
s=
93.
(23,000 )(0.0007 ) = 5.4 ksi
3
The same as 92, except that the gear hub is C1035 steel and class of fit is FN 3.
Solution: D = 3 in
(a) For class FN 3fit, Table 3.2
Tolerances
Hole, in
Shaft, in
+0.0012
+0.0037
-0.0000
+0.0030
Max. interference = 0.0037 in
Min. interference = 0.0030 – 0.0012 = 0.0018 in
Ave. interference = 0.5(0.0037 + 0.0018) = 0.0028 in
(b)
106
SECTION 1– DESIGN FOR SIMPLE STRESSES
(c)
For C1035 steel, E = 30,000 ksi
For maximum interference
Ei
D
(
30,000)(0.0037 )
s=
= 37 ksi
3
s=
For minimum interference
s=
94.
(30,000)(0.0018) = 18 ksi
3
For a No. 7 ball bearing, the New Departure Handbook states that the maximum
bore should be 1.3780 in. and the minimum, 1.3775 in.; for average conditions,
the shaft should have a maximum diameter of 1.3784 in. and a minimum of
1.3779 in. (a) Determine the corresponding tolerances and allowances. (b) What
class of fit is this? (c) New Departure states: “. . . bearing bores are held
uniformly close, . . . averaging within 1.3778 in. to 1.3776 in.” What will be the
maximum and minimum interference of metal with these diameters (if maximum
and minimum sizes are deliberately chosen for assembly)?
Solution:
(a) Tolerances:
For No. 7 ball bearing
Bore, 1.3780 – 1.3775 in = 0.0005 in
+ 0.0005
Therefore, D = 1.3775
in
− 0.0000
107
SECTION 1– DESIGN FOR SIMPLE STRESSES
Shaft, 1.3784 – 1.3775 = 0.0009 in
1.3779 – 1.3775 = 0.0004 in
+ 0.0000
D = 1.37845
in
− 0.0005
Tolerances
Hole, in
+0.0005
-0.0000
Shaft, in
+0.0009
+0.0004
Allowance = 0 – 0.0009 in = - 0.0009 in
(b) Since allowance is < 0.
It is a force and shrink fil class.
(c)
1.3778 – 1.3775 = 0.0003 in
1.3776 – 1.3775 = 0.0001 in
New tolerances
Hole, in
+0.0005
-0.0000
Shaft, in
+0.0003
+0.0001
Maximum interference = 0.0003 in
Minimum interference = 0.0000 in (since 0.0001 – 0.0005 = - 0.0004 < 0)
95.
For a roller bearing having a bore of 65 mm. an SKF catalog states that the largest
diameter should be 2.5591 in. and the smallest, 2.5585 in. If this bearing is to be
used in a gear transmission, it is recommended for the shaft (where the bearing
fits) to have a maximum diameter of 2.5600 in. and a minimum of 2.5595 in. (a)
Determine the tolerances and allowances (or interferences of metal) for this
installation. (b) What class of fit would this be?
Solution:
(a) 65 mm = 2.5591 in
2.5591 – 2.5585 = 0.0006 in
2.5600 – 2.5585 = 0.0015 in
2.5595 – 2.5585 = 0.0010 in
108
SECTION 1– DESIGN FOR SIMPLE STRESSES
Tolerances
Hole, in
+0.0006
-0.0000
Shaft, in
+0.0015
+0.0010
Maximum interference = 0.0015 in
Minimum interference = 0.0010 – 0.0006 = 0.0004 in
(b) Class of fit, Force and shrink fit
TOLERANCES, STATISTICAL CONSIDERATION
96.
(a) A machine tool is capable if machining parts so that the standard deviation of
one critical dimension is 0.0006 in. What minimum tolerance may be specified for
this dimension if it is expected that practically all of the production be acceptable?
Assume that it is possible to “center the process.” (b) The same as (a), except that
it has been decided to tolerate approximately 4.56 % scrap.
Solution:
(a)
T = NS = 6σ = 6(0.0006 ) = 0.0036 in
(b)
A=
0.0456
= 0.0228 in
2
109
SECTION 1– DESIGN FOR SIMPLE STRESSES
From Table 3.3
z
= 2. 0
σ
z = 2σ
T = 2 z = 4σ
T = 4(0.0006 ) = 0.0024 in
97.
A pin and the hole into which it fits have a nominal diameter of 1 ½ in. The pin
tolerance has been set to 0.002 in., the bore tolerance at 0.003 in., and the
allowance at 0.001 in., basic hole system. The parts are to be a natural spread of
0.0015 in. for the pin and 0.002 in. for the hole. Assuming that the processes are
centered, determine the expected minimum clearance and the maximum
clearance. What is the most frequent clearance?
Solution:
1.5015 – 1.4980 = 0.0035 in
NS
0.0015
σ 1 ( pin ) = 1 =
= 0.00025 in
6
6
NS
0.0020
σ 2 (hole ) = 2 =
= 0.00033 in
6
6
σ D2 = σ 12 + σ 22 = (0.00025)2 + (0.00033)2
σ D = 0.00041 in
Natural Spread of Difference = 6σ D = 6(0.00041) = 0.00246 in
110
SECTION 1– DESIGN FOR SIMPLE STRESSES
Expected minimum clearance = 0.00227 in
Expected maximum clearance = 0.00473 in
Most frequent clearance = 0.0035 in
98.
A rod and the hole into which it fits has a nominal diameter of 2 in. The
tolerances are 0.003 in. for both rod and hole, and the allowance as 0..001 in.,
basic hole system. The natural spread of the process of manufacturing the hole is
0.002 in., and for the rod, 0.0015 in. What are the probable maximum and
minimum clearances, provided that the tolerances are met, but assuming that the
processes might simultaneously operate at their extreme permissible position?
Solution:
NS1 0.0015
=
= 0.00025 in
6
6
NS
0.0020
σ 2 ( pin ) = 2 =
= 0.00033 in
6
6
σ D2 = σ 12 + σ 22 = (0.00025)2 + (0.00033)2
σ D = 0.00041 in
σ 1 (rod ) =
111
SECTION 1– DESIGN FOR SIMPLE STRESSES
NS = 6σ D = 6(0.00041) = 0.00246 in
2.0020 – 1.99675 = 0.00525 in
Probable maximum clearance = 0.00648 in
Probable minimum clearance = 0.00402 in
99.
It is desired that the clearance in a 4-in. bearing neither exceed 0.004 in. nor be
less than 0.002 in. Assume that the natural spread of the processes by which the
journal and the bearing surfaces are finished is the same. (a) What should be the
natural spread of these processes? (b) Assuming this natural spread to be equal to
the tolerance, determine the corresponding allowance. (c) If the foregoing
conditions are not practical decide upon practical tolerances and allowances for
the computed natural spread.
Solution:
(a) NS = 6σ D
NS 0.002
σD =
=
= 0.00033 in
6
6
(b) σ D = 2σ
σ
0.00033
σ= D =
= 0.000233 in
2
2
112
SECTION 1– DESIGN FOR SIMPLE STRESSES
NS = 6σ = 6(0.000233) = 0.0014 in
Tolerance = 0.0014 in
Corresponing allowance = 0.0016 in
(c) From Fig. 3.4, T > NS
Tolerance = T = 1.3NS = 1.3(0.0014 ) = 0.0018 in
Allowance = 0.003 – 0.0018 = 0.0012 in
100.
A 4-in, journal-bearing assembly is made for class RC 6 fit. Assume that the
natural spread of the manufacturing process will be about 75 % of the tolerance.
Compute the probable maximum and minimum clearances (which occur when the
processes are not centered) and compare with the allowance. Make a sketch of the
journal and hole properly dimensioned.
113
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
From Table RC 6, D = 4 in
Hole
Shaft
+0.0022
-0.0030
-0.0000
-0.0052
or
+ 0.0022
in
− 0.0000
+ 0.0000
D (shaft ) = 3.997
in
− 0.0022
D(hole ) = 4.000
NS = 6σ
0.00165
σ=
= 0.00028 in
6
σ D = 2σ = 2 (0.00028) = 0.0004 in
NS D = 6σ D = 6(0.0004 ) = 0.0024 in
114
SECTION 1– DESIGN FOR SIMPLE STRESSES
Maximum clearance = 0.00912 in
Minimum clearance = 0.00336 in
Sketch
101.
The same as 100, except that class RC 3 fit is used.
Solution:
From Table RC 3, Table 3.1, D = 4 in
Hole
Shaft
+0.0009
-0.0014
-0.0000
-0.0023
or
+ 0.0009
in
− 0.0000
+ 0.0000
D (shaft ) = 3.9986
in
− 0.0009
D(hole ) = 4.000
115
SECTION 1– DESIGN FOR SIMPLE STRESSES
NS = 6σ
0.000675
σ=
= 0.0001125 in
6
σ D = 2σ = 2 (0.0001125) = 0.00016 in
NS D = 6σ D = 6(0.00016 ) = 0.00096 in
Maximum clearance = 0.001595 in
Minimum clearance = 0.002555 in
Sketch
102.
It is desired that the running clearance for a 3-in. bearing be between
approximately 0.003 in. and 0.007 in. The natural spread of the processes of
finishing the journal and bearing are expected to be virtually the same ( σ 1 = σ 2 ).
Decide upon a suitable tolerance and allowance with a sketch properly
dimensioned (to a ten thousandth). (Suggestion: compute first a theoretical natural
spread for bearing and journal from the given spread of the clearances. Let the
tolerances be approximately equal to this computed NS, and assume that
manufacturing processes are available that produce an actual NS of about 70 % of
this computed NS.) Check for processes being off center but within ± 3σ limits
so that virtually no scrap is manufactured.
Solution:
116
SECTION 1– DESIGN FOR SIMPLE STRESSES
σ D = 0.00067 in
σ 12 + σ 22 = σ D2
2
2σ 12 = (0.00067 )
σ 2 = 0.00047 in
NS = 0.70 NS1
NS = 0.70(6)(0.00047 ) = 0.00197 in
T = NS1 = (6)(0.00047 ) = 0.00282 in
117
SECTION 1– DESIGN FOR SIMPLE STRESSES
For processes off-center
running clearance = 0.00385 in to 0.00785 in
running clearance = 0.00215 in to 0.00615 in
since allowance = 0.00218 in ≈ 0.00215 in, it is checked.
103.
If the tolerances shown are maintained during manufacture, say with the processes
centered, what would be the approximate overall tolerances and limit dimensions?
118
SECTION 1– DESIGN FOR SIMPLE STRESSES
Problem 103.
Solution:
T1 = 4.004 − 4.000 = 0.004 in
T2 = 5.008 − 5.000 = 0.008 in
T3 = 6.707 − 6.700 = 0.007 in
T 2 = T12 + T22 + T32 = (0.004) + (0.008) + (0.007 )
T = 0.0114 in
2
2
2
Limit dimensions
4.000 to 4.0114 in
5.000 to 5.0114 in
6.700 to 6.7114 in
104.
If a cylindrical part needs to have the following tolerances, what process would
you recommend for finishing the surface in each instance? (a) 0.05 in., (b) 0.01
in., (c) 0.005 in., (d) 0.001 in., (e) 0.0001 in., (f) 0.00005 in.?
Solution:
Use fi. 3.9, page 95., Text.
(a) 0.05 in
Surface finishes = 500 or greater
Processes:
1. Flame cutting-machine
2. Rough turning
3. Contour sawing
4. Rough grinding
5. Shaping and planning
6. Drilling
7. Milling – high speed steel
8. Boring
(b) 0.01 in
Surface finishes = 63 to 250
119
SECTION 1– DESIGN FOR SIMPLE STRESSES
Processes:
1. Contour sawing
2. Rough grinding
3. Shaping and planning
4. Drilling
5. Milling – high speed steel
6. Finish turning
7. Broaching
8. Boring
9. Reaming
10. Commercial grinding
11. Barrel finishing
(c) 0.005 in
Surface finishes = 32 to 125
Processes:
1. Shaping and planning
2. Drilling
3. Milling – high speed steel
4. Finish turning
5. Broaching
6. Boring
7. Reaming
8. Commercial grinding
9. Milling – carbides
10. Gear shaping
11. Barrel finishing
12. Honing
(d) 0.001 in
Surface finishes = 8 to 32
Processes:
1. Finish turning
2. Broaching
3. Boring
4. Reaming
5. Commercial griniding
6. Milling – carbides
7. Gear shaping
8. Barrel finishing
9. Roller burnishing
10. Diamond turning
120
SECTION 1– DESIGN FOR SIMPLE STRESSES
11. Diamond and precision boring
12. Precision finish grinding
13. Honing
14. Production lapping
15. Superfinishing
(e) 0.0001 in
Surface finishes = 1 to 8 µin, rms
Processes:
1. Barrel finishing
2. Roller burnishing
3. Diamond turning
4. Diamong and precision boring
5. Precision finish grinding
6. Honing
7. Production lapping
8. Superfinishing
(f) 0.00005 in
Surface finishes = 0 to 2 µin.
Processes:
1. Honing
2. Production lapping
3. Superfinishing
105.
If it cost $100 to finish a certain surface to 500 microinches rms, what would be
the approximate cost to finish it to the following roughness: (a) 125, (b) 32, (c) 8,
(d) 2 µin. rms?
Solution:
From Fig. 3.9
Relative cost of 500 µin rms = 1.75
(a) Relative cost of 125 µin rms = 3
 3 
Cost = $100
 = $171
 1.75 
121
SECTION 1– DESIGN FOR SIMPLE STRESSES
(b) Relative cost of 32 µin rms = 5
 5 
Cost = $100
 = $286
 1.75 
(c) Relative cost of 8 µin rms = 7.75
 7.75 
Cost = $100
 = $443
 1.75 
(d) Relative cost of 2 µin rms = 11.5
 11.5 
Cost = $100
 = $657
 1.75 
DATA LACKING – DESIGNER’S DECISIONS*
*
Properties of rolled structural sections are found in various handbooks.
106-125. Design a bell crank, similar to the one shown, to carry a mild shock load. The
mechanical advantage ( L1 L2 = F2 F1 ), the force F1 , the length L1 , and the
material are given in the accompanying table. (a) Make all significant dimensions,
including tolerances and allowances. One approach could be to compute
dimensions of the yoke connections first; t should be a little less than a . An
assumption for the shaft may be that, on occasion, the torque for F1 is transmitted
through the shaft (ignoring bending for local convenience). (b) Check all
dimensions for good proportion; modify as desirable. (c) Sketch to scale each
part, showing all dimensions with tolerances necessary to manufacture.
Prob. No.
Load F1
L1
AISI No. As Mech,
Rolled
Advantage
106
107
108
109
700
650
600
550
12
14
15
18
C1020
C1020
C1022
C1035
1.5
2
2.5
3
110
111
112
113
500
800
750
750
20
12
14
14
C1040
C1020
C1020
C1020
4
1.5
2
2.5
114
115
116
117
650
600
900
850
18
20
12
14
C1035
C1040
C1020
C1020
3
4
1.5
2
122
SECTION 1– DESIGN FOR SIMPLE STRESSES
118
119
120
121
800
750
700
1000
15
18
20
12
C1022
C1035
C1040
C1020
2.5
3
4
1.5
122
123
124
125
950
900
850
800
14
15
18
20
C1020
C1022
C1035
C1040
2
2.5
3
4
Problems 106 to 125.
Solution:
F1 = 700 lb = 0.7 kip
L1 = 12 in
MA = 1.5
For AISI C1020 as rolled (Table AT 7)
su = 65 ksi
sus = 49 ksi
Designing based on ultimate strength
N = 6 (Table 1.1) mild shock, one direction
su 65
=
= 10 ksi
N
6
s
49
s s = us =
= 8 ksi
N
6
s=
Consider yoke connection A
123
SECTION 1– DESIGN FOR SIMPLE STRESSES
ss =
F1
1

2 π d12 
4

0.7
8=
1
π d12
2
d1 = 0.24 in
1
say d1 = in
4
=
F1
1
π d12
2
F1
a1d1
0. 7
10 =
1
a1  
4
a1 = 0.28 in
5
say a1 = in
16
since t1 < a1
1
say t1 = in
4
F1
s=
D1t1
0. 7
10 =
1
D1  
4
D1 = 0.28 in
5
say D1 = in
16
Consider yoke connection B.
L1
= 1.5
L2
12
L2 =
= 8 in
1.5
F2
= 1.5
F1
F2 = 1.5(0.7 ) = 1.05 kip
F2
F2
ss =
=
1
1

π d 22
2 π d 22 
2
4

s=
124
SECTION 1– DESIGN FOR SIMPLE STRESSES
1.05
1
π d 22
2
d 2 = 0.29 in
5
say d 2 = in
16
8=
F2
a2 d 2
1.05
10 =
5
a2  
 16 
a2 = 0.34 in
3
say a2 = in
8
since t 2 < a2
5
say t 2 = in
16
F1
s=
D1t1
1.05
10 =
5
D2  
 16 
D2 = 0.34 in
3
say D2 = in
8
For shaft diameter
Assume torque, T1 = F1L1 = (0.70 )(12) = 8.4 in − kips
16T1
ss =
π d s3
16(8.4)
8=
π d s3
3
d s = 1.75 in = 1 in
4
Tolerances and allowances, consider RC 4 (Table 3.1)
Hole
Shaft
+ 0.0006
+ 0.0000
d1 = 0.2500
in
d1 = 0.2495
in
− 0.0000
− 0.0006
allowance = 0.0005 in
s=
125
SECTION 1– DESIGN FOR SIMPLE STRESSES
+ 0.0006
+ 0.0000
in
d 2 = 0.3120
in
− 0.0000
− 0.0006
allowance = 0.0005 in
+ 0.0010
+ 0.0000
d s = 1.7490
in
d s = 1.7500
in
− 0.0010
− 0.0000
allowance = 0.0010 in
d 2 = 0.3125
Female
Male
+ 0.0006
+ 0.0000
a1 = 0.3125
in
a1 = 0.3120
in
− 0.0000
− 0.0006
allowance = 0.0005 in
+ 0.0006
+ 0.0000
a2 = 0.3750
in
a2 = 0.3745
in
− 0.0000
− 0.0006
allowance = 0.0005 in
(b) For good proporion use the following dimension
D1 = D2 = 1 in
3
d1 = d 2 = in
4
3
t1 = t 2 = in
4
a1 = a2 = 1 in
d s = 2 in
Tolerances and allowances, consider RC 4 (Table 3.1)
Hole
Shaft
+ 0.0008
+ 0.0000
d1 = 0.7500
in
d1 = 0.7492
in
− 0.0000
− 0.0008
allowance = 0.0008 in
+ 0.0008
+ 0.0000
d 2 = 0.7500
in
d 2 = 0.7492
in
− 0.0000
− 0.0008
allowance = 0.0008 in
+ 0.0012
+ 0.0000
d s = 2.0000
in
d s = 1.9988
in
− 0.0000
− 0.0012
allowance = 0.0012 in
Female
Male
+ 0.0008
+ 0.0000
a1 = 1.0000
in
a1 = 0.9992
in
− 0.0008
− 0.0000
allowance = 0.0008 in
+ 0.0008
+ 0.0000
a2 = 1.0000
in
a2 = 0.9992
in
− 0.0000
− 0.0008
126
SECTION 1– DESIGN FOR SIMPLE STRESSES
allowance = 0.0008 in
(c) Sketch
126.
A simple beam 12 ft. long is to support a concentrated load of 10 kips at the
midpoint with a design factor of at least 2.5 based on yield strength. (a) What is
the size and weight of the lightest steel (C1020, as rolled) I-beam that can be
used? (b) Compute its maximum deflection. (c) What size beam should be used if
the deflection is not to exceed ¼ in.?
Solution:
N = 2. 5
(a) For AISI C1020, as rolled, Table AT 7
s y = 48 ksi
s=
Mc s y
=
I
N
127
SECTION 1– DESIGN FOR SIMPLE STRESSES
48
= 19 ksi
2.5
FL (10 )(144)
M=
=
= 360 lb − in
4
4
M
s=
Z
Z = section modulus
M 360
Z=
=
= 18.95 in 3
s
19
From Table B-3, Strength of Material by F. Singer, 2nd Edition
Select 10I35 Section Index
Unsupported length = 12 ft
Weight per foot = 35 lb
Section Modulus = Z = 29.2 in 3
s=
I = 145.8 in 4 , moment of inertia
Size (Depth) = 10.0 in
Weight of beam = (35)(12) = 420 lb
FL3
48 EI
E = 30,000 ksi
(b) δ =
δ=
(10 )(144 )3 = 0.142 in
48(30,000 )(145.8)
FL3
48 EI
(10)(144 )3
0.25 =
48(30,000)I
I = 82.9 in 4
From Table B-3, Strength of Material by F. Singer, 2nd Edition
Select 10I35 Section Index
Unsupported length = 12 ft
Weight per foot = 35 lb
I = 145.8 in 4 , moment of inertia
Size (Depth) = 10.0 in
(c) δ =
127.
A 10-in., 35-lb. I-beam is used as a simple beam, supported on 18-ft. centers, and
carrying a total uniformly distributed load of 6000 lb. Determine the maximum
stress and the maximum deflection.
Solution:
128
SECTION 1– DESIGN FOR SIMPLE STRESSES
6000 + 630
= 30.7 lb in
(18)(12)
Table B.3, From Strength of Materials, F.L. Singer, 2nd Edition
For 10-in., 35-lb. I-beam
I = 145.8 in 4
Z = 29.2 in 3
w=
M max
Z
2
wL2 (30.7 )(216 )
M max =
=
= 179,042 lb − in = 179 kips − in
8
8
179
s max =
= 6.13 ksi
29.2
5FL3
δ max =
384 EI
E = 30,000 ksi
F = wL = (30.7 )(216) = 6631 lbs = 6.631 kips
s max =
5(6.631)(216 )
= 0.20 in
384(30,000)(145.8)
3
δ max =
128.
The same as 127, except the beam is a cantilever.
Solution:
wL2 (30.7 )(216)
=
= 716,170 lb − in = 716.17 kips − in
8
2
M
716.17
= max =
= 24.53 ksi
Z
29.2
2
M max =
s max
129
SECTION 1– DESIGN FOR SIMPLE STRESSES
w=
6000 + 630
= 30.7 lb in
(18)(12)
(30.7 )(216) = 1.91 in
wL4
=
8EI 8(30,000 )(145.8)
4
δ max =
129.
Two equal angles, placed back to back as shown, act as a simple beam and are to
support a load of F = 2,000 lb .; L = 40 in .; a = 15 in . What size angles should
be used if the maximum stress is not to exceed 20 ksi? The stress due to the
weight of the angles is negligible.
Problems 129, 130.
Solution:
Table AT 2
M
s=
Z
Fab
M=
L
a = 15 in
L = 40 in
b = L − a = 40 − 15 = 25 in
(2 )(15)(25) = 18.75 kips − in
M=
40
M 18.75
Z=
=
= 0.9375 in 3
s
20
1
For each angles, Z = (0.9375) = 0.46875 in 3
2
From Strength of Materials, F.L. Singer, 2nd Edition
Table B-5
Say size 3” x 3”, thickness = ¼ in
I
Z = = 0.58 in3
c
130.
The same as 129, except that a rolled T-section is to be used.
Solution:
From Table AT 1, No. 6
130
SECTION 1– DESIGN FOR SIMPLE STRESSES
aH 2 + bt 2
2(aH + bt )
c2 = H − c1
c1 =
Bt 3
ah 3
+ (Bt )d 2 +
+ (ah )e 2
12
12
say a = t
h=B
B = 4t
H = h+t = B+t
b = B−a
2
t (B + t ) + (B − t )t 2
c1 =
2[t (B + t ) + (B − t )t ]
Ix =
t (B + t ) + (B − t )t 2 t (5t ) + (3t )t 2
=
= 1.75t
2(2 Bt )
4 4t 2
c2 = H − c1
H = B + t = 5t
c2 = 5t − 1.75t = 3.25t
2
2
c1 =
Ix =
( )
Bt 3
ah 3
+ (Bt )d 2 +
+ (ah )e 2
12
12
d = c1 −
t
= 1.75t − 0.5t = 1.25t
2
a=t
h = B = 4t
h
e = c2 − = 3.25t − 0.5(4t ) = 1.25t
2
(4t )t 3 + (4t )t (1.25t )2 + t (4t )3 + t (4t )(1.25t )2 = 18.17t 4
Ix =
12
12
Mc2
s=
I
M = 18.75 kips − in
131
SECTION 1– DESIGN FOR SIMPLE STRESSES
s = 20 =
(18.75)(3.25t )
18.17t 4
t = 0.55 in
9
say t = in
16
1
9
B = 4t = 4  = 2 in
4
 16 
1 9
H = B + t = 2 + = 2.8125 in
4 16
say H = 3 in
″
″
1
9
Size: 2 × 3′′ ×
T section
4
16
- end -
132
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
VARYING STRESSES – NO CONCENTRATION
DESIGN PROBLEMS
141.
The maximum pressure of air in a 20-in. cylinder (double-acting air compressor)
is 125 psig. What should be the diameter of the piston rod if it is made of AISI
3140, OQT at 1000 F, and if there are no stress raisers and no column action? Let
N = 1.75 ; indefinite life desired. How does your answer compare with that
obtained for 4?
Solution:
For AISI 3140, OQT 1000 F
su = 153 ksi
s y = 134 ksi
sn = 0.5su = 0.5(153) = 76.5 ksi
For axial loading, with size factor
sn = 0.5su = (0.8)(0.85)(76.5) = 52 ksi
Soderberg line
1 sm sa
= +
N s y sn
For double-acting
π 
2
Fmax = F = pA = (125) (20 ) = 39,270 lb = 39.27 kips
4
Fmin = − F = −39.27 kips
sm = 0
4 F 4(39.27 ) 50
sa =
=
= 2
π d2
πd2
d
 50 
 2
1
1
d
=
= 0+  
N 1.75
52
d = 1.2972 in
5
say d = 1 in
16
comparative to Problem 4.
142.
A link as shown is to be made of AISI 2330, WQT 1000 F. The load F = 5 kips
is repeated and reversed. For the time being, ignore stress concentrations. (a) If
its surface is machined, what should be its diameter for N = 1.40 . (b) The same
as (a), except that the surface is mirror polished. What would be the percentage
saving in weight? (c) The same as (a), except that the surface is as forged.
Page 1 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Prob. 142 – 144
Solution:
For AISI 2330, WQT 1000 F
su = 105 ksi
s y = 85 ksi
sn = 0.5su = 0.5(105) = 52.5 ksi
sm = 0
4F
4(5)
20
sa =
=
=
2
2
πd
πd
π d2
Soderberg line
1 sm sa
= +
N s y sn
1
s
= 0+ a
N
sn
s
sa = n
N
Size factor = 0.85
Factor for axial loading = 0.80
(a) Machined surface
Surface factor = 0.85 (Fig. AF 5)
sn = 0.5su = (0.80)(0.85)(0.85)(52.5) ksi = 30.345 ksi
20
30.345
sa =
=
2
πD
1 .4
D = 0.542 in
9
say D = in
16
(b) Mirror polished surface
Surface factor = 1.00 (Fig. AF 5)
sn = 0.5su = (0.80)(0.85)(1.00)(52.5) ksi = 35.7 ksi
20
35.7
sa =
=
2
πD
1 .4
Page 2 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
D = 0.5 in
2
2
 9  1
  − 
16
2
Savings in weight =   2  (100% ) = 21%
9
 
 16 
(c) As forged surface
Surface factor = 0.40 (Fig. AF 5)
sn = 0.5su = (0.80)(0.85)(0.40 )(52.5) ksi = 14.28 ksi
20 14.28
sa =
=
π D2
1 .4
D = 0.79 in
3
say D = in
4
143.
The same as 142, except that, because of a corrosive environment, the link is
made from cold-drawn silicon bronze B and the number of reversals of the load
is expected to be less than 3 x 107.
Solution:
For cold-drawn silicon bronze, Type B.
sn = 30 ksi at 3 x 108
s y = 69 ksi
su = 93.75 ksi
0.085
 3 × 108 

sn at 3 x 10 = (30 )
= 36.5 ksi
7 
3
×
10


sn = (0.80)(0.85)(36.5) = 24.82 ksi
20
24.82
sa =
=
2
πD
1 .4
D = 0.60 in
5
say D = in
8
7
144.
The same as 142, except that the link is made of aluminum alloy 2024-T4 with a
minimum life of 107 cycles.
Solution:
For AA 2024-T4
s y = 47 ksi
su = 68 ksi
sn = 20 ksi at 5 x108
Page 3 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
0.085
 5 × 108 

sn at 10 (20 )
= 27.9 ksi
7 
 10 
sn = (0.80)(0.85)(27.9) = 19 ksi
20
19
sa =
=
2
π D 1 .4
D = 0.685 in
11
say D = in
16
7
145.
A shaft supported as a simple beam, 18 in. long, is made of carburized AISI 3120
steel (Table AT 10). With the shaft rotating, a steady load of 2000 lb. is appliled
midway between the bearings. The surfaces are ground. Indefinite life is desired
with N = 1.6 based on endurance strength. What should be its diameter if there
are no surface discontinuities?
Solution:
For AISI 3120 steel, carburized
sn = 90 ksi
s y = 100 ksi
su = 141 ksi
Size Factor = 0.85
Surface factor (ground) = 0.88
sn = (0.85)(0.88)(90) = 67.32 ksi
sm = 0
32 M
sa =
π D3
FL (2000 )(18)
M=
=
= 9000 in − lb = 9.0 in − kips
4
4
Soderberg line
1 sm sa
= +
N s y sn
Page 4 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1
s
= 0+ a
N
sn
s
sa = n
N
32(9 ) 67.32
=
πD3
1 .6
D = 1.2964 in
1
say D = 1 in
4
146.
(a) A lever as shown with a rectangular section is to be designed for indefinite
life and a reversed load of F = 900 lb . Find the dimensions of a section without
discontinuity where b = 2.8t and L = 14 in . for a design factor of N = 2 . The
material is AISI C1020, as rolled, with an as-forged surface. (b) compute the
dimensions at a section where e = 4 in .
Problems 146, 147
Solution:
For AISI C1020, as rolled
su = 65 ksi
s y = 48 ksi
s n = 0.5su = 32.5 ksi
Surface factor (as forged) = 0.55
(a) sm = 0
Mc
sa =
I
3
tb3 t (2.8t )
I=
=
= 1.8293t 4
12
12
b 2.8t
c= =
= 1.4t
2
2
M = FL = (900)(14) = 12,600 in − lb = 12.6 in − kips
(12.6)(1.4t ) = 9.643
sa =
1.8293t 4
t3
sn = (0.85)(0.55)(32.5) = 15.20 ksi
Page 5 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Soderberg line
1 sm sa
= +
N s y sn
1
s
= 0+ a
N
sn
s
sa = n
N
9.643 15.20
=
t3
2
t = 1.08 in
b = 2.8t = 2.8(1.08) = 3.0 in
1
say t = 1 in , b = 3.0 in
16
(b) M = Fe = (900)(4 ) = 3,600 in − lb = 3.6 in − kips
(3.6)(1.4t ) = 2.755
sa =
18293t 4
t3
2.755 15.20
=
t3
2
t = 0.713 in
b = 2.8t = 2.8(0.713) = 1.996 in
23
say t =
in , b = 2 in
32
147.
The same as 146, except that the reversal of the load are not expected to exceed
105 (Table AT 10).
Solution:
sn = 32.5 ksi
0.085
 106 
sn at 10 = (32.5) 5 
= 39.5 ksi
 10 
sn = (0.85)(0.55)(39.5) = 18.5 ksi
5
sn
N
9.643 18.5
=
t3
2
(a) sa =
Page 6 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
t = 1.014 in
b = 2.8t = 2.8(1.014) = 2.839 in
13
say t = 1 in , b = 2 in
16
sn
N
2.755 18.5
=
t3
2
(b) sa =
t = 0.6678 in
b = 2.8t = 2.8(0.6678) = 1.870 in
11
7
say t = in , b = 1 in
16
8
148.
A shaft is to be subjected to a maximum reversed torque of 15,000 in-lb. It is
machined from AISI 3140 steel, OQT 1000 F (Fig. AF 2). What should be its
diameter for N = 1.75 ?
Solution:
For AISI 3140 steel, OQT 1000 F
su = 152 ksi
s y = 134 ksi
sn = 0.5su = 76 ksi
For machined surface,
Surface factor = 0.78
Size factor = 0.85
sns = (0.6)(0.85)(0.78)(134) = 53.3 ksi
s ys = 0.6 s y = 0.6(134 ) = 80.4 ksi
1 sms sas
=
+
N s ys sns
sms = 0
16T
sas =
πD3
T = 15 in − kips
16(15) 240
sas =
=
πD3 πD3
1
s
= 0 + as
N
sns
s
sas = ns
N
Page 7 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
240 53.3
=
πD 3 1.75
D = 1.3587 in
3
say D = 1 in
8
149.
The same as 148, except that the shaft is hollow with the outside diameter twice
the inside diameter.
Solution:
Do = 2 Di
16TDo
16(15)(2 Di )
32
sas =
=
=
4
4
4
4
π (Do − Di ) π (2 Di ) − Di
πDi3
s
sas = ns
N
32 53.3
=
πDi3 1.75
[
]
Di = 0.694 in
11
3
say Di = in , Do = 1 in
16
8
150.
The link shown is machined from AISI 1035 steel, as rolled, and subjected to a
repeated tensile load that varies from zero to 10 kips; h = 2b . (a) Determine these
dimensions for N = 1.40 (Soderberg) at a section without stress concentration.
(b) How much would these dimensions be decreased if the surfaces of the link
were mirror polished?
Problems 150, 151, 158.
Solution:
For AISI 1035, steel as rolled
su = 85 ksi
s y = 55 ksi
sn = 0.5su = 42.5 ksi
Page 8 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1
(10 + 0 ) = 5 kips
2
1
Fa = (10 − 0 ) = 5 kips
2
F
5
10
sm = m =
= 2
2
bh 1.5b
3b
F
5
10
sa = a =
= 2
2
bh 1.5b
3b
(a) Soderberg line
1 sm sa
= +
N s y sn
For machined surface,
Factor = 0.88
Fm =
Size factor = 0.85
sn = (0.80)(0.85)(0.88)(42.5) = 25.4 ksi
1
10
10
= 2
+ 2
1.40 3b (55) 3b (25.4 )
b = 0.5182 in
9
say b = in
16
27
h = 1.5b =
in
32
(b) Mirror polished,
Factor = 1.00
Size factor = 0.85
sn = (0.80)(0.85)(1.00)(42.5) = 28.9 ksi
1
10
10
= 2
+ 2
1.40 3b (55) 3b (28.9 )
b = 0.4963 in
1
say b = in
2
3
h = 1.5b = in
4
151.
The same as 150, except that the link operates in brine solution. (Note: The
corroding effect of the solution takes precedence over surface finish.)
Page 9 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Solution:
Table AT 10, in brine, AISI 1035,
sn = 24.6 ksi
s y = 58 ksi
sn = (0.80)(0.85)(24.6) = 16.73 ksi
1
10
10
= 2
+ 2
1.40 3b (55) 3b (16.73)
b = 0.60 in
5
say b = in
8
15
h = 1.5b = in
16
152.
The simple beam shown, 30-in. long ( = a + L + d ), is made of AISI C1022 steel,
as rolled, left a forged. At a = 10 in , F1 = 3000 lb. is a dead load. At
d = 10 in , F2 = 2400 lb. is repeated, reversed load. For N = 1.5 , indefinite life,
and h = 3b , determine b and h . (Ignore stress concentration).
Problem 152, 153
Solution:
For AISI C1022, as rolled
su = 72 ksi
s y = 52 ksi
sn = 0.5su = 36 ksi
For as forged surface
Figure AF 5, factor = 0.52
Size factor = 0.85
sn = (0.85)(0.52)(36) = 16 ksi
Loading:
Page 10 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
∑M
A
=0
10(3000) + 20(2400 ) = 30 R2
R2 = 2600 lb
∑F
V
=0
R1 + R2 = F1 + F2
R1 + 2600 = 3000 + 2400
R1 = 2800 lb
Shear Diagram
M C1 = (2800 )(10 ) = 28,000 in − lb = 28 in − kips
M D1 = (2600 )(10 ) = 26,000 in − lb = 26 in − kips
Then
Loading
∑M
A
=0
10(3000) + 30 R2 = 20(2400 )
R2 = 600 lb
∑F
V
=0
R1 + F2 = F1 + R2
R1 + 2400 = 3000 + 600
R1 = 1200 lb
Page 11 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Shear Diagram
M C2 = (1200 )(10 ) = 12,000 in − lb = 12 in − kips
M D2 = (600 )(10 ) = 6,000 in − lb = 6 in − kips
Then using
M max = M C1 = 28 in − kips
M min = M C2 = 12 in − kips
1
(M max + M min ) = 1 (28 + 12) = 20 in − kips
2
2
1
1
M a = (M max − M min ) = (28 − 12 ) = 8 in − kips
2
2
M c
M c
sm = m , sa = a
I
I
3
3
bh
b(3b )
I=
=
= 2.25b 4
12
12
h
c = = 1.5b
2
Mm
Ma
sm =
, sa =
3
1.5b
1.5b3
1 sm sa
= +
N s y sn
Mm =
 20   8 

 

1  1.5b3   1.5b3 
=
+
1.5
52
16
b = 0.96 in
say b = 1 in
h = 3b = 3 in
153.
The same as 152, except that the cycles of F2 will not exceed 100,000 and all
surfaces are machined.
Solution:
Page 12 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
0.085
 106 
(
)
sn at 10 cycles = 36  5 
= 43.8 ksi
 10 
su = 72 ksi
Machined surface, factor = 0.90
sn = (0.85)(0.90)(43.8) = 33.5 ksi
5
 20   8 

 

1  1.5b3   1.5b3 
=
+
1.5
52
33.5
b = 0.8543 in
7
say b = in
8
5
h = 3b = 2 in
8
154.
A round shaft, made of cold-finished AISI 1020 steel, is subjected to a variable
torque whose maximum value is 6283 in-lb. For N = 1.5 on the Soderberg
criterion, determine the diameter if (a) the torque is reversed, (b) the torque varies
from zero to a maximum, (c) the torque varies from 3141 in-lb to maximum.
Solution:
For AISI 1020, cold-finished
su = 78 ksi
s y = 66 ksi
sn = 0.5su = 39 ksi
size factor = 0.85
sns = (0.6)(0.85)(39) = 20 ksi
s ys = 0.6 s y = 0.6(66 ) = 40 ksi
1 sms sas
=
+
N s ys sns
(a) Reversed torque
sms = 0
16T
sas =
πD3
T = 6283 in − lb
16(6283) 32,000
32
sas =
=
psi = 3 ksi
3
3
πD
D
D
1
sas
= 0+
N
sns
Page 13 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
 32 
 3
1
D 
= 0+ 
1.5
20
D = 1.34 in
3
say D = 1 in
8
(b) Tmin = 0 , Tmax = 6283 in − lb
1
Tm = (6283) = 3141 in − lb
2
1
Ta = (6283) = 3141 in − lb
2
16(3141) 16,000
16
=
psi = 3 ksi
3
3
πD
D
D
16(3141) 16,000
16
=
sas =
psi = 3 ksi
3
3
πD
D
D
 16   16 

 

1  D3   D 3 
=
+
1.5
40
20
D = 1.22 in
1
say D = 1 in
4
sms =
(c) Tmin = 3141 in − lb , Tmax = 6283 in − lb
1
Tm = (6283 + 3141) = 4712 in − lb
2
1
Ta = (6283 − 3141) = 1571 in − lb
2
16(4712 ) 24,000
24
=
psi = 3 ksi
3
3
πD
D
D
16(1571) 8,000
8
sas =
psi = 3 ksi
=
3
3
πD
D
D
 24   8 

 

1  D3   D 3 
+
=
1.5
40
20
D = 1.145 in
5
say D = 1 in
32
sms =
Page 14 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
CHECK PROBLEMS
155.
A simple beam 2 ft. long is made of AISI C1045 steel, as rolled. The dimensions
of the beam, which is set on edge, are 1 in. x 3 in. At the midpoint is a repeated,
reversed load of 4000 lb. What is the factor of safety?
Solution:
For AISI C1045, as rolled
su = 96 ksi
s y = 59 ksi
sn = 0.5su = 0.5(96) = 48 ksi
size factor = 0.85
sn = (0.85)(48) = 40.8 ksi
1 sm sa
= +
N s y sn
sm = 0
6M
sa = 2
bh
h = 3 in
b = 1 in
FL (4000 )(24 )
M=
=
= 24,000 in − lb = 24 in − kips
4
4
6(24 )
sa =
= 16 ksi
(1)(3)2
1
16
= 0+
N
40.8
N = 2.55
156.
The same as 155, except that the material is normalized and tempered cast steel,
SAE 080.
Solution:
Table AT 6
sn′ = 35 ksi
s y = 40 ksi
sn = (0.85)(35) = 29.75 ksi
1
16
= 0+
N
29.75
N = 1.86
157.
A 1 ½-in. shaft is made of AISI 1045 steel, as rolled. For N = 2 , what repeated
and reversed torque can the shaft sustain indefinitely?
Page 15 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Solution:
For AISI 1045, as rolled
su = 96 ksi
s y = 59 ksi
sn′ = 0.5su = 0.5(96) = 48 ksi
sns = (0.6)(0.85)(48) = 24.48 ksi
s ys = 0.6 s y = (0.6 )(59 ) = 35.4 ksi
1 sms sas
=
+
N s ys sns
sms = 0
1
s
= 0 + as
2
24.48
sas = 12.24 ksi
16T
sas =
= 12.24
πD 3
T = 8 in − kips
VARIABLE STRESSES WITH STRESS CONCENTRATIONS
DESIGN PROBLEMS
158.
The load on the link shown (150) is a maximum of 10 kips, repeated and
reversed. The link is forged from AISI C020, as rolled, and it has a ¼ in-hole
drilled on the center line of the wide side. Let h = 2b and N = 1.5 . Determine b
and h at the hole (no column action) (a) for indefinite life, (b) for 50,000
repetitions (no reversal) of the maximum load, (c) for indefinite life but with a
ground and polished surface. In this case, compute the maximum stress.
Solution:
For AISI C1020, as rolled
su = 65 ksi
s y = 48 ksi
sn = 0.5su = 0.5(65) = 32.5 ksi
For as forged surface
Surface factor = 0.55
Size factor = 0.85
sn = (0.80)(0.85)(0.55)(32.5) = 12.2 ksi
1 sm K f sa
= +
N sy
sn
Page 16 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Fig. AF 8, b h > 1
Assume K t = 3.5
d 1
Figure AF 7, r = = in = 0.125 in
2 8
a = 0.01 in
1
1
q=
=
= 0.926
a
0.01
1+
1+
r
0.125
(
)
K f = q K t − 1 + 1 = 0.926(3.5 − 1) + 1 = 3.3
sm = 0
F
10
=
b(h − d ) b(2b − 0.25)
K s
1
(a)
= 0+ f a
N
sn
1
(3.3)(10)
= 0+
1.5
b(2b − 0.25)(12.2)
2
2b − 0.25b = 4.06
b 2 − 0.125b − 2.03 = 0
b = 1.489 in
1
say b = 1 in , h = 2b = 3 in
2
sa =
(b) For 50,000 repetitions or 50,000 cycles
 106 

sn = (12.2 )
4 
 5 ×10 
(log K ) 3
0.085
= 15.74 ksi
(log 3.3 ) 3
f
n
(5 ×104 )
=
= 2.0
log K
10log 3.3
10 f
1 K fl sa
=
N
sn
1
(2.0)(10)
=
1.5 b(2b − 0.25)(15.74)
2b 2 − 0.25b = 1.906
b 2 − 0.125b − 0.953 = 0
b = 1.04 in
1
1
say b = 1 in , h = 2b = 2 in
16
8
K fl =
(c) For indefinite life, ground and polished surface
Surface factor = 0.90
Page 17 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
sn = (0.80)(0.85)(0.90)(32.5) = 20 ksi
1 K f sa
=
N
sn
1
(3.3)(10)
=
1.5 b(2b − 0.25)(20)
b 2 − 0.125b − 1.2375 = 0
b = 1.18 in
3
3
say b = 1 in , h = 2b = 2 in
16
8
Kf F
Maximum stress =
b(h − d )
b h > 1 , d h = 0.25 2.375 = 0.105
Figure AF 8
K t = 3.5
K f = q (K t − 1) + 1 = 0.926(3.5 − 1) + 1 = 3.315
smax =
159.
(3.315)(10)
= 13.14 ksi
1.1875(2.375 − 0.25)
A connecting link as shown, except that there is a 1/8-in. radial hole drilled
through it at the center section. It is machined from AISI 2330, WQT 1000 F, and
it is subjected to a repeated, reversed axial load whose maximum value is 5 kips.
For N = 1.5 , determine the diameter of the link at the hole (a) for indefinite life;
(b) for a life of 105 repetitions (no column action). (c) In the link found in (a)
what is the maximum tensile stress?
Problem 159
Solution:
For AISI 2330, WQT 1000 F
su = 135 ksi
s y = 126 ksi
sn = 0.5su = 0.5(135) = 67.5 ksi
For machined surface, Fig. AF 7, surface factor = 0.80
Size factor = 0.85
sn = (0.80)(0.85)(0.80)(67.5) = 36.72 ksi
Page 18 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1 sm K f sa
= +
N sy
sn
Fig. AF 8, b h > 1
Assume K t = 2.5
d 1
Figure AF 7, r = = in = 0.0625 in
2 16
a = 0.0025 in
1
1
q=
=
= 0.96
a
0.0025
1+
1+
r
0.0625
K f = q(K t − 1) + 1 = 0.96(2.5 − 1) + 1 = 2.44
(a) Indefinite life, K f = 2.44
sn = 36.72 ksi
sm = 0
F
4F
4(5)
20
sa =
=
=
=
2
2
2
πD
πD − 4 Dd
 1  πD − 0.5 D
πD 2 − 4 D 
− Dd
4
8
K s
1
= 0+ f a
N
sn
1
(2.44)(20)
=
1.5 36.72(πD 2 − 0.5D )
πD 2 − 0.5D = 2.00
D = 0.88 in
7
say D = in
8
(b) For a life of 105 repetitions or cycles
 106 
sn = (36.72 ) 5 
 10 
(log K ) 3
0.085
= 44.66 ksi
(log 2.4 ) 3
f
n
(105 )
K fl = log K f =
10log 2.44
10
1 K fl sa
=
N
sn
1
(1.81)(20)
=
1.5 44.66(πD 2 − 0.5 D )
πD 2 − 0.5D = 1.216
D = 0.71 in
Page 19 of 62
= 1.81
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
say D =
3
in
4
4K f F
πD 2 − 0.5D
7
d 0.125
D = in ,
=
= 0.14
8
D 0.875
Figure AF 8
K t = 2.6
(c) smax =
K f = q(K t − 1) + 1 = 0.96(2.6 − 1) + 1 = 2.54
smax =
160.
4(2.54)(5)
2
7
7
π   − 0.5 
8
8
= 25.82 ksi
A machine part of uniform thickness t = b 2.5 is shaped as shown and machined
all over from AISI C1020, as rolled. The design is for indefinite life for a load
repeated from 1750 lb to 3500 lb. Let d = b . (a) For a design factor of 1.8
(Soderberg), what should be the dimensions of the part? (b) What is the
maximum tensile stress in the part designed?
Problems 160, 161
Solution:
For AISI C1020, as rolled
su = 65 ksi
s y = 48 ksi
sn′ = 0.5su = 0.5(65) = 32.5 ksi
For machined surface
Surface factor = 0.90
Size factor = 0.85
sn = (0.80)(0.85)(0.90)(32.5) = 20 ksi
1 sm K f sa
= +
N sy
sn
(a) For flat plate with fillets
Figure AF 9
b d
r= =
3 3
Page 20 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
r 1
= = 0.333
d 3
h 2b
=
=2
d
b
K t = 1.65
a = 0.01 in
1
q=
≈ 1.0
a
1+
r
K f ≈ K t = 1.65
Fm
bt
Fa
sa =
bt
b
t=
2 .5
1
Fm = (3500 + 1750 ) = 2625 lb
2
1
Fa = (3500 − 1750 ) = 875 lb
2
2625
6562.5
sm =
=
b2
 b 
b

 2 .5 
875
2187.5
sa =
=
 b 
b2
b

 2 .5 
1
6562.5 (1.65)(2187.5)
=
+
1.8 48,000b 2
20,000b 2
b = 0.7556 in
or b = 0.75 in
b
0.75
t=
=
= 0.3 in
2 .5 2 .5
sm =
For flat plate with central hole
Fig. AF 8, b h > 1 , d h = b 2b = 1 2
Assume K f ≈ K t = 2.9
Fm
Fm
F
=
= m
(h − d )t (2b − b )t bt
Fa
Fa
F
sa =
=
= a
(h − d )t (2b − b )t bt
sm =
Page 21 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
2625
6562.5
=
 b 
b2
b

 2 .5 
875
2187.5
sa =
=
 b 
b2
b

 2 .5 
1
6562.5 (2.9)(2187.5)
=
+
1.8 48,000b 2
20,000b 2
b = 0.904 in
15
or b = 0.9375 in = in
16
b
3
t=
= in
2 .5 8
15
d = b = in
16
15
3
15
use b = in , t = in , d = in
16
8
16
sm =
(b) smax = sm + K f sa
d 15
=
in
2 32
1
q=
= 0.98
0.01
1+
 15 
 
 32 
K t = 2.9
r=
K f = q(K t − 1) + 1 = 0.98(2.9 − 1) + 1 = 2.86
Fm 6562.5 6562.5
=
=
= 7467 psi
2
bt
b2
 15 
 
 16 
F 2187.5 2187.5
sa = a =
=
= 2489 psi
2
bt
b2
 15 
 
 16 
smax = sm + K f sa = 7467 + (2.86 )(2489 ) = 14,586 psi
sm =
162.
The beam shown has a circular cross section and supports a load F that
varies from 1000 lb to 3000 lb; it is machined from AISI C1020 steel, as
rolled. Determine the diameter D if r = 0.2 D and N = 2 ; indefinite life.
Page 22 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Problems 162 – 164.
Solution:
For AISI C1020, as rolled
su = 65 ksi
s y = 48 ksi
sn′ = 0.5su = 0.5(65) = 32.5 ksi
For machined surface
Surface factor = 0.90
Size factor = 0.85
sn = (0.85)(0.90 )(32.5) = 24.86 ksi
∑M
A
=0
12 F = 24 B
F = 2B
F
B=
2
F
A= B=
2
At discontinuity
6F
M=
= 3F
2
M max = 3(3000) in − lb = 9000 in − lb = 9 in − kips
M min = 3(1000) in − lb = 3000 in − lb = 3 in − kips
1
M m = (9 + 3) = 6 in − kips
2
1
M a = (9 − 3) = 3 in − kips
2
32 M
s=
πD3
Figure AF 12
D d =1.5d d =1.5
r d = 0.2d d = 0.2
K t = 1.42
assume K f ≈ K t = 1.42
Page 23 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1 sm K f sa
= +
N sy
sn
1 (32 )(6 ) (1.42 )(32 )(3)
=
+
2 48πD 3
24.86πD 3
D = 1.821 in
13
say D = 1 in
16
At maximum moment
12 F
M=
= 6F
2
M max = 6(3000) in − lb = 18000 in − lb = 18 in − kips
M min = 6(1000) in − lb = 6000 in − lb = 6 in − kips
1
M m = (18 + 6 ) = 12 in − kips
2
1
M a = (18 − 6 ) = 6 in − kips
2
32 M
s=
πD3
K f = 1.00
1 sm K f sa
= +
N sy
sn
1 (32 )(12 ) (1.0 )(32 )(6 )
=
+
2 48πD 3
24.86πD 3
D = 1.4368 in
13
Therefore use D = 1 in
16
164. The shaft shown is machined from C1040, OQT 1000 F (Fig. AF 1). It is
subjected to a torque that varies from zero to 10,000 in-lb. ( F = 0 ). Let r = 0.2 D
and N = 2 . Compute D . What is the maximum torsional stress in the shaft?
Solution:
For C1040, OQT 1000 F
su = 104 ksi
s y = 72 ksi
Page 24 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
sn′ = 0.5su = 0.5(104) = 52 ksi
For machined surface
Surface factor = 0.85
Size factor = 0.85
sns = (0.60)(0.85)(0.85)(52) = 22.5 ksi
1
Ta = Tm = (10,000 ) = 5000 in − lb = 5 in − kips
2
s ys = 0.6 s y = 0.6(72 ) = 43.2 ksi
sms = sas =
16T
πD3
K fs sas
1 sms
=
+
N s ys
sns
Figure AF 12
D d =1.5d d =1.5
r d = 0.2d d = 0.2
K ts = 1.2
assume K fs ≈ K ts = 1.2
(16)(5) + (1.2)(16)(5)
1
=
2 43.2πD 3
22.5πD 3
D = 1.5734 in
9
say D = 1 in
16
smax = sm + K f sa
smax =
165.
(16)(5)
 9
π 1 
 16 
3
+
(1.2)(16)(5) = 14.686 ksi
 9
π 1 
 16 
3
An axle (nonrotating) is to be machined from AISI 1144, OQT 1000 F, to the
proportions shown, with a fillet radius r ≈ 0.25 D ; F varies from 400 lb to 1200
lb.; the supports are to the left of BB not shown. Let N = 2 (Soderberg line). (a)
At the fillet, compute D and the maximum tensile stress. (b) Compute D at
section BB. (c) Specify suitable dimensions keeping the given proportions, would
a smaller diameter be permissible if the fillet were shot-peened?
Problems 165 – 167
Page 25 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Solution:
For AISI 1144, OQT 1000 F
su = 118 ksi
s y = 83 ksi
sn′ = 0.5su = 59 ksi
For machined surface
Surface factor = 0.83
Size factor = 0.85
sn = (0.85)(0.83)(59) = 41.62 ksi
(a) At the fillet
D d =1.5d d =1.5
r d = 0.25d d = 0.25
K t = 1.35
assume K f ≈ K t = 1.35
M = 6F
M max = 6(1200) in − lb = 7200 in − lb = 7.2 in − kips
M min = 6(400) in − lb = 2400 in − lb = 2.4 in − kips
1
M m = (7.2 + 2.4 ) = 4.8 in − kips
2
1
M a = (7.2 − 2.4 ) = 2.4 in − kips
2
32 M
s=
πD3
1 sm K f sa
= +
N sy
sn
1 (32 )(4.8) (1.35)(32 )(2.4 )
=
+
2
83πD 3
41.62πD 3
D = 1.4034 in
7
say D = 1 in
16
(b) At section BB,
M = 30 F
M max = 30(1200) in − lb = 36000 in − lb = 36 in − kips
M min = 30(400) in − lb = 12000 in − lb = 12 in − kips
1
M m = (7.2 + 2.4 ) = 4.8 in − kips
2
1
M a = (7.2 − 2.4 ) = 2.4 in − kips
2
Page 26 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
32 M
πD3
K f = 1 .0
s=
1 sm K f sa
= +
N sy
sn
(32)(36) + (1.0)(32)(12)
1
=
2 83π (1.5 D )3 41.62π (1.5 D )3
D = 1.6335 in
11
say D = 1 in
16
(c) Specified dimension:
D = 2 in , 1.5 D = 3 in
A smaller diameter is permissible if the fillet were shot-peened because of increased
fatigue strength.
166.
A pure torque varying from 5 in-kips to 15 in-kips is applied at section C.
( F = 0 ) of the machined shaft shown. The fillet radius r = D 8 and the torque
passes through the profile keyway at C. The material is AISI 1050, OQT 1100 F,
and N = 1.6 . (a) What should be the diameter? (b) If the fillet radius were
increased to D 4 would it be reasonable to use a smaller D ?
Solution:
Tmax = 15 in − kips
Tmin = 5 in − kips
1
Tm = (15 + 5) = 10 in − kips
2
1
Ta = (15 − 5) = 5 in − kips
2
For AISI 1050, OQT 1100 F
su = 101 ksi
s y = 58.5 ksi
sn = 0.5su = 0.5(101) = 50.5 ksi
Page 27 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
For machined surface
Surface factor = 0.85
Size factor = 0.85
sns = (0.60)(0.85)(0.85)(50.5) = 21.9 ksi
(a) At the fillet
r d =r D = =1 8
D d =1.5
K ts = 1.3
assume K fs ≈ K ts = 1.3
At the key profile
K fs = 1.6
use K fs = 1.6
s ys = 0.6 s y = 0.6(58.5) = 35.1 ksi
1 sms K fs sas
=
+
N s ys
sns
(16)(10) + (1.6)(16)(5)
1
=
1.6 35.1πD 3
21.9πD 3
D = 1.7433 in
3
say D = 1 in
4
(b) r = D 4
r D = 0.25
D d =1.5
Figure AF 12
K ts = 1.18
K fs ≈ K ts = 1.18 < 1.6
Therefore, smaller D is not reasonable.
170.
The beam shown is made of AISI C1020 steel, as rolled; e = 8 in . The load F is
repeated from zero to a maximum of 1400 lb. Assume that the stress
concentration at the point of application of F is not decisive. Determine the
depth h and width t if h ≈ 4t ; N = 1.5 ± 0.1 for Soderberg line. Iteration is
necessary because K f depends on the dimensions. Start by assuming a logical
K f for a logical h (Fig. AF 11), with a final check of K f . Considerable
estimation inevitable.
Page 28 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Problem 170
Solution:
1
F
2
At the hole
A= B=
F
M = eB = (8)  = 4 F
2
M max = 4 F
M min = 0
1
M m = (4 F ) = 2 F = 2(1.4 ) = 2.8 in − kips
2
1
M a = (4 F ) = 2 F = 2(1.4 ) = 2.8 in − kips
2
Mc
s=
I
(h − 2d )3 t
I=
12
1
d = in = 0.5 in
2
1 11
c = 1 +   = 1.75 in
2 22
For AISI C1020, as rolled
su = 65 ksi
s y = 48 ksi
sn = 0.5su = 0.5(65) = 32.5 ksi
Size factor = 0.85
sn = (0.85)(32.5) = 27.62 ksi
Fig. AF 7, c d = 1.75 0.5 = 3.5 > 0.5
Page 29 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Assume K t = 3.5
 1  1 
r =    = 0.25 in
 2  2 
a = 0.010 in
1
1
q=
=
= 0.962
a
0.010
1+
1+
r
0.25
K f = q (K t − 1) + 1 = 0.962(3.5 − 1) + 1 = 3.4
1 sm K f sa
= +
N sy
sn
1 12(2.8)(1.75) (3.4 )(12 )(2.8)(1.75)
=
+
3
1.5 48(h − 2d )3 t
27.62(h − 2d ) t
(h − 2d )3 t = 12.70
[h − 2(0.50)]3 t = 12.70
(4t − 1)3 t = 12.70
t = 0.8627 in
7
say t = in
8
h = 4t = 3.5 in
1 1 1
h > 1 + 1 + in
2 2 2
h > 3.5 in
Figure AF 11, h d > 10
h = 10d = 10(0.50) = 5 in
1
d
= 2 = 0 .5
b 5 −1 1
2 2
Therefore K t = 3.5 , K f = 3.4
1
Use h = 5 in , t = 1 in
4
171.
Design a crank similar to that shown with a design factor of 1.6 ± 0.16 based on
the modified Goodman line. The crank is to be forged with certain surfaces
milled as shown and two ¼-in. holes. It is estimated that the material must be of
the order of AISI 8630, WQT 1100 F. The length L = 17 in. , a = 5 in. , and the
load varies form + 15 kips to –9 kips. (a) Compute the dimensions at section AB
with h = 3b . Check the safety of the edges (forged surfaces). (Iteration involves;
one could first make calculations for forged surfaces and then check safety at
holes.) (b) Without redesigning but otherwise considering relevant factors ,
Page 30 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
quantitatively discuss actions that might be taken to reduce the size; holes must
remain as located.
Problems 171-174.
Solution:
(a) AISI 8630, WQT 1100 F
su = 96 ksi
sn = 0.5su = 0.5(96) = 48 ksi
Size factor = 0.85
As-forged surface (Fig. AF I)
Surface factor = 0.4
sn = (0.85)(0.42)(48) = 17 ksi
Milled surface (Machined)
Surface factor = 0.85
sn = (0.85)(0.85)(48) = 34.68 ksi
At AB, machined
1 sm K f sa
= +
N su
sn
Figure AF 11
1
b = in = 0.5 in
2
1
d = in = 0.25 in
4
d 0.25
=
in = 0.5
b 0 .5
Assume K f = 3.50
q = 0.998
K f = q (K t − 1) + 1 = 0.998(3.5 − 1) + 1 = 3.495
Mc
I
3
(
h − 2d ) b
I=
12
s=
Page 31 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1 1
1
h 1  1  1  1 
− +    =  h − 1 +  = (4h − 4 + 1) = (4h − 3)
2 2  2  4  2 
4 8
8
h = 3b
1
M  (4h − 3)
8
s=  
3

 1 
h − 2 4  b
 

12
3
M (12b − 3)
2
s=
(3b − 0.5)3 b
4.5M (4b − 1)
s=
(3b − 0.5)3 b
M = F (L − a )
M max = (15)(17 − 5) = 180 in − kips
c=
M min = (− 9)(17 − 5) = 108 in − kips
1
M m =  (180 − 108) = 36 in − kips
2
1
M a =  (180 + 108) = 144 in − kips
2
1 sm K f sa
= +
N su
sn
1
4.5(36 )(4b − 1) (3.495)(4.5)(144 )(4b − 1)
=
+
3
1.6 96(3b − 0.5)3 b
34.68(3b − 0.5) b
(4b − 1) = 1
(3b − 0.5)3 b 107.2
(3b − 0.5)3 b = 107.2
(4b − 1)
b = 2.6 in
5
say b = 2 in
8
7
h = 3b = 7 in
8
Checking at the edges (as forged)
M max = (15)(17 ) = 255 in − kips
M min = (− 9)(17 ) = −153 in − kips
1
M m =  (255 − 153) = 51 in − kips
2
Page 32 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1
M a =  (255 + 153) = 204 in − kips
2
6M 6M 2 M
s= 2 = 3 = 3
bh
9b
3b
K f ≈ 1 .0
1 sm K f sa
= +
N su
sn
1
2(51) (1.0)(2)(204)
= 3
+
1.6 3b (96)
3b3 (17 )
b = 2.373 in
3
say b = 2 in
8
5
3
since b = 2 in > 2 in , ∴ safe.
8
8
(c) Action: reduce number of repetitions of load.
CHECK PROBLEMS
173.
For the crank shown, L = 15 in , a = 3 in , d = 4.5 in , b = 1.5 in . It is as forged
from AISI 8630, WQT 1100 F, except for machined areas indicated. The load F
varies from +5 kips to –3 kips. The crank has been designed without detailed
attention to factors that affect its endurance strength. In section AB only,
compute the factor of safety by the Soderberg criterion. Suppose it were desired
to improve the margin of safety, with significant changes of dimensions
prohibited, what various steps could be taken? What are your particular
recommendations?
Solution:
For as forged surface
sn = 17 ksi
For machined surface
sn = 34.68 ksi
s n = 72 ksi
In section AB, machined
Page 33 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
M = F (L − a )
M max = (+ 5)(15 − 3) = 60 in − kips
M min = (− 3)(15 − 3) = −36 in − kips
1
M m =  (60 − 36 ) = 12 in − kips
2
1
M a =  (60 + 36 ) = 48 in − kips
2
d = h = 4.5 in , b = 1.5 in
h
=3
b
4.5M (4b − 1)
s=
(3b − 0.5)3 b
4.5(12 )[4(1.5) − 1]
sm =
= 2.8125 ksi
[3(1.5) − 0.5]3 (1.5)
4.5(48)[4(1.5) − 1]
sa =
= 11.25 ksi
[3(1.5) − 0.5]3 (1.5)
1 sm K f sa
= +
N sy
sn
K f = 3.495 from Problem 171.
1 2.8125 (3.495)(11.25)
=
+
N
72
34.68
N = 0.85 < 1 , unsafe
To increase the margin of safety
1. reduce the number of repetitions of loads
2. shot-peening
3. good surface roughness
Recommendation:
No. 1, reducing the number of repetitions of loads.
175.
The link shown is made of AISI C1020, as rolled, machined all over. It is loaded
3
9
5
in tension by pins in the D = in holes in the ends; a = in , t = in ,
8
16
16
1
h = 1 in . Considering sections at A, B, and C, determine the maximum safe
8
axial load for N = 2 and indefinite life (a) if it is repeated and reversed; (b) if it
is repeated varying from zero to maximum; (c) if it is repeatedly varies or
F = −W to F = 3W . (d) Using the results from (a) and (b), determine the ratio of
the endurance strength for a repeated load to that for a reversed load (Soderberg
line).
Page 34 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Problems 175 - 178
Solution:
For AISI C1020, as rolled
su = 65 ksi
s y = 48 ksi
sn = 0.5su = 0.5(65) = 32.5 ksi
Size factor = 0.85
For machined all over
Surface factor = 0.90
sn = (0.85)(0.90)(0.80)(32.5) = 20 ksi
1 sm K f sa
= +
N sy
sn
at A, Figure AF 8
9
b = in
16
1
h = 1 in
8
3
d = D = in
8
5
t = in
16
3
d
= 8 = 0.33
h 11
8
9
b 16
=
= 0.5
1
h
1
8
K tA = 3.6
d 3
r = = in
2 16
a = 0.01 in
Page 35 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1
= 0.95
a
0.01
1+
1+
3
r
16
(
)
k fA = q ktA − 1 + 1 = 0.95(3.6 − 1) + 1 = 3.47
q=
1
=
F
F
64 F
=
=
(h − d )t 1 1 − 3  5  15
 8 8  16 
1 64 Fm 3.47(64)Fa
=
+
2 15(48)
15(20)
8
Fm + 1.48 Fa at A
1=
45
s=
At B Figure AF 9
9
d = a = in
16
1
h = 1 in
8
3
r = in
16
5
t = in
16
3
r 16
=
= 0.33
9
d
16
1
1
h
= 8 =2
9
d
16
K tB = 1.63
a = 0.01 in
1
1
q=
=
= 0.95
a
0.01
1+
1+
3
r
16
k fB = q (ktB − 1) + 1 = 0.95(1.63 − 1) + 1 = 1.6
s=
F
F
256 F
=
=
9
5
dt   
45
  
 16  16 
Page 36 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1 256 Fm 1.6(256)Fa
=
+
2 45(48)
45(20)
32
1=
Fm + 0.455 Fa at B
135
at C, Figure AF 8,
b
>1
h
1
in
8
9
h = a = in
16
1
d
= 8 = 0.22
h 9
16
K tC = 3.5
d 1
r = = in
2 16
a = 0.01 in
1
1
q=
=
= 0.862
a
0.01
1+
1+
1
r
16
k fC = q (ktC − 1) + 1 = 0.862(3.5 − 1) + 1 = 3.2
D=
F
F
256 F
=
=
(h − d )t  9 − 1  5  35
 16 8  16 
1 256 Fm 3.2(256)Fa
=
+
2 35(48)
35(20)
32
1=
Fm + 1.17 Fa at C
105
s=
Equations
8
Fm + 1.48 Fa
45
32
At B, 1 =
Fm + 0.455 Fa
135
32
At C, 1 =
Fm + 1.17 Fa
105
At A, 1 =
Page 37 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
(a) Repeated and reversed load
Fm = 0
Fa = F
use at A
8
1=
Fm + 1.48 Fa
45
8
1 = (0 ) + 1.48 Fa
45
F = 0.676 kip
(b) Fm = Fa = F
8
at A, 1 =
F + 1.48 F
45
F = 0.603 kip
32
F + 0.455 F
at B, 1 =
135
F = 1.480 kips
32
at C, 1 =
F + 1.17 F
105
F = 0.678 kip
use F = 0.603 kip
(c) Fmin = −W , Fmax = 3W
1
Fm = (3W − W ) = W
2
1
Fa = (3W + W ) = 2W
2
8
at A, 1 = W + 1.48(2W )
45
W = 0.319 kip
32
at B, 1 =
W + 0.455(2W )
135
W = 0.884 kip
32
at C, 1 =
W + 1.17(2W )
105
W = 0.378 kip
use W = 0.319 kip
Fmax = 0.957 kip
Page 38 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
(d) Ratio =
179.
F (b ) 0.603
=
= 0.892
F (a ) 0.676
A steel rod shown, AISI 2320, hot rolled, has been machined to the following
3
1
dimensions: D = 1 in. , c = in. , e = in. A semicircular groove at the
4
8
1
1
midsection has r = in. ; for radial hole, a = in. An axial load of 5 kips is
8
4
repeated and reversed ( M = 0 ). Compute the factor of safety (Soderberg) and
make a judgement on its suitability (consider statistical variations of endurance
strength – i4.4). What steps may be taken to improve the design factor?
Problems 179-183
Solution:
AISI 2320 hot-rolled (Table AT 10)
su = 96 ksi
s y = 51 ksi
sn = 48 ksi
Size factor = 0.85
Surface factor = 0.85 (machined)
sn = (0.80)(0.85)(0.85)(48) = 27.74 ksi
1 sm K f sa
= +
N sy
sn
sm = 0 , reversed
sa = s
1 K f sa
=
N
sn
s
sa = n
NK f
at the fillet, Figure AF 12
1
r = e = in
8
Page 39 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
d =c=
3
in
4
D = 1 in
1
r 8
= = 0.17
d 3
4
D 1
= = 1.3
d 3
4
K t = 1.55
a = 0.010 in
1
1
q=
=
= 0.926
a
0.010
1+
1+
1
r
8
K f = q (K t − 1) + 1 = 0.926(1.55 − 1) + 1 = 1.51
sa = s =
N=
4(5)
3
π 
4
2
= 11.32 ksi
sn
27.74
=
= 1.62
sa K f (11.32)(1.51)
At the groove, Figure AF 14
1  3
d = b = D − 2r = 1 in − 2 in  = in
8  4
D = 1 in
1
r = in
8
1
r 8
= = 0.17
d 3
4
D 1
= = 1.3
d 3
4
K t = 1.75
a = 0.010 in
Page 40 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1
= 0.926
a
0.010
1+
1+
1
r
8
(
)
K f = q K t − 1 + 1 = 0.926(1.75 − 1) + 1 = 1.7
q=
1
=
4F
4(5)
=
= 11.32 ksi
2
2
πd
3
π 
4
s
27.74
N= n =
= 1.44
sa K f (11.32)(1.7 )
sa = s =
At the hole, Figure AF8
D = h = 1 in
1
d = a = in
4
1
D 4
= = 0.25
h 1
K t = 2.44
a = 0.010 in
1
1
q=
=
= 0.926
a
0.010
1+
1+
1
r
8
K f = q(K t − 1) + 1 = 0.926(2.44 − 1) + 1 = 2.33
sa = s =
F
πD
4
N=
2
− Dd
=
5
π (1)
2
1
− (1) 
4
4
= 9.34 ksi
27.74
sn
=
= 1.27
sa K f (9.34)(2.33)
Factor of safety is 1.27
From i4.4
s = 0.76sn
sn
N=
= 1.32 min > 1.27
0.76 sn
Therefore, dimensions are not suitable.
Steps to be taken:
1. Reduce number of cycle to failure
Page 41 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
2. Good surface condition
3. Presetting
186.
A stock stud that supports a roller follower on a needle bearing for a cam is
5
7
3
made as shown, where a = in , b = in , c = in . The nature of the junction
8
16
4
of the diameters at B is not defined. Assume that the inside corner is sharp. The
material of the stud is AISI 2317, OQT 1000 F. Estimate the safe, repeated load
F for N = 2 . The radial capacity of the needle bearing is given as 1170 lb. at
2000 rpm for a 2500-hr life. See Fig. 20.9, p. 532, Text.
Problem 186
Solution:
AISI 2317, OQT 1000 F
su = 106 ksi
s y = 71 ksi
sn = 0.5su = 53 ksi
Size factor = 0.85
sn = (0.85)(53) = 45 ksi
Figure AF 12
5
d = a = in
8
3
D = c = in
4
r d ≈ 0 , sharp corner
3
D 4
= = 1.2
d 5
8
Assume K t = 2.7
K f ≈ K t = 2 .7
s=
32 M
π a3
7
M = Fb = F   = 0.4375 F
 16 
Page 42 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
5
in = 0.625 in
8
32(0.4375)F
s=
= 18.25 F
π (0.625)3
sm = sa = s = 18.25F
1 sm K f sa
= +
N sy
sn
a=
1 18.25 F (2.7 )(18.25 F )
=
+
2
71
45
F = 0.370 kip = 370 lb < less than radial capacity of the needle bearing. Ok.
187.
The link shown is made of AISI C1035 steel, as rolled, with the following
3
7
1
1
dimensions a = in. , b = in. , c = 1 in. , d = in. , L = 12 in. , r = in. The
8
8
2
16
axial load F varies from 3000 lb to 5000 lb and is applied by pins in the holes.
(a) What are the factors of safety at points A, B, and C if the link is machined all
over? What are the maximum stresses at these points?
Problems 187, 188
Solution:
AISI C1035, as rolled
su = 85 ksi
s y = 55 ksi
sn = 0.5su = 42.5 ksi
size factor = 0.85
sn = (0.6)(0.85)(42.5) = 21.68 ksi
1 sm K f sa
= +
N sy
sn
1
(5 + 3) = 4 kips
2
1
Fa = (5 − 3) = 1 kip
2
Fm =
Page 43 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
(a) at A, Figure AF 9
1
r = in
16
3
d = a = in
8
7
h = b = in
8
1
r 16
=
= 0.17
3
d
8
7
h 8
= = 2.33
d 3
8
K t = 1.9
a = 0.010 in
1
1
q=
=
= 0.862
a
0.010
1+
1+
1
r
16
K f = q(K t − 1) + 1 = 0.862(1.9 − 1) + 1 = 1.78
s=
F
ac
4
= 10.67 ksi
 3
 (1)
8
1
sa =
= 2.67 ksi
3
 (1)
8
1 10.67 (1.78)(2.67 )
=
+
N
55
21.68
N = 2.42
sm =
At B, same as A, K f = 1.78
F
(b − a )c
4
sm =
= 8 ksi
 7 3
 − (1)
8 8
s=
Page 44 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1
= 2 ksi
 7 3
 − (1)
8 8
1
8 (1.78)(2 )
=
+
N 55
21.68
N = 3.23
sa =
At C, Figure AF 8
1
d = in
2
h = c = 1 in
b h >1
1
d 2
= = 0.5
h 1
K t = 2.2
a = 0.010 in
d 1
r = = in = 0.25 in
2 4
1
1
q=
=
= 0.962
a
0.010
1+
1+
r
0.25
K f = q(K t − 1) + 1 = 0.962(2.2 − 1) + 1 = 2.15
F
(b − a )(c − d )
4
sm =
= 16 ksi
7
3

 1 
 − 1 − 
 8 8  2 
1
sm =
= 4 ksi
 7 3  1 
 − 1 − 
 8 8  2 
1 16 (2.15)(4 )
=
+
N 55
21.68
N = 1.45
s=
(b) Maximum stresses
at A
s A = sm + K f sa = 10.67 + 1.78(2.67 ) = 15.42 ksi
at B
sB = sm + K f sa = 8 + 1.78(2 ) = 11.56 ksi
Page 45 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
at C
sC = sm + K f sa = 16 + 2.15(4 ) = 24.6 ksi
IMPACT PROBLEMS
189.
A wrought-iron bar is 1in. in diameter and 5 ft. long. (a) What will be the stress
and elongation if the bar supports a static load of 5000 lb? Compute the stress
and elongation if a 5000 lb. weight falls freely 0.05 in. and strikes a stop at the
end of the bar. (b) The same as (a), except that the bar is aluminum alloy 3003H14.
Solution:
D = 1 in. , L = 5 ft
For wrought iron,
E = 28× 106 psi
(a) elongation
F = 5000 lb
(5000)(5)(12) = 0.01364 in
FL
=
δ=
π
AE
(1)2 28 ×106
4
Stress and elongation
h = 0.05 in
W = 5000 lb
L = 5 ft = 60 in
(
)
1
W W  2hEA  2
s = + 1 +

A A
LW 
1
2

2
6 π 
2
(
0
.
05
)
28
×
10
(
1
)



5000 5000 
 4   = 24,741 psi
1 +
s=
+
π 2 π 2
(60)(5000)

(1)
(1)


4
4

sL (24,741)(60 )
δ=
=
= 0.053 in
E
28 × 106
(
(b) Aluminum alloy 3003-H14
E = 10× 106 psi
F = 5000 lb
FL
(5000)(5)(12) = 0.038 in
=
δ=
AE π (1)2 10 ×106
4
Stress and elongation
h = 0.05 in
(
Page 46 of 62
)
)
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
W = 5000 lb
L = 5 ft = 60 in
1
W W  2hEA  2
s = + 1 +

A A
LW 
1
2

2
6 π 
2
0
.
05
10
×
10
1
(
)
(
)



5000 5000 
 4   = 18,475 psi
1 +
s=
+
π 2 π 2
(60)(5000)

(1)
(1)


4
4

sL (18,475)(60 )
δ=
=
= 0.111 in
E
10 × 106
(
190.
)
What should be the diameter of a rod 5 ft. long, made of an aluminum alloy
2024-T4, if it is to resist the impact of a weight of W = 500 lb dropped through a
distance of 2 in.? The maximum computed stress is to be 20 ksi.
Solution:
For aluminum alloy, 2024-T4
E = 10.6 × 106 psi
W = 500 lb
h = 2 in
L = 5 ft = 60 in
s = 20 ksi = 20,000 psi
1
W W  2hEA  2
s = + 1 +

A A
LW 
(
)
1
5000 5000  2(2 ) 10.6 × 106 A  2
20,000 =
+
1+
(60)(500) 
A
A 
1
40 A = 1 + (1 + 1413 A)2
A=
πD 2
4
= 0.9332
D = 1.09 in , say D = 1
191.
1
in
16
A rock drill has the heads of the cylinder bolted on by 7/8-in. bolts somewhat as
shown. The grip of the bolt is 4 in. (a) If the shank of the bolt is turned down to
the minor diameter of the coarse-thread screw, 0.7387 in., what energy may each
bolt absorb if the stress is not to exceed 25 ksi? (b) Short bolts used as described
above sometimes fail under repeated shock loads. It was found in one instance
that if long bolts, running from head to head, were used, service failures were
eliminated. How much more energy will the bolt 21 in. long absorb for a stress of
Page 47 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
25 ksi. That the bolt 4 in. long? As before, let the bolt be turned down to the
minor diameter. The effect of the threads on the strength is to be neglected.
Problem 191
Solution:
2
s2
( AL ) = s AL
U=
2E
2E
2
πD
(a) A =
4
L = 4 in
D = 0.7387 in
E = 30×106 psi
s = 25 ksi = 25,000 psi
(25,000)2  π (0.7387 )2 (4)
4
2 30 × 106
U=
(
)
= 17.86 in − lb
(b) L = 21 in
(25,000)2  π (0.7387 )2 (21)
4
= 93.75 in − lb
2 30 × 106
∆U = 93.75 − 17.86 = 75.89 in − lb
U=
192.
(
)
As seen in the figure, an 8.05-lb body A moving down with a constant
acceleration of 12 fps2, having started from rest at point C. If A is attached to a
steel wire, W & M gage 8 (0.162 in. diameter) and if for some reason the sheave
D is instantly stopped, what stress is induced in the wire?
Problems 192, 193
Solution:
s 2 AL
U=
2E
1 2 1
U = mv = m(2ah ) = mah = maL
2
2
Page 48 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
s 2 AL
= maL
2E
2maE 2WaE
s2 =
=
A
gA
W = 8.05 lb
a = 12 fps 2
g = 32 fps 2
b = 12 fps 2
E = 30×106 psi
A=
πD 2
4
8WaE 8(8.05)(12)(30 × 106 )
s2 =
=
πD 2 g
π (0.162)2 (32)
s = 93,741 psi
193.
The hoist A shown, weighing 5000 lb. and moving at a constant v = 4 fps is
attached to a 2 in. wire rope that has a metal area of 1.6 sq. in. and a modulus
E = 12× 106 psi . When h = 100 ft , the sheave D is instantly stopped by a brake
(since this is impossible, it represents the worst conceivable condition).
Assuming that the stretching is elastic, compute the maximum stress in the rope.
Solution:
s 2 AL
2E
1
W 2
v
U = mv 2 =
2
2g
U=
s 2 AL W 2
=
v
2E
2g
Wv 2 E
gAL
W = 5000 lb
s2 =
Page 49 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
v = 4 fps
E = 12× 106 psi
A = 1.6 in 2
L = h = 100 ft
g = 32 fps 2
s2 =
(5000)(4)2 (12 ×106 )
(32)(1.6)(100)
s = 13,693 psi
194.
A coarse-thread steel bolt, ¾ in. in diameter, with 2 in. of threaded and 3 in. of
unthreaded shank, receives an impact caused by a falling 500-lb weight. The area
at the root of the thread is 0.334 sq. in. and the effects of threads are to be
neglected. (a) What amount of energy in in-lb. could be absorbed if the maximum
calculated stress is 10 ksi? (b) From what distance h could the weight be
dropped for this maximum stress? (c) How much energy could be absorbed at the
same maximum stress if the unthreaded shank were turned down to the root
diameter.
Solution:
s 2 AL
U=
2E
(a) U = U1 + U 2
s12 A1L1
2E
2
s AL
U2 = 2 2 2
2E
A1 = 0.334 in 2
U1 =
A2 =
π
(0.75) = 0.442 in 2
4
s1 = 10,000 psi
s A (10,000 )(0.334 )
s2 = 1 1 =
= 7556 psi
A2
0.442
L1 = 2 in
L2 = 3 in
E = 30 × 10 6 psi
U1 =
(10,000)2 (0.334)(2) = 1.113 in − lb
U2 =
(7556)2 (0.442)(3) = 1.262 in − lb
2(30 × 106 )
2(30 ×106 )
Page 50 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
U = U1 + U 2 = 1.113 + 1.262 = 2.375 in − lb
1


W   2hEA  2 
(b) s =
1 + 1 +

A 
LW  


1


2





 

W
2hE
 
s = 1 + 1 +
A1  
 L L 
W  1 + 2   

 
 A1 A2   


1


W 
2hEA1 A2  2 
s = 1 + 1 +

A1   W ( A2 L1 + A1L2 )  


W = 500 lb
A1 = 0.334 in 2
A2 = 0.442 in 2
L1 = 2 in
L2 = 3 in
E = 30×106 psi
s = 10,000 psi
(
)
1
 2h 30 × 106 (0.334 )(0.442 )  2
500
10,000 =
1 + 1 +

0.334
 500[(0.442 )(2 ) + (0.334 )(3)]
h = 0.0033 in
s 2 AL
2E
A = 0.334 in 2
L = 5 in
E = 30×106 psi
s = 10,000 psi
(c) U =
2
(
10,000) (0.334)(5)
U=
= 2.783 in − lb
2(30 ×106 )
196.
A part of a machine that weighs 1000 lb. raised and lowered by 1 ½-in. steel rod
that has Acme threads on one end (see i8.18 Text, for minor diameter). The
length of the rod is 10 ft. and the upper 4 ft are threaded. As the part being
Page 51 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
lowered it sticks, then falls freely a distance of 1/8 in. (a) Compute the maximum
stress in the rod. (b) What would be the maximum stress in the rod if the lower
end had been turned down to the root diameter?
Solution:
1


W   2hEA  2 
s=
1 + 1 +

A 
LW  


1


2 
 

 
W  
2hE
 
s = 1 + 1 +
A1  
 L L 
W  1 + 2   

 
 A1 A2   


1


W 
2hEA1 A2  2 
s = 1 + 1 +

A1   W ( A2 L1 + A1L2 )  


1
see i8.18 , D2 = 1 in , D1 = 1.25 in
2
2
π (1.25)
A1 =
= 1.227 in 2
4
π (1.5)2
A2 =
= 1.767 in 2
4
L1 = 4 in
L2 = 6 in
1
h = in = 0.125 in
8
W = 1000 lb
E = 30×106 psi
(
)
1
 2(0.125) 30 × 106 (1.227 )(1.767 )  2
1000
s=
1 + 1 +
 = 28,186 psi
1.227
1000[(1.767 )(4 ) + (1.227 )(6 )] 

1


W   2hEA  2 
(b) s =
1 + 1 +

A 
LW  


2
A = A1 = 1.227 in
L = L1 + L2 = 10 in
Page 52 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
(
)
1
 2(0.125) 30 × 106 (1.227 )  2
1000
s=
1 + 1 +
 = 25,552 psi
1.227
10(1000 )


197.
A weight W of 50 lb is moving on a smooth horizontal surface with a velocity of
2 fps when it strikes head-on the end of a ¾-in. round steel rod, 6 ft. long.
Compute the maximum stress in the rod. What design factor based on yield
strength is indicated for AISI 1010, cold drawn?
Solution:
1

2
2


Wv E

s=
 g AL1 + We  
 o  W  
W
We = b
3
Wb = ρAL
ρ = 0.284 lb in3
π 3
2
2
  = 0.442 in
4 4
L = 6 ft = 72 in
Wb = (0.284)(0.442)(72) = 9.038 lb
9.038
= 3.013 lb
We =
3
W = 50 lb
v = 2 fps
A=
g o = 32 fps 2
E = 30×106 psi
L = 6 ft
1

2
2
6

(50)(2) (30 ×10 )  = 8166 psi
s=
 (32)(0.442)(6)1 + 3.013  

50  

For AISI 1010, cold drawn
s y = 55 ksi = 55,000 psi
N=
s y 55,000
=
= 6.74
s
8166
Page 53 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
199.
A rigid weight of 100 lb is dropped a distance of 25 in. upon the center of a 12
in., 50-lb. I-beam ( I x = 301.6 in 4 ) that is simply supported on supports 10 ft
apart. Compute the maximum stress in the I-beam both with and without
allowing for the beam’s weight.
Solution:
Without beams weight
y
s = sst
yst
FL3
48EI
F 48EI
k= = 3
y
L
y=
1


W   2hk  2 
y =δ =
1 + 1 +

k  
W  


6
E = 30×10 psi
L = 10 ft = 120 in
I = 301.6 in 4
48(30 × 106 )(301.6)
= 251,333 lb in
(120)3
W = 100 lb
h = 25 in
k=
1


 100   2(25)(251,333)  2 
y=
1 + 1 +
  = 0.1415 in
100
 251,333  


WL3
(100)(120)
yst =
=
= 0.0004 in
48EI 48(30 × 106 )(301.6)
Mc
sst =
I
WL (100 )(120 )
M=
=
= 3000 in − lb
4
4
h 12
c= =
= 6 in
2 2
(3000)(6) = 59.68 psi
sst =
301.6
 0.1415 
s = (59.68)
 = 21,112 psi
 0.0004 
3
Page 54 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
with mass of beam
1
 2h  2

y = yst + yst 1 +
 yst 
h - correction factor =
1
W
1+ e
W
17Wb
35
Wb = (50 lb ft )(10 ft ) = 500 lb
17(500 )
We =
= 243 lb
35
1
h - correction factor =
= 0.292
243
1+
100
1


  2(25)(0.292 )  2 
y = 0.00041 + 1 +
  = 0.0764 in
0
.
0004



y
 0.0764 
s = sst
= (59.68)
 = 11,400 psi
yst
 0.0004 
We =
201.
A 3000 lb. automobile (here considered rigid) strikes the midpoint of a guard rail
that is an 8-in. 23-lb. I-beam, 40 ft. long; I = 64.2 in4 . Made of AISI C1020, as
rolled, the I-beam is simply supported on rigid posts at its ends. (a) What level
velocity of the automobile results in stressing the I-beam to the tensile yield
strength? Compare results observed by including and neglecting the beam’s
mass.
Solution:
For AISI C1020, as rolled
s y = 48 ksi = 48,000 psi
Fδ Wv 2
=
2
2 go
F 48 EI
k= = 3
δ
L
Mc FLc
s=
=
I
4I
4 Is
F=
Lc
Fδ F 2 L3
16 I 2 s 2 L3
s 2 IL
=
= 2 2
=
2
96 EI L c (96 EI ) 6 Ec 2
Page 55 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
neglecting mass of beam
Fδ s 2 IL Wv 2
=
=
2
6 Ec 2 2 go
3Wv 2 Ec 2
2 g o IL
W = 3000 lb
g o = 32 fps 2
h 8
c = = = 4 in
2 2
E = 30×106 psi
I = 64.2 in4
L = 40 ft
s = s y = 48 ksi = 48,000 psi
s2 =
(
)
3Wv 2 Ec 2 3(3000 )v 2 30 × 106 (4 )
s = (48,000 ) =
=
2 g o IL
32(64.2 )(40 )
v = 6.62 fps
Including mass of beam


2
2 
3Wv Ec
1 


s2 =
2 g o IL  1 + We 


 W 
17Wb
We =
35
Wb = (23 lb ft )(40 ft ) = 920 lb
17(920 )
We =
= 447 lb
35
2
2
2


2
2
2
2
6


3
Wv
Ec
3
(
3000
)
v
(
30
×
10
)
(
4
)
1
s 2 = (48,000) =
=

447 
2 go IL
32(64.2)(40)
1 +

 3000 
v = 7.10 fps
2
DATA LACKING – DESIGNER’S DECISIONS
202.
A simple beam is struck midway between supports by a 32.2-lb. weight that has
fallen 20 in. The length of the beam is 12 ft. If the stress is not to exceed 20 ksi,
what size I-beam should be used?
Solution:
Page 56 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1
 2h  2

y = yst + yst 1 +
 yst 
y
s = sst
yst
h = 20 in
s = 20,000 psi
WL3
yst =
48EI
1
y
 96 EIh  2
= 1 + 1 +

yst
WL3 

with correction factor
1


 2


96 EIh
1 
y


= 1 + 1 +
3
We 
yst
WL


1+


 W 
Mc WLd
sst =
=
I
8I
17 wL
We =
35
1


2





 
WLd   96 EIh 
1

 
s=
1 + 1 +
8I  
WL3  1 + 17 wL  


35W  

 


W = 32.2 lb
h = 20 in
L = 12 ft = 144 in
E = 30×106 psi
s=
(32.2)(144)d
8I
1


2




 
1
 96 30 × 106 (I )(20) 
1 + 1 +


(32.2)(144)3 1 + 17(w)(12)  

35(32.2)  




(
)
1


579.6d  
1

 2 
s=
 
1 + 1 + 599 I 
I  
 1 + 0.181w  


From The Engineer’s Manual
By Ralph G. Hudson, S.B.
Page 57 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Use 3”, 5.7 lb, I = 2.5 in4
1


2
579.6(3)
1
s=
1 + 1 + 599(2.5)
  = 19,600 psi < 20,000 psi
2.5
1 + 0.181(5.7 )  

Therefore use 3-in depth, 5.7-lb I-beam ( I = 2.5 in4 )
204.
A 10-in., 25.4-lb.., I-bean, AISI 1020, as rolled, is 10 ft. long and is simply
supported at the ends shown. There is a static load of F1 = 10 kips , 4 ft from the
left end, and a repeated reversed load of F2 = 10 kips , 3 ft from the right end. It is
desired to make two attachments to the beam through holes as shown. No
significant load is supported by these attachments, but the holes cause stress
concentration. Will it be safe to make these attachments as planned? Determine
the factor of safety at the point of maximum moment and at points of stress
concentration.
Problem 204
Solution:
Mass of beam negligible
For AISI C1020, as rolled
s y = 48 ksi
su = 65 ksi
(∑ M
A
= 0)
4 F1 + (10 − 3)F2 = 10 B
1
B = (4 F1 + 7 F2 )
10
(∑ M B = 0)
3F2 + (10 − 4)F1 = 10 A
1
A = (6 F1 + 3F2 )
10
F1 = 10 kips
F2 = −10 to 10 kips
1
Bmin = [4(10 ) + 7(− 10 )] = −3 kips
10
Page 58 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
1
[4(10) + 7(10)] = 11 kips
10
1
Amin = [6(10 ) + 3(− 10 )] = 3 kips
10
1
Amax = [6(10 ) + 7(30 )] = 9 kips
10
Figure AF 11,
1
e = 1 in ,
2
1
d = in
4
1 1
c = 1 + 2  = 1.625 in
2 4
h = 10 in
h
10 1
b = − e = − 1 = 3.5 in
2
2
2
d 0.25
=
= 0.07
b 3 .5
e 1.50
=
= 6 > 0 .5
d 0.25
Use K t = 3.0
1
q=
= 0.926
0.010
1+
1
8
K f = q (K t − 1) + 1 = 0.926(3 − 1) + 1 = 2.85
Bmax =
sn = 0.5su = 0.5(65) = 32.5 ksi
size factor = 0.85
sn = 0.85(32.5) = 27.6 ksi
left hole, M = (2)A
M max = 2(9) = 18 ft − kips
M min = 2(3) = 6 ft − kips
Mc
s=
I
1
M m = (18 + 6 ) = 12 ft − kips = 144 in − kips
2
1
M a = (18 − 6 ) = 6 ft − kips = 72 in − kips
2
c = 1.625 in
I = 122.1 in 4 (Tables)
Page 59 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
sm =
sa
(144)(1.625) = 1.92 ksi
122.1
(72 )(1.625) = 0.96 ksi
=
122.1
1 sm K f sa
= +
N sy
sn
1 1.92 (2.85)(0.96 )
=
+
N
48
27.6
N = 7 .2
right hole , M = (1.5)B
M max = 1.5(11) = 16.5 ft − kips
M min = 1.5(− 3) = −4.5 ft − kips
Mc
s=
I
1
M m = (16.5 − 4.5) = 6 ft − kips = 72 in − kips
2
1
M a = (16.5 + 4.5) = 10.5 ft − kips = 126 in − kips
2
c = 1.625 in
I = 122.1 in 4 (Tables)
(72)(1.625) = 0.96 ksi
sm =
122.1
(126)(1.625) = 1.68 ksi
sa =
122.1
1 sm K f sa
= +
N sy
sn
1 0.96 (2.85)(1.68)
=
+
N
48
27.6
N = 5.67
at maximum moment, or at , F2
M max = 3(11) = 33 ft − kips
M min = 3(− 3) = −9 ft − kips
Mc
s=
I
1
M m = (33 − 9 ) = 12 ft − kips = 144 in − kips
2
1
M a = (33 + 9 ) = 21 ft − kips = 252 in − kips
2
Page 60 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
10
= 5 in
2
I = 122.1 in 4 (Tables)
(144)(5) = 5.90 ksi
sm =
122.1
(
252 )(5)
sa =
= 10.32 ksi
122.1
K f = 1 .0
c=
1 sm K f sa
= +
N sy
sn
1 5.90 (1.0 )(10.32 )
=
+
N
48
27.6
N =2
Since the design factor at the holes is much larger than at the point of maximum moment,
it is safe to make these attachment as planned.
205.
The runway of a crane consists of L = 20 ft. lengths of 15-in., 42.9-lb. I-beams,
as shown, each section being supported at its ends; AISI C1020, as rolled. The
wheels of the crane are 9 ft apart, and the maximum load expected is
F = 10,000 lb on each wheel. Neglecting the weight of the beam, find the design
factor (a) based on variable stresses for 105 cycles, (b) based on the ultimate
strength. (Hint. Since the maximum moment will occur under the wheel, assume
the wheels at some distance x from the point of support, and determine the
dM
reaction, R1 as a function of x ;
= 0 gives position for a maximum bending
dx
moment.)
Problem 205.
Solution:
∑ M R2 = 0
(
)
(L − x )F + (L − x − a )F = LR1
(2 L − 2 x − a )F
R =
1
L
x
M = R1 x = (2 L − 2 x − a )F
L
Page 61 of 62
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
dM F
= [(2 L − 2 x − a ) + x(− 2 )] = 0
dx
L
2L − 2x − a − 2x = 0
1
a
x = L− 
2
2
2
a

a
L−  F
L− 
2
2 2 L −  L − a  − a  F = 
M max =


L 
2 
2L

L = 20 ft = 240 in
a = 9 ft = 108 in
F = 10,000 lb = 10 kips
2
108 

 240 −
 (10 )
2 

M max =
= 720.75 in − kips
2(240 )
For 15-in., 42.9 lb, I-beam
I = 441.8 in4
15
c = = 7.5 in
2
Mc (720.75)(7.5)
smax =
=
= 12.24 ksi
I
441.8
For AISI C1029, as rolled
su = 65 ksi
sn = 0.5su = 32.5 ksi
size factor = 0.85
sn = 0.85(32.5) = 27.6 ksi
(a) at 105 cycles
0.085
 106 
sn = 27.6 5 
= 34 ksi
 10 
s
34
N= n =
=7
s 12.24
(b) N =
su
65
=
= 5.31
s 12.24
- end -
Page 62 of 62
SECTION 3 – SCREW FASTENINGS
SIMPLE TENSION INCLUDING TIGHTENING STRESSES
DESIGN PROBLEMS
221.
A 5000-lb. gear box is provided with a steel (as rolled B1113) eyebolt for use
in moving it. What size bolt should be used: (a) if UNC threads are used? (b)
If UNF threads are used? (c) If the 8-thread series is used? Explain the basis
of your choice of design factor.
Solution:
B1113, as rolled
s y = 45 ksi (Table AT-7)
Fe = 5000 lb
i5.6, sd =
sy
6
1
( As ) 2
3 

 D < in 
4 

3
4
sd = 0.4 sy
For D ≥ in
sd = 0.4(45,000) = 18,000 psi
As =
Fe
5000
=
= 0.2778 sq.in.
sd 18 ,000
Table AT 14 and Table 5.1
(a) UNC Threads
3
4
Use D = in , As = 0.334 sq.in.
(b) UNF Threads
3
4
Use D = in , As = 0.373sq.in.
(c) 8-Thread Series
Use D = 1 in , As = 0.606sq.in.
Page 1 of 42
SECTION 3 – SCREW FASTENINGS
222.
A motor weighing 2 tons is lifted by a wrought-iron eye bolt which is screwed
into the frame. Decide upon a design factor and determine the size of the
eyebolt if (a) UNC threads are used, (b) UNF threads are used. Note: Fine
threads are not recommended for brittle materials.
Solution:
Table AT-7
Wrought iron, s y = 25 ksi
sd = 0.4 sy = 0.4(25) = 10 ksi = 10,000 psi
As =
Fe 2(2000)
=
= 0.4 sq.in.
sd 10,000
Table AT 17
(a) UNC Threads
7
8
Use D = in , As = 0.462 sq.in.
(b) UNF Threads
7
8
Use D = in , As = 0.509 sq.in.
224.
A wall bracket, Fig. 8-13, Text, is loaded so that the two top bolts that fasten it
to the wall are each subjected to a tensile load of 710 lb. The bolts are to be
cold forged from AISI C1020 steel with UNC threads, Neglecting the effect of
shearing stresses, determine the diameter of these bolts if they are well
tightened.
Figure 8-13
Page 2 of 42
SECTION 3 – SCREW FASTENINGS
Solution:
cold forged, AISI C1020
s y = 66 ksi (Table AT-7)
Fe = 710 lb
s
3
3 

Fe = y ( As ) 2  D < in 
4 
6

3
66,000
( As ) 2
710 =
6
3
As = 0.161 sq.in. , D < in
4
Table AT 14 , UNC Threads
Use D =
225.
9
in , As = 0.1820 sq.in.
16
A connection similar to Fig. 5.9, Text, is subjected to an external load Fe of
1250 lb. The bolt is made from cold-finished AISI B1113 steel with UNC
threads. (a) Determine the diameter of the bolt if it is well tightened. (b)
Compute the initial tension and corresponding approximate tightening torque
if si = 0.85s y (i5.8).
Figure 5.9
Page 3 of 42
SECTION 3 – SCREW FASTENINGS
Solution:
Cold-finished AISI B1113
Table A-7, s y = 72 ksi
Fe = 1250 lb
s
3
(a) Fe = y ( As ) 2
6
3
72,000
1250 =
( As ) 2
6
3
in
4
Table AT 14 , UNC Threads
As = 0.2214 sq.in. , D <
Use D =
5
in , As = 0.2260 sq.in.
8
(b) si = 0.85s y = 0.85(72,000 ) = 61,200 psi
Initial Tension
Fi = si As = (61,200 )(0.2260 ) = 13,831 lb
Tightening torque
T = CDFi
5
T = 0.2 DFi = 0.2 (13,831) = 1729 in − lb
8
226.
The cylinder head of a 10 x 18 in. Freon compressor is attached by 10 stud
bolts made of SAE Grade 5. The cylinder pressure is 200 psi. (a) What size
bolts should be used? (b) What approximate tightening torque should be
needed to induce a tightening stress si of 0.9 times the proof stress?
Solution:
Page 4 of 42
SECTION 3 – SCREW FASTENINGS
Table 5.2
SAE Grade 5
Assume s y = 88 ksi
π 
2
200 (10 )
4
 
(a) Fe =
= 1571 lb
10
s
3
3
Fe = y ( As ) 2 , D < in
6
4
3
88,000
1571 =
( As ) 2
6
3
As = 0.2255 sq.in. , D < in
4
Table AT 14 , UNC Threads
Use D =
5
in , As = 0.2260 sq.in.
8
(b) T = CDFi
C = 0. 2
s i = 0. 9 s p
s p = 85 ksi , (Table 5.2)
si = 0.9(85,000 ) = 76,500 psi
Fi = si As = (76,500 )(0.2260 ) = 17,289 lb
Tightening torque
5
T = 0.2 DFi = 0.2 (17,289) = 2161 in − lb
8
227.
The American Steel Flange Standard specifies that 8 bolts are to be used on
flanges for 4-in. pipe where the steam or water pressure is 1500 psi. It is also
specified that, in calculating the bolt load, the outside diameter of the gasket,
which is 6 3/16 in., should be used. Determine (a) the diameter of the UNC bolts
if they are well-tightened and made of ASTM 354 BD (Table 5-2), (b) the
approximate torque to tighten the nuts if the initial stress is 90 % of the proof
stress. The Standard specifies that 1 1/4 –in. bolts with 8 th./in. be used (these
bolts are also subjected to bending). How does your answer compare?
Solution:
Table 5.2, ASTM 354 BD
s p = 120 ksi
s y = 125 ksi
Page 5 of 42
SECTION 3 – SCREW FASTENINGS
2
 π  3 
1500  6 
 4  16  = 5638 lb
Fe =
8
sy
(a) Fe =
3
( As ) 2
, D<
6
3
125,000
5638 =
( As ) 2
6
As = 0.4184 sq.in. ,
Table AT 14 , UNC Threads
Use D =
D>
3
in
4
7
in , As = 0.4620 sq.in.
8
3
in
4
use
sd = 50,000 psi § 5.6, ASTM A354 BD
F
5638
As = e =
= 0.1128 sq.in.
sd 50 ,000
Table AT 14 , UNC Threads
3
4
Use D = in , As = 0.334 sq.in.
(b) T = 0.2 DFi
s i = 0. 9 s p
si = 0.9(120,000) = 108,000 psi
Fi = si As = (108,000)(0.3340) = 36,072 lb
Tightening torque
3
T = 0.2DFi = 0.2 (36,072) = 5411 in − lb
4
1
D < 1 in as specified by the standard.
4
CHECK PROBLEMS
228.
A cap screw, ¾ in.-10-UNC-2, with a hexagonal head that is 9/16 in. thick,
carries a tensile load of 3000 lb. If the material is AISI 1015, cold drawn, find
the factor of safety based on ultimate strengths of (a) the threaded shank, (b)
the head against being sheared off, and (c) the bearing surface under the head.
(d) Is there any need to consider the strength of standard cap-screw heads in
design?
Page 6 of 42
SECTION 3 – SCREW FASTENINGS
Solution:
For ¾ in. UNC, Table AT 14,
As = 0.334 sq.in.
Head:
1
A = 1 in.
8
For AISI 1015, cold drawn
su = 77 ksi , sus = 58 ksi
F
3000
=
= 8982 psi
As 0.334
s
77,000
N= u =
= 8.57
sd
8982
(a) s =
(b) s s =
F
πDt
9
in
16
3000
ss =
= 2264 psi
 3  9 
π   
 4  16 
s
58,000
N = us =
= 25.6
ss
2264
t=
(c) θ =
360 o
= 30 o
12
2
 1
1
 1  A  A

 1  8 
Area = 6(2)   tan θ  = 6(2) 
tan 30 = 1.096 sq.in.
 2  2  2

 2  2 
 
Page 7 of 42
SECTION 3 – SCREW FASTENINGS
sb =
N=
F
=
Area − Ab
3000
1.096 −
π 3
2
= 4586 psi
 
4 4
su 77,000
=
= 16.8
sb
4586
(d)
No need to consider the strength of standard cap-screw heads since its factor of
safety is very much higher than for the threaded shank.
229.
A bolt, 1 1/8 in.-7-UNC-2, is subjected to a tensile load of 10,000 lb. The head
has a thickness of ¾ in. and the nut a thickness of 1 in. If the material is SAE
grade 2 (Table 5.2), find the design factor as based on ultimate stresses (a) of
the threaded shank, (b) of the head against being sheared off, and (c) of the
bearing surface under the head. The bolt head is finished. (d) Is there any need
to consider the strength of standard bolt heads in design?
Solution:
1
For SAE grade 2 (Table 5.2), D = 1 in
8
su = 55 ksi , sus = 0.75su
For 1 1/8 in.-7-UNC-2 (Table AT 14)
As = 0.763 sq.in.
A =1
11
in
16
F = 10,000 lb.
F 10,000
=
= 13,106 psi
As
0.763
s
55,000
N= u =
= 4.2
sd 13,106
(a) s =
(b) s s =
t=
F
πDt
3
in
4
Page 8 of 42
SECTION 3 – SCREW FASTENINGS
10,000
= 3773 psi
 1  3 
π 1  
 8  4 
0.75(55,000 )
s
N = us =
= 11
ss
3773
ss =
(c) θ =
360 o
= 30 o
12
2
 11 
1
 1  A  A

 1  16 
Area = 6(2)   tan θ  = 6(2) 
tan 30 = 2.4661 sq.in.
 2  2  2

 2  2 


F
10,000
sb =
=
= 6793 psi
2
Area − Ab
π  1
2.4661 − 1 
4  8
s
55,000
N= u =
= 8.1
sb
6793
(d)
No need to consider the strength of standard bolt head in design since its factor of
safety is higher than for the threaded shank.
230.
An axial force is applied to a regular nut which of course tends to shear the
threads on the screw. (a) What is the ratio of the force necessary to shear the
threads (all threads initially in intimate contact) to the force necessary to pull
the bolt in two? Use coarse threads, a 1 ½ -in. bolt, and assume that
sus = 0.75su . The head thickness is 1 in. and the nut thickness is 1 5/16 in. (b)
Is failure of the thread by shear likely in this bolt?
Solution:
1 ½ - in. UNC
As = 1.405 sq.in.
(a) Fs = shear force = susπDt
1
D = 1 in.
2
5
t = 1 in.
16
Page 9 of 42
SECTION 3 – SCREW FASTENINGS
sus = 0.75su
 1  5 
Fs = (0.75su )(π )1 1  = 4.6388su
 2  16 
F = su As = 1.405su
4.6388su
= 3.3
Ratio =
1.405su
(b) Ratio > 1, failure by shear is not likely to occur.
231.
For bolted structural joints, specifications suggest that ½-in. bolts (highstrength material) be tightened to an initial tension of Fi = 12,500 lb . What
should be the approximate tightening torque? How does your answer compare
with T = 90 ft − lb ., which is the value in the specification?
Solution:
1
T = 0.2 DFi = 0.2 (12,500) = 1250 in − lb
2
T = 90 ft − lb = 1080 in − lb < 1250 in − lb o.k.
232.
One method of estimating the initial tensile stress in a tightened bolt is to turn
the nut until it is snug, but with no significant stress in the bolt. Then the nut is
turned through a predetermined angle that induces a certain unit strain
corresponding to the desired stress. A ¾ - in. bolt of the type shown in Fig.
5.4, Text, is turned down until, for practical purposes, the diameter of the
entire shank is the minor diameter. The material is AISI 4140, OQT 1200 oF.
The grip is 5 in. and the effective strain length is estimated to be 5.3 in. If the
initial tensile stress at the root diameter is to be about 75 % of the yield
strength, through what angle should the nut be turned after it is just snug? The
threads are UNC and the parts being bolted are assumed to be rigid.
Solution:
For ¾ in., UNC
Dr = 0.6273 in
As = 0.334 sq.in.
Th in. = 10
AISI 4140, OQT 1200 oF
Page 10 of 42
SECTION 3 – SCREW FASTENINGS
s y = 115 ksi
s = 0.75(115) = 86.25 ksi
sL
E
L = 5.3 in
δ=
pitch, p =
1
in = 0.10 in
10
δ
(360o )
p
sL
θ=
(
360 o )
pE
(86,250)(5.3) (360o ) = 55o
θ=
(0.10)(30 ×106 )
θ=
233.
When both ends of a bolt are accessible for micrometer measurements, the
total elongation δ caused by tightening can be determined by measuring
lengths before and after tightening. In order to reduce this total elongation to
unit elongation, thence to stress, the effective strain length for the bolt must be
known. For a 1 ¼-in steel bolt, threaded for its full length, 8-thread series, the
effective strain length has been found by experiment to be
Le = 0.97G + 1.1 in. , where G is the grip (by W.A. McDonald, North
Carolina State College). Let the bolt material be AISI 8742, OQT 1000 oF. (a)
It is desired that the initial tensile stress be about 0.7 s y . What total elongation
should be obtained for a grip length of 4.8 in.? (b) Investigate the approximate
tightening torque for the specified condition. How could this torque be
obtained?
Solution:
1 ¼ in., 8-thread series
Table 5.1
Dr = 1.0966 in
As = 1.000 sq.in.
Th in. = 8
AISI 8742, OQT 1000 oF
s y = 147 ksi
(a) si = 0.70 s y
si = 0.70(147 ) = 102.9 ksi = 102,900 psi
Page 11 of 42
SECTION 3 – SCREW FASTENINGS
si Le
E
Le = 0.97G + 1.1 in.
G = 4.8 in
Le = 0.97(4.8) + 1.1 in. = 5.756 in
δ=
δ=
si Le (102,900 )(5.756 )
=
= 0.01975 in
E
30 × 106
(b) θ =
δ
p
=
64TL
πDr4G
G = 11.5 ×10 6 psi
1
p = in = 0.125 in
8
0.01975
64T (5.756 )
=
θ=
4
0.125
π (1.0966) (11.5 ×106 )
T = 22,408 in − lb
ELASTIC CONSIDERATIONS
235.
The member C shown is part of a swivel connection that is to be clamped by a
1-in. bolt D to the member B, which has large dimensions in the plane
perpendicular to the paper. Both B and C are aluminum alloy 2024-T4, HT
aged. The bolt is made of AISI C1113, cold-drawn steel; consider the
unthreaded shank to be 2 in. long; it is well tightened with a torque of 250 ftlb.; UNC threads, unlubricated. (a) Estimate the initial tension by equation
(5.2), assume elastic action, and compute the bolt elongation and the total
deformation of B and C. Let the effective strain length be 2 in. (b) After
tightening an external axial force Fe of 5000 lb. is applied to member C.
Determine the total normal stresses in the bolt and in B and C. (c) Determine
the load required to “open” the connection. Draw a diagram similar to Fig.
5.6, Text, locating points A, B, D and M.
Page 12 of 42
SECTION 3 – SCREW FASTENINGS
Prob. 235, 236
Solution:
For aluminum alloy, 2024-T4 HT aged,
E = 10.6 ×106 psi
s y = 47 ksi
For AISI C1113, cold-drawn steel,
E = 30×106 psi
s y = 72 ksi
(a) T = 0.2 DFi
D = 1 in.
T = 250 ft − lb = 3000 in − lb
Fi = 15,000 lb
Deformations: L = 2 in.
Table AT 14, 1-in. UNC Bolt,
As = 0.66 sq.in.
π 2
Ab = (1) = 0.785 sq.in.
4
Bolt:
FL
(15,000)(2) = 0.00127 in
δi = i =
Ab Ei (0.785)(30 × 10 6 )
Member B and C
FL
δc = i
Ac Ec
π
π
Ac = De2 − D 2
4
4
h
De = (Nut or head width across flats) +
2
Table AT 14
1
A = 1 in
2
1 2
1
De = 1 + = 2 in.
2 2
2
π 2 π 2
Ac = De − D
4
4
π
2
Ac = (2.5) − (1)2 = 4.1234 sq.in.
4
FL
(15,000)(2) = 0.000686 in.
δc = i =
Ac Ec (4.1234 )(10.6 × 10 6 )
(b) Fe = 5000 lb
[
Page 13 of 42
]
SECTION 3 – SCREW FASTENINGS
 kb 
 δb 
0.000686


 = Fe 
 = 5000
∆Fb = Fe 

 0.000686 + 0.00127 
 kb + k c 
 δb + δc 
∆Fb = 1754 lb
Bolt:
Ft = Fi + ∆Fb = 15,000 + 1754 = 16,754 lb
F 16,754
sb = t =
= 29,132 psi
As 0.606
Member B and C
 kc 

Fc = Fi − Fe 
 kb + k c 
 δc 

Fc = Fi − Fe 
δ
δ
+
b
c


0.00127


Fc = 15,000 − 5000
 = 11,754 lb
 0.00127 + 0.000686 
F 11,754
sc = c =
= 2851 psi
Ac 4.1234
(c) Fo = opening load
 δ + δc
Fo = Fi  i
 δi
Fig. 5.6
237.

 0.00127 + 0.000686 
 = 15,000
 = 23,102 lb
0
.
00127



A 1-in. steel bolt is used to clamp two aluminum (2014-T6, HT aged) plates
together as shown by Fig. 5.9, Text. The aluminum plates have a total
thickness of 2 in. and an equivalent diameter of 2 in. The bolt is heated to a
temperature of 200 oF, the inserted in the aluminum plates, which are at 80 oF,
and tightened so as to have a tensile tightening stress of 30 ksi in the
unthreaded shank while steel at 200 oF. What is the tensile stress in the bolt
after assembly has cooled to 80 oF? The deformations are elastic.
Figure 5.9
Page 14 of 42
SECTION 3 – SCREW FASTENINGS
Solution:
For aluminum 2014-T6
E = 10.6 × 106 psi
s b = 30,000 psi
π  2
Fi = s b Ab = (30,000 ) (1) = 23,562 lb
4
Steel bolt. Eb = 30 × 10 6 psi
s L (30,000 )(2 )
δi = b =
= 0.002 in.
Eb
30 × 10 6
δc =
Ac =
Ac =
Fi L
.
Ac E c
π
4
π
De2 −
[(2)
4
2
π
4
D2
]
− (1) = 2.3562 sq.in.
2
Ec = 10.6 × 10 6 psi
FL
(23,562)(2) = 0.001887 in
δc = i =
Ac Ec (2.3562) 10.6 × 10 6
L′ = L − δ c = 2 − 0.001887 = 1.998113 in.
∆L = αL′∆t
α = 0.000007 in. (in. − F ) for steel
∆L = (0.000007 )(1.998113)(80 − 200 ) = −0.001678 in.
δ i′ = δ i + ∆L = 0.002 − 0.001678 = 0.000322 in.
(
Page 15 of 42
)
SECTION 3 – SCREW FASTENINGS
sb′ L
Eb
δ i′ =
sb′ (2 )
30 × 106
sb′ = 4830 psi
0.000322 =
238.
A 1 1/8-in. steel bolt A passes through a yellow brass (B36-8) tube B as
shown. The length of the tube is 30 in. (virtually the unthreaded bolt length),
the threads on the bolt are UNC, and the tube’s cross-sectional area is 2 sq. in.
After the nut is snug it is tightened ¼ turn. (a) What normal stresses will be
produced in the bolt and in the tube? Assume that washers, nut, and head are
rigid. (b) What are the stresses if an axial load of 5 kips is now applied to the
bolts end? Compute the bolt load that just results in a zero stress in the tube.
Prob. 238
Solution:
For Yellow brass, B36-8,
E = 15× 10 6 psi
Steel bolt
E = 30× 106 psi
Table AT 14, 1 1/8 in., UNC
Dr = 0.9497 in.
As = 0.763 sq.in.
Th in = 7
L = 30 in.
θ=
δi
p
1
p = in.
7
1
θ = turn
4
 1  1  1
δ i =    = in.
 4  7  28
Page 16 of 42
SECTION 3 – SCREW FASTENINGS
δi =
Fi L
Ab Eb
1
Fi (30)
=
28  π  1  2
6
 1  30 × 10
 4  8 
Fi = 35,500 lb
(
(a)
Bolt: sb =
)
Fi 35,500
=
= 46,527 psi
As
0.763
Fi
Ac
Ac = 2 sq.in.
F 35,500
sc = i =
= 17,750 psi
Ac
2
Tube: sc =
(b) Fe = 5000 lb
(
)
Ac Ec (2 ) 15 × 106
kc =
=
= 1,000,000 lb in
L
30
2
 π  1 
6
 1  30 × 10
Ab Eb  4  8 
kb =
=
= 994,000 lb in
L
30
(
)
Bolts:
 kb 
 Fe
Ft = Fi + 
 kb + k c 
994,000


Ft = 35,500 + 
(5000 ) = 38,000 lb
 994,000 + 1,000,000 
F 38,000
st = t =
= 49,800 psi
As
0.763
Tube:
 kc 
 Fe
Fc = Fi − 
 kb + kc 
1,000,000


Fc = 35,500 − 
(5000 ) = 33,000 lb
 994,000 + 1,000,000 
F 33,000
sc = c =
= 16,500 psi
Ac
2
For zero stress in the tube
Page 17 of 42
SECTION 3 – SCREW FASTENINGS
k +k
Fo =  b c
 kc

 994,000 + 1,000,000 
 Fi = 
(35,500) = 70,787 lb
1,000,000



ENDURANCE STRENGTH
DESIGN PROBLEMS
239.
As shown diagrammatically, a bearing is supported in a pillow block attached
to an overhead beam by two cap screws, each of which, it may be assumed,
carried half the total bearing load. This load acts vertically downward, varying
from 0 to 1500 lb. The screws are to be made of AISI C1118, as rolled, and
they are tightened to give an initial stress of about si = 0.5s y . The pillow
block is made of class-20 cast iron. Assume that the effective length of screw
is equal to the thickness t , as shown, and that the head and beam are rigid
(overly conservative?). The equivalent diameter of the compression area may
be taken as twice the bolt diameter. For a design factor of 1.75, determine the
size of the screw: (a) from the Soderberg line, (b) from the modified Goodman
line. (c) What size do you recommend using?
Problem 239
Solution:
For AISI C1118, as rolled
s y = 46 ksi
su = 75 ksi
si = 0.5s y
si = 0.5(46) = 23 ksi = 23,000 psi
Fi = si As = 23 As kip = 23,000 As lb
 kb 
 Fe
∆Fb = 
k
+
k
 b c
Ab Eb
kb =
Lb
Page 18 of 42
SECTION 3 – SCREW FASTENINGS
Eb = 30× 10 6 psi (steel)
Lb = t
Ac Ec
Lc
For cast-iron class 20
Ec = 9.6 × 106 psi
Lc = t
kc =
Ac =
π
4
De2 −
π
4
D2
π
D2
4
De = 2 D
Ab =
Ac =
π
(2 D )2 − π D 2 = 3π D 2 = 3 Ab
4
AE
kb = b b
Lb
(
4
4
)
Ab 30× 10 6
t
3 A 9.6 × 106
kc = b
t
kb
30 ×10 6
300
=
=
6
6
kb + k c 30 × 10 + 3(9.6 × 10 ) 588
∆Fb1 = 0
kb =
(
)
 kb 
 300  1500 
 Fe = 
∆Fb 2 = 

 = 383 lb
 588  2 
 k b + kc 
1
1
Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 23,000 As + (383 + 0) = 23,000 As + 192 lb
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (383 − 0) = 192 lb
2
2
s n = 0.5su = 0.5(75,000 psi ) = 37,500 psi
For axial loading with size factor
s n = (0.8)(0.85)(37,500 psi ) = 25,500 psi
N = 1.75
sm =
Fm 23,000 As 192
192
=
+
= 23,000 +
As
As
As
As
Page 19 of 42
SECTION 3 – SCREW FASTENINGS
sa =
Fa 192
=
As
As
Table AT 12, K f = 1.8
(a) Soderberg line
1 sm K f sa
=
+
N sy
sn

 192 
192 
 23,000 +
 (1.8)

As 
As 
1


=
+
1.75
46,000
25,500
As = 0.2482 sq.in.
Table AT 14, UNC
3
Use D = in. , As = 0.334 sq.in.
4
(b) Modifies Goodman line
1 sm K f sa
=
+
N su
sn

 192 
192 
 23,000 +
 (1.8)

As 
As 
1


+
=
1.75
75,000
25,500
As = 0.0609 sq.in.
Table AT 14, UNC
3
Use D = in. , As = 0.0775 sq.in.
8
(c) Recommended, D =
240.
3
in. − UNC
4
A connection similar to Fig. 5.9, Text, is subjected to an external load that
varied from 0 to 1250 lb. The bolt is cold forged from AISI B1113 steel; UNC
threads.The aluminum parts C (3003 H14) have a total thickness of 1 ½ in.
and an external diameter of 2 D . It is desired that the connection not open for
an external load of 1.5Fe . Determine (a) the initial tensile load on the bolt, (b)
the bolt diameter for N = 2 based on the Soderberg line.
Fig. 5.9
Page 20 of 42
SECTION 3 – SCREW FASTENINGS
Solution:
 kc 
 lb
(a) Fi = QFe 
k
+
k
 b c
Q = 1.5
AE
kb = b b
Lb
Ab =
π
D2
4
Eb = 30 × 10 6 psi
1
Lb = 1 in.
2
Ac Ec
kc =
Lc
Ac =
π
De2 −
4
De = 2 D
Ac =
π
π
4
D2
(2 D )2 − π D 2 = 3π D 2 = 3 Ab
4
4
4
6
Ec = 10 × 10 psi (3003-H14 aluminum)
1
Lc = 1 in.
2
A 30 × 10 6
kb = b
 1
1 
 2
3 A 10 × 10 6
kc = b
 1
1 
 2
kc
3 10 × 10 6
=
= 0.5
kb + k c 30 ×10 6 + 3 10 × 10 6
(
)
(
)
(
Page 21 of 42
(
)
)
SECTION 3 – SCREW FASTENINGS
Fe = 1250 lb
 kc 
 lb
Fi = QFe 
k
+
k
 b c
Fi = (1.5)(1250 )(0.5) = 937.5 lb
(b) For AISI B1113 steel, cold forged
su = 83 ksi
s y = 72 ksi
s n = 0.5su = 0.5(83) = 41.5 ksi = 41,500 psi
For axial loading with size factor
s n = (0.8)(0.85)(41,500 psi ) = 28,220 psi
 kb 
 Fe
∆Fb = 
 kb + k c 
∆Fb1 = 0
 kb 


30 × 106
 Fe = 
(1250 ) = 625 lb
∆Fb 2 = 
6
6 
 30 × 10 + 3(10 × 10 )
 kb + k c 
1
(∆Fb 2 + ∆Fb1 ) = 937.5 + 1 (625 + 0) = 1250 lb
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (625 − 0) = 312.5 lb
2
2
F
1250
sm = m =
As
As
F 312.5
sa = a =
As
As
Fm = Fi +
Soderberg line, K f = 1.8 Table AT 12
1 sm K f sa
=
+
N sy
sn
1
1250
(1.8)(312.5)
=
+
2 72,000 As
28,220 As
As = 0.07459 sq.in.
Table AT 14, UNC
Page 22 of 42
SECTION 3 – SCREW FASTENINGS
3
in. , As = 0.0775 sq.in.
8
Use D =
243.
This problem concerns the Freon compressor of 226: size, 10 x 18 in.; 10
studs, UNC; made of C1118, as rolled; 200 psi gas pressure. The initial
tension in the bolts, assumed to be equally loaded, is such that a cylinder
pressure of 300 psi is required for the joint to be on the opening. The bolted
parts are cast steel and for the first calculations, it will be satisfactorily to
assume the equivalent diameter of the compressed parts to be twice the bolt
size. (a) For N = 2 on the Soderberg criterion, what bolt size is required? (b)
Compute the torque required for the specified initial tension.
Solution:
 π  (10 )
Fo = 300 
= 2356 lb.
 4  10
 kc 

Fi = Fo 
 kb + k c 
2
kb =
Ab Eb
Lb
π
D2
4
Eb = 30 × 10 6 psi
Ab =
Lb = L
AE
kc = c c
Lc
Ac =
π
De2 −
4
De = 2 D
Ac =
π
π
4
D2
(2 D )2 − π D 2 = 3π D 2 = 3 Ab
4
4
4
6
Cast Steel, Ec = 30 × 10 psi
Lc = L
Ab (30× 10 6 )
L
3 Ab 30 × 106
kc =
= 3kb
L
 kc 
 3kb 
 = (2356 )
 = 1767 lb
Fi = Fo 
 kb + k c 
 kb + 3kb 
kb =
(
Page 23 of 42
)
SECTION 3 – SCREW FASTENINGS
 kb 
 Fe
(a) ∆Fb = 
 kb + k c 
∆Fb1 = 0
 kb
∆Fb 2 = 
 k b + kc

 k b  π  (10 )2 (200 )
 Fe = 
 
= 393 lb
10

 kb + 3kb  4 
1
(∆Fb 2 + ∆Fb1 ) = 1767 + 1 (393 + 0 ) = 1964 lb
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (393 − 0) = 196 lb
2
2
F
1964
sm = m =
As
As
F 196
sa = a =
As
As
For C1118, as rolled
su = 75 ksi
Fm = Fi +
s y = 46 ksi
s n = 0.5su = 0.5(75) = 37.5 ksi = 37,500 psi
For axial loading with size factor
s n = (0.8)(0.85)(37,500 psi ) = 25,500 psi
K f = 1.8 Table AT 12
1 sm K f sa
=
+
N sy
sn
1
1964
(1.8)(196)
=
+
2 46,000 As 25,500 As
As = 0.1131 sq.in.
Table AT 14, UNC
1
Use D = in. , As = 0.1419 sq.in.
2
(b) T = 0.2 DFi
1
T = 0.2 (1767 ) = 176.7 in − lb.
2
245.
A cast-iron (class 35) Diesel-engine cylinder head is held on 8 stud bolts with
UNC threads. These bolts are made of AISI 3140 steel, OQT 1000 oF (Fig.
AF2). Assume that the compressed material has an equivalent diameter twice
Page 24 of 42
SECTION 3 – SCREW FASTENINGS
the bolt size. The maximum cylinder pressure is 750 psi and the bore of the
engine is 8 in. Let the initial bolt load be such that a cylinder pressure of 1500
psi brings the joint to the point of opening. For a design factor of 2, determine
the bolt diameter (a) using the Soderberg equation, (b) using the Goodman
equation. (c) What approximate torque will be required to induce the desired
initial stress? (d) Determine the ratio of the initial stress to the yield strength.
Considering the lessons of experience (i5.8), what initial stress would you
recommend? Using this value, what factor of safety is computed from the
Soderberg equation?
Solution:
 π  (8)
Fo = 1500 
= 9425 lb.
4 8
 kc 

Fi = Fo 
 kb + k c 
2
kb =
Ab Eb
Lb
π
D2
4
Eb = 30 × 10 6 psi
Ab =
Lb = L
AE
kc = c c
Lc
Ac =
π
De2 −
4
De = 2 D
Ac =
π
π
4
D2
(2 D )2 − π D 2 = 3π D 2 = 3 Ab
4
4
4
6
Ec = 14.5 × 10 psi , for cast-iron (class 35)
Lc = L
(
)
Ab 30× 10 6
L
3 Ab 14.5 × 106
kc =
L
 kc 


3(14.5 × 10 6 )
 = (9425)
 = 5578 lb
Fi = Fo 
6
6 
 30 × 10 + 3(14.5 × 10 ) 
 kb + k c 
kb =
(
Page 25 of 42
)
SECTION 3 – SCREW FASTENINGS
 kb 
 Fe
∆Fb = 
 kb + k c 
∆Fb1 = 0
 kb 

 π  (8)2 (750 )
30 × 10 6
 Fe = 
∆Fb 2 = 
= 1923 lb

6
6 
8
 30 × 10 + 3(14.5 × 10 ) 4 
 kb + k c 
1
(∆Fb 2 + ∆Fb1 ) = 5578 + 1 (1923 + 0) = 6540 lb
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (1923 − 0) = 962 lb
2
2
Fm = Fi +
Fm 6540
=
As
As
F
962
sa = a =
As
As
(a) For AISI 3140 steel, OQT 1000 oF
su = 153 ksi
s y = 134 ksi
sm =
s n = 0.5su = 0.5(153) = 76.5 ksi = 76,500 psi
For axial loading with size factor
s n = (0.8)(0.85)(76,500 psi ) = 52,000 psi
Table AT 12, K f = 3.3 (hardened)
Soderberg Equation
1 sm K f sa
=
+
N sy
sn
1
6540
(3.3)(962)
=
+
2 134,000 As 52,000 As
As = 0.2197 sq.in.
Table AT 14, UNC
5
Use D = in. , As = 0.226 sq.in.
8
(b) Goodman Equation
1 sm K f sa
=
+
N su
sn
Page 26 of 42
SECTION 3 – SCREW FASTENINGS
(3.3)(962)
1
6540
=
+
2 153,000 As 52,000 As
As = 0.2076 sq.in.
Table AT 14, UNC
5
Use D = in. , As = 0.226 sq.in.
8
(c) T = 0.2 DFi
5
T = 0.2 (5578) = 697 in − lb.
8
Fi 5578
=
= 24,681 psi
As 0.226
s
24,681
Ratio = i =
= 0.184
s y 134,000
(d) si =
i5.8 si = 0.85s y = 0.85(134,000) = 113,900 psi
Factor of safety
Fi = si As = (113,900 )(0.226 ) = 25,742 lb
1
Fm = 25,742 + (1923) = 26,704 lb
2
1
Fa = (1923) = 962 lb
2
Fm 26,704
sm =
=
= 118,159 psi
As
0.226
F
962
sa = a =
= 4257 psi
As 0.226
Soderberg Equation
1 sm K f sa
=
+
N sy
sn
1 118,159 (3.3)(4257 )
=
+
N 134,000
52,000
N = 0.87
246.
A 30,000-lb. body is to be mounted on a shaker (vibrator). The shaker will
exert a harmonic force of F = 30,000 sin 2π t f lb. on the body where f cps is
the frequency and t sec. is the time. The frequency can be varied from 5 to
10,000 cps. The harmonic force will exert a tensile load on the bolts that
Page 27 of 42
SECTION 3 – SCREW FASTENINGS
attach the body to the shaker when F is positive. Determine the minimum
number of ½-in.-UNF bolts that must be used for N = 2 based on Soderberg
line. The material of the bolts is to be AISI 8630, WQT 1100 oF; the material
of the body that is to be vibrated is aluminum alloy, 2014-T6 and the joint is
not to open for an external force that is 1.25 times the maximum force exerted
by the shaker. It may be assumed that the equivalent diameter of the material
in compression is twice the bolt diameter.
Solution:
Fe min = 0
Fe max = 30,000 lb
 kc 

Fi = QFe 
k
+
k
 b c
Q = 1.25
AE
kb = b b
Lb
Ab =
π
D2
4
Eb = 30 × 10 6 psi
Lb = L
AE
kc = c c
Lc
Ac =
π
De2 −
4
De = 2 D
Ac =
π
π
4
D2
(2 D )2 − π D 2 = 3π D 2 = 3 Ab
4
4
4
6
Ec = 10.6 × 10 psi , (Aluminum 2014-T6)
Lc = L
Page 28 of 42
SECTION 3 – SCREW FASTENINGS
Ab Eb Ab (30× 106 )
kb =
=
Lb
L
Ac Ec 3 Ab (10.6 × 106 )
kc =
=
Lc
L
 kc 

Fi = QFe 
 kb + k c 
(
)


3 10.6 × 10 6
Fi = (1.25)(30,000 )
= 19, 296 lb
6
6 
30
×
10
+
3
10
.
6
×
10


(
)
 kb 
 Fe
∆Fb = 
k
+
k
 b c
∆Fb1 = 0
 kb 


30 × 106
 Fe = 
(30,000 ) = 14,563 lb
∆Fb 2 = 
6
6 
k
+
k
30
×
10
+
3
(
10
.
6
×
10
)
b
c




1
(∆Fb 2 + ∆Fb1 ) = 19,296 + 1 (14,563 + 0) = 26,578 lb
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (14,563 − 0 ) = 7282 lb
2
2
Fm = Fi +
Fm
nAs
F
sa = a
nAs
For ½-in.-UNF (Table AT 14)
As = 0.1419 sq.in.
F
26,578 187,300
sm = m =
=
nAs 0.1419n
n
F
7282
51,318
sa = a =
=
nAs 0.1419n
n
sm =
For AISI 8630, WQT 1100 oF
K f = 3.3
su = 137 ksi
s y = 125 ksi
s n = 0.5su = 0.5(137 ) = 68.5 ksi = 68,500 psi
For axial loading with size factor
Page 29 of 42
SECTION 3 – SCREW FASTENINGS
s n = (0.8)(0.85)(68,500 psi ) = 46,580 psi
Soderberg Equation, N = 2
1 sm K f sa
=
+
N sy
sn
1 187,300 (3.3)(51,318)
=
+
2 125,000n
46,580n
n = 10.3
Minimum number of bolts = 10 bolts
248.
The maximum external load on the cap bolts of an automotive connecting rod
end, imposed by inertia forces at top dead center, is taken to be 4000 lb.; the
minimum load is zero at bottom dead center. The material is AISI 4140, OQT
1100oF (qualifying for SAE grade 5); assume that s n′ = 0.45su . The grip for
through bolts is 1.5 in. For design purposed, let each bolt take half the load,
3
and use an equivalent De = 1 in. for the connected parts. The threads extend
8
a negligible amount into the grip. For the initial computation, use an opening
load Fo = 1.75 Fe . Considering the manner in which the bolt is loaded, we
decide that a design factor of 1.4 (Soderberg) should be quite adequate. (a)
Does a 5/16-24 UNF satisfy this situation? If not, what size do you
recommend? (b) Experience suggests that, in situations such as this, an initial
stress of the order suggested in i5.8, Text, is good insurance against fatigue
failure. Decide upon such an si and recomputed N . How does it change?
Would you be concerned about the safety in this case? Consider the variation
of si as a consequences of the use of torque wrench and also the stress
relaxation with time (due to seating and other factors), and discuss. Compute
the required tightening torque for each si .
Solution:
Fo = 1.75Fe = 1.75(4000) = 7000 lb
 kc
Fi = Fe 
 kb + kc
AE
kb = b b
Lb
Ab =
π
D2



4
Eb = 30 × 10 6 psi
Page 30 of 42
SECTION 3 – SCREW FASTENINGS
Lb = 1.5 in.
AE
kc = c c
Lc
Ac =
π
4
De2 −
π  3
π
4
2
D2
π
π
2
2
1  − D = 1.485 − D = 1.485 − Ab
4  8 4
4
Ec = 10.6 × 106 psi , (Aluminum 2014-T6)
Ac =
Lc = 1.5 in.
As ≈ Ab
Ab Eb Ab (30 × 106 )
kb =
=
Lb
1.5
Ac Ec (1.485 − As )(30 × 106 )
kc =
=
Lc
1.5
1.485 − As 
Fi = 7000
= 7000 − 4714 As
 1.485 
 kb 
 Fe
∆Fb = 
 kb + k c 
∆Fb1 = 0
 kb 
 A 
 Fe =  s (4,000 ) = 2694 As
∆Fb 2 = 
 1.485 
 kb + k c 
1
(∆Fb 2 + ∆Fb1 ) = 7000 − 4714 As + 1 (2694 As + 0) = 7000 − 3367 As
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (2694 As − 0 ) = 1347 As
2
2
Fm = Fi +
Table 5.2
su = 120 ksi
s y = 88 ksi
s n′ = 0.45su = 0.45(120 ) = 54 ksi = 54,000 psi
F
7000
sm = m =
− 3367
As
As
F
s a = a = 1347
As
Page 31 of 42
SECTION 3 – SCREW FASTENINGS
K f = 3.3 (hardened, Table AT 12)
Soderberg Equation, N = 1.4
1 sm K f sa
=
+
N sy
sn
1
7000
3367 (3.3)(1347 )
=
−
+
1.4 88,000 As 88,000
54,000
As = 0.1187 sq.in.
7
Table At14, we D = in , As = 0.1187 sq.in.
16
(a) 5/16-24 UNF will not
satisfy the situation. Instead use
D=
7
in ,
16
As = 0.1187 sq.in.
(b) i5.8, Text
si = 0.85s y = 0.85(88) = 74.8 ksi = 74,800 psi
Fi = si As = (74,800)(0.1187 ) = 8879 lb
1
1
Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 8879 + (2694 As + 0 ) = 8879 + 1347 As
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (2694 As − 0 ) = 1347 As
2
2
F
8879
sm = m =
+ 1347
As
As
F
s a = a = 1347
As
1 sm K f sa
=
+
N sy
sn
 8879

+ 1347 

1  0.1187
 + (3.3)(1347 )
=
N
88,000
54,000
N = 1.06 , it decreases
N > 1 , therefore, safe.
Considering variation of si , si tends to exceeds the limiting stress therefore reduces the
factor of safety. While stress relaxation tends to reduce the limiting stress approaching
the si and causing lower design factor.
(c) Fi = 7000 − 4714 As = 7000 − 4714(0.1187 ) = 6440 lb
Page 32 of 42
SECTION 3 – SCREW FASTENINGS
7
T = 0.2 DFi = 0.2 (6440) = 564 in − lb
 16 
at Fi = 8879 lb
7
T = 0.2 DFi = 0.2 (8879 ) = 777 in − lb
 16 
CHECK PROBLEMS
249.
A 1-in. steel bolt A (normalized AISI 1137, cold-rolled threads) passes
through a yellow brass tube B (B36-8, ½ hard) as shown. The tube length is
30 in., its cross-sectional area is 2 sq. in. and the UNC bolt threads extend a
negligible amount below the nut. The steel washers are ¼ in. thick and are
assumed not to bend (clearances are exaggerated). The nut is turned ¼ turn.
(a) If an external tensile axial load, varying from 0 to 5 kips, is repeatedly
applied to the bolt, what is the factor of safety of the bolt by the Soderberg
criterion? (b) What is the external load on the bolt at the instant that the load
on the tube becomes zero.
Problem 249, 250
Solution:
For 1-in. UNC
As = 0.606 sq.in.
Dr = 0.8466 in.
Th in. = 8
1
1
= in.
Th in 8
δ 1
=
p 4
1 1 1
δ =   = in.
4  8  32
p=
Page 33 of 42
SECTION 3 – SCREW FASTENINGS
1
Lb = 30 + 2  = 30.5 in.
4
Eb = 30× 10 6 psi
1
Fi (30.5)
=
δ=
π
32
(1)2 30 ×106
4
Fi = 24,141 lb
(
)
 kb 

∆Fb = Fe 
 kb + k c 
π
Ab Eb 4
=
Lb
AE
kc = c c
Lc
Ac = 2 sq.in.
Lc = 30 in.
(1)2 (30 ×106 )
kb =
30.5
= 772,525
Ec = 15× 106 psi (Yellow Brass)
(
)
2 15 ×10 6
kc =
= 1,000,000
30
(a) ∆Fb1 = 0
772,525


∆Fb 2 = (5000 )
 = 2179 lb
 772,525 + 1,000,000 
1
1
Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 24,141 + (2179 + 0 ) = 25,230 lb
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (2179 − 0) = 1090 lb
2
2
Fm 25,230
sm =
=
= 38,227 psi
As
0.606
F
1090
sa = a =
= 1799 psi
As 0.606
For normalized AISI 1137, cold-rolled thread
su = 98 ksi
s y = 58 ksi
s n = 0.5su = 0.5(98) = 49 ksi = 49,000 psi
K f = 1.4 (Table AT 12)
Page 34 of 42
SECTION 3 – SCREW FASTENINGS
For axial loading, s n = 0.8(49,000 ) = 39,200 psi
1 sm K f sa
=
+
N sy
sn
1 38,227 (1.4 )(1799)
=
+
N 58,000
39,200
N = 1.38
 kc 

(b) Fi = Fo 
 kb + k c 
1,000,000


24,141 = Fo 

 772,525 + 1,000,000 
Fo = 42,790 lb
250.
A ¾-in. fine-thread bolt, made of AISI 1117, cold drawn, with rolled threads,
passes through a yellow brass tube and two steel washers, as shown. The tube
is 4 in. long, 7/8 in. internal diameter, 1 ¼-in. external diameter. The washers
are each ¼-in. thick. The unthreaded part of the bolt is 3 in. long. Assume that
there is no stretching of the bolt inside the nut in finding its k . The
unlubricated bolt is tightened by a torque of 1800 in-lb. The external load,
varying from 0 to 4 kips, is axially applied to the washers an indefinite
number of times. (a) Compute the factor of safety of the bolt by the Soderberg
criterion. Is there any danger of failure of the bolt? (b) What pull must be
exerted by the washers to remove all load from the brass tube?
Solution:
T = 0.2 DFi
3
1800 = 0.2  Fi
4
Fi = 12,000 lb
 kb 

∆Fb = Fe 
 kb + k c 
1
1
1
=
+
kb kb1 kb 2
AE
kb1 = b b
Lb1
Lb1 = 3 in.
Ab =
π 3
2
  = 0.4418 sq.in.
4 4
Page 35 of 42
SECTION 3 – SCREW FASTENINGS
Eb = 30× 10 6 psi
(0.4418)(30 ×106 ) = 4,418,000
kb1 =
3
As Eb
Lb 2
For ¾-in. UNF (Table AT 14)
As = 0.373 sq.in.
kb 2 =
1
Lb 2 = 4 + 2  − 3 = 1.5 in.
2
(0.373) 30 ×106 = 7,460,000
kb 2 =
1.5
1
1
1
=
+
kb kb1 kb 2
1
1
1
=
+
kb 4,418,000 7,460,000
kb = 2,774,733
AE
kc = c c
Lc
(
)
2
2
π  1   7  
1  −    = 0.6259 sq.in.
4  4   8  
Ec = 15× 106 psi
Ac =
Lc = 4 in.
kc =
(0.6259)(15 ×106 ) = 2,347,125
4
∆Fb1 = 0
2,774,733


∆Fb 2 = (4000)
 = 2167 lb
 2,774,733 + 2,347,125 
1
1
Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 12,000 + (2167 + 0) = 13,084 lb
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (2167 − 0) = 1084 lb
2
2
F
13,084
sm = m =
= 35,078 psi
As
0.373
F
1084
sa = a =
= 2906 psi
As 0.373
For AISI 111, cold drawn, rolled threads
s n = 40 ksi
Page 36 of 42
SECTION 3 – SCREW FASTENINGS
s y = 68 ksi
K f = 1.4
s n = 0.8(40) = 32 ksi = 32,000 psi , axial loading
(a)
1 sm K f sa
=
+
N sy
sn
1 35,078 (1.4 )(2906 )
=
+
N 68,000
32,000
N = 1.56
 kc 

(b) Fi = Fo 
 kb + k c 
2,347,125


12,000 = Fo 

 2,774,733 + 2,347,125 
Fo = 26,186 lb
A coupling bolt (i5.13, Text) is used to connect two parts made of cast-iron,
251.
class 35. The diameter of the coarse-thread bolt is ½-in.; its grip is 2 in., which
is also nearly the unthreaded length. The bolt tightened to have an initial
tension of 4000 lb. The parts support an external load Fe that tends to separate
them and it varies from zero to 5000 lb. What is the factor of safety,
(Soderberg)?
Solution:
Fi = 4000 lb
 kb 

∆Fb = Fe 
 kb + k c 
AE
kb = b b
Lb
π 1
2
  = 0.19635 sq.in. (unthreaded length)
4 2
Eb = 30× 10 6 psi
Ab =
Lb = 2 in.
kb =
(0.19635)(30 ×106 ) = 2,945,250
2
Table AT 14, UNC
Page 37 of 42
SECTION 3 – SCREW FASTENINGS
1
in.
2
As = 0.1419 sq.in.
3
A = in.
4
h
De = A +
2
h = 2 in.
3 2
3
De = + = 1 in.
4 2
4
AE
kc = c c
Lc
D=
Ac =
π
4
De2 −
π
4
D2 =
2
2
π  3   1  
1  −    = 2.209 sq.in.
4  4   2  
Ec = 14.5 × 106 psi , (Cast iron, class 35)
Lc = 2 in.
kc
(
2.209 )(14.5 × 106 )
=
= 16,015,250
2
∆Fb1 = 0
2,945,250


∆Fb 2 = (5000 )
 = 777 lb
 2,945,250 + 16,015,250 
1
1
Fm = Fi + (∆Fb 2 + ∆Fb1 ) = 4000 + (777 + 0 ) = 4389 lb
2
2
1
1
Fa = (∆Fb 2 − ∆Fb1 ) = (777 − 0) = 389 lb
2
2
Fm
4389
sm =
=
= 30,930 psi
As 0.1419
F
389
sa = a =
= 2741 psi
As 0.1419
1
For ASTM 354 BC (Table 5.2), D = in.
2
su = 125 ksi
s y = 109 ksi
s n = 0.5su
For axial loading
s n = (0.8)(0.5)(125) = 50 ksi = 50,000 psi
K f = 1.8
Soderberg Line
Page 38 of 42
SECTION 3 – SCREW FASTENINGS
1 sm K f sa
=
+
N sy
sn
1
30,930 (1.8)(2741)
=
+
N 109,000
50,000
N = 2. 6
252.
The cap on the end of a connecting rod (automotive engine) is held on by two
5/16-in. bolts that are forged integrally with the main connecting rod. These
bolts have UNF threads with a 5/8-in. on an unthreaded length of virtually 5/8
in. The nuts are to be tightened with a torque of 20 ft-lb. and the maximum
external load on one bolt is expected to be 2330 lb. Let the equivalent
diameter of the connected parts be ¾ in. (a) Estimate the maximum force on
the bolt. (b) Compute the opening load. Is this satisfactory? (c) If the bolt
material is AISI 4140, OQT 1000 oF, what is the factor of safety based on the
Soderberg criterion?
Solution:
T = 20 ft − lb = 240 in − lb
T = 0.2 DFi
5
240 = 0.2  Fi
 16 
Fi = 3840 lb
 kb 

∆Fb = Fe 
 kb + k c 
AE
kb = b b
Lb
π5
2
  = 0.0767 sq.in. (unthreaded length)
4  16 
Eb = 30× 10 6 psi
5
Lb = in.
8
(0.0767) 30 ×106 = 3,681,600
kb =
5
 
8
AE
kc = c c
Lc
Ab =
(
π
π
)
2
2
π  3   5  
  −    = 0.3651 sq.in.
4  4   16  
Ec = 30× 10 6 psi , (Cast iron, class 35)
Ac =
4
De2 −
4
Page 39 of 42
D2 =
SECTION 3 – SCREW FASTENINGS
5
in.
8
(
0.3651) 30 × 10 6
kc =
= 17,524,800
5
 
8
3,681,600


∆Fb = (2330)
 = 405 lb
 3,681,600 + 17,524,800 
Lc =
(
)
(a) Fmax = Fi + ∆Fb = 3840 + 405 = 4245 lb
 kc 

(b) Fi = Fo 
k
+
k
 b c
17,524,800


3840 = Fo 

 3,681,600 + 17,524,80 
Fo = 4647 lb < Fmax
∆Fb
405
= 3840 +
= 4042 lb
2
2
∆F
405
Fa = b =
= 202 lb
2
2
For AISI 4140, OQT 1000 oF
su = 170 ksi
(c) Fm = Fi +
s y = 155 ksi
Table AT 12, K f = 2.6
s n = 0.5su
For axial loading
s n = (0.8)(0.5)(170) = 68 ksi = 68,000 psi
Soderberg Line
1 sm K f sa
=
+
N sy
sn
For 5/16-in.-UNF, Table AT 14, As = 0.0580 sq.in.
F
4042
sm = m =
= 69,690 psi
As 0.0580
F
202
sa = a =
= 3843 psi
As 0.0580
1 sm K f sa
=
+
N sy
sn
Page 40 of 42
SECTION 3 – SCREW FASTENINGS
1
69,690 (2.6 )(3483)
=
+
N 155,000
68,000
N = 1.72
SET SCREWS
254.
A 6-in. pulley is fastened to a 1 ¼ in. shaft by a set screw. If a net tangential
force of 75 lb, is applied to the surface of the pulley, what size screw should
be used when the load is steady?
Solution:
 6 
Tangential force = (75 lb )
 = 365 lb
 1.25 
Assume tangential force = holding force
Table 5.3, use Screw size 8, Holding force = 385 lb.
255.
An eccentric is to be connected to a 3-in. shaft by a setscrew. The center of the
eccentric is 1 ¼ in. from the center of the shaft when a tensile force of 1000
lb. is applied to the eccentric rod perpendicular to the line of centers. What
size set screw should be used for a design factor of 6?
Solution:
 1.25 
 = 833 lb
Tangential force = (1000 lb )
32
Holding force = (6)(833) = 5000 lb
Table 5.3, use Screw size ¾ in.
256.
A lever 16 in. long is to be fastened to a 2-in. shaft. A load of 40 lb. is to be
applied normal to the lever at its end. What size of set screw should be used
for a design factor of 5?
Solution:
Torque = (16)(40) = 640 in − lb
2(640 )
Tangential force =
= 640 lb
2
Holding force = (5)(640 ) = 3200 lb
Table 5.3, use Screw size 9/16 in.
257.
A 12-in. gear is mounted on a 2-in. shaft and is held in place by a 7/16 in.
setscrew. For a design factor of 3, what would be the tangential load that
could be applied to the teeth and what horsepower could be transmitted by the
screw.
Solution:
Page 41 of 42
SECTION 3 – SCREW FASTENINGS
Table 5.3, 7/16 in.
Holding force = 2500 lb
2500
Tangential force =
= 833 lb
3
2
Tangential load on gear = 833  = 139 lb
 12 
Assume vm = 4500 fpm
(139)(4500) = 19 hp
Hp transmitted =
33,000
- end -
Page 42 of 42
SECTION 4 - SPRINGS
HELICAL COMPRESSION SPRINGS
DESIGN – LIGHT, MEDIUM SERVICE
271.
A solenoid brake (Fig. 18.2, Text) is to be actuated by a helical compression
spring. The spring should have a free length of approximately 18 in. and is to
exert a maximum force of 2850 lb. when compressed to a length of 15 in. The
outside diameter must not exceed 7 in. Using oil-tempered wire, design a
spring for this brake, (wire diameter, coil diameter, number of active coils,
pitch, pitch angle, “solid stress”). General Electric used a spring made of 1 in.
wire, with an outside diameter of 6 in., and 11 ½ free coils for a similar
application.
Solution:
For oil tempered wire, Table AT 17
146
su = 0.19 ksi , [0.032 < Dw < 0.5]
Dw
“solid stress” = 0.6 s u
design stress, (average service)
ssd = 0.324 su
0.324(146 ) 47.304
ssd =
=
ksi
Dw0.19
Dw0.19
Dw + Dm ≤ 7
F = 2850 lb = 2.85 kips
 8FDm  47.304
=
ss = K 
3 
Dw0.19
 πDw 
say K = 1.3
 8(2.85)(7 − Dw )  47.304
ss = 1.3
=
πDw3
Dw0.19


Dw = 1.062 in > 0.5 in
47.304
use ssd =
ksi = 54 ksi
(0.5)0.19
 8(2.85)(7 − Dw ) 
ss = 1.3
 = 54
πDw3


Dw = 1.015 in
say Dw = 1.0 in
 8(2.85)Dm 
ss = 1.3
 = 54
3
 π (1) 
Dm = 5.72 in
Page 1 of 70
SECTION 4 - SPRINGS
say Dm = 5.0 in
OD = Dm + Dw = 5.0 + 1.0 = 6 in < 7 in.
D
5 .0
C= m =
=5
Dw 1.0
δ = Free length – Compressed length = 18 in – 15 in = 3 in.
δ=
8FC 3 N c
GDw
G = 10,500 ksi , Dw >
3
in
8
8(2.85)(5) N c
(10,500)(1)
N c = 11.05
say N c = 11.5
3
δ =3=
8(2.85)(5) (11.5)
= 3.12 in
(10,500)(1)
Free length = 15 + 3.12 = 18.12 in
3
δ=
At 2.85 kips
 8FDm 

ss = K 
3 
 πDw 
C =5
4C − 1 0.615 4(5) − 1 0.615
K=
+
=
+
= 1.3105
4C − 4
C
4(5) − 4
5
 8(2.85)(5) 
ss = 1.3105
 = 47.55 ksi
3
 π (1) 
Permissible solid stress
0.6(146 )
sso = 0.6 su =
ksi = 99.93 ksi
(0.5)0.19
F
k=
δ
using
or let δ T = Free length – Solid height
47.55 99.93
=
3.12
δT
δ T = 6.56 in
δ T = Free length – Solid height = (P − Dw )N c
6.56 = (P − 1)(11.5)
P = 1.570 in
Page 2 of 70
SECTION 4 - SPRINGS
1
use P = 1 in
2
Pitch angle,
 1.5 
P
o
o
λ = tan −1
= tan −1 
 = 5.5 < 12 , o.k.
πD
π
(
5
)


For actual solid stress
δ T = (1.5 − 1)(11.5) = 5.75 in.
47.55 sso
=
3.12 5.75
sso = 87.63 ksi < 99.93 ksi , ok
Summary of answer:
Dw = wire diameter = 1 in.
Dm = coil diameter = 5 in.
N c = no. of active coils = 11 1/2
P = pitch = 1 ½ in.
γ = pitch angle = 5.5o
sso = solid stress = 87.63 ksi
272.
A coil spring is to be used for the front spring of a automobile. The spring is
to have a rate of 400 lb./in., an inside diameter of 4 3/64 in., and a free length
of 14 1/8 in., with squared-and-ground ends. The material is to be oiltempered chrome vanadium steel. Decide upon the diameter of the wire and
the number of free coils for a design load of F = 1500 lb . Be sure “solid
stress” is all right. How much is the pitch angle?
Solution:
Table AT 17 Cr-V steel
168
su = 0.166 ksi , [0.032 < Dw 0.437]
Dw
average service
ssd = 0.324su
0.324(168) 54.432
ssd =
= 0.166 ksi
Dw0.166
Dw
Max “solid stress” = 0.6su
Page 3 of 70
SECTION 4 - SPRINGS
3
in = 4.046875 in
64
Dm = Dw + 4.046875 in
ID = Dm − Dw = 4
 8FDm 
 = ssd
ss = K 
3 
 πDw 
Assume K = 1.3
F = 1500 lb = 1.5 kips
ssd =
 8(1.5)(Dw + 4.046875) 
54.432
= 1.3

0.166
Dw
πDw3


Dw = 0.747 in > 0.437 in
use
54.432
ssd =
ksi = 62.45 ksi
(0.437 )0.19
 8(1.5)(Dw + 4.046875) 
ssd = 1.3
 = 62.45
πDw3


Dw = 0.724 in
3
use Dw = in
4
3
3
51
Dm = + 4 = 4 in
4
64
64
 8FDm 

ss = K 
3 
 πDw 
4C − 1 0.615
K=
+
4C − 4
C
 51 
4 
Dm  64 
C=
=
≈ 6.4
Dw
3
 
4
4(6.4) − 1 0.615
K=
+
= 1.235
4(6.4) − 4
6.4

 51  
 8(1.5) 4  
 64   = 53.64 ksi < 62.45 ksi , (o.k.)
ss = 1.235
3


3
π
 


4


δ=
8FC 3 N c
GDw
Page 4 of 70
SECTION 4 - SPRINGS
3
G = 10,500 ksi , D w > in
8
F 1500
δ= =
= 3.75 in
k
400
3
8(1.5)(6.4 ) N c
δ = 3.75 =
(10,500) 3 
4
N c = 9.4
Table AT 16, Total coils = N c + 2 = 9.4 + 2 = 11.4 for square and grounded end.
Summary of answer:
Dw = wire diameter = ¾ in.
No. of free coils = 11.4
To check for solid stress.
Permissible solid stress =
0.6(168)
= 115.65 ksi
(0.437 )0.166
Free length = PN c + 2 Dw
3
Solid height = Dw ( N c + 2 ) =  (11.4 ) = 8.55 in
4
 1

 14 − 8.55 
 = 78.74 ksi < 115.65 ksi (safe)
Solid stress = (53.64 ) 8
 3.75 




Pitch:
1
PN c + 2 Dw = 14 in
8
1
3
P(9.4 ) + 2  = 14
8
4
11
P = 1.343 in = 1 in
32
Pitch angle,
 11 
 1

−1 P
−1
= tan  32  = 5.1o < 12o , o.k.
λ = tan
πD
 π  4 51  
  64  
Page 5 of 70
SECTION 4 - SPRINGS
273.
A coiled compression spring is to fit inside a cylinder 5/8 in. in diameter. For
one position of the piston, the spring is to exert a pressure on the piston
equivalent to 5 psi of piston area, and in this position, the overall length of the
spring must not exceed (but may be less than) 2 in. A pressure of 46 psi on the
piston is to compress the spring ¾ in. from the position described above.
Design a spring for medium service. Specify the cheapest suitable material,
number of total and active coils for square-and-ground ends, and investigate
the pitch angle, and “solid stress”.
Solution:
 8FDm 

ss = K 
3 
D
π
w 

OD = Dm + Dw =
Dm + 1.5 Dw =
5
D
in − w
8
2
5
in
8
2
 π  5 
F1 = (5)   = 1.534 lb
 4  8 
2
 π  5 
F2 = (46 + 5)   = 15.647 lb
 4  8 
Using hard-drawn spring wire, Cost Index = 1
ssd = 0.324su (0.85)
140
su = 0.19 ksi
Dw
, [0.028 < Dw < 0.625]
70
ksi
Dw0.19
140 38.556
ssd = 0.324(0.85) 0.19 =
Dw
Dw0.19
Max “solid stress” =
 8FC  38.556
38,556
=
ss = K 
ksi = 0.19 psi
2 
0.19
Dw
Dw
 πDw 
 8(15.647 )C 
K
= 38,556 Dw1.81

π


K (39.845)C = 38,556 Dw1.81
Dm + 1.5Dw = 0.625
CDw + 1.5Dw = 0.625
0.625
Dw =
C + 1 .5
Page 6 of 70
SECTION 4 - SPRINGS
K=
4C − 1 0.615
+
4C − 4
C
1.81
 4C − 1 0.615 
 0.625 
+

(39.845)C = 38,556

C 
 4C − 4
 C + 1 .5 
 4C − 1 0.615 
1.81
+

(C + 1.5) C = 413.3
C 
 4C − 4
C = 7.035
0.625
0.625
=
= 0.0732 in
Dw =
C + 1.5 7.035 + 1.5
Table AT 15, Dw = 0.0720 in , W & M 15
Dm = 7.035(0.0720) = 0.5065 in
For N c
δ 2 − δ1 =
8(F2 − F1 )C 3 N c
GDw
G = 11.5 × 106 psi
3 8(15.647 − 1.534)(7.035) N c
δ 2 − δ1 = =
4
11.5 × 106 (0.0720)
N c = 15.8
Table AT 16,
Total coils = N c + 2 = 15.8 + 2 = 17.8
Solid height = ( N c + 2)Dw = (15.8 + 2)(0.0720) = 1.28 in
Free length = PN c + 2 Dw
Free length = 2 + δ1
3
(
δ1 =
)
8(F1 )C 3 N c
GDw
8(1.534 )(7.035) (15.8)
= 0.082 in.
11.5 × 106 (0.0720)
Free length = 2 + 0.082 = 2.082 in
3
δ1 =
δ2 =
(
)
8(F2 )C 3 N c
GDw
8(15.647 )(7.035) (15.8)
= 0.832 in.
11.5 × 106 (0.0720)
Solid Height ≤ Free Length - δ 2
Solid Height ≤ 2.082 − 0.832 in
Solid Height ≤ 1.25 in
But Solid Height > 1.25 in.
3
δ2 =
(
)
Therefore change material to Oil-tempered spring wire, Cost Index = 1.5
Page 7 of 70
SECTION 4 - SPRINGS
Table AT 17
146
su = 0.19 ksi , 0.028 < Dw < 0.5
Dw
Max “solid stress” =
ssd = 0.324
87.5
ksi
D w0.19
146 47.304
=
Dw0.19
Dw0.19
 8FC  47.304
47,304
=
ss = K 
ksi =
psi
2 
0.19
0.19
π
D
D
D
w
w
 w
(
)
 8 15.647 C 
K
= 47,304 Dw1.81

π


K (39.845)C = 47,304 Dw1.81
0.625
Dw =
C + 1 .5
4C − 1 0.615
K=
+
4C − 4
C
1.81
 4C − 1 0.615 
 0.625 
+

(39.845)C = 47,304

C 
 4C − 4
 C + 1 .5 
 4C − 1 0.615 
1.81
+

(C + 1.5) C = 507.1
C 
 4C − 4
C = 7.684
0.625
0.625
Dw =
=
= 0.0680 in
C + 1.5 7.684 + 1.5
Table AT 15, Dw = 0.0625 in , W & M 16
Dm = 7.684(0.0625) = 0.48025 in
15
say Dm =
= 0.46875 in
32
D
0.46875
C= m =
= 7 .5
Dw
0.0625
 8FC 

ss = K 
2 
 πDw 
4C − 1 0.615 4(7.5) − 1 0.615
K=
+
=
+
= 1.1974
4C − 4
C
4(7.5) − 4
7.5
 8(15.647 )(7.5) 
ss = 1.1974
= 91,600 psi = 91.6 ksi
2 
 π (0.0625) 
For N c
Page 8 of 70
SECTION 4 - SPRINGS
δ 2 − δ1 =
8(F2 − F1 )C 3 N c
GDw
G = 11.5 × 106 psi
3 8(15.647 − 1.534)(7.5) N c
=
4
11.5 × 106 (0.0625)
N c = 11.32
Table AT 16, squared and ground ends
Total coils = N c + 2 = 11.32 + 2 = 13.32
Solid height = ( N c + 2)Dw = (11.32 + 2)(0.0625) = 0.8325 in
Free length = PN c + 2 Dw
Free length = 2 + δ1
3
δ 2 − δ1 =
δ1 =
(
)
8(F1 )C 3 N c
GDw
8(1.534 )(7.5) (11.32)
= 0.082 in.
11.5 × 106 (0.0625)
Free length = 2 + 0.082 = 2.082 in = P(11.32) + 2(0.0625)
3
δ1 =
(
)
P = 0.1729 in ≈
11
in
64
Pitch angle,
 0.1729 
P
o
o
= tan −1 
λ = tan −1
 = 6.7 < 12 , o.k.
0
.
46875
πD
π
(
)


Solid stress
 2 − 0.8325 
sso = (91.6 )
 = 142.6 ksi
 0.75 
87.5
Permissible solid stress =
= 148.8 ksi > 137.5 ksi , safe.
(0.0625)0.19
Summary of answer:
Suitable material = Oil-Tempered Spring Wire
Total Coils = 13.32
Active Coils, N c = 11.32
274.
A helical spring is to fit about a 11/16-in. rod with a free length of 2 ¾ in. or
less. A maximum load of 8 lb. is to produce a deflection of 1 ¾ in. The spring
is expected to be compressed less than 5000 times during its life, but is
subjected to relatively high temperatures and corrosive atmosphere. Select a
material and determine the necessary wire size, mean coil diameter, and
number of coils. Meet all conditions advised by Text.
Page 9 of 70
SECTION 4 - SPRINGS
Solution:
For 5000 cycles < 104 cycles, use light service
Use stainless steel, type 302 (Cr-Ni), ASTM A313 – for relative high temperature and
corrosive atmosphere, Table AT 17.
ssd = 0.32su (i)
170
ksi , [0.01 < Dw < 0.13]
Dw0.14
97
su = 0.41 ksi , [0.13 < Dw < 0.375]
Dw
Maximum “solid” so = 0.47 su
su =
 8FDm 

ss = K 
3 
π
D
w 

F = 8 lb
11 D
Dm − Dw = + w
16 2
Dm − 1.5Dw = 0.6875 in
CDw − 1.5 Dw = 0.6875
0.6875
Dw =
C − 1 .5
4C − 1 0.615
K=
+
4C − 4
C
170
assume su = 0.14 ksi
Dw
0.32(170 ) 54.4
54,400
ssd =
= 0.14 ksi =
psi
0.14
Dw
Dw
Dw0.14
 4C − 1 0.615   8(8)C 
+
= 54,400 D1w.86



C  π 
 4C − 4
1.86
 4C − 1 0.615  64C 
 0.6875 
+


 = 54,400

C  π 
 4C − 4
 C − 1 .5 
 4C − 1 0.615 
1.86
+

(C − 1.5) C = 1330
C 
 4C − 4
C = 12.919
0.6875
Dw =
= 0.0602 in
12.919 − 1.5
Use Table AT 15, Dw = 0.0625 in , 16 W & M
Dm = 12.919(0.0602) = 0.8074 in
25
say Dm =
in = 0.78125 in
32
Page 10 of 70
SECTION 4 - SPRINGS
11
16
0.78125 − 0.0625 > 0.6875
0.71875 > 0.6875
Dm − Dw >
0.71875 − 0.6875 = 0.03125 =
C=
Dw
, o.k.
2
Dm 0.71875
=
= 12.5
Dw
0.0625
[0.0625 < 0.13], therefore,
su =
170
ksi is o.k.
Dw0.14
 8FC 

ss = K 
2 
π
D
 w
4(12.5) − 1 0.615
K=
+
= 1.1144
4(12.5) − 4 12.5
 8(8)(12.5) 
ss = 1.1144
= 72,648 psi
2
 π (0.0625) 
8(F )C 3 N c
δ=
GDw
G = 10.6 × 106 psi
3
8(8)(12.5) N c
4 10.6 × 106 (0.0625)
N c = 9.3
To check for solid stress and pitch
Minimum solid height = Dw N c = (0.0625)(9.3) = 0.58125 in
3
δ =1 =
(
)
(72,648) 2 3 − 0.58125 
 4
 = 90,000 psi = 90 ksi
3
1
4
(0.47 )(170) = 117.8 ksi > 90 ksi , o.k.
Permissible solid stress =
(0.0625)0.14
Free length = PN c , minimum
3
P (9.3) = 2
4
P = 0.2957 in
Pitch angle,
 0.2957 
P
= tan −1 
= 7.5o < 12o , o.k.
λ = tan −1

πD
 π (0.71825) 
Summary of answer
Material, Stainless Steel, Cr-Ni. ASTM A313
Solid stress =
Page 11 of 70
SECTION 4 - SPRINGS
Dw = 0.0625 in , 16 W & M
25
Dm =
in
32
N c = 9.3
275.
In order to isolate vibrations, helical compression springs are used to support a
machine. The static load on each spring is 3500 lb., under which the deflection
should be about 0.5 in. The solid deflection should be about 1 in. and the
outside coil diameter should not exceed 6 in. Recommend a spring for this
application; include scale, wire size, static stress, material, number of coils,
solid stress, and pitch of coils.
Solution:
Use Music wire (The best material)
Table AT 17
190
su = 0.154 ksi , [0.004 < Dw < 0.192]
Dw
Maximum “solid” sso = 0.5su
Light service, ssd = 0.405su
0.405(190 ) 76.95
76,950
ssd =
= 0.154 ksi = 0.154 psi
0.154
Dw
Dw
Dw
 8FC 

ss = K 
2 
 πDw 
F = 3500 lb
OD = Dm + Dw = 6 in
(C + 1)Dw = 6
6
Dw =
C +1



76,950
 4C − 1 0.615  8(3500 )C 
=
ss = 
+

C    6    6  0.154
 4C − 4
π
  C + 1    C + 1 


 4C − 1 0.615 
1.846
+
= 235.9

 C (C + 1)
C 
 4C − 4
C = 5.635
6
Dw =
= 0.9043 in > 0.192 in
5.635 + 1
76.950
use ss =
= 99,216 psi
(0.192)0.154
[
Page 12 of 70
]
SECTION 4 - SPRINGS



 4C − 1 0.615  8(3500 )C 
 = 99,216
ss = 
+

C   6  
 4C − 4
π
  C + 1  
 4C − 1 0.615 
2
+

C (C + 1) = 400.8
C 
 4C − 4
C = 6.205
6
= 0.8328 in
Dw =
6.205 + 1
13
Say Dw = in = 0.8125 in
16
Dm = (6.205)(0.8125) = 5.042 in
Say Dm = 5 in
D
5
C= m =
= 6.154
Dw 0.8125
4(6.154) − 1 0.615
K=
+
= 1.2455
4(6.154) − 4 6.154
 8FC 

ss = K 
2 
 πDw 
 8(3500)(6.154) 
ss = 1.2455
 = 103,481 psi > 99,216 psi , not o.k.
2
 π (0.8125) 
Use Dm = 4.5 in
D
4 .5
C= m =
= 5.5385
Dw 0.8125
4(5.5385) − 1 0.615
K=
+
= 1.2763
45.5385 − 4 5.5385
 8(3500)(5.5385) 
ss = 1.2763
 = 95,435 psi > 99,216 psi , o.k.
2
 π (0.8125) 
To check for solid stress
(0.5)(190) = 122.488 ksi = 122,488 psi
Permissible solid stress =
(0.192 )0.154
 1 
Solid stress = 
(95,435) = 190,870 psi > 122,488 psi , not ok
 0 .5 
Use
 0 .5 
ssd = 122,488
 = 61,244 psi
 1 
.
Page 13 of 70
SECTION 4 - SPRINGS



 4C − 1 0.615  8(3500 )C 
 = 61,244
ss = 
+

C   6  
 4C − 4
π
  C + 1  
 4C − 1 0.615 
2
+

C (C + 1) = 247.4
C 
 4C − 4
C = 5 .1
6
= 0.9836 in
Dw =
5 .1 + 1
Say Dw = 1.0 in
Dm = (5.1)(1.0) = 5.1 in
Say Dm = 5 in
D
5
C = m = =5
Dw 1
 8FC 

ss = K 
2 
 πDw 
4(5) − 1 0.615
K=
+
= 1.3105
4(5) − 4
5
 8(3500)(5) 
ss = 1.3105
= 58,400 psi > 61,244 psi o.k.
2 
(
)
1
.
0
π


Use Dw = 1.0 in , Dm = 5 in
 1 
Solid stress = 
(58,400 ) = 116,800 psi < 122,488 psi , o.k.
 0 .5 
8(F )C 3 N c
δ=
GDw
(Table AT 17)
G = 12× 106 psi
8(3500 )(5) N c
δ = 0.5 =
12 ×106 (1.0)
N c = 1.7143
say N c = 1.75
Free length – Solid length = Solid Deflection
PN c − Dw N c = 1 in
P(1.75) − (1)(1.75) = 1
9
P = 1.5714 in ≈ 1 in
16
Pitch angle,
3
(
Page 14 of 70
)
SECTION 4 - SPRINGS
 9 
1 
P
= tan −1  16  = 5.68o < 12o , o.k.
λ = tan −1
πD
 π (5) 


Summary of answer.
F 3500
Scale, k = =
= 7000 lb in
δ
0 .5
Wire size, Dw = 1.0 in
Material = Music Wire
Solid sress = 116,800 psi
9
Pitch of stress = P = 1 in
16
CHECK PROBLEMS – LIGHT, MEDIUM SERVICE
276.
The front spring of an automobile has a total of 9 ½ coils, 7 3/8 active coils
(square-and-ground ends), an inside diameter of 4 3/64 in., and a free length
of 14 ¼ in. It is made of SAE 9255 steel wire, OQT 1000oF, with a diameter
of 43/64 in. Compute (a) the rate (scale) of the spring; (b) the “solid stress”
and compare with a permissible value (is a stop needed to prevent solid
compression?). (c) Can 95 % of the solid stress be repeated 105 times without
danger of failure? Would you advise shot peening of the spring?
Solution:
8FC 3 N c
GDw
43
3
Dw =
in > in
64
8
6
G = 10.5 × 10 psi
D
C= m
Dw
Dm − Dw = ID
43
3
Dm −
= 4 in
64
64
23
Dm = 4 in
32
23
4
C = 32 = 7.0233
43
64
(a) δ =
Page 15 of 70
SECTION 4 - SPRINGS
Nc = 7
3
8
43 
(10.5 ×10 ) 64

6
F
GD
k = rate = = 3 w =
δ 8C N c
  = 345 lb in
3 3 
8(7.0233)  7 
 8
(b) “Solid Stress”
 43  1 
Solid height = N c = (Dw )(Total Coils ) =   9  = 6.3828 in
 64  2 
Solid deflection = Free length – Solid height = 14 ¼ - 6.3828 = 7.8672 in.
Solid Force = Fso = 7.8672(345) = 2714 lb
 8F C 
Solid Stress = K  so2 
 πDw 
4C − 1 0.615
K=
+
4C − 4
C
4(7.0233) − 1 0.615
K=
+
= 1.212
4(7.0233) − 4 7.0233




8(2714)(7.0233) 

ss = 1.212
= 130,322 psi
2


 43 
π 


 64 


Permissible value, ss = s ys = 0.6 sy , [Dw > 0.5 in]
SAE 9255, OQT 1000 oF
s y = 160 ksi su = 180 ksi
,
s ys = 0.6(160 ) = 96 ksi = 96,000 psi < 130,322 psi
Therefore a stop is needed to prevent solid compression.
(c) ssd = 0.324su (105 cycles)
ssd = 0.324(180) = 58.32 ksi
0.95sso = 0.95(130,322) = 123,800 psi = 123.8 ksi > 58.32 ksi
There is a danger of failure, shot peening is advisable
s ys = 1.25(96,000 ) = 120,000 psi ≈ 0.95sso
277.
An oil-tempered steel helical compression spring has a wire size of No. 3 W
& M, a spring index of 4.13, 30 active coils, a pitch of 0.317 in., ground-andsquared ends; medium service. (a) What maximum load is permitted if the
recommended stress is not exceeded (static approach)? Compute (b) the
Page 16 of 70
SECTION 4 - SPRINGS
corresponding deflection, (c) “solid stress,”. (d) pitch angle, (e) scale, (f) the
energy absorbed by the spring from a deflection of 0.25 in. to that of the
working load. (g) Is there any danger of this spring buckling? (h) What
maximum load could be used if the spring were shot peened?
Solution:
Table AT 17, oil-tempered
146
su = 0.19 ksi , [0.032 < Dw < 0.5]
Dw
87.5
Maximum “solid” sso = 0.19 ksi
Dw
ssd = 0.324su (medium service)
Table AT 15, No. 3 W & M
Dw = 0.2437 in
C = 4.13
Dm = CDw = 4.13(0.2437 ) = 1.0 in
 8F C 
(a) ss = K  s 2 
 πDw 
4C − 1 0.615
K=
+
4C − 4
C
4(4.13) − 1 0.615
K=
+
= 1.3885
4(4.13) − 4 4.13
0.324(146 )
ss = ssd =
= 61.858 ksi = 61,858 psi
(0.2437 )0.19
 8(F )(4.13) 
ss = 61,858 = 1.3885
2
 π (0.2437 ) 
F = 252 lb
8FC 3 N c
δ=
GDw
G = 11.5 × 106 psi
N c = 30
8(252)(4.13) (30)
= 1.52 in
11.5 × 106 (0.2437 )
3
δ=
(
)
(c) For solid stress . Square-and-ground end)
Free length = PN c + 2 Dw = (0.317 )(30) + 2(0.2437 ) = 9.9974 in
Page 17 of 70
SECTION 4 - SPRINGS
Solid height = Dw N c + 2 Dw = (30 + 2)(0.2437 ) = 7.7984 in
Solid deflection = 9.9974 – 7.7984 = 2.199 in.
 2.199 
Solid stress = (61,858)
 = 89,491 psi
 1.52 
Maximum “solid” sso =
87.5
87.5
ksi =
ksi = 114.4 ksi > 89.491 ksi , o.k. safe
0.19
Dw
(0.2437 )0.19
 0.317 
P
= tan −1 
= 5.76o < 12o , o.k.

πD
 π (1) 
F 252
(e) scale = k = =
= 166 lb in
δ 1.52
1
(f) U s = k (δ 22 − δ12 )
2
k = 166 lb in
δ1 = 0.25 in
δ 2 = 1.52 in
1
2
2
U s = (166 ) (1.52 ) − (0.25) = 186.6 in − lb
2
(d) λ = tan −1
[
]
(g) i 6.18 Free length = 9.9974 in
Mean Diameter = Dm = 1.0 in
Free length
9.9974
=
= 9.9974 > 4
Mean Diameter
1 .0
There is a danger for spring buckling
(h) Shot peened, Table AT 17
ssd = (61,858)(1.25) = 77,322 psi
 8(F )(4.13) 
ss = 77,322 = 1.3885
2
 π (0.2437 ) 
F = 314 lb
280.
It is desired to isolate a furnace, weighing 47,300 lb., from the surroundings
by mounting it on helical springs. Under the weight, the springs should deflect
approximately 1 in., and at least 2 in. before becoming solid. It has been
decided to use springs having a wire diameter of 1 in., an outside diameter of
5 3/8 in., 4.3 free coils. Determine (a) the number of springs to be used, (b)
the stress caused by the weight, (c) the “solid stress”. (d) What steel should be
used?
Solution:
Page 18 of 70
SECTION 4 - SPRINGS
Dw = 1 in
3
Dm + Dw = 5 in
8
3
Dm = 4 in
8
3
4
D
C = m = 8 = 4.375
Dw
1
8FC 3 N c
(a) δ =
GDw
Assume N c = 4.3
G = 10.5 × 106 psi , Dw >
3
in
8
8F (4.375) (4.3)
10.5 × 106 (1)
F = 3645 lb
W 47,300
No. of springs =
=
= 13
F
3645
3
δ = 1.0 =
(
)
W 47,300
=
= 3638 lb
13
13
 8F C 
ss = K  s 2 
 πDw 
(b) F =
4C − 1 0.615
+
4C − 4
C
4(4.375) − 1 0.615
K=
+
= 1.3628
4(4.375) − 4 4.375
K=
 8(3638)(4.375) 
ss = 1.3628
 = 55,235 psi
π (1.0)2


2
(c) “Solid Stress” = ss = 55,235  = 110,470 psi
1
(d) s ys ≈ 110,470 psi
s ys
110,470
= 184,117 psi = 184.117 ksi
0.6
0.6
From Table AT 7,
Use AISI 8760, OQT 800 oF, s y = 200 ksi
sy =
=
Page 19 of 70
SECTION 4 - SPRINGS
VARYING STRESS APPROACH
DESIGN PROBLEMS
282.
A spring, subjected to a load varying from 100 lb. to 250 lb., is to be made of oiltempered, cold-wound wire. Determine the diameter of the wire and the mean
diameter of the coil for a design factor of 1.25 based on Wahl’s line. The spring
index is to be at least 5. Conform to good practice, showing checks for all
significant parameters. Let the free length be between 6 and 8.
Solution:
Fmax = 250 lb
Fmin = 100 lb
1
1
Fm = (Fmax + Fmin ) = (250 + 100 ) = 175 lb = 0.175 kip
2
2
1
1
Fa = (Fmax − Fmin ) = (250 − 100 ) = 75 lb = 0.075 kip
2
2
Wahl’s line
1 sms − sas 2sas
=
+
N
s ys
sno
8 KFa Dm 8 KFa C
=
sas =
πDw3
πDw2
8 KFm Dm 8 KFmC
sms =
=
K cπDw3
K cπDw2
C =5
4C − 1 0.615
K=
+
4C − 4
C
4(5) − 1 0.615
K=
+
= 1.31
4(5) − 4
5
Fig. AF 15, C = 5
K c = 1.19
For oil-tempered wire,
87.5
s ys = 0.19
Dw , [0.032 < Dw < 0.5]
47
Dw0.1 , [0.041 < Dw < 0.15]
30
sno = 0.34
Dw , [0.15 < Dw < 0.625]
sno =
N = 1.25
Page 20 of 70
SECTION 4 - SPRINGS
8(1.31)(0.075)(5) 1.251
=
πDw2
Dw2
8(1.31)(0.175)(5) 2.453
sms =
=
(1.19)πDw2
Dw2
30
say sno = 0.34 ksi
Dw
1 sms − sas 2sas
=
+
N
s ys
sno
sas =
 2.453 − 1.251 
 1.251 

 2 2 
2
Dw
1
 +  Dw 
=
1.25
 87.5 
 30 
 0.19 
 0.34 
 Dw 
 Dw 
1
1
1
=
+
1.81
1.25 72.8 Dw
11.99 Dw1.66
Dw = 0.2857 in > 0.15 in
Table AT 15, use No. 1, W & M
Dw = 0.2830 in
Dm = CDw = 5(0.2830) = 1.415 in
7
say Dm = 1 in
16
Check for Free length
6 in < Free length < 8 in
 7
Free length = 4 Dm = 41  = 5.75 in
 16 
Increase Dm
1
Dm = 1 in
2
 1
Free length = 4 Dm = 41  = 6 in , o.k.
 2
Summary of answer
Dw = 0.2830 in
1
Dm = 1 in
2
283.
A carbon-steel spring is to be subjected to a load that varies from 500 to 1200 lb.
The outside diameter should be between 3.5 and 4 in., the spring index between 5
to 10; approximate scale of 500 lb./in. Choose a steel and for a design factor of
1.4 by the Wahl line, find the wire diameter. Also determine the number of active
Page 21 of 70
SECTION 4 - SPRINGS
coils and the free length for squared-and-ground ends. Conform to the general
conditions specified in the Text.
Solution:
For carbon steel, Table AT 17
91
s ys = 0.1 ksi , [0.093 < Dw < 0.25]
Dw
49
sno = 0.15 ksi , [0.093 < Dw < 0.25]
Dw
Fmax = 1200 lb
Fmin = 500 lb
1
1
Fm = (Fmax + Fmin ) = (1200 + 500 ) = 850 lb = 0.85 kip
2
2
1
1
Fa = (Fmax − Fmin ) = (1200 − 500 ) = 350 lb = 0.35 kip
2
2
OD = 3.5 ~ 4.0 in
C = 5 ~ 10
Wahl’s line
1 sms − sas 2sas
+
=
N
s ys
sno
Figure AF 15, C = 5 ~ 10
Assume K = 1.2 , K c = 1.125
8 KFa Dm
sas =
πDw3
8 KFm Dm
sms =
K cπDw3
OD ≈ 3.75 in
Dm = 3.75 − Dw
8(1.2 )(0.35)(3.75 − Dw ) 1.0695(3.75 − Dw )
=
sas =
πDw3
Dw3
8(1.2 )(0.85)(3.75 − Dw ) 2.3088(3.75 − Dw )
sms =
=
(1.125)πDw3
Dw3

Dw

 91 
 0.1 
 Dw 
1
3.75 − Dw
3.75 − Dw
=
+
2.9
1.4 73.4285 Dw
22.9079 Dw2.85
Dw = 0.6171 in > 0.25 in
Use
1
=
1.4

 3.75 − Dw 

2(1.0695)
Dw3
+


 49 
 0.15 
 Dw 
(2.3088 − 1.0695) 3.75 −3 Dw 
Page 22 of 70
SECTION 4 - SPRINGS
91
= 104.53 ksi
(0.25)0.1
49
sno =
= 60.33 ksi
(0.25)0.15
s ys =


 3.75 − Dw 

2(1.0695)
Dw3
+


60.33
(2.3088 − 1.0695) 3.75 −3 Dw 
Dw
1

=
1 .4
104.53
1
3.75 − Dw 3.75 − Dw
=
+
1.4 84.346 Dw3 28.205 Dw3
1
3.75 − Dw
=
1.4 21.137 Dw3
Dw = 0.5935 in
use
19
Dw =
in
32
3
Dm + Dw ≈ 3 in
4
19
3
= 3 in
Dm +
32
4
5
Dm = 3 in
32
 5 
3 
Dm  32 
C=
=
= 5.316
Dw
 19 
 
 32 
. o.k.
19
Wire Diameter Dw =
in , Carbon Steel
32
Number of coils:
8FC 3 N c
δ=
GDw
3
G = 10.5 × 106 psi = 10,500 ksi , Dw > in
8
F
GDw
=k = 3
δ
8C N c

(10.5 ×10 ) 19

32
6
 
8(5.316 ) N c
N c = 10.4
Table AT 16, square-and-ground ends
500 =
Page 23 of 70
3
SECTION 4 - SPRINGS
Free length = PN c + 2 Dw
Solid height = Dw N c + 2 Dw
Total Coils = N c + 2
 19 
Solid height = Dw N c + 2 Dw = (10.4 + 2 )  = 7.3625 in
 32 
F 1200
δ= =
= 2.4 in
k
500
Min. Free length = 2.4 + 7.3625 in = 9.7625 in
Use Free length = 10 in
To check for pitch angle.
Free length = PN c + 2 Dw
 19 
P(10.4 ) + 2  = 10
 32 
P = 0.8474 in



P
0.8474 
 = 4.885o < 12o , o.k.
= tan −1 
λ = tan −1
πDm
π  3 5  
  32  
Solid stress:
δ T = solid deflection = 10 − 7.3625 = 2.6375 in
F = kδ T = (500)(2.6375) = 1319 lb
4C − 1 0.615
K=
+
4C − 4
C
4(5.316) − 1 0.615
K=
+
= 1.29
4(5.316) − 4 5.316
 5
8(1.29)(1319) 3 
8 KFDm
 32  = 23,033 psi = 23.033 ksi < s (= 104.53 ksi )
ss =
=
ys
3
3
πDw
 19 
π 
 32 
284.
A helical compression spring, made of oil-tempered, cold-wound carbon steel, is
to be subjected to a working load varying from 100 to 300 lb. for an indefinite
time (severe). A mean coil diameter of 2 in. should be satisfactory. (a) Using the
static approach, compute a wire diameter. (b) For this wire size, compute the
factor of safety as given by the Wahl line.
Solution:
Page 24 of 70
SECTION 4 - SPRINGS
Table AT 16,
For carbon steel,
182
su = 0.1 ksi , [0.093 < Dw < 0.25]
Dw
91
s ys = 0.1 ksi
91
Dw
Max. “solid” s ys = 0.1 ksi
Dw
49
sno = 0.15 ksi , [0.093 < Dw < 0.25]
Dw
Dm = 2 in.
Fmax = 300 lb
Fmin = 100 lb
(a) F = 300 lb = 0.3 kip
severe service, ssd = 0.263su =
(0.263)(182) = 47.866 ksi
Dw0.1
 8F C 
ss = K  s 2 
 πDw 
4C − 1 0.615
K=
+
4C − 4
C
Dm
C=
Dw
D
2
Dw = m =
C C




 4C − 1 0.615   8(0.3)(2 ) 47.866
ss = 
+
=
C    2 3   2  0.1
 4C − 4
 π    
 C   C 
 4C − 1 0.615  2.9
 4C − 4 + C C = 233.84


C = 6.075
2
Dw =
= 0.3292 in > 0.25 in
6.075
47.866
Therefore use ssd =
= 54.984 ksi
(0.25)0.1
Page 25 of 70
Dw0.1
SECTION 4 - SPRINGS




 4C − 1 0.615   8(0.3)(2 )
ss = 
+
= 54.984
C    2 3 
 4C − 4
 π  
 C 
 4C − 1 0.615  3
 4C − 4 + C C = 287.9


C = 6.136
2
= 0.3259 in
Dw =
6.136
21
say Dw =
in
64
91
= 104.53 ksi
(0.25)0.1
49
sno =
= 60.33 ksi
(0.25)0.15
1
1
Fm = (Fmax + Fmin ) = (300 + 100 ) = 200 lb = 0.2 kip
2
2
1
1
Fa = (Fmax − Fmin ) = (300 − 100 ) = 100 lb = 0.1 kip
2
2
Dm
2
C=
=
= 6.095
Dw  21 
 
 64 
Figure AF 15
K c = 1.15
K = 1.25
K  8Fm Dm 
sms =


K c  πDw3 
(b) s ys =
sms




1.25  8(0.2 )(2) 
=
= 31.34 ksi
1.15   21 3 
π  
  64  
sas =
8 KFa Dm
πDw3
Page 26 of 70
SECTION 4 - SPRINGS




8(0.1)(2) 

sas = 1.25
= 18.02 ksi
  21 3 
π  
  64  
Wahl’s line
1 sms − sas 2sas
=
+
N
s ys
sno
1 31.34 − 18.02 2(18.02 )
=
+
N
104.53
60.33
N = 1.38
285.
A helical spring of hard-drawn wire with a mean diameter of 1 ½ in. and squareand-ground ends is to be subjected to a maximum load of 325 lb. (a) Compute the
wire diameter for average service. (b) How many total coils are required if the
scale is 800 lb./in.? (c) For a minimum load of 100 lb., what is the factor of safety
according to Wahl line? Would it be safe for an indefinite life?
Solution:
Table AT 17,
Hard-drawn wire,
140
su = 0.19 ksi , [0.028 < Dw < 0.625]
Dw
70
Maximum “solid” ss = s ys = 0.19 ksi
Dw
(0.9)(47 ) ksi , [0.041 < D < 0.15]
sno =
w
Dw0.1
(0.9)(30) ksi , [0.15 < D < 0.625]
sno =
w
Dw0.34
Average service
(a) ssd = 0.85(0.324 )su = 0.2754 su =
F = 325 lb = 0.325 kip
1
Dm = 1 in
2
 8 FDm 
ss = K 
3 
 πDw 
Page 27 of 70
0.2754(140 ) 38.556
=
ksi
Dw0.19
Dw0.19
SECTION 4 - SPRINGS
4C − 1 0.615
+
4C − 4
C
1 .5
Dw =
C
K=




 4C − 1 0.615   8(0.325)(1.5)  38.556
ss = 
+
=

3
0.19
C 
 4C − 4
 1.5    1.5 
 π    
C   C 

 4C − 1 0.615  2.81
+

C = 97.05
C 
 4C − 4
C = 4.586
1 .5
1 .5
Dw =
=
= 0.3271 in < 0.625 in
C 4.586
21
Dw =
in
64
Dm
1 .5
=
= 4.57
Dw  21 
 
 64 
4(4.57 ) − 1 0.615
K=
+
= 1.345
4(4.57 ) − 4 4.57
(b) C =
8FC 3 N c
GDw
F
GD
=k = 3 w
δ
8C N c
k = 800 lb in = 0.8 kip in
(11,500) 21 
 64 
0.8 =
3
84.57 N c
N c = 6.2
δ=
(c) s ys =
70
0.19
 21 
 
 64 
(0.9)(30)
= 86.5 ksi
= 39.44 ksi , Dw > 0.15 in
0.19
 21 
 
 64 
1
Fm = (325 + 100 ) = 212.5 lb = 0.2125 kip
2
sno =
Page 28 of 70
SECTION 4 - SPRINGS
1
(325 − 100) = 112.5 lb = 0.1125 kip
2
K c = 1.212 , Fig. AF 15
K = 1.345




K  8Fm Dm  1.345  8(0.2125)(1.5)
sms =
=
= 25.5 ksi


3

K c  πDw3  1.212 
 21 
 π 

 64 






 8Fm Dm 
8(0.1125)(1.5) 

sas = K 
= 1.345
= 16.36 ksi
3
3 


 21 
 πDw 
 π 

 64 


1 sms − sas 2sas 25.5 − 16.36 2(16.36)
+
=
+
=
86.5
39.44
N
s ys
sno
Fa =
N = 1.07 < 1.15[N min ]
Not safe for indefinite life.
286.
A helical spring is to be subjected to a maximum load of 200 lb. (a) Determine
the wire size suitable for medium service if the material is carbon steel ASTM
A230; C = 6 . Determine the factor of safety of this spring according to the Wahl
line (b) If the minimum force is 150 lb., (c) if the minimum force is 100 lb., (d) if
the minimum force is 25 lb.
Solution:
For carbon steel ASTM A230
Table AT 17
182
su = 0.1 ksi , [0.093 < Dw < 0.25]
Dw
91
s ys = 0.1 ksi , [0.093 < Dw < 0.25]
Dw
49
sno = 0.15 ksi , [0.093 < Dw < 0.25]
Dw
Medium Service
ssd = 0.324su
 182  58.968
58,968
(a) ssd = 0.324 0.1  =
ksi =
psi
0.1
Dw
Dw0.1
 Dw 
Page 29 of 70
SECTION 4 - SPRINGS
 8 FDm 
ss = K 
3 
 πDw 
4C − 1 0.615
K=
+
4C − 4
C
4(6) − 1 0.615
K=
+
= 1.2525
4(6) − 4
6
F = 200 lb
 8(200)(6)  58,968
ss = 1.2525
=
2
Dw0.1
 πDw 
Dw = 0.2371 in
Table At 15, use Dw = 0.2437 in , No. 3 W & M
Dw = 0.2437 in < 0.25 in , o.k.
Factor of safety.
91
91
s ys = 0.1 ksi =
ksi = 104.8 ksi
Dw
(0.2437 )0.1
49
49
sno = 0.15 ksi =
ksi = 60.56 ksi
Dw
(0.2437 )
(a) Fm =
1
(200 + 150) = 175 lb = 0.175 kip
2
1
(200 − 150) = 25 lb = 0.025 kip
2
Figure AF 15, K c = 1.156
Fa =
K  8FmC  1.2525  8(0.175)(6 ) 

 = 48.8 ksi

=
K c  πDw2  1.156  π (0.2437 )2 
 8(0.025)(6 ) 
 8F C 
sas = K  a 2  = 1.2525
= 8.1 ksi
2
 πDw 
 π (0.2437 ) 
1 sms − sas 2sas
=
+
N
s ys
sno
1 48.8 − 8.1 2(8.1)
=
+
N
104.8
60.56
N = 1.525
sms =
(b) Fm =
Fa =
1
(200 + 100) = 150 lb = 0.15 kip
2
1
(200 − 100) = 50 lb = 0.05 kip
2
Page 30 of 70
SECTION 4 - SPRINGS
Figure AF 15, K c = 1.156
 8FmC  1.2525  8(0.15)(6 ) 
=
= 41.8 ksi


2
2 
 πDw  1.156  π (0.2437 ) 
 8(0.05)(6) 
 8F C 
sas = K  a 2  = 1.2525
= 16.11 ksi
2
 πDw 
 π (0.2437 ) 
1 sms − sas 2sas
+
=
N
s ys
sno
1 41.8 − 16.11 2(16.11)
+
=
N
104.8
60.56
N = 1.287
sms =
K
Kc
(c) Fm =
1
(200 + 25) = 112.5 lb = 0.1125 kip
2
1
(200 − 25) = 87.5 lb = 0.0875 kip
2
Figure AF 15, K c = 1.156
Fa =
K  8FmC  1.2525  8(0.1125)(6) 

 = 31.36 ksi

=
K c  πDw2  1.156  π (0.2437 )2 
 8(0.0875)(6) 
 8F C 
sas = K  a 2  = 1.2525
= 28.2 ksi
2 
D
π
(
)
0
.
2437
π
 w 


1 sms − sas 2sas
=
+
N
s ys
sno
1 31.36 − 28.20 2(28.20 )
=
+
N
104.8
60.56
N = 1.04
sms =
CHECK PROBLEMS
A Diesel valve spring is made of 3/8-in. chrome-vanadium steel wire, shot-peened; inside
diameter is 3 in., 7 active coils, free length is 7 3/8 in., solid length is 4 1/8 in., length
with valve closed, 6 ¼ in., length when open, 5 1/8 in. (a) Compute the spring constant
and the factor of safety as defined by the Wahl criterion (see § 6.13, Text). (b) Is there
any danger of damage to the spring if it is compressed solid? (c) What is the natural
frequency? If this spring is used on a 4-stroke Diesel engine at 450 rpm, is there any
danger of surge? Compute the change of stored energy between working lengths.
Solution:
For chrome-vanadium steel wire, shot-peened, Table AT 17
Page 31 of 70
SECTION 4 - SPRINGS
(1.25)(168) ksi , [0.032 < D
w < 0.437 ]
Dw0.166
(1.25)(100 ) ksi , [0.032 < D < 0.437]
s ys =
w
Dw0.166
(1.25)(49) ksi , [0.028 < D < 0.5]
sno =
w
Dw0.15
3
Dw = in = 0.375 in
8
(1.25)(100) ksi = 147.1 ksi
s ys =
(0.375)0.166
(1.25)(49) ksi = 70.96 ksi
sno =
(0.375)0.15
su =
8FC 3 N c
GDw
F
GD
=k = 3 w
δ
8C N c
(a) δ =
G = 11.5 × 106 psi
Nc = 7
Dw = 0.375 in
Dm − Dw = ID = 3 in
Dm = 3.375 in
D
3.375
C= m =
=9
Dw 0.375
k = spring constant
GDw
(
11.5 × 106 )(0.375)
k= 3 =
= 105.64 lb in
3
8C N c
8(9 ) (7 )
3
1
δ1 = 7 − 4 = 3.25 in
8
8
F1 = kδ1 = (105.64)(3.25) = 343.33 lb
3
1
δ 2 = 7 − 6 = 1.125 in
8
4
F2 = kδ 2 = (105.64)(1.125) = 118.85 lb
1
Fm = (343.33 + 118.85) = 231.09 lb = 0.231 kip
2
1
Fa = (343.33 − 118.85) = 112.24 lb = 0.11224 kip
2
K  8FmC 
sms =


K c  πDw2 
Page 32 of 70
SECTION 4 - SPRINGS
4C − 1 0.615
+
4C − 4
C
4(9) − 1 0.615
K=
+
= 1.162
4(9 ) − 4
9
Figure AF 15, K c = 1.10
K=
 8FmC  1.162  8(0.231)(9) 
=
= 39.8 ksi


2 
2 
 πDw  1.10  π (0.375) 
 8(0.11224 )(9) 
 8F C 
sas = K  a 2  = 1.162
 = 21.3 ksi
2
 πDw 
 π (0.375) 
1 sms − sas 2sas
=
+
N
s ys
sno
1 39.8 − 21.3 2(21.3)
=
+
N
147.1
70.96
N = 1.377
sms =
K
Kc
(b) max. “solid” ss = s ys = 147.1 ksi
Min. Solid Height = Dw N c = (0.375)(7 ) = 2.625 in
3
Solid deflection = 7 − 2.625 = 4.75 in.
8
F = kδ = (105.64)(4.75) = 501.8 lb = 0.5018 kip
 8(0.5018)(9) 
 8 FC 
Solid stress = ss = K  2  = 1.162
= 95 ksi < 147.1 ksi
2 
 πDw 
 π (0.375) 
There is no danger of damage
(c) Natural frequency
For steel
14,050 Dw
φ=
cps
N c Dm2
14,050
φ=
cps
N c C 2 Dw
14,050
φ=
cps = 66 cps
(7 )(9)2 (0.375)
 2π 
For 450 rpm, φ = 450
 = 47 cps
 60 
66
= 1.4 < 12 , there is danger of surging.
47
Page 33 of 70
SECTION 4 - SPRINGS
(d) U s =
[
]
1
1
2
2
k δ12 − δ 22 = (105.64 ) (3.25) − (1.125) = 491 in − lb
2
2
289.
(
)
A helical spring is hot wound from 5/8-in. carbon-steel wire with an outside
diameter of 3 ¼ in. A force of 3060 lb. is required to compress the spring 1 ¾
in to the solid heigh. In service the spring is compressed so that its
deformation varies form ½ in. to1 1/8 in. (a) What is the factor of safety by
the Wahl criterion? (b) Is the “solid stress” safe? Compute (c) the pitch angle,
(d) the change of stored energy between the working lengths, (e) the factor of
safety if the spring is peened?
Solution:
For hot-wound carbon steel wire
5
Dw = in
8
Table AT 17
91
s ys = 0.1 ksi , [0.093 < Dw < 0.25]
Dw
91
s ys =
ksi = 104.5 ksi , Dw > 0.25 in.
(0.25)0.1
49
sno = 0.15 ksi , [0.093 < Dw < 0.25]
Dw
49
sno =
ksi = 60.33 ksi , Dw > 0.25 in.
(0.25)0.15
Permissible solid stress = ss =
ss =
117
ksi , [Dw > 0.375 in.] § 6.3
Dw0.31
117
ksi = 35.4 ksi
(0.625)0.31
3060
= 1748.6 lb in
δ 1.75
1
F1 = kδ1 = (1748.6 )  = 874.3 lb
2
 1
F2 = kδ 2 = (1748.6 )1  = 1967.2 lb
 8
1
Fm = (1967.2 + 874.3) = 1420.7 lb = 1.4207 kip
2
1
Fa = (1967.2 − 874.3) = 546.4 lb = 0.5464 kip
2
(a) k =
F
=
Page 34 of 70
SECTION 4 - SPRINGS
5
in = 0.625 in
8
1
Dm + Dw = 3 in
4
Dm = 2.625 in
D
2.625
C= m =
= 4 .2
Dw 0.625
4C − 1 0.615
K=
+
4C − 4
C
4(4.2) − 1 0.615
K=
+
= 1.3808
4(4.2) − 4
4.2
K c = 1.234
Dw =
K  8FmC  1.3808  8(1.4207 )(4.2) 

 = 43.5 ksi

=
K c  πDw2  1.234  π (0.625)2 
 8(0.5464)(4.2) 
 8F C 
sas = K  a 2  = 1.3808
 = 20.7 ksi
2
 πDw 
 π (0.625) 
1 sms − sas 2sas
=
+
N
s ys
sno
1 43.5 − 20.7 2(20.7 )
+
=
N
104.5
60.33
N = 1.106
sms =
(b) Permissible solid stress = 135.4 ksi
F = 3.060 kip
 8(3.060)(4.2) 
 8 FC 
= 115.7 ksi < 135.4 ksi , safe
Solid stress, ss = K  2  = 1.3808
2 
 πDw 
 π (0.625) 
3
(c) Solid deflection = 1 in
4
(P − Dw )N c = 1.75 in
8FC 3 N c
δ=
GDw
G = 10.5 × 106 psi , hot-wound
F
GD
k= = 3w
δ 8C N c
Page 35 of 70
SECTION 4 - SPRINGS
(10.5 ×10 )(0.625)
6
1748.6 =
8(4.2 ) N c
3
N c = 6.332
(P − 0.625)(6.332) = 1.75
P = 0.9014 in
Pitch angle
P
P
tan λ =
=
πDm πCDw
λ = tan −1
 0.9014 
P
= tan −1 
= 6.24o

πCDw
 π (4.2)(0.625) 
[
]
1
1
2
2
k δ12 − δ 22 = (1748.6 )(1.125) − (0.5) = 888 in − lb
2
2
(e) When peened
s ys = 12.5(104.5) = 130.6 ksi
(
(d) U s =
)
sno = 1.25(60.33) = 75.4 ksi
1 43.5 − 20.7 2(20.7 )
=
+
N
130.6
75.4
N = 1.38
ENERGY STORAGE
293.
A 10-lb. body falls 10 in. and then strikes a helical spring. Design a harddrawn carbon steel spring that will absorb this shock occasionally without
permanent damage. Determine appropriate values of wire diameter, coil
diameter, pitch, free length, closed length, and the maximum stress under the
specified conditions, and scale. Let C = 7 .
Solution:
For hard-drawn carbon steel, Table AT 17
182
su = 0.1 ksi , [0.093 < Dw < 0.25]
Dw
91
Max. “solid” ss = 0.1 ksi , [0.093 < Dw < 0.25]
Dw
36.855
ssd = (0.50 )(0.405)su =
ksi
Dw0.1
Us =
ss2V
4 K 2G
Page 36 of 70
SECTION 4 - SPRINGS
4C − 1 0.615
+
4C − 4
C
4(7 ) − 1 0.615
K=
+
= 1.213
4(7 ) − 4
7
K=
 πD 2 
V ≈  w (πDm )N c
 4 
π 2 Dw2 Dm N c
V=
4
Dm = CDw
V=
π 2CDw3 N c
4
8FC 3 N c
δ=
GDw
U s = W (h + δ )
 8 FC 
ss = K  2 
 πDw 
s πD 2
F= s w
8KC
 s πD 2   8C 3 N c 
δ =  s w 

 8KC   GDw 
s sπDwC 2 N c
KG
2 2
3

ssπDwC 2 N c  s s π CD w N c
U s = W h +
=
KG
16 K 2G


Wh
Nc = 2 2 2
ss π CDw s sπDwC 2W
−
16 K 2G
KG
36.855
when ss =
ksi
Dw0.1
Wh
Nc =
2 2
2.8
(36.855) π CDw − 36.855πD w0.9C 2W
16 K 2G
KG
(
0.010)(10)
Nc =
(36.855)2 π 2 (7)Dw2.8 − 36.855πD w0.9 (7 )2 (0.010)
2
(1.213)(11,500)
16(1.213) (11,500)
0.10
Nc =
2.8
0.3466 Dw − 0.004067 D w0.9
δ=
Page 37 of 70
SECTION 4 - SPRINGS
combination of Dw and N c
Gage No. W & M
12
11
10
9
8
7
6
5
4
3
Dw
0.1055
0.1205
0.1350
0.1483
0.1620
0.1770
0.1920
0.2070
0.2253
0.2437
Nc
991.2
312.1
166.1
108.0
75.2
53.7
40.2
31
23.4
18.1
Dw N c
105
37.6
22.4
16.0
12.2
9.5
7.7
6.4
5.3
4.4
Use Dw = 0.2437 in < 0.25 in , N c = 18.1
45
Dm = 7 Dw = 7(0.2437 ) = 1.7059 in = 1 in
64
0.9 2
0.9
2
s sπDwC 2 N c 36.855πD w C N c 36.855π (0.2437 ) (7 ) (18.1)
=
=
= 2.066 in
(1.213)(11,500)
KG
KG
36.855
ss =
= 42.44 ksi
(0.2437 )0.1
91
91
sso = 0.1 =
= 104.8 ksi
Dw
(0.2437 )0.1
Solid deflection
 104.8 
=
(2.066 ) = 5.1 in
 42.44 
(P − Dw )N c = 5.1
δ=
(P − 0.2437)(18.1) = 5.1
P = 0.5255 in
17
say P =
= 0.53125 in
32
Minimum Solid Height = Dw N c = (0.2437 )(18.1) = 4.41 in
Assume squared and ground end
Solid height = Dw N c + 2 Dw = (0.2437 )(18.1) + 2(0.2437 ) = 5.0 in
Solid deflection = (0.53125 − 0.2437 )(18.1) = 5.2 in
Free length = 5.0 in + 5.2 in = 10.2in
Summary of answer:
Dw = 0.2437 in , No. 3 W & M
Page 38 of 70
SECTION 4 - SPRINGS
Dm = 1
45
in
64
17
in
32
Free length = 10.2 in
Closed length = 5 in
Maximum stress = 42.44 ksi
P=
294.
A helical spring, of hard-drawn steel wire, is to absorb 75 in-lb of energy
without being stressed beyond the recommended value of average service. Let
C = 6 . Decide upon satisfactory dimensions; Dw , Dm , N c , free length, pitch
angle, solid stress, volume of metal, possibility of spring buckling.
Solution:
For hard-drawn steel wire, shock load, average service
140
su = 0.19 ksi , [0.028 < Dw < 0.625]
Dw
70
Max. “solid” ss = 0.19 ksi , [0.028 < Dw < 0.625]
Dw
 140  19.278
ssd = (0.50)(0.85)(0.324)su = (0.50)(0.85)(0.324) 0.19  = 0.19 ksi
 Dw  Dw
s 2V
s 2π 2 Dw3 CN c
Us = s 2 = s
4K G
16 K 2G
C=6
4C − 1 0.615
K=
+
4C − 4
C
4(6) − 1 0.615
K=
+
= 1.2525
4(6) − 4
6
U s = 75 in − lb = 0.075 in − kip
19.278 
π 2 Dw3 (6)N c
U s = 0.075 =  0.19 
2
 Dw  16(1.2525) (11,500)
0.9837 = Dw2.62 N c
Table AT-15
W&M
Page 39 of 70
Dw
Nc
Dw N c
SECTION 4 - SPRINGS
9
8
7
6
5
4
3
2
1
0
2-0
3-0
4-0
5-0
0.1483
0.1620
0.1770
0.1920
0.207
0.2253
0.2437
0.2625
0.2830
0.3065
0.3310
0.3625
0.3938
0.4305
146
116
92
74
61
49
40
32.7
26.9
21.8
17.8
14.0
11.3
8.95
Use Dw = 0.4305 in , 5-0 W & M
Nc ≈ 9
9
Dm = 6(0.4305) = 2.583 in ≈ 2 in
16
19.278
ss =
= 22.63 ksi
(0.4305)0.19
70
Max. Solid Stress = sso =
= 82.16 ksi
(0.4305)0.19
s sπDwC 2 N c (22.63)(π )(0.4305)(6) (9)
=
= 0.6885 in
KG
(1.2525)(11,500)
 82.16 
Solid deflection = 
(0.6885) = 2.5 in
 22.63 
(P − Dw )N c = 2.5
(P − 0.4305)(9) = 2.5
P = 0.7083 in
45
say P =
= 0.703125 in
64
Solid deflection = (0.703125 − 0.4305)(9) = 2.453625 in
 2.453625 
Solid stress = 22.63
 = 80.65 ksi
 0.6885 
2
δ=
7
Minimum Solid Height = Dw N c = (0.4305)(9 ) = 3.8745 in ≈ 3 in
8
45
21
 
Minimum Free Length = PN c =  (9 ) = 6.328125 in ≈ 6 in
64
 64 
Pitch Angle
Page 40 of 70
21.65
18.79
16.28
14.21
12.63
11.04
9.75
8.58
7.61
6.68
5.89
5.075
4.45
3.85
SECTION 4 - SPRINGS
 45 


P
= tan −1  64  = 5o < 12o
λ = tan −1
πDm
π  2 9  
  16  
Volume
 π (0.4305)2    9 
 πDw2 
3
(πDm )N c = 
V ≈ 
 π  2 (9 ) = 10.55 in
4
 4 

   16 
Summary of answer:
Dw = 0.4305 in , No. 5-0 W & M
9
Dm = 2 in
16
Nc = 9
21
Free length = 6 in
64
Pitch Angle = λ = 5o
Solid Stress = 80.65 ksi
Volume of metal = 10.55 in3
Possibility of spring buckling
21
6
64 = 2.47 < 4 , no possibility
9
2
16
CONCENTRIC HELICAL SPRINGS
297.
Two concentric helical springs are to be subjected to a load that varies from a
maximum of 235 lb. to a maximum of 50 lb. They are to fit inside a 1 5/8 in.
cylinder. The maximum deflection is to be ¾ in., and the deflection when
compressed solid is to be approximately 1 in. Using the “static approach” for
severe service (maximum load), determine the wire diameter, mean coil
diameter, number of coils, solid length, and free length of both springs. (Start
with oil-tempered wire and assume a diametral clearance between the outer
D
spring and the cylinder of w , assume a similar clearance between springs.
2
Search for a suitable spring index and wire size.)
Solution:
For oil-tempered wire
Table AT 17
Page 41 of 70
SECTION 4 - SPRINGS
146
ksi , [0.032 < Dw < 0.5]
Dw0.19
87.5
Max. “solid” ss = 0.19 ksi , [0.032 < Dw < 0.5]
Dw
Severe service
0.263(146 ) 38.398
ssd = 0.263su =
=
ksi
Dw0.19
Dw0.19
F = 235 lb = 0.235 kip
δo = δi
su =
8FoCo3 N co 8Fi Ci3 N i
=
GDwo
GDwi
Assume, Co = Ci
3GDwo
Fo =
32C 3 N co
3GDwi
Fi =
32C 3 N ci
 8F C 
sso = K  o2 
 πDwo 
 8F C 
ssi = K  i 2 
 πDwi 
D
D
C = mo = mi
Dwo Dwi
4C − 1 0.615
K=
+
4C − 4
C
D
Dmo − Dwo − wi = Dmi + Dwi
2
Dmo − Dmi = Dwo + 1.5Dwi
D
1.625 − wo = Dmo + Dwo
2
Dmo + 1.5Dwo = 1.625
CDwo + 1.5Dwo = 1.625
1.625
Dwo =
C + 1 .5
1.625C
Dmo =
C + 1 .5
CDwo − CDwi = Dwo + 1.5Dwi
(C − 1)Dwo = (C + 1.5)Dwi
Page 42 of 70
SECTION 4 - SPRINGS
1.625(C − 1)
(C + 1.5)2
1.625C (C − 1)
Dmi =
(C + 1.5)2
Dwi =
 8F C  38.398
sso = K  o2  = 0.19 ksi
Dwo
 πDwo 
1.81
15.08Dwo
Fo =
KC
 8F C  38.398
ssi = K  i 2  = 0.19 ksi
Dwi
 πDwi 
1.81
15.08Dwi
KC
Fo + Fi = F = 0.235 kip
Fi =
1.81
1.81
15.08Dwo
15.08Dwi
+
= 0.235
KC
KC
1.81
1.81
15.08 Dwo
+ 15.08 Dwi
= 0.235 KC
1.81
1.81
 1.625 
1.625(C − 1) 
15.08
= 0.235 KC
 + 15.08
 C + 1 .5 
 C + 1.5 
1.81


(
1
C − 1)
 4C − 1 0.615 
154.52 
+
= 0.235
+
C
1.81
3.62 
C 
(C + 1.5) 
 4C − 4
 (C + 1.5)
C = 5.328
1.625(5.328 − 1)
Dwi =
= 0.1509 in
(5.328 + 1.5)2
1.625
Dwo =
= 0.2380 in
5.328 + 1.5
Table AT 15, use Dwi = 0.1620 in , No. 8 W & M and Dwo = 0.2625 in , No. 2 W & M
13
Dmo = CDwo = (5.328)(0.2625) = 1.3986 in ≈ 1 in
32
7
Dmi = CDwi = (5.328)(0.1620 ) = 0.8631 in ≈ in
8
7
 
Dmi
8
Ci =
=   = 5.401
Dwi 0.1620
Co =
Dmo
Dwo
 13 
1 
32 
=
= 5.357
0.2625
Page 43 of 70
SECTION 4 - SPRINGS
1.81
15.08Dwo
K o Co
4(5.357 ) − 1 0.615
Ko =
+
= 1.287
4(5.357 ) − 1 5.357
Fo =
15.08(0.2625)
= 0.194 kip
(1.287 )(5.357 )
1.81
Fo =
1.81
15.08Dwi
K i Ci
4(5.401) − 1 0.615
Ki =
+
= 1.2843
4(5.401) − 1 5.401
15.08(0.1620)
Fi =
= 0.081 kip
(1.2843)(5.401)
Fo + Fi = 0.194 + 0.071 = 0.275 kip > 0.235 kip , ok
3GDwo
Fo =
32C 3 N co
3(11,500)(0.2625)
0.194 =
3
32(5.357 ) N co
N co = 9.5
3GDwi
Fi =
32C 3 N ci
3(11,500)(0.1620)
0.071 =
3
32(5.401) N ci
N ci = 15.6
87.5
Max. solid stress, sss = 0.19 ksi ,
Dw
87.5
ssso =
= 112.82 ksi
(0.2625)0.19
87.5
sssi =
= 123.65 ksi
(0.1620)0.19
Stress
38.398
ssi =
= 54.26 ksi
(0.1620)0.19
38.398
sso =
= 49.51 ksi
(0.2625)0.19
Fi =
Solid stress
 1 
sso = 49.51
 = 66.01 ksi < 112.82 ksi
 0.75 
Page 44 of 70
SECTION 4 - SPRINGS
 1 
ssi = 54.26
 = 72.35 ksi < 123.65 ksi
 0.75 
Solid length
Dwo N co = (0.2625)(9.5) = 2.5 in
Dwi N ci = (0.1620 )(15.6) = 2.53 in
assume solid length = 3 in
Dwi ( N ci + xi ) = (0.1620)(15.6 + xi ) = 3 in
xi = 2.92
Total coils = 15.6 + 2.92 = 18.52
Dwo ( N co + xo ) = (0.2625)(9.5 + xo ) = 3 in
xo = 1.93
Total coils = 9.5 + 1.93 = 11.43
Free Length = 3 in + 1 in = 4 in
Summary of answer:
Outside wire.
Dwo = 0.2625 in , No. 2 W & M
13
Dmo = 1 in
32
N to = 11.43
Solid length = 3 in
Free length = 4 in
Inside wire.
Dwi = 0.1620 in , No. 8 W & M
7
Dmi = in
8
N ti = 18.52
Solid length = 3 in
Free length = 4 in
298.
Two concentric, helical compression springs are used on a freight car. The
larger spring has an outside diameter of 7 in., a free length of 7 1/8 in., and is
made of a 1 ¾ in. steel bar. The smaller has an outside diameter of 4 1/8 in., a
free length of 6 13/16 in. , and is made of 7/8 in. steel bar. The solid height of
each spring is 5 ¼ in. and the forces required to compress them solid are
15,530 lb. and 7,000 lb., respectively. The working load on the two springs is
11,350 lb. Determine (a) the number of free coils in each spring, (b) the stress
in each spring when compressed solid, (c) the stresses induced by the working
Page 45 of 70
SECTION 4 - SPRINGS
load. Notice that the outer spring deflects 5/16 in. before the inner one takes a
load. (d) What energy is absorbed while changing deflection from that at the
working load to that when the springs are compressed “solid”?
Solution:
ODo = 7 in
3
Dwo = 1 in
8
1
FLo = 7 in
8
1
ODi = 4 in
8
7
Dwi = in
8
13
FLi = 6 in
16
1
(a) Solid height = Dw N T = 5 in
4
5.25
N To =
= 3.82
1.375
5.25
=6
N Ti =
0.875
(b) Fo = 15,530 lb
Fi = 7000 lb
 8 FC 
ss = K  2 
 πDw 
3
Dmo = 7 − 1 = 5.625 in
8
Dmo 5.625
Co =
=
= 4.091
Dwo 1.375
4(4.091) − 1 0.615
Ko =
+
= 1.393
4(4.091) − 4 4.091
1 7
Dmi = 4 − = 3.25 in
8 8
D
3.25
Ci = mi =
= 3.714
Dwi 0.875
4(3.714) − 1 0.615
Ki =
+
= 1.442
4(3.714) − 4 3.714
Page 46 of 70
SECTION 4 - SPRINGS
Solid stress
 8(15,530)(4.091) 
sso = 1.393
 = 119,203 psi
2
(
)
π
1
.
375


 8(7000 )(3.714 ) 
ssi = 1.442
 = 124,689 psi
2
 π (0.875) 
(b) Stresses induced by working load
Fi + Fo = 11,350 lb
15,530
ko =
= 8283 lb in
1
 1
7 − 5 
4
 8
7000
ko =
= 4480 lb in
1
 13
6 − 5 
4
 16
5
δ o − δ i = = 0.3125 in
16
Fi = kiδ i = 4480δ i
Fo = koδ o = 8283(0.3125 + δ i )
Fi + Fo = koδ o = 4480δ i + 8283(0.3125 + δ i ) = 11,350 lb
δ i = 0.6865 in
δ o = 0.3125 + 0.6865 = 0.9990 in
Fi = (4480)(0.6865) = 3076 lb
Fo = (8283)(0.9990) = 8275 lb
Stresses
 8(8275)(4.091) 
sso = 1.393
 = 63,516 psi
2
 π (1.375) 
 8(3076)(3.714) 
ssi = 1.442
 = 54,792 psi
2
 π (0.875) 
(d) Energy
1
U so = ko (δ o22 − δ o21 )
2
1
1
δ o 2 = 7 − 5 = 1.875 in
8
4
δ o1 = 0.9990 in
1
2
2
U so = (8283)(1.875) − (0.999 ) = 10,427 in − lb
2
[
Page 47 of 70
]
SECTION 4 - SPRINGS
1
ki δ i22 − δ i21
2
13
1
δ i 2 = 6 − 5 = 1.5625 in
16
4
δ o1 = 0.6865 in
1
2
2
U si = (4480 )(1.5625) − (0.6865) = 4,413 in − lb
2
U si =
(
)
[
]
TORSION-BAR SPRINGS
299.
A torsion-bar similar to that shown is to be used for the front spring of an
automobile. Its rate should be 400 lb./in. of deflection of the end of the arm
which is e = 10 in. long. It is made of AISI 9261,OQT 900 oF, and the
maximum repeated load is 1500 lb. perpendicular to the centerline of the arm.
The support is such that bending of the bar is negligible. (a) Determine its
diameter and length so that no permanent set occurs due to a 30 % overload
(limited by a stop). Use s ys = 0.6 s y , but check with equation (c) § 6.3, Text, if
appropriate. (b) Determine the factor of safety according to the Soderberg
criterion if the load varies from 1200 lb. to 1500 lb.; minimum r d = 0.1 ,
D d = 3 . (c) The same as (b) except that the bar is shot-peened all over. What
other steps may be taken to improve the fatigue strength?
Problem 299, 300
Solution:
e = 10 in
For AISI 9261, OQT 900 oF
s y = 192 ksi
su = 215 ksi
s ys = 0.6 s y = 115.2 ksi = 115,200 psi
Page 48 of 70
SECTION 4 - SPRINGS
16T
πd3
T = Fe = (1.3)(1500)(10) = 19,500 in − lb
16(19,500 )
115,200 =
π d3
d = 0.95 in
use d = 1 in
117 117
§ 6.3 ss = 0.3 = 0.3 = 117 ksi ≈ s ys
Dw
(1)
(a) ss =
(b) Soderberg Criterion
1 sms K f sas
=
+
N s ys
sns
sns = (0.6)(0.5)(215) = 64.5 ksi
Figure AF 12, r d = 0.1 , D d = 3
K t = 1.45
K f ≈ K t = 1.45
1
(1500 + 1200) = 1350 lb
2
Tm = (1350)(10) = 13,500 in − lb = 13.5 in − kips
16(13.5)
sms =
= 68.8 ksi
π (1)3
Fm =
1
(1500 − 1200) = 150 lb
2
Ta = (150)(10) = 1500 in − lb = 1.5 in − kips
16(1.5)
sas =
= 7.64 ksi
π (1)3
1
68.8 (1.45)(7.64 )
=
+
N 115.2
64.5
N = 1.30
Fa =
(c) Shot-peened
s ys = 1.25(115.2 ) = 144 ksi
sns = 1.25(64.5) = 80.6 ksi
1 68.8 (1.45)(7.64 )
=
+
N 144
80.6
N = 1.625
Page 49 of 70
SECTION 4 - SPRINGS
300.
A solid steel torsion bar is loaded through a 10 in. arm as shown. The load F
perpendicular to the center-line of the arm varies from 500 to 1000 lb.,
7
200,000 cycles. The bar is d = in. in diameter and 30 in. long; let D d = 3 ;
8
r d = 0.1 ; (a) Determine the maximum stress in the bar, the angular
deflection, and the scale (lb./in.) where F is applied. The support is such that
bending of the bar is negligible. (b) Select a material and heat treatment for
this bar for a minimum N = 1.2 , Soderberg criterion.
Problem 299, 300
Solution:
Fig. AF 12, K f = 1.45
1
(1000 + 500) = 750 lb
2
1
Fa = (1000 − 500 ) = 250 lb
2
Tm = (750)(10) = 7500 in − lb = 7.5 in − kips
Fm =
Ta = (250)(10) = 2500 in − lb = 2.5 in − kips
K fl =
n
(log K f ) 3
Kf
=
(200,000)(log1.45) 3 1.33
1.45
16T
πd3
16(7.5)
sms =
= 57 ksi
3
7
π 
8
16(2.5)
sas =
= 19 ksi
3
7
π 
8
ss =
(a) smax = sms + K fl sas = 57 + (1.33)(19 ) = 82.27 ksi
Page 50 of 70
SECTION 4 - SPRINGS
TL 64TL
=
JG π d 4G
64(500 )(10 )(30 )
θ min =
= 0.4533 rad
4
7
6
π   11.5 ×10
8
64(1000 )(10 )(30 )
θ max =
= 0.9066 rad
4
7
6
π   11.5 ×10
8
F
1000
scale =
=
= 110.3 lb in
θ e (0.9066)(10)
θ=
(c)
(
)
(
)
1 sms K f sas
=
+
N s ys
sns
 10 6 

sns = (0.6 )(0.5)su 
 200,000 
s ys = 0.6 s y
0.085
= 0.344 su
1
57
(1.33)(19)
=
+
1.2 0.6s y 0.344su
Use AISI 8760, OQT 800 oF
s y = 200 ksi
su = 220 ksi
N = 1.24
HELICAL SPRINGS – NON CIRCULAR SECTION
301.
A spring is to be designed of square oil-tempered steel wire and subjected to a
repeated maximum load of 325 lb.; mean coil diameter, 1 ½ in.; deflection,
13/32 in. Determine (a) the wire size for average service, (b) the required
number of active coils, (c) the solid height, free length, and pitch (the ends are
squared and ground, the “solid stress” must be satisfactory, and the pitch angle
not excessive). (d) What amount of energy is stored when the load is 325 lb.?
Express in in-lb. and Btu.
Solution:
For oil-tempered wire,
146
su = 0.19 , [0.032 < Dw < 0.5]
Dw
Max. “solid” ss = 0.6su
Page 51 of 70
SECTION 4 - SPRINGS
(a) average service,
b = Dw , t = b
K FD (3b + 1.8t )
2.4 FDm
ss= q m 2 2
= Kq
2b t
b3
s sd = 0.324su , average service
2.4 FDm 0.324(146 )
Kq
=
2b 3
b 0.19
F = 0.325 kip
1
Dm = 1 in
2
K q = 1.25 (assumed)
 2.4(0.325)(1.5) 0.324(146 )
1.25
=
b3
b 0.19


b = 0.2902 in
Table AT 15, use b = 0.313 in , # 1 wire size
D
1 .5
C= m =
= 4 .8
b
0.313
Figure AF 15, K q = 1.275
(b) δ =
2.45 FDm3 N c
2.45FDm3 N c
=
Gt 3 (b − 0.56t )
0.44Gb 4
13 2.45(0.325)(1.5) N c
=
32 0.44(11,500 )(0.313)4
N c = 7.34
3
(c) Solid height = b( N c + 2) = 0.313(7.34 + 2) = 2.92 in
Free length = PN c + 2b
F = 0.325 lb
 2.4(0.325)(1.5)
 2.4 FDm 
ss = Kq 
= 1.275
 = 48.65 ksi
3
3

 b

 (0.313)

0.6(146 )
solid stress = =
= 109.2 ksi
(0.313)0.19
109.2  13 
solid deflection = =
  = 0.91 in
48.65  32 
(P − b )N c = 0.91
(P − 0.313)(7.34) = 0.91
P = 0.437 in
Page 52 of 70
SECTION 4 - SPRINGS
use P =
7
in
16
27
7
Free length = PN c + 2b =  (7.34 ) + 2(0.313) = 3.837 in ≈ 3 in
32
 16 
7
 
P
16
tan λ =
=  
πDm π (1.5)
λ = 5.3o < 10o
(d) U s =
Us =
1 2 1
1
 13 
kδ = Fδ = (0.325)  = 0.066 in − kip = 66 in − lb
2
2
2
 32 
66
= 0.085 Btu
778
302.
A coil spring, of hard-drawn carbon steel, is to deflect 1 in. under a load of
100 lb. The outside coil diameter is to be 1 in. Compute the number of active
coils, (a) if the wire is round, 5/32 in. in diameter, (b) if the wire is square,
5/32 in. on the side, (c) if the wire is rectangular 1/8 x 3/16 in., long
dimension parallel to the axis, (d) If the wire is rectangular 3/16 x 1/8 in.,
short dimension parallel to the axis. (e) What is the maximum stress in each of
the above springs under the 100-lb load? (f) What is the ratio of the
approximate volumes, square- or rectangular-wire to round wire spring?
Solution:
Dm + Dw = 1 in
5
(a) Dw =
in
32
5 27
Dm = 1 −
=
in
32 32
 27 
 
Dm  32 
C=
=
= 5.4
Dw  5 
 
 32 
8FC 3 N c
δ=
GDw
8(100 )(5.4 ) N c
 5 
11.5 × 10 6  
 32 
N c = 14.3
3
1=
(
Page 53 of 70
)
SECTION 4 - SPRINGS
5
in
32
5 27
Dm = 1 −
=
in
32 32
2.45 FD m3 N c
δ=
0.44Gb 4
(b) Square, b =
3
 27 
2.45(100)  N c
 32 
1=
4
6  5 
0.44 11.5 × 10  
 32 
N c = 20.5
(
)
3
1
in , t = in
16
8
1 7
Dm = 1 − t = 1 − = in
8 8
3
2.45FD m N c
δ= 3
Gt (b − 0.56t )
(c) b =
3
7
2.45(100 )  N c
8
1=
3
1  3
 1 
11.5 × 10 6    − 0.56 
 8  16
 8 
(
)
N c = 16.1
1
3
in , t = in
8
16
3 13
Dm = 1 − t = 1 − = in
16 16
3
2.45FD m N c
δ= 3
Gt (b − 0.56t )
(d) b =
3
 13 
2.45(100 )  N c
 16 
1=
3
 3  1
 3 
11.5 × 10 6    − 0.56 
 16   8
 16 
(
)
N c = 11.5
(e) Maximum Stress
Page 54 of 70
SECTION 4 - SPRINGS
8 FC
π Dw3
4(5.4) − 1 0.615
K=
+
= 1.284
4(5.4) − 4
5.4
For (a) ss = K




8(100)(5.4) 
ss = 1.284
= 72,320 psi
  5 3 
 π  
  32  
FD m (3b + 1.8t )
2.4 FD m
= Kq
2 2
2b t
b3
 27 
 
Dm  32 
C=
=
= 5.4
b
 5 
 
 32 
K q = 1.25
For (b) ss = K q

 27  
 2.4(100)  
 32   = 66,355 psi
ss = 1.25
3


 5 
 


 32 


FD m (3b + 1.8t )
For (c) ss = K q
2b 2t 2
7
 
D
8
C = m =   =7
t
1
 
8
K q = 1 .1

7 
 (100)  
 8   3 3  + 1.8 1  = 68,992 psi
ss = 1.1
 
 
2
  3   1  2    16 
 8 
 2    
  16   8  
FD m (3b + 1.8t )
For (d) ss = K q
2b 2t 2
 13 
 
Dm  16 
C=
=
= 4.33
t
3
 
 16 
Page 55 of 70
SECTION 4 - SPRINGS
K q = 1 .2

 13  
 (100)  
 16   3 1  + 1.8 3  = 63,232 psi
ss = 1.2
 
 
2
  1   3  2    8 
 16 
 2    
  8   16  
(e) Ratio of the approximate volumes
For (a) Round wire
π

Va =  Dw2 (π Dm )N c
4

 π  5  2    27 
Va =     π   (14.3) = 0.727 in 3
 4  32     32 
For (b) Square wire
Vb = b 2 (π Dm )N c
2
 5    27 
Vb =   π   (20.5) = 1.327 in 3
 32    32 
For (c) rectangular wire
Vc = bt (π Dm )N c
 3  1    7 
Vc =    π   (16.1) = 1.037 in3
 16  8    8 
For (d) rectangular wire
Vd = bt (π Dm )N c
 1  3    13 
Vd =    π   (11.5) = 0.688 in 3
 8  16    16 
Ratio of volume
Square to round wire
V 1.327
= b =
= 1.825
Va 0.727
Rectangular to round wire (long dimension parallel to the axis)
V 1.037
= c =
= 1.426
Va 0.727
Rectangular to round wire (short dimension parallel to the axis)
V
0.688
= d =
= 0.946
Va 0.727
Page 56 of 70
SECTION 4 - SPRINGS
TENSION SPRINGS
305.
Design two tension springs for a spring balance with a capacity of 200 lb.
Each spring supports a maximum load of 100 lb. The outside diameter must
not exceed 1 ¼ in. and the total length including end loops must not exceed 9
½ in. Select a material and determine the dimension, including wire diameter,
number of coils, and free length.
Solution:
Table AT 17, assume oil tempered wire
146
su = 0.19 ksi
Dw
87.5
s ys = 0.19 ksi
Dw
0.8(87.5)
70
ssd =
= 0.19 ksi , [0.032 < Dw < 0.5]
0.19
Dw
Dw
F = Fi + kδ
8 K cFDm
ss =
π Dw3
2r
D
C= m = m
Dw Dw
8FaC 3 N c
G Dw
GD
k = 3w
8C N
3
 G D  8F C N c 
kδ =  3 w  a
= Fa
 8C N  GDw 
Fa = 100 lb = 0.10 kip
Figure AF 15, assume K c = 1.2
8 K cFi Dm 8 K cFa Dm
ss =
+
π Dw3
π Dw3
8 K cFaC
s s = K c si +
π Dw2
OD = Dm + Dw = 1.25 in
1.25
Dw =
C +1
§ 6.21, assume si = 18 ksi
ssd = ss
δ=
Page 57 of 70
SECTION 4 - SPRINGS
70
8(1.2 )(0.1)C
= (1.2 )(18) +
0.19
Dw
π Dw2
70(C + 1)
(1.25)0.19
0.19
8(1.2 )(0.1)C (C + 1)
π (1.25)2
2
= 21.6 +
67.1(C + 1)
0.19
= 21.6 + 0.1956C (C + 1)
2
67.1(C + 1) − 0.1956C (C + 1) = 21.6
C = 6 .7
1.25
1.25
=
= 0.1623 in
Dw =
C + 1 6 .7 + 1
Table AT 15, use Dw = 0.1620 in , 8 W & M
0.19
2
Dm = CDw = (6.7 )(0.1620 ) = 1.085 in
say Dm = 1.0 in
D
1 .0
C= m =
= 6.17
Dw 0.1620
si = 17.7 ksi
To check, Fig. AF 15, K c = 1.15
8(1.15)(0.10 )(6.17 )
ss = 1.15(17.7 ) +
= 89.20 ksi
π (0.1620)2
70
ssd =
= 98.92 ksi > 89.20 ksi , o.k.
(0.1620)0.19
Total length = Dw N c + 2(Dm + Dw )
9.5 = (0.162)N c + 2(1.0 + 0.162)
N c = 44.3 coils
Free length = Dw N c = (0.1620)(44.3) = 7.18 in
Summary of answer:
Material, oil-tempered wire
Dw = 0.1620 in , 8 W & M
N c = 44.3 coils
Free length = 7.18 in.
306.
Two helical tension springs are to be used in scales for weighing milk. The
capacity of the scales is 30 lb., each spring carries 15 lb. with a deflection of 3
9/16 in. The springs are made of No. 14, W & M steel wire, outside diameter,
29/32 in. (a) how many coils should each spring have? (b) What is the
maximum stress in the wire? What material should be used?
Page 58 of 70
SECTION 4 - SPRINGS
Solution:
kδ = 15 lb
9
δ = 3 in
16
(a) Table AT 15, No. 14 W &M
Dw = 0.0800 in
29
− 0.0800 = 0.82625 in
32
D
0.82625
C= m =
= 10.328
Dw
0.0800
Dm = OD − Dw =
δ=
8(kδ )C 3 N c
GDw
9 8(15)(10.328) N c
=
16 11.5 ×106 (0.080)
N c = 24.8
3
3
(
)
(b) F = Fi + kδ
Fi =
π si Dw3
8Dm
§ 6.21, C = 10.328
si = 11,272 psi
Fi =
π (11,272)(0.08)3
= 2.743 lb
8(0.82625)
F = 2.743 + 15 = 17.743 lb
Figure AF 15, K c = 1.09
8 K c FDm 8(1.09 )(17.743)(0.82625)
ss =
=
= 79,476 psi
πDw3
π (0.080)3
s
79,476
s ys ≈ s =
= 99,345 psi = 99.345 ksi
0 .8
0 .8
Table AT 17, use Hard drawn wire
70
70
s ys = 0.19 =
= 113 ksi > 99.345 ksi
Dw
(0.080)0.19
307.
A tension spring for a gas-control lever is made of Dw = 0.078 in steel wire;
inside diameter, 0.609 in.; number of coils, 55; free length including end
loops, 5 9/16 in. When the spring is extended to a length of 6 5/16 in., it must
exert a force 5 ½ lb.; it must extend to (a) the initial tension, (b) the stress in
the spring caused by the initial tension (compare with the recommended
Page 59 of 70
SECTION 4 - SPRINGS
maximum values), (c) the stress caused by the 5 ½-lb load, (d) the maximum
stress. What material should be used? (e) What energy is absorbed from the
point where the load is the initial tension until the spring’s length is 6 5/16 in.?
(Data courtesy Worthington Corporation.)
Solution:
Dw = 0.078 in
Dm − Dw = 0.609 in
Dm = 0.609 + 0.078 = 0.687 in
D
0.687
C= m =
= 8 .8
Dw 0.078
N c = 55
8FC 3 N c
GDw
1
F = 5 lb
2
5
9
δ = 6 − 5 = 0.75 in
16
16
3
8(kδ )(8.8) (5)
δ = 0.75 =
11.5 × 106 (0.078)
kδ = 2.244 lb
δ=
(
)
(a) Fi = F − kδ = 5.5 − 2.244 = 3.256 lb
8 F C 8(3.256 )(8.8)
(b) si = i 2 =
= 12,000 psi
π Dw
π (0.078)2
§ 6.21, C = 8.8
si = 13,300 psi > 12,000 psi , ok
(c) F = 5.5 lb
8 K c FC
ss =
π Dw2
C = 8 .8
Figure AF 15
K c = 1.1
8(1.1)(5.5)(8.8)
ss =
= 22,284 psi
π (0.078)2
(d) maximum stress
kδ 2.244
k=
=
= 2.992 lb in
δ
0.75
Page 60 of 70
SECTION 4 - SPRINGS
F = kδ ′
5
9
δ ′ = 9 − 5 = 3.75 in
16
16
F = Fi + kδ ′ = 3.256 + (2.992)(3.75) = 14.476 lb
8 K c FC 8(1.1)(14.476 )(8.8)
ss =
=
= 58,651 psi
π Dw2
π (0.078)2
Table AT 16
s
58,651
s ys ≈ s =
= 73,300 psi = 73.3 ksi
0 .8
0 .8
Table AT 17, use Hard drawn wire
70
70
s ys = 0.19 =
= 113.658 ksi > 73.3 ksi
Dw
(0.078)0.19
(e) U s =
1 2 1
2
kδ = (2.992 )(0.75) = 0.8415 in − lb
2
2
TORSION SPRINGS
308.
A carbon-steel (ASTM A230) torsion spring is to resist a force of 55 lb. at a
radius of 2 in.; the mean diameter is to be 2 ½ in. Compute (a) the diameter of
the wire for average service, (b) the number of coils for a deflection of 180o
under the given torque, (c) the energy the spring has absorbed when the force
is 55 lb.
Solution:
T = M = Fa
F = 55 lb
a = 2 in
T = M = (55)(2) = 110 in − lb
Dm = 2.5 in
182
Table AT 17, su = 0.1 ksi , [0.093 < Dw < 0.25]
Dw
Average service
 182  117.936
117,936
sd = (1.6 )(0.405)su = 0.648 0.1  =
ksi =
psi
0.1
Dw
Dw0.1
 Dw 
KMc
I
For round wire, assume K c = K ci = 1.08 , Table AT 18
D
c= w
2
(a) ss =
Page 61 of 70
SECTION 4 - SPRINGS
I π Dw3
=
c
32
(1.08)(110)(32) = 117,936
ss =
π Dw3
Dw0.1
Dw = 0.2060 in < 0.25 in
Table AT 15, use Dw = 0.2070 in , No. 5 W & M
r D
2
= 9.66 > 9 , ok
To check: = m =
c Dw 0.2070
Table AT 18, K = 1.08
(1.08)(110)(32) = 136,430 psi
ss =
π (0.2070 )3
117,936
ssd =
= 138,054 psi > 136,430 psi
(0.2070)0.1
Therefore, use No. 5 W & M, Dw = 0.2070 in
MπDm N c
EI
6
E = 30× 10 psi
(b) θ =
I=
π Dw4
64
θ = 180o = π
64 MDm N c
θ=
EDw4
64(110 )(2 )N c
π=
(30 ×106 )(0.2070)4
N c = 12.29
1
1
(c) U s = Tθ = (110 )(π ) = 172.8 in − lb
2
2
312.
A pivoted roller follower is held in contact with the cam by a torsion spring.
The moment exerted by the spring varies from 20 lb-in to 50 lb-in. as the
follower oscillates through 30o. The spring is made of AISI 6152 steel, OQT
1000 oF. What should be the value of Dw , Dm , and N c if the factor of safety
is 1.75 based on the Soderberg line? Would this be a conservative or risky
approach?
Solution:
AISI 6152, OQT 1000 oF
su = 184 ksi
Page 62 of 70
SECTION 4 - SPRINGS
s y = 173 ksi
sn = 0.5su = 92 ksi
1
M m = (50 + 20 ) = 35 lb − in
2
1
M a = (50 − 20 ) = 15 lb − in
2
assume K = 1.08
32 KM m 32(1.08)(35) 385
sm =
=
= 3 psi
πDw3
πDw3
Dw
32 KM a 32(1.08)(15) 165
sa =
=
= 3 psi
πDw3
πDw3
Dw
1 s m sa
=
+
N s y sn
1
385
165
=
+
3
1.75 173,000 Dw 92,000 Dw3
Dw = 0.1916 in
Table AT 15, use Dw = 0.1920 in , No. 6 W & M
To solve for K
32(35)K
sm =
= 50,369 K psi
π (0.1920)3
32(15)K
sa =
= 21,587 K psi
π (0.1920)3
1
50,369 K 20,587 K
=
+
1.75 173,000
92,000
K = 1.0868
Table AT 18 K ci = K = 1.0868
r Dm
=
= 9.32 > 9 , ok
c Dw
Dm = 9.32(0.1920) = 1.7894 in
7
use Dm = 1 in = 1.875 in
8
∆MπDm N c 64∆MDm N c
∆θ =
=
EI
EDw4
30π 64(50 − 20 )(1.875)N c
=
180
(30 ×106 )(0.1920)4
N c = 5.93
Summary of answer:
Dw = 0.1920 in , No. 6 W & M
Page 63 of 70
SECTION 4 - SPRINGS
7
Dm = 1 in
8
N c = 5.93 , N > 1.4 , therefore conservative.
FLAT AND LEAF SPRINGS
315.
A cantilever flat spring of uniform strength, Fig. 6.20, Text, is to absorb an
energy impact of 500 ft-lb. Let the thickness of the steel, AISI 1095, OQT 900
o
F, be ½ in. and let the maximum stress be half of the yield strength. (a) Find
the width b of the spring at the widest point in terms of the length L .
Determine values of b for lengths of 36 in., 48 in., 60 in., and 72 in. (b)
Determine the deflection of the spring for each set of values found in (a).
Solution.
Fig. 6/20
6 FL
bh 2
6 FL3
δ=
Ebh3
AISI 1095, OQT 900 oF, s y = 104 ksi , Table AT 9
sB =
s B = 0.5s y = 0.5(104 ) = 52 ksi = 52,000 psi
1
Fδ
2
s bh 2
F= B
6L
 s bh 2  L3
s B L2

=
δ = 6 B
3
Eh
 6 L  Ebh
Us =
1  s bh 2  s B L2  1  s B2 bhL 

= 

U s =  B
2  6 L  Eh  12  E 
U s = 500 ft − lb = 6000 in − lb
Page 64 of 70
SECTION 4 - SPRINGS
(52,000)2 b 1  L
1
2
12
30 × 106
bL = 1598 in 2
6000 =
b=
1598 in 2
L
L = 36 in , b =
1598 in 2
= 44.4 in
36 in
L = 48 in , b =
1598 in 2
= 33.3 in
48 in
L = 60 in , b =
1598 in 2
= 26.6 in
60 in
L = 72 in , b =
1598 in 2
= 22.2 in
72 in
(b) δ =
s B L2
Eh
2
(
52,000 )(36 )
L = 36 in , δ =
= 4.4928 in
1
30 × 10  
2
2
(
52,000 )(48)
L = 48 in , δ =
= 7.9872 in
6 1
30 ×10  
2
2
(
52,000 )(60 )
L = 60 in , δ =
= 12.48 in
6 1
30 ×10  
2
2
(
52,000)(72 )
L = 72 in , δ =
= 17.9712 in
6 1
30 × 10  
2
(
317.
6
)
(
)
(
)
(
)
One of the carbon contacts on a circuit breaker is mounted on the free end of
a phosphor-bronze beam ( µ = 0.35 ). This beam has the shape of the beam
9
1
1
shown in Fig. 6.24, Text, with b = 1 in. , b′ = in. , L = 4 in. , and h = in.
16
2
16
When the contacts are closed, the beam deflects ¾ in. Compute (a) the force
on the contacts, (b) the maximum stress.
Solution:
Page 65 of 70
SECTION 4 - SPRINGS
Figure 6.24
6 FL 3WL
s= 2 =
bh
bh 2
K FL3 1 − µ 2
K WL3 1 − µ 2
δ= 1
= 1
3EI
6 EI
9
 
b′  16 
=
= 0.5625
b
1
Figure 6.25, K1 = 1.14
(
)
(
)
(a) Force on contacts = F
K FL3 1 − µ 2
δ= 1
3EI
E = 16× 106 psi (phosphor bronze)
(
)
bh3
12
3
4 K FL 1 − µ 2
δ= 1
Ebh 3
3
2
3 4(1.14 )F (4.5) 1 − (0.35)
=
3
4
(16 ×106 )(1) 161 
 
F = 8 lb
as a beam, I =
(
)
[
(b) s =
318.
]
6 FL 6(8)(4.5)
=
= 55,296 psi
2
bh 2
1
(1) 
 16 
A cantilever leaf spring 26 in. long is to support a load of 175 lb. The
construction is similar to that shown in Fig. 6.22 (a), Text. The leaves are to
be 2 in. wide, 3/16 in. thick; SAE 9255 steel, OQT 1000 oF; 107 cycles (§
6.26). (a) How many leaves should be used if the surfaces are left as rolled?
(b) The same as (a) except that the leaves are machined and the surfaces are
not decarburized. (c) The same as (b), except that the surface is peened all
over. (d) Which of these springs absorbs the most energy? Compute for each:
(e) What are the load and deflection of the spring in (b) when the maximum
stress is the standard-test yields strength?
Solution:
Figure 6.22 (a)
6 FL
sA = 2
bh
Page 66 of 70
SECTION 4 - SPRINGS
6 FL3
Ebh3
b = N1b′
F = 175 lb
b′ = 2 in
3
h = in
16
L = 26 in
§ 6.26, SAE 9255, OQT 1000 oF
su = 180 ksi
s y = 160 ksi
δA =
3
in = 0.1875 in
16
sd = 83.75 ksi
t=
(a) As rolled, Figure AF 5
Surface factor = 0.275
sd = 0.275(83.75) = 23 ksi = 23,000 psi
6 FL
sA = 2
bh
6(175)(26 )
23,000 =
2
3
N 1 (2 ) 
 16 
N1 = 16.88
say N1 = 17
(b) Machined, Figure AF 5
Surface factor = 0.75
sd = 0.75(83.75) = 62.8 ksi = 62,800 psi
6 FL
sA = 2
bh
6(175)(26 )
62,800 =
2
3
N 1 (2 ) 
 16 
N1 = 6.2
say N1 = 7
(c) Peened surface, (b)
Page 67 of 70
SECTION 4 - SPRINGS
sd = 1.25(62.8) = 78.5 ksi = 78,500 psi
6 FL
sA = 2
bh
6(175)(26 )
78,500 =
2
3
N 1 (2 ) 
 16 
N1 = 4.95
say N1 = 5
1
Fδ
2
F = 175 lb
(d) U s =
6 FL3
EN1b′h 3
For (a) N1 = 17
δ=
6(175)(26 )
3
δ=
= 2.745 in
3
3
30 × 10 (17 )(2 ) 
 16 
1
U s = (175)(2.745) = 240 in − lb
2
For (b) N1 = 7
(
6
)
6(175)(26 )
3
δ=
= 6.666 in
3
3


30 ×10 6 (7 )(2 ) 
 16 
1
U s = (175)(6.666 ) = 583 in − lb
2
For (c) N1 = 5
(
)
6(175)(26 )
3
δ=
= 9.332 in
3
3
30 × 10 (5)(2 ) 
 16 
1
U s = (175)(9.332 ) = 817 in − lb
2
(
6
)
answer – spring (c)
(e) sd = s y = 160 ksi , N1 = 7 (b)
Page 68 of 70
SECTION 4 - SPRINGS
6 FL
6 FL
=
2
bh
N1b′h 2
6 F (26 )
160,000 =
2
(7 )(2) 3 
 16 
load F = 505 lb
sd =
6(505)(26 )
3
δ=
3
30 × 10 (7 )(2 ) 
 16 
(
319.
6
)
3
= 19.24 in
The rear spring of an automobile has 9 leaves, each with an average thickness
of 0.242 in. and a width of 2 in.; material is SAE 9261, OQT 1000 oF. The
length of the spring is 56 in. and the total weight on the spring is 1300 lb.
Assume the spring to have the form shown in Fig. 6.22 (b), Text. Determine
(a) the rate of the spring, (b) the maximum stress caused by the dead weight.
(c) What approximate repeated maximum force (0 to Fmax ) would cause
impending fatigue in 105 cycles, the number of applications of the maximum
load expected during the ordinary life of a car? (If the leaves are cold rolled
to induce a residual compressive stress on the surfaces, the endurance limit as
su 2 should be conservative.)
Solution:
Figure 6.22 (b)
3FL
sA =
2bh 2
3FL3
δA =
8Ebh 3
F = 1300 lb
h = 0.242 in
N1 = 9
b′ = 2 in
L = 56 in
(a) Rate , k =
k=
F
δA
=
(
8Ebh3
3L3
)
8 EN1b′h 3 8 30 × 10 6 (9 )(2 )(0.242 )
=
= 116.21lb in
3
3L3
3(56 )
(b) s A =
3
3FL
3(1300 )(56 )
=
= 103,590 psi
2
2
2 N1b′h
2(9 )(2 )(0.242 )
Page 69 of 70
SECTION 4 - SPRINGS
(c) SAE 9261. OQT 1000 oF
su = 192 ksi
192
sn =
= 96 ksi
2
3FL
sA =
2 N1b′h 2
3F (56 )
96,000 =
2
2(9 )(2 )(0.242 )
F = 1200 lb
321.
The front spring of an automobile is made similar to Fig. 6.23, Text. The
average thickness for each of the 6 leaves, 0.213 in.; material is SAE 9255,
OQT 1000 oF. The load caused by the weight of the car is 775 lb. (a) What
stress is caused by a force of twice the dead weight? (b) What load would
stress the spring to the yield strength?
Solution:
Figure 6.23
6 FL 3WL
3WL
s= 2 =
=
2
bh
bh
N1b′h 2
W = 775 lb , N1 = 6 , b′ = 2 in , h = 0.213 in
36 in
= 18 in
L=
2
(a) W = 2(775) = 1550 lb
3(1550 )(18)
s=
= 153,740 psi
(6)(2)(0.213)2
(b) SAE 9255, OQT 1000 oF
s y = 160 ksi
6 FL 3WL
3WL
=
=
2
2
bh
bh
N1b′h 2
3W (18)
160,000 =
(6)(2)(0.213)2
W = 1613 lb
s=
- end -
Page 70 of 70
SECTION 5 – COLUMNS
DESIGN PROBLEMS
334.
A round steel rod made of structural steel, AISI C1020, as rolled, is to be used as
a column, centrally loaded with 10 kips; N = 3 . Determine the diameter for (a)
L = 25 in. , (b) L = 50 in. (c) The same as (a) and (b) except that the material is
AISI 8640, OQT 1000 F. Is there any advantage in using this material rather than
structural steel?
Solution:
For AISI C1020,as rolled
s y = 48 ksi
F = 10 kips
N =3
(a) Le = L = 25 in.
Consider first J.B. Johnson
2

 Le  
 sy   
k
Fc = NF = s y A1 −  2  

4π E 




2
πD
A=
4
D
k=
4
E = 30 × 103 ksi
2

 
 



(48) 25
D


 
 2
(3)(10) = (48) πD 1 − 2  4  3
 4   4π 30 ×10



4 

30 = 12πD 2 1 − 2 2 
 π D 
48
30 = 12πD 2 −
(
π
D = 1.096 in
1
say D = 1 in = 1.0625 in
16
Page 1 of 18










)
SECTION 5 – COLUMNS
Le
25
=
= 94 < 120 ∴ o.k.
k  1.0625 


 4 
(b) Le = L = 50 in.
Consider Euler’s Equation
π 2 EA
Fc = NF =
2
 Le 
 
k 
 πD 2 

π 2 30 ×103 
4 

(3)(10 ) =
2
 
 50 
 
 D 
 4
3 4
30 = 0.1875π D
D = 1.507 in
1
say D = 1 in = 1.5 in
2
Le
50
=
= 133 > 120 ∴ o.k.
k  1 .5 
 
 4 
(
)
(c) For AISI 8640, OQT 1000 F
s y = 150 ksi
1
Le  2π 2 E  2
=
k  s y 
(
)
1
Le  2π 2 30 × 103  2
=
 = 62.83
k 
150

For (a) Le = L = 25 in.
Consider first J.B. Johnson
2

 Le  
 sy   
k
Fc = NF = s y A1 −  2  

4π E 




Page 2 of 18
SECTION 5 – COLUMNS
2

 
 



(150) 25
D


 
 2
(3)(10) = (150) πD  1 − 2  4  3
 4   4π 30 ×10



12.5 

30 = 37.5πD 2 1 − 2 2 
 π D 
468.75
30 = 37.5πD 2 −
(










)
π
D = 1.23 in
say D = 1.25 in
Le
25
=
= 80 > 62.83 ∴ use Euler’s equation
k  1.25 


 4 
π 2 EA
Fc = NF =
2
 Le 
 
k 
 πD 2 

π 2 30 ×103 
4 

(3)(10) =
2
 
 25 
 
 D 
 4 
3 4
30 = 0.75π D
D = 1.0657 in
1
say D = 1 in = 1.0625 in
16
Le
25
=
= 94 > 62.83 ∴ ok
k  1.0625 


 4 
For (b) Le = L = 50 in.
Consider Euler’s Equation
π 2 EA
Fc = NF =
2
 Le 
 
k 
(
Page 3 of 18
)
SECTION 5 – COLUMNS
 πD 2 

 4 
π 2 (30 ×103 )
(3)(10) =
2
 
 50 
 
 D 
 4
3 4
30 = 0.1875π D
D = 1.507 in
1
say D = 1 in = 1.5 in
2
Le
50
=
= 133 > 62.83 ∴ o.k.
k  1 .5 
 
 4 
There is no advantage.
335.
A hollow circular column, made of AISI C1020, structural steel, as rolled, is to
support a load of 10,000 lb. Let L = 40 in , Di = 0.75Do , and N = 3 . Determine
Do by (a) using either Euler’s or the parabolic equation; (b) using the straightline equation. (c) What factor of safety is given by the secant formula for the
dimensions found in (a)?
Solution:
For AISI C1020, as rolled
s y = 48 ksi
Le = L = 40 in
F = 10,000 lb = 10 kips
N =3
Di = 0.75Do
I
A
π (Do4 − Di4 )
4
I=
= π Do4 − (0.75Do ) = 0.033556 Do4
64
π (Do2 − Di2 ) π Do2 − (0.75Do )2
A=
=
= 0.343612 Do2
4
4
k=
[
[
k=
0.033556 Do4
= 0.3125Do
0.343612 Do2
Page 4 of 18
]
]
SECTION 5 – COLUMNS
(a) Consider parabolic equation
2

 Le  
 sy   
k
Fc = NF = s y A1 −  2  

4π E 








25
 (48)
 0.3125Do

(3)(10) = (48)(0.343612)Do2 1 −  2 4 3
4π 30 × 10




30 = 16.493376 Do2 − 10.9519
Do = 1.576 in
9
say Do = 1 in = 1.5625 in
16
Le
40
=
= 82 < 120 ∴ o.k.
k 0.3125(1.5625)
(
(b) Straight-line equation
F
L
= 16,000 − 70
A
k


10,000
40

= 16,000 − 70
2
0.343612 Do
 0.3125 Do 
10,000 = 5498 Do2 − 3078 Do
Do = 1.6574 in
5
say Do = 1 in = 1.625 in
8
Le
40
=
= 78.8 < 120 ∴ o.k.
k 0.3125(1.625)
(c) Secant formula
NF  ec  Le NF
sy =
1 + sec
A  k 2
 2k EA
Do = 1.5625 in
k = 0.3125Do = 0.4883 in
A = 0.343612 Do2 = 0.8389 in 2
Page 5 of 18




)





2










SECTION 5 – COLUMNS
ec
= 0.25 , (i7.8)
k2
48 =

N (10) 
40
1 + 0.25 sec 
0.8389 
 2(0.4883)
[
(
48 = 11.92 N 1 + 0.25 sec 0.81645 N
N = 2.289
336.
)]
A column is to be built up of ½-in., AISI C1020, rolled-steel plates, into a square
box-section. It is 6 ft long and centrally loaded to 80,000 lb. (a) Determine the
size of section for N = 2.74 . (b) Compute N from the secant formula for the
size found and compare with 2.74.
Solution:
For AISI C1020, rolled-steel plate
s y = 48 ksi
4
4
b 4 (b − 1) b 4 − (b − 1)
−
=
12
12
12
2
2
A = b − (b − 1)
I=
4
k=
 
10 N
(30 ×103 )(0.8389)  
I
b 4 − (b − 1)
=
2
A
12 b 2 − (b − 1)
[
]
Le = L = 6 ft = 72 in
F = 80,000 lb = 80 kips
(a) N = 2.74
Consider J.B. Johnson
2

 Le  
 sy   
k
NF = s y A1 −  2  

4π E 




2

 72  
(48)  


(2.74)(80) = (48)A 1 − 2  k  3 
 4π (30 ×10 )




10.085 A
219.2 = 48 A −
k2
Page 6 of 18
SECTION 5 – COLUMNS
try b = 3.23 in
4
4
(
3.23) − (3.23 − 1)
k=
2
2 = 1.1331 in
12[(3.23) − (3.23 − 1) ]
2
2
2
A = b 2 − (b − 1) = (3.23) − (3.23 − 1) = 5.46 in 2
10.085(5.46 )
219.2 = 48(5.46 ) −
= 219.2 ok
(1.1331)2
Therefore use b = 3.23 in
Le
72
=
= 63.54 < 120 ∴ o.k.
k 1.1331
1
b = 3.23 in or b = 3 in
4
(b) s y =
NF  ec  Le
1 + sec
A  k 2
 2k
NF
EA




ec
= 0.25 , (i7.8)
k2
48 =
 72
N (80) 
1 + 0.25 sec 
5.46 
 2(1.1331)
[
(
 
80 N
(30 ×103 )(5.46) 
48 = 14.652 N 1 + 0.25 sec 0.70214 N
N = 2.2 < 2.74
337.
)]
A column is to be made of ½-in structural steel plates (AISI 1020, as rolled),
welded into an I-section as shown in Table AT 1 with G = H . The column, 15 ft
long, is to support a load of 125 kips. (a) Determine the cross-sectional
dimensions from the straight-line equation. (Using either Johnson’s or Euler’s
equation, compute the equivalent stress and the factor of safety. (c) Compute N
from the secant formula.
Solution:
For AISI C1020, as rolled
s y = 48 ksi
Le = L = 15 ft = 180 in
F = 125 kips
Page 7 of 18
SECTION 5 – COLUMNS
Table AT 1.
G=H
A = GH − gh = H 2 − (H − 0.5)(H − 1) = H 2 − H 2 − 1.5H + 0.5 = 1.5H − 0.5 = 0.5(3H − 1)
(
k=
)
3
1  GH 3 − gh3 
1  H 4 − (H − 0.5)(H − 1) 

 =

=
12  GH − gh 
12 
0.5(3H − 1)

(a) Straight-line equation
L

F = 16,000 A1 − 0.0044 
k

(180)

125,000 = 16,000 A1 − 0.0044
k 

 0.792 
7.8125 = A1 −

k 

use H = 7.37 in
(7.37 )4 − (7.37 − 0.5)(7.37 − 1)3
6(3(7.37 ) − 1)
A = 0.5[3(7.37 ) − 1] = 10.555 in
k=
= 3.04527 in
0.792 

7.8125 ≈ 10.5551 −
 = 7.81
 3.04527 
Therefore use H = 7.37 in
3
Or H = 7 in = 7.375 in
8
L
180
= 59 < 120
(b) Consider J.B. Johnson, e =
k 3.04527
s
se = y
N
F
125
se =
=
= 13.8 ksi
2
2


 Le  
 180  
 
 sy   
 (48)
k  
3.04527  




A 1−
10.555 1 −


4π 2 E 
4π 2 30 ×103 








(
Page 8 of 18
)
3
H 4 − (H − 0.5)(H − 1)
6(3H − 1)
SECTION 5 – COLUMNS
N=
sy
48
=
= 3.48
se 13.8
(c) From secant formula
NF  ec  Le NF 
sy =
1 + sec

A  k 2
 2k EA 
ec
= 0.25 , (i7.8)
k2
48 =
 180
N (125) 
1 + 0.25 sec 
10.555 
 2(3.04527 )
[
(
48 = 11.843 N 1 + 0.25 sec 0.5872 N
N = 2 .8
338.
 
125 N
3
(30 ×10 )(10.555) 
)]
The link shown is to be designed for N = 2.5 to support an axial compressive
load that varies from 0 to 15 kips; L = 20 in ; Material AISI 1030, as rolled. (a)
Determine the diameter considering buckling only. (b) Determine the diameter
considering varying stresses and using the Soderberg line (perhaps too
conservative). Estimate an appropriate strength-reduction factor (see Fig. AF 6).
(c) Keeping in mind that the stress is always compressive, do you think that the
answer from (a) will do? Discuss.
Problem 338.
Solution:
For AISI C1030, as rolled
s y = 51 ksi
su = 80 ksi
1
1
Le  2π 2 E  2  2π 2 30 ×103  2
=
=
 = 108
k  s y 
51


L = 20 in
N = 2 .5
(a) F = 15 kips
Consider J.B. Johnson
(
Page 9 of 18
)
SECTION 5 – COLUMNS
2

 Le  
s
 

y
k
NF = s y A1 −  2  

4π E 




D
k=
4
πD 2
A=
4
Le = L = 20 in
2

 20  

(51) D  

 2
(2.5)(15) = (51) πD 1 − 2  4  3 
 4   4π (30 ×10 )






2.72 

37.5 = 12.75π D 2 1 − 2 2 
 π D 
34.68
37.5 = 12.75π D 2 −
π
D = 1.101 in
3
say D = 1 in = 1.1875 in
16
Le
20
=
= 68 < 108 ∴ o.k.
k 1.1875
4
(b) Variable stresses
sn = 0.5su = 0.5(80) = 40 ksi
Size factor = 0.85
sn = 0.85(40) = 34 ksi
K f = 2.8 (Figure AF 6)
1 sm K f sa
= +
N sy
sn
F = 0 to 15 kips
Fm = Fa = 7.5 kips
sem = sea
(2.8)sem
1
s
= em +
2.5 51
34
sem = 3.923 ksi
Page 10 of 18
SECTION 5 – COLUMNS
2

 Le  
s
 

em 
k
Fm = sem A1 −  2  

4π E 




2

 20  

(51) D  
2 
 πD  
 1 − 2  4  3 
7.5 = (3.923)
 4   4π (30 ×10 )






2.72 

7.5 = 0.98π D 2 1 − 2 2 
 π D 
2.67
7.5 = 0.98π D 2 −
π
D = 1.65 in
5
say D = 1 in = 1.625 in
8
Le
20
=
= 49 < 108 ∴ o.k.
k 1.625
4
(c) The answer in (a) will not do because it is lower than (b)
339.
The connecting link for a machine (see figure) is subjected to a load that varies
fro + 450 (tension) to –250 lb. The cross section is to have the proportions
G = 0.4 H , t = 0.1H , fillet radius r ≈ 0.05 H ; L = 10 in ; material, AISI C1020,
as rolled. (a) Considering buckling only, determine the dimensions for a design
factor of 2.5. (b) For the dimension found compute the factor of safety from the
Soderberg criterion.
Problems 339, 340
Solution:
For AISI C1030, as rolled
s y = 48 ksi
su = 65 ksi
Page 11 of 18
SECTION 5 – COLUMNS
Table AT 1
G = 0 .4 H
t = 0.1H
r ≈ 0.05 H
A = GH − gh
g = G − t = 0.4 H − 0.1H = 0.3H
h = H − 2(0.1H ) = 0.8 H
A = (0.4 H )(H ) − (0.3H )(0.8H ) = 0.16 H 2
k=
3
3
1  GH 3 − gh3 
1  (0.4 H )(H ) − (0.3H )(0.8 H ) 

 =

 = 0.35824 H
12  GH − gh 
12 
0.16 H 2

(
)
(a) Consider J.B. Johnson
2

 Le  
s
 

y
k
NF = s y A1 −  2  

4π E 




F = 350 lb = 0.35 kip
Le = 10 in
2


 
10

 48
2  
(
0
.
35824
H
)
2 
 
(2.5)(0.35) = (48) 0.16 H 1 −  2
3

4π 30 × 10




0.875 = 7.68 H 2 − 0.2425
H = 0.3815 in
Le
10
=
= 73 < 120 ∴ ok
k (0.35824)(0.3815)
15
say H =
in = 0.46875 in
32
3
G = 0.4 H = 0.1875 in = in
16
3
t = 0.1H = 0.046875 in =
in
64
(
Page 12 of 18
)
(
)
SECTION 5 – COLUMNS
(b) with H = 0.46875 in
2
A = 0.16(0.46875) = 0.0352 in 2
k = 0.35824(0.46875) = 0.1679 in
− 350
Fmin
0.0352
A
smin = s e =
=
= −11,600 psi = −11.6 ksi
2
2



 Le 
 10  
 
 s y     48
1 −  k   1 −  0.1679  

4π 2 E   4π 2 30 × 103 
 




 
F
+ 450
smax = max =
= +12,800 psi = +12.8 psi
A
0.0352
1
sm = (12.8 − 11.6 ) = 0.6 ksi
2
1
sa = (12.8 + 11.6 ) = 12.1 ksi
2
su = 0.5sn = 0.5(65) = 32.5 ksi
Size factor = 0.85
su = 0.85(32.5) ksi = 27.62 ksi
(
)
Figure AF 9, r = 0.05H = 0.05(0.46875) = 0.023
h = 1.5H = 15.(0.46875) = 0.7031 in
d = H = 0.4688 in
r 0.05 H
=
= 0.05
d
H
h 1 .5 H
=
= 1 .5
d
H
K t = 2.65
1
1
q=
=
= 0.70
0.01
0.01
1+
1+
r
0.023
K f = 0.70(2.65 − 1) + 1 = 2.2
1 sm K f sq
=
+
N sy
sn
1 0.6 (2.2 )(12.1)
=
+
N 48
27.62
N = 1.024
CHECK PROBLEMS
341.
The link shown is subjected to an axial compressive load of 15 kips. Made of
AISI C1030, as rolled, it has sectional length of 20 in. Assume a loose fit with the
Page 13 of 18
SECTION 5 – COLUMNS
pins. What is (a) the critical load for this column, (b) the design factor, (c) the
equivalent stress under a load of 15 kips? What material does the secant formula
indicate as satisfactory for the foregoing load, when (e) ec k 2 = 0.25 , (f)
L
e= e .
400
Problem 341, 342
Solution:
For AISI C1030, as rolled
s y = 51 ksi
b = 0.75 in
h = 1.75 in
A = bh = (0.75)(1.75) = 1.3125 in 2
For loose fit
bh3
I=
12
I
bh 3
h
1.75
k=
=
=
=
= 0.5052 in
A
12bh
12
12
Le
20
=
= 39.6 < 108 for AISI C1030, as rolled
k 0.5052
use J.B. Johnson equation
2
2


 Le  
 20  
 
 sy   
 51
k  
0.5052  




(a) Fc = s y A 1 −
= (51)(1.3125) 1 −
= 62.42 kips

 4π 2 (30,000) 
4π 2 E 








(b) Fc = NF
F 62.42
N= c =
= 4.16
F
15
s
51
(c) se = y =
= 12.26 ksi
N 4.16
F
15
(d) Actual s = =
= 11.43 ksi
A 1.3125
Page 14 of 18
SECTION 5 – COLUMNS
Secant Formula
NF  ec  Le NF 
sy =
1 + sec

A  k 2
 2k EA 
ec
(e) 2 = 0.25
k

 
62.42 
20
62.42
sy =
1 + 0.25 sec 
  = 64.4 ksi
3
1.3125 
 2(0.5052) 30 × 10 (1.3125)  
use AISI C1020, cold drawn, s y = 66 ksi
(
)
Le
20
=
= 0.05 in
400 400
h 1.75
c= =
= 0.875 in
2
2
ec (0.05)(0.875)
=
= 0.1714
k2
(0.5052)2
(f) e =
sy =

62.42 
20
1 + 0.1714 sec 
1.3125 
 2(0.5052 )
 
62.42
  = 59.12 ksi
3
30 × 10 (1.3125)  
(
)
use AISI C1045, cold drawn, s y = 59 ksi
343.
A schedule-40, 4-in. pipe is used as a column. Some of its properties are:
Do = 4.5 in , Di = 4.026 in , I = 3.174 sq.in. , L = 15 ft ; material equivalent to
AISI C1015, as rolled. The total load to be carried is 200 kips. (a) What
minimum number of these columns should be used if a design factor of 2.5 is
desired and the load evenly distributed among them? For the approximately fixed
ends, use Le = 0.65L as recommended by AISC. (b) What is the equivalent stress
in the column?
Solution:
For AISI C1015, as rolled
s y = 45.5 ksi
1
1
Le  2π 2 E  2  2π 2 30 × 103  2
=
=
 = 114
k  s y 
45.5


L = 15 ft = 180 in
Le = 0.65L = 0.65(180) = 117 in
Le
117
=
= 77.5 < 114
k 1.509
(
Page 15 of 18
)
SECTION 5 – COLUMNS
Use J.B. Johnson equation
2

 Le  
sy   
2
s y A 
k   (45.5)(3.174)  45.5(77.5) 

(a) F =
1−
=
1 − 2
 = 44.4 kips
N 
4π 2 E 
2.5
 4π (30,000) 




No. of columns
200
=
= 4.5 say 5 columns
44.4
F
A
(b) se =
2

 Le  
s
 

y
1 −  k  

4π 2 E 




200
F=
= 40 kips
5
40
3.174
se =
= 16.4 ksi
2

 77.5  
 
 45.5
 k  
1 −
 4π 2 (30,000) 




344.
A generally loaded column is a 10-in. x 49 lb., wide-flange I-beam whose
properties are (see figure); k x = 4.35 in , k y = 2.54 in , area A = 14.4 sq.in. ,
I x = 272.9 in 4 , I y = 93.0 in 4 ; length L = 30 ft , material AISI 1022, as rolled. Let
the ends be a “little” fixed with Le = 0.8L and determine the critical load (a)
according to the Johndon or the Euler equation; (b) according to the secant
formula if ec k 2 is assumed to be 0.25.
Page 16 of 18
SECTION 5 – COLUMNS
Solution:
For AISI C1022, as rolled
s y = 52 ksi
1
1
Le  2π 2 E  2  2π 2 (30 ×103 ) 2
=
=
 = 107
k  s y 
52


(a) k = 2.54 in
I = 93.0 in 4
Le = 0.8(30)(12) = 288 in
Le 288
=
= 113.4 > 107
k 2.54
Use Euler’s Equation
π 2 EA π 2 (30,000)(14.4)
Fc =
=
= 332 kips
2
(113.4)2
 Le 
 
k 
(b) Secant formula
NF  ec  Le NF
sy =
1 + sec
A  k 2
 2k EA
113.4
F 
52 = c 1 + 0.25 sec 
14.4 
 2
F
52 = c 1 + 0.25 sec 0.0863
14.4
Fc = 273 kips
{
348.
[




 
Fc
3
(30 ×10 )(14.4)  
Fc
]}
A 4 x 3 x ½-in. angle is used as a flat-ended column, 5 ft. long, with the resultant
load passing through the centroid G (see figure); k x = 1.25 in , k y = 0.86 in ,
ku = 1.37 in , kv = 0.64 in , A = 3.25 sq.in. Find the safe load if N = 2.8 and the
material is (a) structural steel, (b) magnesium alloy AZ 91C (i7.12.\, Text), (c)
magnesium alloy AZ 80A, (d) magnesium alloy AZ 80A as before, but use the
Johnson formula and compare.
Page 17 of 18
SECTION 5 – COLUMNS
Solution:
L (5)(12 )
Le = =
= 30 in
2
2
k = k min = 0.64 in
Le
30
=
= 46.875
k 0.64
(a) Structural steel, s y = 48 ksi
Le
= 46.875 < 120
k
use J.B. Johnson
2

 Le  
sy   
2
s y A 
k   (48)(3.25)  48(46.875) 

F=
1−
=
1 − 2
 = 50.75 kips
N 
4π 2 E 
2.8
 4π (30,000 )




(b) magnesium alloy AZ 91C
NF
C
=
2
A
 Le 
C 
k
1+   6
64.4 × 10
C = 57,000
(2.8)(F ) =
57,000
psi
2
3.25
57,000(46.875)
1+
64.4 ×106
F = 22,467 lb = 22.467 kips
(c) magnesium alloy AZ 80A
C = 82,900
(2.8)(F ) =
82,900
psi
2
3.25
82,900(46.875)
1+
64.4 × 106
F = 25,134 lb = 25.134 kips
(d) By J.B. Johnson
For magnesium alloy AZ 80A, s y = 36 ksi
2

 Le  
s
 
y
s y A 
(36)(3.25) 1 − 36(46.875)2  = 39 kips > 25.134 kips
k
F=
1−  2   =


2
N 
4π E 
2.8
 4π (30,000) 




- end -
Page 18 of 18
SECTION 6 – COMBINED STRESSES
ECCENTRIC LOADING (NORMAL STRESSES)
DESIGN PROBLEM
361.
It is necessary to shape a certain link as shown in order to prevent interference with
another part of the machine. It is to support a steady tensile load of 2500 lb. with a
design factor of 2 based on the yield strength. The bottom edge of the midsection is
displaced upward a distance a = 2 ½ in. above the line of action of the load. For AISI
C1022, as rolled, and h ≈ 3b, what should be h and b?
Solution:
F = 2500 lb
Ny = 2
a = 2.5 in
F Fec
σ= ±
A
I
For AISI C1022, as rolled, sy = 52 ksi (Table AT7).
σ=
sy
Ny
=
52
= 26 ksi = 26,000 psi
2
A = bh = 3b2
bh 3 b(3b )2
I=
=
= 2.25b 4
12
12
h
c = = 1.5b
2
h
e = a + = a + 1.5b = 2.5 + 1.5b
2
F Fec
σ= ±
A
I
(2.5 + 1.5b)(1.5b) 
 1
26,000 = 2500 2 +

2.25b 4
 3b
2.5 + 1.5b 
 1
26,000 = 2500 2 +

1.5b 3 
 3b
By trial and error method:
5
b = 0.625 in = in
8
7
h = 3b = 3(0.625) in = 1.875 in = 1 in
8
Page 1 of 133
SECTION 6 – COMBINED STRESSES
362.
A tensile load on a link as described in 361 varies from 0 to 3000 lb.; it is machined from
AISI 1045, as rolled, and the lower edge of the link is a = 0.5 in. above the center line of
the pins; h ≈ 3b. Determine the dimensions of the link for N = 2 based on the Soderberg
line.
Solution:
Soderberg Line:
1 sm sa
=
+
N s y sn
For AISI 1045, as rolled (Table AT 7).
sy = 59 ksi
su = 96 ksi
sn′ = 0.5su = 48 ksi
Size factor = 0.85
Load factor (axial) = 0.80
sn = 0.85(0.80)(48) = 32.64 ksi
1
(3000 lb + 0 ) = 1500 lb
2
1
Fa = (3000 lb − 0 ) = 1500 lb
2
h
e = a + = a + 1.5b = 0.5 + 1.5b
2
 1 ec 
sm = Fm  + 
A I 
(0.5 + 1.5b )(1.5b) 
 1
sm = 1500 2 +

2.24b 3
 3b
0.5 + 1.5b 
 1
sm = 1500 2 +

1.5b 3 
 3b
 1 ec 
sa = Fa  + 
A I 
(0.5 + 1.5b)(1.5b ) 
 1
sa = 1500 2 +

3
b
2.24b 3

0.5 + 1.5b 
 1
sa = 1500 2 +

1.5b 3 
 3b
1 sm sa
=
+
N s y sn
Fm =
Page 2 of 133
SECTION 6 – COMBINED STRESSES
1  1500
1500  1 0.5 + 1.5b 
=
+
+


2  59,000 32,640  3b 2
1.5b3 
1
0.5 + 1.5b
+
= 7.00
2
3b
1.5b 3
By trial and error method:
b = 0.53 in
h = 3b = 3(0.53) in = 1.59 in
Use b x h = 9/16 in x 1 5/8 in
363.
The same as 362, except that the load continuously reverses, 3 kips to -3 kips.
Solution:
1
Fm = (3 − 3) = 0 lb
2
1
Fa = (3 + 3) = 3 kips
2
 1 ec 
sm = Fm  + 
A I 
(0.5 + 1.5b )(1.5b) 
 1
sm = 1500 2 +

2.24b 3
 3b
0.5 + 1.5b 
 1
sm = (0) 2 +
=0
1.5b 3 
 3b
 1 ec 
sa = Fa  + 
A I 
(0.5 + 1.5b)(1.5b) 
 1
s a = 3 2 +

2.24b 3
 3b
0.5 + 1.5b 
 1
sa = 3 2 +

1.5b 3 
 3b
1 sm sa
=
+
N s y sn
1 
3  1
0.5 + 1.5b 
=0+
 2 +

2 
32.64  3b
1.5b 3 
1
0.5 + 1.5b
+
= 5.44
2
3b
1.5b 3
By trial and error method:
b = 0.5905 in
Page 3 of 133
SECTION 6 – COMBINED STRESSES
h = 3b = 3(0.5905) in = 1.7715 in
Use b x h = 5/8 in x 1 13/16 in
364.
A circular column (See Fig. 8.3, Text), the material of which is SAE 1020, as rolled, is to
have a length of 9 ft. and support an eccentric load of 16 kips at a distance of 3 in. from
the center line. Let N = 3. (a) What should be the outside diameter Do if the column is
hollow and Di = 0.75Do? (b) What should be the diameter if the column is solid?
Solution:
a. Try J.B. Johnson
2

 Le  
 sy   
F
k
= se 1 −  2  

A
4π E 




For SAE 1020, as rolled, s y = 48 ksi
E = 30×10 6 psi
Transition point
1
(
)
1
Le  2π 2 E  2  2π 2 30 ×10 6  2
=
=
 = 111
k  sy 
 48000 
Le = 2L
L = 9 ft = 108 in
Le = 2L = 2(108) = 216 in
I
A
π Do4 − Di4 π Do4 − (0.75Do )4 0.6836πDo4
I=
=
=
64
64
64
k=
) [
(
A=
π
(D
4
2
o
)
− Di2 =
Page 4 of 133
π
4
]
[D
2
o
]
− (0.75Do )2 = 0.109375πDo2
SECTION 6 – COMBINED STRESSES
0.6836πDo4
64
k=
= 0.3125Do
0.109375πDo2
16
46.5642
F
2
0.109375πDo
Do2
A
se =
=
=
2
2

 Le   
 216   1 − 19.363 
s
   48
  

y
Do2 
0.3125Do   
1 −  k   


4π 2 E  1 − 4π 2 30 ×103 

 


 



(
σ = se +
Fec
I
Do
= 0.5Do
2
e = 3 in
F = 16 kips
c=
sy
48
= 16 ksi
N 3
46.5642
Do2
16(3)(0.5Do )
σ = 16 =
+
 19.363   0.6836Do4 
1 −
 

64
Do2  


σ=
=
σ = 16 =
46.5642
Do2
 19.363 
1 −

Do2 

+
715.22
Do3
By trial and error method
Do = 3.23 in
k = 0.3125(3.23) = 1.0094 in
Le 2L
216
= =
= 214 > 111
k
k 1.0094
Therefore use Euler’s equation
π 2E
se =
2
L 
N e 
k
Page 5 of 133
)
SECTION 6 – COMBINED STRESSES
se =
π 2 (30,000)
 216
3
 0.3125Do
Fec
σ = se +
I



16 = 0.20658Do2 +
2
= 0.20658Do2 ksi
16(3)(0.5Do )
715.22
= 0.20658Do2 +
4
 0.6836πDo 
Do3




64


Do = 3.802 in
Di = 0.75Do = 0.75(3.802) = 2.8515 in
To check:
k = 0.3125Do = 0.3125(3.802) in = 1.188125 in
Le
216 in
=
= 182 > 111
k 1.188125 in
Use Do = 3 13/16 in, Di = 2 13/16 in
b. For solid, also using Euler’s equation.
I
A
πD 4
I=
64
πD 2
A=
4
k=
πD 4
k=
se =
se =
64 = 1 D = 0.25D
πD 2 4
4
π 2E
2
L 
N e 
k
π 2 (30,000)
 216 
3

 0.25D 
Fec
σ = se +
I
2
= 0.1322D 2 ksi
16(3)(0.5D )
489
= 0.1322D 2 + 3
4
 πD 
D


 64 
By trial and error method.
16 = 0.1322D 2 +
Page 6 of 133
SECTION 6 – COMBINED STRESSES
D = 3.221 in
k = 0.25D = 0.25(3.221) in = 0.80525 in
Le
216 in
=
= 268 > 111
k 0.80525 in
Use D = 3 ¼ in.
365.
The same as 364, except that the length is 15 ft.
Solution:
Euler’s Equation:
Le = 2L = 2(15)(12 ) = 360 in
σ = 16 ksi
e = 3 in
0.6836πDo4
64
A = 0.109375πDo2
k = 0.3125Do
a. I =
se =
se =
π 2E
L 
N e 
k
2
π 2 (30,000)
 360
3
 0.3125Do
Fec
σ = se +
I



16 = 0.07437Do2 +
2
= 0.07437Do2 ksi
16(3)(0.5Do )
715.22
= 0.07437Do2 +
4
 0.6836πDo 
Do3




64


Do = 3.624 in
Di = 0.75Do = 0.75(3.624 ) = 2.718 in
To check:
k = 0.3125Do = 0.3125(3.624) in = 1.1325 in
Le
216 in
=
= 191 > 111
k 1.1325 in
Use Do = 3 5/8 in, Di = 2 5/8 in
Page 7 of 133
SECTION 6 – COMBINED STRESSES
b. I =
A=
πD 4
64
πD 2
4
k = 0.25D
se =
se =
π 2E
2
L 
N e 
k 
2
π (30,000)
 360 
3

 0.25D 
Fec
σ = se +
I
2
16 = 0.0476D 2 +
= 0.0476D 2 ksi
16(3)(0.5D )
489
= 0.0476D 2 + 3
4
 πD 
D


 64 
By trial and error method.
D = 3.158 in use 3 3/16 iin
k = 0.25D = 0.25(3.158) in = 0.7895 in
Le
216 in
=
= 274 > 111
k 0.7895 in
Use D = 3 3/16 in.
366.
A link similar to one shown is to be designed for: steady load F = 8 kips, L = 20 in. θ = 30o;
aluminum alloy 2024-T4; N = 2.6 on the yield strength. It seems desirable for the
dimension b not to exceed 1 3/8 in. Determine b and h and check their proportions for
reasonableness. The support is made so that the pin at B carries the entire horizontal
component of F.
Page 8 of 133
SECTION 6 – COMBINED STRESSES
Solution:
Aluminum alloy (2024-T4), sy = 47 ksi
RBH = F sin 30o
FL cos 30o
d
F (L + d )cos 30o
RBV =
d
M = RAd = FL cos 30o
Mc
s2 =
I
h
c=
2
bh 3
I=
12
6M 6FL cos 30o
s2 = 2 =
bh
bh 2
F sin 30o
s1 =
bh
σ t = s1 + s2
RA =
F sin 30o 6FL cos 30o
+
bh
bh 2
s
47
σt = y =
= 18 ksi
N 2. 6
F = 8 kips
3
b = 1 in = 1.375 in
8
L = 20 in
σt =
σt =
F sin 30o 6FL cos 30o
+
bh
bh 2
Page 9 of 133
SECTION 6 – COMBINED STRESSES
6(8)(20)cos 30o
1.375h
1.375h 2
24.75h 2 − 4h − 831.4 = 0
18 =
(8)sin 30o
+
7
h = 5.877 in ≈ 5 in
8
7
5 in
h
= 8 = 4.27
b 1 3 in
8
7
3
Therefore, use h = 5 in , b = 1 in
8
8
367.
A column 15 ft. long is to support a load F2 = 50,000 lb. Acting at a distance of e = 8 in.
from the axis of the column as shown (with F1 = 0). Select a suitable I-beam for a design
factor of 3 based on yield strength. The upper end of the column is free. See handbook
for the properties of rolled sections.
Solution:
Use C1020, structural steel, sy = 48 ksi
Secant Formula
NF  ec
L NF 
1 + 2 sec e

sy =
A  k
2 EI 
F = F2 = 50,000 lbs = 50 kips
e = 8 in
N =3
E = 30,000 ksi
depth
c=
2
Le = 2L = 2(15)(12 ) = 360 in
Page 10 of 133
SECTION 6 – COMBINED STRESSES


(8) depth  


(
)(
)
3(50) 
360
3
50
2

 sec
48 ≥
1 +



A 
k2
 2 (30,000)(I )  


rd
From Strength of Materials, 3 Edition by F.S. Singer and A. Pytel, Table B-2, pg. 640, select
Wide-Flange Sections by trial and error.
Then selecting W360 x 51, properties are
A = 6450 mm2 = 10 in2
Depth = 355 mm = 14 in
k = 148 mm = 5.83 in
I = 1.41 x 108 mm4 = 338.8 in4
Substitute,


(8) 14  



(3)(50)
3(50) 
 2  sec 360
48 ≥
1 +


2
10  (5.83)
 2 (30,000)(338.8)  


48 ≥ 47.08
Therefore suitable wide flange I-beam is W14 x 34 lb. (English units)
A = 10 in2
Depth = 14 in
k = 5.83 in
I = 338.8 in4
368.
The same as 367, except that F1 = 50,000 lb.
Solution:
Use C1020, structural steel, sy = 48 ksi
N =3
E = 30,000 ksi
Le = 2L = 2(15)(12 ) = 360 in
Transition Point
1
1
Le  2π 2E  2  2π 2 (30,000)  2
=
=
 = 111
k  sy 
48


F = F1 = 50,000 lbs = 50 kips
Page 11 of 133
SECTION 6 – COMBINED STRESSES
Check J.B. Johnson Formula
2

 Le  
 sy   
k
Fc = NF1 = sy A 1 −  2  

4π E 




NF1
sy =
2

 Le  
 sy   
k
A 1 −  2  

4π E 




3(50 )
48 ≥
2

 360  
48
 

k  
A 1 − 2
 4π (30,000) 




150
48 ≥
 5.2549 
A1 −

k2 

From Strength of Materials, 3rd Edition by F.S. Singer and A. Pytel, Table B-2, pg. 640, select
Wide-Flange Sections by trial and error.
Then selecting W310 x 21, properties are
A = 2690 mm2 = 4.17 in2
k = 117 mm = 4.61 in
Substitute,
150
48 ≥
 5.2549 
4.17 1 −
2
 (4.61) 
48 ≥
150
 5.2549 
4.17 1 −
2
 (4.61) 
48 ≥ 47.8
Check for validity of JB Johnson Formula
Le 360
=
= 78.1 < 111
k 4.61
Therefore, JB Johnson formula is valid and suitable wide flange I-beam is W12 x 14 lb. (English
units)
A = 4.17 in2
k = 4.61 in
Page 12 of 133
SECTION 6 – COMBINED STRESSES
CHECK PROBLEMS
369.
A cam press, similar to that of Fig. 19-1, Text, exerts a force of 10 kips at a distance of 7
in. from the inside edge of the plates that make up the frame. If these plates are 1 in.
thick and the horizontal section has a depth of 6 in., what will be the maximum stress in
this section?
Solution:
σ=
F Fec
+
A
I
F = 10 kips
6
e = 7 in + in = 10 in
2
6
c = in = 3 in
2
A = 2(1)(6 ) = 12 in 2
I=2
(1)(6 )3
12
= 36 in 4
F Fec
+
A
I
10 (10)(10)(3)
σ= +
= 9.2 ksi
12
36
σ=
370.
A manufacturer decides to market a line of aluminum alloy (6061-T6) C-clamps, (see Fig.
8.4, Text). One frame has a T-section with the following dimensions (letters as in Table
AT 1): H = 1 1/16 , B = 17/32, a = 1/8, and t = 1/8. The center line of the screw is 2 3/8 in.
from the inside face of the frame. (a) For N = 3 on the yield strength, what is the
Page 13 of 133
SECTION 6 – COMBINED STRESSES
capacity of the clamp (gripping force)? (b) Above what approximate load will a
permanent deformation of the clamp occur?
Solution:
AA 6061-T6, sy = 40 ksi (Table AT 3)
See Fig. 8.4, (C-clamp) T-section
F Fe′c
+
A
I
 1 e′c 
σ = F +

A I 
From Table AT 1 (T-section)
A = Bt + a(H − t )
σ=
I=
Bt 2
ah 2
+ (Bt )d 2 +
+ (ah )e 2
12
12
H = 1.0625 in
B = 0.53125 in
a = 0.125 in
t = 0.125 in
h = H − t = 1.0625 − 0.125 = 0.9375 in
For c1 :
 h
t 
c1 (Bt + ha ) =  t + (ha ) +  (Bt )
 2
2
0.9375 

c1 [(0.53125)(0.125) + (0.9375)(0.125)] =  0.125 +
(0.9375)(0.125)
2 

 0.125 
+
(0.53125)(0.125)
 2 
c1 = 0.4016 in
c 2 = H − c1 = 1.0625 − 0.4016 = 0.6609 in
Page 14 of 133
SECTION 6 – COMBINED STRESSES
t
0.125
d = c1 − = 0.4016 −
= 0.3391 in
2
2
h
0.9375
e = t + − c1 = 0.125 +
0.4016 = 0.19215 in
2
2
A = Bt + a(H − t )
A = (0.53125)(0.125) + (0.125)(1.0625 − 0.125) = 0.1836 in 2
Bt 2
ah 2
+ (Bt )d 2 +
+ (ah )e 2
12
12
(0.53125)(0.125)2
(0.125)(0.9375)2
I=
+ (0.53125)(0.125)(0.3391)2 +
12
12
2
4
+ (0.125)(0.9375)(0.19215) = 0.02063 in
I=
c = c1 = 0.4016 in
e′ = 2.375 + 0.4016 = 2.7766 in
sy
40
= 13 ksi
N 3
(2.7766)(0.4016) 
 1
σ = 13 = F 
+

0.02063
 0.1836
F = 0.218 kips = 218 lbs
a. σ =
=
b. σ = sy = 40 ksi
(2.7766)(0.4016) 
 1
+

0.02063
 0.1836
F = 0.672 kips = 672 lbs
σ = 40 = F 
371.
A C-frame (Fig. 8.5 Text) of a hand-screw press is made of annealed cast steel, ASTM
A27-58 and has a section similar to that shown. The force F acts normal to the plane of
the section at a distance of 12 in. from the inside face. The various dimensions of the
sections are: a = 3 in., b = 6 in., h = 5 in., d = e= f = 1 in. Determine the force F for N = 6
based on the ultimate strength.
Page 15 of 133
SECTION 6 – COMBINED STRESSES
Solution:
a = 3 in
b = 6 in
h = 5 in
d = e = f = 1 in
A = fa + dh + (b − f − d )e
A = (1)(3) + (1)(5) + (6 − 1 − 1)(1) = 12 in 2
For c1 :
f
d
b

c1 A = hd   + (b − f − d )  + af  b − 
2
2
2

1
1
6

c1 (12) = (5)(1)  + (6 − 1 − 1)  + (3)(1) 6 − 
2
 2
2

c1 = 2.583 in
c 2 = b − c1 = 6 − 2.583 = 3.417 in
Page 16 of 133
SECTION 6 – COMBINED STRESSES
j=
6 −1 + 1
b − f −d
b− f +d
− 2.583 = 0.417 in
− (c1 − d ) =
− c1 =
2
2
2
f
1
= 3.417 − = 2.917 in
2
2
d
1
m = c1 − = 2.583 − = 2.083 in
2
2
k = c2 −
af 3
e(b − f − d )3
hd 3
+ afk 2 +
+ e(b − f − d ) j 2 +
+ hdm 2
12
12
12
(3)(1)3
(1)(6 − 1 − 1)3
(5)(1)3
I=
+ (3)(1)(2.917 )2 +
+ (1)(6 − 1 − 1)(0.417 )2 +
+ (5)(1)(2.083)2
12
12
12
I = 53.92 in 4
I=
For ASTM A27-58 Annealed Cast Steel
su = 60 ksi
σ=
su 60
=
= 10 ksi
N 6
Page 17 of 133
SECTION 6 – COMBINED STRESSES
 1 e′c 
+

A I 
c = c1 = 2.583 in
e′ = 12 + 2.583 = 14.583 in
σ = F
 1 (14.583)(2.583) 
10 = F  +

53.92
12
F = 12.789 kips = 12,789 lbs
In the link shown (366), let b = ½ in., h = 2 in., d = 2 in., L = 18 in., and θ = 60o. The
clearance at the pins A and B are such that B resists the entire horizontal component of
F; material is AISI C1020, as rolled. What may be the value of F for N = 3 based on the
yield strength?
372.
Solution:
Refer to Prob. 366.
F sinθ 6FL cos θ
+
σt =
bh
bh 2
 sinθ 6L cos θ 
+
σ t = F

bh 2 
 bh
For AISI C1020, as rolled, sy = 48 ksi .
σ=
sy
N
=
48
= 16 ksi
3
b = 0.5 in
h = 2 in
L = 18 in
θ = 60o
 sin 60 6(18)cos 60 
16 = F 
+
(0.5)(2)2 
 (0.5)(2)
F = 0.574 kips = 574 lbs
373.
The link shown is subjected to a steady load F1 = 2.1 kips; b = 0.5 in., h = a = d = 2 in., L =
18 in.; material AISI 1040, cold drawn (10% work). The dimensions are such that all of
the horizontal reaction from F2 occurs at A; and F2 varies from 0 to a maximum, acting
Page 18 of 133
SECTION 6 – COMBINED STRESSES
towards the right. For N = 1.5 based on the Soderberg line, what is the maximum value
of F2? Assume that the stress concentration at the holes can be neglected.
Solution:
b = 0.5 in
h = a = d = 2 in
L = 18 in
F1 = 2.1 kips
∑M
A
=0
(a + L + d )E = aF1 + hF2
E=
aF1 + hF2
a+L+d
∑F
V
=0
Ay = F1 − E
∑F
H
=0
Ax = F2
For F2 = 0 :
(2 )(2.1) + 0 = 0.191 kip
E=
2 + 18 + 2
Ay = F1 − E = 2.1 − 0.191 = 1.909 kips
Ax = F2 = 0
M = aAy = (L + d )E
M=
(L + d )(aF1 + hF2 )
a+L+d
F2 = 0
Page 19 of 133
SECTION 6 – COMBINED STRESSES
M=
(18 + 2)[(2)(2.1) + 0] = 3.82 in − kips
2 + 18 + 2
Let F2 = max F2
Mmax + 3.82
2
(
18 + 2)[2(2.1) + 2F2 ]
Mmax =
= 1.82(2.1 + F2 )
2 + 18 + 2
1.82(2.1 + F2 ) + 3.82
Mm =
= 3.82 + 0.91F2
2
Mm =
Mmax − 3.82
2
1.82(2.1 + F2 ) − 3.82
Ma =
= 0.91F2
2
Ma =
Fm = Fa = 0.5F2
For SAE AISI 1040 Cold Drawn (10% Work)
sy = 85 ksi
sn′ = 54 ksi
sn (bending ) = sn′ × size factor = (54)(0.85) = 45.9 ksi
se = sm +
se =
sy
sn
sa
M c F
= m + m
N  I
A
sy
 sy  Ma c Fa 
+ 
+ 
A
 sn  I




85  (1.82 )(2.1 + F2 )(1) 0.5F2  85  (0.91F2 )(1) 0.5F2 
=
+
+
+
1
1.5 
(0.5)(2)  45.9  1
(0.5)(2) 
3
3




F2 = 3.785 kips
376.
A free-end column as shown, L = 12 ft. long, is made of 10-in. pipe, schedule 40, (Do =
10.75 in., Di = 10.02 in., k = 3.67 in., Am = 11.908 in2., I = 160.7 in4., Z = 29.9 in3.). The load
completely reverses and e = 15 in.; N = 3; material is similar to AISI C1015, as rolled. (a)
Using the equivalent-stress approach, compute the safe (static) load as a column only.
(b) Judging the varying loading by the Soderberg criterion, compute the safe maximum
load. (c) Determine the safe load from the secant formula. (d) Specify what you consider
to be a reasonable safe loading.
Page 20 of 133
SECTION 6 – COMBINED STRESSES
Solution:
For AISI C1015, as rolled.
sy = 45.5 ksi
su = 61 ksi
sn = 0.5su = 0.5(61) = 30.5 ksi
Do = 10.75 in
Di = 10.02 in
L = 12 ft
k = 3.67 in
Am = 11.908 in 2
I = 160.07 in 4
Z = 29.9 in3
N =3
a. As a column only (static)
Le = 2L = 2(12)(12) = 288 in
Le 288 in
=
= 78.5 < 120
k 3.67 in
E = 30,000 ksi
Use J.B. Johnson Formula:
F
s1 =
2

 Le  
 sy   
k
A 1 −  2  

4π E 




Page 21 of 133
SECTION 6 – COMBINED STRESSES
F
s1 =
 (45.5)(78.5)2 
11.9081 −

2
 4π (30,000) 
F
s1 =
9.09
Fe
s2 =
Z
e = 15 in
F (15) F
s2 =
=
29.9 2
σ = s1 + s2
45.5
F
F
=
+
3
9.09 2
F = 24.863 kips = 24,863 lbs
b. Varying load: Fmax = F , Fmin = −F
F −F
=0
2
F − (− F )
Fa =
=F
2
s
se = sm + y sa
sn
Axial load factor = 0.80
sn = sn′ × size factor × load factor = (30.5)(0.85)(0.80) = 20.74 ksi
Fm =
F e F  s F e F 
se =  m + m  + y  a + a 
A  sn  Z
A
 Z
sy
45.5  F (15)
F 
se = = (0) +
+

 = 2.25F
N
20.74  29.9 1.908 
45.5
= 2.25F
3
F = 6.74 kips = 6740 lbs
c. Secant Formula
sy =
Z=
NF  ec
L
1 + 2 sec e

A  k
2
I
c
Page 22 of 133
NF
EI




SECTION 6 – COMBINED STRESSES
I 160.07
=
= 5.354 in
Z
29.9
ec (15)(5.354)
=
= 5.963
k2
(3.67 )2
c=
Le NF 288
3F
=
= 0.11382 F
2 EI
2 (30,000)(160.07 )
NF
3F
=
= 0.252F
A 11.908
NF  ec
L NF 
1 + 2 sec e

sy =

A  k
2 EI 
[
(
sy = 45.5 = 0.252F 1 + 5.963 sec 0.11382 F
)]
F = 22.5 kips = 22,500 lbs
d. 6740 lbs.
377.
A bracket is attached as shown (367) onto a 14-in. x 193-lb., wide flange I-beam (A =
56.73 sq. in., depth = 15.5 in., flange width = 15.710 in., Imax = 2402.4 in4., Imin = 930.1
in4., kmin = 4.05 in.). The member is an eccentrically loaded column, 40 ft. long, with no
central load (F1 = 0) and no restraint at the top. For e = 12 in. and N = 4, what may be
the value of F2?
Solution:
Using secant formula:
NF  ec
L NF 
1 + 2 sec e

sy =

A  k
2 EI 
I = Imax = 2402.4 in 4
E = 30,000 ksi
e = 12 in
k = kmax = 6.50 in
Le = 2L = 2(40)(12) = 960 in
Page 23 of 133
SECTION 6 – COMBINED STRESSES
Le
= 148
k
N=4
A = 56.73 in 2
depth 15.5
c=
=
= 7.75 in
2
2
For C1020, as rolled, structural steel,
sy = 48 ksi
sy = 48 =
4F  (12)(7.75)
460
sec
1 +
2
56.73 
2
(6.50)
[
(
48 = 0.0705F 1 + 2.2 sec 0.11308 F
F = F2 = 104.9 kips = 104,900 lbs
378.

4F
(30,000)(2402.4) 
)]
A 14-in. x 193-lb., wide flange I-beam is used as a column with one end free (A = 56.73
sq. in., depth = 15.5 in., Imax = 2402.4 in.4, Imin = 930.1 in.4, kmin = 4.05 in., length L = 40
ft.). If a load F2 is supported as shown on a bracket at an eccentricity e = 4 in. (with F1 =
0), what may be its value for a design factor of 4? Flange width = 15.71 in.
Solution:
Using secant formula:
L NF 
NF  ec
1 + 2 sec e

sy =
A  k
2 EI 
F = F2
A = 56.73 in 2
e = 4 in
flange width 15.71
c=
=
= 7.855 in
2
2
k = kmin = 4.05 in
E = 30,000 ksi
Page 24 of 133
SECTION 6 – COMBINED STRESSES
I = Imin = 930.1 in 4
Le = 2L = 2(40 )(12) = 960 in
N=4
s y = 48 ksi
s y = 48 =
4F2  (4)(7.855)
960
1 +
sec
2

56.73 
2
(4.05)
(
48 = 0.0705F2 1 + 0.554 sec 0.18174 F2

4F2

(30,000)(930.1) 
)
F2 = 68.88 kips = 68,880 lbs
379. The same as 378, except that F1 = 0.5 F2.
Solution:
s
L
F  ec
σ = y = 2 1 + 2 sec e
N A k
2
NF2
EI
 F1
+
 A

F1 = 0.5F2
F
0.5F2
48
= 2 1 + 0.554 sec 0.18174 F2 +
4 56.73
56.73
12 = 0.01763F2 1 + 0.554 sec 0.18174 F2 + 0.008814F2
(
)
(
)
12 = 0.026444F2 + 0.009767F2 sec 0.18174 F2
By Trial and error:
F2 = 68.56 kips
F1 = 0.5F2 = 34.28 kips
380.
The cast-steel link (SAE 080) shown (solid lines) is subjected to a steady axial tensile load
and was originally made with a rectangular cross section, h = 2 in., b = ½ in., but was
found to be too weak. Someone decided to strengthen it by using a T-section (dotted
addition), with h and b as given above. (a) Will this change increase the strength?
Explain. (b) What tensile load could each link carry with N = 3 based on yield?
Page 25 of 133
SECTION 6 – COMBINED STRESSES
Solution:
For SAE 080, s y = 40 ksi
(a) This change will not increase the strength because of increased bending action that
tends to add additional stress.
s y 40
(b) σ = =
= 13.3 ksi
N
3
Rectangular cross section:
F F
σ= =
A bh
F
13.3 =
(0.5)(2 )
F = 13.3 kips
T-section:
h = 2 in , b = 0.5 in
A = b(h − b ) + bh = 0.5(2 − 0.5) + 0.5(2) = 1.75 in 2
b
1

c1 A = b(h − b ) (h − b ) + b + bh 
 2
2

 0. 5 
1

c1 (1.75) = 0.5(2 − 0.5) (2 − 0.5) + 0.5 + 0.5(2)

 2 
2

c1 = 0.6786 in
c 2 = h − c1 = 2 − 0.6786 = 1.3214 in
1
(h − b) = 1.3214 − 1 (2 − 0.5) = 0.5714 in
2
2
b
0.5
f = c1 − = 0.6786 −
= 0.4286 in
2
2
d = c2 −
b(h − b )3
hb 3
+ b(h − b )d 2 +
+ hbf 2
12
12
0.5(2 − 0.5)3
2(0.5)3
I=
+ 0.5(2 − 0.5)(0.5714)2 +
+ 2(0.5)(0.4286)2
12
12
I=
I = 0.59 in 4
Page 26 of 133
SECTION 6 – COMBINED STRESSES
F Fec
+
A
I
c = c1 = 0.6786 in
e = c = 0.6786 in
σ=
(0.6786)(0.6786) 
 1
13.3 = F 
+

0.59
1.75
F = 6.941 kips
COPLANAR SHEAR STRESSES
381.
The figure shows a plate riveted to a vertical surface by 5 rivets. The material of the
plate and rivets is SAE 1020, as rolled. The load F = 5000 lb., b = 3 in., θ = 0, and c = 5 in.;
let a = 3D. Determine the diameter D of the rivets and the thickness of plate for a design
factor of 3 based in yield strengths.
Solution:
θ =0
For SAE 1020, as rolled.
s y = 48 ksi
s sy = 0.6s y
N=3
Page 27 of 133
SECTION 6 – COMBINED STRESSES
1
2

F  2
R = F12 +   
 5  

F2 F1
=
a 2a
F1 = 2F2
[∑ M
C
=0
]
2F1 (2a ) + 2F2 (a) = F (c ) + F sinθ (2a − b)
θ =0
a = 3D
b = 3 in
c = 5 in
F = 5 kips
F1 = 2F2
4(2F2 )(3D ) + 2F2 (3D ) = (5)(5) + 5 sin 0 o (6D − 3)
30F2 D = 25
5
F2 =
6D
 5  5
F1 = 2F2 = 2  =
 6D  3D
A=
τ=
τ=
π
4
D2
R
A
s sy
N
=
0.6s y
N
Page 28 of 133
SECTION 6 – COMBINED STRESSES
0.6 sy
N
=
R
A
1
 2  F 2  2
F1 +   
 5  
0.6 sy 
=
N
A
1
0.6(48)
=
3
 5  2  5  2  2
  +   
 3D   5  
π
4
D2
1
 2.778  2
 2 + 1
D

9. 6 = 
0.7854D 2
By trial and error method.
5
D = 0.625 in = in
8
For thickness of plate, t .
A = Dt
R
σ=
A
1
1
2
2
 5  2  5  2  2  5
  5   2
R =   +    =  
+
    = 2.85 kips
 3D   5  
 3(0.625)   5  
s
R
σ= y =
N Dt
48
2.85
=
3 0.625t
1
t = 0.285 in = in
4
382.
The same as 383, except that θ = 30o.
Page 29 of 133
SECTION 6 – COMBINED STRESSES
Solution:
1
2

2
F 
F 
R = F12 +   − 2F1   cos120o 
5
5


1
 2  F 2
2
F 
R = F1 +   + 2F1   cos 60o 
5
5


MC = 0
[∑
]
2F1 (2a ) + 2F2 (a) = F (c ) + F sinθ (2a − b)
(
)
4(2F2 )(3D ) + 2F2 (3D ) = 5 cos 30 o (5) + 5 sin 30 o (6D − 3)
30F2 D = 21.65 + 7.5(2D − 1) = 15D + 14.15
15D + 14.15
F2 =
30D
0.472
F2 = 0.5 +
D
0.472 
0.944

F1 = 2F2 = 2 0.5 +
 =1+
D 
D

R
τ=
A
s
0.6s y
τ = sy =
N
N
0.6 sy
N
=
R
A
1
 2  F 2
2
F 
o
F1 +   + 2F1   cos 60 
0.6 sy 
5
5

=
N
A
Page 30 of 133
SECTION 6 – COMBINED STRESSES
1
0.6(48)
=
3
 0.944  2  5  2  0.944  5 
2
o
 +   + 21 +
  cos 60 
1 +
D  5
D  5 



π 2
D
4
1
 0.944  2
2
 0.944 
o
1
+
+
1
+
2
1
+
cos
60






D 
D 



9. 6 =
2
0.7854 D
By trial and error method. D = 0.641 in
5
Say D = 0.625 in = in (same as 381).
8
For t .
1
 0.944  2
2
 0.944 
o
R = 1 +
 + 1 + 21 +
 cos 60 
D 
D 



1
 0.944  2
2
 0.944 
o
R = 1 +
 + 1 + 21 +
 cos 60  = 3.1325 in
 0.625 
 0.625 

R
σ=
A
s
R
σ= y =
N Dt
48 3.1325
=
3 0.625t
5
t = 0.31325 in = in
16
383.
Design a riveted connection, similar to that shown, to support a steady vertical load of F
= 1500 lb. when L = 18 in. and θ = 0o. Let the maximum spacing of the rivets, horizontally
and vertically, be 6D, where D is the diameter of the rivet; SAE 1020, as rolled, is used
for all parts; N = 2.5 based on yield. The assembly will be such that there is virtually no
twisting of the channel. The dimensions to determine at this time are: rivet diameter
and minimum thickness of the plate.
Page 31 of 133
SECTION 6 – COMBINED STRESSES
Solution:
θ = 0 o , a = b = 6D
1
(6D ) = 3D
2
F
R = F1 +
4
R
τ=
A
For SAE 1020, as rolled, s y = 48 ksi .
c=
N = 2.5
s sy = 0.6s y
τ=
s sy
N
=
0.6s y
N
=
0.6(48)
= 11.52 ksi
2.5
4F1c = F (L + c )
4F1 (3D ) = 1.5(18 + 3D )
0.375(6 + D )
F1 =
D
0.375(6 + D ) 1.5 2.25
R=
+
=
+ 0.75
D
4
D
1
A = πD 2
4
Page 32 of 133
SECTION 6 – COMBINED STRESSES
τ=
R
A
2.25
+ 0.75
11.52 = D
1 2
πD
4
11
D = 0.6875 in = in
16
sy
R
N A′
2.25
+ 0.75
48
= D
2.5
Dt
2.25
+ 0.75
0
.
6875
19.2 =
0.6875t
5
t = in
16
384.
The same as 383, except that θ = 45o.
σ=
=
Solution:
τ = 11.52 ksi , σ = 19.2 ksi
Page 33 of 133
SECTION 6 – COMBINED STRESSES
2
F 
F 
R
+   + 2F1   cos 45 o
4
4
4F1 (3D ) = F cos θ (L + 3D ) = 1.5 cos 45 o (18 + 3D )
0.2652(6 + D )
F1 =
d
2
= F12
2
2
 0.2652(6 + D )   1.5 
 0.2652(6 + D )  1.5 
o
R =
+
 + 2

 cos 45


D
D

  4 

 4 
2
2
 0.2652(6 + D ) 
6+D
R2 = 
+ 0.140625 + 0.140625


D


 D 
2
0.84375
6+D
R = 0.0703
+ 0.28125
 +
D
 D 
R
τ=
A
2
1
2

2
0.84375
6+D
+ 0.28125
0.0703
 +
D
 D 


11.52 = 
1 2
πD
4
D = 0.594 in
19
say D = in = 0.59375 in
32
1
2

2
0.84375
 6 + 0.59375 
R = 0.0703
+ 0.28125 = 3.221 kips
 +
0.59375
 0.59375 


R
σ=
Dt
R
19.2 =
Dt
3.221
19.2 =
0.59375t
t = 0.2815 in
say t =
385.
1
in .
4
The plate shown (381) is made of SAE 1020 steel, as rolled, and held in place by five ¾
in. rivets that are made of SAE 1022 steel, as rolled. The thickness of the plate is ½ in., a
= 2 ½ in., c = 6 in., b = 4 in., and θ = 0. Find the value of F for a design factor of 5 based
on the ultimate strength.
Page 34 of 133
SECTION 6 – COMBINED STRESSES
Solution:
Plate, SAE 1020, as rolled (Table AT 7)
su = 65 ksi
su 65
=
= 13 ksi
N
5
R = σπDt
 3  1 
R = (13)(π )   = 15.3 kips
 4  2 
σ=
Rivets, SAE 1022, as rolled (Table AT 7)
s su = 54 ksi
s
54
τ = su = = 10.8 ksi
N
5
  3 2 
π   
 πD 2 
4


R =τ
= 10.8    = 4.77 ksi


4 
 4 




use R = 4.77 ksi
From 381.
1
2

F  2
R = F12 +   
 5  

F1 = 2F2
2F1 (2a ) + 2F2 (a ) = Fc
2(2F2 )(2 )(2.5) + 2F2 (2.5) = F (6)
25F2 = 6F
F2 = 0.24F
F1 = 2(0.24F ) = 0.48F
2
R = (0.48F )
2
2
F 
+   = (4.77 )2
5
F = 9.173 kips = 9,173 lbs
Page 35 of 133
SECTION 6 – COMBINED STRESSES
386.
The same as 385, except that θ = 90o.
Solution:
R = 4.77 kips
F
R = F1 +
5
F1 = 2F2
2F1 (2a) + 2F2 (a ) = F (2a − b)
2(2F2 )(2 )(2.5) + 2F2 (2.5) = F [2(2.5) − 4]
25F2 = F
F2 = 0.04F
F1 = 0.08F
F
R = 4.77 = 0.08F +
5
F = 17 ,000 lbs
Page 36 of 133
SECTION 6 – COMBINED STRESSES
387.
The plate shown is made of AISI 1020 steel, as rolled, and is fastened to an I-beam (AISI
1020, as rolled) by three rivets that are made of a steel equivalent to AISI C1015, cold
drawn. The thickness of the plate and of the flanges of the I-beam is ½ in., the diameter
of the rivets is ¾ in., a = 8.5 in., b = 11.5 in. and c = 4.5 in., d = 4 in. For F2 = 0, calculate
the value of F1 for N = 2.5 based on yield strength.
Solution:
3
D = in
4
1
t = in
2
a = 8.5 in
b = 11.5 in
c = 4.5 in
d = 4 in
Plate, AISI 1020 Steel, as rolled, s y = 48 ksi
Rivet, AISI C1015, cols drawn, s sy = 0.6(63) = 37.8 ksi
sy
48
= 19.2 ksi
N 2.5
s
37.8
τ = sy =
= 15.12 ksi
N
2.5
σ=
=
 3  1 
R = σ (πDt ) = 19.2(π )   = 22.6 kips
 4  2 
 πD 2
R = τ 
 4
2

π 3
 = 15.12   = 6.68 kips

 4  4 

 πD 2
Use R = τ 
 4
2

π 3
 = 15.12   = 6.68 kips

 4  4 

F2 = 0 ,
Page 37 of 133
SECTION 6 – COMBINED STRESSES
d
2
2
 4
 2
ρ = 2.5 in
2
c
 3
2
ρ2 =  + 
 4. 5 

 3 
2
ρ 2 =   +
F′
F
=
 2c  ρ
 
 3
F  2c 
F′ =  
ρ 3 
c

 2c 
F1  a +  = 2Fρ + F ′ 
3


 3
 2(4.5) 
F
4. 5 

 3 
F1  8.5 +
 = 2F (2.5) +
3 
2. 5

F1 = 0.86F
F
F= 1
0.86
c3
cos θ =
2
ρ
4. 5 3
cos θ =
= 0.60
2. 5
2
F 
F 
R 2 = F 2 +  1  + 2F  1  cos θ
3
3
2
2
 F  F 
 F  F 
R =  1  +  1  + 2 1  1 (0.60)
 0.86   3 
 0.86  3 
2
Page 38 of 133
SECTION 6 – COMBINED STRESSES
R = 1.389F1
R = 6.68 kips = 1.389F1
F1 = 4.8 kips .
388.
The same as 387, except that F1 = 0, and the value of F2 is calculated.
Solution:
R = 6.68 kips
R=F+
F2
3
Fρ
(2c 3)
c   2c 

F2  b −  = F   + 2F ′ρ
3  3 

F′ =
4.5   2(4.5)  2F (2.5)2

F2 11.5 −
+
=F
3   3   2(4.5) 

 3 
F2 = 0.7167F
Page 39 of 133
SECTION 6 – COMBINED STRESSES
0.7167F
= 1.24F
3
6.68 = 1.24F
F = 5.387 kips
R =F +
NORMAL STRESSES WITH SHEAR
DESIGN PROBLEMS
The bracket shown is held in place by three bolts as shown. Let a = 5 ¼ in., θ = 30o, F =
1500 lb.; bolt material is equivalent to C1022, as rolled. (a) Compute the size of the bolts
by equation (5.1), Text. (b) Assuming that the connecting parts are virtually rigid and
that the initial stress in the bolts is about 0.7sy, compute the factor of safety by (i) the
maximum shear stress theory, (ii) the octahedral shear theory. (c) Compute the
maximum normal stress.
389.
Solution:
3
(a) Eq. 5-1, Fe =
[∑ M
corner
=0
s y As 2
6
3 

, D < in
4 

]
2FA (9 ) + FB (3) = F sinθ (3) + F cos θ (a )
(
)
(
)
18FA + 3FB = (1500) sin 30 o (3) + (1500) cos 30 o (5.25)
18FA + 3FB = 9070 lbs
FA FB
=
9
3
FA = 3FB
Page 40 of 133
SECTION 6 – COMBINED STRESSES
18FA + FA = 9070 lbs
FA = 477.4 lbs
Fe = FA
For C1022, as rolled, s y = 52,000 psi
3
Fe =
s y As 2
6
3
52,000 As 2
Fe = 477.4 =
6
2
As = 0.1448 in
Select
Say D =
1
in , UNC, As = 0.1419 in 2
2
Fe
+ si
A
477.4
st =
+ 0.7(52,000) = 39,764 psi
0.1419
(b) st =
ss =
(
)
F cos θ 3 (1500) cos 30 o 3
=
= 3052 psi
As
0.1419
(i)
Maximum shear theory
1

1  s
=
N  s y

2
  ss 
 +

  s 2
  y 
2
2
2 

 =  39,764  +  3052

 52,000   52,000

1


2 
2
2


N = 1.293
(ii)
Octahedral shear theory
1

1  s
=
N  s y

2
  ss 

 +
 s

3
  y

2
1
2
2
2 

 2
 =  39,764  +  3052  

 52,000   52,000 3  



N = 1.296
(c) Maximum normal stress = 39,764 psi.
390.
For the mounted bracket shown, determine the rivet diameter (all same size) for N = 3,
the design being for the external loading (initial stress ignored); F = 2.3 kips, θ = 0, c = 17
in., a = 1 ½ in., b = 14 ½ in.; rivet material is AISI 1015, as rolled. Compute for (a) the
maximum shear theory, (b) the maximum normal stress theory, (c) the octahedral shear
theory.
Page 41 of 133
SECTION 6 – COMBINED STRESSES
Solution:
F2
F
= 1
a a+b
F2
F1
=
1.5 1.5 + 14.5
F2 = 0.09375F1
2F1 (a + b) + F2 (a ) = Fc
2F1 (1.5 + 14.5) + 0.09375F1 (1.5) = (2.3)(17 )
F1 = 1.2165 kips
F1 1.2165
=
ksi
A
A
F
2.3 0.7667
ss =
=
=
ksi
3 A 3A
A
For AISI 4015, as rolled. s y = 45.5 ksi
s=
(a) Maximum shear theory
1

1  s
=
N  sy

  ss   2
 +  
 s  
  ys  
s ys = 0.5sy = 0.5(45.5) = 22.75 ksi
2
Page 42 of 133
2
SECTION 6 – COMBINED STRESSES
1
2
2
1  1.2165   0.7667   2
= 
 +
 
3  45.5A   22.75A  
A = 0.1291 in 2
πD 2
= 0.1291 in 2
4
D = 0.4054 in
3
say D = in
8
A=
1
1
2
 2 1.2165  1.2165  2  0.7667  2 1.5869
s  s 
(b) σ = +   + s s2  =
+ 
 +
 =
2  2 
2A
A

 2 A   A 
1 σ
=
N sy
σ=
sy
N
1.5869 45.5
=
A
3
A = 0.1046 in 2
A=
πD 2
= 0.1046 in 2
4
D = 0.365 in
3
say D = in
8
(c) s ys =
sy
3
=
45.5
3
= 26.27 ksi
1
2
1  1.2165   0.7667  2
= 
 +

3  45.5A   26.27 A 
A = 0.11874 in 2
A=
πD 2
= 0.11874 in 2
4
D = 0.3888 in
3
say D = in
8
Page 43 of 133
SECTION 6 – COMBINED STRESSES
392.
The same as 390, except that the two top rivets are 2 in. long and the bottom rivet is 1 ¼
in. long.
Solution:
δ2
a
δ2
1.5
=
=
δ1
a+b
δ1
1.5 + 14.5
δ 2 = 0.09375δ1
F2 (1.25) = 0.09375(2)F1
F2 = 0.15F1
2F1 (a + b) + F2 a = Fc
2F1 (16) + 0.15F1 (1.5) = (2.3)(17 )
F1 = 1.2133 kips
F1 1.2133
=
ksi
A
A
F
2.3 0.7667
ss =
=
=
ksi
3 A 3A
A
For AISI 4015, as rolled. s y = 45.5 ksi
s=
Page 44 of 133
SECTION 6 – COMBINED STRESSES
(b) Maximum shear theory
1

1  s
=
N  sy

  ss   2
 +  
 s  
  ys  
s ys = 0.5sy = 0.5(45.5) = 22.75 ksi
2
2
1
2
2
1  1.2133   0.7667   2
= 
+
 
 
3  45.5 A   22.75A  
A = 0.1289 in 2
πD 2
= 0.1289 in 2
4
D = 0.4051 in
3
say D = in
8
A=
1
1
2
 2 1.2133  1.2133  2  0.7667   2 1.5843
s  s 
+ 
(b) σ = +   + s s2  =
 +
 =
2  2 
2A
A

 2 A   A  
1 σ
=
N sy
σ=
sy
N
1.5843 45.5
=
A
3
A = 0.1045 in 2
πD 2
= 0.1045 in 2
4
D = 0.3648 in
3
say D = in
8
A=
(c) s ys =
sy
3
=
45.5
3
= 26.27 ksi
1
2
1  1.2133   0.7667   2
= 
 +

3  45.5 A   26.27 A  
A = 0.1186 in 2
Page 45 of 133
SECTION 6 – COMBINED STRESSES
πD 2
= 0.1186 in 2
4
D = 0.3886 in
3
say D = in
8
A=
393.
The same as 390, except that the load is applied vertically at B instead of at A; let AB = 8
in. The two top rivets are 12 in. apart.
Solution:
b
ρ = 6 + 
 3
2
2
2
 14.5 

 3 
ρ = 7.705 in
2
ρ 2 = 62 + 
F1
ρ
=
F2
2b 3
 2b  F   2(14.5)  F1 
F2 =   1  = 

 = 1.2546F1
 3  ρ   3  7.705 
Page 46 of 133
SECTION 6 – COMBINED STRESSES
 2b 
F (8) = 2F1 ρ + F2  
 3 
2(14.5) 
 3 
(2.3)(8) = 2F1 (7.705) + 1.2546F1 
F1 = 0.6682 kips
cos θ =
6
ρ
=
6
= 0.7787
7.705
2
F 
F 
R 2 = F12 +   + 2F1   cos θ
3
3
2
 2.3 
 2.3 
R = (0.6682) + 
 + 2(0.6682)
(0.7787 )
 3 
 3 
R = 1.3536 kips
R 1.3536
ss = =
ksi
A
A
2
2
1

1
s
(a) = 

N  sy

2
  ss 
 + 
 s 
  ys 
2
2



From Problem 390.
1.2165
s=
ksi
A
s y = 45.5 ksi
s ys = 22.75 ksi
N=3
1
2
2
1  1.2165   1.3536   2
= 
 +
 
3  45.5A   22.75A  
A = 0.1957 in 2
πD 2
= 0.1957 in 2
4
D = 0.5 in
1
say D = in
2
A=
1
1
2
 2 1.2165  1.2165  2  1.3536   2 2.10
s  s 
(b) σ = +   + s s2  =
+ 
 +
 =
2  2 
2A
A

 2 A   A  
1 σ
=
N sy
Page 47 of 133
SECTION 6 – COMBINED STRESSES
sy
σ=
N
2.10 45.5
=
A
3
A = 0.1385 in 2
A=
πD 2
= 0.1385 in 2
4
D = 0.42 in
7
say D = in
16
(c) s ys = 26.27 ksi
1
2
1  1.2165   1.3536  2
= 
 +

3  45.5A   26.27 A 
A = 0.1742 in 2
πD 2
= 0.1742 in 2
4
D = 0.471 in
1
say D = in
2
A=
394.
The bracket shown is made of SAE 1020, as rolled, and the rivets are SAE 1015, cold
drawn. The force F = 20 kips, L = 7 in., and θ = 60o. Let the design factor (on yield) be 2.
(a) Determine the thickness t of the arm. (b) Compute the rivet diameter by both
maximum shear and octahedral shear theories and specify a standard size. (c) Decide
upon a proper spacing of rivets and sketch the bracket approximately to scale. Is some
adjustment of dimensions desirable? Give suggestions, if any. (No additional calculations
unless your instructor asks for a complete design.)
Solution:
Bracket: SAE 1020, as rolled, s y = 48 ksi
Rivets: SAE 1015, cold drawn, s y = 63 ksi
Page 48 of 133
SECTION 6 – COMBINED STRESSES
N=2
(a) Bracket.
(F cosθ )(L ) 4 
F sinθ
2
s=
+
A
I
A = 4t
t (4)3
I=
= 5.333t
12
(
)
48 20 sin 60 o
20 cos 60 o (7 )(2)
=
+
N
2
4t
5.333t
t = 1.275 in
1
say t = 1 in
4
s=
sy
=
(b)
F2 F1
=
2 6
F1 = 3F2
(
)
(
)
3(3F )(6) + 2F (2 ) = (20)(cos 60 )(7 ) + (20)(sin 60 )(3)
3F1 (6) + 2F2 (2) = F cos 60o (L ) + F sin 60 o (3)
o
2
2
F2 = 2.10 kips
F1 = 3F2 = 3(2.10) = 6.31 kips
F cos θ 20 cos 60 2
=
=
5A
5A
A
F1 F sin 60 6.31 20 sin 60 9.774
s= +
=
+
=
A
5A
A
5A
A
ss =
1

1  s
=
N  sy

2
  ss
 +
 s
  ys




2
2



s y = 63 ksi
Max. shear: s ys = 0.5sy = 0.5(63) = 31.5 ksi
1
2
2
1  9.774   2   2
= 
+
 
 
2  63A   31.5A  
A = 0.3353 in 2
Page 49 of 133
o
SECTION 6 – COMBINED STRESSES
A=
πD 2
= 0.3353 in 2
4
D = 0.653 in
3
say D = in
4
Octahedral shear, s ys =
sy
3
63
=
3
1
2
2
1  9.774   2   2
= 
 +
 
2  63A   36.37 A  
A = 0.3292 in 2
A=
πD 2
= 0.3292 in 2
4
D = 0.6474 in
3
say D = in
4
(c) Spacing
F cos θ
5(S − D )t
t = 1.5 in , s y = 48 ksi
σ=
D = 0.75 in
F = 20 kips
sy
F cos θ
N 5(S − D )t
48
20 cos 60
=
2 5(S − 0.75)(1.5)
S = 0.806 in
7
use S = in adjust to 2 in
8
σ=
=
Page 50 of 133
= 36.37 ksi
SECTION 6 – COMBINED STRESSES
Adjust spacing to 2 in from 7/8 in as shown.
CHECK PROBLEMS
396.
(a) If the rivets supporting the brackets of 390 are 5/8 in. in diameter, θ = 0, c = 14 in. a =
2 in., and b = 18 in., what are the maximum tensile and shear stresses in the rivets
induced by a load of F = 10 kips. (b) For rivets of naval brass, ¼ hard, compute the factor
of safety by maximum shear and octahedral shear theories (initial tension ignored).
Solution:
Page 51 of 133
SECTION 6 – COMBINED STRESSES
F2
F
= 1
a a+b
F2
F
= 1
2 2 + 18
F2 = 0.1F1
2F1 (a + b) + F2 (a ) = Fc
2F1 (2 + 18) + 0.1F1 (2) = (10)(14)
F1 = 3.8425 kips
F1
F
3.8425
= 1 =
= 12.525 ksi
π
π
A
(0.625)2
D2
4
4
F
F
10
ss =
=
=
= 10.865 ksi
3A
π  2
π 
2
3 D
3 (0.625)
4
4
s=
1
1
 s  2 2  2  12.525  2
2
2
(a) τ =   + s s  = 
 + (10.865)  = 12.541 ksi
 2 

 2 

1
1
2
 2 12.525  12.525  2
2
s  s
2
σ = +   + s s2  =
+ 
 + (10.865)  = 18.804 ksi
2  2 
2

 2 

(b) Naval Brass, ¼ hard, s y = 48 ksi .
1

1  s
=
N  sy

  ss   2
 +  
 s  
  ys  
Max. shear theory; s ys = 0.5sy = 0.5(48) = 24 ksi
2
Page 52 of 133
2
SECTION 6 – COMBINED STRESSES
1
2
2
1  12.525   10.865   2
= 
 +
 
N  48   24  
N = 1.914
Octahedral shear theory; s ys =
sy
3
=
48
3
= 27.71 ksi
1
2
2
1  12.525   10.865   2
= 
 +
 
N  48   27.71  
N = 2.123
397.
The same as 396, except that the two top rivets are ¾ in. in diameter and the bottom
one is ½ in. in diameter.
Solution:
δ2
=
δ1
a+b
a
δ2 =
δ1
a+b
F2
a F1
=
A2 a + b A1
a
Page 53 of 133
SECTION 6 – COMBINED STRESSES
2
 a  D2 
F2 = 
F1

 a + b  D1 
2
 2   0. 5 
F2 = 

 F1 = 0.0444F1
 2 + 18  0.75 
2F1 (a + b) + F2 (a ) = Fc
2F1 (2 + 18) + 0.0444F1 (2) = (10)(14)
F1 = 3.492 kips
F1
3.492
=
= 7.9043 ksi
π
A1
(0.75)2
4
F
10
ss =
=
= 9.26 ksi
2 A1 + A2
π 
π 
2
2
2 (0.75) +  (0.50)
4
4
s=
1
1
 s  2 2  2  7.9043  2
2
2
(a) τ =   + s s  = 
 + (9.26)  = 10.068 ksi
 2 
 2 


1
1
2
 2 7.9043  7.9043  2
2
s  s
2
σ = +   + s s2  =
+ 
 + (9.26)  = 14.02 ksi
2  2 
2

 2 

1

1
s
(b) = 
N  sy

  ss   2
 +  
 s  
  ys  
Max. shear theory; s ys = 0.5sy = 0.5(48) = 24 ksi
2
2
1
2
2
1  7.9043   9.26   2
= 
 +
 
N  48   24  
N = 2.384
Octahedral shear theory; s ys =
1
2
2
1  7.9043   9.26   2
= 
 +
 
N  48   27.71  
N = 2.684
Page 54 of 133
sy
3
=
48
3
= 27.71 ksi
SECTION 6 – COMBINED STRESSES
398.
What static load F may be supported by the ¾-in. rivets shown, made of cold-finished
C1015, with N = 3; θ = 0, a = 1 ½, b = 9, c = 14, f = 9, g = 12 in.? Count on no help from
friction and ignore the initial tension. Check by both maximum shear and octahedral
shear theories.
Solution:
For cold-finished, C1015, s y = 63 ksi .
F1
F
= 2
a+b a
F1
F
= 2
1.5 + 9 1.5
F1 = 7F2
θ = 0o
3F1 (a + b) + F2 a = Fc
3(7F2 )(1.5 + 9 ) + F2 (1.5) = F (14)
F2 = 0.0631F
F1 = 7F2 = 0.4417F
s=
F1 0.4417F
=
=F
A π (0.75)2
4
Page 55 of 133
SECTION 6 – COMBINED STRESSES
ss =
F
F
=
= 0.5659F
4A
π 
2
(
)
4  0.75
 4
1

1  s
=
N  sy

2
2
  ss   2
 +  
 s  
  ys  
Max. shear theory; s ys = 0.5sy = 0.5(63) = 31.5 ksi
1
2
2
1  F   0.5659F   2
=   + 
 
3  63   31.5  
F = 13.9 kips
Octahedral shear theory; s ys =
sy
3
=
63
3
= 36.37 ksi
1
2
2
1  F   0.5659F   2
=   + 
 
3  63   36.37  
F = 15 kips
399.
The 2-in., UNC cap screw shown has been subjected to a tightening torque of 20 in-kips.
The force F = 12 kips, θ = 60o, and Q = 0; L = 24 in., a = 20 in., b = 15 in.; screw material is
AISI C1137 as rolled. (a) What is the approximate initial tightening load? (b) What is the
increase in this load caused by the external force F if the bar is 8 in. wide and 2 in. thick
and the unthreaded shank of the screw is 2 in. long? (See §5.9, Text.) (c) What are the
maximum tensile and shear stresses in the bolt? (d) Compute the factor of safety from
maximum normal stress, maximum shear, and octahedral shear theories.
Solution:
(a) T = CDFi
C = 0.2
D = 2 in
T = 20 in − kips
20 = 0.2(2)Fi
Page 56 of 133
SECTION 6 – COMBINED STRESSES
Fi = 50 kips
(b) θ = 60 o
Fe (L − a ) = F cos 60 o (a )
(
)
(
)
Fe (24 − 20) = 12 cos 60 o (20)
Fe = 30 kips
 kb 

∆Fb = Fe 
 kb + kc 
AE
kb = s
Lb
AE
kc = c
Lc
For 2”-UNC
As = 2.50 in 2
Width across flat = 3 in.
2.5E
kb =
= 1.25E
2
π
kc = 4
(3)2 E
2
= 3.5343E
1.25E


∆Fb = (30)
 = 7.84 kips
 1.25E + 3.5343E 
F + ∆Fb 50 + 7.84
(c) s = i
=
= 23.14 ksi
As
2.50
ss =
F sinθ 12 sin 60o
=
= 4.16 ksi
As
2.5
1
1
 s  2
 2  23.14  2
2
2
τ =   + ss2  = 
+
(
4
.
16
)

 = 12.3 ksi
2
2








1
1
2
 2 23.14  23.14  2
2
s  s 
2
2
σ = +   + s s  =
+ 
 + (4.16)  = 23.87 ksi
2  2 
2

 2 

(d) For C1137, as rolled, s y = 55 ksi
σ=
sy
N
23.87 =
55
N
Page 57 of 133
SECTION 6 – COMBINED STRESSES
N = 2.304
Maximum Shear:
0.5sy
τ=
N
0.5(55)
12.3 =
N
N = 2.236
Octahedral shear, s ys =
sy
3
=
55
3
= 31.75 ksi
1

1  s
=
N  sy

2
2
  ss   2
 +  
 s  
  ys  
1
2
2
1  23.14   4.16   2
= 
 +
 
N  55   31.75  
N = 2.269
400.
The plate shown is attached by three ½-in., UNC cap screws that are made of ASTM
A325, heat-treated bolt material; L = 26 in., a = 6 in., b = 4 in., θ = 0. The shear on the
screws is across the threads and they have been tightened to an initial tension of 0.6sp
(sp = proof stress, §5.8, Text). Which screw is subjected to (a) the largest force, (b) the
largest stress? What safe static load can be supported by the screws for N = 1.5 based
on the Hencky-Mises criterion?
Solution:
For ASTM A325, Heat-Treated, ½-in. UNC
s y = 88 ksi
s p = 85 ksi
si = 0.6s p = 0.6(85) = 51 ksi
Page 58 of 133
SECTION 6 – COMBINED STRESSES
2
b a
 2  3
ρ = 2.83 in
2
 4
 2
2
 6
 3
2
ρ 2 =   +  =   + 
(a) Largest force, at A
(b) Largest stress, at A
F2
F1
ρ (2a 3)
ρF1
2.83F1
F2 =
=
= 0.7075F1
(2a 3) 2(6 ) 3
=
2a 
 2a  
2F2 ρ + F1   = F  L + 
3 
 3  
2(6) 
 2(6 ) 

2(0.7075)(F1 )(2.83) + F1 
= F 26 +

3 
 3 

F1 = 3.75F
F
F
R = F1 + = 3.75F + = 4.1F
3
3
For ½ in UNC, As = 0.1419 in 2
R
4.1F
ss = =
= 29F
A 0.1419
s = si = 51 ksi
By Hencky-Mises Criterion.
1

1− µ
(s x + s y )+ (1 + µ ) s x − sy
σ=
2
 2
Page 59 of 133



2
2
+ s s2 

SECTION 6 – COMBINED STRESSES
s x = s = 51 ksi
sy = 0
s s = 29F
µ = 0.3
1

1− µ
(s x + sy ) + (1 + µ ) s x − sy
σ= =
N
2
 2
sy



2
2
2
+ ss 

1
2

2
88 1 − 0.3
=
(51) + (1 + 0.3) 51  + (29F )2 
1.5
2
 2 

F = 0.63 kips = 630 lbs
401.
The same as 400, except that the cap screw A is ¾ in. in diameter.
Solution:
A1 = As1 = 0.3340 in 2 (3/4 in UNC)
A2 = As2 = 0.1419 in 2 (1/2 in UNC)
F2
F1
=
A2 ρ A1 (2a 3)
F2
F1
=
(0.1419 )(2.83) (0.3340)[2(6) 3]
F2 = 0.3006F1
2a 
 2a  
2F2 ρ + F1   = F  L + 
3 
 3  
2(6) 
 2(6) 

2(0.3006F1 )(2.83) + F1 
= F  26 +

3 
 3 

F1 = 5.262F
F2 = 0.3006F1 = 0.3006(5.262F ) = 1.582F
For ¾ in UNC.
F
F
R = F1 + = 5.262F + = 5.6F
3
3
Page 60 of 133
SECTION 6 – COMBINED STRESSES
ss =
5.6F
= 16.8F ksi
0.3340
For ½ in. UNC,
1
2

2
F 
F 
R = F22 +   − 2F2   cos θ 
3
3


a3 63
cos θ =
=
= 0.7067
ρ 2.83
1
2

2
F 
F 
R = (1.582F )2 +   − 2(1.582F ) (0.7067 ) = 1.367F
3
3


1.367F
ss =
= 9.6F ksi
0.1419
(a) Max. force, at ¾ in.
(b) Max. stress, at ¾ in.
s s = 16.8F ksi
s = si = 51 ksi
1
2

2
88 1 − 0.3
σ= =
=
(51) + (1 + 0.3) 51  + (16.8F )2 
N 1.5
2
 2 

sy
F = 1.09 kips = 1090 lbs
NORMAL STRESSES WITH TORSION
DESIGN PROBLEMS
402.
A section of a machined shaft is subjected to a maximum bending moment of 70,000 inlb., a torque of 50,000 in-lb., and an end thrust of 25,000 lb. The unsupported length is 3
ft. and the material is AISI C1030, normalized. Since the computations are to be as
though the stresses were steady, use N = 3.3. Compute the diameter from both the
maximum-shear and the octahedral-shear theories and specify a standard size.
Solution:
wL2
2
M1 = 70,000 in − lb
L = 3 ft = 36 in
M = M1 +
Page 61 of 133
SECTION 6 – COMBINED STRESSES
w = ρA
ρ = 0.284 lb in 3
πD 2
A=
4
0.284πD 2
w=
= 0.223D 2 lb in
4
0.223D 2 (36 )2
M = 70,000 +
= 70,000 + 144.5D 2
2
F = 25,000 lb
F Mc
s= +
A
I
D
c=
2
πD 2
A=
4
πD 4
I=
64
4F 32M
s= 2 +
πD
πD 3
4(25,000) 32 70,000 + 144.5D 2
s=
+
πD 2
πD 3
(
)
31,831 713,014 1472
+
+
D
D2
D3
713,014 31,831 1472
s=
+
+
D
D3
D2
s=
ss =
16T 16(50,000) 254,6548
=
=
πD 3
πD 3
D3
For AISI C1030, normalized, s y = 47 ksi = 47 ,000 psi
1

1  s
=
N  sy

2
  ss
 +
 s
  ys




2
2



Maximum shear
s ys = 0.5sy = 0.5(47 ) = 23.5 ksi = 23,500 psi
1
 713,014 31,831 1472  2
2
2
+
+



1  D 3
D  +  254,648  
D2
=
  23,500D 3  
3.3 
47 ,000







By trial and error,
Page 62 of 133
SECTION 6 – COMBINED STRESSES
D = 4.125 in
Octahedral Shear
sy
47
s ys =
=
= 27.14 ksi = 27 ,140 psi
3
3
1
 713,014 31,831 1472  2
2
2
+
+



1  D 3
D  +  254,648  
D2
=
  27 ,140D 3  
3.3 
47 ,000






By trial and error,
D = 4.125 in
1
use D = 4 in standard
4
403. The same as 402, except that the unsupported length is 15 ft. Do not overlook the
moment due to the weight of the shaft, which acts in the same sense as the given
bending moment.
Solution:
L = 15 ft = 180 in
M = 70,000 +
0.223D 2 (180)2
= 70,000 + 3612.6D 2
2
F = 25,000 lb
F Mc
s= +
A
I
4F 32M
s= 2 +
πD
πD 3
4(25,000) 32 70,000 + 3612.6D 2
s=
+
πD 2
πD 3
(
)
31,831 713,014 36,798
+
+
D
D2
D3
713,014 31,831 36,798
s=
+
+
D
D3
D2
s=
Maximum shear
1
 713,014 31,831 36,798  2
2
2
+
+



1  D 3
D  +  254,648  
D2
=
  23,500D 3  
3.3 
47 ,000






By trial and error,
D = 5.125 in
Page 63 of 133
SECTION 6 – COMBINED STRESSES
Octahedral Shear
1
 713,014 31,831 36,798  2
2
2
+
+



3
2
1  D
D  +  254,648  
D
=
  27 ,140D 3  
3.3 
47 ,000






By trial and error,
D = 5.0625 in
1
use D = 5 in standard
4
404.
A shaft is to be made in two sections, I and II, of diameters D1 and D2, somewhat as
shown, machined from AISI 1045, annealed. It is expected that a = 8 in., b = 24 in., L = 20
in., and the load Q = 2 kips, so seldom repeated that the design is for steady load. The
factor of safety is to be 2.2 on the basis of the octahedral-shear theory and closely the
same in each section. The ends A and B are restrained from twisting, but they are
designed to support the balancing reactions from Q without other moments. Decide
upon standard size for D1 and D2.
Solution:
T = QL = (2)(20 ) = 40 in − kips
T1 + T2 = T = 40 in − kips
θ1 = θ 2
T1L1 T2 L2
=
J1G J 2 G
T1a
Tb
= 24
4
πD1 πD2
32
32
8T1 24T2
= 4
D14
D2
Page 64 of 133
SECTION 6 – COMBINED STRESSES
4
D 
T1 = 3T2  1 
 D2 
T1 + T2 = T = 40 in − kips
 D
T2 3 1
  D 2
4


 + 1 = 40


40
T2 =
  D 4 
3 1  + 1
  D 2 

4
4
D 
D 
40(3) 1 
120 1 
 D2 
 D2 
T1 =
=
4
 D 
   D 4 
1
3  + 1 3 1  + 1
  D 2 
   D2 

A + B = Q = 2 kips
aA = bB
8A = 24B
A = 3B
3B + B = 2
B = 0.5 kips
A = 1.5 kips
4
  D 4 
D 
16120 1  
1920 1 
  D 2  
16T
 D2 
s s1 = 31 =
=
πD1
  D 4 
  D 4 
3
3
1




πD1 3  + 1 πD1 3 1  + 1
  D 2 

  D 2 

16T2
16(40)
640
s s2 =
=
=
3
4
πD 2
 D 

  D 4 
3
3
1




πD2 3  + 1 πD2 3 1  + 1
  D 2 

  D 2 

32M1 32 Aa 32(1.5)(8) 384
s1 =
=
=
= 3
πD13
πD13
πD13
πD1
32M 2 32Bb 32(0.5)(24) 384
s2 =
=
=
= 3
πD 23
πD23
πD 23
πD2
For AISI 1045, annealed, s y = 55 ksi ,
N = 2.2
Octahedral Shear
Page 65 of 133
SECTION 6 – COMBINED STRESSES
1

1  s
=
N  sy

s ys =
sy
3
2
  ss 
 + 
 s 
  ys 
=
55
3
2
2



= 31.75 ksi
1

1  s1
=
N  sy

2
  s s1
 +
 s
  ys




2
2



1
2


 2
4

D 

 
1920 1 
2

 
D2 
1  384  



=
+
 
2.2  55πD13  
  D 4   
3
1

 31.75πD1 3  + 1  
D

  


  2
 

1
2


 2
4

 19.25 D1   
2

D   
1  2.2224  
 2  

= 
+

 
2.2  D13     D  4   
3
1

 D1 3  + 1  
  
   D 2 

 

1

1  s2
=
N  sy

2
  s s2 
 +

 s 
  ys 
2
2



1




2




1  384
640
 + 
= 
3 
2.2  55πD 2  
 D
3
1

 31.75πD 2 3
  D 2







4


 + 1 
 


1
2


 2


 
2


 
1  2.2224  
6.42
 +
= 
 
2.2  D 23     D  4   
3
1

 D 2 3  + 1  
D

  
   2 

 

Page 66 of 133
2
2








SECTION 6 – COMBINED STRESSES
By trial and error,
D
Trial 1
D2
1
1.25
1.37
1.40
D1
D2
2.268
2.372
2.398
2.403
1.820
1.730
1.714
1.712
D1
D2
1.25
1.37
1.40
1.40
Actual
1
3
Use standard D1 = 2 in , D 2 = 1 in
2
4
1
But D1 > D 2 , use D1 = 2 in
2
 1
D 2 = D1 + 2r = 1.3D1 = 1.3 2  = 3.25 in
 2
1
say D 2 = 3 in
2
405.
The shaft shown overhangs a bearing on the right and has the following dimensions: a =
5 in., b = ½ in., and e = 10 in. The material is AISI C1040, annealed. This shaft is subjected
to a torque T = 10,000 in-lb., forces F1 = 10,000 lb., and F2 = 20,000 lb. Using a staticdesign approach, determine the diameter D for N = 2.5, with computations from the
maximum-shear and octahedral-shear theories.
Solution:
ρ = 0.284 lb in 3
π

w = 0.284 D 2  = 0.223D 2
4


2
we
0.223D 2 (10 )2
M=
=
= 11.15D 2
2
2
Bending due to F1 and load
[
]
32(F1a + M ) 32 (10,000)(5) + 11.15D 2 509,296 114
=
=
+
πD 3
πD 3
D3
D
Bending due to F2
s1 =
Page 67 of 133
SECTION 6 – COMBINED STRESSES
1
32(20,000 ) 
32F2 b
 2  = 101,859
s2 =
=
3
3
πD
πD
D3
Tension,
4(20,000 ) 25,465
s3 =
=
πD 2
D2
since s1 > s 2
s = s3 + s1 − s2
25,465 509,296 114 101,859
+
+
−
s=
D
D2
D3
D3
407,437 25,465 114
s=
+
+
D
D3
D2
16T 16(10,000) 50,930
=
=
πD 3
πD 3
D3
N = 2.5
ss =
1

1  s
=
N  sy

  ss   2
 +  
 s  
  ys  
For AISI C1040, annealed (Fig. AF 1) s y = 48 ksi
2
2
Maximum shear, s ys = 0.5sy = 0.5(48) = 24 ksi
1
 407,437 25,465 114  2
2
2
+
+



1  D 3
D  +  50,930  
D2
=

  24,000D 3  
2.5 
48,000







D = 2.95 in
Octahedral Shear
sy
48
s ys =
=
= 27.71 ksi
3
3
1
 407,437 25,465 114  2
2
2
+
+



1  D 3
D  +  50,930  
D2
=
  27,710D 3  
2.5 
48,000







D = 2.95 in
use D = 3.0 in
Page 68 of 133
SECTION 6 – COMBINED STRESSES
406.
The same as 405, except that F2 = 0.
Solution
F2 = 0
s 2 = 0 , s3 = 0
509,296 114
s = s1 =
+
D
D3
50,930
ss =
D3
N = 2.5
Maximum shear, s ys = 0.5sy = 0.5(48) = 24 ksi
1
 509,296 114  2
+


1  D 3
D
 +  50,930
=
  24,000D 3
2.5 
48,000




2
2
 

 


D = 3 in
Octahedral Shear
sy
48
s ys =
=
= 27.71 ksi
3
3
1
 509,296 114  2
2
2
+



1  D 3
D  +  50,930  
=
  27 ,710D 3  
2.5 
48,000







D = 2.997 in
use D = 3.0 in
CHECK PROBLEMS
407.
The shaft shown overhangs a bearing at the right and has the following dimensions: D =
2 in., a = 4 in., b = ¾ in., c = 2 in., d = 6 in., e = 8 in., r = ¼ in. This shaft is subjected to a
torque T = 8000 in-lb. and forces F1 = 8000 lb., and F2 = 16,000 lb. Determine the
maximum-shear and normal stresses, and the octahedral-shear stress: (a) at points A
and B (θ = 45o), (b) at points M and N, (c) at point G.
Page 69 of 133
SECTION 6 – COMBINED STRESSES
Solution:
Bending due to F2 :
32F2 b 32(16,000 )(0.75)
s2 =
=
= 15,279 psi
πD 3
π (2)3
Tension
4F
4(16,000)
s3 = 22 =
= 5093 psi
πD
π (2)2
Bending due to F1 and weight of beam

wx 2 

32 F1a +
2 

s1 =
πD 3
x from the free end.
ss =
16T 16(8000)
=
= 5093 psi
πD 3
π (2)3
w = 0.223D 2 = 0.223(2)2 = 0.892 lb in
(A) x = e − c = 8 − 2 = 6 in

0.892(6)2 
32 (8000)(4) +

2


s1 =
= 40,764 psi
3
π (2 )
at A
s = s3 + (s1 − s 2 ) sin 45o = 5093 + (40,764 − 15,279) sin 45o = 23,114 psi
Max. Shear
1
1
 s 2
 2  23,114  2
2
2
τ =   + s s2  = 
 + (5093)  = 12,629 psi
 2 

 2 

Max. Normal
1
1
2
 2 23,114  23,114  2
2
s  s 
2
2
τ = +   + s s  =
+ 
 + (5093)  = 24,186 psi
2  2 
2

 2 

Octahedral Shear Stress
Page 70 of 133
SECTION 6 – COMBINED STRESSES
τo =
2 2
s + 3s s2
3
(
1
2
)
=
[
2
(23,114)2 + 3(5093)2
3
1
2
]
= 11,662 psi
At B:
s = (s1 − s2 )sin 45o − s3 = (40,764 − 15,279) sin 45o − 5093 = 12,928 psi
Max. Shear
1
1
 s 2
 2  12,928  2
2
2
τ =   + s s2  = 
 + (5093)  = 8229 psi
 2 

 2 

Max. Normal
1
1
2
 2 12,928  12,928  2
2
s  s
2
σ = +   + s s2  =
+ 
 + (5093)  = 14,693 psi
2  2 
2

 2 

Octahedral Shear Stress
1
2 2
2
τo =
s + 3s s2 2 =
(12,928)2 + 3(5093)2
3
3
(
[
)
1
2
]
= 7 ,378 psi
(B)
1
= 7.75 in
4

0.892(7.75)2 
32(8000)(4 ) +

2


x =e −c =8−
s1 =
π (2)3
= 40,778 psi
at M
s = s3 + s1 − s2 = 5093 + 40,778 − 15,279 = 30,592 psi
Max. Shear
1
1
 s 2
 2  30,592  2
2
2
τ =   + s s2  = 
 + (5093)  = 16,122 psi
 2 
 2 


Max. Normal
1
1
2
 2 30,592  30,592  2
2
s  s 
2
2
σ = +   + s s  =
+ 
 + (5093)  = 31,418 psi
2  2 
2

 2 

Octahedral Shear Stress
1
2 2
2
τo =
s + 3s s2 2 =
(30,592)2 + 3(5093)2
3
3
(
)
[
1
2
]
= 15,009 psi
At N:
s = s1 − s 2 − s3 = 40,778 − 15,279 − 5093 = 20,406 psi
Max. Shear
Page 71 of 133
SECTION 6 – COMBINED STRESSES
1
1
 s  2
 2  20,406  2
2
2
2
τ =   + s s  = 
 + (5093)  = 11,403 psi
 2 

 2 

Max. Normal
1
1
2
 2 20,406  20,406  2
2
s  s
2
τ = +   + s s2  =
+ 
 + (5093)  = 21,606 psi
2  2 
2

 2 

Octahedral Shear Stress
1
1
2 2
2
τo =
s + 3s s2 2 =
(20,406)2 + 3(5093)2 2 = 10,480 psi
3
3
(
[
)
]
(C) At G.
x = e − c = 8 − 2 = 6 in

0.892(2 )2 
32 (8000)(4) +

2


s1 =
= 40,746 psi
3
π (2)
s = s3 + (s1 − s 2 ) cos 30 o = 5093 + (40,746 − 15,279) cos 30 o = 17 ,826 psi
Max. Shear
1
1
 s 2
 2  17 ,826  2
2
2
τ =   + s s2  = 
 + (5093)  = 10,265 psi
 2 

 2 

Max. Normal
1
1
2
 2 17 ,826  17 ,826  2
2
s  s
2
σ = +   + s s2  =
+ 
 + (5093)  = 19,178 psi
2  2 
2

 2 

Octahedral Shear Stress
1
2 2
2
τo =
s + 3s s2 2 =
(17,826)2 + 3(5093)2
3
3
(
411.
)
[
1
2
]
= 9376 psi
A 4-in. shaft carries an axial thrust of 20 kips. The maximum bending moment is 2/3 of
the twisting moment; material is AISI 8630, WQT 1100 F, and N =3. Use the steady stress
approach and compute the horsepower that may be transmitted at 2000 rpm?
Solution:
For AISI 8630, WQT 1100oF, s y = 125 ksi
32M 4F
+
πD 3 πD 2
16T
ss = 3
πD
2
M= T
3
F = 20 kips
s=
Page 72 of 133
SECTION 6 – COMBINED STRESSES
D = 4 in
2 
32 T 
 3  4(20 )
s=
+
= 0.1061T + 1.59155
π (4 )3
π (4)2
16T
ss =
= 0.0796T
π (4)3
By maximum shear.
s ys = 0.5s y = 0.5(125) = 62.5 ksi
1

1  s
=
N  sy

2
2
  ss   2
 +  
 s  
  ys  
1
2
2
1  0.1061T + 1.59155   0.0796T   2
= 
 +
 
3 
125
  62.5  
T = 213 in − kips = 213,000 in − lb
Tn
(213,000 )(2000)
hp =
=
= 6762 hp
63,000
63,000
412.
The same as 411, except that the shaft is hollow with an inside diameter of 2 ½ in.
Solution:
32MDo
4F
s=
+
4
4
2
π Do − Di
π D o − D i2
(
) (
)
2 
32 T (4)
4(20)
3 
s=
+
= 0.1252T + 2.612
4
4
π (4) − (2.5)
π (4)2 − (2.5)2
16TDo
16T (4)
ss =
=
= 0.0939T
4
4
π Do − Di
π (4)4 − (2.5)4
[
] [
(
)
[
]
]
1

1  s
=
N  sy

2
2
  ss   2
 +  
 s  
  ys  
1
2
2
1  0.1252T + 2.612   0.0939T   2
= 
 +
 
3 
125
  62.5  
T = 177 in − kips = 177,000 in − lb
Tn
(177,000)(2000)
hp =
=
= 5619 hp
63,000
63,000
413.
A hollow, alloyed-steel shaft, AISI 4130, OQT 1100 F, has an OD of 3 ¼ in. and an ID of 2
½ in. It is transmitting 1500 hp at 1200 rpm, and at the same time is withstanding a
maximum bending moment of 40,000 in-lb. and an axial compressive force F = 10 kips.
The length of the shaft between bearings is 10 ft. Using a steady stress approach,
Page 73 of 133
SECTION 6 – COMBINED STRESSES
determine (a) the maximum shearing stress in the shaft, (b) the maximum normal stress,
(c) the factor of safety in each case. (d) Also compute N from the octahedral –shear
theory.
Solution:
For alloy-steel shaft, AISI 4130, WQT 1100oF, s y = 114 ksi (Table AT 7)
63,000hp 63,000(1500 )
=
= 78,750 in − lb
n
1200
1
Do = 3 in
4
1
Di = 2 in
2
π 
π 
w = 0.284  Do2 − Di2 = 0.284  (3.25)2 − (2.5)2 = 0.962 lb in
4
 
4
2
wL
M = M1 +
(Table AT 8)
8
L = 10 ft = 120 in
T=
(
[
)
]
0.962(120)2
= 41,732 in − lb
8
32MDo
4F
s=
+
4
4
2
π Do − Di
π Do − Di2
32(41,732)(3.25)
4(10,000)
s=
+
= 22,000 psi
4
4
π (3.25) − (2.5)
π (3.25)2 − (2.5)2
16TDo
16(78,750)(3.25)
ss =
=
= 17 ,978 psi
4
4
π Do − Di
π (3.25)4 − (2.5)4
M = 40,000 +
(
) (
[
)
] [
) [
(
]
]
1
2
1
 s  2
 22,000  2
2
2
(A) τ =   + s s2  = 
 + (17 ,978)  = 21,076 psi
 2 

 2 

1
1
2
 2 22,000  22,000  2
2
s  s 
2
2
(B) σ = +   + s s  =
+ 
 + (17 ,978)  = 32,076 psi
2  2 
2

 2 

sys
(C) N =
τ
=
0. 5s y
τ
=
0.5(114)
= 2.704
21.076
sy
114
N= =
= 3.554
σ 32.076
1

1
s
(D) = 
N  sy

s ys =
sy
3
=
2
  ss
 +
 s
  ys
114
3




2
2



= 65.82 ksi
Page 74 of 133
SECTION 6 – COMBINED STRESSES
1
2
2
1  22.0   17.978   2
= 
 +
 
N  114   65.82  
N = 2.99
VARYING STRESSES COMBINED
DESIGN PROBLEMS
414.
The force F on the lever in the illustration (in the plane of the lever) varies from a
maximum of 424.2 lb. to a minimum of -141.4 lb.; L = 20 in., a = 15 in., D2 = 1.2D1, r =
0.125D1, θ = 45o; the material is cold-drawn SAE 1040, 10% worked, the design factor N
= 1.5. Compute the diameter D1 using the Soderberg-line approach with both the
maximum-shear and octahedral-shear theories; indefinite life.
Solution:
For SAE 040, 10% Worked
s n′ = 54 ksi
s y = 85 ksi
s n 54
1
=
=
sy 85 1.574
s ns s n
1
=
=
s ys s y 1.574
SF = 0.85
Strength Reduction Factors
With r d = r D1 = 0.125
D d = D 2 D1 = 1.2
Fig. AF 12
K t = 1.54
K ts = 1.27
Page 75 of 133
SECTION 6 – COMBINED STRESSES
Assume q = 1
K f = 1 + q (K t − 1) = 1 + (1)(K t − 1) = K t
K f = 1.54
K fs = 1.27
Forces:
Fmax = 424.2 lb
Fmin = −141.4 lb
1
(Fmax + Fmax ) = 1 (424.2 − 141.4) = 141.4 lb
2
2
1
1
Fa = (Fmax − Fmax ) = (424.2 + 141.4) = 282.8 lb
2
2
Fm =
 πD 2 
 = 0.223D 2
w = 0.284

 4 
wL2
wL2
M = M1 +
= FL +
2
2
0.233D 2 (20 )2
M m = (141.4)(20) +
= 2828 + 44.6D 2
2
0.233D 2 (20)2
Ma = (282.8)(20) +
= 5656 + 44.6D 2
2
T = (F cos θ )a
(
)
= (282.8)(cos 45 )(15) = 3000 in − lb
Tm = (141.4) cos 45o (15) = 1500 in − lb
Ta
o
32M
πD 3
32 2828 + 44.6D 2
28,806 454
sm =
=
+
3
D
πD
D3
32 5656 + 44.6D 2 57 ,612 454
sa =
=
+
D
πD 3
D3
s=
(
)
(
)
16T
πD 3
16(1500) 7640
sms =
= 3
πD 3
D
16(3000 ) 15,280
sas =
=
πD 3
D3
ss =
Page 76 of 133
SECTION 6 – COMBINED STRESSES
s es =
K fs sas
sns
sms +
s ys
SF
1  7640   1.27  15,280  27,684

+

=
1.574  D 3   0.85  D 3 
D3
K f sa
s
se = n s m +
sy
SF
s es =
se =
1  28,806 454   1.54  57,612 454  122,681 1111
+
+
+

+

=
1.574  D 3
D   0.85  D 3
D 
D
D3
1
2
2 2
1  se   ses  
=   +   
N  sn   s ns  


Maximum shear, sns = 0.5sn = 0.5(54,000) = 27,000 psi
1
 122,681 1111  2
2
2
+



1  D 3
D  +  27 ,684  
=
1.5 
54,000
  27,000D 3  







D = 1.5625 in = 1
9
in
16
Octahedral Shear. sns =
sn
3
=
54,000
3
= 31,177 psi
1
 122,681 1111  2
2
2
+



1  D 3
D  +  27,684  
=
1.5 
54,000
  31,177D 3  







D = 1.55 in
9
say D = 1 in
16
5
say D = 1 in
8
417.
A hollow steel shaft, SAE 1045, as rolled, has an inside diameter of one half of the
outside diameter and is transmitting 1600 hp at 600 rpm. The maximum bending
moment is 40,000in-lb. Determine the diameter for N = 3 by both the maximum-shear
and octahedral shear theories. Specify a standard size. Use the Soderberg line for
obtaining the equivalent stresses.
Page 77 of 133
SECTION 6 – COMBINED STRESSES
Solution:
For SAE 1045, as rolled, s y = 59 ksi , su = 96 ksi
sn = 0.5su = 48 ksi
sn sns 48
1
=
=
=
sy sys 59 1.229
Assume K f = K fs = 1
Do = 2Di
For bending:
sm = 0
sa = s
s
1
se = n sm + K f sa =
(0) + (1)s = s
sy
1.229
32MDo
32(40,000)(2Di ) 54,325
=
=
4
4
Di3
π Do − D i
π (2Di )4 − Di4
For torsion:
63,000hp 63,000(1600)
T=
=
= 168,000 in − lb
n
600
s
s es = ns sms + K fs sas
sys
se = s =
(
[
)
]
sas = 0
s ms = s
sns
1  16TDo 
1 16(168,000)(2Di )  92,825
=
sms =

=

4
4 
s ys
1.229  π Do − Di  1.229  π (2Di )4 − Di4 
D i3
Maximum shear, sns = 0.5sn = 0.5(48,000) = 24,000 psi
ses =
(
[
)
1
2
2 2
1  54,325   92,825  






=
+
1.5  48,000Di3   24,000D i3  


Di = 2.295 in
say Di = 2.25 in , Do = 2Di = 4.5 in
Octahedral Shear. sns =
sn
3
=
48,000
3
= 27,713 psi
1
2
2 2
1  54,325   92,825  






=
+
1.5  48,000Di3   27,713Di3  


Di = 2.20 in
say Di = 2.25 in , Do = 2Di = 4.5 in
Page 78 of 133
]
SECTION 6 – COMBINED STRESSES
1
1
Standard Size Di = 2 in , Do = 4 in
4
2
418.
A section of a shaft without a keyway is subjected to a bending moment that varies
sinusoidally from 30 to 15 then to 30 in-kips during two revolutions, and to a torque that
varies similarly and in phase from 25 to 15 to 25 in-kips; there is also a constant axial
force of 40 kips; the material is AISI 2340, OQT 1000 F; N = 1.5. Determine the diameter
by (a) the maximum-shear-stress theory; (b) the octahedral-shear-stress theory.
Solution:
For AISI 2340, OQT 1000oF. s y = 120 ksi , su = 137 ksi
s n = 0.5su = 0.5(137 ) = 68.5 ksi
sn sns 68.5
1
=
=
=
sy sys 120 1.752
Assume K f = K fs = 1
1
(30 + 15) = 22.5 in − kips
2
1
Ma = (30 − 15) = 7.5 in − kips
2
1
Tm = (25 + 15) = 20 in − kips
2
1
Ta = (25 − 15) = 5 in − kips
2
Mm =
32M m
4F 32(22.5) 4(40) 229 51
+
=
+
= 3 + 2
3
D
D
πD
πD 2
πD 3
πD 2
32Ma 32(7.5) 76
sa =
=
= 3
D
πD 3
πD 3
sm =
16Tm 16(20) 102
=
= 3
D
πD 3
πD 3
16Ta 16(5) 25.5
sas =
=
= 3
πD 3 πD 3
D
sms =
se =
Kf
sn
1  229 51   1  76  220 29
sm +
sa =
+
+

+
  =
sy
SF
1.752  D 3 D 2   0.85  D 3  D 3 D 2
ses =
K fs
sns
1  102   1  25.5  88
sms +
sas =

+

=
s ys
SF
1.752  D 3   0.85  D 3  D 3
(a) Maximum shear, sns = 0.5sn = 0.5(68.5) = 34.25 psi
Page 79 of 133
SECTION 6 – COMBINED STRESSES
1
 220 29
+
1  D 3 D 2
=
1.5  68.5








2
2
2
 88  
+
3 
 34.25D  


D = 1.93 in
say D = 2 in ,
Octahedral Shear. sns =
sn
3
=
68.5
3
= 39.55 psi
1
 220 29
+
1  D 3 D 2
=
1.5  68.5








2
2
2
 88  
+
3 
 39.55D  


D = 1.909 in
say D = 2 in
419.
The same as 418, except that the shaft has a profile keyway at the point of maximum
moment.
Solution:
K f = 1.6
K fs = 1.3
Kf
sn
1  229 51   1.6  76  274 29
sm +
sa =
+
+

+
  =
sy
SF
1.752  D 3 D 2   0.85  D 3  D 3 D 2
K fs
s
1  102   1.3  25.5  97
ses = ns sms +
sas =

+

=
s ys
SF
1.752  D 3   0.85  D 3  D 3
se =
(a) Maximum shear, sns = 0.5sn = 0.5(68.5) = 34.25 psi
1
 274 29
+
1  D 3 D 2
=
1.5  68.5


2
2

2

 +  97  
  34.25D 3  




D = 2.04 in
say D = 2 in
(b) Octahedral Shear. sns =
Page 80 of 133
sn
3
=
68.5
3
= 39.55 psi
SECTION 6 – COMBINED STRESSES
1
 274 29
+
1  D 3 D 2
=
1.5  68.5








2
2
2
 97  
+
3 
 39.55D  


D = 2.02 in
say D = 2 in
CHECK PROBLEMS
420.
A 2-in. shaft made from AISI 1144, elevated temperature drawn, transmits 200 hp at 600
rpm. In addition to the data on the figure, the reactions are B = 4.62 kips and E = 1.68
kips. Compute the factor of safety by the maximum-shear and octahedral-shear
theories.
Solution:
For AISI 1144, Elevated Temperature, drawn, s y = 83 ksi , su = 118 ksi
sn = 0.5su = 0.5(118) = 59 ksi
sn 59
1
=
=
sy 83 1.407
MB = (2.1)(10) = 21 in − kips , MC = (1.68)(10) = 16.8 in − kips
63,000hp 63,000(200)
T=
=
= 21,000 in − lbs = 21 in − kips
n
600
Table AT 13
K f = 2.0 , K fs = 1.6
se =
sn
sm + K f sa
sy
Mm = 0 , s m = 0
Ma = Mm
 32(16.8) 
 32MC 
s e = K f sa = K f 
= 2.0
= 42.8 ksi
3 
3 
 πD 
 π (2) 
s
s es = ns sms + K fs sas
sys
Tm = T ,
Ta = 0
Page 81 of 133
SECTION 6 – COMBINED STRESSES
ses =
sns
s  16T 
1 16(21) 
sms = ns  3  =

 = 9.5 ksi
s ys
sys  πD  1.407  π (2)3 
Maximum shear, sns = 0.5sn = 0.5(59 ) = 29.5 psi
1
1  se
= 
N  sn

2
  ses
 + 
  sns



2
2


1
2
2
1  42.8   9.5   2
= 
+
 
 
N  59   29.5  
N = 1.26
Octahedral Shear. s ns = 0.577 sn = 0.577(59 ) = 34.05 psi
1
2
2
1  42.8   9.5   2
= 
 +
 
N  59   34.05  
N = 1.26
421.
In the figure (399), the bar supports a static load Q = 3000 lb. acting down; L = 16 in., a =
12 in., b = 7 in. The force F = 2500 lb. is produced by a rotating unbalanced weight and is
therefore repeated and reversed in both the horizontal and the vertical directions. The
1-in. cap screw, with cut UNC threads, is made of AISI C1137, annealed, and it has been
subjected to a tightening torque of 4600 in-lb. The thickness of the bar is 2 in. (a)
Compute the factor of safety for the load reversing in the vertical direction, and (b) in
the horizontal direction (maximum-shear theory), with the conservative assumption
that friction offers no resistance.
Solution:
For AISI C1137, annealed, s y = 50 ksi , su = 85 ksi
sn = 0.5su = 0.5(85) = 42.5 ksi
K f = 2.8 (Table AT 12)
sn sns 42.5
1
=
=
=
sy sny
50 1.1765
T = 0.2DFi
4600 = 0.2(1)(Fi )
Page 82 of 133
SECTION 6 – COMBINED STRESSES
Fi = 23,000 lbs = 23 kips
For 1-in cap screws, UNC
As = 0.606 in 2
Nut: A = 1.5 in
 kb 

∆Fb = Fe 
 kb + kc 
AE
AE
kb = s , kc = c
Le
Le
kb
As
0.606
=
=
= 0.2554
k b + k c As + Ac 0.606 + π (1.5)2
4
Q = 3000 lb = 3 kips
F = 2500 lb = 2.5 kips
(a) Vertical, Moment at Edge = 0, Q > F
(a − b)Fe max = (Q + F )(b)
(12 − 7 )Fe max = (3.0 + 2.5)(7 )
Fe max = 7.7 kips
(a − b)Fe min = (Q − F )(b)
(12 − 7 )Fe min = (3.0 − 2.5)(7 )
Fe min = 0.7 kips
∆Fb max = (7.7 )(0.2554) = 1.97 kips
∆Fb min = (0.7 )(0.2554) = 0.18 kips
Fb max = Fi + ∆Fbmax = 23 + 1.97 = 24.97 kips
Fb min = Fi + ∆Fb min = 23 + 0.18 = 23.18 kips
1
(Fb max + Fb min ) = 1 (24.97 + 23.18) = 24.1 kips
2
2
1
1
Fa = (Fb max − Fb min ) = (24.97 − 23.18) = 0.9 kip
2
2
Fm =
Fm
24.1
=
= 39.8 ksi
As 0.606
F
0.9
sa = a =
= 1.5 ksi
As 0.606
sm =
Page 83 of 133
SECTION 6 – COMBINED STRESSES
1 sm K f sa
=
+
N sy
sn
say K f = 2.8 , SF = 0.85 , Factor for tension = 0.80
sn = (0.85)(0.80)(42.5) = 28.9 ksi
1 39.8 2.8(1.5)
=
+
N
50
28.9
N = 1.06
(b) Horizontal:
(a − b)Fe = Qb
(12 − 7 )Fe = (3)(7 )
Fe = 4.2 kips
∆Fb = (4.2 )(0.2554 ) = 1.073 kips
Fb = Fi + ∆Fb = 23 + 1.073 = 24.1 kips
F
24.1
s= b =
= 39.8 ksi
A 0.606
sm = s
sa = 0
K f sa
s
1
se = n sm +
=
(39.8) + 0 = 33.83 ksi
sy
SF
1.1765
Shear:
Fm = 0
1
Fa = (Fmax − Fmin ) = 2.5 kips
2
s ms = 0
Fa
2.5
=
= 4.13 ksi
As 0.606
K fs sas
s
(1.0)(4.13) = 4.86 ksi
ses = ns sms +
=0+
s ys
SF
0.85
sas =
1
1  se
= 
N  sn

  ses   2
 + 
 
  sns  
sns = 0.5sn = 0.5(42.5) = 21.25 ksi , maximum shear
2
2
1
2
2
1  33.83   4.86   2
= 
+
 
 
N  42.5   21.25  
N = 1.21
Page 84 of 133
SECTION 6 – COMBINED STRESSES
422.
The load Q, as seen (404), acts on the arm C and varies from 0 to 3 kips. The ends A and
B of the shaft are restrained from turning through an angle but are supported to take
the reactions A and B without other moments. The shaft is machined from AISI 1045, as
rolled; D1 = 2, D2 = 2.5, L = 15, a = 10, b = 20 in. For calculation purposes, assume that
the shaft size changes at the section of application of Q. Determine the factor of safety
in accordance with the maximum-shear and octahedral-shear theories. Investigate both
sections I and II. Would you judge the design to be 100% reliable?
Solution:
T = QL
Tmax = (3)(15) = 45 in − kips
T1 + T2 = 45 in − kips
T1a T2 b
=
J
J
T1a T2 b
=
D14 D 24
T1 (10) T2 (20)
=
(2 )4 (2.5)4
T1 = 0.8192T2
0.8192T2 + T2 = 45 in − kips
T2 = 24.74 in − kips
T1 = 0.8192T2 = 0.8192(24.74) = 20.27 in − kips
1
1
Ta1 = Tm1 = T1 = (20.27 ) = 10.14 in − kips
2
2
1
1
Ta 2 = Tm 2 = T2 = (24.74 ) = 12.37 in − kips
2
2
16T
πD 3
16T 16(10.14)
sms1 = sas1 = 31 =
= 6.46 ksi
πD1
π (2)3
sms = sas =
Page 85 of 133
SECTION 6 – COMBINED STRESSES
sms 2 = sas 2 =
16T2 16(12.37 )
=
= 4.03 ksi
πD 23
π (2.5)3
A + B = Q = 3 kips
Aa = Bb
A(10) = B(20)
A = 2B
2B + B = 3 kips
B = 1 kip
A = 2 kips
M = Aa = Bb = (2 )(10) = 20 in − kips
Mmax = M = 20 in − kips
1
M m = Ma = Mmax = 10 in − kips
2
32M
sm = sa =
πD 3
32(10 )
sm1 = sa1 =
= 12.73 ksi
π (2)3
32(10)
sm 2 = sa 2 =
= 6.52 ksi
π (2.5)3
Use (1)
sms = 6.46 ksi , sm = 12.73 ksi
sas = 6.46 ksi , sa = 12.73 ksi
r = 0.15D1
r
D D
2.5
= 0.15 , = 2 =
= 1.25
D1
d D1 2.0
K t = 1.5 , K ts = 1.25 (Figure AF 12)
r = 0.15D1 = 0.15(2 ) = 0.30
1
=
= 0.968
a
0.01
1+
1+
r
0. 3
K f = q (K t − 1) + 1 = 0.968(1.5 − 1) + 1 = 1.484
q=
1
K fs = q (K ts − 1) + 1 = 0.968(1.25 − 1) + 1 = 1.242
Profile Keyway
K f = 1.6 , K fs = 1.3
Net
Page 86 of 133
SECTION 6 – COMBINED STRESSES
K f = (1.484)(1.6 )(1 − 0.20) = 1.9
K fs = (1.242)(1.3)(1 − 0.20 ) = 1.3
For AISI 1045, as rolled, s y = 59 ksi , su = 96 ksi
sn = 0.5su = 48 ksi
sns sn 48
1
= =
=
s ys sy 59 1.229
SF = 0.85 , RF = 0.85
se =
K f sa
sn
1
sm +
=
(12.73) + 1.9(12.73) = 43.84 ksi
sy
(SF )(RF ) 1.229
(0.85)(0.85)
ses =
K fs sas
sns
1
s ms +
=
(6.46) + 1.3(6.46) = 16.88 ksi
s ys
(SF )(RF ) 1.229
(0.85)(0.85)
Maximum shear, sns = 0.5sn = 0.5(48) = 24 psi
1
1  se
= 
N  sn

2
  ses
 + 
  sns



2
2


1
2
2
1  43.84   16.88   2
= 
 +
 
N  48   24  
N = 0.87
Octahedral Shear. sns = 0.577 sn = 0.577(48) = 27.7 psi
1
2
2
1  43.84   16.88   2
= 
 +
 
N  48   27.7  
N = 0.91
Not 100% reliable, N < 100.
423.
A rotating shaft overhangs a bearing, as seen in the illustration. A ¼-in. hole is drilled at
AB. The horizontal force F2 varies in phase with the shaft rotation from 0 to 5 kips, but
its line of action does not move. A steady torque T = 8 in-kips is applied at the end of the
shaft; D = 2, D2 = 2.5, a = 2, b = 5, e = 0.5, r = ¼ in. The material is AISI C1040, annealed.
What steady vertical load F1 can be added as shown if the design factor is to be 2.5 from
the octahedral-shear theory? Assume that the cycling of F2 may be such that the worst
stress condition occurs at the hole.
Page 87 of 133
SECTION 6 – COMBINED STRESSES
Solution:
AISI C1040, annealed, s y = 48 ksi (Fig. AF 1), su = 80 ksi
sn = 0.5su = 40 ksi
sn 40 1
=
=
sy 48 1.2
For hole: d D = 0.25 2 = 0.125
K t = 2.2 , K ts = 1.6
a = 0.01 (annealed)
0.25
r=
= 0.125
2
1
1
q=
=
= 0.926
a
0.01
1+
1+
r
0.125
K f = q (K t − 1) + 1 = 0.926(2.2 − 1) + 1 = 2.11
K fs = q (K ts − 1) + 1 = 0.926(1.6 − 1) + 1 = 1.56
At hole
s = s3 + s1 − s2
Bending F2 :
Mc M
s2 =
=
I
I c
F2 e
(5)(0.5)
s2 =
=
= 4.04 ksi
3
2
3
πD
dD
π (2) (0.25)(2 )2
−
−
32
6
32
6
F2
(
5)
s3 =
=
= 1.89 ksi
πD 2
π (2)2
− dD
− (0.25)(2)
4
4
F1b
(F1 )(5)
s1 =
=
= 8.08F1
3
2
3
πD
dD
π (2) (0.25)(2)2
−
−
32
6
32
6
Page 88 of 133
SECTION 6 – COMBINED STRESSES
ss =
T
πD
3
dD
−
16
6
2
=
π (2 )
3
16
−
(8)
(0.25)(2 )2
= 5.70 ksi
6
smin = s3 + s1 − s2 = 1.89 + 8.08F1 − 4.04 = 8.08F1 − 2.15
smax = s1 = 8.08F1
1
1
sm = (smax + smin ) = (8.08F1 + 8.08F1 − 2.15) = 8.08F1 − 1.08 ksi
2
2
1
1
sa = (smax − smin ) = (8.08F1 − 8.08F1 + 2.15) = 1.08 ksi
2
2
se = 6.74F1 − 1.78
s
s es = ns sms + K fs sas
sys
s ms = s s
sas = 0
1
(5.7 ) = 4.75 ksi
1.2
N = 2.5
ses =
Octahedral Shear Theory
s
sns = n = 0.577 sn = 0.577(40) = 23.08 ksi
3
1
1  se
= 
N  sn

2
  ses
 + 
  sns



2
2


1
2
2
1  6.74F1 − 1.78   4.75   2
= 
 +
 
2.5 
40
  23.08  
F1 = 2.3 kips
POWER SCREWS
424.
Design a square-thread screw for a screw jack, similar to that shown, which is to raise
and support a load of 5 tons. The maximum lift is to be 18 in. The material is AISI C1035,
as rolled, and N ≈3.3 based on the yield strength.
Page 89 of 133
SECTION 6 – COMBINED STRESSES
Solution:
AISI C1035, as rolled, s y = 55 ksi
sy
55
= 16.6 ksi
N 3.3
F = (5)(2) = 10 kips
F 10
A= =
= 0.6034 in 2
s 16.6
πD 2
A = r = 0.6034 in 2
4
Dr = 0.876 in
say 1 ¼ in, Dr = 1.000 in
L = 18 in
Le = 2L = 36 in
1
1
k = Dr = (1.000) = 0.125 in
8
8
Le
36
=
= 288 > 40
k 0.125
Transition:
s=
=
1
1
Le  2π 2E  2  2π 2 (30,000)  2
=
=
 = 104
k  sy 
55


L
Use column formula, Eulers e > 104
k
π 2EI
F= 2
NLe
10,000 =
I=
π 2 (30 × 10 6 )I
3.3(36)2
πDr4
= 0.14444
64
Dr = 1.31 in
use 1 ¾ in, Dr = 1.400 in
Page 90 of 133
SECTION 6 – COMBINED STRESSES
425.
(a) For the screw of 424, what length of threads h will be needed for a bearing pressure
of 1800 psi? (b) Complete the design of the jack. Let the base be cast iron and the
threads integral with the base. Devise a method of turning the screw with a round steel
rod as a lever and fix the details of a nonrotating cap on which the load rests. (c) What
should be the diameter of the rod used to turn the screw? If a man exerts a pull of 150
lb. at the end, how long must the rod be?
Solution:
(a)
Th/in = 2.5
Dr = 1.40 in
1
= 0.4 in
2.5
Lead
λ = tan −1
πD m
1
Dm = (1.75 + 1.40 ) = 1.575 in
2
 0.40 
o
λ = tan −1 
 = 4.62
(
)
π
1
.
575


Lead Pitch =
f = 0.15
tan β = f = 0.15
β = 8.53o
F cos(β + λ )
s=
(Do − Dr )L
10,000 cos(8.53 + 4.62 )
(1.75 − 1.40)L
L = 16.30 in
1800 =
h = L tan λ = 16.30 tan 4.62 o = 1.32 in
say h = 1.5 in
Page 91 of 133
SECTION 6 – COMBINED STRESSES
(b) Assume ASTM 20. sus = 32 ksi , su = 20 ksi , N = 5
32
= 6.4 ksi
5
20
s=
= 4 ksi
5
ss =
F
πDh
10
6.4 =
πD(1.5)
D = 0.33 in
ss =
Dr = 1.4 in > 0.33 in
3
Do = 1 in
4
Use proportions from figure based on diameter.
Method: Manual, normal pull.
7
(c) D = in (Based on proportion)
8
FDm
(10 )(1.575) tan(8.53 + 4.62) = 1.84 in − kips
T=
tan(β + λ ) =
2
2
T = F ′a
F ′ = 150 lb = 0.15 kips
1.84 = 0.15a
a = 12.3 in
426.
A screw jack, with a 1 ¼-in. square thread, supports a load of 6000 lb. The material of
the screw is AISI C1022, as rolled, and the coefficient of friction for the threads is about
0.15. The maximum extension of the screw from the base is 15 in. (a) Considering the
ends of the screw restrained so that Le = L, find the equivalent stress and the design
factor. (b) If the load on the jack is such that it may sway, the screw probably acts as a
column with one end free and the other fixe. What is the equivalent stress and the
factor of safety in this instance? (c) What force must be exerted at the end of a 20-in.
lever to raise the load? (d) Find the number of threads and the length h of the threaded
portion in the cast-iron base for a pressure of 500 psi on the threads. (e) What torque is
necessary to lower the load?
Page 92 of 133
SECTION 6 – COMBINED STRESSES
Solution:
From Table AT 7,
AISI C1022, as rolled, sy = 52 ksi
F = 6000 lb = 6 kips
For 1 ¼ in square thread, Dr = 1.0 in, Th/in. = 3.5
f = 0.15
(a) With Le = L = 15 in
1
1
k = Dr = (1.0) = 0.125 in
8
8
Le
15
=
= 120
k 0.125
Transition for AISI C1020;
1
1
Le  2π 2E  2  2π 2 (30,000)  2
=
=
 = 107
k  sy 
52


Le
Use column formula, Eulers
> 107
k
π 2EA
Fc = NF =
(Le k )2
sd =
F
π 2E
=
A N (Le k )2
Equivalent stress
F
4F
sd = = 2
A πDr
4(6)
sd =
= 7.64 ksi
π (1.0)2
Design factor
π 2E
sd =
N (Le k )2
Page 93 of 133
SECTION 6 – COMBINED STRESSES
7.64 =
π 2 (30,000)
N (120)2
N = 2.69
(b) With Le = 2L = 30 in
Le
30
=
= 240
k 0.125
Transition for AISI C1020;
1
1
Le  2π 2E  2  2π 2 (30,000)  2
=
=
 = 107
k  sy 
52


Le
Use column formula, Eulers
> 107
k
π 2EA
Fc = NF =
(Le k )2
sd =
F
π 2E
=
A N (Le k )2
Equivalent stress
F
4F
sd = = 2
A πDr
4(6)
sd =
= 7.64 ksi
π (1.0)2
Design factor
π 2E
sd =
N (Le k )2
7.64 =
π 2 (30,000)
N (240)2
N = 0.673 not safe
(c) For force exerted at the end of 20-in. lever to raise the load = Fa
WDm
T=
tan(β + λ )
2
1
Lead = Pc =
= 0.2857 in
3. 5
1
Dm = (1.25 + 1.00) = 1.125 in
2
Lead
0.2857
λ = tan−1
= tan−1
= 4.62o
πDm
π (1.125)
f = tan β = 0.15
β = 8.53o
W = 6000 lb
Page 94 of 133
SECTION 6 – COMBINED STRESSES
WDm
tan(β + λ )
2
6000(1.125)
Fa (20) =
tan(8.53 + 4.62)
2
Fa = 39.43 lb
T = Fa a =
(d) Let p = pressure = 500 psi, W = 6000 lb, Do = 1.25 in, Di = 1.00 in.
Nt = number of threads, h = length of threaded portion.
4W
p=
2
π Do − Dr2 Nt
4(6000 )
500 =
π (1.25)2 − (1.00 )2 Nt
Nt = 27
Then
h = Nt Pc = (27 )(0.2857 ) = 7.7 in
(
)
[
]
(e) Torque necessary to lower the load.
WDm
T=
tan(β − λ )
2
(6000)(1.125) tan(8.53 − 4.62)
T=
2
T = 230.7 in − lb.
427.
A square-thread screw, 2 in. in diameter, is used to exert a force of 24,000 lb. in a shaftstraightening press. The maximum unsupported length of the screw is 16 in. and the
material is AISI C1040, annealed. (a) What is the equivalent compressive stress in the
screw? Is this a satisfactory value? (b) What torque is necessary to turn the screw
against the load for f = 0.15? (c) What is the efficiency of the screw? (d) What torque is
necessary to lower the load?
Solution:
For 2 in. square thread screw, Do = 2 in, Dr = 1.612 in, Th/in. = 2.25 from Table 8.1
W = 24,000 lb = 24 kips, L = 16 in
(a) For unsupported length, Le = L = 16 in.
For AISI C1040, annealed, Figure AF-1, sy = 47.5 ksi
Transition,
1
1
k = Dr = (1.612 ) = 0.2015 in
8
8
1
1
Le  2π 2E  2  2π 2 (30,000)  2
=
=
 = 112
k  sy 
47.5


Then
Le
16
=
= 79.4 < 112
k 0.2015
Use column formula, JB Johnson Formula,
Page 95 of 133
Le
< 112
k
SECTION 6 – COMBINED STRESSES
 sy (Le k )2 
F
= se 1 −

A
4π 2E 

se =
4W

sy (Le k )2 

4π 2E 

4(24)
se =
 47.5(79.4)2 
π (1.612)2 1 − 2

 4π (30,000) 
πDr2 1 −
se = 15.74 ksi
s
47.5
= 3.0 satisfactory
N= y =
se 15.74
(b) Torque to turn the screw against the load
WDm
tan(β + λ )
2
1
Lead = Pc =
= 0.4445 in
2.25
1
Dm = (2.00 + 1.612 ) = 1.806 in
2
Lead
0.4445
λ = tan−1
= tan−1
= 4.48o
πDm
π (1.806)
f = tan β = 0.15
T=
β = 8.53o
W = 24,000 lb
WDm
tan(β + λ )
2
24,000(1.806)
T=
tan(8.53 + 4.48)
2
T = 5008 in − lb
T=
(c) Torque necessary to lower the load.
WDm
T=
tan(β − λ )
2
24,000(1.806)
T=
tan(8.53 − 4.48)
2
T =1535 in − lb.
428.
(a) A jack with a 2-in., square-thread screw is supporting a load of 20 kips. A single
thread is used and the coefficient of friction may be as low as 0.10 or as high as 0.15.
Page 96 of 133
SECTION 6 – COMBINED STRESSES
Will this screw always be self-locking? What torque is necessary to raise the load? What
torque is necessary to lower the load? (b) The same as (a) except that a double thread is
used. (c) The same as (a) except that a triple thread is used.
Solution:
Table 8.1, 2 in. square thread, Do = 2 in, Dr = 1.612 in, Th/in = 2.25
(a) Self-locking? And Torque necessary to raise the load.
1
Dm = (2.00 + 1.612 ) = 1.806 in
2
1
Lead = Pc =
= 0.4445 in
2.25
Lead
0.4445
λ = tan−1
= tan−1
= 4.48o
πDm
π (1.806)
If f = 0.10
f = tan β = 0.10
β = 5.71o
If f = 0.15
f = tan β = 0.15
β = 8.53o
Since β is always greater than λ, the screw is always self-locking.
WDm
tan(β + λ )
2
W = 20 kips
WDm
tan(β + λ )
T=
2
20(1.806)
T=
tan(8.53 + 4.48)
2
T = 4.173 in − kips
Torque necessary to lower the load.
WDm
T=
tan(β − λ )
2
20(1.806)
T=
tan(8.53 − 4.48)
2
T = 1.279 in − kips.
T=
(b) Self-locking? And Torque necessary to raise the load.
2
Lead = 2Pc =
= 0.8889 in
2.25
Lead
0.8889
λ = tan−1
= tan−1
= 8.904o
πDm
π (1.806)
If f = 0.10
f = tan β = 0.10
β = 5.71o
If f = 0.15
Page 97 of 133
SECTION 6 – COMBINED STRESSES
f = tan β = 0.15
β = 8.53o
Since β is always less than λ, the screw is always not self-locking.
WDm
tan(β + λ )
2
W = 20 kips
WDm
tan(β + λ )
T=
2
20(1.806)
T=
tan(8.53 + 8.904)
2
T = 5.671 in − kips
Torque necessary to lower the load = 0
T=
(c) Self-locking? And Torque necessary to raise the load.
3
Lead = 3Pc =
= 1.3333 in
2.25
Lead
1.3333
λ = tan−1
= tan−1
= 13.224o
πDm
π (1.806 )
If f = 0.10
f = tan β = 0.10
β = 5.71o
If f = 0.15
f = tan β = 0.15
β = 8.53o
Since β is always less than λ, the screw is always not self-locking.
WDm
tan(β + λ )
2
W = 20 kips
WDm
T=
tan(β + λ )
2
20(1.806)
T=
tan(8.53 + 13.224)
2
T = 7.207 in − kips
Torque necessary to lower the load = 0
T=
429.
The conditions for a self-locking screw are given in §8.23, Text. Assume that the
coefficient of friction is equal to the tangent of the lead angle and show that the
efficiency of a self-locking screw is always less than 50%.
Solution:
tan λ
λ
e=
≈
tan(β + λ ) β + λ
For self-locking, β > λ, then β + λ > 2λ
Then,
Page 98 of 133
SECTION 6 – COMBINED STRESSES
e<
λ
2λ
e < 0.50
e < 50%
CURVED BEAMS
430.
It is necessary to bend a certain link somewhat as shown in order to prevent
interference with another part of the machine. It is estimated that sufficient clearance
will be provided if the center line of the link is displaced e = 3 in. from the line of action
of F, with a radius of curvature of R ≈ 5.5 in., L = 10 in., material is wrought aluminum
alloy 2014 T6; N = 2 on the basis of the maximum shear stress; F = 2500 lb. with the
number of repetitions not exceeding 106. (a) If the section is round, what should be its
diameter D? (b) If the link is bend to form cold, will the residual stresses be helpful or
damaging? Discuss.
Solution:
(a) Table AT3. Wrought aluminum alloy 2014 T6
sn = 18 ksi @5 ×108 cycles
sy = 60 ksi
At 106 cycles
 106 

sn = sn′ 

 nc 
0.09
0.09
 10 6 

18 = sn′ 
8 
 5 ×10 
sn′ = 31.49 ksi
With size factor.
sn = 0.85sn′ = 0.85(31.49) = 26.77 ksi
sn 26.77
=
= 13.38 ksi
N
2
Equation:
F K Mc
s= + c
A
I
s=
Page 99 of 133
SECTION 6 – COMBINED STRESSES
A=
I=
πD 2
4
πD 4
64
D
c=
2
M = Fe
D
K c (Fe ) 
4F
2
s= 2 +
πD
πD 4
64
s=
4F 32K c Fe
+
πD 2
πD 3
Using Trial and error and Table AT 18:
r 2R 2(5.5) 11
=
=
=
c D
D
D
4(2.5) 32K c (2.5)(3)
+
13.38 =
πD 2
πD 3
By trial and error D = 1.92 in
r 11
=
= 6.0908
c 1.92
Table AT 18: Kc = 1.152
4(2.5) 32(1.152)(2.5)(3)
s=
+
= 13.30 ksi ≈ 13.38 ksi
π (1.92)2
π (1.92)3
Use D = 2 in.
(b) Residual stress is helpful due to a decrease in total stress on tension side.
431.
The same as 430, except that the section is rectangular with h ≈ 3b; see figure.
Solution:
(a) Table AT3. Wrought aluminum alloy 2014 T6
sn = 18 ksi @5 ×108 cycles
sy = 60 ksi
At 106 cycles
Page 100 of 133
SECTION 6 – COMBINED STRESSES
 106 

sn = sn′ 

 nc 
0.09
0.09
 10 6 

18 = sn′ 
8 
5
×
10


sn′ = 31.49 ksi
With size factor.
sn = 0.85sn′ = 0.85(31.49) = 26.77 ksi
s
26.77
s= n =
= 13.38 ksi
N
2
Equation:
F K Mc
s= + c
A
I
A = bh = b(3b) = 3b 2
bh 3 b(3b)3
=
= 2.25b 4
12
12
h
c = = 0.5h = 1.5b
2
M = Fe
F
K (Fe )(1.5b)
s= 2 + c
3b
2.25b 4
I=
s=
F
K Fe
+ c 3
2
3b 1.5b
Using Trial and error and Table AT 18:
r 2R 2(5.5) 11 11
=
=
= =
c h
h
h 3b
2.5 K (2.5)(3)
13.38 = 2 + c
3b
1.5b3
By trial and error b = 0.787 in
r
11
=
= 4.66
c 3(0.787 )
Table AT 18: Kc = 1.1736
2.5
(1.1736 )(2.5)(3)
s=
+
= 13.38 ksi
2
3(0.787 )
1.5(0.787 )3
Use b = 7/8 in. h = 3b = 2 5/8 in
(b) Residual stress is helpful due to a decrease in total stress on tension side.
432.
A hook is to be designed similar to that shown to support a maximum load F = 2500 lb.
that will be repeated an indefinite number of times; the horizontal section is to be
circular of radius c and the inside radius a is 1 ½ in. (a) Determine the diameter of the
Page 101 of 133
SECTION 6 – COMBINED STRESSES
horizontal section for N = 2 based on the Soderberg line, if the material is AISI 4130,
WQT 1100 F. (b) Calculate the value of the static load that produces incipient yielding.
Solution:
(a) For AISI 4130, WQT 1100 F, Table AT 7
sy = 114 ksi, su = 127 ksi, sn’ = su/2 for reversed bending
sn = SFsn′ = 0.85sn′ = 0.85(su 2)
Soderberg line:
1 sm K f sa
=
+
N sy
sn
sm = sa =
s
repeated load
2
K f = 1.0
s2
1
s
=
+
N 2 sy 0.85(su 2 )
1  1
1 
=
+
s
N  2sy 0.85su 

1  1
1
s
= 
+
2  2(114) 0.85(127 ) 
sd = 36.63 ksi
For curved beam
F K Mc
s= + c
A
I
a = 1.5 in
A=
π
(2c )2 = πc 2
4
F = 2500 lb = 2.5 kips
M = F (a + c )
I=
π (2c )4
64
Page 102 of 133
=
πc 4
4
SECTION 6 – COMBINED STRESSES
Table AT 18,
r =a+c
r a + c 1. 5 + c
=
=
c
c
c
Substitute:
2.5 K (2.5)(1.5 + c )c
36.63 = 2 + c
πc
πc 4 4
2.5 10K c (1.5 + c )
36.63 = 2 +
πc
πc 3
By trial and error: c = 0.633
r 1.5 + 0.633
=
= 3.37 , K c = 1.293
c
0.633
2.5
10(1.293)(1.5 + 0.633)
36.63 = s =
+
2
π (0.633)
π (0.633)3
36.63 = s ≈ 36.60 ksi
Use c = 11/16 = 0.6875 in
Diameter = 2c = 1.375 in = 1 3/8 in
(b) Static load that produces incipient yielding.
sd = sy = 114 ksi
F K Mc
s= + c
A
I
K (F )(1.5 + c )c
F
114 = 2 + c
πc
πc 4 4
F
K (F )(1.5 + c )
114 = 2 + c
πc
πc 3
r 1.5 + 0.6875
=
= 3.18 , K c = 1.312
c
0.6875
1.312(F )(1.5 + 0.6875)
F
114 =
+
2
π (0.6875)
π (0.6875)3
F = 32.71 kips
433.
The same as 432, except that the hook is expected to be subjected to 100,000
repetitions of the maximum load.
Page 103 of 133
SECTION 6 – COMBINED STRESSES
Solution:
(a) For AISI 4130, WQT 1100 F, Table AT 7
sy = 114 ksi, su = 127 ksi, sn’ = su/2 for reversed bending
At 100,000 repetitions
 10 6 

sn = 0.85(su 2 )

n
 c 
Soderberg line:
1 sm K f sa
=
+
N sy
sn
sm = sa =
0.085
 106 

= 0.85(su 2)

100
,
000


s
repeated load
2
K f = 1.0
s2
1
s
=
+
N 2 sy 0.5169su

1  1
1
s
=
+

N  2sy 1.0338su 

1  1
1
s
= 
+
2  2(114 ) 1.0338(127 ) 
sd = 41.66 ksi
For curved beam
F K Mc
s= + c
A
I
a = 1.5 in
π
(2c )2 = πc 2
4
F = 2500 lb = 2.5 kips
M = F (a + c )
A=
I=
π (2c )4
=
πc 4
64
4
Table AT 18,
r =a+c
r a + c 1. 5 + c
=
=
c
c
c
Substitute:
2.5 K (2.5)(1.5 + c )c
41.66 = 2 + c
πc
πc 4 4
2.5 10K c (1.5 + c )
41.66 = 2 +
πc
πc 3
By trial and error: c = 0.601
r 1.5 + 0.601
=
= 3.5 , K c = 1.28
c
0.601
Page 104 of 133
0.085
= 0.5169su
SECTION 6 – COMBINED STRESSES
2. 5
10(1.28)(1.5 + 0.601)
+
2
π (0.601)
π (0.601)3
41.66 = s ≈ 41.64 ksi
Use c =5/8
Diameter = 2c = 1.25 in = 1 1/4 in
41.66 = s =
(b) Static load that produces incipient yielding.
sd = sy = 114 ksi
F K Mc
s= + c
A
I
K (F )(1.5 + c )c
F
114 = 2 + c
πc
πc 4 4
F
K (F )(1.5 + c )
114 = 2 + c
πc
πc 3
r 1.5 + 0.625
=
= 3.4 , K c = 1.29
c
0.625
1.29(F )(1.5 + 0.625)
F
114 =
+
2
(
)
π 0.625
π (0.625)3
F = 25.97 kips
434.
A hook, similar to that shown with a horizontal circular section of diameter 2c, is to be
designed for a capacity of 2000 lb. maximum, a load that may be applied an indefinite
number of times. A value of a = 2 in. should be satisfactory for the radius of curvature of
the inside of the hook. Let N = 1.8 based on the modified Goodman line. At the outset of
design, the engineer decided to try AISI C1040, OQT 1100 F. (a) Compute the diameter
of the horizontal section, (b) If the 45o circular section is made the same diameter, what
is its design factor (modified Goodman)? Could this section be made smaller or should it
be larger?
Solution:
(a) For AISI C1040, OQT 1100 F, Figure AF 1
su = 100 ksi, sn’ = su/2 for reversed bending
sn = SF x sn’ = 0.85(0.5)(100) = 42.5 ksi
Kf = 1.0
Modified Goodman line:
Page 105 of 133
SECTION 6 – COMBINED STRESSES
1 sm K f sa
=
+
N su
sn
s
sm = sa = repeated load
2
K f = 1.0
s2 s2
1
=
+
1.8 100 42.5
sd = 33.14 ksi
For curved beam
F K Mc
s= + c
A
I
a = 2.0 in
A=
π
(2c )2 = πc 2
4
F = 2000 lb = 2.0 kips
M = F (a + c )
I=
π (2c )4
=
πc 4
64
4
Table AT 18,
r =a+c
r a + c 2. 0 + c
=
=
c
c
c
Substitute:
2.0 K (2.0)(2.0 + c )c
33.14 = 2 + c
πc
πc 4 4
2.0 8K (2.0 + c )
33.14 = 2 + c 3
πc
πc
By trial and error: c = 0.639
r 2.0 + 0.639
= 4.13 , K c = 1.224
=
c
0.639
2. 0
8(1.224)(2.0 + 0.639)
33.14 = s =
+
2
π (0.639)
π (0.639)3
33.14 = s ≈ 33.08 ksi
Use c = 11/16 in
Diameter = 2c = 1.375 in = 1 3/8 in
(b) sus = 0.6su = 0.6 x 100 ksi = 60 ksi
sns = 0.6sn = 0.6 x 42.5 ksi = 25.5 ksi
Equivalent stress (Modified Goodman Line)
s s
se = m n + K f sa
su
s s
ses = ms ns + K f sas
sus
Page 106 of 133
SECTION 6 – COMBINED STRESSES
1
2
2 2
1  se   ses  
=   +   
N  sn   sns  


s
sm = sa =
2
s
sms = sas = s
2
F cos 45 K c Mc F cos 45 K c F (a + c )(cos 45)c
s=
+
=
+
A
I
πc 2
πc 4 4
F cos 45 4K c F (a + c )(cos 45)
+
s=
πc 2
πc 3
F sin 45 F sin 45
ss =
=
A
πc 2
11
c = in = 0.6875 in (assuming constant diameter)
16
r a + c 2.0 + 0.6875
=
=
= 3.91
c
c
0.6875
Table AT 18,
K c = 1.239
(2.0 )cos 45 4(1.239)(2.0)(2.0 + 0.6875)(cos 45)
s=
+
π (0.6875)2
π (0.6875)3
s = 19.40 ksi
(2.0 )sin 45
ss =
= 0.95 ksi
π (0.6875)2
Then
s s
se = m n + K f sa
su
 19.40  42.5 
ss
se =  n + K f  =
+ 1 = 13.82 ksi

2  su
2  100


s s
ses = ms ns + K f sas
sus
ses =
 0.85  25.5 
ss  sns

+ K f  =
+ 1 = 0.68 ksi


2  sus
2  60


2
2
1  se   ses  
=   +   
N  sn   sns  


1
2
2
1  13.82   0.68   2
= 
 +
 
N  42.5   25.5  
N = 3.06
Since N > 1.8, this section could be made smaller.
Page 107 of 133
SECTION 6 – COMBINED STRESSES
435.
A C-frame hand press is made of annealed cast steel (A27-58) and has a modified Isection, as shown. The dimensions of a 45o section CD are: a = 3, b = 6, h = 4, t = 1 in.,
radius r = 1 in.; also g = 12 in.; and the maximum force is F = 17 kips, repeated a
relatively few times in the life of the press. (a) Applying the straight-beam formula to
the 45o section, compute the maximum and minimum normal stresses. (b) Do the same,
applying the curved-beam formula. (c) By what theory would you judge this section to
have been designed? If the radius r were increased several times over, as it could have
been done, would the stress have been materially reduced? Give reasons for your
conclusions.
Solution:
(a) Straight-beam formula
Consider only normal stresses, relatively static.
F cos 45 Mc
s=
±
A
I


c

M = F g − r +  2 + r  cos 45
2




A = ht + at + (b − 2t )t
t
t 
 b − 2t 

ht   + (b − 2t )(t )
+ t  + at  b − t + 
2
2
 2


c2 =  
ht + (b − 2t )t + at
Page 108 of 133
SECTION 6 – COMBINED STRESSES
ht 2
t
 bt 

+ (b − 2t )  + at  b − 
2
2
2
 


c2 =
ht + (b − 2t )t + at
c1 = b − c 2
(4)(1)2
1
 (6 )(1) 

+ [6 − 2(1)]
+ (3)(1) 6 − 

2
2
 2 

c2 =
= 2.77273 in
(4 )(1) + [6 − 2(1)](1) + (3)(1)
c1 = b − c 2 = 6 − 2.7723 = 3.22727 in
I = I + Ad 2
A1 = ht
A2 = (b − 2t )t
A3 = at
ht 3
t

I1 =
+ ht  c 2 − 
12
2

I2 =
t (b − 2t )3
b

+ (b − 2t )(t ) − c2 
12
2


at 3
t

I3 =
+ at  c1 − 
12
2

I1 =
2
2
2
2
(4)(1)3 ( )( )
1
+ 4 1  2.77273 −  = 21 in 4

12
2
2
(1)[6 − 2(1)]3 [ ( )]( ) 6

I2 =
+ 6 − 2 1 1  − 2.77273 = 5.54 in 4
12
(3)(1)
3
2

2
1

+ (3)(1) 3.22727 −  = 22.564 in 4
12
2

I = 21 + 5.54 + 22.564 = 49.104 in 4
Then
F cos 45 Mc 2
smax =
+
A
I
F cos 45 Mc1
smin =
−
A
I
A = (4 )(1) + (3)(1) + [6 − 2(1)](1) = 11 in 2
I3 =


 2.77273 
M = 17 12 − 1 + 
+ 1 cos 45 = 215.686 in − kips
2




17 cos 45 (215.686)(2.77273)
smax =
+
= 13.27 ksi in tension
11
49.104
17 cos 45 (215.686)(2.77273)
smin =
−
= −13.08 ksi = 13.08 ksi in compression
11
49.104
(b) Curved-beam formula
Page 109 of 133
SECTION 6 – COMBINED STRESSES
F cos 45 K ci Mc 2
+
A
I
F cos 45 K co Mc1
smin =
−
A
I
Using Table AT18
r
Z = −1 + [b1 log e (r + c1 ) − (t − b1 )log e (r + c 4 ) + (b − t )log e (r − c3 ) − blog e (r − c 2 )]
A
smax =
r = 1 + 2.77273 = 3.77273 in
c 2 = 2.77273 in
c1 = 3.22727 in
c 4 = 3.22727 − 1 = 2.22727 in
c3 = 2.77273 − 1 = 1.77273 in
b1 = 3 in
t = 1 in
b = 4 in

c 
1 + Z (r + c ) I

Kc = 
Arc
3.77273 3log e (3.77273 + 3.22727 ) − (1 − 3)log e (3.77273 + 2.22727) 
Z = −1 +


11 + (4 − 1)log e (3.77273 − 1.77273) − 4 log e (3.77273 − 2.77273)
Z = 2.944455
c = −c2

c2  

2.77273
(49.104)
1 +
I 1 +
Z (r − c 2 )   2.944455(3.77273 − 2.77273) 

K ci =
=
(11)(3.77273)(2.77273)
Arc 2
K ci = 0.8286
c = c1

c1  

3.22727
(49.104)
1 +
I 1 +
Z (r + c1 )   2.944455(3.77273 + 3.22727 ) 

K co =
=
(11)(3.77273)(3.22727 )
Arc1
K co = 0.424
F cos 45 K ci Mc 2
smax =
+
A
I
F cos 45 K co Mc1
smin =
−
A
I
17 cos 45 (0.8286)(215.686)(2.77273)
smax =
+
= 11.18 ksi in tension
11
49.104
Page 110 of 133
SECTION 6 – COMBINED STRESSES
smin =
17 cos 45 (0.424)(215.686)(2.77273)
−
= −4.07 ksi = 4.07 ksi in compression
11
49.104
(c) This section must be designed based on straight beam formula. Maximum stress is
higher.
Increasing the radius r.
Table A-18.
r = 2 + 2.77273 = 4.77273 in
c 2 = 2.77273 in
c1 = 3.22727 in
c 4 = 3.22727 − 1 = 2.22727 in
c3 = 2.77273 − 1 = 1.77273 in
b1 = 3 in
t = 1 in
b = 4 in

c 
1 + Z (r + c ) I

Kc = 
Arc
4.77273 3log e (4.77273 + 3.22727 ) − (1 − 3)log e (4.77273 + 2.22727 ) 
Z = −1 +


11 + (4 − 1)log e (4.77273 − 1.77273) − 4 log e (4.77273 − 2.77273)
Z = 3.622343
c = −c2

c2  

2.77273
1 +
I 1 +
(49.104)
(
)
Z
r
−
c
3
.
622343
(
4
.
77273
−
2
.
77273
)
2 



K ci =
=
(
)(
)(
)
Arc 2
11 4.77273 2.77273
K ci = 0.4664
c = c1

c1  

3.22727
(49.104)
1 +
I 1 +
Z
(
r
+
c
)
3.622343(4.77273 + 3.22727 ) 

1 

K co =
=
(11)(4.77273)(3.22727)
Arc1
K co = 0.3221
F cos 45 K ci Mc 2
smax =
+
A
I
F cos 45 K co Mc1
smin =
−
A
I
17 cos 45 (0.4664)(215.686)(2.77273)
smax =
+
= 6.77 ksi in tension
11
49.104
17 cos 45 (0.3221)(215.686)(2.77273)
smin =
−
= −2.83 ksi = 2.83 ksi in compression
11
49.104
Page 111 of 133
SECTION 6 – COMBINED STRESSES
436.
The stress is reduced using by increasing the radius r in Curved Beam Formula.
Reason: As the radius r increased the stress factor for curved beam decreases thence
the maximum stress is reduced.
A heavy C-clamp, similar to the figure, is made of normalized cast steel (A27-58) and has
a T-section where t= 7/16 in.; q= 2 ¾ , a =1 ¾ in. What is the safe capacity if N = 2 based
on yield?
Solution:
F K Mc
s = + ci i
A
I
Table AT 1
2
 3t 
7
A = 4t   + t (4.5t ) = 10.5t 2 = 10.5  = 2.009766 in 2
2
 16 
(t ) 4.5t + 3 t 
2
2
3 
+ (4t − t ) t 
2 
 2  = 2.035714t
c1 = 

3 
 3 
2t  4.5t + t  + (4t − t ) t 
2

 2 

7
c1 = 2.035714  = 0.890625 in
 16 
c 2 = 4.5t + 1.5t − c1 = 6t − 2.035714t = 3.964286t
7
c 2 = 3.964286  = 1.734375 in
 16 
Table AT 18
r = a + c1 = 1.75 + 0.890625 = 2.640625 in
r a + c1 2.640625
=
=
= 2.965
c
c1
0.890625
K ci = 1.4212
M = F (q + ci ) = F (2.75 + 0.890625) = 3.640625F
For Normalized cast steel, A27-58,
sy = 36 ksi
Moment of Inertia
Page 112 of 133
SECTION 6 – COMBINED STRESSES
3
(4t ) 3 t 
2
2
3
 2  + (4t ) 3 t  2.035714t − 3 t  + (t )(4.5t ) + (t )(4.5t ) 3.964286t − 4.5 t 
I=
 



12
4 
12
2 
 2 

4
7
I = 31.861607t = 31.861607  = 1.167293 in 4
 16 
4
F K ci Mc i
+
A
I
sy 36
F
(1.4212 )(3.640625)(F )(0.890625)
s= = =
+
N 2 2.009766
1.167293
F = 4.049 kips = 4049 lb
s=
437.
The same as 436, except that the section is trapezoidal with b = ¾ in. (see figure). Ignore
the effect of resounding off the corners.
Solution:
F K Mc
s = + ci i
A
I
From other sources.
2
1
(b + 2b)(3b ) = 4.5b 2 = 4.5 3  = 2.53125 in 2
2
 4
3b  2b + 2b  4
43
c1 = 
 = b =   = 1 in
3  b + 2b  3
34
4
5
5 3
c 2 = 3b − b = b =   = 1.25 in
3
3
3 4
A=
4
(3b)3 [b 2 + 4b(2b) + (2b)2 ]
3
4
I=
= 3.25b = 3.25  = 1.02832 in 4
36(b + 2b)
4
Table AT 18
3.25b4
Z = −1 +



2r  b − a
(r + c 2 ) × log e  r + c 2  − (b − a )
a +
(a + b)c 
c


 r − c1 
Page 113 of 133
SECTION 6 – COMBINED STRESSES
r = a + c1 = 1.75 + 1 = 2.75 in
a = b = 0.75 in
b = 2b = 2(0.75) = 1.50 in
c = 3b = 3(0.75) = 2.25 in
Z = −1 +


2(2.75)

 1.50 − 0.75 
 2.75 + 1.25 
(2.75 + 1.25) × log e 
 − (1.50 − 0.75)
0.75 + 
(0.75 + 1.5)(2.25) 
 2.25 
 2.75 − 1 


Z = 0.05627

c1  

1
(1.02832)
1 +
I 1 +
Z (r − c1 )   0.05627(2.75 − 1) 

K ci =
=
= 1.6479
(2.53125)(2.75)(1)
Arc1
M = F (q + c1 ) = F (2.75 + 1) = 3.75F
For Normalized cast steel, A27-58,
sy = 36 ksi
F K ci Mc i
+
A
I
sy 36
F
(1.6479)(3.75)(F )(1)
s= = =
+
N 2 2.53125
1.02832
F = 2.810 kips = 2810 lb
s=
THICK-SHELL CYLINDERS; INTERFERENCE FITS
438.
Special welded steel pipe, equivalent in strength to SAE 1022, as rolled, is subjected to
an internal pressure of 8000 psi. The internal diameter is to be 4 ½ in. and the factor of
safety is to be 3, including an allowance for the weld. (a) Find the thickness of the pipe
according to the distortion-energy theory. (b) Using this thickness find the maximum
normal and shear stresses and the corresponding safety factors. (c) Compute the
thickness from the thin-shell formula and from the Barlow formula.
Solution:
4.5
ri =
= 2.25 in , N = 3 , pi = 8000 psi
2
SAE 1022, as rolled, sy = 52 ksi
(a) Distortion-Energy Theory
1


2 



1
 − 1 in
t = ri 



 1 − 3pi 



s 


sy 52
s = = = 17.333 ksi = 17 ,333 psi
N 3
Page 114 of 133
SECTION 6 – COMBINED STRESSES
1


2 



1
 − 1 = 2.774 in
t = 2.25



 1 − 3 × 8000 



17 ,333 


(b) Maximum normal stress
p r 2 + r 2 − 2p r 2
σ ti = i o 2 i 2 o o
ro − ri
(
)
(
)
pi ro2 + ri2
ro2 − ri 2
ri = 2.25 in
ro = 2.25 + 2.774 = 5.024 in
σ ti =
(
)
8000 5.024 2 + 2.25 2
= 12,014 psi
5.0242 − 2.252
s
52,000
= 4.33
N= y =
σ ti 12,014
Maximum shear stress
r 2 (p − p )
τ = o 2 i 2o
ro − ri
σ ti =
τ=
τ=
ro2 pi
ro2 − ri 2
(5.024)2 (8000)
= 10,007 psi
5.0242 − 2.252
sy
52,000
N=
=
= 2.60
2τ 2(10,007 )
(c) From thin-shell formula
p r (8000)(2.25)
t= i i =
= 1.0385 in
st
17,333
From Barlow formula
pr
p (r + t )
t= i o = i i
st
st
pr
(8000)(2.25)
t= i i =
= 1.929 in
st − pi 17,333 − 8000
439.
The internal diameter of the cast-steel cylinder, SAE 0030, of a hydraulic press is 12 in.
The internal working pressure is 6000 psi, N = 2.5. Find the thickness of the cylinder
walls (a) from the maximum-shear-stress theory, (b) from the octahedral-shear theory.
(c) Compute the thickness from the thin-shell and Barlow formulas. What do you
recommend?
Solution:
Page 115 of 133
SECTION 6 – COMBINED STRESSES
Table AT 6. SAE 0030 = A27-58, sy = 35 ksi
(a) Maximum shear theory
r 2 (p − p ) s
τ = o 2 i 2o = y
ro − ri
2N
12
ri = = 6 in
2
pi = 6,000 psi = 6 ksi
po = 0 ksi
ro2 (6 − 0 )
35
=
2
2
2(2.5)
ro − (6)
ro = 15.8745 in
t = ro − ri = 15.8745 − 6 = 9.8745 in
(b) Octahedral Sheat Theory
12






1
 − 1
t = ri 


3pi 

 1 −

s 


sy 35
= 14 ksi
s= =
N 2.5
12






1



t = (6 )
− 1 = 5.8195 in


3 (6) 

 1 −

14 


(c) Thin shell formula
p r sy
st = i i =
t
N
(6)(6) = 35
t
2.5
t = 2.5714 in
Barlow formula
sy
pr
st = i o =
t
N
6(6 + t ) 35
=
t
2.5
t = 4.5 in
Recommended: Maximum shear theory , t = 9.8745 in thick.
Page 116 of 133
SECTION 6 – COMBINED STRESSES
440.
The same as 439, except a higher-strength material is selected. Try cast-steel SAE 0105.
Solution:
Table AT 6. SAE 0105 = A148-58, sy = 85 ksi
(a) Maximum shear theory
r 2 (p − p ) s
τ = o 2 i 2o = y
ro − ri
2N
12
ri = = 6 in
2
pi = 6,000 psi = 6 ksi
po = 0 ksi
ro2 (6 − 0 )
85
=
2
2
2(2.5)
ro − (6)
ro = 7.459 in
t = ro − ri = 7.459 − 6 = 1.459 in
(b) Octahedral Sheat Theory
12






1
 − 1
t = ri 


3pi 

 1 −

s 


sy 85
s= =
= 34 ksi
N 2.5
12






1
 − 1 = 1.2005 in
t = (6)


3 (6 ) 

 1 −

34 


(c) Thin shell formula
p r sy
st = i i =
t
N
(6)(6) = 85
t
2.5
t = 1.0588 in
Barlow formula
sy
pr
st = i o =
t
N
6(6 + t ) 85
=
t
2.5
t = 1.2857 in
Recommended: Maximum shear theory , t = 1.459 in thick.
Page 117 of 133
SECTION 6 – COMBINED STRESSES
441.
A 2 ½ in. heavy-wall pipe has the following dimensions: OD = 2.875, ID = 1.771, t = 0.552
in.; inside surface area per foot of length = 66.82 in.2, outside surface area per foot of
length = 108.43 in.2. The material is chromium-molybdenum alloy, for which the
permissible tangential tensile stress is 15 ksi at temperatures between 700 – 800 F. (a)
Compute the maximum internal working pressure for this pipe from Lame’s formula, by
the maximum-shear and octahedral-shear theories. (b) What is the stress at an external
fiber? (c) A higher design stress would be permitted for an external pressure alone.
Nevertheless, compute the external pressure corresponding to a maximum tangential
stress of 15 ksi.
Solution:
OD 2.875
ro =
=
= 1.4375 in
2
2
ID 1.771
ri = =
= 0.8855 in
2
2
t = 0.552 in
(a) Lame’s Equation
p r 2 + r 2 − 2p r 2
σ ti = i o 2 i 2 o o = s
ro − ri
(
)
[
]
pi (1.4375)2 + (0.8855)2 − 0
(1.4375)2 − (0.8855)2
pi = 6.7477 ksi
Maximum shear theory
r 2 (p − p ) s
τ = o 2 i 2o =
ro − ri
2
15 =
(1.4375)2 (pi )
15
=
2
2
(1.4375) − (0.8855) 2
pi = 4.654 ksi
Octahedral shear theory
12






1



t = ri
− 1


3pi 

 1 −

s 


12






1



0.552 = (0.8855)
− 1


3pi 

 1 −

15 


pi = 5.374 ksi
(b) Stress at external fiber, pi = 4.654 ksi
σ to =
(
2pi ri 2 − po ro2 + ri2
ro2 − ri2
Page 118 of 133
)
SECTION 6 – COMBINED STRESSES
2(4.654)(0.8855)2 − 0
σ to =
= 5.592 ksi
(1.4375)2 − (0.8855)2
(c) External pressure alone.
p r 2 + r 2 − 2p r 2
σ ti = i o 2 i 2 o o = s
ro − ri
(
)
0 − 2po (1.4375)2
(1.4375)2 − (0.8855)2
po = 4.654 ksi
− 15 =
442.
A cast-steel hub is to be shrunk on a 1.5-in., SAE 1035, as-rolled, steel shaft. The
equivalent diameter of the hub is 2.5 in., its length is 4 in. (a) What must be the
interference of metal if the holding power of this fit is equal to the torsional yield
strength of the shaft? Use Baugher’s recommendations. (b) What are the corresponding
tangential and radial stresses in the hub?
Solution:
Table AT 7, SAE 1035, as rolled, sy = 55 ksi.
sys = 0.6 sy = 33 ksi
Es = 30,000 ksi
µs = 0.3
For hub, Cast steel, Eh = 30,000 ksi, µh ~ 0.3
(a) Interference of metal
For solid shaft, same E and µ.
2
Ei   Di  
1 −   
pi =
2Di   Do  


Di = 1.5 in
Do = 2.5 in
L = 4 in
For pi:
fp πD 2L
T= i i
2
But
πDi3 sys
T=
16
Then
πDi3 sys fpiπDi2L
=
16
f = 0.1 as per Baugher’s recommendation
Di sys (1.5)(33)
pi =
=
= 15.46875 ksi
8 fL 8(0.1)(4 )
Then
Page 119 of 133
SECTION 6 – COMBINED STRESSES
pi =
2
Ei   Di  
1 −   
2Di   Do  


(30,000)(i )   1.5  2 
15.46875 =
 
1 − 
2(1.5)   2.5  
i = 0.002417 in - answer.
(b) Tangential and radial stresses in the hub
Tangential stress
Ei   Di
1 + 
σ th =
2Di   Do

σ th =



2



(30,000)(0.002417 )   1.5  2 
  = 32.87 ksi
1 + 
2(1.5)
  2.5  
Radial stress
σ rh = − pi = −15.46875 ksi
443.
The same as 442, except that the hub is ASTM 20, cast iron. Will the resulting tensile
stresses be safe for cast iron?
Solution:
Table AT 6, ASTM 20, cast iron, suc = 83 ksi, su = 20 ksi (hub)
Table AT 7, SAE 1035, as rolled, sy = 55 ksi.
sys = 0.6 sy = 33 ksi
(a) Interference of metal
For hub of cast iron and the shaft is steel.
  D 2 
Ei 1 −  i  
  Do  
pi =
2

D  
Di 3 + µ + (1 − µ ) i  

 Do  
Di = 1.5 in
Do = 2.5 in
L = 4 in
E = 30,000 ksi
µ = 0.27
For pi:
fp πD 2L
T= i i
2
But
πDi3 sys
T=
16
Page 120 of 133
SECTION 6 – COMBINED STRESSES
Then
πDi3 sys
=
fpiπDi2L
16
f = 0.1 as per Baugher’s recommendation
Di sys (1.5)(33)
pi =
=
= 15.46875 ksi
8 fL 8(0.1)(4 )
Then
  D 2 
Ei 1 −  i  
  Do  
pi =
2

 Di  



Di 3 + µ + (1 − µ )  

 Do  
2


(30,000)(i )1 −  1.5  
  2.5  
15.46875 =
2


(1.5)3 + 0.27 + (1 − 0.27 ) 1.5  
 2.5  

i = 0.004269 in - answer.
(b) Tangential and radial stresses in the hub
Tangential stress
  D 2 
Ei 1 +  i  
  Do  
σ th =
2

D  
Di 3 + µ + (1 − µ ) i  

 Do  
2


(30,000)(0.004269)1 +  1.5  
  2.5   = 32.87 ksi
σ th =
2


(1.5)3 + 0.27 + (1 − 0.27 ) 1.5  
 2.5  

(30,000)(0.002417 )   1.5  2 
σ th =
  = 32.87 ksi
1 + 
2(1.5)
  2.5  
> 20 ksi.
Not safe for cast iron ASTM 20.
Radial stress
σ rh = − pi = −15.46875 ksi
444.
A cast-steel gear is pressed onto a 2-in. shaft made of AISI 3140, OQT 1000 F. The
equivalent hub diameter is 4 in., and the hub length is 4 in. (a) What are the maximum
tangential and radial stresses in the hub caused by a class FN 2 interference fit?
Compute for the apparent maximum value of i (but recall the probability of this event).
Page 121 of 133
SECTION 6 – COMBINED STRESSES
(b) What axial force F in tons will be required to press the gear on the shaft if f1 is
assumed to be 0.2? (c) What torque may the force fit safely transmit? (d) Is the holding
capacity of this fit large enough to transmit a torque that produces a simple torsional
stress of 0.6sys in the shaft?
Solution:
Cast steel, E = 30 x 106 psi, µ = 0.27 or approximately 0.3
AISI 3140, OQT 1000 F, E = 30 x 106 psi, µ = 0.3, sy = 133 ksi (Fig. AF 2).
Di = 2 in, Do = 4 in, L = 4 in.
For Class FN 2 interference fit.
Table 3.2, page 85, 2 in diameter.
Maximum value of i = 0.0027 – 0.0000 = 0.0027 in
(a) For same material and same Poisson’s ratio
Tangential stress
Ei   Di
1 + 
σ th =
2Di   Do




2



(30 ×10 )(0.0027) 1 +  2 
=
6
σ th
2(2)


 = 25,313 psi
 4  
2
Radial stress
2
Ei   Di  
1 −   
σ rh = − pi = −
2Di   Do  


2
6
30 ×10 (0.0027 )   2  
σ rh = −
1 −    = −15,188 psi
2(2)
  4  
(
)
(b) Axial force F in tons.
f p πD L
F = 1 i i tons
2000
(
0.2 )(15,188)(π )(2)(4)
= 38.17 tons
F=
2000
(c) Torque safely transmit.
fp πD 2L
T= i i
2
f = 0.1 as recommended by Baugher.
(0.1)(15,188)(π )(2)2 (4)
T=
= 38,172 in − lb
2
(d) With simple torsional stress of 0.6sys.
ss = 0.6sys = 0.6(0.6sy ) = 0.6(0.6 )(133) = 47.88 ksi = 47 ,880 psi
ssπDi3 (47 ,880)(π )(2)3
=
= 72,210 psi
16
16
No. The holding capacity of this fit is not large enough to transmit a torque that
produces a simple torsional stress of 0.6sys in the shaft.
T=
Page 122 of 133
SECTION 6 – COMBINED STRESSES
445.
The same as 444, except that a class FN 4 fit is investigated and the computation is
made for the average i.
Solution:
Cast steel, E = 30 x 106 psi, µ = 0.27 or approximately 0.3
AISI 3140, OQT 1000 F, E = 30 x 106 psi, µ = 0.3, sy = 133 ksi (Fig. AF 2).
Di = 2 in, Do = 4 in, L = 4 in.
For Class FN 4 interference fit.
Table 3.2, page 85, 2 in diameter.
Maximum value of i = 0.0042 – 0.0000 = 0.0042 in
Minimum value of i = 0.0035 – 0.0012 = 0.0023 in
Average value of i = 0.5 (0.0042 + 0.0023) = 0.00325 in
(a) For same material and same Poisson’s ratio
Tangential stress
Ei   Di
1 + 
σ th =
2Di   Do




2



(30 ×10 )(0.00325) 1+  2 
=
6
σ th
2(2 )


 = 30,469 psi
 4  
2
Radial stress
2
Ei   Di  
1 −   
σ rh = − pi = −
2Di   Do  


2
6
30 ×10 (0.00325)   2  
σ rh = −
1 −    = −18,281 psi
2(2 )
  4  
(
)
(b) Axial force F in tons.
f p πD L
F = 1 i i tons
2000
(
0.2)(18,281)(π )(2 )(4)
= 45.95 tons
F=
2000
(c) Torque safely transmit.
fp πD 2L
T= i i
2
f = 0.1 as recommended by Baugher.
(0.1)(18,281)(π )(2)2 (4)
T=
= 45,945 in − lb
2
(d) With simple torsional stress of 0.6sys.
ss = 0.6sys = 0.6(0.6sy ) = 0.6(0.6 )(133) = 47.88 ksi = 47 ,880 psi
ssπDi3 (47 ,880)(π )(2)3
=
= 72,210 psi
16
16
No. The holding capacity of this fit is not large enough to transmit a torque that
produces a simple torsional stress of 0.6sys in the shaft.
T=
Page 123 of 133
SECTION 6 – COMBINED STRESSES
446.
A No. 217 ball bearing has a bore of 3.3465 in., a width of 1.1024 in., and the inner race
is approximately 3/8 in. thick. This bearing is to be mounted on a solid shaft with i =
0.0014. (a) Calculate the maximum radial and tangential stresses in the race. (b)
Estimate the force required to press the bearing onto the shaft.
Solution:
Di = 3.3465 in, Do = 3.3465 + 2(3/8) = 4.0965 in, i = 0.0014 in.
(a) Maximum radial stress in the race
2
Ei   Di  
1 −   
σ rh = − pi = −
2Di   Do  


2
6
30 ×10 (0.0014)   3.3465  
σ rh = −
  = −2,087 psi
1 − 
2(3.3465)
  4.0965  
Tangential stress
2
Ei   Di  
1 +   
σ th =
2Di   Do  


2
6
30 ×10 (0.0014)   3.3465  
σ th =
1
+
  = 10,463 psi
 
2(3.3465)
  4.0965  
(b) Force required to press the bearing onto the shaft
f p πD L
F = 1 i i tons , use f1 = 0.175 on the average
2000
(0.175)(2,087 )(π )(2)(4 ) = 4.59 tons
F=
2000
(
(
)
)
447.
A steel disk of diameter Do and thickness L = 4 in. is to be pressed onto a 2-in. steel
shaft. The parts are manufactured with class FN 5 fit, but assembled parts are selected
so as to give approximately the average interference. What will be the maximum radial
and tangential stresses in the disk if (a) Do is infinitely large; (b) Do = 10 in.; (c) Do = 4 in.;
(d) Do = 2.5 in.?
Solution:
(a) Maximum radial stress if Do → ∞ .
2
Ei   Di  
1 −   
σ rh = − pi = −
2Di   Do  


Ei
σ rh = − pi = −
2Di
(30 ×10 )(0.005) = −37,500 psi
6
σ rh = −
2(2 )
Maximum tangential stress if Do → ∞ .
σ th =
Ei   Di
1 + 
2Di   Do

Page 124 of 133



2



SECTION 6 – COMBINED STRESSES
σ th =
Ei
2Di
(30 ×10 )(0.005) = 37,500 psi
6
σ th =
2(2 )
(b) Maximum radial stress if Do = 10 in .
2
Ei   Di  
1 −   
σ rh = − pi = −
2Di   Do  


2
30 ×10 6 (0.005)   2  
σ rh = −
1 −    = −36,000 psi
2(2 )
  10  
Maximum tangential stress if Do = 10 in .
(
)
2
Ei   Di  
1 +   
σ th =
2Di   Do  


6
30 ×10 (0.005)   2  
σ th =
1 +  10   = 39,000 psi
2(2 )
  
(c) Maximum radial stress if Do = 2.5 in .
(
)
2
Ei   Di  
1 −   
σ rh = − pi = −
2Di   Do  


2
30 ×106 (0.005)   2  
σ rh = −
  = −13,500 psi
1 − 
2(2 )
  2.5  
Maximum tangential stress if Do = 2.5 in .
(
)
2
Ei   Di  
1 +   
σ th =
2Di   Do  


6
30 ×10 (0.005)   2 
σ th =
1 +  2.5  = 61,500 psi
2(2 )

 
(
448.
)
A steel cylinder is to have an inside diameter of 3 in. and pi = 30,000 psi. (a) Calculate
the tangential stresses at the inner and outer surfaces if the outside diameter is 6 in. (b)
It was decided to make the cylinder in two parts, the inner cylinder with D1 = 3 in. and Di
= 4.5 in., the outer cylinder with Di = 4.5 in. and Do = 6 in. (see figure). The two cylinders
were shrunk together with i = 0.003 in. Calculate the pressure at the interface and the
tangential stresses at the inner and outer surfaces of each cylinder. (Suggestion: first
derive an equation for the interface pressure).
Page 125 of 133
SECTION 6 – COMBINED STRESSES
Solution:
(a) Tangential stresses at the inner and outer surface.
Di = 3 in, ri = 1.5 in, pi = 30,000 psi
Do = 6 in, ro = 3 in, po = 0
p r 2 + r 2 − 2p r 2
σ ti = i o 2 i 2 o o
ro − ri
(
)
σ ti =
(30,000)[(3)2 + (1.5)2 ]− 0
= 50,000 ksi
(3)2 − (1.5)2
σ to =
2pi ri 2 − po ro2 + ri2
ro2 − ri2
(
)
2(30,000 )(1.5)2 − 0
= 20,000 ksi
(3)2 − (1.5)2
(b) Pressure at the interface, tangential stresses at the inner and outer surface of each
cylinder.
 σ + µ h pi σ ts + µ s pi 

i = 2( δ h + δ s ) = Di  th
−

E
E
h
s


σ to =
Eh = E s , µh = µs
(
)
σ th =
pi ro2 + ri2
ro2 − ri 2
σ ts =
2 p1r12 − pi ri2 + r12
ri 2 − r12
(
)
µp σ
µp  D
σ
i = Di  th + i − ts − i  = i (σ th − σ ts )
E
E
E  E
 E
D p r2 +r2 p r2 +r2
2p r 2 
i = i  i 2o 2i + i 2i 21 − 2 1 1 2 
E  ro − ri
ri − r1
ri − r1 
p1 = 30,000 psi, ro = 3 in, ri = 2.25 in, r1 = 1.5 in
(
Page 126 of 133
)
(
)
SECTION 6 – COMBINED STRESSES
Pressure at the interface, pi.
Ei 2 p1r12
+
Di ri2 − r12
pi = 2 2
ro + ri
r2 +r2
+ i2 12
2
2
ro − ri
ri − r1
(30 ×10 )(0.003) + 2(30,000)(1.5)
2
6
4.5
(2.25)2 − (1.5)2 = 20,000 + 48,000
3.571429 + 2.6
(3)2 + (2.25)2 + (2.25)2 + (1.5)2
2
2
2
2
(3) − (2.25) (2.25) − (1.5)
pi = 11,018.5 psi
Tangential stresses:
Inner cylinder:
Inner surface:
p r 2 + r 2 − 2p r 2
σ ti = 1 i 2 1 2 i i
ri − r1
pi =
(
)
[
]
30,000 (2.25)2 + (1.5)2 − 2(11,018.5)(2.25)2
= 38,333.4 psi
(2.25)2 − (1.5)2
Outer surface:
2p r 2 − p r 2 + r 2
σ to = 1 1 2 i 2i 1
ri − r1
σ ti =
(
σ to =
)
[
]
2(30,000)(2.25)2 − 11,018.5 (2.25)2 + (1.5)2
= 79,351.9 psi
(2.25)2 − (1.5)2
Outer cylinder:
Inner surface:
p r 2 + r 2 − 2p r 2
σ ti = i o 2 i 2 o o
ro − ri
(
)
[
]
11,018.5 (3)2 + (2.25)2 − 2(0 )(3)2
= 39,351.8 psi
(3)2 − (2.25)2
Outer surface:
2p r 2 − p r 2 + r 2
σ to = i i 2 o 2o i
ro − ri
σ ti =
(
σ to =
449.
)
[
]
2(11,018.5)(2.25)2 − 0 (3)2 + (2.25)2
= 28,333.3 psi
(3)2 − (2.25)2
A phosphor-bronze (B139C) bushing has an ID = ¾ in., an OD = 1 ¼ in., and a length of 2
in. It is to be pressed into a cast-steel cylinder that has an outside diameter of 2 ½ in. An
ASA class FN 2 fit is to be used with selective assembly to give approximately the
interference i = 0.0016 in. Calculate (a) pi, (b) the maximum tangential stress in the steel
Page 127 of 133
SECTION 6 – COMBINED STRESSES
cylinder, (c) the force required to press bushing into the cylinder, (d) the decrease of the
inside diameter of the bushing.
Solution:
Phosphor Bronze B139C, Es = 16 x 106 psi (Table AT3), µs = 0.36 (other reference).
Cast steel, Eh = 30 x 106 psi , µh = 0.27 (Table AT 6)
 σ + µ h pi σ ts + µ s pi 

i = 2( δ h + δ s ) = Di  th
−

Eh
Es


(
)
σ th =
pi ro2 + ri2
ro2 − ri 2
σ ts =
− pi ri2 + r12
ri2 − r12
(
)
σ
µ p σ
µp 
i = Di  th + h i − ts − s i 
Eh
Es
Es 
 Eh
 p r2 +r2 µ p p r2 + r2 µ p 
i = Di  i o2 i 2 + h i + i i 2 12 − s i 
Eh
Es 
E s ri − r1
 E h ro − ri
(a) pi
i
Di
pi =
2
2
ro + ri
ri2 + r12
µ µ
+
+ h− s
2
2
2
2
E h ro − ri
E s ri − r1
Eh Es
(
(
)
)
(
(
(
)
2.25
= 1.125 in
2
1.25
ri =
= 0.625 in
2
0.75
r1 =
= 0.375 in
2
ro =
Page 128 of 133
(
)
)
)
SECTION 6 – COMBINED STRESSES
L = 2 in
Di = 1.25 in
pi =
(1.125)2 + (0.625)2
30 ×10 6 [(1.125)2 − (0.625)2 ]
0.0016
1.25
(
0.625)2 + (0.375)2
0.27
0.36
+
+
−
2
2
6
6
30 ×10 16 ×10 6
16 ×10 (0.625) − (0.375)
[
]
−3
1.28 ×10
6.309524 ×10 + 1.328125 ×10 −7 + 0.9 ×10 −8 − 2.25 ×10−8
pi = 7,017 psi
(b) Maximum tangential stress in the steel cylinder.
p r2 + r2
σ th = i 2o 2i
ro − ri
pi =
−8
(
)
(7,017 )[(1.125)2 + (0.625)2 ]
= 13,282 psi
(1.125)2 − (0.625)2
σ th =
(c) F
f1piπDi L
tons , use f1 = 0.175 on the average
2000
(0.175)(7,017 )(π )(1.25)(4) = 4.82 tons
F=
2000
(d) Decrease of the inside diameter of the bushing. The bushing is phosphor bronze.
Subscript is “s” as in shaft.
 σ + µ s pi 

δ s = − ts

Es


F=
(
)
σ ts =
− pi ri2 + r12
ri2 − r12
σ ts =
− 7,017 (0.625)2 + (0.375)2
= −14,911 psi
(0.625)2 − (0.375)2
[
]
 − 14,911 + 0.36(7,017 ) 
 = 0.000774 in
16 ×10 6


δ s = −
DESIGN PROJECTS
DESIGN PROJECTS
450.
A jib crane similar to the one shown is to be designed for a capacity of F = ___ (say, 1 to
3 tons). The load F can be swung through 360o; L ≈ 10 ft., b ≈ 8.5 ft., c ≈ 2 ft. The
moment on the jib is balanced by a couple QQ on the post, the forces Q acting at
supporting bearings. The crane will be fastened to the floor by 6 equally spaced bolts on
a D1 = 30-in. bolt circle; outside diameter of base D2 = 36 in. (a) Choose a pipe size
(handbooks) for the column such that the maximum equivalent stress does not exceed
12 ksi. (b) Choose an I-beam for the jib such that the maximum stress does not exceed
12 ksi. (c) Compute the maximum external load on a base bolt and decide upon the size.
Page 129 of 133
SECTION 6 – COMBINED STRESSES
(d) Complete other details as required by the instructor, such as: computing Q and
choosing bearings (ball or roller?), the internal construction and assembly in this vicinity,
detail sketches giving full information.
451.
Design an air-operated punch press similar to the one shown. Let the force at the punch
be 12 tons, (or other capacity as specified by the instructor), the depth of throat to the
inside edge of the frame be 25 in., the diameter and stroke of the piston about 8 in. by 8
in., the mechanical advantage of the lever about 7, and the diameter of the punch 3/16
in. Determine first the horizontal section of the frame, and locate and design the
cylinder. Then determine the relative arrangement of the various links and make a force
analysis, from which the design of certain parts follows. Determine the actual distance
of movement of the punch (not less than about 1 in.). The illustration will assist the
student in settling upon the proportions of parts for which strength calculations cannot
be made.
Page 130 of 133
SECTION 6 – COMBINED STRESSES
452.
Design a screw press similar to that shown for a load of ___ (say, 3) tons on the screw.
The depth of the throat g is to be ___ (10) in. and the height of the throat h is to be ___
(15) in. (The instructor will assign the data.) The order of procedure may be as follows:
(a) Find the diameter of the screw. If Le/k > 40, check as a column. If the top of the screw
is squared off for a handwheel or handle, check this section for twisting. The equation
for pivot friction, if desired, is in §18.10, Text. (b) Decide upon the diameter of the
handwheel or the length of handle (if one is needed to obtain the maximum pressure),
assuming that the maximum force to be exerted by a man is about 150 lb. Dimensions
of handwheels may be found in handbooks. The handle may be designed by the flexure
formula. (c) Design the frame. The shape of the section of the frame will depend on the
material used. A T-section is suitable for cast iron (say N = 6 on the ultimate strength), a
hollow box or modified I-section is suitable for cast steel. The 45o section CD of the
frame should be safe as a curved beam. See Table AT 18. In this connection, it will be
well to make the radius r as large as practicable, since the larger r the less the stresses
from a given load. Compute the dimensions of the vertical section. It is a good plan to
keep t and a the same in all sections. (d) Design the bushing if one is used. The height b
depends upon the number of threads in contact, which in turn depends upon the
bearing pressure used in design. (Say half-hard yellow brass?) Compute the outside
Page 131 of 133
SECTION 6 – COMBINED STRESSES
bushing diameter, the diameter and thickness of the collar, and decide upon dimensions
to be used. (e) Fix the location and number of bolts to be used to fasten the frame to
the base plate, and determine their size. Use a common bolt material. (f) Decide upon
all other details of design. Make a separate sketch of each part of the machine showing
thereon all dimensions necessary for manufacture. It is suggested that, first, all
materials be tentatively decided upon, after which design stresses may be chosen. See
that design stresses for the various parts bear a logical relation to one another. It is not
necessary to follow this procedure in detail. It is likely that one will have to leave certain
details unfinished from time to time, because these details depend on parts of the
design not yet completed. Make sure that all parts can be assembled after they are
made. Notice that the plate on the lower end of the screw must be connected in such a
manner that the screw may turn while the plate does not.
453.
Design a jib crane, as suggested by the illustration, to lift a load of W of ___ tons. The
maximum radius of swing is to be about ___ ft. (The instructor will assign data).
Suggested procedure: (a) From catalogues, select a hoist to suit the purpose, giving
reasons for your choice, and noting significant dimensions. Of course, in the end, the
hoist trolley has to match the size of I-beam used. (b) Let the angle that the diagonal
tension rod makes with the horizontal be about 20o to 25o, and decide upon the
dimensions H and L. Note that the point G does not necessarily have to be at the
extreme position of the load. As a matter of fact, some advantage may result from
having G inside the outermost position of the load. Make the force analyses (including
weight of hoist as part of load) for (1) the condition of maximum column action, (2) the
condition of maximum bending moment on the beam, and (3) the condition for
maximum force on the hinge B (to be used for the design of this hinge). (c) Find the size
of I-beam such that the maximum stress for any position of the load falls within the
limits of 12 and 15 ksi, usually by assuming a standard beam and checking the stress.
According to the arrangement of parts, it may be necessary to design the connection at
G between the rod and the beam first. With the details of this connection known and
Page 132 of 133
SECTION 6 – COMBINED STRESSES
with the details of beam assumed, the location of point G, the point of application of the
force T, can be determined. The bending moment of a section a minute distance to the
right of G is W(x – dx). A minute distance to the left of G, the bending moment is W(x +
dx) – Txe – Tydx; that is, the moment changes suddenly at G by the amount Txe. (d)
Determine the size of diagonal support, including details of connections. (e) Design the
connections at each end of the diagonal and the hinge at C. Settle upon the details
including the method of attaching the hinge to the vertical surface, which may be wideflange beam. (f) Design the hinge at B and the connection to the I-beam; also the details
of the method of attaching the hinge to the vertical surface. Where material is not
specified, make your choice clear. There should be no doubt as to your design stresses
or design factor. Show a neat large sketch, fully dimensioned, of each part separately. It
is unlikely that too much detail will be shown.
-
E
n
d
-
Page 133 of 133
SECTION 7 – SHAFT DESIGN
471.
A short stub shaft, made of SAE 1035, as rolled, receivers 30 hp at 300 rpm via a
12-in. spur gear, the power being delivered to another shaft through a flexible
coupling. The gear is keyed (profile keyway) midway between the bearings. The
pressure angle of the gear teeth φ = 20 o ; N = 1.5 based on the octahedral shear
stress theory with varying stresses. (a) Neglecting the radial component R of the
tooth load W , determine the shaft diameter. (b) Considering both the tangential
and the radial components, compute the shaft diameters. (c) Is the difference in
the results of the parts (a) and (b) enough to change your choice of the shaft size?
Problem 471.
Solution:
For SAE 1035, as rolled
s y = 55 ksi
su = 85 ksi
sn = 0.5su = 0.5(85) = 42.5 ksi
A = W cos φ
63,000hp 63,000(30 )
T=
=
= 6300 in − lb
n
300
AD
T=
2
A(12 )
6300 =
2
A = 1050 lb
A = W cos φ
1050 = W cos 20
W = 1118 lb
Shear stress
16T 16(6300 )
ss =
=
π d3
π d3
100,800
ss = sms =
π d3
sas = 0
Page 1 of 76
SECTION 7 – SHAFT DESIGN
bending stress
From Table AT 2
FL
M=
4
(a) Negligible R :
AL (1050 )(16 )
M=
=
= 4200 in − lb
4
4
32 M 32(4200 ) 134,400
s=
=
=
π d3
π d3
π d3
sm = 0
134,400
sa = s =
π d3
K f sa
s
se = n s m +
sy
SF
For profile keyway
K f = 2 .0
K fs = 1.6
SF = 0.85
K s
(2.0)(134,400) = 100,661
se = f a =
SF
(0.85) π d 3
d3
K fs sas
s
ses = ns sms +
s ys
SF
(
)
sns sn 42.5
1
= =
=
s ys s y
55 1.294
sns
 1  100,800  24,796
 =
sms = 

3
s ys
d3
 1.294  π d 
Octahedral-shear theory
ses =
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


2
1  100,661  
24,796
= 
 + 
3
1.5  42,500d   0.577 42,500d 3
d = 1.569 in
(
Page 2 of 76
1
)



2
2


SECTION 7 – SHAFT DESIGN
use d = 1
11
in
16
(b) Considering both radial and tangential component.
WL (1118)(16 )
M=
=
= 4472 in − lb
4
4
32 M 32(4472 ) 143,104
s=
=
=
π d3
π d3
π d3
sm = 0
143,104
sa = s =
π d3
K s
(2.0)(143,104) = 107,180
se = f a =
SF
(0.85)(π d 3 )
d3
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


2
1  107,180  
24,796
= 
 + 
3
1.5  42,500d   0.577 42,500d 3
d = 1.597 in
11
use d = 1 in
16
(
1
)



2
2


(c) The difference in the results of the parts (a) and (b) is not enough to change the choice
of the shaft size.
472.
A cold-finished shaft, AISI 1141, is to transmit power that varies from 200 to 100
and back to 200 hp in each revolution at a speed of 600 rpm. The power is
received by a 20-in. spur gear A and delivered by a 10-in. spur gear C. The
tangential forces have each been converted into a force ( A and C shown) and a
couple (not shown). The radial component R of the tooth load is to be ignored in
the initial design. Let 2 and, considering varying stresses with the maximum
shear theory, compute the shaft diameter.
Problems 472 – 474
Page 3 of 76
SECTION 7 – SHAFT DESIGN
Solution:
For AISI 1141, cold-finished
s y = 90 ksi
sn = 50 ksi
sn
1
=
s y 1.8
SF = 0.85
63,000hp
T=
n
63,000(200 )
Tmax =
= 21,000 in − lb
600
63,000(100 )
Tmin =
= 10,500 in − lb
600
1
1
Tm = (Tmax + Tmin ) = (21,000 + 10,500 ) = 15,750 in − lb
2
2
1
1
Ta = (Tmax − Tmin ) = (21,000 − 10,500 ) = 5,250 in − lb
2
2
16T
ss =
πd3
16(15,750 ) 252,000
sms =
=
π d3
π d3
16(5250 ) 24,000
=
sas =
π d3
π d3
K fs sas
s
ses = ns sms +
s ys
SF
For profile keyway
K f = 2 .0
K fs = 1.6
sns sn
1
= =
s ys s y 1.8
 1  252,000  (1.6)(84,000) 94,894
 +
ses =  
=
3
0.85π d 3
d3
 1.8  π d 
Bending stress, negligible radial load
T = 21,000 in − lb at 200 hp
For A:
 20 
A  = T
 2 
A(10 ) = 21,000
Page 4 of 76
SECTION 7 – SHAFT DESIGN
A = 2100 lb at 200 hp
For C:
 10 
C  = T
2
C (5) = 21,000
C = 4200 lb at 200 hp
[∑ M
]
= 0 A(10 ) + D(25) = C (15)
at 200 hp
(2100)(10) + D(25) = (4200)(15)
D = 1680 lb
[∑ F
B
]
=0
A+C = B + D
at 200 hp
2100 + 4200 = B + 1680
B = 4620 lb
At 200 hp: A = 2100 lb , B = 4620 lb , C = 4200 lb , D = 1680 lb
Shear Diagram
V
Maximum moment at B
M = (2100)(10 ) = 21,000 in − lb
32 M 32(21,000 ) 672,000
s=
=
=
π d3
π d3
π d3
sm = 0
672,000
sa = s =
π d3
K f sa
(2.0)(672,000) = 503,304
s
se = n s m +
= 0+
sy
SF
0.85π d 3
d3
Page 5 of 76
SECTION 7 – SHAFT DESIGN
94,894
d3
Maximum Shear Theory
ses =
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.5sn  


2
1  503,304  
94,894
= 
 + 
3
2  50,000d   0.5 50,000d 3
d = 2.78 in
3
use d = 2 in
4
(
475.
1
)



2
2


A shaft S, of cold-drawn AISI 1137, is to transmit power received from shaft W,
which turns at 2000 rpm, through the 5-in. gear E and 15-in. gear A. The power
is delivered by the 10-in. gear C to gear G, and it varies from 10 hp to 100 hp and
back to 10 hp during each revolution of S. The design is to account for the
varying stresses, with calculations based on the octahedral shear stress theory.
Let N = 1.8 and compute the shaft diameter, using only the tangential driving
loads for the first design.
Problem 475 – 477
Solution.
For AISI 1137, cold drawn
s y = 93 ksi
su = 103 ksi
sn = 0.5su = 0.5(103) = 51.5 ksi
sn sns 51.5
1
=
=
=
s y s ys
93 1.806
63,000hp
T=
n
5 in. E
n=
(2000 rpm) = 667 rpm
15 in. A
Page 6 of 76
SECTION 7 – SHAFT DESIGN
63,000(100 )
= 9450 in − lb
667
63,000(10 )
Tmin =
= 945 in − lb
667
1
1
Tm = (Tmax + Tmin ) = (9450 + 945) = 5197.5 in − lb
2
2
1
1
Ta = (Tmax − Tmin ) = (9450 − 945) = 4252.5 in − lb
2
2
16T
ss =
πd3
16(5197.5) 83,160
sms =
=
π d3
π d3
16(4252.5) 68,040
=
sas =
π d3
π d3
K fs sas
s
ses = ns sms +
s ys
SF
For profile keyway
K f = 2 .0
Tmax =
K fs = 1.6
SF = 0.85
 1  83,160  (1.6)(68,040) 55,425
+
ses = 
=

3 
0.85π d 3
d3
 1.806  π d 
Bending stress, using only tangential loads
For 100 hp:
T = 9450 in − lb
 15 
A  = T
2
A(7.5) = 9450
A = 1260 lb
For C:
 10 
C  = T
2
Page 7 of 76
SECTION 7 – SHAFT DESIGN
C (5) = 9450
C = 1890 lb
[∑ M
B
]
= 0 6 A + 20 D = 14C
6(1260) + 20 D = 14(1890)
D = 945 lb
[∑ F
]
=0
A+C = B + D
1260 + 1890 = B + 945
B = 2205 lb
Shear diagram
V
Maximum moment at B
M = (1260)(6) = 7560 in − lb
32 M 32(7560 ) 241,920
s=
=
=
π d3
π d3
π d3
sm = 0
241,920
sa = s =
π d3
K f sa (2.0 )(241,920 ) 181,189
s
se = n s m +
=
=
sy
SF
0.85π d 3
d3
55,425
d3
Octahedral Shear Theory
ses =
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


2
1  181,189  
55,425
= 
 + 
3
2  51,500d   0.577 51,500d 3
(
d = 1.997 in
use d = 2 in
Page 8 of 76
1
)



2
2


SECTION 7 – SHAFT DESIGN
478.
A shaft made of AISI 1137, cold rolled, for a forage harvester is shown.
Power is supplied to the shaft by a vertical flat belt on the pulley A. At B, the
roller chain to the cutter exerts a force vertically upwards, and the V-belt to
the blower at C exerts a force vertically upwards. At maximum operating
conditions, the flat belt supplies 35 hp at 425 rpm, of which 25 hp is delivered
to the cutter and 10 hp to the blower. The two sections of the shaft are joined
by a flexible coupling at D and the various wheels are keyed (sled-runner
keyway) to the shafts. Allowing for the varying stresses on the basis of the
von Mises-Hencky theory of failure, decide upon the diameters of the shafts.
Choose a design factor that would include an allowance for rough loading.
Problem 478.
Solution:
For AISI 1137, cold rolled
s y = 93 ksi
su = 103 ksi
sn = 0.5su = 0.5(103) = 51.5 ksi
sn sns 51.5
1
=
=
=
s y s ys
93 1.806
Pulley,
63,000hp 63,000(35)
TA =
=
= 5188 in − lb
n
425
For flat-belt
 2T  4(5188)
FA = F1 + F2 = 2(F2 − F1 ) = 2 A  =
= 692 lb
30
 DA 
Sprocket,
63,000hp 63,000(25)
TB =
=
= 3706 in − lb
n
425
For chain,
2T
2(3706 )
FB = B =
= 741 lb
DB
10
Sheave,
63,000hp 63,000(10 )
TC =
=
= 1482 in − lb
n
425
Page 9 of 76
SECTION 7 – SHAFT DESIGN
For V-belt,
 2T
FC = F1 + F2 = 1.5(F2 − F1 ) = 1.5 C
 DC
Consider shaft ABD.
35 hp
Shaft ABD
[∑ M
D'
=0
]
(6 + 8 + 4)FA = (8 + 4)A'+4FB
18(692) = 12 A'+4(741)
A' = 791 lb
[∑ F
V
=0
]
FA + D′ = FB + A′
692 + D′ = 741 + 791
D′ = 840 lb
Shear Diagram
Maximum M at A’.
M = (6)(692) = 4152 in − lb.
32 M 32(4152 ) 132,864
s=
=
=
π d3
π d3
π d3
sm = 0
sa = s =
132,864
π d3
Page 10 of 76
 3(1482 )
 =
= 445 lb
10

SECTION 7 – SHAFT DESIGN
K f sa
sn
sm +
sy
SF
For sled-runner keyway (Table AT 13)
K f = 1 .6
se =
K fs = 1.6
SF = 0.85
K f sa
(1.60)(132,864) = 79,610
s
se = n s m +
= 0+
sy
SF
0.85π d 3
d3
at A’ T = TA = 5188 in − lb
16T 16(5188) 83,008
ss =
=
=
π d3
π d3
π d3
sms = ss
sas = 0
K fs sas
s
ses = ns sms +
s ys
SF
 1  83,000  14,630
=
ses = 

3 
d3
 1.806  π d 
Choose a design factor of 2.0
N = 2 .0
von Mises-Hencky theory of failure (Octahedral shear theory)
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


2
1  79,610  
14,630
= 
 + 
3
2  51,500d   0.577 51,500d 3
d = 1.48 in
1
use d = 1 in
2
(
1
)



2
2


Consider shaft D-C
63,000hp 63,000(10 )
TC =
=
= 1482 in − lb
n
425
For V-belt,
 2T  3(1482 )
FC = F1 + F2 = 1.5(F2 − F1 ) = 1.5 C  =
= 445 lb
10
 DC 
Page 11 of 76
SECTION 7 – SHAFT DESIGN
[∑ M
C'
=0
]
8 D′′ = 3FC
8 D′′ = 3(445)
D′′ = 167 lb
[∑ F
V
=0
]
C ′ = D′′ + FC
C ′ = 167 + 445
C ′ = 612 lb
Shear Diagram
M = (167 )(8) = 1336 in − lb
32 M 32(1336 ) 42,752
s=
=
=
π d3
π d3
π d3
sm = 0 , sa = s
K f sa
(1.60)(42,752) = 25,616
s
se = n s m +
= 0+
sy
SF
0.85π d 3
d3
at C’, TC = 1482 in − lb
16T 16(1482 ) 23,712
ss =
=
=
π d3
π d3
π d3
sms = ss
sas = 0
K fs sas
s
ses = ns sms +
s ys
SF
4180
 1  23,712 
+0 = 3
ses = 

3 
d
 1.806  π d 
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


2
1  25,616  
4180
= 
 + 
3
2  51,500d   0.577 51,500d 3
(
Page 12 of 76
1
)



2
2


SECTION 7 – SHAFT DESIGN
d = 1.011 in
use d = 1 in
479.
A shaft for a punch press is supported by bearings D and E (with L = 24 in.)
and receives 25 hp while rotating at 250 rpm, from a flat-belt drive on a 44in. pulley at B, the belt being at 45o with the vertical. An 8-in. gear at A
delivers the power horizontally to the right for punching operation. A 1500-lb
flywheel at C has a radius of gyration of 18 in. During punching, the shaft
slows and energy for punching comes from the loss of kinetic energy of the
flywheel in addition to the 25 hp constantly received via the belt. A
reasonable assumption for design purposes would be that the power to A
doubles during punching, 25 hp from the belt, 25 hp from the flywheel. The
phase relations are such that a particular point in the section where the
maximum moment occurs is subjected to alternating tension and
compression. Sled-runner keyways are used for A, B, and C; material is colddrawn AISI 1137, use a design factor of N = 2.5 with the octahedral shear
theory and account for the varying stresses. Determine the shaft diameters.
Problems 479-480
Solution:
Flat-Belt Drive (B)
63,000hp 63,000(25)
TB =
=
= 6300 in − lb
n
250
 2T  4(6300)
FB = F1 + F2 = 2(F1 − F2 ) = 2 B  =
= 573 lb
44
 DB 
Gear A, Doubled hp
63,000hp 63,000(25 + 25)
TA =
=
= 12,600 in − lb
n
250
2T
2(12,600 )
FA = A =
= 3150 lb
DA
8
Loading:
Page 13 of 76
SECTION 7 – SHAFT DESIGN
Vertical:
BV = FB cos 45 = 573 cos 45 = 405 lb
[∑ M
D
=0
]
6(1500) + 8BV = 24 EV
6(1500) + 8(405) = 24 EV
EV = 510 lb
[∑ F
V
=0
]
1500 + EV = DV + BV
1500 + 510 = DV + 405
DV = 1605 lb
Shear Diagram
Page 14 of 76
SECTION 7 – SHAFT DESIGN
M DV = (6 )(1500 ) = 9000 in − lb
M BV = (16 )(510 ) = 8160 in − lb
M AV = (5)(510 ) = 2550 in − lb
Horizontal:
Bh = FB sin 45 = 573 sin 45 = 405 lb
[∑ M
D
=0
]
8 Bh + 24 Eh = 19 FA
8(405) + 24 Eh = 19(3150)
Eh = 2359 lb
[∑ F
h
=0
]
Dh + Bh + Eh = FA
Dh + 405 + 2359 = 3150
Dh = 386 lb
Shear Diagram
Page 15 of 76
SECTION 7 – SHAFT DESIGN
M Dh = 0 in − lb
M Bh = (8)(386 ) = 3088 in − lb
M Ah = (5)(2359 ) = 11,795 in − lb
MA =
MB =
(M ) + (M )
(M ) + (M )
2
Ah
2
Bh
2
=
(11,795)2 + (2550)2
2
=
(3088)2 + (8160)2
AV
BV
= 12,068 in − lb
= 8725 in − lb
M D = 9000 in − lb
Therefore
M max = 12,068 in − lb
32 M 32(12,068) 386,176
s=
=
=
π d3
π d3
π d3
Maximum moment subjected to alternating tension and compression
sm = 0
386,176
sa = s =
π d3
K f sa
sn
sm +
sy
SF
For AISI 1137, cold-drawn,
s y = 93 ksi
se =
su = 103 ksi
sn = 0.5su = 0.5(103) = 51.5 ksi
For sled-runner keyway (Table AT 13)
K f = 1 .6
K fs = 1.6
SF = 0.85
(1.60)(386,176 ) = 231,386
se = 0 +
0.85π d 3
d3
Page 16 of 76
SECTION 7 – SHAFT DESIGN
At A, 50 hp max. and 25 hp min.
50 hp
63,000hp 63,000(25 + 25)
TA =
=
= 12,600 in − lb
n
250
2T
2(12,600 )
FA = A =
= 3150 lb
DA
8
16T 16(12,600 ) 201,600
ss max =
=
=
π d3
π d3
π d3
25 hp
63,000hp 63,000(25)
TA =
=
= 6,300 in − lb
n
250
2T
2(6,300 )
FA = A =
= 1575 lb
DA
8
16T 16(6,300 ) 100,800
ss min =
=
=
π d3
π d3
π d3
sms =

1
,800  151,200
 =
(ss max + ss min ) = 1  201,600 + 100
3
2
2
πd
π d3

sas =

1
,800  50,400
 =
(ss max − ss min ) = 1  201,600 − 100
3
2
2
πd
π d3

ses =
K fs sas
sns
sms +
s ys
SF
 1  151,200  (1.6)(50,400) 56,848
 +
ses = 
=

3
0.85π d 3
d3
 1.806  π d 
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


2
1  231,386  
56,848
= 
 + 
3
2  51,500d   0.577 51,500d 3
d = 2.14 in
3
say d = 2 in
16
(
THRUST LOADS
Page 17 of 76
1
)



2
2


SECTION 7 – SHAFT DESIGN
481.
A cold-drawn monel propeller shaft for a launch is to transmit 400 hp at 1500
rpm without being subjected to a significant bending moment; and Le k < 40 .
The efficiency of the propeller is 70 % at 30 knots (1.152 mph/knot). Consider
that the number of repetitions of the maximum power at the given speed is 2x
105. Let N = 2 based on the maximum shear theory with varying stress.
Compute the shaft diameter.
Solution:
For cold-drawn monel shaft, Table AT 10
s y = 75 ksi
sn = 42 ksi at 108
at 2 x 105
0.085
 108 

sn ≈ 42
= 71.23 ksi
5 
2
×
10


sn sns 71.23
1
=
=
=
s y s ys
75
1.053
63,000hp 63,000(400 )
=
= 16,800 in − lb
n
1500
16T 16(16,800 ) 268,800
ss =
=
=
π D3
π D3
π D3
s ms = ss
T=
s as = 0
K fs sas
sns
sms +
s ys
SF
SF = 0.85
assume K f = K fs = 1.0
ses =
81,255
 1  268,800 
 + 0 =
ses = 

3
D3
 1.053  π D 
Fvm
= ηhp
33,000
vm = (30 knots )(1.152 mph knot )(5280 ft mi )(1 hr 60 min ) = 3041 fpm
F (3041)
= (0.70)(400)
33,000
F = 3040 lb
4F
4(3040 ) 12,160
s=
=
=
2
πD
π D2
π D2
sm= s
sa = 0
Page 18 of 76
SECTION 7 – SHAFT DESIGN
se =
K f sa
sn
sm +
sy
SF
3676
 1  12,160 
+0 = 2
se = 

2 
D
 1.053  π D 
Maximum Shear Theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.5sn  


2
1  3676  
81,255
= 
 + 
2
2  71,230 D   0.5 71,230 D 3
(
1
)



2
2


1
2
2
1 
1
  2.2815   2
= 
 +
 
2  19.377 D 2   D 3  
By trial and error
11
D = 1.66 in = 1 in
16
482.
A shaft receives 300 hp while rotating at 600 rpm, through a pair of bevel gears,
and it delivers this power via a flexible coupling at the other end. The shaft is
designed with the average forces ( at the midpoint of the bevel-gear face); the
tangential driving force is F , G = 580 lb , Q = 926 lb ; which are the rectangular
components of the total reaction between the teeth; Dm = 24 in , L = 36 in ,
a = 10 in . Let the material be AISI C1045, cold drawn; N = 2 . Considering
varying stresses and using the octahedral shear theory, determine the shaft
diameter.
Problems 482, 485, 486.
Solution:
For AISI C1045, cold drawn
s y = 85 ksi
su = 100 ksi
Page 19 of 76
SECTION 7 – SHAFT DESIGN
sn = 0.5su = 0.5(100) = 50 ksi
SF = 0.85
sn sns 50 1
=
=
=
s y s ys 85 1.7
63,000hp 63,000(300 )
T=
=
= 31,500 in − lb
600
600
16T 16(31,500 ) 504,000
ss =
=
=
π D3
π D3
π D3
s ms = ss
s as = 0
K fs sas
s
ses = ns sms +
s ys
SF
94,370
 1  504,000 
 + 0 =
ses = 

3
D3
 1.7  π D 
D 
F m  = T
 2 
 24 
F   = 31,500
 2 
F = 2625 lb
Vertical:
D 
 24 
Q m  = 926  = 11,112 in − lb
 2 
 2 
G = 580 lb
[∑ M
B
]
=0 − Av (36 ) +
QDm
− G (10 ) = 0
2
QDm
= G (10 ) + Av (36 )
2
11,112 = 580(10) + Av (36)
Av = 148 lb
Page 20 of 76
SECTION 7 – SHAFT DESIGN
[∑ F =0]
v
Av + Bv = 580 lb
148 + Bv = 580 lb
Bv = 432 lb
Shear Diagram
Moment Diagram
M Cv = 11,112 in − lb
M Bv = 5328 in − lb
Horizontal:
[∑ M
B
=0
]
Ah (36) = (2625)(10)
Ah = 729 lb
[∑ F
h
=0
]
Bh = Ah + F
Bh = 725 + 2625
Page 21 of 76
SECTION 7 – SHAFT DESIGN
Bh = 3354 lb
Shear Diagram
M Ch = 0
M Bh = (36 )(729 ) = 26,244 in − lb
Maximum M
M = MB =
(M ) + (M )
2
Bh
(26,244)2 + (5328)2 = 26,780 in − lb
32(26,780 ) 4(926 ) 856,960 3,704
=
+
=
+
2
BV
=
32 M
4Q
+
3
π D π D2
π D3
π D2
π D3
4Q 32 M 3704 856,960
smin =
−
=
−
π D 2 π D3 π D 2
π D3
1
sm = (smax + smin )
2
1  856,960 3704 3704 856,960  3704
=
sm = 
+
+
−
π D2 π D2
π D 3  π D 2
2  π D3
1
sa = (smax − smin )
2
856,960
sa =
π D3
K f sa
s
se = n s m +
sy
SF
smax =
assume K f = 1.0 at B
 1  3704   1.0  856,960  964 320,916
+
 = 2 +
se = 


2  
3
D3
 1.7  π D   0.85  π D  D
Octahedral Shear Theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


Page 22 of 76
π D2
SECTION 7 – SHAFT DESIGN
1
2
 694 320,916 

1
2
2
2 2
+






2
3


1
94,370
6.42   3.27 
 1
D
 + 
  = 
=  D
+ 3  + 3  
3 
2
2 
50,000
D   D  
  0.577(50,000)D  
 72 D





By trial and error, use
1
D = 2 in
2
2
483.
The worm shown is to deliver 65.5 hp steadily at 1750 rpm. It will be integral
with the shaft if the shaft size needed permits, and its pitch diameter 3 in. The 12in. pulley receives the power from a horizontal belt in which the tight tension
F1 = 2.5F2 . The forces (in kips) on the worm are as shown, with the axial force
taken by bearing B. The strength reduction factor for the thread roots may be
taken as K f = 1.5 , shear or bending. The shaft is machined from AISI 1045, as
rolled. (a) For N = 2.2 (Soderberg criterion) by the octahedral-shear theory,
compute the required minimum diameter at the root of the worm thread (a first
approximation). (b) What should be the diameter of the shaft 2.5 in. to the left of
the centerline of the worm? (c) Select a shaft size D and check it at the pulley A.
Problem 483.
Solution:
For AISI 1045, as rolled
s y = 59 ksi
su = 96 ksi
sn = 0.5su = 48 ksi
sn sns 48
1
=
=
=
s y s ys 59 1.229
63,000hp 63,000(65.5)
T=
=
= 2358 in − lb
1750
1750
(F1 − F2 ) 12  = T
2
(2.5F2 − F2 )(6) = 2358
F2 = 262 lb
F1 = 2.5F2 = 655 lb
Page 23 of 76
SECTION 7 – SHAFT DESIGN
FA = F1 + F2 = 655 + 262 = 917 lb
Horizontal
[∑ M
B
=0
]
(917 )(6) + (1570)(6.5) = 13Eh
Eh = 1208 lb
[∑ F
h
=0
]
917 + Eh = Bh + 1570
917 + 1208 = Bh + 1570
Bh = 555 lb
Shear Diagram
M Ah = 0
M Bh = (917 )(6 ) = 5502 in − lb
M Ch = (1208)(6.5) = 7852 in − lb
Vertical:
3
M ′ = (2540 )  = 3810 in − lb
2
∑ M E =0
[
]
M ′ + (1170)(6.5) = 13Bv
3810 + (1170)(6.5) = 13Bv
Bv = 878 lb
Page 24 of 76
SECTION 7 – SHAFT DESIGN
[∑ F =0]
v
Ev + Bv = 1170
Ev + 878 = 1170
Ev = 292 lb
Shear Diagram
Moment Diagram
M Av = 0
M Bv = 0
M Cv = 5707 in − lb
M=
MA =
MB =
MC =
(M h )2 + (M v )2
(0)2 + (0)2 = 0
(5502)2 + (0)2 = 5502 in − lb
(7852)2 + (5707)2 = 9707 in − lb
(a) Minimum diameter at the root of the warm thread.
K f = K fs = 1.5
M = M C = 9707 in − lb
F = 2540 lb
smax =
32 M
4F
32(9707 ) 4(2540 ) 310,624 10,160
+
=
+
=
+
3
2
π Dr π Dr
π Dr3
π Dr2
π Dr3
π Dr2
Page 25 of 76
SECTION 7 – SHAFT DESIGN
smin = −
310,624 10,160
+
π Dr3
π Dr2
1
(smax + smin )
2
10,160
sm =
π Dr2
1
sa = (smax − smin )
2
310,624
sa =
π Dr3
K f sa
s
se = n s m +
sy
SF
sm =
 1  10,160   1.5  310,624  2632 174,485
se = 
+


2  
3
 = D2 + D3
 1.229  π Dr   0.85  π Dr 
r
r
16T 16(2358) 12,000
ss =
=
=
π Dr3
π Dr3
Dr3
sms = ss
sas = 0
K fs sas
s
ses = ns sms +
s ys
SF
9764
 1  12,000 
ses = 
+0= 3

3

Dr
 1.229  Dr 
N = 2.2 , Octahedral shear theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
 2632 174,485  2
2
1
+
2
2
2 2

 2

3

 
1  Dr
Dr  
9764
1
3.635   1  


 +
 

=
+ 
=
+
2

  0.577(48,000)Dr3  
2.2 
48,000
Dr3   2.84 Dr3  
 18.24 Dr







By trial and error
Dr = 2.023 in
1
say Dr = 2 in
16
(b) D – shaft diameter 2.5 in. to the left of the center line of worm
Page 26 of 76
SECTION 7 – SHAFT DESIGN
3
in
16
Figure AF 12
3
r
16
=
≈ 0.1
d
 3
2.023 − 2 2 
 16 
D
2.023
=
= 1 .2
d
 3
2.023 − 2 2 
 16 
K f = K t = 1.65
r=
K fs = K ts = 1.34
at 2.5 in to the shaft
M h = (917 )(6) + (362)(6.5 − 2.5) = 6950 in − lb
M v = (878)(6.5 − 2.5) = 3512 in − lb
M=
(6950)2 + (3512)2
= 7787 in − lb
10,160
π D2
32 M 32(7787 ) 249,184
sa =
=
=
π D3
π D3
π D3
K f sa
s
se = n s m +
sy
SF
sm =
 1  10,160   1.65  249,184  2632 153,970
+
 =
se = 
+


2  
3
D2
D3
 1.229  π D   0.85  π D 
9764
ses =
D3
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
 2632 153,970  2
2
1
2
2
2 2
+






2
3


1
9764
1
3.21   1 

D
 + 
  = 
=  D
+ 3  +
 
3 
2
2.2 
48,000
18.24 D
D   2.84 D 3  
  0.577(48,000)D  







By trial and error
D = 1.9432 in
15
say D = 1 in
16
Page 27 of 76
SECTION 7 – SHAFT DESIGN
15
in = 1.9375 in
16
At the pulley A, or 3 in. right of centerline
M h = (917 )(3) = 2751 in − lb
(c) Selecting D = 1
Mv = 0
M = 2751 in − lb
For sled runner keyway
K f = 1 .6
K fs = 1.6
sm = 0
32 M
32(2751)
sa =
=
= 3853 psi
3
πD
π (1.9375)3
K f sa
s
se = n s m +
sy
SF
 1 .6 
se = 0 + 
(3853) = 7253 psi
 0.85 
9764
ses =
= 1343 psi
(1.9375)3
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
2 2
2
 
1  7253  
1343
 
= 
 +
N  48,000   0.577(48,000)  
N = 6.30 > 2.2 , therefore o.k.
484.
A propeller shaft as shown is to receive 300 hp at 315 rpm from the right through
a flexible coupling. A 16-in. pulley is used to drive an auxiliary, taking 25 hp.
The belt pull FB is vertically upward. The remainder of the power is delivered to
a propeller that is expected to convert 60% of it into work driving the boat, at
which time the boat speed is 1500 fpm. The thrust is to be taken by the right-hand
bearing. Let N = 2 ; material cold-worked stainless 410. Use the octahedral shear
theory with varying stresses. (a) Determine the shaft size needed assuming no
buckling. (b) Compute the equivalent column stress. Is this different enough to
call for another shaft size? Compute N by the maximum shear stress theory,
from both equations (8.4) and (8.11).
Page 28 of 76
SECTION 7 – SHAFT DESIGN
Problem 484.
Solution:
For stainless 410, cold-worked
s y = 85 ksi
sn = 53 ksi
SF = 0.85
Belt drive
63,000hp 63,000(25)
TB =
=
= 5000 in − lb
n
315
 2T  4(5000)
FB = F1 + F2 = 2(F1 − F2 ) = 2 B  =
= 1250 lb
16
 DB 
Propeller
63,000hp 63,000(300 − 25)
TP =
=
= 55,000 in − lb
n
315
Thrust
Fvm = ηhp(33,000)
F (1500) = (0.60)(300 − 25)(33,000)
F = 3630 lb
Vertical loading
[∑ M
E
=0
]
(20)(1250) = 60C
C = 417 lb
[∑ F =0]
v
A + C = FB
A + 417 = 1250
Page 29 of 76
SECTION 7 – SHAFT DESIGN
A = 833 lb
Shear Diagram
M B = (20)(833) = 16,660 in − lb
Maximum T at B
T = TB + TP = 60,000 in − lb
(a) Shaft size assuming no buckling
M = 16,660 in − lb
F = 3630 lb
4F
4(3630 ) 14,520
=
=
sm =
2
πD
π D2
π D2
32 M 32(16,660 ) 533,120
sa =
=
=
π D3
π D3
π D3
For sled-runner keyway
K f = 1 .6
K fs = 1.6
sn sns 53
1
=
=
=
s y s ys 85 1.604
se =
K f sa
sn
sm +
sy
SF
 1  14,520   1.6  533,120  2882 319,430
+
=
se = 
+


2  
3 
D2
D3
 1.604  π D   0.85  π D 
16
16(60,000 ) 960,000
ss = sms =
=
=
3
πD
π D3
π D3
sas = 0
K fs sas
s
ses = ns sms +
s ys
SF
190,510
 1  960,000 
 + 0 =
ses = 

2
D3
 1.604  π D 
Page 30 of 76
SECTION 7 – SHAFT DESIGN
1
2
1  se   ses  
=   +   
N  sn   sns  


N = 2 , Octahedral Shear Theory, sns = 0.577 sn
2
2
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
2
 2882 319,430 

1
2
2
2 2
+






2
3


1
190,510
1
6.027   6.230 

D
 + 
  = 
=  D
+
 +
 
3 
2
2 
53,000
18.39 D
D 3   D 3  
  0.577(53,000 )D  







By trial and error
D = 2.6 in
5
say D = 2 in = 2.625 in
8
2
(b) Equivalent Column Stress
4F
s=
α
π D2
Le = 12 + 60 + 10 = 82 in
1
1
k = D = (2.625) = 0.65625 in
4
4
Le
82
=
= 125 > 120
k 0.65625
Use Euler’s equation
2
L 
sy  e 
2
85(125)
k 

α=
= 2
= 4.486
π 2E
π 30 ×103
4F
4(3630 )
s=
(4.486) = 3000 psi
α=
2
πD
π (2.625)2
Since α > 1 , it is different enough to call for another shaft size.
(
)
Solving for N by maximum shear theory.
2882 319,430
2882
319,430
se =
+
=
+
= 18,078 psi
2
2
3
D
D
(2.625) (2.625)3
190,510
ses =
= 10,533 psi
(2.625)3
Equation (8.4)
Page 31 of 76
SECTION 7 – SHAFT DESIGN
1
1
2
 2  s 2  2 
 18,078   2
2
τ =  ss +    = (10,533) + 
  = 13,880 psi
 2  
 2  


0.5sn 0.5(53,000)
N=
=
= 1.91
τ
13,880
Equation (8.11) sns = 0.5sn
1
1
2
2 2
 18,078  2  10,533  2  2
1  s   ss  
 
=   +    = 
 + 
(
)
N  sn   sns  
53
,
000
0
.
5
53
,
000




 



N = 1.91
CHECK PROBLEMS
485.
A 3-in. rotating shaft somewhat as shown (482) carries a bevel gear whose mean
diameter is Dm = 10 in and which is keyed (profile) to the left end. Acting on the
gear are a radial force G = 1570.8 lb , a driving force Q = 3141.6 lb . The thrust
force is taken by the right-hand bearing. Let a = 5 in and L = 15 in ; material,
AISI C1040, annealed. Base calculations on the maximum shearing stress theory
with variable stress. Compute the indicated design factor N . With the use of a
sketch, indicate the exact point of which maximum normal stress occurs.
Solution:
For AISI C1040, annealed, Figure AF 1
s y = 48 ksi
su = 80 ksi
sn = 0.5su = 40 ksi
sn sns 40 1
=
=
=
s y s ys 48 1.2
FDm (6283.2 )(10 )
T=
=
= 31,416 in − lb
2
2
16T 16(31,416 )
ss =
=
= 5926 psi
π D3
π (3)3
sms = ss
sas = 0
K fs sas
s
ses = ns sms +
s ys
SF
 1 
ses = 
(5926 ) + 0 = 4940 psi
 1 .2 
Vertical
Page 32 of 76
SECTION 7 – SHAFT DESIGN
QDm (3141.6 )(10 )
=
= 15,708 in − lb
2
2
∑ M E =0
[
]
QDm
= 5G + 15 AV
2
15,708 = 5(1570.8) + 15 AV
AV = 523.6 lb
[∑ F =0]
v
AV + BV = G
523.6 + BV = 1570.8
BV = 1047.2 lb
Shear Diagram
Moment Diagram
M CV = 15,708 in − lb
M BV = 7854 in − lb
Page 33 of 76
SECTION 7 – SHAFT DESIGN
Horizontal
[∑ M
B
=0
]
15 Ah = 5(6283.2)
Ah = 2094.4 lb
[∑ F
h
=0
]
Bh = Ah + F
Bh = 2094.4 + 6283.2
Bh = 8377.6 lb
Shear Diagram
M Ch = 0
M Bh = (15)(2094.4 ) = 31,416 in − lb
Maximum Moment
2
M = M B2h + M Bv
=
(31,416)2 + (7854)2
= 32,383 in − lb
Since thrust force is taken by the right-hand bearing
sms = 0
32 M 32(32,383)
sas =
=
= 12,217 psi
π D3
π (3)3
K f sa
s
se = n s m +
sy
SF
Assume K f = 1.0 at the bearing B
 1 .0 
se = 0 + 
(12,217 ) = 14,373 psi
 0.85 
Page 34 of 76
SECTION 7 – SHAFT DESIGN
Maximum shear theory sns = 0.5sn
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.5sn  


1
2 2
2
1  14,373   4940  
 
= 
 +
N  40,000   0.5(40,000)  
N = 2 .3
Location of maximum normal stress
487.
A 2 7/16-in. countershaft in a machine shop transmits 52 hp at 315 rpm. It is
made of AISI 1117, as rolled, and supported upon bearing A and B, 59-in. apart.
Pulley C receives the power via a horizontal belt, and pulley D delivers it
vertically downward, as shown. Calculate N based on the octahedral-shearstress theory considering varying stresses.
Problem 487, 488
Solution:
For AISI 1117, as rolled
s y = 44.3 ksi
Page 35 of 76
SECTION 7 – SHAFT DESIGN
su = 70.6 ksi
sn = 0.5su = 35.3 ksi
sn sns 35.3
1
=
=
=
s y s ys 44.3 1.255
SF = 0.85
63,000(52 )
T=
= 10,400 in − lb
315
Pulley C
 2T  4(10,400)
 =
FC = F1 + F2 = 2(F2 − F1 ) = 2
= 2311 lb
D
18
 C
Pulley D
 2T  4(10,400)
 =
FD = F1 + F2 = 2(F2 − F1 ) = 2
= 1664 lb
D
25
 D
Horizontal
[∑ M
A
=0
]
15(2311) = 59 Bh
Bh = 588 lb
[∑ F
h
=0
]
Ah + Bh = 2311
Ah + 588 = 2311
Ah = 1723 lb
Shear Diagram
M Ch = (1723)(15) = 25,845 in − lb
Page 36 of 76
SECTION 7 – SHAFT DESIGN
M Dh = (1723)(15) − (588)(26 ) = 10,557 in − lb
Vertical
[∑ M
B
=0
]
18(1664) = 59 Av
Av = 508 lb
[∑ F =0]
v
Av + Bv = 1664
508 + Bv = 1664
Bv = 1156 lb
Shear Diagram
M Cv = (508)(15) = 7620 in − lb
M Dv = (1156 )(18) = 20,808 in − lb
M C = M C2h + M C2v =
(25,845)2 + (7620 )2
M D = M D2 h + M D2 v =
(10,557 )2 + (20,808)2
Maximum M at C
M = M C = 26,945 in − lb
sm = 0
32 M
sa =
π D3
7
D = 2 in = 2.4375 in
16
Page 37 of 76
= 26,945 in − lb
= 23,333 in − lb
SECTION 7 – SHAFT DESIGN
32(26,945)
= 18,952 psi
π (2.4375)3
assume K f = K fs = 1.0
sa =
se =
K f sa
sn
sm +
sy
SF
(1.0)(18,952) = 22,300 psi
 1 
se = 
(0 ) +
0.85
 1.255 
16T 16(10,400 )
ss =
=
= 3658 psi
π D 3 π (2.4375)3
sms = s s = 3658 psi
sas = 0
ses =
K fs sas
sns
sms +
s ys
SF
 1 
ses = 
(3658) + 0 = 2915 psi
 1.255 
Octahedral shear theory sns = 0.577 sn
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
2 2
2
 
1  22,300  
2915
 
= 
 +
N  35,300   0.577(35,300)  
N = 1.544
489.
A shaft for a general-purpose gear-reduction unit supports two gears as shown.
The 5.75-in. gear B receives 7 hp at 250 rpm. The 2.25-in. gear A delivers the
power, with the forces on the shaft acting as shown; the gear teeth have a
o
1
A
B
pressure angle of φ = 14
( tan φ = h = h ). Both gears are keyed (profile) to
2
Av Bv
the shaft of AISI 1141, cold rolled. (a) If the fillet radius is 1/8 in. at bearing D,
where the diameter is 1 3/8 in., compute N based on the octahedral-shear-stress
theory (Soderberg line). The shaft diameter at A is 1 11/16 in. What is N here?
Page 38 of 76
SECTION 7 – SHAFT DESIGN
Problem 489, 490
Solution:
For AISI 1141, cold rolled
s y = 90 ksi
sn = 50 ksi
sn sns 50 1
=
=
=
s y s ys 90 1.8
SF = 0.85
63,000(7 )
T=
= 1764 in − lb
250
16T
sms =
πD 3
sas = 0
Gear B:
 5.75 
Bv 
 = T = 1764 in − lb
 2 
Bv = 614 lb
Bh = Bv tan φ = 614 tan 14.5 = 159 lb
Gear A:
 2.25 
Av 
 = T = 1764 in − lb
 2 
Av = 1568 lb
Ah = Av tan φ = 1568 tan 14.5 = 406 lb
Vertical
Page 39 of 76
SECTION 7 – SHAFT DESIGN
[∑ M
D
=0
]
8Cv = 4(1568) − 3(614)
Cv = 554 lb
[∑ F =0]
v
Cv + Dv = Av + Bv
554 + Dv = 1568 + 614
Dv = 1628 lb
Shear Diagram
M Av = (554 )(4 ) = 2216 in − lb
M Dv = (614 )(3) = 1842 in − lb
Horizontal
[∑ M
C
=0
]
4(406) + 8Dh = 11(159)
Dh = 16 lb
[∑ F
h
=0
]
Ch + Bh = Ah + Dh
Ch + 159 = 406 + 16
Ch = 263 lb
Shear Diagram
Page 40 of 76
SECTION 7 – SHAFT DESIGN
M Ah = (263)(4 ) = 1052 in − lb
M Dh = (159 )(3) = 477 in − lb
M A = M A2h + M A2v =
(1052)2 + (2216)2
= 2453 in − lb
M D = M D2 h + M D2 v =
(477 )2 + (1842)2
= 1903 in − lb
(a) At bearing D
1
r = in
8
3
d = 1 in
8
r 0.125
=
≈ 0.10
d 1.375
D 1.375 + 0.25
=
≈ 1 .2
d
1.375
K t ≈ K f = 1 .6
K ts ≈ K fs = 1.34
M = MD
sm = 0
32 M 32(1903)
sa =
=
= 7456 psi
π d 3 π (1.375)3
K f sa
s
se = n s m +
sy
SF
se = 0 +
sms
(1.6)(7456) = 14,035 psi
0.85
16T
16(1764 )
=
=
= 3456 psi
3
π D π (1.375)3
sas = 0
ses =
K fs sas
sns
sms +
s ys
SF
 1 
ses =  (3456 ) + 0 = 1920 psi
 1 .8 
Octahedral shear theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


Page 41 of 76
SECTION 7 – SHAFT DESIGN
1
2 2
2
 
1  14,035  
3456


= 
 +

N  50,000   0.577(50,000)  
N = 3.28
(b) At A
For profile keyway
K f = 2.0 , K fs = 1.6
11
in = 1.6875 in
16
M = M A = 2453 in − lb
sm = 0
32 M
32(2453)
sa =
=
= 5200 psi
3
πd
π (1.6875)3
K f sa
s
se = n s m +
sy
SF
d =1
se = 0 +
sms
(2.0)(5200) = 12,235 psi
0.85
16T
16(1764 )
=
=
= 1870 psi
3
π D π (1.6875)3
sas = 0
ses =
K fs sas
sns
sms +
s ys
SF
 1 
ses =  (1870 ) + 0 = 1040 psi
 1 .8 
Octahedral shear theory
1  se   ses 

=   + 
N  sn   0.577 sn 

2
2



1
2
1
2 2
2
 
1  12,235  
1040
 
= 
 +
N  50,000   0.577(50,000)  
N = 4.043
THRUST LOADS
491.
The high-speed shaft of a worm-gear speed reducer, made of carburized AISI
8620, SOQT 450 F, is subjected to a torque of 21,400 in-lb. Applied to the right
Page 42 of 76
SECTION 7 – SHAFT DESIGN
end with no bending. The force on the worm has three components: a horizontal
force opposing rotation of W = 6180 lb , a vertical radial force S = 1940 lb , and a
rightward thrust of F = 6580 lb . The shaft has the following dimensions: a = 6 ,
7
9
3
9
b = 4 , c = 10 , d = 4 , e = 2 , f = 13 , g = 11.646 , h = 10.370 ,
8
16
4
16
13
D2 = 4 ,
D3 = 4 ,
D4 = 3.3469 , D5 = 3.253 ,
r1 = 0.098 ,
D1 = 3.740 ,
16
3
1
r2 = r3 = , r4 = 0.098 , r5 = , all in inches. The pitch diameter of the worm,
4
16
6.923 in., is the effective diameter for the point of application of the forces. The
root diameter, 5.701 in. is used for stress calculations. The left-hand bearing
takes the thrust load. Calculate N based on the octahedral-shear-stress theory
with varying stresses. (Data courtesy of Cleveland Worm and Gear Company.)
Problem 491
Solution:
Table AT 11n For AISI 8620, SOQT 450 F
s y = 120 ksi
su = 167 ksi
sn = 0.5su = 83.5 ksi
sn sns 83.5
1
=
=
=
s y s ys 120 1.437
SF = 0.85
T = 21,400 in − lb
Vertical
Page 43 of 76
SECTION 7 – SHAFT DESIGN
 6.923 
 6.923 
M ′ = F
 = 6580
 = 22,777 in − lb
 2 
 2 
∑ M A =0
[
]
22,777 + (11.646)(1940) = (11.646 + 10.370)Gv
Gv = 2061 lb
[∑ F =0]
v
S + Av = Gv
1940 + Av = 2061
Av = 121 lb
Shear Diagram
Moment Diagram
M Av = 0
M Bv = −(121)(1.2035) = −146 in − lb
M Cv = −(121)(1.2035 + 4.875) = −736 in − lb
M Dv = −(121)(1.2035 + 4.875 + 5.5675) = −1409 in − lb at left side
M Dv = −1409 + M ′ = −1409 + 22,777 = 21,368 in − lb at right side
M Ev = 21,368 − (2061)(4.4325) = 12,233 in − lb
M Fv = 12,233 − (2061)(4.5625) = 2830 in − lb
M Gv = 2830 − (2061)(1.375) = 0
Horizontal
Page 44 of 76
SECTION 7 – SHAFT DESIGN
[∑ M
A
=0
]
(11.646)(6180) = (11.646 + 10.370)Gh
Gh = 3269 lb
[∑ F
h
=0
]
Ah + Gv = W
Ah + 3269 = 6180
Ah = 2911 lb
Shear Diagram
Moment Diagram
M Ah = 0
M Bh = (2911)(1.2035) = 3500 in − lb
M Ch = (2911)(1.2035 + 4.875) = 17,695 in − lb
M Dh = 33,900 in − lb
M Eh = 33,900 − (3269 )(4.4325) = 19,410 in − lb
M Fh = 19,410 − (3269 )(4.5625) = 4495 in − lb
M Fh = 4495 − (3269 )(1.375) = 0
Combined
M = M h2 + M v2
MA =
MB =
MC =
MD =
MD =
ME =
(0)2 + (0)2 = 0 in − lb
(3500)2 + (146)2 = 3503 in − lb
(17,695)2 + (736)2 = 17,710 in − lb
(33,900)2 + (1409)2 = 33,930 in − lb (left)
(33,900)2 + (21,368)2 = 40,073 in − lb (right)
(19,410)2 + (12,233)2 = 22,944 in − lb
Page 45 of 76
SECTION 7 – SHAFT DESIGN
MF =
MG =
(2830)2 + (4495)2 = 5312 in − lb
(0)2 + (0)2 = 0 in − lb
Bending stresses (Maximum)
At A, s A = 0
32 M B 32(3503)
At B, s B =
=
= 682 psi
π D13 π (3.740)3
32 M C 32(17,710 )
=
= 1618 psi
At C, sC =
π D23 π (4.8125)3
32 M D 32(40,073)
At D, s D =
=
= 2203 psi
π Dr3
π (5.701)3
32 M E 32(22,944 )
At E, s E =
=
= 3652 psi
π D33
π (4)3
32 M F
32(5312 )
At F, s F =
=
= 1443 psi
3
π D4
π (3.3469 )3
At G, sG = 0
Shear Stresses:
16T 16(21,400 )
ssA = ssB =
=
= 2083 psi
π D13 π (3.740)3
16T 16(21,400 )
ssC =
=
= 978 psi
π D23 π (4.8125)3
16T 16(21,400 )
ssD =
=
= 588 psi
π Dr3 π (5.701)3
16T 16(21,400 )
ssE =
=
= 1703 psi
π D33
π (4)3
16T 16(21,400 )
ssF = ssG =
=
= 2907 psi
π D43 π (3.3469)3
Tensile stresses: F = 6580 lb
4F
4(6580 )
s′A = s′B =
=
= 599 psi
2
π D1 π (3.740)2
4F
4(6580 )
sC′ =
=
= 362 psi
2
π D2 π (4.8125)2
4F
4(6580 )
s′D =
=
= 258 psi
2
π Dr π (5.701)2
4F
4(6580 )
s′E =
=
= 524 psi
2
π D3
π (4 )2
Page 46 of 76
SECTION 7 – SHAFT DESIGN
s′E = s′F =
4F
4(6580 )
=
= 748 psi
2
π D4 π (3.3469 )2
r1 0.098
=
= 0.03
D1 3.740
D2 4.8125
=
= 1 .3
D1
3.740
Figure AF 12
K f ≈ K t = 2 .3
At B:
K fs ≈ K ts = 1.7
K f sa
sn
sm +
sy
SF
sm = s′B = 599 psi
se =
sa = s B = 682 psi
(2.3)(682 ) = 2262 psi
 1 
se = 
(599 ) +
0.85
 1.437 
K fs sas
s
ses = ns sms +
s ys
SF
sms = ssB = 2083 psi
sas = 0
 1 
ses = 
(2083) + 0 = 1450 psi
 1.437 
Octahedral shear theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
2 2
2
 
1  2262  
1450
 
= 
 +
N  83,500   0.577(83,500)  
N = 24.7
r2
0.75
=
= 0.16
D2 4.8125
Dr
5.701
=
= 1 .2
D2 4.8125
Figure AF 12
K f ≈ K t = 1 .5
At C:
K fs ≈ K ts = 1.2
Page 47 of 76
SECTION 7 – SHAFT DESIGN
se =
K f sa
sn
sm +
sy
SF
sm = 362 psi
sa = 1618 psi
(1.5)(1618) = 3107 psi
 1 
se = 
(362 ) +
0.85
 1.437 
K fs sas
s
ses = ns sms +
s ys
SF
sms = ssC = 978 psi
sas = 0
 1 
ses = 
(978) + 0 = 681 psi
 1.437 
Octahedral shear theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
2 2
2
 
1  3107  
681
 
= 
 +
N  83,500   0.577(83,500)  
N = 25.1
At D:
Assume K f = 1.5 as in Prob. 483
se =
K f sa
sn
sm +
sy
SF
sm = 258 psi
sa = 2203 psi
(1.5)(2203) = 4067 psi
 1 
se = 
(258) +
0.85
 1.437 
K fs sas
s
ses = ns sms +
s ys
SF
sms = s sD = 588 psi
sas = 0
 1 
ses = 
(588) + 0 = 409 psi
 1.437 
Octahedral shear theory
Page 48 of 76
SECTION 7 – SHAFT DESIGN
1  se   ses 

=   + 
N  sn   0.577 sn 

2
2



1
2
1
2 2
2
 
1  4067  
409
 
= 
 +
N  83,500   0.577(83,500)  
N = 20.2
r3 0.75
=
= 0.19
D3
4
Dr 5.701
=
= 1.43
D3
4
Figure AF 12
K f ≈ K t = 1.45
At E:
K fs ≈ K ts = 1.25
K f sa
sn
sm +
sy
SF
sm = s′E = 524 psi
se =
sa = s E = 3652 psi
(1.45)(3652) = 6595 psi
 1 
se = 
(524 ) +
0.85
 1.437 
K fs sas
s
ses = ns sms +
s ys
SF
sms = s sE = 1703 psi
sas = 0
 1 
ses = 
(1703) + 0 = 1185 psi
 1.437 
Octahedral shear theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
2 2
2
 
1  6595  
1185
 
= 
 +
N  83,500   0.577(83,500)  
N = 12
At F:
r4
0.098
=
= 0.03
D4 3.3469
Page 49 of 76
SECTION 7 – SHAFT DESIGN
D3
4
=
= 1 .2
D4 3.3469
Figure AF 12
K f ≈ K t = 2 .3
K fs ≈ K ts = 1.7
K f sa
sn
sm +
sy
SF
sm = s′F = 748 psi
se =
sa = s F = 1443 psi
(2.3)(1443) = 4425 psi
 1 
se = 
(748) +
0.85
 1.437 
K fs sas
s
ses = ns sms +
s ys
SF
sms = s sF = 2907 psi
sas = 0
 1 
ses = 
(2907 ) + 0 = 2023 psi
 1.437 
Octahedral shear theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
2 2
2
 
1  4425  
2023
 
= 
 +
N  83,500   0.577(83,500)  
N = 14.8
Then N = 12 at r3 =
492.
3
in (E)
4
The slow-speed shaft of a speed reducer shown, made of AISI 4140, OQT 1200
F, transmits 100 hp at a speed of 388 rpm. It receives power through a 13.6 in.
gear B. The force on this gear has three components: a horizontal tangential
driving force Ft = 2390 lb , a vertical radial force S = 870 lb , and a thrust force
Q = 598 lb taken by the right-hand bearing. The power is delivered to a belt at
F that exerts a downward vertical force of 1620 lb.; sled runner keyways. Use
the octahedral shear theory with the Soderberg line and compute N at sections C
and D. (Data courtesy of Twin Disc Clutch Company.)
Page 50 of 76
SECTION 7 – SHAFT DESIGN
Problem 492, 493
Solution:
For AISI 4140, OQT 1200 F
s y = 83 ksi
su = 112 ksi
sn = 0.5su = 56 ksi
sn sns 56
1
=
=
=
s y s ys 83 1.482
SF = 0.85
63,000(100 )
T=
= 16,237 in − lb
388
Vertical
 13.6 
 13.6 
M ′ = Q
 = (598)
 = 4066.4 in − lb
 2 
 2 
∑ M A =0
[
]
Page 51 of 76
SECTION 7 – SHAFT DESIGN
5
5
3
7
11 13
3
 3
 3
1 + 1 (870 ) + 1 + 1 + 3 + 1 + 1 + + 2 (1620 ) + 4066.4
8 32 32 16
4
 16 8 
 16 8
5
3
7 
 3
= 1 + 1 + 3 + 1 Gv
8 32 
 16 8
Gv = 3573 lb
[∑ F =0]
v
Av + S + F = Gv
Av + 870 + 1620 = 3573
Av = 1083 lb
Shear Diagram
Moment Diagram
M Av = 0
 3
M Pv = −(1083)1  = −1286 in − lb
 16 
 5
M Bv = −1286 + (− 1083)1  = −3046 in − lb at the left
 8
M Bv = −3046 + 4066.4 = 1021 in − lb at the right
 3
M Cv = 1021 − (1953) 3  = −5570 in − lb
 8
 7 
M Gv = −5570 − (1953)1  = −7950 in − lb
 32 
Page 52 of 76
SECTION 7 – SHAFT DESIGN
 11 
M Dv = −7950 + (1620 )1  = −5773 in − lb
 32 
 13 
M Ev = −5773 + (1620 )  = −4457 in − lb
 16 
 3
M Fv = −4457 + (1620 ) 2  = 0 in − lb
 4
Horizontal
[∑ M
A
=0
]
19 
 13 
 13
 2 (2390 ) +  2 + 4 Gh
32 
 16 
 16
Gh = 908 lb
[∑ F
h
=0
]
Ah + Gh = Ft
Ah + 908 = 2390
Ah = 1482 lb
Shear Diagram
M Ah = 0
 3
M Ph = (1482 )1  = 1760 in − lb
 16 
 5
M Bh = 1760 + (1482 )1  = 4168 in − lb
 8
Page 53 of 76
SECTION 7 – SHAFT DESIGN
 3
M Ch = 4168 − (908) 3  = 1104 in − lb
 8
 7 
M Ch = 1104 − (908)1  = 0 in − lb
 32 
M Dh = 0 in − lb
M Eh = 0 in − lb
M Fh = 0 in − lb
Combined
M = M h2 + M v2
M A = 0 in − lb
MP =
MB =
MC =
MD =
ME =
MF =
(1760)2 + (1286)2 = 2180 in − lb
(4168)2 + (3046)2 = 5163 in − lb
(1104)2 + (5570)2 = 5678 in − lb
(0)2 + (5773)2 = 5773 in − lb
(0)2 + (4457 )2 = 4457 in − lb
(0)2 + (0)2 = 0 in − lb
1
in = 0.125 in
8
d = 2.750 in
D = 2.953 in
r 0.125
=
= 0.05
d 2.750
D 2.953
=
= 1.10
d 2.750
Figure AF 12
K f 1 ≈ K t = 1 .9
at C: r =
K fs1 ≈ K ts = 1.3
For sled runner keyway
K f 2 = 1 .6
K fs 2 = 1.6
K f = 0.8 K f 1 K f 2 = 0.8(1.9 )(1.6 ) = 2.4
K fs = 0.8 K fs1 K fs 2 = 0.8(1.3)(1.6 ) = 1.7
Page 54 of 76
SECTION 7 – SHAFT DESIGN
se =
K f sa
sn
sm +
sy
SF
4Q
4(598)
=
= 101 psi
2
πd
π (2.750)2
32 M C 32(5678)
sa =
=
= 2781 psi
π d3
π (2.750)3
(2.4)(2781) = 7920 psi
 1 
se = 
(101) +
0.85
 1.482 
K fs sas
s
ses = ns sms +
s ys
SF
sm =
16T 16(16,237 )
=
= 3976 psi
π d 3 π (2.750)3
sas = 0
sms =
 1 
ses = 
(3976 ) + 0 = 2683 psi
 1.482 
Octahedral shear theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
2 2
2
 
1  7920  
2683
 
= 
 +
N  56,000   0.577(56,000)  
N =6
1
in = 0.0625 in
16
d = 2.953 in
3
D = 3 in = 3.375 in
8
r 0.0625
=
= 0.02
d
2.953
D 3.375
=
= 1.14
d 2.953
Figure AF 12
K f ≈ K t = 2 .4
at D: r =
K fs ≈ K ts = 1.6
se =
K f sa
sn
sm +
sy
SF
Page 55 of 76
SECTION 7 – SHAFT DESIGN
4Q
4(598)
=
= 87.3 psi
2
πd
π (2.953)2
32 M C 32(5773)
sa =
=
= 2284 psi
π d3
π (2.953)3
(2.4)(2284) = 6508 psi
 1 
se = 
(87.3) +
0.85
 1.482 
K fs sas
s
ses = ns sms +
s ys
SF
sm =
16T 16(16,237 )
=
= 3211 psi
π d 3 π (2.953)3
sas = 0
sms =
 1 
ses = 
(3211) + 0 = 2167 psi
 1.482 
Octahedral shear theory
1
2
2 2
1  se   ses  
 
=   + 
N  sn   0.577 sn  


1
2 2
2
 
1  6508  
2167
 
= 
 +
N  56,000   0.577(56,000)  
N = 7 .5
TRANSVERSE DEFLECTIONS
494.
The forces on the 2-in. steel shaft shown are A = 2 kips , C = 4 kips . Determine
the maximum deflection and the shaft’s slope at D.
Problems 494-496
Solution:
Page 56 of 76
SECTION 7 – SHAFT DESIGN
[M B = 0]
2(10) + 25D = 4(15)
D = 1.6 kips
[Fv = 0]
A+C = B + D
2 + 4 = B + 1 .6
B = 4.4 kips
Shear Diagram
Moment Diagram
M
64 M
=
EI Eπ D 4
M (in − kip )
M  4 4
  D 10
 EI 
Page 57 of 76
A
0
B
-20
C
16
D
0
0
-135.8
108.6
0
SECTION 7 – SHAFT DESIGN
Scale ss = 10 in in
M
200 × 10−4
, Scale sM =
per in
EI
EI
D4
Slope θ , Scale sθ = 0.2 D 4 rad in
y deflection, Scale s y = 2.0 D 4 in in
Deflection:
At A: y A =
0.625
in
D4
Page 58 of 76
SECTION 7 – SHAFT DESIGN
At C: yC =
0.375
in
D4
Slope:
0.075
rad
D4
0.0125
At B: θ =
rad
D4
0.05625
At D: θ =
rad
D4
At A: θ =
Maximum deflection:
0.625
y = yA =
= 0.04 in
(2 )4
Shaft’s slope at D
0.05625
θ=
= 0.0035 rad
(2)4
495.
The forces on the steel shaft shown are A = 2 kips , C = 4 kips . Determine the
constant shaft diameter that corresponds to a maximum deflection of 0.006 in. at
section C.
Solution:
(see Problem 494)
0.375
yC =
= 0.006
D4
D = 2.812 in
7
say D = 2 in
8
496. The forces on the steel shaft shown are A = 2 kips , C = 4 kips . Determine a
constant shaft diameter that would limit the maximum deflection at section A to
0.003 in.
Solution:
(see Problem 494)
0.625
yA =
= 0.003
D4
D = 3.80 in
7
say D = 3 in
8
497.
A steel shaft is loaded as shown and supported in bearings at R1 and R2 .
Determine (a) the slopes at the bearings and (b) the maximum deflection.
Page 59 of 76
SECTION 7 – SHAFT DESIGN
Problem 497
Solution:
∑ M R1 = 0
[
]
(3000 ) 7 + 1 1  − (2100) 7 + 2 1 + 1 =  7 + 2 1 + 2 + 7  R2
8 8
R2 = −444 lb
[∑ F = 0]
R1 + R2 + 2100 = 3000
R1 − 444 + 2100 = 3000
R1 = 1344 lb
Loading
Shear Diagram
Moment Diagram
Page 60 of 76
8
4
 8
4
8
SECTION 7 – SHAFT DESIGN
M A = 0 in − lb
7
M B = (1344 )  = 1176 in − lb
8
7 1
M C = (1134 ) + 1  = 2688 in − lb
8 8
 1
M D = 2688 − (1656 )1  = 825 in − lb
 8
M E = 825 − (1656)(1) = −831 in − lb
M F = −831 + (444)(1) = −387 in − lb
7
M G = −387 + (444 )  = 0 in − lb
8
M (in − kips )
D(in )
M
10 4
EI
(
)( )
A
0
1½
B1
1.18
1½
B2
1.18
2
C
2.69
2
D1
0.83
2
D2
0.83
1¾
E
-0.83
1¾
F1
-0.39
1¾
F2
-0.39
1½
G
0
1½
0
1.58
0.50
1.14
0.35
0.60
-0.60
-0.28
-0.52
0
Scale ss = 2 in in
Page 61 of 76
SECTION 7 – SHAFT DESIGN
M
2 × 10−4
, Scale sM =
per in
EI
EI
D4
Slope θ , Scale sθ = 4 × 10 −4 D 4 rad in
y deflection, Scale s y = 8 × 10 −4 D 4 in in
(a) Slopes at the bearings
(
)
at R1 , θ A = 0.375 4 × 10 −4 = 1.5 × 10 −4 rad
at R2 , θ G = 0 rad
(b) Maximum deflection
at C, yC = 0.1875(8 × 10 −4 ) = 1.5 × 10 −4 in
Page 62 of 76
SECTION 7 – SHAFT DESIGN
498.
(a) Determine the diameter of the steel shaft shown if the maximum deflection is
to be 0.01 in.; C = 1.5 kips , A = 1.58 kips , L = 24 in . (b) What is the slope of the
shaft at bearing D? See 479.
Problems 498, 505, 506.
Solution:
Vertical
[∑ M
D
=0
]
6(1.5) + 8(0.424) = 24 Ev
Ev = 0.516 kip
[∑ F
v
=0
]
Dv + 0.424 = 1.5 + Ev
Dv + 0.424 = 1.5 + 0.516
Dv = 1.592 kip
Shear Diagram
M C = 0 ; M D = −6(1.5) = −9 in − kips
M B = −9 + 8(0.092) = −8.264 in − kips
Page 63 of 76
SECTION 7 – SHAFT DESIGN
M A = −8.264 + 11(0.516) = −2.588 in − kips
M E = −2.588 + 5(0.516) = 0
M (in − kips )
M
D 4 ×10 4
EI
(
)
C
0
0
D
-9
-61.1
Scale ss = 8 in in
M
120 × 10 −4
, Scale sM =
per in
EI
EI
D4
Slope θ , Scale sθ = 0.096 D 4 rad in
Page 64 of 76
B
-8.264
-56.1
A
-2.588
-17.6
E
0
0
SECTION 7 – SHAFT DESIGN
y deflection, Scale s y = 0.768 D 4 in in
Deflections.
0.384
yCv =
in
D4
0.288
y Bv =
in
D4
0.168
y Av =
in
D4
Slope
0.057
θ Dv =
rad
D4
Horizontal
[∑ M
D
=0
]
8(0.424 ) + 24 Eh = 19(1.58)
Eh = 1.1095 kip
[∑ F
h
=0
]
Dh + Eh + 0.424 = 1.58
Dh + 1.1095 + 0.424 = 1.58
Page 65 of 76
SECTION 7 – SHAFT DESIGN
Dh = 0.0465 kip
Shear Diagram
Moments
MC = 0
MD = 0
M B = 8(− 0.0465) = −0.372 in − kip
M A = −0.372 + 11(− 0.4705) = −5.5475 in − kips
M E = −5.5475 + 5(1.1095) = 0
M (in − kips )
M
D 4 ×10 4
EI
(
)
C
0
0
D
0
0
Scale ss = 8 in in
M
4 × 10−4
, Scale sM =
per in
EI
EI
D4
Slope θ , Scale sθ = 0.032 D 4 rad in
Page 66 of 76
B
-0.372
-2.53
A
-5.5475
-37.7
E
0
0
SECTION 7 – SHAFT DESIGN
y deflection, Scale s y = 0.256 D 4 in in
Deflections.
0.064
yC h =
in
D4
0.072
y Bh =
in
D4
0.096
y Ah =
in
D4
Slope
0.012
θ Dh =
rad
D4
Resultant deflection:
(
y = yh2 + yv2
1
2
)
1
2 2
[(0.064) + (0.384) ]
2
yC =
D
4
[(0.072) + (0.288) ]
D
4
[(0.096) + (0.168) ]
Slope:
Page 67 of 76
=
0.297
D4
=
0.194
D4
1
2 2
2
yA =
0.390
D4
1
2 2
2
yB =
=
D
4
SECTION 7 – SHAFT DESIGN
θ = (θ + θ
2
h
1
2 2
v
)
1
2 2
[(0.012) + (0.057) ]
=
2
θD
D
4
=
0.05823
rad
D4
(a) Diameter D.
Maximum deflection = yC =
0.390
= 0.01 in
D4
D = 2.50 in
(b) slope of the shaft at bearing D
0.05823 0.05823
θD =
=
= 0.0015 rad
D4
(2.5)4
CRITICAL SPEED
499.
A small, high-speed turbine has a single disk, weighing 0.85 lb., mounted at the
midpoint of a 0.178-in. shaft, whose length between bearings is 6 ½ in. What is
the critical speed if the shaft is considered as simply supported?
Solution:
Table AT 2
3
WL3
(
0.85)(6.5)
y=
=
= 0.052634 in
4
3EI
6  π (0.178 ) 
3 30 × 10 

64


(
)
1
1
1
30  g o (∑ Wy ) 2 30  g o  2 30  386  2
nc =
=
= 818 rpm

 =
π  ∑ Wy 2 
π  y 
π  0.052634 
500.
The bearings on a 1 ½-in. shaft are 30 in. apart. On the shaft are three 300-lb
disks, symmetrically placed 7.5 in. apart. What is the critical speed of the shaft?
Solution:
Page 68 of 76
SECTION 7 – SHAFT DESIGN
Table AT 2
Deflection of B.
y B = y B1 + y B2 + y B3
y B1 =
(300)(22.5)(7.5)([ 30)2 − (22.5)2 − (7.5)2 ] = 0.01273 in
4
6  π (1.5 ) 
6(30 × 10 )
(30)


 64 
(300)(15)(7.5)(30)2 − (15)2 − (7.5)2 = 0.01556 in
y B2 =
 π (1.5)4 
6 30 × 106 
 (30)
 64 
2
2
2
(
300)(7.5)(7.5)(30) − (7.5) − (7.5)
y B3 =
= 0.00990 in
4
6  π (1.5 ) 
6 30 × 10 
 (30)
64


y B = 0.01273 + 0.00990 + 0.01556 = 0.03819 in
[
(
]
)
[
(
]
)
Deflection of C.
yC = yC1 + yC2 + yC3
yC1
2
2
2
(
300)(7.5)(30 − 15)(
[
30) − (7.5) − (30 − 15) ]
=
= 0.01556 in
4
6  π (1.5 ) 
6(30 × 10 )
(30)


 64 
(300)(15)(30 − 15)(30)2 − (15)2 − (30 − 15)2 = 0.02264 in
yC2 =
4
6  π (1.5 ) 
6 30 ×10 
 (30)
 64 
(300)(7.5)(15)(30)2 − (7.5)2 − (15)2 = 0.01556 in
yC3 =
 π (1.5)4 
6 30 × 106 
 (30)
 64 
yC = 0.01556 + 0.02264 + 0.01556 = 0.05376 in
Deflection of D.
y D = y D1 + y D2 + y D3
[
(
)
[
(
Page 69 of 76
]
]
)
SECTION 7 – SHAFT DESIGN
y D1 =
(300)(7.5)(30 − 22.5)([ 30)2 − (7.5)2 − (30 − 22.5)2 ] = 0.00990 in
4
6  π (1.5 ) 
6(30 × 10 )
(30)


 64 
(300)(15)(30 − 22.5)(30)2 − (15)2 − (30 − 22.5)2 = 0.01556 in
y D2 =
4
6  π (1.5 ) 
6 30 ×10 
 (30)
 64 
(300)(7.5)(22.5)(30)2 − (7.5)2 − (22.5)2 = 0.01273 in
y D3 =
 π (1.5)4 
6 30 × 106 
 (30 )
 64 
y D = 0.00990 + 0.01556 + 0.01273 = 0.03819 in
[
(
]
)
[
(
]
)
1
2
1
30  g o (∑ Wy )
30  g o ( y B + yC + y D )  2
nc =
=


π  ∑ Wy 2 
π  y B2 + yC2 + y D2 
1
30  386(0.03819 + 0.05376 + 0.03819)  2
nc = 
 = 888 rpm
π  (0.03819)2 + (0.05376)2 + (0.03819)2 
501.
A fan for an air-conditioning unit has two 50-lb. rotors mounted on a 3-in. steel
shaft, each being 22 in. from an end of the shaft which is 80 in. long and simply
supported at the ends. Determine (a) the deflection curve of the shaft considering
its weight as well as the weight of the rotors, (b) its critical speed.
Solution:
W1 = 50 lb
W3 = 50 lb
π  2
W2 = (0.284 ) (3) (80 ) = 160 lb weight of shaft
4
160
w2 =
= 2 lb in
80
Deflection of B.
y B = y B1 + y B2 + y B3
Page 70 of 76
SECTION 7 – SHAFT DESIGN
y B1 =
(50)(50)(22)([ 80)2 − (58)2 − (22)2 ] = 0.002844 in
4
6  π (35 ) 
6(30 × 10 )
(80)


 64 
(50)(22)(22)(80)2 − (22)2 − (22)2 = 0.002296 in
y B3 =
4
6  π (3) 
6 30 × 10 
 (80)
 64 
(2)(22)(80)3 − 2(80)(22)2 − (22)3 = 0.006843 in
y B2 =
 π (3)4 
6 30 × 106 

 64 
y B = 0.002844 + 0.006843 + 0.002296 = 0.011983 in
[
(
]
)
[
]
(
)
Deflection of C.
yC = yC1 + yC2 + yC3
yC1
2
2
2
(
50)(22)(80 − 40)(
[
80) − (22) − (80 − 40) ]
=
= 0.003317 in
4
6  π (35) 
6(30 ×10 )
(80)


 64 
2
2
2
(
50)(22 )(40)(80) − (22) − (40)
yC3 =
= 0.003317 in
4
6  π (3) 
6 30 × 10 
 (80)
 64 
(2)(40)(80)3 − 2(80)(40)2 − (40)3 = 0.008942 in
yC2 =
4
6  π (3) 
6 30 × 10 

 64 
yC = 0.003317 + 0.008942 + 0.003317 = 0.015576 in
[
(
]
)
[
]
(
By symmetry
y D = y B = 0.011983 in
(a) Deflection curve
Page 71 of 76
)
SECTION 7 – SHAFT DESIGN
(b) Critical speed
1
30  g o (∑ Wy ) 2
nc =


π  ∑ Wy 2 
∑Wy = (50)(0.011983) + (160)(0.015576) + (50)(0.011983) = 3.69046
∑Wy = (50)(0.011983) + (160)(0.015576) + (50)(0.011983) = 0.053177
2
2
2
2
1
30  386(3.69046)  2
nc =
= 1563 rpm
π  0.053177 
ASME CODE
502.
A cold-rolled transmission shaft, made of annealed AISI C1050, is to transmit a
torque of 27 in-kips with a maximum bending moment of 43 in-kips. What
should be the diameter according to the Code for a mild shock load?
Solution:
For AISI C1050, annealed
s y = 53 ksi
su = 92 ksi
0.3s y = 15.9 ksi
0.18su = 16.56 ksi
use τ d = 0.3s y = 15.9 ksi
M = 43 in − kips
T = 27 in − kips
1
2 2


16
α FD 1 + B 2  
2
3
 
(K sT ) +  K m M +
D =
8
πτ d 1 − B 4 

 

Reduce to
1
16
2
2 2
3
D =
(K sT ) + ( K m M )
πτ d 1 − B 4
For mild shock load, rotating shafts
K m = 1.75
K s = 1.25
B=0
1
16
2
2 2
3
D =
[(1.25)(27,000)] + [(1.75)(43,000)]
π (15,900)
D = 2.98 in
say D = 3 in
(
)
(
)[
{
Page 72 of 76
(
)
]
}
SECTION 7 – SHAFT DESIGN
503.
A machinery shaft is to transmit 82 hp at a speed of 1150 rpm with mild shock.
The shaft is subjected to a maximum bending moment of 7500 in-lb. and an axial
thrust load of 15,000 lb. The material is AISI 3150, OQT 1000 F. (a) What
should be the diameter when designed according to the Code? (b) Determine the
corresponding conventional factor of safety (static-approach and maximum-shear
theory).
Solution:
For AISI 3150, OQT 1000 F
s y = 130 ksi
su = 151 ksi
0.3s y = 39 ksi
0.18su = 27.18 ksi
use τ d = 0.18su = 27.18 ksi
63000(82 )
T=
= 4492 in − lb
1150
M = 7500 in − lb
F = 15,000 lb


16
α FD 1 + B 2
2
3
(K sT ) +  K m M +
(a) D =
πτ d 1 − B 4 
8


(
(
)
1
) 


2
2


For mild shock load
K m = 1.75
K s = 1.25
B=0
α =1
1
2

(
16
1)(15,000)D   2

2
3
D =
[(1.25)(4492)] + (1.75)(7500) +
 
π (27180) 
8


{
1
2 2
D 3 = 0.1874 31.53 + [13.125 + 1.875D ]
D = 1.4668 in
say D = 1.5 in
(b) s =
}
32 M
4F
32(7500 ) 4(15,000 )
+
=
+
= 31,124 psi = 31.124 ksi
3
2
πD πD
π (1.5)3
π (1.5)2
Page 73 of 76
SECTION 7 – SHAFT DESIGN
ss =
16T 16(4492 )
=
= 6778.5 psi = 6.7785 ksi
π D 3 π (1.5)3
2
2
1  s   ss  
=
+
N  s y   s ys  


Maximum shear theory
s ys = 0.5s y
2
2
1  31.124   6.7785  
 
= 
 +
N  130   0.5(130)  
N = 3.83
504.
short stub shaft, made of SAE 1035, as rolled, receives 30 hp at 300 rpm via a
12-in. spur gear, the power being delivered to another shaft through a flexible
coupling. The gear is keyed midway between the bearings and its pressure angle
φ = 20o . See the figure for 471. (a) Neglecting the radial component of the tooth
load, determine the shaft diameter for a mild shock load. (b) Considering both
tangential and radial components, compute the shaft diameter. (c) Is the
difference in the foregoing results enough to change your choice of the shaft
size?
Solution:
Figure for 471.
For SAE 1035, as rolled
s y = 55 ksi
su = 85 ksi
0.3s y = 16.5 ksi
0.18su = 15.3 ksi
use τ d = 0.18su = 15.3 ksi
Page 74 of 76
SECTION 7 – SHAFT DESIGN
Data are the same as 471.
From Problem 471.
(a) M = 4200 in − lb = 4.2 in − kips
T = 6300 in − lb = 6.3 in − kips


16
α FD 1 + B 2
2


(
)
D3 =
K
T
+
K
M
+
s
 m
πτ d 1 − B 4 
8


Reduce to
1
16
2
2 2
D3 =
K
T
+
K
M
(
)
(
)
s
m
πτ d 1 − B 4
For mild shock load, rotating shafts
K m = 1.75
K s = 1.25
B=0
1
16
D3 =
[(1.25)(6.3)]2 + [(1.75)(4.2)]2 2
π (15.3)
D = 1.5306 in
9
say D = 1 in
16
(
)
(
)[
(
1
) 


2
2


]
{
}
(b) M = 4472 in − lb = 4.472 in − kips
T = 6300 in − lb = 6.3 in − kips
16
D =
[(1.25)(6.3)]2 + [(1.75)(4.472)]2
π (15.3)
D = 1.5461 in
9
say D = 1 in
16
{
3
1
2
}
(c) Not enough to change the shaft size.
505.
Two bearings D and E, a distance D = 24 in . Apart, support a shaft for a punch
press on which are an 8-in. gear A, a 44-in. pulley B, and a flywheel C, as
indicated (498). Weight of flywheel is 1500 lb.; pulley B receives the power at an
angle of 45o to the right of the vertical; gear A delivers it horizontally to the right.
The maximum power is 25 hp at 250 rpm is delivered, with heavy shock. For
cold-finish AISI 1137, find the diameter by the ASME Code.
Solution:
Page 75 of 76
SECTION 7 – SHAFT DESIGN
Data and figure is the same as in Problem 479. Also figure is the same as in Problem 498.
For AISI 1137, cold-finished
s y = 93 ksi
su = 103 ksi
0.3s y = 27.9 ksi
0.18su = 18.54 ksi
use τ d = 0.18s u = 18.54 ksi
From Problem 479
M = M B = 14,343 in − lb = 14.343 in − kips
T = TA = 12,600 in − lb = 12.6 in − kips
For heavy shock load
K m = 2.5
K s = 1.75
B=0


16
α FD 1 + B
2
(K sT ) +  K m M +
4
πτ d 1 − B 
8


1
16
2
2 2
(
)
(
)
D3 =
K
T
+
K
M
s
m
πτ d 1 − B 4
D3 =
(
)
(
)[
(
]
16
[(1.75)(12.6)]2 + [(2.5)(14.343)]2
π (18.54)
D = 2.2613 in
5
say D = 2 in
16
D3 =
{
2
1
2
}
- end -
Page 76 of 76
)  
2

 
1
2
SECTION 8 – KEYS AND COUPLINGS
FLAT AND SQUARE KEYS
DESIGN PROBLEMS
521.
A 2-in. shaft, of cold-drawn AISI 1137, has a pulley keyed to it. (a) Compute the length of square
key and the length of flat key such that a key made of cold-drawn C1020 has the same yield
strength as the shaft does in pure torsion. (b) The same as (a), except that the key material is
AISI 2317, OQT 1000 F. (c) Would you discard either of these keys? Explain.
Solution:
For AISI 1137 shaft, Table AT 8, sy = 93 ksi
Yield Strength of Shaft
sys = 0.6sy = 0.6(93) = 55.8 ksi
sysπD 3 (55.8)(π )(2)3
T=
=
= 87.65 in − kips
16
16
(a) Key Material, cold-drawn, C1020, Table AT 7
sy = 66 ksi
sys = 0.6sy = 0.6(66) = 39.6 ksi
Table AT 19, use b = ½ in , t = 3/8 in.
For square key, b = t = ½ in.
By shear, ss = sys
2T
2(87.65)
L=
=
= 4.43 in
ss bD 39.6(1 2 )(2 )
By compression, sc = sy. Key has lowest yield strength.
4T
4(87.65)
L=
=
= 5.32 in
sc tD 66(1 2)(2)
Use L = 5.32 in
For Flat key, b = ½ in , t = 3/8 in.
By shear, ss = sys
2T
2(87.65)
L=
=
= 4.43 in
ss bD 39.6(1 2 )(2 )
By compression, sc = sy. Key has lowest yield strength.
4T
4(87.65)
L=
=
= 7.09 in
sc tD 66(3 8)(2)
Use L = 7.09 in
(b) Key Material, AISI 2317, QOT 1000 F. Table AT 8
sy = 71 ksi
sys = 0.6sy = 0.6(71) = 42.6 ksi
Table AT 19, use b = ½ in , t = 3/8 in.
For square key, b = t = ½ in.
By shear, ss = sys
2T
2(87.65)
L=
=
= 4.12 in
ss bD 42.6(1 2)(2)
By compression, sc = sy. Key has lowest yield strength.
1
SECTION 8 – KEYS AND COUPLINGS
4T
4(87.65)
=
= 4.94 in
sc tD 71(1 2 )(2 )
Use L = 4.94 in
L=
For Flat key, b = ½ in , t = 3/8 in.
By shear, ss = sys
2T
2(87.65)
L=
=
= 4.12 in
ss bD 42.6(1 2)(2)
By compression, sc = sy. Key has lowest yield strength.
4T
4(87.65)
L=
=
= 6.59 in
sc tD 71(3 8)(2 )
Use L = 6.59 in
(c) Either of the above is not to be discarded since they are designed based on yield strength with
the same factor of safety.
522.
A cast-iron pulley transmits 65.5 hp at 1750 rpm. The 1045 as-rolled shaft to which it is to be
keyed is 1 ¾ in. in diameter; key material, cold-drawn 1020. Compute the length of flat key and
of square key needed.
Solution:
For shaft: 1045 as-rolled, Table AT 7, sy = 59 ksi
For key: Cold-drawn 1020, sy = 66 ksi
D = 1 ¾ in = 1.75 in
hp = 65.5 hp, n = 1750 rpm
63,000hp 63,000(65.5)
=
= 2358 in − lb = 2.358 in − kips
n
1750
Table AT 19, use b = 3/8 in, t = 1/4 in for D = 1 ¾ in
Assume smooth load, N = 1.5
T=
For Flat key, b = 3/8 in, t = 1/4 in.
By shear,
sys = 0.6sy = 0.6(66) = 39.6 ksi
sys 39.6
ss =
=
= 26.4 ksi
N
1. 5
2T
2(2.358)
L=
=
= 0.272 in
ss bD 26.4(3 8)(1.75)
By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron).
sy 59
sc = =
= 39.3 ksi
N 1. 5
4T
4(2.358)
L=
=
= 0.549 in
sc tD 39.3(1 4)(1.75)
Use L = 0.549 in - answer
2
SECTION 8 – KEYS AND COUPLINGS
For Square key, b = 3/8 in, t = 3/8 in.
By shear, ss = sys
sys = 0.6sy = 0.6(66) = 39.6 ksi
sys 39.6
ss =
=
= 26.4 ksi
N
1. 5
2T
2(2.358)
L=
=
= 0.272 in
ss bD 26.4(3 8)(1.75)
By compression, use sy of shaft the lowest. Pulley has the highest compressive strength (Cast iron).
sy 59
sc = =
= 39.3 ksi
N 1. 5
4T
4(2.358)
L=
=
= 0.366 in
sc tD 39.3(3 8)(1.75)
Use L = 0.366 in - answer
523.
A 3 ¼-in. shaft transmits with medium shock 85 hp at 100 rpm. Power is received through a
sprocket (annealed nodular iron 60-45-10) keyed to the shaft of cold-rolled AISI 1040 (10%
work), with a key of cold-finished B1113. What should be the length of (a) a square key? (b) a
flat key?
Solution:
For sprocket, annealed nodular iron, 60-45-10, Table AT 6, sy = 55 ksi
For shaft, cold-rolled AISI 1040 (10% work), Table AT 10, sy = 85 ksi
For key, cold-finished B1113, Table AT 7, sy = 72 ksi
D = 3 ¼ in = 3.25 in
hp = 85 hp
n = 100 rpm
63,000hp 63,000(85)
=
= 53,550 in − lb = 53.55 in − kips
n
100
Table AT 19, use b = ¾ in, t = 1/2 in for D = 3 ¼ in
For medium shock, N = 2.25
(a) Square key,
b = ¾ in, t = ¾ in.
By shear,
sys = 0.6sy = 0.6(72) = 43.2 ksi
sys 43.2
ss =
=
= 19.2 ksi
N 2.25
2T
2(53.55)
L=
=
= 2.29 in
ss bD 19.2(3 4)(3.25)
By compression, use sy of sprocket the lowest.
sy
55
sc = =
= 24.4 ksi
N 2.25
4T
4(53.55)
L=
=
= 3.60 in
sc tD 24.4(3 4)(3.25)
T=
3
SECTION 8 – KEYS AND COUPLINGS
Use L = 3.60 in. - answer
(b) Flat key,
b = ¾ in, t = ½ in.
By shear, ss = sys
sys = 0.6sy = 0.6(72) = 43.2 ksi
sys 43.2
ss =
=
= 19.2 ksi
N 2.25
2T
2(53.55)
L=
=
= 2.29 in
ss bD 19.2(3 4)(3.25)
By compression, use sy of sprocket the lowest.
sy
55
sc = =
= 24.4 ksi
N 2.25
4T
4(53.55)
L=
=
= 5.40 in
sc tD 24.4(1 2)(3.25)
Use L = 5.40 in. - answer
524.
A cast-steel gear (SAE 0030), with a pitch diameter of 36 in., is transmitting 75 hp at 210 rpm to
a rock crusher, and is keyed to a 3-in. shaft (AISI 1045, as rolled); the key is made of AISI C1020,
cold drawn. For a design factor of 4 based on yield strength, what should be the length of (a) a
square key, (b) flat key? (c) Would either of these keys be satisfactory?
Solution:
For cast-steel gear (SAE 0030), Table AT 6, sy = 35 ksi
For shaft, AISI 1045, as rolled, Table AT 7, sy = 59 ksi
For key, AISI C1020, cold-drawn, Table AT 7, sy = 66 ksi
D = 3 in
hp = 75 hp
n = 210 rpm
63,000hp 63,000(75)
=
= 22,500 in − lb = 22.5 in − kips
n
210
Table AT 19, use b = ¾ in, t = 1/2 in for D = 3 in
Design factor, N = 4
(a) Square key,
b = ¾ in, t = ¾ in.
By shear,
sys = 0.6sy = 0.6(66) = 39.6 ksi
sys 39.6
ss =
=
= 9.9 ksi
N
4
2T
2(22.5)
L=
=
= 2.02 in
ss bD 9.9(3 4)(3)
By compression, use sy of cast-steel gear the lowest.
sy 35
sc = =
= 8.75 ksi
N 4
T=
4
SECTION 8 – KEYS AND COUPLINGS
4T
4(22.5)
=
= 4.57 in
sc tD 8.75(3 4)(3)
Use L = 4.57 in. - answer
(b) Flat key,
b = ¾ in, t = ½ in.
By shear, ss = sys
sys = 0.6sy = 0.6(66) = 39.6 ksi
sys 39.6
ss =
=
= 9.9 ksi
N
4
2T
2(22.5)
L=
=
= 2.02 in
ss bD 9.9(3 4)(3)
By compression, use sy of cast-steel gear the lowest.
sy 35
sc = =
= 8.75 ksi
N 4
4T
4(22.5)
L=
=
= 6.86 in
sc tD 8.75(1 2)(3)
Use L = 6.86 in. - answer
L=
525.
An electric motor delivers 50 hp at 1160 rpm to a 1 5/8 in. shaft (AISI 13B45, OQT 1100 F). Keyed
to this shaft is a cast-steel (SAE 080, N & T) pulley whose hub is 2 in. long. The loading may be
classified as mild shock. Decide upon a key for this pulley (material), investigating both flat and
square keys.
Solution:
hp = 50 hp
n = 1160 rpm
D = 1 5/8 in = 1.625 in
Shaft material – AISI 13B45, OQT 1100 F, Table AT 10, sy = 112 ksi
Pulley material – SAE 080, N & T, Table AT 6, sy = 40 ksi
L = 2 in
N = 2.0 to 2.25 for mild shock
From Table AT 19 for D = 1 5/8 in
b = 3/8 in, t = ¼ in
63,000hp 63,000(50)
T=
=
= 2,716 in − lb = 2.716 in − kips
n
1160
For flat key: b = 3/8 in, t = ¼ in
Check for compression:
4T
4(2.716)
sc =
=
= 13.37 ksi
LtD (2 )(1 4)(1.625)
sy
40
= 2.99 > 2.25
Based on pulley material, N = =
sc 13.37
Therefore safe in compression.
Determine the yield stress on the key
2T
2(2.716)
ss =
=
= 4.457 ksi
LbD (2)(3 8)(1.625)
5
SECTION 8 – KEYS AND COUPLINGS
N = 2.25
s ys = 0.6 sy = Nss
0.6 sy = (2.25 )(4.457 )
s y = 16.7 ksi
Min. s y = (13.37 )(2.25) = 30 ksi - Minimum yield strength of key material required.
Select SAE 003, Table AT 6, sy = 35 ksi – Answer.
For square key: b = t = 3/8 in
Check for compression:
4T
4(2.716)
sc =
=
= 8.914 ksi
LtD (2 )(3 8)(1.625)
sy
40
= 4.49 > 2.25
Based on pulley material, N = =
sc 8.914
Therefore safe in compression.
Determine the yield stress on the key
2T
2(2.716)
ss =
=
= 4.457 ksi
LbD (2)(3 8)(1.625)
N = 2.25
s ys = 0.6 sy = Nss
0.6 sy = (2.25 )(4.457 )
s y = 16.7 ksi
Min. s y = (8.914)(2.25) = 20 ksi - Minimum yield strength of key material required.
Select SAE 003, Table AT 6, sy = 35 ksi – Answer.
CHECK PROBLEMS
526. A cast-steel (SAE 080, N & T) pulley, attached to a 2-in. shaft, is transmitting 40 hp at 200 rpm,
and is keyed by a standard square key, 3 in. long, made of SAE 1015, cold drawn; shaft material,
C1144, OQT 1000 F. (a) What is the factor of safety of the key? (b) The same as (a) except a flat
key is used.
Solution:
Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi
Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi
Shaft, C1144, OQT 1000 F, sy = 83 ksi
hp = 40 hp
N = 200 rpm
D = 2 in
L = 3 in
63,000hp 63,000(40)
=
= 12,600 in − lb = 12.6 in − kips
n
200
Table AT 19, D = 2 in
b = ½ in, t = 3/8 in
T=
6
SECTION 8 – KEYS AND COUPLINGS
a) Square key, b = ½ in, t = ½ in
By shear:
2T
2(12.6 )
ss =
=
= 8.4 ksi
LbD (3)(1 2)(2)
sys 0.6sy 0.6(63)
N=
=
=
= 4.50 < 6.21
ss
ss
8.4
By compression:
4T
4(12.6)
sc =
=
= 16.8 ksi
LtD (3)(1 2)(2)
sy (pulley ) 40
N=
=
= 2.38 < 6.21
sc
16.8
Answer N = 2.38
b) Flat key, b = ½ in, t = 3/8 in
By shear:
2T
2(12.6 )
ss =
=
= 8.4 ksi
LbD (3)(1 2)(2)
sys 0.6sy 0.6(63)
N=
=
=
= 4.50 < 6.21
ss
ss
8.4
By compression:
4T
4(12.6)
sc =
=
= 22.4 ksi
LtD (3)(3 8)(2)
sy (pulley ) 40
N=
=
= 1.78 < 6.21
sc
22.4
Answer N = 1.78
527.
A cast-steel (SAE 080, N & T) pulley is keyed to a 2 1/2-in. shaft by means of a standard square
key, 3 ½-in. long, made of cold-drawn SAE 1015. The shaft is made of cold-drawn AISI 1045. If
the shaft is in virtually pure torsion, and turns at 420 rpm, what horsepower could the assembly
safely transmit (steady loading)?
Solution:
Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi
Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi
Shaft, AISI 1045, cold drawn, Table AT 7, sy = 85 ksi
N = 420 rpm
L = 3 ½ in = 3.5 in
D = 2 ½ in
Table AT 19, D = 2 ½ in
b = 5/8 in, t = 7/16 in
Square Key, b = t = 5/8 in
N = 1.5 for steady loading (smooth)
For shaft:
7
SECTION 8 – KEYS AND COUPLINGS
3
ssπD 3 0.6 syπD
0.6(85)(π )(2.5)3
T=
=
=
= 104.31 in − kips
16
N(16)
1.5(16)
Key:
By shear:
s LbD
T= s
2
sys 0.6 sy 0.6(63)
ss =
=
=
= 25.2 ksi
N
N
1. 5
(25.2 )(3.5)(5 8)(2.5)
T=
= 68.9 in − kips < 104.31 in-kips
2
By compression:
s LtD
T= c
4
sy (pulley ) 40
sc =
=
= 26.67 ksi
N
1.5
(26.67 )(3.5)(5 8)(2.5)
T=
= 36.46 in − kips < 104.31 in-kips
4
Use T = 36.46 in − kips = 36,460 in − kips
Tn
36,460(420)
hp =
=
= 243 hp
63,000
63,000
528.
The same as 527, except that the diameter is 3 in. and the length of the key is 5 in.
Solution:
Pulley, Cast steel, SAE 080, N & T, Table AT 6, sy = 40 ksi
Key, SAE 1015, cold drawn, Table AT 7, sy = 63 ksi
Shaft, AISI 1045, cold drawn, Table AT 7, sy = 85 ksi
N = 420 rpm
L = 5 in
D = 3 in
Table AT 19, D = 3 in
b = 3/4 in, t = 1/2 in
Square Key, b = t = 3/4 in
N = 1.5 for steady loading (smooth)
For shaft:
3
s πD 3 0.6 sy πD
0.6(85)(π )(3)3
T= s
=
=
= 180.25 in − kips
16
N(16 )
1.5(16 )
Key:
By shear:
s LbD
T= s
2
8
SECTION 8 – KEYS AND COUPLINGS
0.6(63)
= 25.2 ksi
N
N
1. 5
(25.2)(5)(3 4)(3)
T=
= 141.75 in − kips < 180.25 in-kips
2
By compression:
s LtD
T= c
4
sy (pulley ) 40
=
= 26.67 ksi
sc =
N
1.5
(26.67 )(5)(3 4)(3)
T=
= 75 in − kips < 180.25 in-kips
4
ss =
sys
=
0.6 sy
=
Use T = 75 in − kips = 75,000 in − kips
Tn
75,000(420)
hp =
=
= 500 hp
63,000
63,000
MISCELLANEOUS KEYS
529. Two assemblies, one with one feather keys, are shown, with the assumed positions of the
normal forces N. Each assembly is transmitting a torque T. Derive an equation for each case
giving the axial force needed to slide the hub along the shaft (f = coefficient of friction). Does
either have an advantage in this respect?
Solution:
ND
2
2T
N=
D
a) T =
Axial force = F = fN =
2 fT
D
2ND
= ND
2
T
N=
D
b) T =
fT
D
Assembly (b) is stronger than assembly (a) which has an axial force half that of assembly (b).
Axial force = F = fN =
9
SECTION 8 – KEYS AND COUPLINGS
530.
A 1 11/16-in. shaft rotating at 200 rpm, carries a cast-iron gear keyed to it by a ¼ x 1 ¼-in.
Woodruff key; shaft material is cold-finished SAE 1045. The power is transmitted with mild
shock. What horsepower may be safely transmitted by the key, (a) if it is made of cold-drawn
SAE 1118? (b) if it is made of SAE 2317, OQT 1000 F? (c) How many keys of each material are
needed to give a capacity of 25 hp? Specify a choice.
Solution:
Only shear is used.
D = 1 11/16 in
n = 200 rpm
Woodruff key = ¼ x 1 ¼ in
N = 2 for mild shock
Shear force for key
2T
F=
= ss As
D
s AD
T= s s
2
Table 10.1, ¼ x 1 ¼ in Woodruff Key is Key No. 810
Shear area, As = 0.296 sq. in.
(a) Key, cold-drawn 1118, Table AT 7, sy = 75 ksi
sys 0.6 sy 0.6(75)
=
=
= 22.5 ksi < 24.06 ksi
ss =
N
N
2
(22.5)(0.296)1 11 
 16  = 5.62 in − kips = 5620 in − lb
T=
2
(5620)(200)
Tn
hp =
=
= 17.84 hp
63,000
63,000
(b) Key, SAE 2317, OQT 1000F, Table AT 7, sy = 79 ksi
sys 0.6 sy 0.6(79)
ss =
=
=
= 23.7 ksi < 24.06 ksi
N
N
2
(23.7 )(0.296 )1 11 
 16  = 5.92 in − kips = 5920 in − lb
T=
2
(5920)(200)
Tn
hp =
=
= 18.79 hp
63,000
63,000
(c) Number of keys for (a) = 25 / 17.84 = 1.4 or 2 keys
Number of keys for (b) = 25 / 18.79 = 1.33 or 2 keys
Select (b) which is stronger.
531.
A 3/16 x 1-in. Woodruff key is used in a 1 3/16-in. shaft (cold-drawn SAE 1045). (a) If the key is
made of the same material, will it be weaker or stronger than the shaft in pure torsion? (b) If the
key is made of SAE 4130, WQT 1100 F, will it be weaker or stronger? For the purposes here, the
weakening of the shaft by the keyway is ignored.
10
SECTION 8 – KEYS AND COUPLINGS
Solution:
Woodruff key, 3/16 x 1 in.
D = 1 3/16 in
Shaft: Cold drawn, SAE 1045 (Table AT 8) sy = 85 ksi
s AD
T= s s
2
Table 10.1, 3/16 x 1 in., Woodruff key is Key no. 608.
Shear area = As = 0.178 sq. in.
(a) Key material = Shaft Material
In yield: For key
 3
0.6(85)(0.178)1 
ss As D 0.6sy As D
 16  = 5.39 in − kips
T=
=
=
2
2
2
For shaft:
3
 3
0.6(85)(π )1 
3
ssπD 3 0.6 syπD
 16  = 16.77 in − kips
T=
=
=
16
16
16
Therefore the key is weaker.
(b) Key material = SAE 4130, WQT 1100, Table AT 7, sy = 114 ksi
In yield: For key
 3
0.6(114)(0.178)1 
ss As D 0.6 sy As D
 16  = 7.23 in − kips
T=
=
=
2
2
2
For shaft:
3
 3
0.6(85)(π )1 
3
ssπD 3 0.6 syπD
 16  = 16.77 in − kips
T=
=
=
16
16
16
Therefore the key is weaker.
532.
A 2-in. shaft (cold-finished SAE 1137) is connected to a hub by a 3/8-in. radial taper pin made of
4150, OQT 1000 F. (a) What horsepower at 1800 rpm would be transmitted when the pin is
about to be sheared off? (b) For this horsepower, what peak torsional stress may be repeated in
the shaft? Is the shaft safe from fatigue at this stress?
Solution:
D = 2 in
d = 3/8 in
n = 1800 rpm
(a) For pin material , 4150, OQT 1000 F, Table AT9, su = 193.5 ksi
 πd 2 
 πd 2 
 = ss 

F = ss (2 As ) = ss (2)
 4 
 2 
 πd 2  D  1
FD
  = ssπd 2D
T=
= ss 
2
 2  2  4
s s = sus = 0.75 su = 0.75(193.5 ) = 145.1 ksi
11
SECTION 8 – KEYS AND COUPLINGS
2
1
 3
T = (145.1)(π )  (2 ) = 32.05 in − kips = 32,050 in − lb
4
8
(32,050 )(1800)
Tn
hp =
=
= 915.7 hp
63,000
63,000
(b) For shaft, cold-finished, SAE 1137, Table AT 8, su = 103 ksi
s 
 103 
sns = 0.6 sn′ = 0.6 u  = 0.6
 = 30.96 ksi
 2 
2
But,
16T 16(32,050)
ss = 3 =
= 20,404 psi = 20.4 ksi < 30.96 ksi
πD
π (2)3
s
30.96
N = ns =
= 1.52 > 1.5, therefore safe from fatigue at this stress.
ss
20.4
533.
A 20-in. lever is keyed to a 1 7/8-in. shaft (cold-finished SAE 1141) by a radial taper pin whose
mean diameter is 0.5 in.; pin material, C1095, OQT 800 F. The load on the lever is repeatedly
reversed; N = 2 on endurance strength. What is the safe lever load (a) for the shaft, (b) for the
pin key (shear only), (c) for the combination?
Solution:
T = FL where F = safe lever load.
L = 20 in
D = 1 7/8 in = 1.875 in
Shaft Material, cold finished, SAE 1141, Table AT 10.
sy = 90 ksi
sy
sn
= 1.8
90
= 50 ksi
1.8
Pin Material, C1095, OQT 800 F, Table AT 9.
su = 176 ksi
sn = 0.5su = 0.5(176 ) = 88 ksi
(a) For the shaft.
0.6sy 0.6(50)
ss =
=
= 15 ksi
N
2
s πD 3 15(π )(1.875)3
T= s
=
= 19.414 in − kips = 19,414 in − lbs
16
16
T = FL
19,414 = F (20 )
F = 970.7 lb
(b) For the pin.
0.6 sy 0.6(88)
ss =
=
= 26.4 ksi
N
2
sn =
12
SECTION 8 – KEYS AND COUPLINGS
2T d
T
4T
=
= 3
π 2
2 As
d (d ) πd
4
s πd 3 26.4(π )(0.5)3
T= s
=
= 2.592 in − kips = 2592 in − lbs
4
4
T = FL
2592 = F (20 )
F = 129.6 lb
(c) For the combination.
Use F = 129.6 lb
ss =
534.
A lever is keyed to a 2 ½-in. shaft made of SAE 1035, as rolled, by a radial taper pin, made of SAE
1020, as rolled. A load of 200 lb. is applied to the lever 22 in. from the center of the shaft. (a)
What size pin should be used for N = 3 based on the yield strength in shear? (b) Let the hub
diameter be 5 in. and assume that the part of the pin in the hub is uniformly loaded cantilever
beam. Compute the bending stress and comment on the bending strength (especially if the
loading varies).
Solution:
Shaft material, SAE 1035, as rolled, Table AT 7, sy = 55 ksi
Pin material, SAE 1020, as rolled, Table AT 7, sy = 48 ksi
F = 200 lb, L = 22 in, N = 3
D = 2 ½ in, Dh = 5 in
T = FL = (200)(22) = 4400 in-lb = 4.4 in-kips
(a) For the pin:
0.6 sy 0.6(48)
ss =
=
= 9.6 ksi
N
3
2T d
T
4T
ss =
=
= 3
π 2
2 As
d (d ) πd
4
4(4.4)
9.6 =
πd 3
d = 0.836 in
7
Use d = in = 0.875 in
8
(b) For the bending stress.
As cantilever beam let
1
L = (Dh − D )
2
1
L = (5 − 2.5) = 1.25 in
2
From Table AT 2.
wL2 FL
M=
=
2
2
Where F is the uniform load.
13
SECTION 8 – KEYS AND COUPLINGS
2T
4T
4(4400)
=
=
= 2347 lb
Dm Dh + D 5 + 2.5
F = 1174 lb
FL (1174)(1.25)
M= =
= 734 lb
2
2
Bending stress
32M 32(734)
sb = 3 =
= 11,160 psi = 11.16 ksi
πd
π (0.875)3
If the loading varies and factor of safety of 3.
sn = Nsb = 3(11.16 ) = 33.48 ksi
Pin material, SAE 1020, as rolled, Table AT 7, sy = 48 ksi, su = 65 ksu.
sn′ = 0.5 su = 0.5(65 ) = 32.5 ksi
The bending stress is nearly safe as the load varies.
2F =
535.
A sprocket, transmitting 10 hp at 100 rpm, is attached to a 1 7/16 in. shaft as shown in Fig.
10.15, p. 290., Text; E = 3-1/2 in. What should be the minimum shear pin diameter if the
computed stress is 85% of the breaking stress mentioned in the Text?
Solution:
hp = 10 hp, n = 100 rpm, D = 1 7/16 in = 1.4375 in, E = 3 ½ in = 3.5 in
63,000hp 63,000(10)
T=
=
= 6300 in − lb
n
100
T 6300
F= =
= 1800 lb
E
3.5
From text, Page 290. Breaking stress = 50,000 psi
s s = 0.85(50 ,000 ) = 42 ,500 psi
4F
ss = 2
πd
4(1800 )
42,500 =
πd 2
d = 0.2322 in use ¼ in.
536.
A gear is attached to a 2-in. shaft somewhat as shown in Fig. 10-15, p. 290, Text; E = 3 5/16 in.;
minimum shear-pin diameter = 3/8 in. with a rated torque of 22 in-kips. (a) For this torque,
compute the stress in the shear pin. (b) From the ferrous metals given in the Appendix, select
those that would give a resisting torque of about 110% of the rated value. Choose one,
specifying its heat treatments or other conditions.
Solution:
D = 2 in
E = 3 5/16 in = 3.3125 in
d = 3/8 in
T = 22 in-kips
(a) Stress in shear-pin
T
22
F= =
= 6.64 lb
E 3.3125
14
SECTION 8 – KEYS AND COUPLINGS
4(6.64)
= 60.12 ksi
π (3 8)2
(b) Select material. sus = 1.1(60.12 ) = 66.13 ksi
From appendix, Table AT 7, select Cold drawn, C1020 with sus = 66 ksi
ss =
SPLINES
537.
A shaft for an automobile transmission has 10 splines with the following dimensions: D = 1.25
in., d = 1.087 in., and L = 1.000 in. (see Table 10.2, p. 287, Text). Determine the safe torque
capacity and horsepower at 3600 rpm of this sliding connection.
Solution:
D = 1.25 in, d = 1.087 in, L = 1.000 in, Nt = 10, n = 3600 rpm
T = (1000 )(hL )(rm )(Nt ) in − lb
D +d
rm =
4
Table 10.2 for 10 splines, sliding connection
h = 0.095D
Then
h = 0.095(1.25 ) = 0.11875 in
But
D − d 1.25 − 1.087
h=
=
= 0.0815 in actual
2
2
1.25 + 1.087
rm =
= 0.58425 in
4
T = (1000 )(0.0815)(1.000 )(0.58425 )(10 ) = 476.2 in − lb - ans
Tn
(476.2 )(3600)
hp =
=
= 27.2 hp - ans
63000
63000
538.
The rear axle of an automobile has one end splined. For this fitting there are ten splines, and D =
1.31 in., d = 1.122 in., and L = 1 15/16 in. The minimum shaft diameter is 1 3/16 in. (a)
Determine the safe torque capacity of the splined connection, sliding under load. (b) Determine
the torque that would have the splines on the point of yielding if the shaft is AISI 8640, OQT
1000 F, if one fourth of the splines are in contact. (c) Determine the torsional stress in the shaft
corresponding to each of these torques.
Solution:
D = 1.31 in, d = 1.122 in, L = 1 15/16 = 1.9375 in, Nt = 10
Dr = 1 3/16 in = 1.1875 in
T = (1000 )(hL )(rm )(Nt ) in − lb
D +d
rm =
4
Table 10.2 for 10 splines, sliding connection
h = 0.095D
Then
h = 0.095(1.31) = 0.12445 in
15
SECTION 8 – KEYS AND COUPLINGS
But
D − d 1.31 − 1.122
=
= 0.094 in (actual)
2
2
1.31 + 1.122
rm =
= 0.608 in
4
(a) Safe Torque
T = (1000 )(0.094 )(1.9375)(0.608 )(10 ) = 1107.32 in − lb - ans
(b) Torque by splines required on the point of yielding with one fourth of splines in contact (Page
288).
From Table AT-7, AISI 8640, OQT 1000 F. sy = 150 ksi, ss = sys = 0.6sy = 0.6(150) = 90 ksi
 s πDL  D   (90)(π )(1.31)(1.9375)  1.31 
T = s
  =
 2  = 58.76 in − kips
8
 8  2  
T = 58,760 in − lbs
(c) Torsional stress in the shaft
From safe torque of 1107.32 in-lb
16T 16(1107.32 )
ss = 3 =
= 3368 psi - ans
πDr π (1.1875)3
From torque at the point of yield
16T 16(58,760)
ss = 3 =
= 178,711 psi - ans (too high)
πDr π (1.1875)3
h=
539.
An involute splined connection has 10 splines with a pitch Pd of 12/24 (a) Determine the
dimension of this connection. (b) Compute the length of spline to have the same torsional
strength as the shaft when one fourth the splines carry the load; minimum shaft diameter is
9/16 in. (no sliding). Check for compression.
Solution:
Nt = 10, Pd = 12, Dr = 9/16 in = 0.5625 in
(a) Dimension D
N 10
D = t = = 0.8333 in
Pd 12
(b) Length of spline (same torsional strength as the shaft when one fourth the splines carry the load
(Page 288).
D 3 (0.5625)3
L= r =
= 0.2136 in
D
0.8333
Check for compression.
Failure in compression is not likely (Page 289) and can be checked by using the projected
contact area.
Projected contact area:
Ac = 0.2LD = 0.2(0.2136 )(0.8333) = 0.0356 in 2 based on one-fourth of the teeth being under load.
COUPLINGS
540.
A flange coupling has the following dimensions (Fig. 10.19, p. 291, Text): d = 5, D = 8 5/8, H = 12
¼, g = 1 ½, h = 1, L = 7 ¼ in.; number of bolts = 6; 1 ¼ x 1 ¼-in. square key. Materials: key, colddrawn AISI 1113; shaft, cold-rolled, AISI 1045; bolts, SAE grade 5 (§5.8). Using the static
16
SECTION 8 – KEYS AND COUPLINGS
approach with N = 3.3 on yield strengths, determine the safe horsepower that this connection
may transmit at 630 rpm.
Solution:
d = 5 in
D = 8 5/8 in = 8.625 in
H = 12 ¼ in = 12.25 in
g = 1 ½ in = 1.5 in
h = 1 in
L = 7 ¼ in = 7.25 in
N = 3.3
nb = 630 rpm
Square key = 1 ¼ in x 1 ¼ in
Materials:
Key: cold-drawn AISI 1113, Table AT 7, sy = 72 ksi, sys = 0.6sy = 0.6(72) = 43.2 ksi
Shaft: cold-rolled, AISI 1045, Table AT 8, sy = 85 ksi, sys = 0.6sy = 0.6(85) = 51 ksi
Bolt: SAE Grade 5, h = 1 in. sy = 81 ksi, sys = 0.6sy = 0.6(81) = 48.6 ksi
No given material for the flange.
Bolts in shear:
sys 48.6
ss =
=
= 14.73 ksi
N
3. 3
πh 2
F=
N b ss
4
FH πh 2 Nb ss H
T=
=
2
8
2
π (1) (6 )(14.73)(12.25)
T=
= 425.158 in − kips
8
T = 425,158 in − lbs
(425,158)(630)
Tn
hp =
=
= 4252 hp
63,000
63,000
Bolts in compression:
sy 81
sc = =
= 24.55 ksi
N 3.3
F = Nb hgs c
17
SECTION 8 – KEYS AND COUPLINGS
FH Nb hgsc H
=
2
2
(6)(1)(1.5)(24.55)(12.25)
T=
= 1353.319 in − kips
2
T = 1,353,319 in − lbs
(1,353,319)(630)
Tn
hp =
=
= 13,533 hp
63,000
63,000
T=
Key in shear:
sys 43.2
ss =
=
= 13.09 ksi
N
3. 3
s bdL (13.09 )(1.25)(5)(7.25)
T= s
=
= 296.570 in − kips
2
2
T = 296,570 in − lbs
(296,570)(630)
Tn
hp =
=
= 2966 hp
63,000
63,000
Key in compression:
sy 72
sc = =
= 21.82 ksi
N 3.3
s tdL (21.82)(1.25)(5)(7.25)
T= c
=
= 247.180 in − kips
4
4
T = 247 ,180 in − lbs
(247,180 )(630)
Tn
hp =
=
= 2472 hp
63,000
63,000
Shaft in shear:
sys 51
ss =
=
= 15.45 ksi
N 3.3
πd 3 ss π (5)3 (15.45)
T=
=
= 379.200 in − kips
16
16
T = 379,200 in − lbs
(379,200)(630)
Tn
hp =
=
= 3792 hp
63,000
63,000
The safest horsepower is the lowest which is 2472 hp.
541.
A cast-iron (ASTM 25) jaw clutch with 4 jaws transmits 50 hp at 60 rpm. The inside diameter of
the jaws is 3 in. Considering rough handling, choose N = 8 on ultimate strengths. Make
reasonable and conservative assumptions and compute (a) the outside diameter of the jaws, (b)
the length of jaws h.
18
SECTION 8 – KEYS AND COUPLINGS
Solution:
For ASTM 25, suc = 97 ksi, in shear sus = 35 ksi (Table AT 6)
63,000hp 63,000(50)
T=
=
= 52,500 in − lbs
n
60
T = 52.5 in − kips
(a) The outside diameter of the jaws
s
35
ss = us = = 4.375 ksi
N
8
Assume Dm as the average diameter, t = thickness = Do – Di , Nj = number of jaws
Shear area,
1  πD 
1  π  D + Di  Do − Di 
As =  m (t ) =   o




2  Nj 
2  N j  2  2 
1  π  D 2 − Di2  π 2
=
As =   o
Do − Di2
2  4  4  32
2T
4T
F=
=
Dm Do + Di
F
4T
32
128T
ss = =
⋅
=
2
2
π (Do + Di ) Do2 − Di2
As Do + Di π Do − Di
128(52.5)
4.375 =
π (Do + 3) Do2 − 9
By trial and error.
Do = 7.466 in or D o = 7.5 in
(b) The length of jaws h.
s
97
sc = uc =
= 12.125 ksi
N
8
N j h(Do − Di )
Ac =
2
F
2F
2F
F
sc =
=
=
=
Ac N j h(Do − Di ) 4h(Do − Di ) 2h(Do − Di )
(
(
(
sc =
)
)
(
)
)
4T (Do + Di )
2T
=
2
2h(Do − Di ) h Do − Di2
(
)
19
SECTION 8 – KEYS AND COUPLINGS
2(52.5)
h (7.5)2 − (3)2
3
h = 0.1833 in or h = in
16
12.125 =
542.
[
]
The universal joint shown is made of AISI 3150, OQT 1000 F; a = 2 7/16 in., D = 9/16; n = 400
rpm. (a) What torque may be transmitted for shear of the pin (N = 5 on ultimate)? (b)
Considering the pin as a simply supported beam of length a with the load distributed from a
maximum at the outer (triangular), compute the safe transmitted torque (Same N). (c) In order
not to have excessive wear on the pin, the average bearing pressure should not excced 3 ksi.
Compute this transmitted torque. (d) What is the safe power?
Solution:
For AISI 3150, OQT 1000 F, Table AT 7, su = 151 ksi, sus = 113 ksi
N=5
pb = 3 ksi
a = 2 7/16 in = 2.4375 in
D = 9/16 in = 0.5625 in
n = 400 rpm
(a) Torque transmitted for shear of the pin.
s
113
ss = us =
= 22.6 ksi
N
5
Each shear area
 πD 2  (22.6)(π )(0.5625)2
 =
F = ss 
= 5.616 kips
4
 4 
Fa
T = 2 = Fa = (5.616)(2.4375) = 13.687 in − kips
2
T = 13,687 in − lbs
(b) Torque transmitted for shear of the pin (simply supported beam)
Fa
T =M=
3
(5.616)(2.4375)
T=
= 4.563 in − kips
3
T = 4,563 in − lbs
20
SECTION 8 – KEYS AND COUPLINGS
(c) Torque transmitted for shear of the pin (pb = 3 ksi)
F
F
F
pb =
=
=
Ab D(a 2) 2Da
F
3=
2(0.5625)(2.4375)
F = 8.23 ksi
Fa
T =M=
3
(8.23)(2.4375)
T=
= 6.687 in − kips
3
T = 6,687 in − lbs
(d) Safe power
(4,563)(400)
Tn
hp =
=
= 28.97 hp
63,000
63,000
544.
A diagrammatic representation of a universal joint is shown, two yoke parts, the type being
similar to Figs. 10.28 and 12.10, Text. The pin extensions have a diameter D = ¾ in.; a = 11/16 in.,
material of all parts is 4340, OQT 800. Let N = 4 on ultimate stresses; n = 2400 rpm. Compute the
safe torque for (a) shear of pins, (b) the pin extensions in bending, assuming that the load
distribution is from zero at the outside pin ends to a maximum at the inside yoke surfaces, (c) an
average bearing pressure on pins of 4 ksi. (d) What is the corresponding horsepower capacity?
Solution:
For AISI 4340, OQT 800 F, Table AT 7, su = 221 ksi, sus = 0.75su = 0.75(221) = 166 ksi
N=4
pb = 4 ksi
a = 11/16 in = 0.6875 in
D = 3/4 in = 0.75 in
n = 2400 rpm
(a) Torque transmitted for shear of the pin.
s
166
ss = us =
= 41.5 ksi
N
4
Each shear area
 πD 2  (41.5)(π )(0.75)2
 =
F = ss 
= 18.33 kips
4
 4 
T = M = Fa = (18.33)(0.6875) = 12.602 in − kips
T = 12,602 in − lbs
21
SECTION 8 – KEYS AND COUPLINGS
(b) Torque transmitted for shear of the pin (simply supported beam)
2Fa
T =M=
3
2(18.33)(0.6875)
T=
= 8.401 in − kips
3
T = 8,401 in − lbs
(c) Torque transmitted for shear of the pin (pb = 4 ksi)
F
F
pb =
=
Ab Da
F
4=
(0.75)(0.6875)
F = 2.063 ksi
2Fa
T =M=
3
2(2.063)(0.6875)
T=
= 0.9455 in − kips
3
T = 945.5 in − lbs
(d) Safe power
(945.5)(2400)
Tn
hp =
=
= 36.02 hp
63,000
63,000
-
End -
22
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
LIGHTLY LOADED BEARINGS
551.
(a) A 3 x 3 – in. full bearing supports a load of 900 lb., c d D = 0.0015 ,
n = 400 rpm . The temperature of the SAE 40 oil is maintained at 140 oF.
Considering the bearing lightly loaded (Petroff), compute the frictional torque,
fhp, and the coefficient of friction. (b) The same as (a) except that the oil is
SAE 10W.
Solution.
(a) T f =
µπ DLvips  D 
 
2
(cd 2)
L = 3 in
D = 3 in
πDn π (3)(400)
vips =
=
= 20π ips
60
60
c d D = 0.0015
SAE 40 oil, 140 oF, Figure A16.
µ = 7.25 µreyns
µπ DLvips 7.25 ×10 −6 (π )(3)(3)(20π )
F=
=
= 17.173 lb
(cd 2)
(0.0015 2)
(
)
D
3
T f = F   = (17.173)  = 25.76 in − lb
2
2
Fvm
fhp =
33,000
π Dn π (3)(400 )
vm =
=
= 314.16 fpm
12
12
Fvm
(17.173)(314.16) = 0.1635 hp
fhp =
=
33,000
33,000
F 17.173
f =
=
= 0.0191
W
900
(b) SAE 10W oil, 140 oF, Figure A16.
µ = 2.2 µreyns = 2.2 ×10 −6 reyn
µπ DLvips 2.2 ×10 −6 (π )(3)(3)(20π )
F=
=
= 5.211 lb
(cd 2)
(0.0015 2)
(
)
D
3
T f = F   = (5.211)  = 7.817 in − lb
2
2
Fvm
fhp =
33,000
Page 1 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
vm =
π Dn
π (3)(400 )
= 314.16 fpm
12
12
Fvm
(5.211)(314.16) = 0.0496 hp
fhp =
=
33,000
33,000
F 5.211
f =
=
= 0.00579
W
900
553.
=
The average pressure on a 6-in. full bearing is 50 psi, cd = 0.003 in. , L D = 1 .
While the average oil temperature is maintained at 160 oF with n = 300 rpm ,
the frictional force is found to be 13 lb. Compute the coefficient of friction
and the average viscosity of the oil. To what grade of oil does this
correspond?
Solution:
W
p=
LD
D = 6 in.
L D =1
L = 6 in.
W = pLD = (50)(6)(6 ) = 1800 lb
F = 13 lb
Coefficient of Friction
F
13
f =
=
= 0.0072
W 1800
µπ DLvips
F=
(cd 2)
πDn π (6)(300)
vips =
=
= 30π ips
60
60
µπ DLvips µ (π )(6)(6)(30π )
F=
=
= 13 lb
(cd 2)
(0.003 2)
µ = 1.8 ×10 −6 reyn = 1.8 µreyns
Figure AF 16, 160 oF use SAE 10W or SAE 20W
FULL BEARINGS
554.
The
load
on
a
4-in.
full
bearing
is
2000
lb.,
n = 320 rpm ; L D = 1 ; cd D = 0.0011 ; operating temperature = 150 oF;
ho = 0.00088 in . (a) Select an oil that will closely accord with the started
conditions. For the selected oil determine (b) the frictional loss (ft-lb/min), (c)
the hydrodynamic oil flow through the bearing, (d) the amount of end leakage,
Page 2 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
(e) the temperature rise as the oil passes through the bearing, (f) the maximum
pressure.
Solution:
(a) D = 4 in
L D =1
L = 4 in
cd = 0.0011D = 0.0011(4) = 0.0044 in
ho = 0.00088 in
2h
2(0.00088)
ε = 1− o = 1−
= 0 .6
cd
0.0044
Table AT 20
ε = 0.6 , L D = 1
Sommerfield Number
µns  D 
2
 
p  cd 
320
ns =
= 5.333 rps
60
W
2000
p=
=
= 125 psi
LD (4)(4)
cd D = 0.0011
S=
µ (5.333) 
2
1 
0.121 =


125  0.0011 
µ = 3.4 ×10 −6 reyn = 3.4 µreyns
Figure AF-16, 150 oF, use SAE 30 or SAE 20 W
Select SAE 30, the nearest
µ = 3.9 ×10−6 reyn
(b) Table AT 20, L D = 1 , ε = 0.6
r
f = 3.22
cr
r
D
1
=
=
cr cd 0.0011
 1 

 f = 3.22
 0.0011 
f = 0.003542
F = f W = (0.003542)(2000 ) = 7.084 lb
Page 3 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
vm =
π Dn
=
π (4)(320)
= 335.1 fpm
12
12
Frictional loss = Fvm = (7.084 )(335.1)2374 ft − lb min
(c) Table AT 20, L D = 1 , ε = 0.6
q
= 4.33
rcr ns L
D
r = = 2.0 in
2
c
0.0044
cr = d =
= 0.0022 in
2
2
ns = 5.333 rps
L = 4 in
q = 4.33rcr ns L = 4.33(2.0 )(0.0022 )(5.333)(4 ) = 0.4064 in 3 sec
(d) Table AT 20, L D = 1 , ε = 0.6
qs
= 0.680
q
qs = 0.680q = 0.680(0.4064 ) = 0.2764 in 3 sec
(e) Table AT 20, L D = 1 , ε = 0.6
ρ c∆to
= 14.2
p
ρ c = 112 , p = 125 psi
14.2 p 14.2(125)
∆to =
=
= 15.85 o F
ρc
112
(f) Table AT 20, L D = 1 , ε = 0.6
p
= 0.415
pmax
125
pmax =
= 301.2 psi
0.415
555.
A 4-in., 360o bearing, with L D = 1.1 (use table and chart values for 1), is to
support 5 kips with a minimum film thickness 0.0008 in.; cd = 0.004 in. ,
n = 600 rpm . Determine (a) the needed absolute viscosity of the oil .(b)
Suitable oil if the average film temperature is 160 F, (c) the frictional loss in
hp. (d) Adjusting only ho to the optimum value for minimum friction,
determine the fhp and compare. (e) This load varies. What could be the
magnitude of the maximum impulsive load if the eccentricity ration ε
becomes 0.8? Ignore “squeeze” effect.
Page 4 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
Solution:
D = 4 in
L = 1.1D = 1.1(4) = 4.4 in
W
5000
p=
=
= 284 psi
LD (4.4)(4)
ho = 0.0008 in
cd = 0.004 in.
2h
2(0.0008)
ε = 1− o = 1−
= 0 .6
cd
0.004
600
η=
= 10 rps
60
(a)
Table AT20, L D = 1 , ε = 0.6
S = 0.121
r
S = 
 cr
2
 µ ns µ n s  D 
 

=
p  cd 
 p
µ (10) 
4 
0.121 =


284  0.004 
µ = 3.4 ×10−6 reyn
2
2
(b) Figure AF16, 160 F
Use SAE 30, µ = 3.2 ×10−6 reyn
(c) Table AT 20, L D = 1 , ε = 0.6
r
f = 3.22
cr
D
f = 3.22
cd
 4 

 f = 3.22
 0.004 
f = 0.00322
F = f W = (0.00322)(5000 lb ) = 16.1 lb
π Dn π (4)(600)
vm =
=
= 628.3 fpm
12
12
Page 5 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
fhp =
(16.1)(628.3) = 0.3065 hp
Fvm
=
33,000
33,000
(d) adjusting ho , cd = 0.004 in.
Table AT 20, L D = 1
ho cr = 0.30 optimum value for minimum friction
r
f = 2.46
cr
D
f = 2.46
cd
 4 

 f = 2.46
 0.004 
f = 0.00246
F = f W = (0.00246)(5000 lb ) = 12.3 lb
π Dn π (4)(600)
vm =
=
= 628.3 fpm
12
12
Fvm
(12.3)(628.3) = 0.234 hp < fhp (c )
fhp =
=
33,000
33,000
(e) ε = 0.8 , Table AT 20, L D = 1
S = 0.0446
r
S = 
 cr
2
 µ ns µ n s

=
p
 p
D
 
 cd 
(3.2 ×10 )(10) 
0.0446 =
−6
p
2
4 

 0.004 
2
p = 717.5 psi
W = pDL = (717.5)(4)(4.4) = 12,628 lb
556.
For an 8 x 4 – in. full bearing, cr = 0.0075 in. , n = 2700 rpm , average
µ = 4 ×10 −6 reyn . (a) What load may this bearing safely carry if the minimum
film thickness is not to be less than that given by Norton, i11.14, Text? (b)
Compute the corresponding frictional loss (fhp). (c) Complete calculations for
the other quantities in Table AT 20, φ , q , qs , ∆to , pmax . Compute the
maximum load for an optimum (load) bearing (d) if cr remains the same, (e)
if ho remains the same.
Page 6 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
Solution:
D × L = 8× 4
L D =1 2
cr = 0.0075 in
r = D 2 = 4 in
µ = 4 ×10 −6 reyn
(a) by Norton, ho = 0.00025D = 0.00025(8) = 0.002 in
ho
0.002
=
= 0.27
cr 0.0075
h
Table AT 20, L D = 1 2 , o = 0.27
cr
S = 0.172
2
 r  µns
S =  
 cr  p
2700
ns =
= 45 rps
60
2
(
)
−6
 4  4 × 10 (45)
S = 0.172 = 

p
 0.0075 
p = 298 psi
W = pDL = (298)(8)(4) = 9536 lb
(b) Table AT 20, L D = 1 2 ,
ho
= 0.27
cr
φ = 38.5o
r
f = 4.954
cr
D
f = 4.954
cd
 4 

 f = 4.954
 0.004 
f = 0.0093
F = f W = (0.0093)(9536 lb ) = 88.7 lb
π Dn π (8)(2700)
vm =
=
= 5655 fpm
12
12
Fvm
(88.7 )(5655) = 15.2 hp
fhp =
=
33,000
33,000
Page 7 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
(c) Table AT 20, L D = 1 2 ,
ho
= 0.27
cr
φ = 38.5o
q
= 5.214
rcr ns L
q = 5.214rcr ns L = 5.214(4 )(0.0075)(45)(4 ) = 28.2 in 3 sec
qs
= 0.824
q
qs = 0.824(28.2 ) = 23.2 in 3 sec
ρ c∆t
= 20.26
p
20.26(298)
∆t =
= 54 o F
112
p
= 0.3013
pmax
298
pmax =
= 989 psi
0.3013
h
To solve for maximum load, Table AT 20, L D = 1 2 , o = 0.43
cr
r
S = 
 cr
2
 µns

= 0.388
 p
(d) cr = 0.0075 in
2
(
)
−6
 4  4 × 10 (45)
S = 0.388 = 

p
 0.0075 
p = 132 psi
W = pDL = (132)(8)(4) = 4224 lb
(e) ho = 0.002 in
ho
= 0.43
cr
0.002
cr =
= 0.00465 in
0.43
2
(
)
−6
4

 4 × 10 (45)
S = 0.388 = 

p
 0.00465 
p = 343.3 psi
W = pDL = (343.3)(8)(4) = 10,986 lb
Page 8 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
557.
A 6 x 6 – in full bearing has a frictional loss of fhp = 11 when the load is
68,500 lb. and n = 1600 rpm ; cr r = 0.001 . (a) Compute the minimum film
thickness. Is this in the vicinity of that for an optimum bearing? (b) What is
the viscosity of the oil and a proper grade for an operating temperature of 160
F? (c) For the same ho , but for the maximum-load optimum, determine the
permissible load and the fhp.
Solution:
L = 6 in
D = 6 in
L D =1
r = D 2 = 3 in
cr r = 0.001
n = 1600 rpm
π Dn π (3)(1600)
vm =
=
= 2513 fpm
12
12
Fvm
fhp =
33,000
33,000(11)
F=
= 144.45 lb
2513
F 144.45
f =
=
= 0.00211
W 68,500
r
 1 
(a)
f =
(0.00211) = 2.11
cr
 0.001 
r
Table AT 20, L D = 1 ,
f = 2.11
cr
Near the vicinity of optimum bearing
cr = 0.001r = 0.001(3) = 0.003 in
ho = 0.254cr = 0.254(0.003) = 0.0008 in
(b) Table AT 20, L D = 1 ,
r
f = 2.11
cr
S = 0.0652
2
 r  µns
S =  
= 0.388
 cr  p
1600
ns =
= 26.67 rps
60
W
68,500
p=
=
= 1902.8 psi
LD (6)(6)
Page 9 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
2
 1  (µ )(26.67 )
S = 0.0652 = 

 0.001  1902.8
µ = 4.7 ×10 −6 reyn
Figure AF 16, 160 F, use SAE 40.
(c) Table AT 20, L D = 1
optimum bearing, maximum load,
ho
= 0.53
cr
ho
0.0008
=
= 0.0015 in
0.53
0.53
ho
r
= 0.53 , S = 0.214 , f = 4.89
cr
cr
ho the same, cr =
r
S = 
 cr
2
 µns

 p
2
(
)
−6
 3  4.7 × 10 (26.67 )
S = 0.214 = 

p
 0.0015 
p = 2343 psi
W = pDL = (2343)(6)(6) = 84,348 lb
r
f = 4.89
cr
 3 

 f = 4.89
 0.0015 
f = 0.00245
F = f W = 0.00245(84,348) = 206.65 lb
vm = 2513 fpm
(206.65)(2513) = 15.74 hp
Fvm
fhp =
=
33,000
33,000
558.
The maximum load on a 2.25 x 1.6875 in. main bearing of an automobile is
3140 lb. with wide-open throttle at 1000 rpm. If the oil is SAE 20W at 210 F,
compute the minimum film thickness for a bearing clearance of (a) 0.0008 in.
and (b) 0.0005 in. Which clearance results in the safer operating conditions?
Note: Since a load of this order exists for only 20-25o of rotation, the actual
ho does not reach this computed minimum (squeeze effect).
Solution:
D × L = 2.25 × 1.6875 in
L 1.6875
=
= 0.75
D
2.25
Page 10 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
SAE 20 W at 210 oF
µ = 0.96 ×10 −6 reyn
W = 3140 lb
n = 1000 rpm
W
3140
p=
=
= 827 psi
DL (2.25)(1.6875)
1000
ns =
= 16.67 rps
60
D
r = = 1.125 in
2
µ ns  r 
2
 
p  cr 
(a) cr = 0.0008 in
S=
(0.96 ×10 )(16.67 )  1.125 
S=
−6
2
= 0.038
827
 0.0008 
Table AT 20, L D = 3 4 , S = 0.038
L D
1
½
ho cr
0.2
0.2
S
0.0446
0.0923
¾
0.2
0.0685
L D
1
½
ho cr
0.1
0.1
S
0.0188
0.0313
¾
0.1
0.0251
At L D = 3 4
ho  0.038 − 0.0251 
=
(0.2 − 0.1) + 0.1 = 0.13
cr  0.0685 − 0.0251 
ho = 0.13cr = 0.13(0.0008) = 0.0001 in
(b) cr = 0.0005 in
(0.96 ×10 )(16.67 )  1.125 
S=
−6
827
 0.0005 
2
= 0.098
Table AT 20, L D = 3 4 , S = 0.098
Page 11 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
L D
1
½
ho cr
0.2
0.2
S
0.0446
0.0923
¾
0.2
0.0685
L D
1
½
ho cr
0.4
0.4
S
0.121
0.319
¾
0.4
0.220
At L D = 3 4
ho  0.098 − 0.0685 
=
(0.4 − 0.2) + 0.2 = 0.239
cr  0.220 − 0.0685 
ho = 0.239cr = 0.239(0.0005) = 0.00012 in
use cr = 0.0005 in , ho = 0.00012 in
A 360o bearing supports a load of 2500 lb.; D = 5 in. , L = 2.5 in. ,
cr = 0.003 in. , n = 1800 rpm ; SAE 20 W oil entering at 100 F. (a) Compute
the average temperature t av of the oil through the bearing. (An iteration
procedure. Assume µ ; compute S and the corresponding ∆to ; then the
average oil temperature t av = ti + ∆to 2 . If this t av and the assumed µ do not
locate a point in Fig. AF 16 on the line for SAE 20 W oil, try again.) Calculate
(b) the minimum film thickness, (c) the fhp, (d) the amount of oil to be
supplied and the end leakage.
561.
Solution:
D = 5 in
L = 2.5 in
L 2 .5
=
= 0 .5
D
5
cr = 0.003 in
(a) Table AT 20
ρ c∆to
Parameter,
, ρ c = 112
p
r
S = 
 cr
2
 µns

 p
Page 12 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
W
2500
=
= 200 psi
DL (5)(2.5)
1800
ns =
= 30 rps
60
D
r = = 2.5 in
2
cr = 0.003 in
Fig. AF 16, SAE 20 W, Table AT 20, L D = 0.5 , ti = 100 o F
ρ c∆to
∆to o F
Trial µ ( t o F ), reyns
S
p
3.5 x 10-6 (130 F)
0.365
36.56
65
-6
3.2 x 10 (134 F)
0.333
34.08
61
-6
3.4 x 10 (132 F)
0.354
35.71
64
p=
Therefore, use t av = 132 o F , S = 0.354
(b) Table AT 20, L D = 0.5 , S = 0.354
ho
= 0.415
cr
ho = 0.415(0.003) = 0.00125 in
(c) Table AT 20, L D = 0.5 , S = 0.354
r
f = 8.777
cr
 2 .5 

 f = 8.777
 0.003 
f = 0.0105
F = f W = 0.0105(2500 ) = 26.25 lb
π Dn π (5)(1800)
vm =
=
= 2356 fpm
12
12
Fvm
(26.25)(2356) = 1.874 hp
fhp =
=
33,000
33,000
(d) Table AT 20, L D = 0.5 , S = 0.354
q
= 4.807
rcr ns L
q = 4.807 rcr ns L = 4.807(2.5)(0.003)(30 )(2.5) = 2.704 in 3 sec
qs
= 0.7165
q
Page 13 of 63
t av = ti + ∆to 2 o F
132.5
130.5
132.0
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
qs = 0.7165(2.704 ) = 1.937 in 3 sec
PARTIAL BEARINGS
562.
A 2 x 2-in. bearing has a clearance cr = 0.001 in , and ho = 0.0004 in. ,
n = 2400 rpm , and for the oil, µ = 3 × 10−6 reyn . Determine the load, frictional
horsepower, the amount of oil to enter, the end leakage of oil, and the
temperature rise of the oil as it passes through for : (a) a full bearing, partial
bearings of (b) 180o, (c) 120o, (d) 90o, (e) 60o.
Solution:
D = L = 2 in
L D =1
cr = 0.001 in
r = D 2 = 1 in
n = 2400 rpm
ns = 40 rps
µ = 3 ×10−6 reyn
ho = 0.004 in.
ho 0.0004
=
= 0. 4
0.001
cr
πDn π (2)(2400)
vm =
=
= 1257 fpm
12
12
(a) Full bearing
Table AT 20, L D = 1 , ho cr = 0.4
S = 0.121
rf
= 3.22
cr
q
= 4.33
rcr ns L
qs
= 0.680
q
ρ c∆to
= 14.2
p
p
= 0.415
pmax
Load W
Page 14 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
r
S = 
 cr
2
 µ ns

 p
2
(
)
−6
 1  3 ×10 (40 )
0.121 = 

p
 0.001 
p = 992 psi
W = pDL = (992)(2)(2) = 3968 lb
fhp:
F = fW
rf
= 3.22
cr
 1 

 f = 3.22
 0.001 
f = 0.00322
F = f W = (0.00322)(3968) = 12.78 lb
(12.78)(1257 ) = 0.4868 hp
Fvm
fhp =
=
33,000
33,000
Oil flow, q
q
= 4.33
rcr ns L
q
= 4.33
(0.1)(0.001)(40)(2)
q = 0.3464 in3 sec
End leakage
qs
= 0.680
q
qs = 0.68(0.3464 ) = 0.2356 in3 sec
Temperature rise, ∆to
ρ c∆to
= 14.2
p
(112)∆to = 14.2
992
∆to = 126 o F
(b) 180o Bearing
Table AT 21, L D = 1 , ho cr = 0.4
S = 0.128
Page 15 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
rf
= 2.28
cr
q
= 3.25
rcr ns L
qs
= 0.572
q
ρ c∆to
= 12.4
p
Load W
r
S = 
 cr
2
 µ ns

 p
2
(
)
−6
 1  3 × 10 (40 )
0.128 = 

p
 0.001 
p = 937.5 psi
W = pDL = (937.5)(2)(2) = 3750 lb
fhp:
F = fW
rf
= 2.28
cr
 1 

 f = 2.28
 0.001 
f = 0.00228
F = f W = (0.00228)(3750) = 8.55 lb
(8.55)(1257) = 0.3257 hp
Fvm
fhp =
=
33,000
33,000
Oil flow, q
q
= 3.25
rcr ns L
q
= 3.25
(0.1)(0.001)(40)(2)
q = 0.26 in3 sec
End leakage
qs
= 0.572
q
qs = 0.572(0.26 ) = 0.1487 in3 sec
Temperature rise, ∆to
Page 16 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
ρ c∆to
= 12.4
p
(112)∆to = 12.4
937.5
∆to = 104 o F
(c) 12o Bearing
Table AT 22, L D = 1 , ho cr = 0.4
S = 0.162
rf
= 2.16
cr
q
= 2.24
rcr ns L
qs
= 0.384
q
ρ c∆to
= 15
p
Load W
r
S = 
 cr
2
 µ ns

 p
2
(
)
−6
 1  3 × 10 (40 )
0.162 = 

p
 0.001 
p = 741 psi
W = pDL = (741)(2 )(2) = 2964 lb
fhp:
F = fW
rf
= 2.16
cr
 1 

 f = 2.16
 0.001 
f = 0.00216
F = f W = (0.00216)(2964) = 6.4 lb
Fvm
(6.4)(1257 ) = 0.2438 hp
fhp =
=
33,000
33,000
Oil flow, q
Page 17 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
q
= 2.24
rcr ns L
q
= 2.24
(0.1)(0.001)(40)(2)
q = 0.1792 in3 sec
End leakage
qs
= 0.384
q
q s = 0.384(0.1792 ) = 0.0688 in 3 sec
Temperature rise, ∆to
ρ c∆to
= 15
p
(112)∆to = 15
741
∆to = 99 o F
(d) 60o Bearing
L D = 1 , ho cr = 0.4
S = 0.450
rf
= 3.29
cr
q
= 1.56
rcr ns L
qs
= 0.127
q
ρ c∆to
= 28.2
p
Load W
r
S = 
 cr
2
 µ ns

 p
2
(
)
−6
 1  3 × 10 (40 )
0.450 = 

p
 0.001 
p = 267 psi
W = pDL = (267 )(2)(2) = 1068 lb
fhp:
F = fW
Page 18 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
rf
= 3.29
cr
 1 

 f = 3.29
 0.001 
f = 0.00329
F = f W = (0.00329)(1068) = 3.514 lb
Fvm
(3.514)(1257) = 0.1339 hp
fhp =
=
33,000
33,000
Oil flow, q
q
= 1.56
rcr ns L
q
= 1.56
(0.1)(0.001)(40)(2)
q = 0.1248 in3 sec
End leakage
qs
= 0.127
q
qs = 0.127(0.1248) = 0.0158 in3 sec
Temperature rise, ∆to
ρ c∆to
= 28.2
p
(112)∆to = 28.2
267
∆to = 67 o F
563.
A 2 x 2 in. bearing sustains a load of W = 5000 lb. ; cr = 0.001 in. ;
n = 2400 rpm ; µ = 3 × 10−6 reyn . Using Figs. AF 17 and AF 18, determine the
minimum film thickness and the frictional loss (ft-lb/min.) for (a) a full
bearing, and for partial bearings of (b) 180o, (c) 120o, (d) 90o, (e) 60o.
Solution:
L = 2 in
D = 2 in
W = 5000 lb
cr = 0.001 in.
n = 2400 rpm
ns = 40 rps
µ = 3 ×10−6 reyn
r = D 2 = 1 in
Page 19 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
p=
W
5000
=
= 1250 psi
LD (2)(2)
2
2
(
)
 r  µns  1  3 × 10 −6 (40 )
S =  
=
= 0.10

1250
 cr  p  0.001 
πDn π (2)(2400)
vm =
=
= 1257 fpm
12
12
Using Fig. AF 17 and AF 18
(a) Full Bearing
ho
= 0.346
cr
r
f = 2 .8
cr
ho = 0.346(0.001) = 0.000346 in
 1 

 f = 2 .8
 0.001 
f = 0.0028
F = f W = (0.0028)(5000) = 14 lb
Fvm = (14)(1257 ) = 17,600 ft − lb min
(b) 180o Bearing
ho
= 0.344
cr
r
f = 2 .0
cr
ho = 0.344(0.001) = 0.000344 in
 1 

 f = 2 .0
 0.001 
f = 0.0020
F = f W = (0.0020)(5000) = 10 lb
Fvm = (10)(1257 ) = 12,570 ft − lb min
(c) 120o Bearing
ho
= 0.302
cr
Page 20 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
r
f = 1 .7
cr
ho = 0.302(0.001) = 0.000302 in
 1 

 f = 1 .7
 0.001 
f = 0.0017
F = f W = (0.0017 )(5000) = 8.5 lb
Fvm = (8.5)(1257 ) = 10,685 ft − lb min
(d) 60o Bearing
ho
= 0.20
cr
r
f = 1 .4
cr
ho = 0.20(0.001) = 0.0002 in
 1 

 f = 1 .4
 0.001 
f = 0.0014
F = f W = (0.0014)(5000) = 7 lb
Fvm = (7 )(1257 ) = 8,800 ft − lb min
564.
A 120o partial bearing is to support 4500 lb. with ho = 0.002 in. ; L D = 1 ;
D = 4 in. ; cd = 0.010 in. ; n = 3600 rpm . Determine (a) the oil’s viscosity,(b)
the frictional loss (ft-lb/min), (c) the eccentricity angle, (d) the needed oil
flow, (e) the end leakage, (f) the temperature rise of the oil as it passes
through, (g) the maximum pressure. (h) If the clearance given is the average,
what approximate class of fit (Table 3.1) is it? (i) What maximum impulsive
load would be on the bearing if the eccentricity ratio suddenly went to 0.8?
Ignore “squeeze” effect.
Solution:
W = 4500 lb
ho = 0.002 in
L D =1
D = 4 in
L = 4 in
r = D 2 = 2 in
cd = 0.010 in.
n = 3600 rpm
Page 21 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
3600
= 60 rps
60
πDn π (2)(3600)
vm =
=
= 3770 fpm
12
12
W
4500
p=
=
= 281.25 psi
LD (4 )(4)
ho 2ho 2(0.002 )
=
=
= 0 .4
cr
cr
0.010
Table AT 22, L D = 1 , ho cr = 0.4
S = 0.162
φ = 35.65o
r
f = 2.16
cr
q
= 2.24
rcr ns L
qs
= 0.384
q
ρ c∆to
= 15.0
p
p
= 0.356
pmax
ns =
r
(a) S = 
 cr
2
 µns

 p
2
 D  µns
S =  
 cd  p
2
 4  µ (60 )
0.162 = 

 0.010  281.25
µ = 4.75 ×10−6 reyn
(b)
r
f = 2.16
cr
D
f = 2.16
cd
 4 

 f = 2.16
 0.010 
f = 0.0054
F = f W = 0.0054(4500) = 24.30 lb
Page 22 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
Fvm = (24.30)(3770) = 91,611 ft − lb min
(c) φ = 35.65o
q
4q
=
= 2.24
rcr ns L Dcd ns L
4q
= 2.24
(4)(0.010)(60)(4)
q = 5.4 in3 sec
(d)
qs
= 0.384
q
qs = 0.384(5.4 ) = 2.07 in3 sec
(e)
(f)
ρ c∆to
= 15.0
p
(112)∆to = 15.0
281.25
∆to = 38 o F
(g)
p
= 0.356
pmax
281.25
pmax =
= 790 psi
0.356
(h) cd = 0.010 in , D = 4 in
Table 3.1
RC 8, Hole, average = + 0.0025
Shaft, average = - 0.00875
cd = 0.0025 + 0.00875 = 0.01125 ≈ 0.010 in
Class of fit = RC 9
(i) ε = 0.80
Table AT 22, , L D = 1
S = 0.162
2
 D  µ ns
S =  
 cd  p
2
(
)
−6
 4  3 × 10 (60 )
0.0531 = 

p
 0.010 
p = 542 psi
Page 23 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
W = pDL = (542)(4)(4) = 8672 lb
565.
A 120o partial bearing is to support 4500 lb., D = 3 in. , cd = 0.003 in. ;
n = 3600 rpm ; SAE 20W entering at 110 F. Calculate (a) the average
temperature of the oil as it passes through,(b) the minimum film thickness, (c)
the fhp, (d) the quantity of oil to be supplied. HINT: In (a) assume µ and
determine the corresponding values of S and ∆to ; then tav = ti + ∆to 2 . If
assumed µ and tav do not locate a point in Fig. AF 16 that falls on line for
SAE 20W, iterate.
Solution:
W = 4500 lb
D = 3 in
L = 3 in
L D =1
cd = 0.003 in.
2
 D  µns
S =  
 cd  p
3600
ns =
= 60 rps
60
W
4500
p=
=
= 500 psi
DL (3)(3)
ρ c∆to
, (SAE 20W)
p
(a) Using Table AT22, L D = 1 , ρ c = 112 , ti = 110o F
Trial µ
t , oF
S
3.5 x 10-6
2.0 x 10-6
2.6 x 10-6
2.35 x 10-6
2.4 x 10-6
130
160
145
150
149
0.42
0.24
0.312
0.282
0.288
∴ Use tav = 149 o F
(b) Table AT 22, L D = 1 , S = 0.288
Page 24 of 63
ρ c∆to
p
19.8
15.4
17.7
17.2
17.3
∆to
tav = ti + ∆to 2
88
68
79
76
78
154
144
149.5
148
149
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
ho
= 0.513
cr
2ho
= 0.513
cd
2ho = 0.513(0.003)
ho = 0.00077 in
(c) Table At 22, L D = 1 , S = 0.288
r
f = 2.974
cr
D
3
f =
f = 2.974
cr
0.003
f = 0.002974
F = f W = (0.002974)(4500) = 13.383 lb
Fvm
fhp =
33,000
πDn π (3)(3600)
vm =
=
= 2827 fpm
12
12
Fvm
(13.383)(2827 ) = 1.15 hp
fhp =
=
33,000
33,000
(d) Table At 22, L D = 1 , S = 0.288
q
= 2.528
rcr ns L
4q
= 2.528
Dcd ns L
4q
= 2.528
(3)(0.003)(60)(3)
q = 1.024 in3 sec
566.
The 6000-lb. reaction on an 8 x 4 –in., 180o partial bearing is centrally
applied; n = 1000 rpm ; ho = 0.002 in . For an optimum bearing with minimum
friction determine (a) the clearance, (b) the oil’s viscosity, (c) the frictional
horsepower. (d) Choose a cd D ratio either smaller or larger than that
obtained in (a) and show that the friction loss is greater than that in the
optimum bearing. Other data remain the same.
Solution:
W = 6000 lb
Page 25 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
D = 8 in
L = 4 in
n = 1000 rpm
1000
ns =
= 16.67 rps
60
L D =1 2
ho = 0.002 in
(a) Table AT 21, L D = 1 2
Optimum value (minimum friction)
ho cr = 0.23
0.002
cr =
= 0.0087 in
0.23
(b) Table AT 21, L D = 1 2 , ho cr = 0.23
S = 0.126
2
 r  µns
S =  
 cr  p
W
6000
p=
=
= 187.5 psi
DL (4)(8)
D
r = = 4 in
2
2
 4  µ (16.67 )
S = 0.126 = 

 0.0087  187.5
µ = 6.70 ×10−6 reyn
(c) Table AT 21, L D = 1 2 , ho cr = 0.23
r
f = 2.97
cr
 4 

 f = 2.97
 0.0087 
f = 0.00646
F = f W = (0.00646 )(6000) = 38.76 lb
πDn π (8)(1000)
vm =
=
= 2094 fpm
12
12
Page 26 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
fhp =
Fvm
(38.76)(2094) = 2.46 hp
=
33,000
33,000
For (a)
cd 2cr 2(0.0087 )
=
=
= 0.0022
D
D
8
cd
> 0.0022
D
cd
= 0.0030
D
cd = 0.0030(8) = 0.0240 in
cr = 0.0120 in
ho 0.002
=
= 0.1667
cr 0.012
Table AT 21, L D = 1 2
r
f = 1.67
cr
 4 

 f = 1.67
 0.0016 
f = 0.00668
F = f W = (0.00668)(6000 ) = 40.08 lb
πDn π (8)(1000)
vm =
=
= 2094 fpm
12
12
(40.08)(2094) = 2.54 hp > 2.46 hp
Fvm
fhp =
=
33,000
33,000
cd
< 0.0022
D
cd
= 0.0020
D
cd = 0.0020(8) = 0.0160 in
cr = 0.0080 in
ho 0.002
=
= 0.25
cr 0.008
Table AT 21, L D = 1 2
r
f = 3.26
cr
 4 

 f = 3.26
 0.0016 
f = 0.00652
Page 27 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
F = f W = (0.00652 )(6000) = 39.12 lb
πDn π (8)(1000)
vm =
=
= 2094 fpm
12
12
Fvm
(39.12)(2094) = 2.48 hp > 2.46 hp
fhp =
=
33,000
33,000
A 120o partial bearing supports 3500 lb. when n = 250 rpm ; D = 5 in. ,
L = 5 in. ; µ = 3 × 10−6 reyn . What are the clearance and minimum film
thickness for an optimum bearing (a) for maximum load, (b) for minimum
friction? (c) On the basis of the average clearance in Table 3.1, about what
class fit is involved? Would this fit be on the expensive or inexpensive side?
(d) Find the fhp for each optimum bearing.
567.
Solution:
D = 5 in.
L = 5 in.
L
=1
D
n = 250 rpm
250
ns =
= 4.17 rps
60
µ = 3 ×10−6 reyn
W = 3500 lb
W
3500
p=
=
= 140 psi
DL (5)(5)
(a)
Table AT 22,
h
L
= 1 , max. load o = 0.46
D
cr
S = 0.229
2
 r  µns
S =  
 cr  p
D
r = = 2.5 in
2
2
(
)
 2.5  3.0 × 10−6 (4.17 )

S = 0.229 = 
140
 cr 
cr = 0.00156 in
ho = 0.46cr = 0.46(0.00156) = 0.00072 in
Page 28 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
(b)
Table AT 22,
L
h
= 1 , min. friction o = 0.40
D
cr
S = 0.162
2
 r  µns
S =  
 cr  p
D
r = = 2.5 in
2
2
(
)
 2.5  3.0 × 10−6 (4.17 )

S = 0.162 = 
140
 cr 
cr = 0.00186 in
ho = 0.46cr = 0.40(0.00186) = 0.00074 in
(c) cd 1 = 2(0.00156) = 0.00312 in
cd 2 = 2(0.00186) = 0.00372 in
Use Class RC4, ave. cd = 0.00320 in , expensive side
(d) Table AT 22,
L
h
= 1 , max. load o = 0.46
D
cr
r
f = 2.592
cr
 2 .5 

 f = 2.592
 0.00156 
f = 0.00162
F = f W = (0.00162 )(3500 ) = 5.67 lb
πDn π (5)(250)
vm =
=
= 327.25 fpm
12
12
Fvm
(5.67)(327.25) = 0.0562 hp
fhp =
=
33,000
33,000
For minimum friction,
ho
= 0.40
cr
r
f = 2.16
cr
 2 .5 

 f = 2.16
 0.00186 
f = 0.00161
F = f W = (0.00161)(3500) = 5.635 lb
Page 29 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
πDn
π (5)(250)
= 327.25 fpm
12
12
Fvm
(5.635)(327.25) = 0.0559 hp
fhp =
=
33,000
33,000
vm =
570.
=
A 180o partial bearing is to support 17,000 lb. with p = 200 psi ,
n = 1500 rpm , ho = 0.003 in , L D = 1 . (a) Determine the clearance for an
optimum bearing with minimum friction. (b) Taking this clearance as the
average, choose a fit (Table 3.1) that is approximately suitable. (c) Select an
oil for an average temperature of 150 F. (d) Compute fhp.
Solution:
W = 17,000 lb
p = 200 psi
n = 1500 rpm
1500
ns =
= 25 rps
60
L D =1
L=D
W
p=
DL
17,000
200 =
D2
D = L = 9.22 in
D 9.22
r= =
= 4.61 in
2
2
(a) For optimum bearing with minimum friction
Table AT 21, L D = 1 , ho cr = 0.44
ho cr = 0.44
0.003
= 0.44
cr
cr = 0.00682 in
(b) Table 3.1, D = 9.22 in
cd = 2cr = 2(0.00682 ) = 0.01364 in
Use Class RC7, average cd = 0.01065 in
Or use Class RC8, average cd = 0.01575 in
(c) Table AT 21, L D = 1 , ho cr = 0.44
S = 0.158
Page 30 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
r
S = 
 cr
2
 µns

 p
2
 4.61  µ (25)
0.158 = 

 0.00682  200
µ = 2.8 ×10−6 reyn
Fig. AF 16, at 150 F
Use Either SAE 20W or SAE 30.
(d)
Table AT 21, L D = 1 , ho cr = 0.44
r
f = 2.546
cr
 4.61 

 f = 2.546
 0.00682 
f = 0.00377
πDn π (9.22)(1500)
vm =
=
= 3621 fpm
12
12
F = f W = (0.00377 )(17,000) = 64.09 lb
Fvm
(64.09)(3621) = 7.0 hp
fhp =
=
33,000
33,000
571.
The reaction on a 120o partial bearing is 2000 lb. The 3-in journal turns at
1140 rpm; cd = 0.003 in. ; the oil is SAE 20W at an average operating
temperature of 150 F. Plot curves for the minimum film thickness and the
frictional loss in the bearing against the ratio L D , using L D = 0.25, 0.5, 1,
and 2. (Note: This problem may be worked as a class problem with each
student being responsible for a particular L D ratio.)
Solution:
W = 2000 lb
D = 3 in.
n = 1140 rpm
1140
ns =
= 19 rps
60
cd = 0.003 in
cr = 0.0015 in
For SAE 20W, 150 F
µ = 2.75 ×10−6 reyn
(a)
L
= 0.25
D
Page 31 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
L = 0.25D = 0.25(3) = 0.75 in
W
2000
p=
=
= 889 psi
DL (3)(0.75)
L
Table AT 22,
= 0.25
D
D
r = = 1.5 in
2
2
2
(
)
 r  µns  1.5  2.75 ×10 −6 (19 )
S =  
=
= 0.0588

c
p
0
.
0015
889


 r
ho
= 0.083
cr
ho = 0.083(0.0015) = 0.000125 in
r
f = 2.193
cr
 1 .5 

 f = 2.193
 0.0015 
f = 0.002193
F = f W = (0.002193)(2000) = 4.386 lb
πDn π (3)(1140)
vm =
=
= 895 fpm
12
12
Fvm
(4.386)(895) = 0.119 hp
fhp =
=
33,000
33,000
L
(b)
= 0 .5
D
L = 0.5 D = 0.5(3) = 1.5 in
W
2000
p=
=
= 444 psi
DL (3)(1.5)
L
Table AT 22,
= 0 .5
D
D
r = = 1.5 in
2
2
2
(
)
 r  µns  1.5  2.75 × 10−6 (19 )
S =  
=
= 0.1177

444
 cr  p  0.0015 
ho
= 0.2159
cr
ho = 0.2159(0.0015) = 0.000324 in
r
f = 2.35
cr
Page 32 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
 1 .5 

 f = 2.35
 0.0015 
f = 0.00235
F = f W = (0.00235)(2000) = 4.7 lb
πDn π (3)(1140)
vm =
=
= 895 fpm
12
12
Fvm
(4.7 )(895) = 0.1275 hp
fhp =
=
33,000
33,000
L
(c)
=1
D
L = D = 3 in
W
2000
p=
=
= 222 psi
DL (3)(3)
L
Table AT 22,
=1
D
D
r = = 1.5 in
2
2
2
(
)
 r  µns  1.5  2.75 × 10−6 (19 )
S =  
=
= 0.2354

222
 cr  p  0.0015 
ho
= 0.4658
cr
ho = 0.4658(0.0015) = 0.000699 in
r
f = 2.634
cr
 1 .5 

 f = 2.634
 0.0015 
f = 0.002634
F = f W = (0.002634)(2000 ) = 5.268 lb
πDn π (3)(1140)
vm =
=
= 895 fpm
12
12
Fvm
(5.268)(895) = 0.1429 hp
fhp =
=
33,000
33,000
L
(d)
=2
D
L = 2 D = 2(3) = 6 in
W
2000
p=
=
= 111 psi
DL (3)(6)
L
Table AT 22,
=2
D
Page 33 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
r=
D
= 1.5 in
2
2
2
(
)
 r  µns  1.5  2.75 × 10−6 (19 )
S =  
=
= 0.47

111
 cr  p  0.0015 
ho
= 0.718
cr
ho = 0.718(0.0015) = 0.00108 in
r
f = 3.8118
cr
 1 .5 

 f = 3.8118
 0.0015 
f = 0.003812
F = f W = (0.003812)(2000) = 7.624 lb
πDn π (3)(1140)
vm =
=
= 895 fpm
12
12
Fvm
(7.624)(895) = 0.2068 hp
fhp =
=
33,000
33,000
L
D
0.25
0.5
1.0
2.0
Page 34 of 63
ho , in
fhp
0.000125
0.000324
0.000699
0.001080
0.119
0.128
0.143
0.207
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
STEADY-STATE TEMPERATURE
A 180o partial bearing is subjected to a load of 12,000 lb.; D × L = 8 × 8 in. ,
cr r = 0.0015 , ho ≈ 0.0024 in. , n = 500 rpm . The air speed about the bearing
is expected to be in excess of 1000 fpm (on moving vehicle) and the effective
radiating area is 20 DL . Determine: (a) the eccentricity factor, (b) µreyns, (c)
the frictional loss (ft-lb/min), (d) the estimated temperature of oil and bearing
( a self-contained oil-bath unit) for steady-state operation, and a suitable
oil.(e) Compute ∆to of the oil passing through the load-carrying area, remark
on its reasonableness, and decide upon whether some redesign is desirable.
572.
Solution:
D = 8 in.
L = 8 in.
L D =1
W = 12,000 lb
D
r = = 4 in
2
cr = 0.0015r = 0.0015(4) = 0.0060 in
ho 0.0024
=
= 0. 4
cr 0.0060
n = 500 rpm
500
ns =
= 8.33 rps
60
Table AT 21, ho cr = 0.4 , L D = 1
S = 0.128
r
f = 2.28
cr
ρ c∆to
= 12.4
p
W 12,000
p=
=
= 187.5 psi
DL (8)(8)
(a) ε = 1 −
ho
= 1 − 0 .4 = 0 .6
cr
r
(b) S = 
 cr
2
 µns

 p
2
 4  µ (8.33)
S = 0.128 = 

 0.0060  187.5
µ = 6.5 ×10−6 reyn
Page 35 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
(c)
r
f = 2.28
cr
 4 

 f = 2.28
 0.0060 
f = 0.00342
F = f W = (0.00342 )(12,000) = 41.04 lb
πDn π (8)(500)
vm =
=
= 1047 fpm
12
12
Fvm
(41.04)(1047) = 1.302 hp
fhp =
=
33,000
33,000
Frictional loss = 43,000 ft-lb/min
(d) Q = hcr Ab ∆tb ft-lb/min
Q = 43,000 ft − lb min
hcr = hc + hr
hr = 0.108 ft − lb min − sq.in. − F
va0.6
, va ≥ 1000 fpm
D 0.4
0.6
(
1000 )
hc = 0.017
= 0.467 ft − lb min − sq.in. − F
(8)0.4
hcr = 0.467 + 0.108 = 0.575 ft − lb min − sq.in. − F
Ab = 20 DL = 20(8)(8) = 1280 sq.in.
Q = hcr Ab ∆tb
43,000 = (0.575)(1280)(∆tb )
∆tb = 58.42 F
Oil-bath, 1000 fpm
∆toa ≈ (1.2)(1.3)(∆tb )
∆toa = (1.2)(1.3)(58.42) = 91.1 F
assume 100 F ambient temperature
tb = 100 + 58.42 F = 158.42 F
tb = 100 + 91.1 F = 191.1 F
hc = 0.017
(c)
ρ c∆to
= 12.4
p
(112)∆to = 12.4
187.5
∆to = 20.8 F
Solve for to 2
Page 36 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
to1 + to 2 = 2(191.1) = 382.2 F
to1 = 382.2 − to 2
to 2 − to1 = 20.8 F
to 2 − 362.2 + to 2 = 20.8
to 2 = 201.5 F ≈ 200 F
∴ not reasonable since the oil oxidizes more rapidly above 200 F, a redesign is
desireable.
573.
A 2 x 2-in. full bearing (ring-oiled) has a clearance ratio cd D = 0.001 . The
journal speed is 500 rpm, µ = 3.4 × 10−6 reyn , and ho = 0.0005 in. The ambient
temperature is 100 F; Ab = 25DL , and the transmittance is taken as
hcr = 2 Btu hr − sq. ft. − F . Calculate (a) the total load for this condition; (b)
the frictional loss, (c) the average temperature of the oil for steady-state
operation. Is this temperature satisfactory? (d) For the temperature found,
what oil do you recommend? For this oil will ho be less or greater than the
specified value? (e) Compute the temperature rise of the oil as it passes
through the bearing. Is this compatible with other temperatures found? (f)
What minimum quantity of oil should the ring deliver to the bearing?
Solution:
L = 2 in.
D = 2 in.
cd D = 0.001
cd = (0.001)(2) = 0.0020 in
µ = 3.4 ×10−6 reyn
ho = 0.0005 in.
cr = 0.0010 in
ho cr = 0.0005 0.0010 = 0.5
Table AT 20, L D = 1 , ho cr = 0.5 , Full Bearing
S = 0.1925
r
f = 4.505
cr
q
= 4.16
rcr ns L
ρ c∆to
= 19.25
p
(a) S = 0.1925
Page 37 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
2
 r  µns
S =  
 cr  p
D
r = = 1 in
2
500
ns =
= 8.33 rps
60
2
(
)
−6
 1  3.4 × 10 (8.330 )
S = 0.1925 = 

p
 0.0010 
p = 147 psi
W = pDL = (147 )(2)(2) = 588 lb
(b)
r
f = 4.505
cr
 1 

 f = 4.505
 0.001 
f = 0.004505
F = f W = (0.004505)(588) = 2.649 lb
πDn π (2)(500)
vm =
=
= 261.8 fpm
12
12
U f = Fvm = (2.649 )(261.8) = 693.5 ft − lb min
(c) Q = hcr Ab ∆tb
hcr = 2 Btu hr − sq. ft. − F = 0.18 ft − lb min − sq.in. − F
Ab = 25DL = 25(2)(2) = 100 sq.in.
Q =Uf
(0.18)(100)(∆tb ) = 693.5
∆tb = 38.53 F
∆toa = 2∆tb = 2(38.53) = 77 F
to = 77 + 100 = 177 F , near 160 F
∴ satisfactory.
(d) to = 177 F , µ = 3.4 × 10−6 reyn
Figure AF 16
Use SAE 40 oil, µ = 3.3 × 10−6 reyn
r
S = 
 cr
2
 µns

 p
Page 38 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
2
(
)
−6
 1  3.3 × 10 (8.33)
S =
= 0.187

147
 0.0010 
Table AT 20, L D = 1 , S = 0.187
ho cr = 0.4923
ho = 0.4923(0.0010) = 0.00049 in < ho (= 0.0005 in )
ρ c∆to
(e)
= 19.25
p
(112)∆to = 19.25
147
∆to = 25.3 F
∆to1 + ∆to 2 = 2(177 ) = 354 F
∆t o 2 − ∆t o1 = 25.3 F
2∆to 2 = 354 + 25.3
∆to 2 = 190 F < 200 F
∴
compatible.
q
= 4.16
rcr ns L
q
= 4.16
(1)(0.001)(8.33)(2)
q = 0.0693 in3 sec
(f)
574.
An 8 x 9-in. full bearing (consider L D = 1 for table and chart use only)
supports 15 kips with n = 1200 rpm ; cr r = 0.0012 ; construction is medium
heavy with a radiating-and-convecting area of about 18 DL ; air flow about the
bearing of 80 fpm may be counted on (nearby) pulley; ambient temperature is
90 F. Decide upon a suitable minimum film thickness. (a) Compute the
frictional loss and the steady state temperature. Is additional cooling needed
for a reasonable temperature? Determine (b) the temperature rise of the oil as
it passes through the load-carrying area and the grade of oil to be used if it
enters the bearing at 130 F, (c) the quantity of oil needed.
Solution:
D = 8 in.
L = 9 in.
W = 15,000 lb.
n = 1200 rpm.
1200
ns =
= 20 rps
60
cr r = 0.0012
Page 39 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
r = D 2 = 4 in
cr = 0.0012(4) = 0.0048 in
By Norton: ho = 0.00025 D = 0.00025(8) = 0.002 in
ho
0.002
=
= 0 .4
cr 0.0048
Table AT 20, L D = 1 , ho cr = 0.4
S = 0.121
r
f = 3.22
cr
q
= 4.33
rcr ns L
ρ c∆to
= 14.2
p
(a)
r
f = 3.22
cr
 4 

 f = 3.22
 0.0048 
f = 0.003864
F = f W = (0.003864)(15,000) = 57.96 lb
πDn π (8)(1200)
vm =
=
= 2513 fpm
12
12
U f = Fvm = (57.96 )(2513) = 145,654 ft − lb min
Q = hcr Ab ∆tb
hr = 0.108 ft − lb min − sq.in. − F
va0.6
ft − lb min − sq.in. − F
D 0.4
(80)0.6 = 0.103 ft − lb min − sq.in. − F
hc = 0.017
(8)0.4
hcr = hc + hr = 0.103 + 0.108 = 0.211 ft − lb min − sq.in. − F
Ab = 18DL = 18(8)(9 ) = 1296 sq.in.
Uf =Q
hc = 0.017
145,654 = (0.211)(1296)∆tb
∆tb = 533 F , very high, additional cooling is necessary.
(b)
ρ c∆to
p
= 14.2
Page 40 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
W 15,000
=
= 208 psi
DL (8)(9)
(112)∆to = 14.2
208
∆to = 26 F
ti = 130 F
to = 156 F
tave = 12 (130 + 156) = 143 F
p=
r
S = 
 cr
2
 µns

 p
2
 4  µ (20 )
S = 0.121 = 

 0.0048  208
µ = 1.8 ×10−6 reyn
Figure AF 16, µ = 1.8 µreyns , 143 F
Use SAE 10W
q
= 4.33
rcr ns L
q
= 4.33
(4)(0.0048)(20)(9)
q = 14.96 in3 sec
(c)
575.
A 3.5 x 3.5-in., 360o bearing has cr r = 0.0012 ; n = 300 rpm ; desired
minimum ho ≈ 0.0007 in . It is desired that the bearing be self-contained (oilring); air-circulation of 80 fpm is expected; heavy construction, so that
Ab ≈ 25DL . For the first look at the bearing, assume µ = 2.8 × 10−6 reyn and
compute (a) the frictional loss (ft-lb/min), (b) the average temperature of the
bearing and oil as obtained for steady-state operation, (c) ∆to as the oil passes
through the load-carrying area (noting whether comparative values are
reasonable). (d) Select an oil for the steady-state temperature and decide
whether there will be any overheating troubles.
Solution:
D = 3.5 in.
L = 3.5 in.
cr r = 0.0012
r = D 2 = 1.75 in.
cr = (0.0012)(1.75) = 0.0021 in
ho ≈ 0.0007 in
Page 41 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
ho cr = 0.0007 0.0021 = 0.333
Table AT 20, 360o Bearing, L D = 1 , ho cr = 0.333
S = 0.0954
r
f = 2.71
cr
ρ c∆to
= 12.12
p
2
 r  µns
(a) S =  
 cr  p
300
ns =
= 5 rps
60
µ = 2.8 ×10−6 reyn
2
(
)
−6
 1.75  2.8 × 10 (5)
S = 0.0954 = 

p
 0.0021 
p = 102 psi
W = pDL = (102)(3.5)(3.5) = 1250 lb
r
f = 2.71
cr
 1.75 

 f = 2.71
 0.0021 
f = 0.00325
F = f W = (0.00325)(1250) = 4.0625 lb
πDn π (3.5)(300 )
vm =
=
= 275 fpm
12
12
U f = Fvm = (4.0625)(275) = 1117 ft − lb min
(b) Q = hcr Ab ∆tb
hr = 0.108 ft − lb min − sq.in. − F
va0.6
ft − lb min − sq.in. − F
D 0.4
0.6
(
80 )
hc = 0.017
= 0.143 ft − lb min − sq.in. − F
(3.5)0.4
hcr = hc + hr = 0.143 + 0.108 = 0.251 ft − lb min − sq.in. − F
Ab = 25DL = 25(3.5)(3.5) = 306.25 sq.in.
Uf =Q
hc = 0.017
1117 = (0.251)(306.25)∆tb
Page 42 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
∆tb = 14.5 F
∆toa = 2∆tb = 2(14.5) = 29 F
assume ambient temperature of 100 F
tb = 114.5 F
to = 129 F
(c)
ρ c∆to
= 12.12
p
(112)∆to = 12.12
102
∆to = 11 F
to1 + to 2 = 2(129) = 258 F
to 2 − to1 = 11 F
2to 2 = 269 F
to 2 = 135 F < 140 F
∴ reasonable
(d) to = 129 F , µ = 2.8 × 10−6 reyn
use SAE 10W
Figure AF 16, to = 126 F
∆toa = 126 − 100 = 26 F
∆toa = 2∆tb
26
∆tb =
= 13 F
2
Q = hcr Ab ∆tb = (0.251)(306.25)(13) = 999 ft − lb min < U f
∴ there is an overheating problem.
576.
A 10-in. full journal for a steam-turbine rotor that turns 3600 rpm supports a
20-kip load with p = 200 psi ; cr r = 0.00133 . The oil is to have
µ = 2.06 ×10−6 reyn at an average oil temperature of 130 F. Compute (a) the
minimum film thickness (comment on its adequacy), (b) the fhp, (c) the
altitude angle, the maximum pressure, and the quantity of oil that passes
through the load-carrying area (gpm).(d) At what temperature must the oil be
introduced in order to have 130 F average? (e) Estimate the amount of heat
lost by natural means from the bearing (considered oil bath) with air speed of
300 fpm. If the amount of oil flow computed above is cooled back to the
entering temperature, how much heat is removed? Is this total amount of heat
enough to care for frictional loss? If not, what can be done (i11.21)?
Page 43 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
Solution:
D = 10 in.
n = 3600 rpm
3600
ns =
= 60 rps
60
W = 20,000 lb
p = 200 psi
W
p=
DL
20,000
200 =
10 L
L = 10 in
L D =1
D
r = = 5 in
2
cr r = 0.00133
cr = 0.00133(5) = 0.00665 in
µ = 2.06 ×10−6 reyn
tave = 130 F
r
S = 
 cr
2
 µns

 p
2
(
)
−6
5

 2.06 × 10 (60 )
S =
= 0.35

200
 0.00665 
Table AT 20, L D = 1 , S = 0.35
ho cr = 0.647
φ = 65.66o
r
f = 7.433
cr
q
= 3.90
rcr ns L
p
= 0.495
pmax
ρ c∆to
= 30.8
p
qs
= 0.446
q
(a) ho = 0.647cr = 0.647(0.00665) = 0.00430 in
Page 44 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
Norton’s recommendation = 0.00025D = 0.00025(10) = 0.00250 in < 0.00430 in
∴ adequate
(b)
r
f = 7.433
cr
5



 f = 7.433
0
.
00665


f = 0.0099
F = f W = (0.0099)(20,000) = 198 lb
πDn π (10)(3600)
vm =
=
= 9425 fpm
12
12
Fvm
(198)(9425) = 56.55 hp
fhp =
=
33,000
33,000
(c) φ = 65.66o
p
200
pmax =
=
= 404 psi
0.495 0.495
q = 3.90rcr ns L
q = 3.90(5)(0.00665)(60)(10) = 77.805 in3 sec
q = (77.805 in3 sec )(1 gpm 231 in3 )(60 sec min ) = 0.21 gpm
(d)
ρ c∆to
= 30.8
p
(112)∆to = 30.8
200
∆to = 55 F
∆t
tave = ti + o
2
55
130 = ti +
2
ti = 102.5 F
(e) Q = hcr Ab ∆tb
hr = 0.108 ft − lb min − sq.in. − F
va0.6
ft − lb min − sq.in. − F
D 0.4
0.6
(
300 )
hc = 0.017
= 0.207 ft − lb min − sq.in. − F
(3.5)0.4
hcr = hc + hr = 0.207 + 0.108 = 0.315 ft − lb min − sq.in. − F
Assume
hc = 0.017
Page 45 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
Ab = 25DL = 25(10 )(10) = 2500 sq.in.
∆toa = 130 − 100 = 30 F
∆toa = 1.3∆tb
30
∆tb =
= 23 F
1 .3
Q = (0.315)(2500)(23) = 18,113 ft − lb min
Qr = ρ c(q − qs )∆to in − lb sec
Qr = (112 )(1 − 0.446)(77.805)(55)(1 12)(60) = 1,327,602 ft − lb min
QT = Q + Qr = 18,113 + 1,327,602 = 1,345,735 ft − lb min
U f = Fvm = (198)(9425) = 1,866,150 ft − lb min > QT
not enough to care for frictional loss, use pressure feed (i11.21).
DESIGN PROBLEMS
578.
A 3.5-in. full bearing on an air compressor is to be designed for a load of 1500
lb.; n = 300 rpm ; let L D = 1 . Probably a medium running for would be
satisfactory. Design for an average clearance that is decided by considering both
Table 3.1 and 11.1. Choose a reasonable ho , say one that gives ho cr ≈ 0.5 .
Compute all parameters that are available via the Text after you have decided on
details. It is desired that the bearing operate at a reasonable steady-state
temperature (perhaps ring-oiled medium construction), without special cooling.
Specify the oil to be used and show all calculations to support your conclusions.
What could be the magnitude of the maximum impulsive load if the eccentricity
ration ε becomes 0.8, “squeeze” effect ignored?
Solution:
L D =1
D = 3.5 in
L = 3.5 in
W = 1500 lb
n = 300 rpm
300
ns =
= 5 rps
60
W
1500
p=
=
= 122.45 psi
DL (3.5)(3.5)
Table 3.1, medium running fit,
D = 3.5 in
RC 5 or RC 6
Use RC 6
Average cd = 0.0052 in
Page 46 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
Table 11.1, air-compressor
General Machine Practice
Average cd = 0.0055 in
Using cd = 0.0055 in
cr = 0.00275 in
ho = 0.5cr = 0.5(0.00275) = 0.001375 in
Table AT 20, L D = 1 , ho cr = 0.5
ε = 0.5
S = 0.1925
φ = 56.84o
r
f = 4.505
cr
q
= 4.16
rcr ns L
ρ c∆to
= 19.25
p
p
= 0.4995
pmax
Specifying oil:
Q = hcr Ab ∆tb
U f = Fvm
r
f = 4.505
cr
 1.75 

 f = 4.505
 0.00275 
f = 0.00708
F = f W = (0.00708)(1500) = 10.62 lb
πDn π (3.5)(300 )
vm =
=
= 275 fpm
12
12
U f = Fvm = (10.62 )(275) = 2921 ft − lb min
Q = hcr Ab ∆tb
Assume hcr = 0.516 ft − lb min − sq.in. − F
Medium construction
Ab = 15.5DL = 15.5(3.5)(3.5) = 189.875 sq.in.
Oil-ring bearing
∆toa = 2∆tb
Q =Uf
(0.516)(189.875)(∆tb ) = 2921
Page 47 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
∆tb = 30 F
∆toa = 2∆tb = 2(30) = 60 F
assume ambient temperature = 90 F
t o = 150 F
r
S = 
 cr
2
 µns

 p
2
 1.75  µ (5)
S = 0.1925 = 

 0.00275  122.45
µ = 11.6 ×10−6 reyn
Figure AF 16, 150 F, µ ≈ 11.6 × 10−6 reyn
Use SAE 70 oil
Maximum load, W with ε = 0.8
Table AT 20, L D = 1
S = 0.0446
r
S = 
 cr
2
 µns

 p
2
(
)
−6
 1.75  11.6 × 10 (5)
S = 0.0446 = 

p
 0.00275 
p = 527 psi
W = pDL = (527 )(3.5)(3.5) = 6456
580.
A 2500-kva generator, driven by a water wheel, operates at 900 rpm. The weight
of the rotor and shaft is 15,100 lb. The left-hand, 5 –in, full bearing supports the
larger load, R = 8920 lb . The bearing should be above medium-heavy
construction (for estimating Ab ). (a) Decide upon an average clearance
considering both Table 3.1 and 11.1, and upon a minimum film thickness
( ho cr ≈ 0.5 is on the safer side). (b) Investigate first the possibility of the
bearing being a self-contained unit without need of special cooling. Not much air
movement about the bearing is expected. Then make final decisions concerning
oil-clearance, and film thickness and compute all the parameters given in the text,
being sure that everything is reasonable.
Solution:
n = 900 rpm
900
ns =
= 15 rps
60
D = 5 in
W = R = 8920 lb
Page 48 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
(a) Table 3.1, D = 5 in
RC 5, average cd = 0.0051 in
cr = 0.00255 in
ho = 0.5cr = 0.5(0.00255) = 0.00128 in
(b) Use L D = 1
L = 5 in
D
r = = 2.5 in
2
W
8920
p=
=
= 356.8 psi
DL (5)(5)
Table AT 20, L D = 1 , ho cr = 0.5
S = 0.1925
r
f = 4.505
cr
q
= 4.16
rcr ns L
ρ c∆to
= 19.25
p
r
S = 
 cr
2
 µns

 p
2
 2.5  µ (15)
S = 0.1925 = 

 0.00255  356.8
µ = 4.8 ×10−6 reyn
r
f = 4.505
cr
 2 .5 

 f = 4.505
 0.00255 
f = 0.00460
F = f W = (0.00460 )(8920) = 41.032 lb
πDn π (5)(900)
vm =
=
= 1178 fpm
12
12
U f = Fvm = (41.032 )(1178) = 48,336 ft − lb min
Q = hcr Ab ∆tb
Medium-Heavy
Ab = 20.25DL = 20.25(5)(5) = 506.25 sq.in.
Assume hcr = 0.516 ft − lb min − sq.in. − F
Q =Uf
Page 49 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
(0.516)(506.25)(∆tb ) = 48,336
∆tb = 185 F , very high
Therefore, special cooling is needed.
ρ c∆to
= 19.25
p
(112)∆to = 19.25
356.8
∆to = 61 F
Assume ti = 100 F
61
tave = 100 + ≈ 130 F
2
Figure AF 16, µ = 4.8 µreyns , 130 F
Select SAE 30 oil. µ = 6.0 µreyns
r
S = 
 cr
2
 µns

 p
2
(
)
−6
 2.5  6.0 × 10 (15)
S =
= 0.242

356.8
 0.00255 
Table AT 20, L D = 1 , S = 0.242
SAE 30 oil at 130 F
ho
= 0.569
cr
φ = 61.17o
r
f = 5.395
cr
q
= 4.04
rcr ns L
ρ c∆to
= 22.75
p
p
= 0.4734
pmax
Oil, SAE 30
cr = 0.00255 in
ho = 0.569(0.00255) = 0.00145 in
PRESSURE FEED
581.
An 8 x 8-in. full bearing supports 5 kips at 600 rpm of the journal; cr = 0.006 in. ;
let the average µ = 2.5 × 10−6 reyn . (a) Compute the frictional loss U f . (b) The
Page 50 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
oil is supplied under a 40-psi gage pressure with a longitudinal groove at the
point of entry. Assuming that other factors, including U f , remain the same and
that the heat loss to the surrounding is negligible, determine the average
temperature rise of the circulating oil.
Solution:
L = 5 in
D = 5 in
W = 5000 lb
n = 600 rpm
600
ns =
= 10 rps
60
cr = 0.006 in
µ = 2.5 ×10−6 reyn
L D =1
W
5000
p=
=
= 78.125 psi
DL (8)(8)
r
S = 
 cr
2
 µns

 p
2
(
)
−6
 4  2.5 ×10 (10 )
S =
= 0.1422

78.125
 0.006 
(a) Table AT 20, L D = 1 , S = 0.1422
r
f = 3.6 , ε = 0.57
cr
 4 

 f = 3 .6
 0.006 
f = 0.0054
F = f W = (0.0054)(5000) = 27 lb
πDn π (8)(600)
vm =
=
= 1257 fpm
12
12
U f = Fvm = (27 )(1257 ) = 33,940 ft − lb min
(b) Longitudinal Groove.
c3 p 
2π r 
2
3
q = 2.5 r i  tan −1
 1 + 1.5ε in sec
3µ 
L 
pi = 40 psi
(
)
3
(
0.006) (40)  −1 2π (4) 
q = 2.5
tan
[1 + 1.5(0.57)2 ]in3
3(2.5 × 10−6 ) 
Page 51 of 63
8

sec
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
q = 5.41 in3 sec
U f = ρ cq∆to
(33,940 ft − lb
min )(12 in ft )(1min 60sec ) = (12)(5.41)∆to
∆to = 11.2 F
583.
A 4-in. 360o bearing, with L D = 1 , supports 2.5 kips with a minimum film of
ho = 0.0008 in. , cd = 0.01 in. , n = 600 rpm. The average temperature rise of the oil
is to be about 25 F. Compute the pressure at which oil should be pumped into the
bearing if (a) all bearing surfaces are smooth, (b) there is a longitudinal groove at
the oil-hole inlet. (c) same as (a) except that there is a 360o circumferential
groove dividing the bearing into 2-in. lengths.
Solution:
D = 4 in
L = 4 in
r = 2 in
W = 2500 lb
cd = 0.010 in
cr = 0.005 in
n = 600 rpm
600
ns =
= 10 rps
60
∆to = 25 F
W
2500
p=
=
= 156.25 psi
DL (4)(4)
ho = 0.00080 in
ho 0.0008
=
= 0.16
cr
0.005
Table AT 20, L D = 1 , ho cr = 0.16
r
f = 1.44 , ε = 0.84
cr
 2 

 f = 1.44
 0.005 
f = 0.0036
F = f W = (0.0036)(2500) = 9 lb
πDn π (4)(600)
vm =
=
= 628 fpm
12
12
U f = Fvm = (9 )(628) = 5652 ft − lb min = 1130 in − lb sec
S = 0.0343
Page 52 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
r
S = 
 cr
2
 µns

 p
2
 2  µ (10 )
S = 0.0343 = 

 0.005  156.25
µ = 3.35 ×10−6 reyn
U f = ρ cq∆to
1130 = (112)(q )(25)
q = 0.404 in3 sec
(a) Smooth
q=
cr3 pi  −1 2π r 
2
3
 tan
 1 + 1.5ε in sec
3µ 
L 
(
0.404 =
)
(0.005)3 ( pi ) tan −1 2π (2) [1 + 1.5(0.84)2 ] in3
3(3.35 × 10−6 ) 
pi = 12.5 psi
(b) Longitudinal groove
q=

4
sec
2.5cr3 pi  −1 2π r 
2
3
 tan
 1 + 1.5ε in sec
3µ 
L 
(
3
)
2.5(0.005) ( pi )  −1 2π (2) 
2
tan
1 + 1.5(0.84) in3 sec
−6


3 3.35 × 10
4 

pi = 5 psi
(c) Circumferential groove
0.404 =
(
)
[
]
2π rcr3 pi
q=
1 + 1.5ε 2 in3 sec
3µL
(
)
3
2π (2)(0.005) ( pi )
2
0.404 =
1 + 1.5(0.84) in3 sec
−6
3 3.35 × 10 (4 )
pi = 5 psi
(
) [
]
BEARING CAPS
584.
An 8-in. journal, supported on a 150o partial bearing, is turning at 500 rpm;
bearing length = 10.5 in., c d = 0.0035 in ., ho = 0.00106 in . The average
temperature of the SAE 20 oil is 170 F. Estimate the frictional loss in a 160o cap
for this bearing.
Solution:
Page 53 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
ho = 0.00106 in
c d = 0.0035 in
c r = 0.00175 in
2

 ho  
hav = cr 1 + 0.741 −   in

 cr  
2

 0.00106  
hav = (0.00175)1 + 0.741 −
  = 0.00195 in
 0.00175  

For SAE 20, 170 F
µ = 1.7 ×10−6 reyn
µAvips
F=
hav
1
A = θDL
2
D = 8 in
L = 10.5 in
160
8π
θ = 160o =
π=
180
9
1  8π 
A =  (8)(10.5) = 117.3 sq.in.
2 9 
 500 
vips = π Dns = π (8)
 = 209.5 ips
 60 
1.7 × 10−6 (117.3)(209.5)
= 21.424 lb
F=
0.00195
πDn π (8)(500)
vm =
=
= 1047 fpm
12
12
U f = Fvm = (21.424 )(1047 ) = 22,430 ft − lb min = 1130 in − lb sec
(
585.
)
A
partial
160o
bearing
has
a
160o
L = 2 in ., cd = 0.002 in ., ho = 0.0007 in ., n = 500 rpm , and
For the cap only, what is the frictional loss?
Solution:
c d = 0.002 in
c r = 0.001 in
ho = 0.0007 in
ho 0.0007
=
= 0 .7
cr
0.001
Page 54 of 63
cap;
D = 2 in .,
µ = 2.5 ×10−6 reyn .
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
ε = 1−
ho
= 1 − 0 .7 = 0 .3
cr
(
[
)
2
]
hav = cr 1 + 0.74ε 2 = (0.001) 1 + 0.74(0.3) = 0.001067 in
µAvips
F=
hav
πDn π (2)(500)
vm =
=
= 261.8 fpm
12
12
 12 
vips = (261.8)  = 52.36 ips
 60 
1  160 
1  160 
A= 
πDL = 
(π )(2 )(2 ) = 5.585 sq.in.
2  180 
2  180 
(2.5 ×10−6 )(5.585)(52.36) = 0.685 lb
F=
0.001067
U f = Fvm = (0.685)(261.8) = 179.3 ft − lb min
586.
The central reaction on a 120o partial bearing is 10 kips; D = 8 in .,
L D = 1 ., cr r = 0.001 . Let n = 400 rpm and µ = 3.4 ×10−6 reyn . The bearing has
a 150o cap. (a) For the bearing and the cap, compute the total frictional loss by
adding the loss in the cap to that in the bearing. (b) If the bearing were 360o,
instead of partial, calculate the frictional loss and compare.
Solution:
2
 r  µns
S =  
 cr  p
400
ns =
= 6.67 rps
60
W 10,000
p=
=
= 156.25 psi
DL (8)(8)
2
(
)
−6
 1  3.4 × 10 (6.67 )
S =
= 0.145

156.25
 0.001 
(a) Table AT 22, L D = 1 , S = 0.145
r
f = 2.021
cr
ε = 0.6367
r
f = 2.021
cr
 1 

 f = 2.021
 0.001 
f = 0.002021
Page 55 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
F = f W = (0.002021)(10,000) = 20.21 lb
πDn π (8)(400)
vm =
=
= 838 fpm
12
12
U f 1 = Fvm = (20.21)(838) = 16,936 ft − lb min
CAP:
hav = cr (1 + 0.74ε 2 )
cr = 0.001r
D
r = = 4 in
2
cr = 0.001(4) = 0.004 in
(
)
[
2
]
hav = cr 1 + 0.74ε 2 = (0.004) 1 + 0.74(0.6367 ) = 0.0052 in
µAvips
F=
hav
 12 
vips = (838)  = 167.6 ips
 60 
1  150 
1  150 
A= 
πDL = 
(π )(8)(8) = 83.78 sq.in.
2  180 
2  180 
3.4 ×10−6 (83.78)(167.6)
F=
= 9.18 lb
0.0052
U f 2 = Fvm = (9.18)(838) = 7693 ft − lb min
Total Frictional Loss
= U f 1 + U f 2 = 16,936 + 7693 = 24,629 ft − lb min
(
)
(b) 360o Bearing, L D = 1 , S = 0.145
r
f = 3.65
cr
ε = 0.5664
BEARING:
 1 

 f = 3.65
 0.001 
f = 0.00365
F = f W = (0.00365)(10,000) = 36.5 lb
πDn π (8)(400)
vm =
=
= 838 fpm
12
12
U f 1 = Fvm = (36.5)(838) = 30,587 ft − lb min
CAP:
hav = cr (1 + 0.74ε 2 )
[
]
hav = cr (1 + 0.74ε 2 ) = (0.004) 1 + 0.74(0.5664) = 0.00495 in
Page 56 of 63
2
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
F=
µAvips
hav
(3.4 ×10 )(83.78)(167.6) = 9.645 lb
F=
−6
0.00495
U f 2 = Fvm = (9.645)(838) = 8083 ft − lb min
Total Frictional Loss
= U f 1 + U f 2 = 30,587 + 8083 = 38,670 ft − lb min
587.
The central reaction on a 120o partial bearing is a 10 kips; D = 8 in. , L D = 1 ,
cr r = 0.001 ; n = 1200 rpm . Let µ = 2.5 ×10−6 reyn . The bearing has a 160o cap.
(a) Compute ho and fhp for the bearing and for the cap to get the total fhp. (b)
Calculate the fhp for a full bearing of the same dimensions and compare.
Determine (c) the needed rate of flow into the bearing, (d) the side leakage qs .
(e) the temperature rise of the oil in the bearing both by equation (o), i11.13,
Text, and by Table AT 22. (f) What is the heat loss from the bearing if the oil
temperature is 180 F? Is the natural heat loss enough to cool the bearing? (g) It is
desired to pump oil through the bearing with a temperature rise of 12 F. How
much oil is required? (h) For the oil temperature in (f), what is a suitable oil to
use?
Solution:
2
 r  µns
S =  
 cr  p
1200
ns =
= 20 rps
60
W 10,000
p=
=
= 156.25 psi
DL (8)(8)
−6
 1  (2.5 ×10 )(20 )
S =
= 0.32

156.25
 0.001 
(a) Table AT 22, L D = 1 , S = 0.32
ε = 0.5417
ho
= 0.4583
cr
r
f = 3.18
cr
q
= 2.60
rcr ns L
2
Page 57 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
qs
= 0.305
q
ρ c∆to
= 17.834
p
p
= 0.38434
pmax
ho = 0.4583cr = 0.4583(0.001)(4) = 0.00183 in
BEARING:
r
f = 3.18
cr
 1 

 f = 3.18
 0.001 
f = 0.00318
F = f W = (0.00318)(10,000) = 31.8 lb
πDn π (8)(1200)
vm =
=
= 2513 fpm
12
12
U f 1 = Fvm = (31.8)(2513) = 79,913 ft − lb min , 2.42 hp
CAP:
hav = cr (1 + 0.74ε 2 )
cr = 0.001r
D
r = = 4 in
2
cr = 0.001(4) = 0.004 in
[
]
hav = cr (1 + 0.74ε 2 ) = (0.004) 1 + 0.74(0.5417 ) = 0.00487 in
µAvips
F=
hav
2
 12 
vips = (2513)  = 503 ips
 60 
1  160 
1  160 
A= 
πDL = 
(π )(8)(8) = 89.36 sq.in.
2  180 
2  180 
2.5 ×10−6 (89.36)(5036)
F=
= 23.1 lb
0.00487
U f 2 = Fvm = (23.1)(2513) = 58,050 ft − lb min , 1.76 hp
Total Frictional Loss
= U f 1 + U f 2 = 79,913 + 58,050 = 137,963 ft − lb min
(
)
Uf
137,963
= 4.18 hp
33,000 33,000
(b) Full Bearing, L D = 1 , S = 0.32
fhp =
Page 58 of 63
=
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
Table AT 20
ho
= 0.6305
cr
r
f = 6.86
cr
ε = 0.3695
ho = 0.6305(0.004 ) = 0.002522 in
BEARING:
r
f = 6.86
cr
 1 

 f = 6.86
 0.001 
f = 0.00686
F = f W = (0.00686)(10,000) = 68.6 lb
U f 1 = Fvm = (68.6 )(2513) = 172,392 ft − lb min , 5.224 hp
CAP:
hav = cr (1 + 0.74ε 2 )
[
]
hav = cr (1 + 0.74ε 2 ) = (0.004) 1 + 0.74(0.3695) = 0.00440 in
µAvips
F=
hav
(2.5 ×10 )(89.36)(503) = 25.54 lb
F=
2
−6
0.00440
U f 2 = Fvm = (25.54 )(2513) = 64,182 ft − lb min , 1.946 hp
Total Frictional Loss
= U f 1 + U f 2 = 172,392 + 64,182 = 236,574 ft − lb min
fhp =
Uf
33,000
=
236,574
= 7.17 hp
33,000
(c) 120o Bearing
q
= 2.60
rcr ns L
q
= 2.60
(4)(0.004)(20)(8)
q = 6.656 in3 sec
q
(d) s = 0.305
q
qs
= 0.305
6.656
Page 59 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
qs = 2.03 in 3 sec
(e) Equation(o)
U f 1 = ρ cq∆to
 12 
U f 1 = 79,913 ft − lb min = 79,913  in − lb sec = 15,983 in − lb sec
 60 
U f 1 = 15,983 = (112 )(6.656 )∆to
∆to = 21.4 F
Table 22.
ρ c∆to
= 17.834
p
112∆to
= 17.834
156.25
∆to = 24.9 F
(f) Q = hcr Ab ∆tb
assume hcr = 0.516 ft − lb min − sq.in. − F
Ab = 25DL = 25(8)(8) = 1600 sq.in.
∆t
∆tb = oa
2
assume ambient = 100 F
180 − 100
∆tb =
= 40 F
2
Q = (0.516 )(1600 )(40 ) = 33,024 ft − lb min < U f 1
Therefore not enough to cool the bearing.
(g) Qr + Q = U f 1 + U f 2
Qr + 33,024 = 137,963
Qr = 104,939 ft − lb min
Qr = 20,988 in − lb sec
Qr = ρ cq∆to
20,988 = (112)q(12)
q = 15.62 in3 sec
(h) Fig. AF 16, 180 F, µ = 2.5 ×10−6 reyn
use SAE 30 oil
Page 60 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
IMPERFECT LUBRICATION:
588.
A 0.5 x 0.75-in. journal turns at 1140 rpm. What maximum load may be
supported and what is the frictional loss if the bearing is (a) SAE Type I, bronze
base, sintered bearing, (b) nylon (Zytel) water lubricated, (c) Teflon, with
intermittent use, (d) one with carbon graphite inserts.
Solution:
(a) f = 0.12
πDn π (0.5)(1140)
vm =
=
= 149.23 fpm
12
12
pvm = 50,000
p(149.23) = 50,000
p = 335 psi
W = pDL = (335)(0.5)(0.75) = 126 lb
F = f W = (0.12)(126) = 15.12 lb
U f = Fvm = (15.12 )(149.23) = 2256 ft − lb min
(b) f = 0.14 ~ 0.18 , use f = 0.16
pvm = 2500 , water
p(149.23) = 2500
p = 16.75 psi
W = pDL = (16.75)(0.5)(0.75) = 6.28 lb
F = f W = (0.16)(6.28) = 1.005 lb
U f = Fvm = (1.005)(149.23) = 150 ft − lb min
(c) vm > 100 fpm
f = 0.25
pvm = 20,000 , intermittent
p(149.23) = 20,000
p = 134 psi
W = pDL = (134)(0.5)(0.75) = 50.25 lb
F = f W = (0.25)(50.25) = 12.5625 lb
U f = Fvm = (12.5625)(149.23) = 1875 ft − lb min
(d) pvm = 15,000
p(149.23) = 15,000
p = 100.5 psi
W = pDL = (100.5)(0.5)(0.75) = 37.69 lb
assume f = 0.20
Page 61 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
F = f W = (0.20)(37.69) = 7.54 lb
U f = Fvm = (7.54 )(149.23) = 1125 ft − lb min
590.
A bearing to support a load of 150 lb at 800 rpm is needed; D = 1 in. ; semilubricated. Decide upon a material and length of bearing, considering sintered
metals, Zytel, Teflon, and graphite inserts.
Solution:
πDn π (1)(800)
vm =
=
= 209.44 fpm
12
12
assume, L = D = 1 in
W
150
p=
=
= 150 psi
DL (1)(1)
pvm = (150)(209.44) = 31,416
Use sintered metal, limit pvm = 50,000
THRUST BEARINGS
592.
A 4-in. shaft has on it an axial load of 8000 lb., taken by a collar thrust
bearing made up of five collars, each with an outside diameter of 6 in. The
shaft turns 150 rpm. Compute (a) the average bearing pressure, (b) the
approximate work of friction.
Solution:
(a) p =
4W 4(8000 )
=
= 283 psi
πDo2
π (6)2
(b) assume f = 0.065 , average
F = f W = (0.065)(8000) = 520 lb
πDn π (3)(150)
vm =
=
= 117.81 fpm
12
12
U f = nFvm = (5)(520 )(117.81) = 306,306 ft − lb min
593.
A 4-in. shaft, turning at 175 rpm, is supported on a step bearing. The bearing
area is annular, with a 4-in. outside diameter and a 3/4 –in. inside diameter.
Take the allowable average bearing pressure as 180 psi. (a) What axial load
may be supported? (b) What is the approximate work of friction?
Solution:
πDn
vm =
12
Page 62 of 63
SECTION 9 – JOURNAL AND PLANE-SURFACE BEARINGS
1
(4 + 0.75) = 2.375 in
2
πDn π (2.375)(175)
vm =
=
= 108.81 fpm
12
12
assume f = 0.065 , average
D=
(a) p =
4W
π Do2 − Di2
(
)
π  2 3 
W = (4) −    (180) = 2182 lb
4 
 4  
(b) U f = f Wvm = (0.065)(2182 )(108.81) = 15,433 ft − lb min
2
- end -
Page 63 of 63
SECTION 10 - BALL AND ROLLER BEARINGS
601.
The radial reaction on a bearing is 1500 lb.; it also carries a thrust of 1000 lb.;
shaft rotates 1500 rpm; outer ring stationary; smooth load, 8-hr./day service, say
15,000 hr. (a) Select a deep-groove ball bearing. (b) What is the rated 90 % life
of the selected bearing? (c) For b = 1.34 , compute the probability of the selected
bearing surviving 15,000 hr.
Solution:
Fx = 1500 lb
Fy = 1000 lb
( )
B10 = (15,000 )(60 )(1500 ) 10−6 = 1350 mr
F e= 0.56Cr Fx + Ct Fz
C r = 1.2 , outer ring stationary
assume Ct = 1.8
F e = 0.56(1.2)(1500) + (1.8 )(1000) = 2808 lb
1
1
 B 3
Fr =  10  Fe = (1350) 3 (2808) = 31,034 lb
 Br 
(a) Table 12.3
use 321, Fr = 31,800 lb
Fs = 32,200 lb
To check:
Fz
1000
=
= 0.03125
Fs 32,000
Table 12.2, Ct = 1.96 , Q = 0.2246
Fz
1000
=
= 0.556 > Q
C r Fx (1.2)(1500)
F e= 0.56Cr Fx + Ct Fz
F e = 0.56(1.2)(1500) + (1.96)(1000) = 2968 lb
1
1
 B 3
Fr =  10  Fe = (1350 ) 3 (2968 ) = 32,803 lb
 Br 
3.2 % higher than 31,800 lb. Safe.
Therefore use Bearing 321, Deep-Groove Ball Bearing.
(b) Fr = 31,800 lb
F e = 2968 lb
1
 B 3
31,800 =  10  (2968 )
 1 mr 
B10 = 1230 mr
(
)
B10 = (HR )(60 )(1500 ) 10 −6 = 1230
HR ≈ 13,700 hr
628
SECTION 10 - BALL AND ROLLER BEARINGS
1
  1  b
 ln   
B  P 
(c)
=
B10   1  
 ln   
  P10  
 1 
 1 
ln   = ln 

 0.9 
 P10 
B10 = 1230 mr
B = 1350 mr
1
  1   1.34
ln 
1350   P  

=
1230   1  
ln
  0.9  
P = 0.8875
602.
A certain bearing is to carry a radial load of 500 lb. and a thrust of 300 lb. The
load imposes light shock; the desired 90 % life is 10 hr./day for 5 years at
n = 3000 rpm . (a) Select a deep-groove ball bearing. What is its bore? Consider
all bearings that may serve. (b) What is the computed rated 90 % life of the
selected bearing? (c) What is the computed probability of the bearing surviving
the specified life? (d) If the loads were changed to 400 and 240 lb., respectively,
determine the probability of the bearing surviving the specified life, and the 90 %
life under the new load.
Solution:
Fx = 500 lb
Fz = 300 lb
Assume Cr = 1
Fz
300
=
= 0.6
Cr Fx (1.0)(500 )
Fz
Table 12.2,
>Q
Cr Fx
(a) F e= 0.56Cr Fx + Ct Fz
Cr = 1
Assume Ct = 1.8
F e= 0.56(1)(500) + (1.93)(300) = 820 lb
For light shock, service factor ~ 1.1
F e= (1.1)(820 ) = 902 lb
629
SECTION 10 - BALL AND ROLLER BEARINGS
1
1
 B 3
Fr =  10  Fe = (1350 )3 (2770) = 30,614 lb
 Br 
B10 = (5)(365)(10)(60 )(3000)(10−6 ) = 3285 mr
1
1
 B 3
Fr =  10  Fe = (3285)3 (902 ) = 13,409 lb
 Br 
Table 12.3,
Bearing No. Fr , lb
Fs , lb
217
14,400
12,000
312
14,100
10,900
Bore
85 mm
60 mm
Select, Bearing No. 312
Fr = 14,100 lb
Fs = 10,900 lb
(b) Table 12.2
Fz
300
=
= 0.0285
Fs 10,900
Ct = 1.99
Q = 0.22
F e= 0.56Cr Fx + Ct Fz
F e= 0.56(1)(500) + (1.99)(300 ) = 877 lb
F e= (1.1)(877 ) = 965 lb
1
 B 3
Fr =  10  Fe
 Br 
1
 B 3
14,100 =  10  (965)
 1 
B10 = 3119 mr
( )
B10 = (YR )(365)(10 )(60)(3000 ) 10−6 = 3119
YR = 4.75 years
1
  1  b
 ln   
B  P 
(c)
=
B10   1  
 ln   
  P10  
use b = 1.125
B10 = 3119 mr
630
SECTION 10 - BALL AND ROLLER BEARINGS
B = 3285 mr
 1 
ln  
3285   P  

=
3119   1  
ln
  0.9  
P = 0.8943
1
1.125
(d) Fx = 400 lb
Fz = 240 lb
Cr = 1
Fz
240
=
= 0.6
Cr Fx (1.0)(400)
Table 12.2
Ct = 2.15
Q = 0.21 < 0.6
F e= 0.56Cr Fx + Ct Fz
F e= 0.56(1)(400 ) + (2.15)(240) = 740 lb
F e= (1.1)(740) = 814 lb
1
 B 3
Fr =  10  Fe
 Br 
1
 B 3
14,100 =  10  (814)
 1 
B10 = 5197 mr
1
  1  b
 ln   
B
P
=  
B10   1  
 
 ln 
  P10  
1
  1   1.125
ln  
3285   P  

=
5197   1  
ln
  0.9  
P = 0.939
Life:
B10 = (YR )(365)(10 )(60)(3000 )(10−6 ) = 5197
YR = 8 years
631
SECTION 10 - BALL AND ROLLER BEARINGS
603.
The smooth loading on a bearing is 500-lb radial, 100 lb. thrust; n = 300 rpm . An
electric motor drives through gears; 8 hr./day, fully utilized. (a) Considering
deep-groove ball bearings that may serve, choose one end specify its bore. For
the bearing chosen, determine (b) the rated 90 % life and (c) the probability of
survival for the design lufe.
Solution:
Fx = 500 lb
Fz = 100 lb
Table 12.1, 8 hr./day fully utilized, assume 25,000 hr
B10 = (25,000 )(60)(300 )(10−6 ) = 450 mr
(a) assume Cr = 1
Fz
100
=
= 0.2
Cr Fx (1.0)(500 )
F
consider Q > z
Cr Fx
Fe = Cr Fx = (1.0 )(500) = 500 lb
1
1
 B 3
Fr =  10  Fe = (450)3 (500) = 3832 lb
 Br 
Table 12.3
Bearing No. Fr , lb
Fs , lb
207
4440
3070
306
4850
3340
305
3660
2390
Select 305, Fr = 3660 lb , Fs = 2390 lb
Bore (Table 12.4) = 25 mm
Fz 100
=
= 0.0418
Fs 2390
Table 12.2, 0.22 < Q0.26
F
Q> z
Cr Fx
Fe = Cr Fx = (1.0 )(500) = 500 lb
(a)
1
 B 3
3660 =  10  (500)
 1 
B10 = 392 mr
Rated Life:
B10 = (HR )(60)(300 )(10−6 ) = 392
HR ≈ 22,000 hr
632
SECTION 10 - BALL AND ROLLER BEARINGS
1
  1  b
 ln   
B
P
(c)
=  
B10   1  
 
 ln 
  P10  
b = 1.125
1
  1   1.125
ln  
450   P  

=
392   1  
ln
  0.9  
P = 0.884
605.
A No. 311, single-row, deep-groove ball bearing is used to carry a radial load of
1500 lb. at a speed of 500 rpm. (a) What is the 90 % life of the bearing in hours?
What is the approximate median life? What is the probability of survival if the
actual life is desired to be (b) 105 hr., (c) 104 hr.?
Solution:
Table 12.3, No. 311
Fs = 9400 lb
Fr = 12400 lb
Fx = 1500 lb
assume Cr = 1
Fe = Cr Fx = (1)(1500 ) = 1500 lb
1
 B 3
(a) Fr =  10  Fe
 Br 
1
 B 3
12400 =  10  (1500 )
 1 
B10 = 565 mr
( )
B10 = (HR )(60)(500 ) 10−6 = 565
HR ≈ 18,800 hr
For median life = 5( 90 % life) = 5(18,800) = 94,000 hr
( )
( )
(b) B = 105 (60 )(500) 10 −6 = 3000 mr
633
SECTION 10 - BALL AND ROLLER BEARINGS
 1 
 ln   
B
P
=  

B10
 1 
 
 ln 
  P10  
b = 1.125
1
b
1
  1   1.125
ln  
3000   P  

=
565   1  
ln
  0.9  
P = 0.502
(c) 104 hr
( )
( )
B = 10 4 (60 )(500 ) 10−6 = 300 mr
 1 
 ln   
B
P
=  

B10
 1 
 
 ln 
  P10  
b = 1.125
1
b
1
  1   1.125
ln  
300   P  

=
565   1  
ln
  0.9  
P = 0.950
606.
The load on an electric-motor bearing is 350 lb., radial; 24 hr. service,
n = 1200 rpm ; compressor drive; outer race stationary. (a) Decide upon a deepgroove ball bearing, giving its significant dimensions. Then compute the selected
bearing’s 90 % life, and the probable percentage of failures that would occur
during the design life. What is the approximate median life of this bearing? (b)
The same as (a), except that a 200 series roller bearing is to be selected.
Solution:
Fx = 350 lb
Fe = Cr Fx
outer race stationary, Cr = 1
Fe = (1)(350 ) = 350 lb
Table 12.1
634
SECTION 10 - BALL AND ROLLER BEARINGS
90 % Life, hrs = 50,000 hrs
B = (50,000 )(60 )(1200 ) 10−6 = 3600 mr
( )
1
1
 B 3
(a) Fr =  10  Fe = (3600 )3 (350 ) = 5364 lb
 Br 
Table AT 12.3
earing No.
Fr , lb
Fs , lb
208
5040
3520
209
5660
4010
306
4850
3340
307
5750
4020
Use No. 209 Fr = 5660 lb
Table 12.4, Dimension
Bore = 45 mm
O.D. = 85 mm
Width of Races = 19 mm
Max. Fillet r = 0.039 mm
90 % Life:
1
 B 3
Fr =  10  Fe
 Br 
1
 B 3
5660 =  10  (350)
 1 
B10 = 4229 mr
( )
B10 = (HR )(60 )(1200 ) 10−6 = 4229
HR ≈ 58,740 hr
Probability.
1
  1  b
 ln   
B
P
=  
B10   1  
 
 ln 
  P10  
b = 1.125
1
  1   1.125
ln  
3600   P  

=
4229   1  
ln
  0.9  
P = 0.916
% failures = 1 – 0.916 = 0.084 = 8.4 %
Median Life = 5(58,740) = 293,700 hrs
635
SECTION 10 - BALL AND ROLLER BEARINGS
(b) Table 12.3, Fr = 5364 lb
use No. 207, Fr = 5900 lb
Bore = 35 mm
O.D. = 72 mm
Width of Races = 17 mm
90 % life:
1
 B 3
Fr =  10  Fe
 Br 
1
 B 3
5900 =  10  (350)
 1 
B10 = 4790 mr
( )
B10 = (HR )(60 )(1200 ) 10−6 = 4790
HR ≈ 66,530 hr
Probability.
1
  1  b
 ln   
B
P
=  
B10   1  
 
 ln 
  P10  
b = 1.125
1
  1   1.125
ln  
3600   P  

=
4790   1  
ln
  0.9  
P = 0.926
% failures = 1 – 0.926 = 0.074 = 7.4 %
Median Life = 5(66,530) = 332,650 hrs
608.
A deep-groove ball bearing on a missile, supporting a radial load of 200 lb., is to
have a design life of 20 hr.; with only a 0.5 % probability of failure while
n = 4000 rpm . Using a service factor of 1.2 , choose a bearing. ( A 5- or 6- place
log table is desirable.)
Solution: No need to use log table.
Fx = 200 lb
assume Cr = 1
Fe = Cr Fx = (1.0 )(200 ) = 200 lb
636
SECTION 10 - BALL AND ROLLER BEARINGS
Fe = (1.2 )(200 ) = 240 lb
( )
B10 = (20 )(60)(4000) 10 −6 = 4.8 mr
P = 1 − 0.005 = 0.995
1
  1  b
 ln   
B
P
=  
B10   1  
 
 ln 
  P10  
b = 1.125
  1 
ln 

4.8   0.995  

=
B10   1  
ln
  0.9  
B10 = 72 mr
1
1.125
1
1
 B 3
Fr =  10  Fe = (72 )3 (240) = 998.4 lb
 Br 
Table 12.3
Select No. 201, Fr = 1180 lb
VARIABLE LOADS
610.
A certain bearing is to carry a radial load of 10 kip at a speed of 10 rpm for 20 %
of the time, a load of 8 kips at a speed of 50 rpm for 50 % of the time, and a load
of 5 kips at 100 rpm during 30 % of the time, with a desired life of 3000 hr.; no
thrust. (a) What is the cubic mean load? (b) What ball bearings may be used?
What roller bearings?
Solution:
1
 F13n1 + F23n2 + F33n3 + L  3
(a) Fm = 

∑n


∑n = n + n
1
2
+ n3
For 1 min.
n1 = (0.2 )(10 ) = 2 rev
n 2 = (0.5)(50 ) = 25 rev
n3 = (0.3)(100 ) = 30 rev
∑ n = 2 + 25 + 30 = 57 rev
F1 = 10 kips
F2 = 8 kips
637
SECTION 10 - BALL AND ROLLER BEARINGS
F3 = 5 kips
1
 (10 )3 (2 ) + (8)3 (25) + (5)3 (30)  3
Fm = 
 = 6.88 kips
57


(b) Fx = 6.88 kips = 6880 lb
assume Cr = 1
Fe = (1.0 )(6880 ) = 6880 lb
1 min = 57 rev
B10 = (3000 )(60 )(57 )(10−6 ) = 10.26 mr
1
3
1
B 
Fr =  10  Fe = (10.26 )3 (6880) = 14,950 lb
 Br 
Table 12.3, Ball Bearing
Use Bearing No. 217, Fr = 14,400 lb
(c) Table 12.3 (Roller Bearing)
Use Bearing No. 213, Fr = 14,900 lb
612.
A deep-groove ball bearing No. 215 is to operate 30 % of the time at 500 rpm
with Fx = 1200 lb and Fz = 600 lb , 55 % of the time at 800 rpm with
Fx = 1000 lb and Fz = 500 lb , and 15 % of the time at 1200 rpm with
Fx = 800 lb and Fz = 400 lb . Determine (a) the cubic mean load; (b) the 90 % life
of this bearing in hours, (c) the average life in hours.
Solution:
Bearing No. 215, Fr = 11,400 lb , Fs = 9,250 lb
Table 12.2, Fz Fs
At 30 % of the time, 500 rpm
Fz
600
=
= 0.065
Fs 9250
Ct = 1.66
Q = 0.266
Fz
600
=
= 0.5 > Q
Cr Fx (1)(1200 )
Fe1 = 0.56Cr Fc + Ct Fz = 0.56(1)(1200 ) + (1.66)(600 ) = 1668 lb
At 55 % of the time, 800 rpm
Fz
500
=
= 0.054
Fs 9250
Ct = 1.73
638
SECTION 10 - BALL AND ROLLER BEARINGS
Q = 0.257
Fz
500
=
= 0.5 > Q
Cr Fx (1)(1000 )
Fe 2 = 0.56Cr Fc + Ct Fz = 0.56(1)(1000 ) + (1.73)(500 ) = 1425 lb
At 15 % of the time, 1200 rpm
Fz
400
=
= 0.043
Fs 9250
Ct = 1.84
Q = 0.242
Fz
400
=
= 0.5 > Q
Cr Fx (1)(800 )
Fe1 = 0.56Cr Fc + Ct Fz = 0.56(1)(800) + (1.84)(400) = 1184 lb
1
 F 3n + F23n2 + F33n3 + L  3
(a) Fm =  1 1

∑n


∑n = n + n
1
2
+ n3
F1 = 1668 lb
F2 = 1425 lb
F3 = 1184 lb
For 1 min.
n1 = (0.3)(500 ) = 150 rev
n 2 = (0.55)(800) = 440 rev
n3 = (0.15)(1200) = 180 rev
∑ n = 150 + 440 + 180 = 770 rev
1
 (1668)3 (150 ) + (1425)3 (440) + (1184)3 (180 )  3
Fm = 
 = 1434 kips
770


(b) Fe = Fm = 1434 lb
1
 B 3
Fr =  10  Fe
 Br 
1
 B 3
11,400 =  10  (1434 )
 1 
B10 = 503 mr
For 1 min = 770 rev
639
SECTION 10 - BALL AND ROLLER BEARINGS
( )
B10 = (HR )(60)(770) 10−6 = 503
HR ≈ 11,000 hr
(c) Average life = 5(11,000) = 55,000 hrs
MANUFACTURER’S CATALOG NEEDED
614.
A shaft for the general-purpose gear-reduction unit described in 489 has radial
bearing reactions of RC = 613 lb and RD = 1629 lb ; n = 250 rpm . Assume that
the unit will be fully utilized for at least 8 hr./day, with the likelihood of the same
uses involving minor shock. (a) Select ball bearings for this shaft. (b) Select
roller bearings. (c) What is the probability of both bearings C and D surviving for
the design life?
Solution:
3
Problem 489, D = 1 in = 1.375 in
8
Ref: Design of Machine Members, Doughtie and Vallance
Fc = (K a K l )K o K p K s K t Fr
at C. Fr = RC = 613 lb
K t = 1.0
K p = 1. 0
K o = 1.0
Ks = 3
Kr Na
Nc
N a = 250 rpm
N c = 500 rpm
K r = 1.5
Ks = 3
(1.5)(250) = 0.90856
500
K a = 1.0
Ha
H c K rel
Table 12.1, 8 hr/day, fully utilized, Text
H a = 25,000 hr
Kl = 3
H c = 10,000 hr
assume K rel = 1.0 for 90 % reliability
Kl = 3
25,000
= 1.3572
10,000
640
SECTION 10 - BALL AND ROLLER BEARINGS
Fc = (K a K l )K o K p K s K t Fr
Fc = (1.0 )(1.3572)(1.0)(1.0 )(0.90856)(1.0 )(613) = 756 lb
Table 9-7, Doughtie and Vallance,
Two-row spherical Type, No. 207
Bore = 1.3780 in, Fc = 880 lb
At D. Fr = RD = 1629 lb
Fc = (K a K l )K o K p K s K t Fr
Fc = (1.0 )(1.3572)(1.0)(1.0 )(0.90856)(1.0 )(1629) = 2009 lb
Table 9-7, Doughtie and Vallance,
Two-row spherical Type, No. 407
Bore = 1.3780 in, Fc = 2290 lb
(b) at C, Fc = 756 lb
Table 9.8, Doughtie and Vallance
Use No. 207, Bore = 1.3780 in, Fc = 1540 lb
at C, Fc = 2009 lb
Table 9.8, Doughtie and Vallance
Use No. 307, Bore = 1.3780 in, Fc = 2660 lb
(c) For probability:
(c.1) at C, Bearing No. 207, Two-row spherical bearing, Fc = 880 lb
Fc = 880 lb = (1)K l (1)(1)(0.90856)(1)(613)
K l = 1.58
Kl = 3
Ha
H c K rel
1.58 = 3
25,000
10,000 K rel
K rel = 0.634
Table 9-3, Reference
Probability = 95.8 %
at D, Bearing No. 407, Deep-groove bearing, Fc = 2290 lb
Fc = 2290 lb = (1)K l (1)(1)(0.90856)(1)(1627 )
K l = 1.547
Kl = 3
Ha
H c K rel
641
SECTION 10 - BALL AND ROLLER BEARINGS
1.547 = 3
25,000
10,000 K rel
K rel = 0.675
Table 9-3, Reference
Probability = 93.3 %
(c.2) at C, Roller Bearing No. 207, Fc = 1540 lb
Fc = 1540 lb = (1)K l (1)(1)(0.90856)(1)(613)
K l = 2.765
Kl = 3
Ha
H c K rel
2.765 = 3
25,000
10,000 K rel
K rel = 0.118
Table 9-3, Reference
Probability = 98.8 %
at D, Roller Bearing No. 407, Fc = 2660 lb
Fc = 2660 lb = (1)K l (1)(1)(0.90856 )(1)(1627 )
K l = 1.80
Kl = 3
Ha
H c K rel
1.80 = 3
25,000
10,000 K rel
K rel = 0.43
Table 9-3, Reference
Probability = 95.7 %
615.
A shaft similar to that in 478 has the following radial loads on the bearings, left
to right: 803 lb, 988 lb, 84 lb, and 307 lb; no thrust. The minimum shaft diameter
at the bearings are 1.250 in, 1.125 in, 1.000 in, and 1.0625 in. Assume that the
service will not be particularly gentle; intermittently used, with n = 425 rpm . (a)
Select ball bearing for this shaft. (b) Select roller bearings.
Solution:
Ref: Design of Machine Members by Doughtie and Vallance
Fc = (K a K l )K o K p K s K t Fr
K a = 1.0
642
SECTION 10 - BALL AND ROLLER BEARINGS
Kl = 3
Ha
H c K rel
H c = 10,000 hr
Table 12.1, Text, H a = 10,000 hr (intermittent)
90 % reliability, K rel = 1.0
Kl = 3
10,000
= 1. 0
10,000
K o = 1.0
K p = 1. 0
K r = 1.5 assumed
Ks = 3
Kr Na
Nc
N a = 425 rpm
N c = 500 rpm
Ks = 3
(1.5)(425) = 1.0844
500
K t = 1.0
(a) Ball Bearing
(a.1) 803 lb, D = 1.250 in
Fc = (1.0 )(1.0 )(1.0)(1.0)(1.0844)(1.0)(803) = 870 lb
Table 9-7, Ref.
Two-row spherical type, 207
Fc = 880 lb
Bore = 1.3780 in
(a.2) 988 lb, D = 1.125 in
Fc = (1.0 )(1.0)(1.0)(1.0)(1.0844)(1.0 )(988) = 1071 lb
Table 9-7, Ref.
Two-row spherical type, 306
Fc = 1050 lb
Bore = 1.1811 in
(a.3) 84 lb, D = 1.000 in
Fc = (1.0 )(1.0 )(1.0)(1.0)(1.0844)(1.0)(84) = 91 lb
643
SECTION 10 - BALL AND ROLLER BEARINGS
Table 9-7, Ref.
Deep-groove type, 106
Fc = 544 lb
Bore = 1.1811 in
(a.4) 307 lb, D = 1.0625 in
Fc = (1.0 )(1.0 )(1.0)(1.0)(1.0844)(1.0)(307 ) = 333 lb
Table 9-7, Ref.
Deep-groove type, 106
Fc = 544 lb
Bore = 1.1811 in
(b) Roller Bearing
(b.1) 803 lb, D = 1.250 in
Fc = 870 lb , Bore = 1.3780 in
use No. 207, Fc = 1540 lb
(b.2) 988 lb, D = 1.125 in
Fc = 1071 lb , Bore = 1.1811 in
use No. 206, Fc = 1320 lb
(b.3) 84 lb, D = 1.000 in
Fc = 91 lb , Bore = 1.1811 in
use No. 206, Fc = 1320 lb
(b.4) 307 lb, D = 1.0625 in
Fc = 333 lb , Bore = 1.1811 in
use No. 206, Fc = 1320 lb
- end -
644
SECTION 11 – SPUR GEARS
INTERMITTENT SERVICE
DESIGN PROBLEMS
631.
A pair of gears with 20o full-depth teeth are to transmit 10 hp at 1750 rpm of the
3-in. pinion; velocity ratio desired is about 3.8; intermittent service. Use a
strength reduction factor of about 1.4, with theload at the tip and teeth
commercially cut. Determine the pitch, face width, and tooth numbers if the
material is cast iron, class 20.
Solution:
πD p n p π (3)(1750)
vm =
=
= 1374 fpm < 2000 fpm
12
12
10 hp < 20 hp
Commercially cut gears, vm < 2000 fpm
600 + vm
Fd =
Ft
600
33,000hp 33,000(10 )
Ft =
=
= 240 lb
vm
1374
 600 + 1374 
Fd = 
(240 ) = 790 lb
600


Fs =
sbY
K f Pd
For cast-iron, class 20, sn = 0.4 su
s = 0.4(20 ) = 8 ksi = 8000 psi
K f = 1 .4
10
Pd
Table AT 24, Load at tip, 20o F.D.
Assume Y = 0.33
assume b =
Fs = Fd
(8000)(10)(0.33) = 790
(1.4)Pd2
Pd = 4.89
use Pd = 5
N p = Pd D p = (5)(3) = 15
Y = 0.289
Page 1 of 57
SECTION 11 – SPUR GEARS
Fs = Fd
(8000)(b )(0.33) = 790
(1.4)(5)
b = 2.4 in
8
12.5
<b<
Pd
Pd
use
12.5 12.5
b=
=
= 2.5 in
Pd
5
1
b = 2 in
2
Summary of answers:
Pd = 5
1
b = 2 in
2
N p = 15
N g = 3.8(15) = 57
633.
A pair of gears with 20o full-depth teeth are to transmit 5 hp at 1800 rpm of the
pinion; mw = 2.5 ; N p = 18 teeth; commercially cut teeth; intermittent service;
K f ≈ 1.45 . (a) Determine the pitch, face width, and tooth numbers if the material
is cast iron, class 25. (b) The same as (a) except that the pinion is to be made of
phosphor gear bronze (SAE 65, Table AT 3).
Solution:
Load at tip, Table AT 24, 20o F.D.
N p = 18 , Y = 0.308
N g = (2.5)(18) = 45 , Y = 0.399
vm =
πD p n p
12
N p 18
Dp =
=
Pd
Pd
n p = 1800 rpm
π (18)(1800 )
2700π
12 Pd
Pd
Commercially cut teeth
vm =
Page 2 of 57
=
SECTION 11 – SPUR GEARS
 600 + vm 
Fd = 
 Ft
 600 
33,000(5) 550 Pd
Ft =
=
9π
 2700π 


 Pd 
2700π 
8482.3 


 600 +

 600 +

Pd  550 Pd  
Pd 
(19.4523Pd )
Fd = 
=



 9π  

600
600








 14.1372 

Fd = 19.4523Pd 1 +
Pd 

Fs =
sbY
K f Pd
10
Pd
(a) Cast iron, class 25,
sn = 0.4su = 0.4(25) = 10 ksi = 10,000 psi
assume b =
Fs = Fd
Pinion is weaker
(10,000)(10)(0.308) = 19.4523P 1 + 14.1372 
d
Pd 
(1.45)(Pd )2

 14.1372 
21,241

= 19.4523Pd 1 +
2
Pd
Pd 

Pd = 7.16
use Pd = 7
Face width b
(10,000 )(b )(0.308) = 19.4523(7 )1 + 14.1372 


(1.45)(7 )
7


b = 1.355 in
8
12.5
<b<
Pd
Pd
1.143 < b < 1.786
say b = 1.5 in
Summary of answers:
Pd = 7
1
b = 1 in
2
Page 3 of 57
SECTION 11 – SPUR GEARS
N p = 18
N g = 45
(b) Phosphor gear bronze , (SAE 65, Table AT 3)
su = 80 ksi
sn = 0.4su = 0.4(80) = 32 ksi
or sn = 31 ksi (Table AT 3)
use sn = 31 ksi
Pinion, bronze
sY = (31,000)(0.308) = 9548
Gear, cast iron
sY = (10,000)(0.399) = 3990
Therefore gear is weaker
(10,000)(10)(0.399) = 19.4523P 1 + 14.1372 
d
Pd 
(1.45)(Pd )2

 14.1372 
27,517

= 19.4523Pd 1 +
2
Pd
Pd 

Pd = 8
use Pd = 8
Face width b
(10,000 )(b )(0.399) = 19.4523(8)1 + 14.1372 


(1.45)(8)
8


b = 1.252 in
8
12.5
<b<
Pd
Pd
1 < b < 1.5625
1
say b = 1.25 in = 1 in
4
Summary of answers:
Pd = 8
1
b = 1 in
4
N p = 18
N g = 45
Page 4 of 57
SECTION 11 – SPUR GEARS
634.
It is desired to transmit 120 hp at 1800 rpm of the pinion; intermittent service;
with light shock (§13.18); preferably not less than 18, 20o-full-depth teeth on the
pinion; K f = 1.5 should be conservative; mw = 1.5 . Decisions must be made
concerning the material and quality of cutting the teeth. Since the design is for
strength only, it will be convenient to express Ft , Fd , vm , b in terms of Pd and
arrange an equation containing s and Pd convenient for iteration. Weak material
results in a relatively large pinion with high peripheral speed. A very strong
material may be unnecessarily expensive. On a production basesm carefully cut
teeth should have a reasonable cost. Specify material, accuracy of cutting, pitch,
face width, and tooth numbers.
Solution:
N
18
Dp = p =
Pd
Pd
 18 
(1800 )
πD p n p
Pd 
8482.3

vm =
=
=
12
12
Pd
33,000hp 33,000(120 )
Ft =
=
= 466.85Pd
vm
 8482.3 


 Pd 
120 hp > 20 hp
use Buckingham’s equation
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
Assume a material of steel, as rolled AISI 1050,
su = 102 ksi
For carefully cut teeth, Fig. AF 20, e = 0.001 in (min.)
Table AT 25, steel on steel, 20o F.D.
C = 1660
π 
Try b =
10
Pd
Fd = 466.85Pd +

 8482.3  
 10 
 1660  + 466.85 Pd 
0.05
 Pd  
 Pd 

1
2
 8482.3  
 10 
 + 1660  + 466.85Pd 
0.05
 Pd  
 Pd 

Page 5 of 57
SECTION 11 – SPUR GEARS

424  16,600

+ 466.85Pd 
Pd  Pd

Fd = 466.85Pd +

424  16,600
+ 
+ 466.85Pd 
Pd  Pd

1
2
sbY
K f Pd
Weak pinion
Table AT 24, N p = 18 , Y = 0.522 (Load near middle, 20o F.D.)
Fs =
s = 0.5su = 0.5(102) = 51 ksi = 51,000 psi
K f = 1 .5


(51,000) 10 (0.522)
 Pd 
(1.5)Pd
Fs =
=
177,480
Pd2
Fs = N sf Fd
(§13.18), light shock, N sf = 1.25
Iteration:
Pd
5
4
Use Pd = 5
10 10
b=
=
= 2 in
Pd
5
N p = 18
Ng
Np
= mw
N g = (1.5)(18) = 27
Summary of first computation
Material, AISI 1050, as rolled
Accurately cut gears
Pd = 5
b = 2.0 in
N p = 18
N g = 27
Page 6 of 57
Fd
5331
5342
Fs
7099
11093
N sf (≥ 1.25)
1.33
2.08
SECTION 11 – SPUR GEARS
CHECK PROBLEMS
A pair of carefully cut, full depth, 20o involute gears, made of cast iron, ASTM
30, is transmitting 5 hp at 1150 rpm of the pinion; N p = 24 , N g = 32 , Pd = 8 ,
636.
1
b = 1 in . For the teeth, determine (a) the endurance , (b) the dynamic load, (c)
2
the service factor (§13.18).
Solution: 5 hp < 20 hp
N p = 24 , 20o F.D.
Y = 0.337 , Load at tip
assume K f = 1.45 average
(a) Fs =
sbY
K f Pd
cast-iron, s = 0.4su = 0.4(30) = 12 ksi = 12,000 psi
(12,000)(1.5)(0.337 ) = 523 lb
Fs =
(1.45)(8)
 1200 + vm 
(b) Fd = 
 Ft
 1200 
For carefully cut gears
πD p n p
vm =
12
N
24
Dp = p =
= 3 in
Pd
8
n p = 1150 rpm
vm =
π (3)(1150)
= 903.2 fpm
12
33,000hp 33,000(5)
Ft =
=
= 182.7 lb
vm
903.2
 1200 + vm 
 1200 + 903.2 
Fd = 
 Ft = 
(182.7 ) = 320 lb
1200
 1200 


(c) N sf =
637.
Fs 523
=
= 1.63
Fd 320
A manufacturer’s catalog for cut-tooth spur gears rates a 25-tooth, cast-iron
(ASTM 25) pinion with 5-pitch, 20o full-depth involute teeth at 16.5 hp at 900
Page 7 of 57
SECTION 11 – SPUR GEARS
1
rpm; b = 2 in and mg = 2 ; let K f = 1.5 ; intermittent service; smooth load. (a)
2
What horsepower may these gears transmit? Do you consider the catalog rating
too high or too low? (b) The same as (a) except that the teeth are carfully cut. (c)
The same as (a) except that the pinion is to be made of phosphor bronze (SAE 65,
Table AT 3).
Solution:
16.5 hp < 20 hp
πD p n p
vm =
12
N p 25
Dp =
=
= 5 in
Pd
5
n p = 900 rpm
vm =
π (5)(900)
12
= 1178 fpm
(a) Using commercially cut.
 600 + vm 
Fd = 
 Ft , vm ≤ 2000 fpm
 600 
Table AT 24, Load at tip, 20o F.D.
N p = 25 , Y = 0.340
N g = mg N p = 2(25) = 50 , Y = 0.408
use Y = 0.340 , weaker pinion
Cast iron, s = 0.4su = 0.4(30) = 12 ksi = 12,000 psi
K f = 1 .5
Pd = 5
1
b = 2 in
2
sbY
Fs =
K f Pd
(12,000)(2.5)(0.340) = 1360 lb
Fs =
(1.5)(5)
Fs = N sf Fd
smooth load, N sf = 1
 600 + 1178 
1360 = (1)
 Ft
600


Ft = 459 lb
Page 8 of 57
SECTION 11 – SPUR GEARS
Ft vm
(459)(1178) = 16.4 hp
=
33,000
33,000
The rating is not too low or too high.
hp =
(b) Carefully cut, [1000 fpm < vm < 4000 fpm]
 1200 + vm 
Fd = 
 Ft
 1200 
Fs = N sf Fd
 1200 + 1178 
1360 = (1)
 Ft
 1200

Ft = 686 lb
hp =
Ft vm
(686 )(1178) = 24.5 hp
=
33,000
33,000
(c) Phosphor bronze, (SAE 65, Table AT 3)
sn = 24 ksi = 24,000 psi
Pinion
sY = (24,000 )(0.340) = 8160
Gear
sY = (12,000)(0.408) = 4896
Therefore , gear is weaker
sbY
Fs =
K f Pd
(12,000)(2.5)(0.408) = 1632 lb
Fs =
(1.5)(5)
Fs = N sf Fd
 600 + 1178 
1632 = (1)
 Ft
600


Ft = 551 lb
hp =
638.
Ft vm
(551)(1178) = 19.7 hp
=
33,000
33,000
A pair of commercially cut spur gears transmits 10 hp at 1750 rpm of the 25tooth pinion. The teeth are 20o full depth with 6 pitch; material , cast iron, class
9
30; face width is 1 in .; N g = 40 . Allow for stress concentration. (a) Compute
16
Page 9 of 57
SECTION 11 – SPUR GEARS
the service factor for the teeth (§13.18). (b) If the drive is for a single-cylinder
compressor, would carefully cut teeth be advisable? Show calculations.
Solution:
10 hp < 20 hp
Dp =
vm =
Np
=
Pd
πD p n p
25
= 4.167 in
6
12
π (4.167 )(1750)
vm =
= 1909 fpm
12
For commercially cut gears
 600 + vm 
Fd = 
 Ft
 600 
33,000hp 33,000(10 )
Ft =
=
= 173 lb
vm
1909
 600 + 1909 
Fd = 
(173) = 723.4 lb
600


sbY
Fs =
K f Pd
Assume K f = 1.45 , average, load near tip
Table AT 24, 20o F.D.
N p = 25 , Y = 0.340
Cast iron, s = 0.4su = 0.4(30) = 12 ksi = 12,000 psi
(12,000)1 9 (0.340)
Fs =
 16 
(1.45)(6)
= 732.8 lb
Fs 732.8
=
≈ 1 .0
Fd 723.4
(b) Carefully cut gears
 1200 + vm 
 1200 + 1909 
Fd = 
 Ft = 
(173) = 448.2 lb
 1200 
 1200

F
732.8
N sf = s =
= 1.635
Fd 448.2
(§13.18) Single-cylinder compressore
1.5 < N sf < 1.75
(a) N sf =
Page 10 of 57
SECTION 11 – SPUR GEARS
1.5 < 1.635 < 1.75
Therefore advisable.
CONTINUOUS SERVICE
DESIGN PROBLEMS
NOTE: When using Buckingham’s Fd equation and a K f is used, as intended in design
for continuous service, use Y for load near middle.
639.
The pinion of a pair of steel gears, transmitting 110 hp at 2300 rpm, is to have a
1
diameter of about 4 in .; mg ≈ 2.3 ; 20o full-depth teeth; the drive is to a
3
centrifugal pump, continuous service. (a) Decide upon Pd , b , N p , N g , and the
material to be used. Consider the strength with the load near the middle of the
profile. (b) The same as (a) except that it is not expected that the maximum
loading will occur for more than 107 cycles, can you justify changes in your
previous answers?
Solution:
110 hp > 20 hp
1
D p = 4 in
3
n p = 2300 rpm
 1
πD p n p
3
vm =
=  
= 2609 fpm
12
12
33,000hp 33,000(110 )
Ft =
=
= 1391 lb
vm
2609
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
Fig. AF 19, vm = 2609 fpm
Maximum permissible error = 0.0012 in
Fig. AF 20
Use carefully cut gears
Expected errors = 0.001 in
Table AT 25, steel on steel, 20o F.D.
C = 1660
10
Try b =
Pd
π  4 (2300)
Page 11 of 57
SECTION 11 – SPUR GEARS


 10 
0.05(2609)1660  + 1391
 Pd 


Fd = 1391 +
1

2
 10 
0.05(2609) + 1660  + 1391
 Pd 


 16600

130.45
+ 1391
 Pd

Fd = 1391 +
1
 16600
2
130.45 + 
+ 1391
 Pd

Wear Load
Fw = D pbQK g
Q=
2m g
mg + 1
=
2(2.3)
= 1.394
2 .3 + 1
with Pd = 5 , Fd = 4478 lb
Pd = 4 , Fd = 4919 lb
 1  10 
Pd = 5 , Fw =  4  (1.394 )K g = 4478
 3  5 
K g = 371
 1  10 
Pd = 4 , Fw =  4  (1.394 )K g = 4919
 3  4 
K g = 326
Table AT 26, 20o F.D.
Use sum of BHN = 800. K g = 366
22,109
 1  10 
Fw =  4  (1.394)(366) =
Pd
 3  Pd 
Iteration: Fw ≥ Fd
Pd
5
4
8
12.5
<b<
Pd
Pd
1.6 in < b < 2.5 in
use b = 2.5 in
Use Pd = 5 ,
Page 12 of 57
Fd
4478
4919
Fw
4422
5527
SECTION 11 – SPUR GEARS
To check for strength
sbY
Fs =
K f Pd
 1
Load near middle, N p = Pd D p = (5) 4  = 21.67
 3
N g = mg N p = (2.3)(21.67 ) = 49.84
use N g = 50 , N p = 22
D p = 4.4 in
50
= 10 in
5
For s , average BHN = 400
Dg =
Use
8630 WQT 800 F, BHN = 375 gear
2330 WQT 600 F, BHN = 429 pinion
Sum of BHN = 375 + 429 = 804 ≈ 800
Table AT 24, Load near middle, 20o F.D.
N p = 22
Y = 0.559
N g = 50
Y = 0.694
Pinion:
s = 250(429) = 107,250 psi
sY = (107,250)(0.559) = 59,953 psi
s = 250(375) = 93,750 psi
sY = (93,750)(0.694) = 65,062 psi
Gear:
Pinion is weaker
sbY
(59,953)(2.5) = 17,633 lb > F
Fs =
=
d
(1.7)(5)
K f Pd
Summary of answer:
Pd = 5
b = 2.5 in
N p = 22
N g = 50
Material:
Pinion:
Gear:
2330 WQT 600 F, BHN = 429
8630 WQT 800 F, BHN = 375
(b) For 107 cycles
Page 13 of 57
SECTION 11 – SPUR GEARS
Tables AT 26, K g =< 306 for the less stronger material available, therefore can’t be
justify to change the previous answer. ( N p will become less than minimum). Using
stronger material is expensive.
640.
A pinion with 20o full-depth teeth, transmitting 60 hp at 2400 rpm, is part of a
gear reduction for a lobeblower. It is to be about 3.2 in. in diameter; mg ≈ 1.56 .
(a) Decide upon a material for the mating gears (and its treatment), Pd , b , N p ,
and N g . Determine the strength with the load near the middle of the profile. (b)
The same as (a), except that the maximum loading will occur for no more than
107 cycles.
Solution:
D p = 3.2 in
n p = 2400 rpm
vm =
πD p n p
=
π (3.2)(2400)
= 2011 fpm
12
12
33,000hp 33,000(160 )
Ft =
=
= 2626 lb
vm
2011
160 hp > 20 hp
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
Fig. AF 19, vm = 2011 fpm
Maximum permissible error = 0.0015 in
Fig. AF 20
carefully cut gears
Expected errors = 0.001 in
Table AT 25, steel on steel, 20o F.D.
C = 1660
10
Try b =
Pd


 10 
0.05(2011)1660  + 2626
 Pd 


Fd = 2626 +
1

2
 10 
0.05(2011) + 1660  + 2626
 Pd 


Page 14 of 57
SECTION 11 – SPUR GEARS
 16600

100.55
+ 2626 
 Pd

Fd = 2626 +
1
 16600
2
100.55 + 
+ 2626 
 Pd

Fw = D pbQK g
Q=
2(1.56 )
= 1.22
m g + 1 1.56 + 1
2m g
=
39.04 K g
 10 
Fw = (3.2) (1.22)K g =
Pd
 Pd 
Fw = Fd
For Pd = 5
Fd = 5991 lb
39.04 K g
Fw =
= 5991
5
K g = 767
For Pd = 4
Fd = 6532 lb
39.04 K g
Fw =
= 6532
4
K g = 651
Table AT 26,
Use Steel (600), carburized case hardened, and same 1010 cycles, 20o F.D.
K g = 750
Using Pd = 5 , b =
10
= 2 in
5
1
Say b = 2 in
2
To check for strength
sn = 250 BHN = 250(600) = 150,000 psi
Table AT 24, Load near middle
N p = Pd D p = (5)(3.2 ) = 16
Y = 0.503
N g = mg N p = (1.56 )(16 ) = 25 Y = 0.580
Page 15 of 57
SECTION 11 – SPUR GEARS
assume K f = 1.7
Fs =
(150,000)(2.5)(0.503) = 22,191 lb > F
sbY
=
d
(1.7)(5)
K f Pd
Summary of answer:
Pd = 5
b = 2.5 in
N p = 16
N g = 25
(b) 107 cycles
Table AT 26, K g = 1680
Fw =
39.04(1680 ) 65,587
=
Pd
Pd
Iteration: Fw ≥ Fd
Pd
7
8
9
10
11
12
13
Fd
5560
5421
5311
5222
5148
5086
5033
Use Pd = 13
10
b = in
13
N p = D p Pd = (3.2 )(13) = 42 , Y = 0.667
To check for strength
sbY
(150,000)(10)(0.667) = 3482 lb < F
Fs =
=
d
K f Pd
(1.7)(13)2
Therefore use Fs ≥ N sf Fd
For lobe blower, 1.25 < N sf < 1.5
Assume Y = 0.50
Page 16 of 57
Fw
9370
8198
7287
6559
5962
5466
5045
SECTION 11 – SPUR GEARS
(150,000)(10)(0.50) = 441,177
sbY
=
K f Pd
Pd2
(1.7 )(Pd )2
Iteration:
Pd
Fs
7
9004
8
6894
Fs =
Use Pd = 8
10 10
b=
=
= 1.25 in
Pd
8
N p = Pd D p = (8)(3.2 ) = 25.6
Fd
5560
5421
N sf
1.619
1.272>1.25
say 26
N g = mg N p = (1.56 )(25.6 ) = 40
Summary of answer:
Pd = 8
1
b = 1 in
4
N p = 26
N g = 40
641.
Gears with 20o full-depth teeth are to transmit 100 hp continuously at 5000 rpm
with mg = 4 ; pinion D p = 3 in. ; the drive is subjected to minor shocks with
frequent starts. First calculations are to be made for carburized pinion teeth of
AISI E3310, SOQT 450F, and the gear of cast steel, SAE 0175, WQT. Decide
upon Pd , b , N p , and N g .
Solution:
πD p n p π (3)(5000)
vm =
=
= 3927 fpm
12
12
Fig. AF 19, max. per/ error. e = 0.00075 in
Use precision gears, error, e = 0.0005 in
Table AT 25, steel on steel, 20o F.D.
C = 0.5(1660) = 830
33,000hp 33,000(100 )
Ft =
=
= 840 lb
vm
3927
10
b=
Pd
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
Page 17 of 57
SECTION 11 – SPUR GEARS
  10 

0.05(3927 )830  + 840
  Pd 

Fd = 840 +
1
  10 
2
0.05(3927 ) + 830  + 840
  Pd 

 8300

196.35
+ 840 
 Pd

Fd = 840 +
1
 8300
2
196.35 + 
+ 840 
 Pd

For minor shock with frequent start
1.25 < N sf < 1.5
sbY
K f Pd
Pinion: AISI E3310, SOQT 450 F, Rc = C 57.5 , BHN = 600
sn = 250(600) = 150,000 psi
Gear: Cast Steel, SAE 0175, WQT
sn = 77 ksi = 77,000 psi (Table AT 6)
Fs =
use s = sn = 77,000 psi
sbY
(77,000)(10)(0.50) = 226,471
Fs =
=
K f Pd
Pd2
(1.7 )(Pd )2
F
Iteration: N sf = s , 1.25 < N sf < 1.5
Fd
Pd
5
6
7
8
Fd
2833
2633
2488
2378
Use Pd = 8
Check for wear
Fw = Fd
Fw = D pbQK g
Q=
2m g
mg + 1
=
2(4 )
= 1 .6
4 +1
Page 18 of 57
Fs
9059
6291
4622
3539
N sf
3.2
2.4
1.9
1.49
SECTION 11 – SPUR GEARS
10
= 1.25 in
8
Fw = (3)(1.25)(1.6 )K g = 2378
b=
K g = 396
K g = 396 < for carburized teeth
Therefore
Pd = 8
1
b = 1 in
4
N p = D p Pd = (3)(8) = 24
N g = mg N p = (4 )(24 ) = 96
Summary of answer:
Pd = 8
1
b = 1 in
4
N p = 24
N g = 96
A 20-tooth (20o F.D.) pinion is to transmit 50 hp at 600 rpm, the service being
indefinitely continuous in a conveyor drive; mw = 2.5 . The original pplan is to
use a nodular-iron casting, 80-60-03, for each gear. Determine suitable values for
the pitch, face width, and diameters. (Warning: compute C .)
642.
Solution:
N p = 20
Dp =
Np
Pd
=
20
Pd
 20 
(600 )
πD p n p
Pd 
3142

=
=
vm =
12
12
Pd
33,000hp 33,000(50)
Ft =
=
= 525Pd
vm
 3142 


 Pd 
π 
b=
10
Pd
Fd = Ft +
0.05vm (Cb + Ft )
1
0.05vm + (Cb + Ft )2
Page 19 of 57
SECTION 11 – SPUR GEARS
Fd = 525Pd +
Fd = 525Pd +

 3142    10 
 C   + 525Pd 
0.05
 Pd    Pd 

1
2
 3142    10 
 + C   + 525Pd 
0.05
 Pd    Pd 


157.1  10C

+ 525Pd 
Pd  Pd


157.1  10C
+ 
+ 525Pd 
Pd
 Pd

1
2
Fs ≥ N sf Fd
Conveyor drive, 1 < N sf < 1.25
sbY
K f Pd
assume Y = 0.50 load near middle
K f = 1 .7
For nodular-iron, 80-60-03
sn = 40,000 psi
Fs ≥
E = 23 × 106 psi
(
(
)
)
2
k 23 × 106
= k 11.5 × 106
C=
=
6
Eg + E p
2 23 ×10
k = 0.111e for 20o full depth
kE p E g
Fs =
(
)
(40,000)(10)(0.50) = 117,647
(1.7 )Pd2
Pd2
Fs
Fd
Iteration: Use Fig. AF 19 and Fig AF 20.
3142
vm =
Pd
e
C
Pd
5
628.4
0.00225
2872
4
785.5
0.002625
3351
N sf =
Use Pd = 4 , commercially cut
8
12.5
<b<
Pd
Pd
2 < b < 3.125
use b = 3.0 in
Page 20 of 57
Fd
Fs
4765
5005
4706
7353
Fs
Fd
0.99
1.46
N sf =
SECTION 11 – SPUR GEARS
Fw = D pbQK g
Table AT 26, K g = 248 20o F.D.
Np
20
= 5 in
Pd
4
2mw
2(2.5)
Q=
=
= 1.43
m w + 1 2 .5 + 1
Fw = (5)(3.0)(1.43)(248) = 5320 lb > Fd
Summary of answer:
Pd = 4 , commercially cut gears
b = 3 in
D p = 5 in
Dp =
=
Dg = mw D p = (2.5)(5) = 12.5 in
643.
A 4.8-in. (approximate) pinion with 20o full-depth teeth is to transmit 40 hp at
1000 rpm; indefinitely continuous service with smooth load; mg = 3.5 ; carefully
cut teeth to reduce the chance of an explosive spark, the use of a phosphor0gearbronze (SAE 65, Table AT 3) pinion and a cast-iron (class 35) gear is a tentative
decision. Decide upon an appropriate Pd and b , using Buckingham’s average
dynamic load.
Solution:
Dd = 4.8 in
n p = 1000 rpm
vm =
πD p n p
=
π (4.8)(1000)
= 1257 fpm
12
12
Fig. AF 19, max. permissible error, e = 0.00225 in
Carefully-cut, e = 0.001 in
kE g E p
C=
Eg + E p
k = 0.111e for 20o full depth
Phosphor-bronze pinion SAE 65
sn = 24,000 psi
E p = 16×106 psi
Cast iron gear, class 35
sn = 0.4su = 0.4(35,000) = 14,000 psi
E g = 14.5 ×106 psi
Page 21 of 57
SECTION 11 – SPUR GEARS
C=
(0.111)(0.001)(14.5 ×106 )(16 ×106 ) = 844
14.5 ×106 + 16 ×106
Assume Pd = 4
e = 0.00125 in
C = 1.25(844) = 1055
33,000hp 33,000(40 )
Ft =
=
= 1050 lb
vm
1257
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
10 10
b=
=
= 2.5 in
Pd 4 d
0.05(1257 )[1055(2.5) + 1050]
Fd = 1050 +
= 2926 lb
1
2
0.05(1257 ) + [1055(2.5) + 1050]
sbY
Fs =
K f Pd
assume K f = 1.7 , load near middle, gear, 20o F.D.
N p = D p Pd = (4.8)(4 ) = 19 , Y = 0.534
use s = 14,000 psi , gear
N g = mg N p = (3.5)(19 ) = 66 , Y = 0.7224
Fs =
(14,000)(2.5)(0.7224) = 3718 lb
(1.7 )(4)
Fs ≥ N sf Fd
smooth load, N sf = 1.0
Fs > Fd
3718 lb > 2926 lb
Summary
Pd = 4
b = 2.5 in
644.
The 20o full-depth teeth for a pair of steel gears are to transmit 40 hp at 1200 rpm
of the 20-tooth pinion; mg = 3 ; continuous service and indefinite life: The driven
machine is an off-and-on reciprocating compressor. (a) Determine the pitch, face
width, and steel (with treatment), considering at least three alternatives, including
carefully cut teeth. For the gear teeth decided on, what would be the power
capacity if only intermittent service (wear not considered) were required? (c) If a
limited life of 107 cycles were satisfactory?
Solution:
Page 22 of 57
SECTION 11 – SPUR GEARS
N p = 20
Np
20
Pd
Pd
n p = 1200 rpm
Dp =
=
 20 
(1200 )
πD p n p
P
6283
vm =
=  d
=
12
12
Pd
π 
Ft =
33,000hp 33,000(40)
=
= 210 Pd
vm
 6283 


P
d


Fd = Ft +
0.05vm (Cb + Ft )
1
0.05vm + (Cb + Ft )2
For carefully cut, e = 0.001 in
Table AT 25, steel on steel, 20o F.D.
C = 1660
Fw = Fd
Fw = D pbQK g
assume b =
10
Pd
Fd = 210 Pd +
Fd = 210 Pd +

 6283  
 10 
 1660  + 210 Pd 
0.05
 Pd  
 Pd 


314  16600
+ 
+ 210 Pd 
Pd  Pd

2mw
2(3)
Q=
=
= 1 .5
mw + 1 3 + 1
 20  10 
300
Fw =   (1.5)(K g ) = 2 K g
Pd
 Pd  Pd 
Pd = 5
Page 23 of 57
1
2
 6283  
 10 
 + 1660  + 210 Pd 
0.05
 Pd  
 Pd 


314  16600

+ 210 Pd 
Pd  Pd

1
2
SECTION 11 – SPUR GEARS
Fd = 3179 lb
300
Fw = 2 K g = 3179
(5)
K g = 265
Pd = 6
Fd = 3079 lb
300
Fw = 2 K g = 3079
(6)
K g = 369
(a) For indefinite life, Table AT 26
use Sum of BHN = 700, K g = 270 , Pd = 5
10 10
=
= 2 in
Pd
5
N
20
Dp = p =
= 4 in
Pd
5
Dg = m g D p = (3)(4 ) = 12 in
b=
Material combination. (Sum of BHN = 700)
Alternatives:
(1)
4150 OQT 1200 F, gear, BHN = 331
6152 OQT 1000 F, pinion, BHN = 375
(2)
5150 OQT 1000 F, gear, BHN = 321
8620 OQT 800 F, pinion, BHN = 375
(3)
4150 OQT 1200 F, gear, BHN = 331
C1095 OQT 800 F, pinion, BHN = 363
Using Pinion: C1095, OQT 800 F, BHN = 363
sn = 250(363) = 90,750 psi
Gear: 4150, OQT 1200 F, BHN = 331
sn = 250(331) = 82,750 psi
Table AT 24, Load near middle, 20o F.D.
N p = 20
Y = 0.544
N g = 3(20 ) = 60 , Y = 0.713
Pinion: sY = (90,750)(0.544) = 49,368
Gear: sY = (82,750)(0.713) = 59,000
Therefore, pinion is weaker
Page 24 of 57
SECTION 11 – SPUR GEARS
K f = 1 .7
Fs =
(90,750)(2)(0.544) = 11,616 lb
sbY
=
(1.7)(5)
K f Pd
Fs = N sf Fd
Reciprocating compressor, N sf = 1.4
Fd = Ft +
0.05vm (Cb + Ft )
1
0.05vm + (Cb + Ft )2
πD p n p π (4)(1200)
vm =
=
= 1257 fpm
12
12
C = 1660
b = 2 in
0.05(1257 )[1660(2) + Ft ]
Fd = Ft +
1
0.05(1257 ) + [1660(2) + Ft ]2
62.85(3320 + Ft )
Fd = Ft +
1
62.85 + (3320 + Ft )2
Fs = N sf Fd
11,616 = 1.4 Fd
Fd = 8297 lb
By trial and error method
Ft = 4900 lb
Fv
(4900)(1257 ) = 186.6 hp
hp = t m =
33,000
33,000
(b) 107 cycles, Table AT 26
Use Pd = 4 , K g = 252
Sum of BHN = 500
e = 0.00125 in
C = 1.25(1660)
 314  1.25(1660)(10)

 
0.05
+ 210 Pd 
Pd
 Pd  

Fd = 210 Pd +
1
 314  1.25(1660)(10)
2
 + 
0.05
+ 210 Pd 
Pd
 Pd  

Pd = 4
Fd = 3870 lb
Page 25 of 57
SECTION 11 – SPUR GEARS
300
(252) = 4725 > Fd
(4)2
Therefore Pd = 4
Fw =
10 10
=
= 2.5 in
Pd
4
N
20
Dp = p =
= 5 in
Pd
4
Dg = mg D p = (3)(5) = 15 in
b=
Material combination. (Sum of BHN = 600)
(1) 8630 WQT 1100 F, gear, BHN = 285
9261 OQT 1200 F, pinion, BHN = 311
(2) 6152 OQT 1200 F, gear, BHN = 293
9840 OQT 1000 F, pinion, BHN = 302
(3) 5150 OQT 1200 F, gear, BHN = 269
4150 OQT 1200 F, pinion, BHN = 331
Using Pinion: 4150, OQT 1200 F, BHN = 331
sn = 250(331) = 82,750 psi
Gear: 5150, OQT 1200 F, BHN = 269
sn = 250(269 ) = 62,250 psi
Table AT 24, Load near middle, 20o F.D.
Y = 0.544
N p = 20
N g = 3(20 ) = 60 , Y = 0.713
Pinion: sY = (82,750)(0.544) = 45,016
Gear: sY = (62,750)(0.713) = 47,949
Therefore, pinion is weaker
K f = 1 .7
sbY
(82,750)(2.5)(0.544) = 16,550 lb
=
(1.7 )(4)
K f Pd
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
πD p n p π (5)(1200)
vm =
=
= 1571 fpm
12
12
C = 1.25(1660) = 2075
b = 2.5 in
Fs =
Page 26 of 57
SECTION 11 – SPUR GEARS
Fd = Ft +
0.05(1571)[2075(2.5) + Ft ]
1
0.05(1571) + [2075(2.5) + Ft ]2
78.55(5187.5 + Ft )
Fd = Ft +
1
78.55 + (5187.5 + Ft )2
Fs = N sf Fd
16,550 = 1.4 Fd
Fd = 11,821 lb
By trial and error method
Ft = 6800 lb
Fv
(6800)(1571) = 324 hp
hp = t m =
33,000
33,000
645.
A pair of spur gears, delivering 100 hp to a reciprocating pump at a pinion speed
of 600 rpm, is to serve continuously with indefinite life; minimum number of 20o
full-depth teeth is 18; mw = 2.5 . Since low weight is highly important, it is
decided that the initial design be for carburized case-hardened teeth. (a)
Determine a suitable pitch, face width, diameters, and specify the material and its
heat treatment. (b) Use the same size teeth as determined in (a), but let the
material be flame-hardened 4150, OQT 1100 F. Compute Fs and Fw . If it were
decided that the maximum (specified) loading would be imposed only
occasionally, would these gears transmit more or less power than the carburized
teeth? Explain.
Solution:
N p = 18
Dp =
Np
Pd
=
18
Pd
 18 
(600 )
πD p n p
Pd 
2827

vm =
=
=
12
12
Pd
33,000hp 33,000(100)
Ft =
=
= 1167 Pd
vm
 2827 


P
 d 
π 
Fd = Ft +
0.05vm (Cb + Ft )
1
0.05vm + (Cb + Ft )2
Page 27 of 57
SECTION 11 – SPUR GEARS
10
Pd
assume first class commercial gears
e = 0.002 in
Table AT 25,
C = 2(1660) = 3320
b=

 2827  
 10 
 3320  + 1167 Pd 
0.05
 Pd  
 Pd 

Fd = 1167 Pd +
Fd = 1167 Pd +
1
2
 2827  
 10 
 + 3320  + 1167 Pd 
0.05
 Pd  
 Pd 


141.35  33200

+ 1167 Pd 
Pd  Pd

1
2
141.32  33200
+ 
+ 1167 Pd 
Pd
 Pd

Fw = D pbQK g
2(2.5)
= 1.43
m g + 1 2 .5 + 1
Table AT 26, 20o F.D., carburized, indefinite
K g = 750
Q=
2m g
=
 18  10 
193,050
Fw =   (1.43)(750) =
Pd2
 Pd  Pd 
Fw ≥ Fd
Iteration:
Pd
5
6
Will not equal
Using precision cut gears
C = 0.5(1660) = 830
Page 28 of 57
Fd
8355
9181
Fw
7722
5363
SECTION 11 – SPUR GEARS

141.35  8300

+ 1167 Pd 
Pd  Pd

Fd = 1167 Pd +

141.32  8300
+ 
+ 1167 Pd 
Pd
 Pd

Pd
5
1
2
Fd
7680
Use Pd = 5 , precision cut
To check for strength
sbY
Fs =
K f Pd
BHN = 600
sn = 250 BHN = 250(600) = 150,000 psi
10 10
b=
=
= 2 in
Pd
5
Table AT 24m Load near middle, 20o F.D.
N p = 18 ,
Y = 0.522
K f = 1 .7
Fs =
(150,000)(2)(0.522) = 18,424 lb > F
d
(1.7 )(5)
BHN = 600 , Rc = 57.5
E3310, SOQT 450 F
Pd = 5 , precision cut
b = 2 in
N
18
D p = p = = 3.6 in
Pd
5
Dg = mw D p = (2.5)(3.6 ) = 9 in
Material, E3310, SOQT 450 F
(b) Flame hardened, 4150 OQT 1100 F
BHN ≈ 359
Sum of BHN = 718
K g = 287
Fw = D p bQK g = (3.6 )(2 )(1.43)(287 ) = 2955 lb
Page 29 of 57
Fw
7722
SECTION 11 – SPUR GEARS
Fs =
sbY
K f Pd
s = 250(359) = 89,750 psi
Y = 0.522
(89,750)(2)(0.522) = 11,023 lb
Fs =
(1.7 )(5)
For occasional loading, Fs = Fd
For (a) Fd = 7680 lb < 11,023 lb
Therefore, these gears would transmit more power than carburized teeth for occasional
loading with continuous loading for carburized teeth.
CHECK PROBLEMS
646.
A 6-ft. ball mill runs at 24.4 rpm, the drive being through 14 1/2o involute spur
gears; Pd = 2 , N p = 15 , N g = 176 , b = 5 in ., and hp = 75 . The material of the
pinion is SAE 1040, BHN = 180; of the gear, 0.35% C cast steel, BHN = 180. (a)
Check for strength and wear and give your decision as to the service to be
expected. (b) The foregoing pinion wore out. Actually, the first step was to
replace it with one made of SAE 3140, OQT 1000 F. Would you expect this to
cure the trouble? (c) The drive in (b) also wore out. The following solution which
maintained the same gear diameters, pitch and face, was proposed: 20o full-depth
teeth; pinion of SAE 3140 with BHN = 350; gear of SAE 1045 with BHN = 280.
Would you predict that these gears will give long service? What are the
approximate tempering temperatures to get the specified hardness?
Solution:
n g = 24.4 rpm
vm =
πDg ng
12
N
176
Dg = g =
= 88 in
Pd
2
π (88)(24.4)
vm =
= 3373 fpm
12
33,000hp 33,000(75)
Ft =
=
= 734 lb
vm
3373
Fig. AF 19, max. per. Error = 0.0008 in for vm = 3373 fpm
Fig. AF 20, precision cut, Pd = 2 , e = 0.0010 in ≈ 0.0008 in
Table AT 25, e = 0.0010 in
C = 1600 , 14 1/2o F.D.
Page 30 of 57
SECTION 11 – SPUR GEARS
Fd = Ft +
0.05vm (Cb + Ft )
1
0.05vm + (Cb + Ft )2
b = 5 in
Fd = 734 +
0.05(3373)[1600(5) + 734]
1
2
0.05(3373) + [1600(5) + 734]
= 6354 lb
(a) Wear load
Fw = D pbQK g
Dp =
Np
=
15
= 7.5 in
2
Pd
b = 5 in
2N g
2(176 )
Q=
=
= 1.843
N p + N g 15 + 176
Sum of BHN = 180 + 180 = 360, 14 1/2o F.D.
Table At 26, K g = 46.3
Fw = (7.5)(5)(1.843)(46.3) = 3200 lb
Strength:
sbY
Fs =
K f Pd
s = 250 BHN = 250(180) = 45,000 psi
Pinion is weaker, N p = 15 , 14 1/2o F.D. (involute)
Y = 0.415 , Load near middle
K f = 1 .7
Fs =
(45,000)(5)(0.415) = 27,463 lb
(1.7 )(2)
Fw < Fd , service is intermittent.
(b) Pinion, SAE 3140, OQT 1000 F, BHN = 311
sn = 250 BHN = 250(311) = 77,750 psi
sY = (77,750)(0.415) = 32,266 psi
Gear, s n = 250 BHN = 250(180) = 45,000 psi
N g = 176 , Y = 0.6376 , 14 1/2o F.D.
sY = (45,000)(0.6376) = 28,692 psi
Strength,
Page 31 of 57
SECTION 11 – SPUR GEARS
Fs =
(45,000)(5)(0.6376) = 42,194 lb
sbY
=
(1.7)(2)
K f Pd
Wear, Fw = D pbQK g
Sum of BHN = 180 + 311 = 491, 14 1/2o F.D.
Table AT 26, K g = 92.4
Fw = (7.5)(5)(1.843)(92.4) = 6386 lb
Fw ≈ Fd , this will cure the trouble.
(c) Pinion, SAE 3140, BHN = 350
sn = 250 BHN = 250(350) = 87,500 psi
Gear, SAE 1045, BHN = 280
s n = 250 BHN = 250(280) = 70,000 psi
Table AT 25, 20o F.D., e = 0.001 in
C = 1660
Fd = 734 +
0.05(3373)[1660(5) + 734]
1
2
0.05(3373) + [1660(5) + 734]
= 6512 lb
Wear load, sum of BHN = 350 + 280 = 630
Table At 26, K g = 218.8 , 20o F.D.
Fw = D pbQK g
Fw = (7.5)(5)(1.843)(218.8) = 15,122 lb > Fd
Tempering temperatures:
Pinion: SAE 3140, BHN = 350
Fig. AF 2, OQT 995 F
Gear: SAE 1045, BHN = 280
Table AT 8, WQT 900 F, rod. Diameter = ½ in.
647.
A 22-tooth pinion, transmitting 110 hp at 2300 rpm, drives a 45-tooth gear, both
steel; 20o full depth; Pd = 5 , b = 1.5 in .; The manufacturing process is expected
to result in a maximum effective error of e = 0.0016 in . (a) Compute
Buckingham’s average dynamic load. Compute Fs and Fw if the material is (b)
case-carburized AISI 8620, DOQT 300 F, (c) AISI 8742, OQT 950 F, (d)
induction-hardened AISI 8742. (e) Suppose your company carries a stock of the
foregoing materials. For a minimum-service factor of 1.2, which material do you
Page 32 of 57
SECTION 11 – SPUR GEARS
recommend for (i) intermittent service, (ii) indefinitely continuous service, (iii)
cycles of loading not to exceed 107?
Solution:
N p = 22
N g = 45
Dp =
vm =
Np
=
Pd
πD p n p
22
= 4.4 in
5
=
π (4.4)(2300)
= 2649 fpm
12
12
33,000hp 33,000(110 )
Ft =
=
= 1370 lb
vm
2649
(a) Dynamic Load
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
For e = 0.0016 in , Table AT 25, steel, 20o F.D.
C = 1.6(1660) = 2656
b = 1.5 in
0.05(2649)[2656(1.5) + 1370]
Fd = 1370 +
= 4819 lb
1
0.05(2649) + [2656(1.5) + 1370]2
(b) AISI 8620, DOQT 300 F carborized
Table AT 26, 20o F.D.
K g = 750 , 1010 cycles ≈ indefinite
Table AT 11, Rc = C 64
Figure AF 4, BHN = 700
sn = 250(700) = 175,000 psi
N p = 22
Table AT 24, Load near middle, 20o F.D.
Y = 0.559
K f = 1 .7
Strength
sbY
(175,000)(1.5)(0.559) = 17,263 lb
Fs =
=
(1.7)(5)
K f Pd
Wear
Fw = D pbQK g
Page 33 of 57
SECTION 11 – SPUR GEARS
Q=
2N g
N p + Ng
=
2(45)
= 1.3433
22 + 45
Fw = (4.4)(1.5)(1.3433)(750) = 6649 lb
(c) AISI 8742, OQT 950 F
BHN = 358.5
sum of BHN = 2(358.5) = 717
Table AT 26, 20o F.D.
K g = 286
sn = 250(358.5) = 89,625 psi
Strength
sbY
(89,625)(1.5)(0.559) = 8841 lb
Fs =
=
(1.7 )(5)
K f Pd
Wear
Fw = D pbQK g
Fw = (4.4)(1.5)(1.3433)(286) = 2536 lb
(d) Induction hardened, AISI 8742
Table AT 26, 20o F.D.
K g = 555 at 1010 cycles
sn = 250(500) = 125,000 psi
Strength
sbY
(125,000)(1.5)(0.559) = 12,331 lb
Fs =
=
(1.7 )(5)
K f Pd
Wear
Fw = D pbQK g
Fw = (4.4)(1.5)(1.3433)(555) = 4921 lb
(e)
(i) intermittent service
N sf F d = (1.2 )(4819 ) = 5783 lb
use AISI 8742, OQT 950 F
FS ≥ N sf F d
FS = 8841 lb
(ii) indefinitely continuous service
Fw ≥ F d
Fd = 4819 lb
use AISI 8620, DOQT 300 F
Page 34 of 57
SECTION 11 – SPUR GEARS
Fw = 6649 lb
(iii) 107 cycles
Use AISI 8742, induction hardened
K g = 1190
Fw = (4.4)(1.5)(1.3433)(1190) = 10,550 lb > Fd
Two mating steel gears have 16 and 25 teeth, respectively, 20o F.D.; b = 2 in ,
Pd = 5 ; pinion speed, 2400 rpm. The maximum effective error in the profiles is
planned to be 0.0012 in. The drive is for heavy-duty conveyer, continuous
service. Compute and specify a reasonable rated horsepower if the gear teeth are:
(a) Case carburized AISI 8620, SOQT 300 F, (b) AISI 8742, OQT 950 F and the
flame-hardened, (c) AISI 8742, OQT 800 F.
648.
Solution:
N p = 16
N g = 25
Dp =
vm =
Np
=
Pd
πD p n p
12
Fd = Ft +
16
= 3.2 in
5
=
π (3.2)(2400)
12
0.05vm (Cb + Ft )
= 2011 fpm
1
0.05vm + (Cb + Ft )2
error, e = 0.0012 in
Table AT 25, 20o F.D.
C = 1.2(1660) = 1992
0.05(2011)[1992(1.5) + Ft ]
Fd = Ft +
1
0.05(2011) + [1992(1.5) + Ft ]2
100.55(2988 + Ft )
Fd = Ft +
1
100.55 + (2988 + Ft )2
(a) Case carburized AISI 8620, SOQT 300 F
K g = 750
BHN = 700
sn = 250(700) = 175,000 psi
N p = 16 , Y = 0.503 , Table AT 24, 20o F.D., Load near middle
Page 35 of 57
SECTION 11 – SPUR GEARS
Q=
2N g
N p + Ng
Wear:
Fw = D pbQK g
=
2(25)
= 1.22
16 + 25
Fw = (3.2 )(2)(1.22)(750) = 5856 lb
Strength
sbY
(175,000)(2)(0.503) = 20,712 lb
Fs =
=
(1.7 )(5)
K f Pd
Use Fw = Fd
5856 = Ft +
100.55(2988 + Ft )
1
100.55 + (2988 + Ft )2
Ft = 2635 lb
(2635)(2011) = 160 hp
Fv
hp = t m =
33,000
33,000
(b) AISI 8742, OQT 950 F, flame-hardened
K g = 555
sn = 250(500) = 125,000 psi
Strength
sbY
(125,000)(2)(0.503) = 14,794 lb
Fs =
=
(1.7)(5)
K f Pd
Wear:
Fw = D pbQK g
Fw = (3.2 )(2)(1.22)(555) = 4333 lb
Use Fw = Fd
100.55(2988 + Ft )
4333 = Ft +
1
100.55 + (2988 + Ft )2
Ft = 1590 lb
Fv
(1590)(2011) = 97 hp
hp = t m =
33,000
33,000
(c) AISI 8742, OQT 800 F, BHN = 416.4 (Table AT 9)
Sum of BHN = 2(416.4) = 832.8
Table AT 26, K g = 397.5
sn = 250 BHN = 250(416.4) = 104,100 psi
Strength
(104,100)(2)(0.503) = 12,320 lb
sbY
Fs =
=
(1.7)(5)
K f Pd
Page 36 of 57
SECTION 11 – SPUR GEARS
Wear:
Fw = D pbQK g
Fw = (3.2 )(2)(1.22)(397.5) = 3104 lb
Use Fw = Fd
100.55(2988 + Ft )
3104 = Ft +
1
100.55 + (2988 + Ft )2
Ft = 770 lb
Fv
(770)(2011) = 47 hp
hp = t m =
33,000
33,000
649.
3
The data for a pair of gears are: 20o F.D. teeth, b = 1 in , Pd = 6 , N p = 26 ,
4
N g = 60 , n p = 2300 rpm ; as-rolled AISI 1050; carefully cut teeth; N sf = 1.2 . (a)
Strength alone considered, find the horsepower that may be transmitted. (b)
Determine the required surface hardness in order for Fw = Fd , and specify a
treatment that would make the gears long lasting in continuous service.
Solution:
N
26
Dp = p =
= 4.333 in
Pd
6
πD p n p π (4.333)(2300)
vm =
=
= 2609 fpm
12
12
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
For carefully cut teeth, Pd = 6 , 20o F.D.
C = 1660
0.05(2609)[1660(1.75) + Ft ]
Fd = Ft +
1
0.05(2609 ) + [1660(1.75) + Ft ]2
130.45(2905 + Ft )
Fd = Ft +
1
130.45 + (2905 + Ft )2
(a) Strength
sbY
Fs =
K f Pd
For as rolled AISI 1050, BHN = 229
sn = 250(229 ) = 57,250 psi
or sn = 0.5su = 0.5(102.000) = 51,000 psi
use s = 51,000 psi
Page 37 of 57
SECTION 11 – SPUR GEARS
N p = 26
Table AT 26, 20 o F.D., load near middle
Y = 0.588
K f = 1 .7
Fs =
(51,000)(1.75)(0.588) = 5145 lb
(1.7)(6)
Fs = N sf Fd
5145 = 1.2 Fd
Fd = 4288 lb
Fd = 4288 = Ft +
130.45(2905 + Ft )
1
130.45 + (2905 + Ft )2
Ft = 1400 lb
Fv
(1400)(2609) = 110 hp
hp = t m =
33,000
33,000
(b) Fw = Fd = 1400 lb
Wear
Fw = D pbQK g
Q=
2N g
N p + Ng
=
2(60 )
= 1.395
26 + 60
Fw = (4.333)(1.75)(1.395)K g = 4288 lb
K g = 405
Table of BHN = 838
BHN1 = 229
BHN 2 = 838 − 229 = 609
Therefore use carburized teeth.
650.
Gears with carefully cut, 20o F.D. teeth have Pd = 5 , b = 2 , mg = 5 , N p = 24 .
Pinion material is manganeses gear bronze (heading of Table AT 3); gear is cast
iron, class 25. Gear speed ng = 200 rpm ; smooth load. They are to transmit 18
hp. (a) Are the teeth strong enough for intermittent service? (b) Does the limiting
wear load indicate long life? Suggestion: Compute C for equation (13.7).
Solution:
N
24
Dp = p =
= 4.8 in
Pd
5
Page 38 of 57
SECTION 11 – SPUR GEARS
n p = mg ng = (5)(200 ) = 1000 rpm
vm =
πD p n p
π (4.8)(1000)
= 1257 fpm
12
12
33,000hp 33,000(18)
Ft =
=
= 473 lb
vm
1257
Carefully cut:
kE g E p
C=
Eg + E p
k = 0.111e for 20o full-depth
Fig. AF 20, e = 0.001 in , Pd = 5
C=
=
(0.111)(0.001)(11.5 ×106 )(16 ×106 ) = 743
Fd = Ft +
11.5 ×106 + 16 ×106
0.05vm (Cb + Ft )
1
0.05vm + (Cb + Ft )2
0.05(1257 )[743(2) + 473]
Fd = 473 +
= 1622 lb
1
0.05(1257 ) + [743(2) + 473]2
(a) Fs ≥ N sf Fd
N sf = 1.0 , smooth load
sbY
K f Pd
Gear: cast-iron, class 25, 20o F.D.
N g = mg N p = (5)(24 ) = 120
Fs =
Y = 0.7646 , load near middle
sn = 0.4(25) = 10 ksi = 10,000 psi
snY = (10,000)(0.7646) = 7646 psi
Pinion: manganese gear bronze, 20o F.D.
N p = 24
Y = 0.572 , load near middle
sn = 17 ksi = 17,000 psi
snY = (17,000)(0.572) = 9724 psi
Gear is weaker
K f = 1 .7
Fs =
(10,000)(2)(0.7646) = 1799 lb > N F
sf d
(1.7 )(5)
Page 39 of 57
SECTION 11 – SPUR GEARS
Therefore, enough for intermittent service
(b) Fw = D pbQK g
Q=
2m g
mg + 1
=
2(5)
= 1.667
5 +1
 s 2 sin φ  1
1 

+
K g = 


 1.4  E p E g 
Pinion, manganese gear bronze
s = su = 75 ksi
φ = 20o
 (75,000 )2 sin 20  1
1

Kg = 
+
= 205

6
6 
1 .4

 16 × 10 11.5 × 10 
Fw = (4.8)(2)(1.667 )(205) = 3281 lb > Fd
Therefore, indicates long life.
A 20-tooth pinion. 20o F.D., drives a 100-tooth gear. The pinion is made of SAE
1035, heat treated to Rockwell C15; the gear is cast iron class 35, HT; Pd = 3 ,
b = 2.5 in .; carefully cut teeth; pinion speed n p = 870 rpm , smooth load. (a) For
a continuous service, indefinite life, what is a safe horsepower? (b) For
intermittent service (wear unimportant), compute the safe horsepower.
651.
Solution:
N p = 20
N g = 100
Dp =
vm =
Np
=
Pd
πD p n p
12
Fd = Ft +
20
= 6.667 in
3
=
π (6.667 )(870)
12
0.05vm (Cb + Ft )
1
= 1519 fpm
0.05vm + (Cb + Ft )2
Carefully cut gears, steel and cast iron, 20o F.D.
Pd = 3 , Fig. AF 20, Table AT 25
e = 0.0016 in
C = 1.6(1140) = 1824
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
Page 40 of 57
SECTION 11 – SPUR GEARS
Fd = Ft +
Fd = Ft +
0.05(1519)[1824(2.5) + Ft ]
1
0.05(1519) + [1824(2.5) + Ft ]2
75.95(4560 + Ft )
1
75.95 + (4560 + Ft )2
(a) Continuous ser vice
Strength:
sbY
Fs =
K f Pd
Pinion: SAE 1035, Rc = C15
Fig. AF 4, BHN = 200
sn = (250)(200) = 50,000 psi
N p = 20
Table AT 24, 20o F.D., load near middle
Y = 0.544
snY = (50,000)(0.544) = 27,200 psi
Gear: Cast iron, class 35
sn = 0.4su = 0.4(35,000) = 14,000 psi
N g = 100
Table AT 24, 20o F.D., load near middle
Y = 0.755
snY = (14,000)(0.755) = 10,570 psi
Gear is weaker
K f = 1 .7
Fs =
(14,000)(2.5)(0.755) = 5181 lb
(1.7)(3)
Wear:
Fw = D pbQK g
Q=
2N g
N p + Ng
=
2(100 )
= 1.667
20 + 100
 s 2 sin φ  1
1 

K g = 
+


 1.4  E p E g 
φ = 20o
s = 0.4 BHN − 10 = 0.4(200) − 10 = 70 ksi = 70,000 psi
E p = 30×106 psi
Page 41 of 57
SECTION 11 – SPUR GEARS
E g = 14.5 ×106 psi
 (70,000 )2 sin 20  1
1

Kg = 
+
= 122

6
6 
1 .4

 30 ×10 14.5 × 10 
Fw = (6.667 )(2.5)(1.667 )(122) = 3390 lb
Fw = Fd
3390 = Ft +
75.95(4560 + Ft )
1
75.95 + (4560 + Ft )2
Ft = 700 lb
Fv
(700)(1519) = 32 hp
hp = t m =
33,000
33,000
(b) Fs = N sf Fd
N sf = 1.0 , smooth load
5181 = Ft +
75.95(4560 + Ft )
1
75.95 + (4560 + Ft )2
Ft = 2000 lb
Fv
(2000)(1519) = 92 hp
hp = t m =
33,000
33,000
652.
A pair of steel gears is defined by Pd = 8 , b = 1.5 in , N p = 25 , N g = 75 ,
e = 0.001 in , 20o F.D. If these gears may transmit continuously and without
failure 75 hp at 1140 rpm of the pinion, what horsepower would be satisfactory
for n p = 1750 rpm ?
Solution:
N
25
Dp = p =
= 3.125 in
Pd
8
πD p n p π (3.125)(1140 )
vm =
=
= 933 fpm
12
12
33,000hp 33,000(75)
Ft =
=
= 2653 lb
vm
933
Table AT 25, e = 0.001 in , 20o F.D.
C = 1660
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
Page 42 of 57
SECTION 11 – SPUR GEARS
Fd = 2653 +
0.05(933)[1660(1.5) + 2653]
1
2
0.05(933) + [1660(1.5) + 2653]
= 4680 lb
For n p = 1750 rpm
vm =
πD p n p
π (3.125)(1750 )
= 1432 fpm
12
0.05(1432)[1660(1.5) + Ft ]
Fd = Ft +
= 4680 lb
1
0.05(1432 ) + [1660(1.5) + Ft ]2
Ft = 2260 lb
Fv
(2260)(1432) = 98 hp
hp = t m =
33,000
33,000
654.
12
=
A gear manufacturer recommends that the following gears can transmit 25 hp at
600 rpm of the pinion during continuous 24-hr. service, indefinite life, moderate
shock: N p = 31 , N g = 70 , b = 3.25 in , Pd = 6 , 20o F.D.; pinion material is SAE
2335 with BHN = 300; gear material is SAE 1040 with BHN = 250. At what
horsepower would you rate them?
Solution:
N
31
D p = p = = 5.167 in
Pd
6
πD p n p π (5.167 )(600)
vm =
=
= 812 fpm
12
12
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
Commercial cut, e = 0.001 in , 20o F.D., Pd = 6
Table AT 25
C = (2)(1660) = 3320
0.05(812)[3220(3.25) + Ft ]
Fd = Ft +
1
0.05(812) + [3220(3.25) + Ft ]2
40.6(10,465 + Ft )
Fd = Ft +
1
40.6 + (10,465 + Ft )2
Strength:
Pinion: SAE 2335, bhn = 300
sn = 250(300) = 75,000 psi
N p = 31 , Table AT 24, Load near middle, 20o F.D.
Y = 0.6115
Page 43 of 57
SECTION 11 – SPUR GEARS
snY = (75,000)(0.6115) = 45,862 psi
Gear: SAE 1040, BHN = 250
sn = 250(250 ) = 62,500 psi
N g = 70 , Table AT 24, Load near middle, 20o F.D.
Y = 0.728
snY = (62,500)(0.728) = 45,500 psi
Gear is weaker, K f = 1.7
Fs =
(62,500)(3.25)(0.728) = 14,498 lb
sbY
=
(1.7 )(6)
K f Pd
Wear load:
Fw = D pbQK g
Q=
2N g
N p + Ng
=
2(70 )
= 1.386
31 + 70
Sum of BHN = 300+250 = 550
Table AT 26, 20o F.D.
K g = 162
Fw = (5.167 )(3.25)(1.386)(162) = 3770 lb
Fw = Fd
40.6(10465 + Ft )
3770 = Ft +
1
40.6 + (10465 + Ft )2
Ft = 675 lb
(675)(812) = 16.6 hp
Fv
hp = t m =
33,000
33,000
For carefully cut, C = 1660
Cb = (1660)(3.25) = 5395
40.6(5395 + Ft )
3770 = Ft +
1
40.6 + (5395 + Ft )2
Ft = 1500 lb
Fv
(1500)(812) = 36.9 hp carefully cut
hp = t m =
33,000
33,000
use hp = 36.9 hp , carefully cut.
Page 44 of 57
SECTION 11 – SPUR GEARS
NONMETALLIC GEARS
655.
A 4-in. Bakelite pinion meshing with a cast-iron gear, is to be on the shaft of a 12
–hp induction motor that turns at 850 rpm; mw = 5 , 20o F.D. teeth. (a) Determine
Pd , b , N p , N g for indefinitely continuous service with a smooth load. Is there
serious interference in the gears you have designed? (b) Are the teeth of your
design strong enough to taje without damage an occasional 60 % overload?
Solution:
Pinion as weaker
(200 + vm )Ft
Fd =
v
200 + m
4
πD p n p π (4)(850)
vm =
=
= 890 fpm
12
12
33,000hp 33,000(12 )
Ft =
=
= 445 lb
vm
890
(200 + 890)(445) = 1148 lb
Fd =
890
200 +
4
Strength: s = 6000 psi
sbY
Fs =
Pd
10
and Y = 0.33 , 20o F.D. Load at tip.
assume b =
Pd
Fs = Fd (smooth load)
(6000)(10)(0.33) = 1148 lb
Fs =
Pd2
Pd = 4.15
use Pd = 4
10 10
b=
=
= 2.5 in
Pd
4
say b = 3 in
N p = D p Pd = (4 )(4 ) = 16
Table At 24, Y = 0.295 , 20o F.D. Load at tip.
(6000)(3)(0.295) = 1328 lb > 1148 lb
Fs =
4
Check:
Page 45 of 57
SECTION 11 – SPUR GEARS
Wear:
Fw = D pbQK g
2mw
2(5)
=
= 1.667
mw + 1 5 + 1
Table AT 26,
Use K g = 64 , 20o F.D.
Q=
Fw = (4)(3)(1.667 )(64) = 1280 lb > Fd , o.k.
Then
Pd = 4
b = 3 in
N p = 16
N g = mw N p = (5)(16 ) = 80
N p < 18 , there is interference.
(b) Ft = (1.6)(445) = 712 lb
Fd > 712 lb , strong enough.
656.
A Zytel pinion with molded teeth is to transmit 0.75 hp to a hardened-steel gear;
n p = 1750 rpm , D p ≈ 1.25 in . Determine the pitch, face, and number of teeth on
the pinion for intermittent service.
Solution:
πD p n p π (1.25)(1750)
vm =
=
= 573 fpm
12
12
33,000hp 33,000(0.75)
Ft =
=
= 43 lb
vm
573
Fd = (VF )Ft
For molded teeth, vm < 4000 fpm
VF = 1
Fd = (1)(43) = 43 lb
Load near middle, assume Y = 0.50
Fs = Fd
say s = 4.6 ksi = 4600 psi , §13.27
10
b=
Pd
sbY (4600 )(10 )(0.50 )
Fs =
=
= 43
Pd
Pd2
Page 46 of 57
SECTION 11 – SPUR GEARS
Pd = 23
use Pd = 20
10
= 0.5 in
b=
20
N p = D p Pd = (1.25)(20 ) = 25 teeth
657.
A 10-in. Textolite pinion, driving a hardened steel gear, transmits power at 400
rpm; Pd = 2.5 , b = 5 in ., 14 1/2o F.D. teeth. Determine the safe horsepower (a)
for smooth, continuous, indefinite service, and also (b) for limited-life
intermittent service.
Solution:
D p = 10 in
n p = 400 rpm
vm =
πD p n p
=
π (10)(400)
= 1047 fpm
12
12
(200 + vm )Ft
Fd =
v
200 + m
4
Strength
sbY
Fs =
Pd
s = 6000 psi
b = 5 in
Pd = 2.5
N p = Pd D p = (2.5)(10 ) = 25
Table At 24, 14 1/2o F.D. Load at tip.
Y = 0.305
(6000)(5)(0.305) = 3660 lb
Fs =
2 .5
Wear load, Table At 26, 14 1/2o F.D.
K g = 46
Fw = D pbQK g
Q ≈ 1 .5
Fw = (10)(5)(1.5)(46) = 3450 lb
(a) Continuous service
Fw = Fd (smooth)
Page 47 of 57
SECTION 11 – SPUR GEARS
3450 =
(200 + 1047 )Ft
200 +
1047
4
Fs = 1277 lb
Fv
(1277)(1047) = 40 hp
hp = t m =
33,000
33,000
(b) Intermittent service
Fs = Fd
(200 + 1047 )Ft
3660 =
1047
200 +
4
Fs = 1355 lb
Fv
(1355)(1047) = 43 hp
hp = t m =
33,000
33,000
658.
5
in .
8
(a) What safe horsepower may be transmitted for long-life? (b) For 107 cycles?
A 16-pitch Zytel pinion, with 26, 20o F.D. cut teeth, rotates at 600 rpm; b =
Solution:
πD p n p
vm =
12
N
26
Dp = p =
= 1.625 in
Pd 16
π (1.625)(600)
vm =
= 255 fpm
12
Cut-teeth, vm < 4000 fpm
VF = 1.2
Fd = (VF )Ft
Fd = Ft
(a) Long life, 5× 108 cycles
Pd = 16
s = 2.3 ksi = 2300 psi
sbY
Fs =
Pd
For N p = 26 , 20o F.D., Load near middle
Y = 0.588
Page 48 of 57
SECTION 11 – SPUR GEARS
(2300) 5 (0.588)
8
= 53 lb
16
Fs = 53 = 1.2 Ft
Ft = 44 lb
(44)(255) = 0.34 hp
Fv
hp = t m =
33,000
33,000
Fs =
(b) 107 cycles. s = 4.2 ksi = 4200 psi , Pd = 16
(4200) 5 (0.588)
8
= 96.5 lb
16
Fs = 96.5 = 1.2 Ft
Ft = 80 lb
Fv
(44)(255) = 0.34 hp
hp = t m =
33,000
33,000
Fs =
CAST-TOOTH GEARS
659.
A pair of cast-iron spur gears, ASTM 20, with cast teeth, transmits 10 hp at 125
rpm of the pinion; mw = 5 , D p ≈ 8 in . Determine Pc , b , N p , N g .
Solution:
πD p n p π (8)(125)
vm =
=
= 262 fpm
12
12
600 + vm
Fd =
Ft
600
33,000hp 33,000(10 )
Ft =
=
= 1260 lb
vm
262
600 + 262
Fd =
(1260) = 1810 lb
600
Fs = 0.054sbPc
say b = 2.5Pc
s = 0.4su = 0.4(20) = 8 ksi = 8000 psi
Fs = 0.054(8000 )(2.5)Pc2 = 1080 Pc2
Fs = Fd
1080 Pc2 = 1810
Pc = 1.29 in
Page 49 of 57
SECTION 11 – SPUR GEARS
1
say Pc = 1 in
4
b = 2.4 Pc = 2.4(1.25) = 3 in
πD p π (8)
Np =
=
= 20
Pc
1.25
N g = mw N p = (5)(20 ) = 100
660.
Design the cast teeth for a pair of cast-iron spur gears to transmit 35 hp at 50 rpm
of the pinion; mw ≈ 2.5 . Decide upon a suitable grade of cast iron and find Pc , b ,
D p , Dg , and center distance.
Solution:
Using cast-iron, class 35
sn = 0.4su = 0.4(35) = 14 ksi = 14,000 psi
33,000hp 33,000(35) 1,155,000
Ft =
=
=
vm
vm
vm
600 + vm
Fd =
Ft
600
Try N p = 20
Dp =
vm =
Pc N p
π
πD p n p
12
=
=
20 Pc
= 6.366 Pc
π
π (6.366 Pc )(50)
12
Ft =
1,155,000 13,860
=
83.33Pc
Pc
Fd =
600 + 83.33Pc
600
= 83.33Pc
 13,860  23.1(600 + 83.33Pc )

 =
P
Pc
c


Fs = 0.054sbPc
b = 2.5Pc
Fs = 0.054(14,000 )(2.5)Pc2 = 1890 Pc2
Fs = Fd
23.1(600 + 83.33Pc )
Pc
Pc = 2.117 in
1890 Pc2 =
use Pc = 2 in
PN
20(2)
Dp = c p =
= 12.73 in
π
Page 50 of 57
π
SECTION 11 – SPUR GEARS
Dg = mw D p = (2.5)(12.73) = 31.83 in
b = 2.5Pc = 2.5(2) = 5 in
1
1
C = (D p + Dg ) = (12.73 + 31.83) = 22.28 in
2
2
Summary:
Pc = 2 in
D p = 12.73 in
Dg = 31.83 in
b = 5 in
C = 22.28 in
661.
A manufacturer’s catalog specifies that a pair of gray cast-iron spur gears with
cast teeth will transmit 7.01 hp at a pitch-line speed of 100 fpm; N p = 20 ,
Pc = 1.5 in ., b = 4 in . Compute the stress and specify the grade of cast iron that
should be used.
Solution:
vm = 100 fpm
33,000hp 33,000(7.01)
Ft =
=
= 2313.3 lb
vm
100
600 + vm
600 + 100
Fd =
Ft =
(2313.3) = 2699 lb
600
600
Fs = 0.054sbPc
Fs = Fd
0.054s (4)(1.5) = 2699
s = 8330 psi
s ≈ 0.4su
8330 ≈ 0.4su
su = 20,825 psi = 21 ksi
use Cast Iron, ASTM 25, su = 25 ksi
ARMS AND RIMS
662.
A 24-in. cast-iron gear transmits 30 hp at 240 rpm; 14 1/2o F.D. teeth, Pd = 4 ,
b = 2.5 in . The gear is on a 2 ¼ -in. shaft. Determine (a) the hub diameter, rim
thickness, and bead, (b) the dimensions of the arms at the hub and at the pitch
circle for an elliptical-shaped section, (c) the arm dimensions for a cross shape.
Page 51 of 57
SECTION 11 – SPUR GEARS
Solution:
1
 1
(a) Hub diameter = 2 Ds = 2 2  = 4 in , for cast iron
2
 4
7
π 
Rim thickness = 0.56 Pc = 0.56  = 0.44 in ≈ in
16
4
7
π 
Bead = 0.56 Pc = 0.56  = 0.44 in ≈ in
16
4
(b) Elliptical section
use no. of arms = N a = 4 , D = 24 in < 120 in
FL
M=
Na
1
24 − 4
D − Dh
2 = 9.75 in
L=
=
2
2
F = Fd
M
s=
Z
πh3
Z=
64
0.05vm (Cb + Ft )
Fd = Ft +
1
0.05vm + (Cb + Ft )2
πD p n p π (24)(240)
vm =
=
= 1506 fpm
12
12
assume C = 800 , 14 1/2o F.D.
33,000hp 33,000(30 )
Ft =
=
= 660 lb
vm
1506
0.05(1506)[800(2.5) + 660]
Fd = 660 +
= 2238 lb
1
0.05(1506 ) + [1800(2.5) + 660]2
M
FL
64 FL
s=
=
= 3
Z N a Z πh N a
use s = 8000 psi
Page 52 of 57
SECTION 11 – SPUR GEARS
64(2238)(9.75)
πh3 (4)
h = 2.4 in
8000 =
At the hub
h = 2.4 in
h
h1 = = 1.2 in
2
At the pitch circle
h
h′ = = 1.2 in
2
h
h1 = = 1.2 in
2
(c) Cross shape
6Z
h2
G1 = 0.75G
h = 2.4 in
G=
Z=
πh3
, §13.32
64
π (2.4)3
Z=
= 0.6786 in 3
64
At the hub
6(0.6786 )
G=
= 0.71 in
(2.4)2
G1 = 0.75(0.71) = 0.53 in
h = 2.4 in
h
At the pitch circle, h′ = = 1.2 in
2
G = 0.71 in
G1 = 0.53 in
Page 53 of 57
SECTION 11 – SPUR GEARS
INTERNAL GEARS
A 20-tooth pinion, with 20o F.D. teeth , drives a 75-tooth internal gear ( Pd = 8 ,
b = 1.5 in , n p = 1150 rpm ; material, cast iron, class 20). What horsepower may
be transmitted continuously (a) if the teeth are commercially hobbed (AGMA
equation, §13.15), (b) if they are precision cut? There are minor irregularities in
the loading.
664.
Solution:
(a) Commecially hobbed
1
2
m
50 + v
Ft
50
Fw = D pbQK g
Fd =
Q=
2N g
Ng − N p
=
2(75)
= 2.727
75 − 20
From AT 26, K g = 112
Np
20
= 2.5 in
Pd
8
Fw = (2.5)(1.5)(2.727 )(112) = 1145 lb
Dp =
=
For strength:
sbY
Fs =
K f Pd
N p = 20 , Y = 0.320 at the tip
Let K f = 1.3 , s = 8000 psi
(8000)(1.5)(0.320) = 369 lb
(1.3)(8)
πD p n p π (2.5)(1150 )
=
=
= 753 fpm
Fs =
vm
12
Fs = Fd
12
1
50 + (753)2
Fd = 369 =
Ft
50
Ft = 238 lb
hp =
Ft vm
(238)(753) = 5.43 hp
=
33,000
33,000
(b) Precision cut
Page 54 of 57
SECTION 11 – SPUR GEARS
1
78 + vm2
Fd =
Ft
78
1
2
Fd = 369 =
78 + (753)
Ft
78
Ft = 273 lb
hp =
666.
Ft vm
(273)(753) = 6.23 hp
=
33,000
33,000
A planetary gear train is composed of four gears – the sun gear B , two planet
gears C , as shown. Gears B and C have 20 teeth each, gear D has 60; Pd = 10 ,
1
b = 1 − in ., 20o F.D. teeth, cast iron, class 20, nB = 1750 rpm . (a) Determine the
4
speed of the arm. (b) What horsepower may be transmitted continuously? Note
that the dynamic load (AGMA equation , §13.15) depends on the speed of tooth
engagement, which is not the absolute pitch line speed of B (pitch-line speed
relative to arm for B − C ). Check the speed of tooth engagement of both B − C
and C − D . (c) If the designer wishes to increase the power transmitted by using
three planet gears, instead of two, what changes must be made in tooth numbers
so that the gears can be assembled with the planets 120o apart?
Solution:
nL = enF + na (1 − e )
− (20)(20)
1
e=
=−
(20)(60)
3
nF = nB = 1750 rpm
nL = 0 = nD
1
 1
0 = − (1750 ) + na 1 + 
3
 3
na = 437.5 rpm
(assumed)
(a) Speed pf arm = 437.5 rpm
(b) For BC
vm =
πD p nB A
12
Page 55 of 57
 Np 
(nB − n A ) π  20 (1750 − 437.5)
 
D p 
10

=
=  
= 687 fpm
12
12
π 
SECTION 11 – SPUR GEARS
For BC
vm =
− πDg nD A
12
N
−π g
 Dp

=

(nD − n A ) − π  60 (0 − 437.5)
 


 10 
=
= 687 fpm
12
12
vm = 687 fpm < 1000 fpm
Use commercial teeth
600 + vm
Ft
600
For cast-iron, continuous service
Fs ≥ N sf Fd
Fd =
For cast iron, assume N sf = 1
Fs =
sbY
K f Pd
Let K f = 1.2
s = 0.4su = 0.4(20) = 8 ksi = 8000 psi
For N = 20 , Y = 0.32 , load at tip, 20o F.D.
(8000)1 1 (0.32)
Fs =
 4
(1.2)(10)
= 267 lb
Fs = Fd
600 + 687
267 =
Ft
600
Ft = 124 lb
hp =
Ft vm
(124)(687) = 2.6 hp
=
33,000
33,000
(c) For each planet gear,
 2 .6 
hp = 
 (1.15) = 1.495 hp
 2 
For 3 planet gears
 1.495 
hp = 
 (3) = 3.9 hp
 1.15 
sbY
600 + vm
Fs =
=
Ft
K f Pd
600
Page 56 of 57
SECTION 11 – SPUR GEARS
33,000(3.9 )
= 187 lb
687
assume Y = 0.3
(8000)1 1 (0.30)
600 + 687
 4
Fs =
=
(187)
600
(1.2)(Pd )
Pd = 6.65
Ft =
use Pd = 7
N = DPd = (2)(7 ) = 14 teeth
- end -
Page 57 of 57
SECTION 12 – HELICAL GEARS
DESIGN PROBLEMS
701.
For continuous duty in a speed reducer, two helical gears are to be rated at 7.4 hp
at a pinion speed of 1750 rpm; m w ≈ 2.75 ; the helix angle 15o ; 20o F.D. teeth in
the normal plane; let N p = 21 teeth, and keep b < 2 D p . Determine the pitch, face,
N g , and the material and heat treatment. Use through-hardened teeth with a
maximum of 250 BHM (teeth may be cut after heat treatment).
Solution:
ψ = 15o
φn = 20o
πD p n p
vm =
12
N p 21
Dp =
=
Pd
Pd
n p = 1750 rpm
 21 
(1750)
Pd 
9621

vm =
=
12
Pd
33,000hp (33,000 )(7.4 )
Ft =
=
= 25.38 Pd
vm
 9621 


P
 d 
π 
b ≤ 2 Dp
 21  42
b = 2  =
 Pd  Pd
0.05vm (Ft + Cb cos 2 ψ )cosψ
Fd = Ft +
lb
1
2
2
0.05vm + (Ft + Cb cos ψ )
Table AT 25
Assume C = 1660
ψ = 15o
Fd = 25.38Pd +

 9621  
 42 
 25.38Pd + 1660  cos 2 15 cos15
0.05
 Pd  
 Pd 


 9621  
 42 
 + 25.38Pd + 1660  cos 2 15
0.05
 Pd  
 Pd 

Page 1 of 14
1
2
lb
SECTION 12 – HELICAL GEARS
Fd = 25.38 Pd +
465 
65050 
 25.38 Pd +

Pd 
Pd 
481 
65050 

+  25.38 Pd +
Pd 
Pd 
For continuous service: Fw ≥ Fd
bD pQK g
Fw =
cos 2 ψ
2mg
2(2.75)
Q=
=
= 1.467
mg + 1 2.75 + 1
Table At 26, Bhn = 250
Sum of BHN = 500, φn = 20o
K g = 131
 42  21  (1.467 )(131) 181,670
Fw =   
=
2
Pd2
 Pd  Pd  cos 15
Fw ≥ Fd
By trial and error method
Pd
7
6
Fd
3967
4758
Fw
3708
5046
use Pd = 6
21 21
Dp =
=
= 3.5 in
Pd
6
42 42
b=
=
= 7 in
Pd
6
9621 9621
vm =
=
= 1604 fpm
Pd
6
Fig. AF 19, permissible error = 0.0018 in
Fig. AF 20
Use carefully cut gears, Pd = 6
Error = 0.001 in is o.k.
For material
Strength
sbY cosψ
Fs =
K f Pd
Page 2 of 14
1
2
lb
SECTION 12 – HELICAL GEARS
Np
21
= 23
cos ψ cos3 15
Table AT 24, Load near middle
N ep = 23 , φn = 20o FD
N ep =
3
=
Y = 0.565
assume K f = 2.0
Fs = N sf Fd
assume N sf = 2.0
s (7 )(0.565) cos15
= (4758)(2)
(2 )(6)
s = 29,892 psi
s
use sn = u
3
su = 3(29,892) = 89,676 psi
Use C1050, OQT 1100 F,
su = 122 ksi , BHN = 248 < 250
Ans.
Pd = 6
b = 7 in
N g = mw N p = (2.75)(21) = 58
Material. C1050, OQT 1100 F
703.
A pair of helical gears, subjected to heavy shock loading, is to transmit 50 hp at
3
1750 rpm of the pinion.; mg = 4.25 ; ψ = 15o ; minimum D p = 4 in. ; continuous
4
o
service, 24 hr/day; 20 F.D. teeth in the normal plane, carefully cut; throughhardened to a maximum BHN = 350. Decide upon the pitch, face width, material
and its treatment.
Solution:
π (4.75)(1750)
vm =
= 2176 fpm
12
33,000hp (33,000 )(50 )
Ft =
=
= 758 lb
vm
(2176)
Dynamic load:
0.05vm Ft + Cb cos 2 ψ cosψ
Fd = Ft +
lb
1
2
2
0.05vm + Ft + Cb cos ψ
Fig. AF 19, vm = 2176 fpm
Permissible error = 0.0014 in
(
)
(
Page 3 of 14
)
SECTION 12 – HELICAL GEARS
Use carefully cut gears, e = 0.001 in , Pd = 5 as standard
Table AT 25,
Steel and steel, 20o FD
C = 1660
Fd = 758 +
(
)
0.05(2176) 758 + 1660b cos 2 15 cos15
(
0.05(2176) + 758 + 1660b cos 15
105.1(758 + 1548.8b )
Fd = 758 +
1 lb
108.8 + (758 + 1548.8b )2
Wear load:
bD pQK g
Fw =
cos 2 ψ
2mg
2(4.25)
Q=
=
= 1.619
mg + 1 4.25 + 1
Table At 26, 20o FD,
Sum of BHN =2(350)=700
K g = 270
2
b(4.75)(1.619 )(270 )
= 2225b
cos 2 15
2π
Fw ≥ Fd , bmin = 2 Pa =
= 4.69 in.
Pd tanψ
By trial and error method
Fw =
b
5
6
Fd
5203
5811
Fw
11125
13350
use b = 5 in
Material:
Strength:
sbY
sbY cosψ
Fs =
=
K f Pdn
K f Pd
Np
N ep =
cos3ψ
N p = Pd D p = (5)(4.375) = 22
22
= 25
cos3 15
Table AT 24, Load near middle
N ep =
Page 4 of 14
1
2
)
lb
SECTION 12 – HELICAL GEARS
N ep = 25 , φn = 20o FD
Y = 0.580
assume K f = 1.7
s (5)(0.580) cos15
= 0.32955s
(1.7 )(5)
Fs = N sf Fd
for 24 hr/day service, heavy shock loading
N sf = 1.75
Fs =
0.32955s = (1.75)(5203)
s = 27,629 psi
s
use sn = u
3
su = 3(27,629) = 82,887 psi
Table AT 9
Use 4150, OQT 1200 F,
su = 159ksi , BHN = 331 < 350
Ans.
Pd = 5
b = 5 in
Material. 4150, OQT 1200 F
705.
Design the teeth for two herringbone gears for a single-reduction speed reducer
with mw = 3.80 . The capacity is 36 hp at 3000 rpm of the pinion; ψ = 30o ; F.D.
teeth with φn = 20o . Since space is at a premium, the initial design is for N p = 15
teeth and carburized teeth of AISI 8620; preferably b < 2 D p .
Solution:
N
15
Dp = p =
Pd
Pd
b ≈ 2Dp
b = 2 Dp =
vm =
30
Pd
πD p n p
12
 15 
π  (3000)
P
11,781
vm =  d 
=
12
Pd
Page 5 of 14
SECTION 12 – HELICAL GEARS
33,000hp (33,000 )(36 )
=
= 101Pd
vm
 11,781 


 Pd 
Dynamic load
0.05vm Ft + Cb cos 2 ψ cosψ
Fd = Ft +
lb
1
2
0.05vm + Ft + Cb cos ψ 2
Ft =
(
)
(
φn = 20
)
o
ψ = 30 o
Assume C = 1660 , Table AT 25, 20o FD

 11,781  
 30 
 101Pd + 1660  cos 2 30 cos 30
0.05
 Pd  
 Pd 

Fd = 101Pd +
lb
1
2

 11,781  
 30 
 + 101Pd + 1660  cos 2 30
0.05
 Pd  
 Pd 

510 
37,350 
101Pd +

Pd 
Pd 
Fd = 101Pd +
lb
1
2
589 
37,350 

+ 101Pd +
Pd 
Pd 
Wear load
bD pQK g
Fw =
cos 2 ψ
2 mg
2(3.80)
Q=
=
= 1.583
mg + 1 3.80 + 1
For AISI 8620, carburized, 20o FD
K g = 750 for 1010 cycles
 30  15  (1.583)(750) 712,350
Fw =   
=
2
Pd2
 Pd  Pd  cos 30
By trial and error, Fw ≥ Fd
Pd
Fd
Fw
5
4433
28,494
4
5454
44,522
6
3817
19,788
8
3173
11,130
9
3008
8794
For carefully cut gears, e = 0.001
vmax = 1400 fpm (Fig. AF 9)
Page 6 of 14
SECTION 12 – HELICAL GEARS
Pd
5
4
6
8
9
11,781
Pd
2356.2
1963.5
1683
1473
1309 fpm
vm =
use Pd = 9
Fd = 3008 lb
Fw = 5794 lb > Fd
30 30
b=
=
= 3.3 in
Pd
9
use b = 3.0 in
To check for strength
sbY
sbY cosψ
Fs =
=
K f Pdn
K f Pd
Np
N ep =
cos3ψ
N p = 15
15
= 23
cos3 30
Table AT 24, Load near middle
N ep = 23 , φn = 20o FD
N ep =
Y = 0.565
assume K f = 1.7
8620, SOQT 450, su = 167 ksi
s
sn = u
3
su 167
sn = =
= 83.5
2
2
(83,500)(3.0)(0.565)cos 30 = 8011 lb > F (= 3008 lb )
Fs =
d
(1.7)(9)
Designed Data:
Pd = 9
b = 3.0 in
N p = 15
N g = mw N p = (3.8)(15) = 57
Page 7 of 14
SECTION 12 – HELICAL GEARS
N p 15
= = 1.67 in
Pd
9
N
57
Dg = g =
= 6.33 in
Pd
9
Dp =
CHECK PROBLEMS
707.
The data for a pair of carefully cut gears are: Pdn = 5 , φn = 20o ,ψ = 12o ,
b = 3.5 in. , N p = 18 , N g = 108 teeth; pinion turns 1750 rpm. Materials: pinion,
SAE 4150, OQT to BHN = 350; gear, SAE 3150, OQT to BHN = 300. Operation
is with moderate shock for 8 to 10 hr./day. What horsepower may be transmitted
continuously?
Solution:
N
Dp = p
Pd
Pd = Pdn cosψ = (5)cos15 = 4.89
18
Dp =
= 3.681 in
4.89
Wear load
bD pQK g
Fw =
cos 2 ψ
b = 3.5 in.
2Ng
2(108)
Q=
=
= 1.7143
N p + N g 18 + 108
Table AT 26, φn = 20o
Sum of BHN = 350 + 300 = 650
K g = 233
Fw =
(3.5)(3.681)(1.7143)(233) = 5379 lb
cos 2 12
Strength of gear
sbY
Fs =
lb
K f Pdn
For gear: SAE 3150, OQT to BHN = 300
su = 151 ksi
sn = 0.5su = 0.5(151) = 75.5 ksi
Ng
108
N eg =
=
= 116
3
cos ψ cos3 12
Page 8 of 14
SECTION 12 – HELICAL GEARS
Table AT 24, Load near middle, φn = 20o
Y = 0.763
snY = (75.5)(0.763) = 57.6
For pinion: SAE 4150, OQT to BHN = 350
su = 0.5BHN = 0.5(350) = 175 ksi
sn = 0.5su = 0.5(175) = 87.5 ksi
Np
18
N ep =
=
= 19
3
cos ψ cos3 12
Table AT 24, Load near middle, φn = 20o
Y = 0.534
snY = (87.5)(0.534) = 46.7
Therefore use pinion as weak
Assume K f = 1.7
Fs =
(87,500)(3.5)(0.534) = 19,240 lb
(1.7 )(5)
For moderate shock, 8 to 10 hr./day
Use N sf = 1.5
Fs ≥ N sf Fd
19,240 = 1.5Fd
Fd ≤ 12,827 lb
Therefore use Fd = Fw = 5379 lb
Fd = Ft +
0.05vm (Ft + Cb cos 2 ψ )cosψ
lb
1
0.05vm + (Ft + Cb cos 2 ψ )2
Fig. AF 20, carefully cut gears, Pdn = 5 , e = 0.001 in
Table AT 25, steel and steel, 20o FD
C = 1660
πD p n p π (3.681)(1750)
vm =
=
= 1686 fpm
12
12
0.05(1686) Ft + 1660(3.5) cos 2 12 cos12
Fd = Ft +
lb
1
2
0.05(1686) + Ft + 1660(3.5) cos 12 2
82.46[Ft + 5559]
Fd = Ft +
= 5379 lb
1
84.3 + [Ft + 5559]2
Trial and error
Ft = 1800 lb
Fv
(1800)(1686) = 92 hp
hp = t m =
33,000
33,000
[
]
[
Page 9 of 14
]
SECTION 12 – HELICAL GEARS
708.
Two helical gears are used in a single reduction speed reducer rated at 27.4 hp at
a motor speed of 1750 rpm; continuous duty. The rating allows an occasional 100
% momentary overload. The pinion has 33 teeth. Pdn = 10 , b = 2 in. , φn = 20o ,
ψ = 20o , mw = 2.82 . For both gears, the teeth are carefully cut from SAE 1045
with BHN = 180. Compute (a) the dynamic load, (b) the endurance strength;
estimate K f = 1.7 . Also decide whether or not the 100 % overload is damaging.
(c) Are these teeth suitable for continuous service? If they are not suitable
suggest a cure. (The gears are already cut.)
Solution:
N
Dp = p
Pd
Pd = Pdn cosψ = (10) cos15 = 9.66
33
Dp =
= 3.42 in
9.66
πD p n p π (3.42)(1750)
vm =
=
= 1567 fpm
12
12
33,000hp 33,000(27.4 )
Ft =
=
= 577 lb
vm
1567
(a) Dynamic load
0.05vm (Ft + Cb cos 2 ψ )cosψ
Fd = Ft +
lb
1
0.05vm + (Ft + Cb cos 2 ψ )2
Fig. AF 20, carefully cut gears, Pdn = 10 , e = 0.001 in
Table AT 25, steel and steel, 20o FD
C = 1660
b = 2 in
Fd = 577 +
[
]
0.05(1567 ) 577 + 1660(2)cos 2 15 cos15
[
(b) Endurance strength
sbY
Fs =
lb
K f Pdn
For SAE 1045, BHN = 180
su = 0.5BHN = 0.5(180) = 90 ksi
sn = 0.5su = 0.5(90) = 45 ksi
Np
33
N ep =
=
= 37
3
cos ψ cos3 15
Table AT 24, Load near middle, φn = 20o
Page 10 of 14
1
2
]
0.05(1567 ) + 577 + 1660(2)cos 15
2
= 2578 lb
SECTION 12 – HELICAL GEARS
Y = 0.645
K f = 1 .7
sbY
(45,000)(2)(0.645) = 3415 lb
=
(1.7 )(10)
K f Pdn
For 100 % overload
Ft = 2(577 ) = 1154 lb
Fs =
Fd = Ft +
0.05vm (Ft + Cb cos 2 ψ )cosψ
1
2
lb
0.05vm + (Ft + Cb cos ψ )
2
Fd = 1154 +
[
]
0.05(1567 ) 1154 + 1660(2)cos 2 15 cos15
[
(c) Fw =
1
2
]
0.05(1567 ) + 1154 + 1660(2) cos 15
Since Fs ≈ Fd , 100 % overload is not damaging
2
= 3475 lb
bD pQK g
cos 2 ψ
b = 2 in.
2mw
2(2.82 )
Q=
=
= 1.476
mw + 1 2.82 + 1
Table AT 26, φn = 20o
Sum of BHN = 2(180) = 360
K g = 62.5
Fw =
(2)(3.42)(1.476)(62.5) = 676 lb < F (= 2578 lb )
d
cos 2 15
Therefore not suitable for continuous service.
Cure: Through hardened teeth
For Bhn
2578
Kg =
(62.5) = 238
676
min Bhn = 0.5(650) = 325
709.
Two helical gears are used in a speed reducer whose input is 100 hp at 1200 rpm,
from an internal combustion engine. Both gears are made of SAE 4140, with the
pinion heat treated to a BHN 363 – 415, and the gear to 321 – 363; let the teeth
be F.D.; 20o pressure angle in the normal plane; carefully cut; helix angle
ψ = 15o ; N p = 22 , N g = 68 teeth; Pd = 5 , b = 4 in . Calculate the dynamic load,
the endurance strength load, and the limiting wear load for the teeth. Should these
gears have long life if they operate continuously? (Data courtesy of the Twin
Disc Clutch Co.)
Solution:
Page 11 of 14
SECTION 12 – HELICAL GEARS
N p 22
=
= 4.4 in
Pd
5
πD p n p π (4.4)(1200)
vm =
=
= 1382 fpm
12
12
33,000hp 33,000(100 )
Ft =
=
= 2388 lb
vm
1382
Dynamic load
0.05vm (Ft + Cb cos 2 ψ )cosψ
Fd = Ft +
lb
1
2
0.05vm + (Ft + Cb cos ψ )2
Fig. AF 20, carefully cut gears, Pdn = 5 , e = 0.001 in
Table AT 25, steel and steel, 20o FD
C = 1660
b = 4 in
Dp =
Fd = 2388 +
[
]
0.05(1382) 2388 + 1660(4) cos 2 15 cos15
[
1
2
]
0.05(1382 ) + 2388 + 1660(4)cos 15
Endurance strength load
sbY cosψ
Fs =
lb
K f Pd
2
= 5930 lb
Assume K f = 1.7
Pinion
sn = 0.25BHN = 0.25(363) = 90.75 ksi
Np
22
N ep =
=
= 25
3
cos ψ cos3 15
Table AT 24, Load near middle, φn = 20o
Y = 0.580
sbY cosψ (90,750)(4 )(0.580) cos15
Fs =
=
= 23,925 lb
(1.7 )(5)
K f Pd
Gear
sn = 0.25BHN = 0.25(321) = 80.25 ksi
Np
68
N ep =
=
= 75
3
cos ψ cos3 15
Table AT 24, Load near middle, φn = 20o
Y = 0.735
sbY cosψ (80,250 )(4)(0.735)cos15
Fs =
=
= 26,811 lb
(1.7 )(5)
K f Pd
use Fs = 23,925 lb
Page 12 of 14
SECTION 12 – HELICAL GEARS
Limiting Wear Load
bD pQK g
Fw =
cos 2 ψ
Table AT 26, φn = 20o
Sum of BHN = 684 to 778 use 700
K g = 270
Q=
2Ng
2(68)
=
= 1.511
N p + N g 22 + 68
Fw =
(4)(4.4)(1.511)(270) = 7696 lb
cos 2 15
Since Fw (= 7696 lb ) > Fd (= 5930 lb ) these gears have long life if they operate
continuously.
CROSSED HELICAL
710.
Helical gears are to connect two shafts that are at right angles
( N1 = 20 , N 2 = 40 , Pdn = 10 ,ψ 1 = ψ 2 = 45o ). Determine the center distance.
Solution:
πD cosψ 1
N1 = 1
= Pdn D1 cosψ 1
Pcn
20 = (10)(D1 )cos 45
D1 = 2.83 in
N 2 = Pdn D2 cosψ 2
40 = (10)(D2 )cos 45
D2 = 5.66 in
C = 12 (D1 + D2 ) = 12 (2.83 + 5.66) = 4.25 in
712.
Two shafts that are at right angles are to be connected by helical gears. A
tentative design is to use N1 = 20 , N 2 = 60 , Pdn = 10 , and a center distance of 6
in. What must be the helix angles?
Solution:
Σ = ψ 1 +ψ 2 = 90o
N1
D1 =
Pdn cosψ 1
N2
D2 =
Pdn cosψ 2
1
C = 2 (D1 + D2 )
Page 13 of 14
SECTION 12 – HELICAL GEARS
N1
N2
+
Pdn cosψ 1 Pdn cosψ 2
20
60
+
2(6 ) =
10 cosψ 1 10 cosψ 2
2
6
12 =
+
cosψ 1 cosψ 2
1
3
6=
+
cosψ 1 cosψ 2
By trial and error method
1
3
+
6=
cosψ 1 sinψ 1
2C =
ψ 1 = 39.5o
ψ 2 = 50.5o
- end -
Page 14 of 14
SECTION 13 – BEVEL GEARS
DESIGN PROBLEMS
751.
Decide upon the pitch, face, N g , material, and heat treatment of a pair of straight
bevel gears to transmit continuously and indefinitely a uniform loading of 5 hp at
900 rpm of the pinion, reasonable operating temperature, high reliability;
mg ≈ 1.75 ; D p ≈ 3.333 in . Pinion overhangs, gear is straddle mounted.
Solution:
(
L = rp2 + rg2
tan γ p =
1
2
)
1
1
=
mg 1.75
γ p = 29.75o
L sin γ p = rp
L sin 29.75 =
3.333
2
L = 3.358 in
33,000hp
Ft =
lb
vm
πD p n p π (3.333)(900)
vm =
=
= 785.4 fpm
12
12
33,000(5)
Ft =
= 210 lb
785.4
Fd = (VF )N sf K m Ft
1
1
50 + vm2 50 + (785.4 ) 2
VF =
=
= 1.56
50
50
One gear straddle, one not
K m = 1.2
Table 15.2, uniform
N sf = 1.0
Fd = (1.56)(1.0)(1.2 )(210) = 393 lb
Wear load
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
D p = 3.333 in
b = 0.3L = 0.3(3.358) = 1.0 in
Temperature factor
K t = 1.0 , reasonable operating temperature
Life factor for wear
Page 1 of 17
SECTION 13 – BEVEL GEARS
Cl = 1.0 for indefinite life
Reliability factor for wear
Cr = 1.25 high reliability
Geometry factor for wear, Fig. 15.7
Assume I = 0.080
Elastic coefficient (Table 15.4)
Steel on steel , Ce = 2800
Fw = Fd
2
(3.333)(1.0)(0.08) (scd ) 2  1.0 
(2800)  (1.0)(1.25)
2
= 393
scd = 134,370 psi
Table 15.3, use Steel, (300)
scd = 135 ksi
Strength of bevel gears
s bJ K l
Fs = d
Pd K s K t K r
Size factor, assume K s = 0.71
Life factor for strength
K l = 1 for indefinite life
Temperature factor,
K t = 1 good operating condition
Reliability factor
K r = 1.5 high reliability
Geometry factor for strength (Fig. 15.5)
Assume J = 0.240
b = 1.0 in
sd = design flexural stress
Min. BHN = 300
sd = 19 ksi
Fs = F d
(19,000)(1.0)(0.240) 
Pd

1
 (0.71)(1)(1.5)  = 393


Pd = 11
say Pd = 10
10 10
so that b =
=
= 1.0 in
Pd 10
Dg = D p mg = (3.333)(1.75) = 5.833 in
Page 2 of 17
SECTION 13 – BEVEL GEARS
N g = Pd Dg = (10 )(5.833) = 58.33
say N g = 58
Use Pd = 10 , b = 1.0 in , N g = 58
Material = steel, min. Bhn = 300
752.
A pair of steel Zerol bevel gears to transmit 25 hp at 600 rpm of the pinion;
mg = 3 ; let N p ≈ 20 teeth; highest reliability; the pinion is overhung, the gear
straddle mounted. An electric motor drives a multi-cylinder pump. (a) Decide
upon the pitch, face width, diameters, and steel (with treatment) for intermittent
service. (b) The same as (a) except that indefinite life is desired.
Solution:
N
20
Dp = p =
Pd
Pd
 20 
(600)
πD p n p
Pd 
1000π

vm =
=
=
fpm
12
12
Pd
10
Let b =
Pd
Dynamic load
Fd = (VF )N sf K m Ft
π 
33,000hp
lb
vm
33,000(25)
Ft =
= 262.6 Pd
 1000π 


 Pd 
Ft =
1
 1000π 

50 + 
1
2
Pd 
50 + vm
1.121
1.121

VF =
=
= 1 + 12 = 1 +
50
50
Pd
Pd
Table 15.2, electric motor drives a multi-cylinder pump
Service factor, N sf = 1.25
2
One gear straddle, one not, K m = 1.2
 1.121 


(1.25)(1.2 )(262.6 )Pd = 394 Pd 1 + 1.121 
Fd = 1 +


Pd 
Pd 


(a) Strength of Bevel Gears
s bJ K l
Fs = d
Pd K s K t K r
Size factor, assume K s = 0.71
Page 3 of 17
SECTION 13 – BEVEL GEARS
Life factor for strength
Intermittent service, use K l = 4.6
Temperature factor, say K t = 1.0
Reliability factor, highest reliability
K r = 3.0
Geometry factor for strength
N
mg = g
Np
N p = 20
N p = 3(20 ) = 60
Fig. 15.5, J = 0.205
10
b=
Pd
Design flexural stress, steel
Assume sd = 15 ksi
Fs = Fd


(15,000) 10 (0.205)(4.6)
 1.121 

= 394 Pd 1 +


P
d 

 1.121 
66,408


=
394
P
d 1 +

Pd2
P
d 

Pd = 4.814
say Pd = 5
10 10
b=
=
= 2.0 in
Pd
5
 Pd 
Pd (0.71)(1.0 )(3)
Wear load for bevel gears
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
N p 20
=
= 4 in
Pd
5
K t = 1.0
Dp =
Life factor for wear, intermittent service
Cl = 1.5
Reliability factor for wear, highest reliability
Cr = 1.25
Geometry factor for wear, Fig. 15.7
Page 4 of 17
SECTION 13 – BEVEL GEARS
N p = 20 , N g = 60
I = 0.083
Elastic coefficient, steel on steel (Table 15.4)
Ce = 2800
Pd = 5
Fw = Fd
2
2
(4)(2)(0.083) scd 2  1.5  = 394(5)1 + 1.121 
(2800)  (1.0)(1.25)
5 

scd = 155,730 psi
Table 15.3
Use steel, min. BHN = 360, scd = 160 ksi
Pd = 5
b = 2 in
D p = 4 in
Dg = mg D p = (3)(4 ) = 12 in
steel, min. BHN = 360
(b) For indefinite life,
K l = 1.0 , life factor for strength
Cl = 1.0 , life factor for wear
Strength:
s bJ K l
Fs = d
Pd K s K t K r
Fs = Fd


(15,000) 10 (0.205)(1.0)
 1.121 

= 394 Pd 1 +


P
d 

 1.121 
14,437

= 394 Pd 1 +
2


Pd
P
d 

Pd = 2.799
say Pd = 3
10 10
b=
=
= 3.33 in
Pd
3
 Pd 
Pd (0.71)(1.0)(3)
Wear load
Page 5 of 17
SECTION 13 – BEVEL GEARS
scd2  Cl 

Fw = D pbI 2 
C e  K t Cr 
2
N p 20
=
= 6.67 in
Pd
3
Fw = Fd
Dp =
2
2
(6.67 )(3.33)(0.083) scd 2  1.0  = 394(3)1 + 1.121 
(2800)  (1.0)(1.25)
3 

scd = 113,744 psi
Table 15.3
Use steel, min. BHN = 240, scd = 115 ksi
Pd = 3
b = 3.33 in
D p = 6.67 in
Dg = mg D p = (3)(6.67 ) = 20 in
steel, min. BHN = 240
753.
Decide upon the pitch, face, and number of teeth for two spiral-bevel gears for a
speed reducer. The input to the pinion is 20 hp at 1750 rpm; mg ≈ 1.9 ; pinion
overhung, gear-straddle mounted. It is hoped not to exceed a maximum D p of 4
3/8-in.; steel gears with minimum 245 BHN on pinion and 210 BHN on gear.
The gear is motor-driven, subject to miscellaneous drives involving moderate
shock; indefinite life against breakage and wear with high reliability. If the gears
designed for the foregoing data are to be subjected to intermittent service only,
how much power could they be expected to transmit?
Solution:
(a) vm =
πD p n p
π (4.375)(1750)
= 2000 fpm
12
12
33,000hp 33,000(20 )
Ft =
=
= 330 lb
vm
2000
Dynamic load
Fd = (VF )N sf K m Ft
One gear straddle, one not
K m = 1.2
Table 15.2
Motor-driven, moderate shock
Page 6 of 17
=
SECTION 13 – BEVEL GEARS
N sf = 1.25
1
 70 + vm2  2
VF = 
 , spiral
 70 
1
1
 70 + (2000) 2 
VF = 
 = 1.254
70


Fd = (1.254)(1.25)(1.2)(330) = 621 lb
1
2
Wear load
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
D p = 4.375 in
Temperature Factor, K t = 1.0
Design contact stresses,
BHN = 245 , pinion
scd = 116 ksi
Life factor for wear
Cl = 1.0 , indefinite life
Reliability factor for wear
Cr = 1.25 , high reliability
Geometry factor for wear, Fig. 15.8
Assume I = 0.12
Elastic coefficient, steel on steel (Table 15.4)
Ce = 2800
(116,000)2  1.0 
Fw = (4.375)(b )(0.12)
(2800)2  (1.0)(1.25)
Fw = Fd
721b = 621
b = 0.8613 in
7
say b = in = 0.875 in
8
Strength of gear
s bJ K l
Fs = d
Pd K s K t K r
sd = design flexural stress
min. BHN = 210
sd = 15.4 ksi
Size factor, assume K s = 0.71
Page 7 of 17
2
= 721b
SECTION 13 – BEVEL GEARS
Life factor for strength
K l = 1 for indefinite life
Temperature factor,
Kt = 1
Reliability factor
K r = 1.5 high reliability
Geometry factor Fig. 15.6
Assume J = 0.28
 3543
(15,400)(0.875)(0.28) 
1
Fs =

= P
Pd
 (0.71)(1)(1.5) 
d
Fs = Fd
3543
= 621
Pd
Pd = 5.7
say Pd = 6
7
Then, Pd = 6 , b = in , N p = D p Pd = (4.375)(6 ) = 26
8
N g = mw N p = (1.9 )(26 ) = 50
(b) Intermittent service only
Strength
s bJ K l
Fs = d
Pd K s K t K r
sd = 15,400 psi (Gear)
For Pd = 6 , K s = 0.64
For indefinite service, K l = 4.6
K t = 1.0 , K r = 1.5
Geometry factor, Fig. 15.6, N p = 26 , N g = 50
J = 0.292
(15,400)(0.875)(0.292)  4.6  = 3142 lb
Fs =
 (0.71)(1)(1.5) 
6


Wear load
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
D p = 4.375 in
K t = 1.0
scd = 116 ksi
Ce = 2800
Page 8 of 17
2
SECTION 13 – BEVEL GEARS
Cl = 1.5 intermittent service
Cr = 1.25
Geometry factor for wear, Fig. 15.8
N p = 26 , N g = 50
I = 0.116
(116,000)2  1.5 
Fw = (4.375)(0.875)(0.116 )
(2800)2  (1.0)(1.25)
2
= 1098 lb
use Fw = Fd
Fd = (VF )N sf K m Ft
1098 = (1.254)(125)(1.2)Ft
Ft = 584 lb
Fv
(584)(2000) = 35 hp
hp = t m =
33,000
33,000
CHECK PROBLEMS
755.
A pair of straight-bevel gears are to transmit a smooth load of 45 hp at 500 rpm
3
of the pinion; mg = 3 . A proposed design is Dg = 15 in. , b = 2 in. , Pd = 4 .
8
Teeth are carburized AISI 8620, SOQT 450 F. The pinion overhangs, the gear is
straddle-mounted. Would these gears be expected to perform with high reliability
in continuous service? If not would you expect more than 1 failure in 100?
Solution:
D 15
D p = g = = 5 in
mg
3
πD p n p π (5)(500)
vm =
=
= 655 fpm
12
12
33,000hp 33,000(45)
Ft =
=
= 2267 lb
vm
655
Dynamic load
Fd = (VF )N sf K m Ft
1
1
50 + vm2 50 + (655) 2
VF =
=
= 1.512
50
50
One gear straddle, one not
K m = 1.2
Smooth load, N sf = 1.0
Fd = (1.512)(1.0)(1.2 )(2267 ) = 4113 lb
Page 9 of 17
SECTION 13 – BEVEL GEARS
Strength of bevel gears
s bJ K l
Fs = d
Pd K s K t K r
Size factor, for Pd = 4 ,
K s = 0.71
Life factor for strength
Kl = 1
Temperature factor,
Kt = 1
Geometry factor for strength (Fig. 15.5)
N p = Pd D p = (4 )(5) = 20
N g = Pd Dg = (4 )(15) = 60
J = 0.205
sd = 30 ksi (55 – 63 Rc) for carburized teeth
Fs =
(30,000)(2.375)(0.205) 
 5143
1
 (0.71)(1)(K )  = K
r 
r

4
Fs = Fd
5143
= 4113
Kr
K r = 1.25 < 1.5 will not perform high reliability.
Wear load
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
D p = 5 in
b = 2.375 in
Table 15.3, scd = 225 ksi
Table 15.4, Ce = 2800
Geometry factor for wear, Figure 15.7
N p = 20 , N g = 60
I = 0.083
Kt = 1
life factor for wear Cl = 1
(225,000)2  1 
Fw = (5)(2.375)(0.083)
(2800)2  (1)(Cr )
Fw = Fd
Page 10 of 17
2
=
6364
Cr2
SECTION 13 – BEVEL GEARS
6364
= 4113
Cr2
Cr = 1.244 ≈ 1.25 , high reliability
Since K r < 1.5 ,this will not perform high reliability but 1 in 100, K r ≈ 1.12 < 1.25
756.
A gear catalog rates a pair of cast-iron, straight-bevel gears at 15.26 hp at 800
rpm of the 16-tooth pinion; mg = 3.5 , b = 3 in. , Pd = 3 ; pinion overhangs,
straddle-mounted gear. Assume the cast iron to be class 30. If the load is smooth
is this rating satisfactory, judging by the design approach of the Text for good
reliability (a) when strength alone is considered, (b) when long continuous
service is desired?
Solution:
N
16
Dp = p =
= 5.333 in
Pd
3
πD p n p π (5.333)(800)
vm =
=
= 1117 fpm
12
12
33,000hp 33,000(15.26 )
Ft =
=
= 451 lb
vm
1117
Dynamic load
Fd = (VF )N sf K m Ft
1
1
50 + vm2 50 + (1117 ) 2
VF =
=
= 1.668
50
50
One gear straddle, one not
K m = 1.2
Smooth load, N sf = 1.0
Fd = (1.668)(1.0)(1.2)(451) = 903 lb
(a) Strength
s bJ K l
Fs = d
Pd K s K t K r
Pd = 3 ,
K s = 0.76
Kl = 1
Kt = 1
K r = 1.5
sd = 4.6 ksi , cast-iron class 30
N p = 16
Page 11 of 17
SECTION 13 – BEVEL GEARS
N g = mw N p = (3.5)(16 ) = 56
J = 0.184

(4,600)(3)(0.184) 
1
Fs =

 = 742 lb < 903 lb(= Fd )
3
 (0.76)(1)(1.5) 
with
K l = 1.4 for 106 cycles
Fs = (1.4)(742) = 1040 lb > 903 lb(= Fd )
Therefore satisfactory for 106 cycles.
(b) Continuous service
Wear load
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
D p = 5.333 in
b = 3 in
Table 15.3, scd = 50 ksi , cast-iron class 30
Table 15.4, cast-iron and cast-iron Ce = 2250
Cl = 1
Kt = 1
Cr = 1.25
Geometry factor for wear, Figure 15.7
N p = 16 , N g = 56
I = 0.077
(50,000)2  1 
Fw = (5.333)(3)(0.077 )
(2250)2  (1)(1.25)
2
= 389 lb < Fd (= 903 lb )
Therefore, not satisfactory for long continuous service.
757.
An 870-rpm motor drives a belt conveyor through bevel gears having 18 and 72
3
teeth; Pd = 6 , b = 1 in . Both gears are straddle-mounted. What horsepower may
4
these gears transmit for an indefinite life with high reliability if both gears are (a)
cast-iron, class 40; (b) AISI 5140, OQT 1000 F; (c) AISI 5140, OQT 1000 F,
flame hardened (d) AISI 8620, SOQT 450 F?
Solution:
N
18
D p = p = = 3 in
Pd
6
πD p n p π (3)(870)
vm =
=
= 683 fpm
12
12
Dynamic load
Page 12 of 17
SECTION 13 – BEVEL GEARS
Fd = (VF )N sf K m Ft
Both gears straddle mounted
K m = 1.0
Table 15.2, N sf = 1.0
1
1
50 + vm2 50 + (683) 2
=
= 1.523
50
50
Fd = (1.523)(1.0)(1.0)Ft = 1.523Ft
VF =
(a) Cast-iron, class 40
Strength
s bJ K l
Fs = d
Pd K s K t K r
sd = 7 ksi , cast-iron class 40
3
b = 1 in
4
K l = 1 , indefinite life
Pd = 6
K s = 0.64
Kt = 1
K r = 1.5 , high reliability
Figure 15.5, N p = 18 , N g = 72
J = 0.204

(7000)(1.75)(0.204) 
1
Fs =

 = 434 lb
6
 (0.64)(1)(1.5) 
Wear:
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
D p = 3 in
3
b = 1 in
4
Table 15.3, scd = 65 ksi , cast-iron class 40
Table 15.4, cast-iron and cast-iron Ce = 2250
Cl = 1 , indefinite life
Kt = 1
Cr = 1.25 , high reliability
Geometry factor for wear, Figure 15.7
N p = 18 , N g = 72
I = 0.082
Page 13 of 17
SECTION 13 – BEVEL GEARS
2
(
65,000)  1 
Fw = (3)(1.75)(0.082 )
(2250)2  (1)(1.25)
2
= 230 lb
Fd = Fw
1.523Ft = 230
Ft = 151 lb
Fv
(151)(683) = 3 hp
hp = t m =
33,000
33,000
(b) AISI 5140, OQT 1000 F, BHN = 300
Strength
sd = 19,000 psi
Fs =
(19,000)(1.75)(0.204) 

1
 (0.64)(1)(1.5)  = 1178 lb


6
Wear:
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
Table 15.3, scd = 135 ksi
Table 15.4, steel and steel, Ce = 2800
2
(
135,000)  1 
Fw = (3)(1.75)(0.082)
(2800)2  (1)(1.25)
2
= 640 lb
Fd = Fw
1.523Ft = 640
Ft = 420 lb
Fv
(420)(683) = 8.7 hp
hp = t m =
33,000
33,000
(c) AISI 5140, OQT 1000 F, Flame Hardened
Strength
sd = 13.5 ksi
Fs =
(13,500)(1.75)(0.204) 

1
 (0.64 )(1)(1.5)  = 837 lb


6
Wear:
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
Table 15.3, scd = 190 ksi
Table 15.4, steel and steel, Ce = 2800
Page 14 of 17
SECTION 13 – BEVEL GEARS
2
(
190,000)  1 
Fw = (3)(1.75)(0.082)
(2800)2  (1)(1.25)
2
= 1269 lb
Fd = Fs
1.523Ft = 837
Ft = 550 lb
Fv
(550)(683) = 11.4 hp
hp = t m =
33,000
33,000
(d) AISI 86200, SOQT 450 F, carburized
Strength
sd = 30 ksi (55 – 63 Rc)
Fs =
(30,000)(1.75)(0.204) 

1
 (0.64)(1)(1.5)  = 1859 lb


6
Wear:
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
Table 15.3, scd = 225 ksi
Table 15.4, steel and steel, Ce = 2800
2
(
225,000)  1 
Fw = (3)(1.75)(0.082)
(2800)2  (1)(1.25)
2
= 1779 lb
Fd = Fw
1.523Ft = 1779
Ft = 1168 lb
Fv
(1168)(683) = 24.2 hp
hp = t m =
33,000
33,000
758.
A pair of straight-bevel gears transmits 15 hp at a pinion speed of 800 rpm;
Pd = 5 , N p = 20 , N p = 60 , b = 2 in . Both gears are made of AISI 4140 steel,
OQT 800 F. What reliability factor is indicated for these gears for strength and
for wear (a) for smooth loads, (b) for light shock load from the power source and
heavy shock on the driven machine?
Solution:
N
20
Dp = p =
= 4 in
Pd
5
πD p n p π (4)(800)
vm =
=
= 838 fpm
12
12
Page 15 of 17
SECTION 13 – BEVEL GEARS
33,000hp 33,000(15)
=
= 591 lb
vm
838
Fd = (VF )N sf K m Ft
Ft =
1
1
50 + vm2 50 + (838) 2
VF =
=
= 1.579
50
50
assume K m = 1.0
Fd = (1.579 )(N sf )(1.0 )(591) = 933 N sf
Strength of bevel gear
s bJ K l
Fs = d
Pd K s K t K r
For AISI 4140, OQT 800 F, BHN = 429
sd = 24 ksi
assume K l = 1
Kt = 1
Pd = 5
K s = 0.675
Figure 15.5, N p = 20 , N g = 60
J = 0.205
 2916
1
(24,000)(2)(0.205) 
Fs =

= K
5
r
 (0.675)(1)(K r ) 
Fs = Fd
2916
= 933 N sf
Kr
3.1254
Kr =
N sf
Wear load:
s2  C 
Fw = D pbI cd2  l 
C e  K t Cr 
2
BHN = 429
Table 15.3, scd = 190 ksi
Table 15.4, steel and steel, Ce = 2800
D p = 4 in
b = 2 in
Assume Cl = 1.0 , K t = 1.0
Fig. 15.7, N p = 20 , N g = 60
I = 0.083
Page 16 of 17
SECTION 13 – BEVEL GEARS
2
(
190,000 )  1 
(
)(
)(
)
Fw = 4 2 0.083
(2800 )2  (1)(Cr )
2
=
3058
Cr2
Fd = Fw
3058
Cr2
1.810
Cr =
N sf
933 N sf =
(a) Table 15.2, smooth load
N sf = 1.0
For strength, K r =
For wear, Cr =
3.1254 3.1254
=
= 3.1254
N sf
1
1.810 1.810
=
= 1.810
N sf
1
(b) Table 15.2, light shock source, heavy shock driven
N sf = 2.0
3.1254 3.1254
=
= 1.5627
N sf
2
1.810 1.810
For wear, Cr =
=
= 1.2799
N sf
2
For strength, K r =
- end -
Page 17 of 17
SECTION 14 – WORM GEARS
DESIGN PROBLEMS
791.
(a) Determine a standard circular pitch and face width for a worm gear drive with
an input of 2 hp at 1200 rpm of the triple-threaded worm; the 1.58-in. ( Dw ) is
steel with a minimum BHN = 250; gear is manganese bronze (Table AT 3);
mw = 12 . Consider wear and strength only. Use a φn to match the lead angle λ .
(See i16.13, Text.) (b) compute the efficiency.
Solution:
 1200 + vmg 
 Ft lb
a) Fd = 
 1200 
33,000hp
Ft =
vmg
πDg ng
vmg =
12
n
1200
ng = w =
= 100 rpm
mw
12
Dg = mw Dw tan λ
tan λ =
N t Pc
πDw
 N P  N m P (3)(12)Pc
Dg = mw Dw  t c  = t w c =
= 11.46 Pc
π
π
 πDw 
π (11.45Pc )(100)
vmg =
= 300 Pc
12
33,000(2 ) 220
Ft =
=
300 Pc
Pc
 1200 + 300 Pc  220  55(4 + Pc )
Fd = 
=
lb

1200
Pc

 Pc 
Wear load
Fw = Dg bK w
say b = 2 Pc ,
Dg = 11.46 Pc
Fw = Fd
(11.46 Pc )(2 Pc )(K w ) = 55(4 + Pc )
Pc
22.92 Pc2 K w =
Page 1 of 19
55(4 + Pc )
Pc
SECTION 14 – WORM GEARS
Dg
11.46 Pc
=
= 0.60443Pc
mw Dw (12)(1.58)
tan λ =
By trial and error and using Table AT 27 ( φn ≈ λ )
Kw
Pc
Pc (std)
λ
36
50
0.678
0.605
¾
5/8
24.4
20.7
Use φn = 20o , λ = 20.7o , Pc =
5
in
8
Fw = Fd
(11.46 Pc )(b )(K w ) = 55(4 + Pc )
Pc
5

55 4 + 
(11.46) 5 (b)(50) =  5 8 
8
8
b = 1.1365 in
5
say b = 1 in
32
To check for strength.
sYbPcn sYbPc cos λ
Fs =
=
π
π
For manganese bronze,
s = 30,000 psi
φn = 20o
Y = 0.392
λ = 20.7o
5
Pc = in
8
5
b = 1 in
32
(30,000)(0.392)1 5
 5 
  cos 20.7
 32  8 
= 2530 lb > Fd
Fs =
π
use Pc =
b =1
5
in
8
5
in
32
Page 2 of 19
λmax
i16.11
16
25
φn
14 ½
20
SECTION 14 – WORM GEARS
 cos φn − f tan λ 

(b) e = tan λ 
 cos φn tan λ + f 
φn = 20o
λ = 20.7o
πDwnw π (1.58)(1200 )
vr =
=
= 531 fpm > 70 fpm
12 cos λ
12 cos 20.7
0.32
0.32
=
= 0.0334
0.36
vr
(531)0.36
 cos 20 − 0.0334 tan 20.7 
e = tan 20.7
 = 0.902 = 90.2%
 cos 20 tan 20.7 + 0.0334 
f =
792.
A high-efficiency worm-gear speed reducer is desired, to accept 20 hp from a
1750-rpm motor. The diameter Dw of the integral worm has been estimated to be
7
1 in. ; the next computations are to be for a steel worm with a minimum BHN =
8
250; phosphor-bronze gear (Table AT 3); mw = 11 . Probably, the worm should
not have less than 4 threads. (a) Considering wear and strength only (i16.13),
decide upon a pitch and face width that satisfies these requirements (i16.11,
Text); specifying the pressure angle, diameters, and center distance. How does
Dw used compare with that from equation (m), i16.11, Text? What addendum
and dedendum are recommended by Dudley? Compute a face length for the
worm. (b) Compute the efficiency. What do you recommend as the next trial for
a “better” reducer?
Solution:
33,000hp
Ft =
vmg
πDg ng
vmg =
12
n
1750
ng = w =
= 159.1 rpm
mw
11
PN
Pm N
P (11)(4 )
Dg = c g = c w t = c
= 14 Pc
π
vmg =
π
π (14 Pc )(159.1)
π
= 583Pc
12
33,000(20 ) 1132
Ft =
=
583Pc
Pc
Page 3 of 19
SECTION 14 – WORM GEARS
 1200 + vmg 
 Ft lb
Fd = 
 1200 
 1200 + 583Pc  1132  1132(1 + 0.4858 Pc )
=
Fd = 
lb

1200
Pc

 Pc 
(a) Wear
Fw = Dg bK w
b = 2 Pc ,
Dg = 14 Pc
Fw = Fd
(14 Pc )(2 Pc )(K w ) = 1132(1 + 0.4858Pc )
Pc
1132(1 + 0.4858 Pc )
28 Pc2 K w =
Pc
Table AT 27, steel, min. BHN = 250, and bronze
And by trial and error ethod
NP
4(Pc )
tan λ = t c =
= 0.6791Pc
πDw (π )(1.875)
By trial and error and using Table AT 27
Kw
Pc
Pc (std)
λ
36
50
60
1.213
1.071
1.000
1¼
1¼
1.0
40.33
40.33
34.18
Use φn = 25o , λ = 34.18o , Pc = 1 in
Fw = Fd
(14 Pc )(b )(K w ) = 1132(1 + 0.4858Pc )
Pc
(14)(1)(b )(60) = 1132(1 + 0.4858)
1
b = 2 in
To check for strength
sYbPcn sYbPc cos λ
Fs =
=
π
π
For phosphor-bronze,
s = sn = 31,000 psi
Page 4 of 19
λmax
i16.11
16
25
35
φn
14 ½
20
25
SECTION 14 – WORM GEARS
For φn = 25o , Y = 0.470
(31,000)(0.470 )(2.0)(1.0)cos 34.18 = 7674 lb > F , ok
Fs =
d
π
use Pc = 1.0 in
b = 2.0 in
φn = 25o
7
Dw = 1 in
8
 7
Dg = mw Dw tan λ = (11)1  tan 34.18 = 14.0 in
 8
1
1 7

C = (Dw + Dg ) = 1 + 14  = 7.9375 in
2
2 8

Equation (m)
(7.9375)0.875 = 2.785 in > 1.875 in , ok
C 0.875
Dw =
in =
2 .2
2 .2
Addendum and dedendum (by Dudley)
Addendum = a = 0.3183Pcn = 0.3183Pc cos λ = 0.3183(1.0) cos 34.18 = 0.2633 in
Whole depth = 0.7 Pcn = 0.7 Pc cos λ = 0.7(1.0)cos 34.18 = 0.5791 in
Dedendum = whole depth – addendum = 0.5791 in – 0.2633 in = 0.3158 in
N 

Face length = Pc  4.5 + g 
50 

N g = mw N p = (11)(4 ) = 44
44 

Face length = 1.0 4.5 +  = 5.38 in
50 

Or
[
]
Face length = 2 2a(Dg − 2a )
1
2
Dg = 14 in
a = 0.2633 in
1
Face length = 2{2(0.2633)[14 − 2(0.2633)]} 2 = 5.33 in
Use Face length = 5.38 in
 cos φn − f tan λ 

(b) e = tan λ 
 cos φn tan λ + f 
πDwnw π (1.875)(1750 )
vr =
=
= 1038 fpm > 70 fpm
12 cos λ
12 cos 34.18
Page 5 of 19
SECTION 14 – WORM GEARS
f =
0.32
0.32
=
= 0.0263 ( 70 < vr < 3000 fpm )
0.36
vr
(1038)0.36
φn = 25o , λ = 34.18o ,
 cos 25 − 0.0263 tan 34.18 
e = tan 34.18
 = 0.94 = 94%
 cos 25 tan 34.18 + 0.0263 
recommendation for next trial
φn = 30o
λmax = 45o
793.
The input to a worm-gear set is to be 25 hp at 600 rpm of the worm with
m w = 20 . The hardened-steel worm is to be the shell type with a diameter
approximately as given in i16.11, Text, and a minimum of 4 threads; the gear is
to be chilled phosphor bronze (Table AT 3). (a) Considering wear and strength
only determine suitable values of the pitch and face width. Let φn be appropriate
to the value of λ . (b) Compute the efficiency. (c) Estimate the radiating area of
the case and compute the temperature rise of lubricant. Is special cooling needed?
Solution:
33,000hp
Ft =
vmg
πDg ng
vmg =
12
n
600
ng = w =
= 30 rpm
mw
20
PN
Pm N
P (20 )(4 ) 80 Pc
Dg = c g = c w t = c
=
π
π
π
π
 80 Pc 
(30)
π 

vmg =
= 200 Pc
12
33,000(25) 4125
Ft =
=
200 Pc
Pc
π
 1200 + vmg 
 Ft lb
Fd = 
 1200 
 1200 + 200 Pc  4125  687.5(6 + Pc )
Fd = 
=
lb

1200
Pc

 Pc 
shell type: Dw = 2.4 Pc + 1.1 in
(4)Pc
NP
4 Pc
tan λ = t c =
=
πDw (π )(2.4 Pc + 1.1) π (2.4 Pc + 1.1)
Page 6 of 19
SECTION 14 – WORM GEARS
(a) Wear load
Fw = Dg bK w
b = 2 Pc ,
80 Pc
Dg =
π
Fw = Fd
687.5(6 + Pc )
 80 Pc 

(2 Pc )(K w ) =
Pc
 π 
687.5(6 + Pc )
50.93Pc2 K w =
Pc
Table AT 27, Hardened steel and chilled bronze
By trial and error method
By trial and error and using Table AT 27 ( φn ≈ λ )
Kw
Pc
Pc (std)
λ
90
125
1.017
0.907
1.0
1.0
20
20
Use φn = 20o , λ = 20o , Pc = 1 in
Fw = Fd
1687.5(6 + 1)
 80 
 (b )(125) =
1
π 
b = 1.512 in
5
say b = 1 in
8
 cos φn − f tan λ 

(b) e = tan λ 
 cos φn tan λ + f 
φn = 20o
λ = 20o
Dw = 2.4 Pc + 1.1 = 2.4 + 1.1 = 3.5 in
πDwnw π (3.5)(600 )
vr =
=
= 585 fpm
12 cos λ
12 cos 20
0.32
0.32
f = 0.36 =
= 0.0323 ( 70 < vr < 3000 fpm )
vr
(585)0.36
 cos 20 − 0.0323 tan 20 
e = tan 20
 = 0.9023 = 90.23%
 cos 20 tan 20 + 0.0323 
Page 7 of 19
λmax
i16.11
16
25
φn
14 ½
20
SECTION 14 – WORM GEARS
(c) Radiating area ≈ Amin = 43.2C 1.7 sq. in.
1
(Dw + Dg )
2
Dw = 3.5 in
80 Pc 80(1)
Dg =
=
= 25.5 in
C=
π
π
1
C = (3.5 + 25.5) = 14.5 in
2
1.7
Amin = 43.2(14.5) = 4072 sq.in.
Temperature rise = ∆t
Qc = hcr A∆t ft − lb min
Q = (1 − e )(hpi ) = (1 − 0.9023)(25) = 2.4425 hp(33,000 ft − lb min − hp ) = 80,600 ft − lb min
Figure AF 21, A = 4072 sq.in. = 28.3 sq. ft.
hcr = 0.42 ft − lb min − sq.in. − F
Q = Qc
80,600 = (0.42)(4072)(∆t )
∆t = 47 F
with t1 = 100 F
t2 = 147 F < 150 F
Therefore, no special cooling needed.
794.
A 50-hp motor turning at 1750 rpm is to deliver its power to a worm-gear
reducer, whose velocity ratio is to be 20. The shell-type worm is to be made of
high-test cast iron; since a reasonably good efficiency is desired, use at least 4
threads; manganese –bronze gear (Table AT 3). (a) Decide upon Dw and φn , and
determine suitable values of the pitch and face width. Compute (b) the efficiency,
(c) the temperature rise of the lubricant. Estimate the radiating area of the case. Is
special cooling needed?
Solution:
33,000hp
Ft =
vmg
πDg ng
vmg =
12
n
1750
ng = w =
= 87.5 rpm
mw
20
PN
Pm N
P (20 )(4 ) 80 Pc
Dg = c g = c w t = c
=
π
Page 8 of 19
π
π
π
SECTION 14 – WORM GEARS
 80 Pc 
(87.5)
π 

vmg =
= 583Pc
12
33,000(50 ) 2830
Ft =
=
583Pc
Pc
π
 1200 + vmg 
 Ft lb
(a) Fd = 
 1200 
 1200 + 583Pc  2830  1375(2.06 + Pc )
Fd = 
=
lb

1200
Pc

 Pc 
Wear load
Fw = Dg bK w
b = 2 Pc ,
80 Pc
Dg =
π
Fw = Fd
1375(2.06 + Pc )
 80 Pc 
(2 Pc )(K w ) =

Pc
 π 
1375(2.06 + Pc )
50.93Pc2 K w =
Pc
NP
tan λ = t c
πDw
Shell-type
Dw = 2.4 Pc + 1.1 in
4 Pc
tan λ =
π (2.4 Pc + 1.1)
Table AT 27, high-test cast-iron and manganese bronze
By trial and error and using Table AT 27 ( φn ≈ λ )
Kw
Pc
Pc (std)
λ
80
115
1.012
0.885
1.0
7/8
20
19.2
Use λ = 19.2o , φn = 20o , Pc =
7
in
8
7
Dw = 2.4 Pc + 1.1 = 2.4  + 1.1 = 3.2 in
8
Fw = Fd
Page 9 of 19
λmax
i16.11
16
25
φn
14 ½
20
SECTION 14 – WORM GEARS
7

1375 2.06 + 
 80  7 
8

  (b )(115) =
7
 π  8 
8
b = 1.80 in
7
say b = 1 in
8
 cos φn − f tan λ 

(b) e = tan λ 
 cos φn tan λ + f 
λ = 19.2o
φn = 20o
πDwnw
vr =
12 cos λ
nw = 1750 rpm
Dw = 3.2 in
πDwnw π (3.2 )(1750 )
vr =
=
= 1552 fpm
12 cos λ
12 cos19.2
0.32
0.32
f = 0.36 =
= 0.0227 ( 70 < vr < 3000 fpm )
vr
(1552)0.36
 cos 20 − 0.0227 tan 19.2 
e = tan 19.2
 = 0.9273 = 92.73%
 cos 20 tan 19.2 + 0.0227 
(c) Q = (1 − e )(hpi ) = (1 − 0.9273)(50 ) = 3.635 hp = 119,955 ft − lb min
Qc = hcr A∆t ft − lb min
A = Amin = 43.2C1.7 sq.in.
1
C = (Dw + Dg )
2
Dw = 3.2 in
7
80 
80 Pc
8
Dg =
=   = 22.3 in
π
π
1
C = (3.2 + 22.35) = 12.75 in
2
1.7
A = 43.2(12.75) = 3272 sq.in.
Figure AF 1
3272
A=
= 22.7 ft 2
144
hcr = 0.43 ft − lb min − sq.in. − F
Page 10 of 19
SECTION 14 – WORM GEARS
Q = Qc
119,955 = (0.43)(3272)(∆t )
∆t = 85 F
with t1 = 100 F
t2 = 185 F > 150 F
Therefore, special cooling is needed.
CHECK PROBLEMS
795.
A worm-gear speed reducer has a hardened-steel worm and a manganese-bronze
gear (Table AT 3); triple-threaded worm with Pc = 1.15278 in. , Dw = 3.136 in. ,
1
φn = 25o , b = 2 in. , mw = 12 , nw = 580 rpm . The output is 16 hp. Compute (a)
4
the dynamic load, (b) the endurance strength of the teeth and the indicated
service factor on strength, (c) the limiting wear load (is it good for indefinitely
continuous service?), (d) the efficiency and input hp, (e) the temperature rise of
the oil (estimate case area as Amin , i16.6). (f) Determine the tangential and radial
components of the tooth load. (g) Is this drive self-locking?
Solution:
33,000hp
Ft =
vmg
πDg ng
vmg =
12
nw 580
ng =
=
= 48.3 rpm
mw 20
PN
P m N (1.15278)(12 )(3)
Dg = c g = c w t =
= 13.21 in
π
vmg =
π
π (13.21)(48.3)
12
π
= 167 fpm
 1200 + vmg 
 Ft
(a) Fd = 
 1200 
 1200 + 167 
Fd = 
 Ft
 1200 
33,000(16 )
Ft =
= 3162 lb
167
 1200 + 167 
Fd = 
(3162 ) = 3602 lb
 1200 
Page 11 of 19
SECTION 14 – WORM GEARS
(b) Fs =
sYbPcn
π
=
sYbPc cos λ
π
N P (3)(1.15278)
tan λ = t c =
πDw
π (3.136 )
λ = 19.34o
For manganese-bronze, s = sn = 30,000 psi
For φn = 25o , Y = 0.470
(30,000)(0.470) 2 1 (1.15278)cos19.34
Fs =
 4
π
= 10,984 lb
Service factor
F 10,984
= 3.05
N sf = s =
3602
Fd
(c) Fw = Dg bK w
Dg = 13.21 in
b = 2.25 in
Table AT 27, hardened-steel worn and manganese bronze gear
φn = 25o
K w = 100
Fw = (13.21)(2.25)(100) = 2972 lb < Fd (= 3602 lb )
Therefore, not good for indefinitely continuous service
 cos φn − f tan λ 

(d) e = tan λ 
 cos φn tan λ + f 
πDwnw π (1.15278)(580 )
vr =
=
= 185.5 fpm
12 cos λ
12 cos19.34
0.32
0.32
f = 0.36 =
= 0.0488 ( 70 < vr < 3000 fpm )
vr
(185.5)0.36
 cos 25 − 0.0488 tan 19.34 
e = tan 19.34
 = 0.85 = 85%
 cos 25 tan 19.34 + 0.0488 
hp 16 hp
hpi = o =
= 18.82 hp
e
0.85
(e) Temperature rise, ∆t
Q = (1 − e )(hpi ) = (1 − 0.85)(18.82)(33,000 ) = 93,159 ft − lb min
Qc = hcr A∆t ft − lb min
A = Amin = 43.2C1.7 sq.in.
Page 12 of 19
SECTION 14 – WORM GEARS
1
(Dw + Dg )
2
1
C = (1.15278 + 13.21) = 7.18 in
2
1.7
A = 43.2(7.18) = 1233 sq.in.
Figure AF 1
1233
A=
= 8.6 ft 2
144
hcr = 0.47 ft − lb min − sq.in. − F
C=
Q = Qc
93,159 = (0.47 )(1233)(∆t )
∆t = 161 F
(f) Tangential components on the worm
 cos φn sin λ + f cos λ 
cos 25 sin 19.34 + 0.0488 cos19.34 
 = 3162
Wt = Ft 
 = 1305 lb
 cos 25 cos19.34 − 0.0488 sin 19.34 
 cos φn cos λ − f sin λ 
on the gear
Ft = 3162 lb
radial components
Ft sin φn
3162 sin 25
S=
=
= 1593 lb
cos φn cos λ − f sin λ cos 25 cos19.34 − 0.0488 sin 19.34
(g) λ = 19.34o > 5o , not self-locking
797.
A worm-gear speed reducer has a hardened-steel worm and a phosphor-bronze
gear. The lead angle of the 5-threaded worm λ = 28o57' , Pc = 1.2812 in. ,
1
φn = 25o , b = 2 in. , mw = 8 ; worm speed = 1750 rpm. The gear case is 35 3/8
2
in. high, 22 in. wide, 14 in. deep. Compute (a) the efficiency, (b) the limiting
wear load, the strength load, and the corresponding safe input and output
horsepowers. (c) The manufacturer rates this reducer at 53-hp input. Is this rating
conservative or risky? (d) What is the calculated temperature rise of the oil with
no special cooling? (e) The manufacturer specifies that for continuous service
power should not exceed 36.5 hp if there is to be no artificial cooling and if ∆t is
to be less than 90 F. Make calculations and decide whether the vendor is on the
safe side. (Data courtesy of the Cleveland Worm Gear Co.)
Solution:
Page 13 of 19
SECTION 14 – WORM GEARS
 cos φn − f tan λ 

(a) e = tan λ 
 cos φn tan λ + f 
λ = 28o57' = 28.95o
φn = 25o
πDwnw
vr =
12 cos λ
nw = 1750 rpm
N P
Dg = g c
π
N g = mw N t = (8)(5) = 40
Dg =
(40)(1.2812) = 16.31 in
π
NP
tan λ = t c
πDw
tan 28.95 =
(5)(1.2812)
πDw
Dw = 3.686 in
π (3.686 )(150 )
vr =
= 1923 fpm
12 cos 28o57'
0.32
0.32
f = 0.36 =
= 0.0210 ( 70 < vr < 3000 fpm )
vr
(1923)0.36
 cos 25 − 0.0210 tan 28.95 
e = tan 28.95
 = 0.9475 = 94.75%
 cos 25 tan 28.95 + 0.0210 
(b) Fw = Dg bK w
Dg = 16.31 in
b = 2.5 in
Table AT 27, hardened-steel worn and phosphor bronze gear
φn = 25o
K w = 100
Fw = (16.31)(2.5)(100) = 4078 lb
Fs =
sYbPcn
π
=
sYbPc cos λ
π
For phosphor-bronze, s = sn = 31,000 psi
For φn = 25o , Y = 0.470
(31,000)(0.470 )(2.5)(1.2812 )cos 28.95 = 13,000 lb
Fs =
π
Page 14 of 19
SECTION 14 – WORM GEARS
For safe input and output
 1200 + vmg 
 Ft
Fd = 
 1200 
πDg ng
vmg =
12
n
1750
ng = w =
= 218.75 rpm
mw
8
π (16.31)(218.75)
vmg =
= 934 fpm
12
Fw = Fd
 1200 + 934 
4078 = 
 Ft
 1200 
Ft = 2293 lb
Fv
(2293)(934) = 64.9 hp
safe output = hpo = t mg =
33,000
33,000
hp
64.9
safe input = hpi = o =
= 68.5 hp
e
0.9475
(c) 53-hp input < 68.5 hp. ∴ conservative.
(d) Q = (1 − e )(hpi ) = (1 − 0.9475)(68.5)(33,000) = 118,676 ft − lb min
Qc = hcr A∆t ft − lb min
A = 2[(22)(14) + (22)(35.375)] = 2172.5 sq.in.
Figure AF 1
2172.5
= 15 ft 2
A=
144
hcr = 0.45 ft − lb min − sq.in. − F
Q = Qc
118,676 = (0.45)(2172.5)(∆t )
∆t = 121.4 F
(e) ∆t ′ = 90 F
hpi′ ∆t ′
=
hpi ∆t
hpi′
90
=
68.5 124
hpi′ = 50.8 hp
Since 36.5 hp < 50.8 hp, therefore on the safe side.
Page 15 of 19
SECTION 14 – WORM GEARS
HEATING
799.
The input to a worm-gear reducer is 50.5 hp at 580 rpm of the 4-threaded worm.
The gear case is 22 x 31 x 45 in. in size; φn = 25o , Pc = 1.5 in , Dw = 4.432 in ,
f = 0.035 , room temperature = 80 F. Compute the steady-state temperature for
average cooling.
Solution:
NP
4(1.5)
tan λ = t c =
πDw π (4.432 )
λ = 23.3o
 cos φn − f tan λ 

e = tan λ 
 cos φn tan λ + f 
 cos 25 − 0.035 tan 23.3 
e = tan 23.3
 = 0.9025
 cos 25 tan 23.3 + 0.035 
Q = (1 − e )(hpi ) = (1 − 0.9025)(50.5)(33,000) = 162,484 ft − lb min
Qc = hcr A∆t ft − lb min
A = 2[(22)(31) + (31)(45)] = 4154 sq.in.
Figure AF 1
4154
A=
= 28.85 ft 2
144
hcr = 0.42 ft − lb min − sq.in. − F
Q = Qc
162,484 = (0.42)(4154 )(∆t )
∆t = 93 F
t1 = 80 F
t2 = 173 F
801.
A hardened-steel, 4-threaded worm drives a bronze gear; Dw = 1.875 in ,
Dg ≈ 14 in , Pc = 1.0 in , φn = 25o , area of case ≈ 1500 sq.in. , vr ≈ 1037 fpm ;
input = 20 hp at 1750 rpm of the worm; room temperature = 80 F. Compute the
steady-state temperature of the lubricant for average ventilation.
Solution:
NP
4(1.0 )
tan λ = t c =
πDw π (1.875)
λ = 34.2o
f =
0.32
0.32
=
= 0.0263 ( 70 < vr < 3000 fpm )
0.36
vr
(1037 )0.36
Page 16 of 19
SECTION 14 – WORM GEARS
 cos φn − f tan λ 

e = tan λ 
 cos φn tan λ + f 
 cos 25 − 0.0263 tan 34.2 
e = tan 34.2
 = 0.94
 cos 25 tan 34.2 + 0.0263 
Q = (1 − e )(hpi ) = (1 − 0.94)(20)(33,000) = 39,600 ft − lb min
Qc = hcr A∆t ft − lb min
A = 1500 sq.in.
Figure AF 1
1500
= 10.4 ft 2
A=
144
hcr = 0.46 ft − lb min − sq.in. − F
Q = Qc
39,600 = (0.46 )(1500)(∆t )
∆t = 57 F
t1 = 80 F
t2 = 137 F
802.
The input to a 4-threaded worm is measured to be 20.8 hp; Pc = 1.0 in , Dw = 2 in ,
φn = 25o . The area of the case is closely 1800 sq. in.; ambient temperature = 100
F; oil temperature = 180 F. Operation is at a steady thermal state. Compute the
indicated coefficient of friction.
Solution:
Qc = hcr A∆t ft − lb min
Figure AF 1
1800
A=
= 12.5 ft 2
144
hcr = 0.46 ft − lb min − sq.in. − F
A = 1800 sq.in.
∆t = 180 − 100 = 80 F
Qc = hcr A∆t = (0.46)(1800)(80) = 66,240 ft − lb min
Q = (1 − e )(hpi )(33,000) ft − lb min
Q = Qc
(1 − e)(hpi )(33,000) = 66,240
e = 0.9035
 cos φn − f tan λ 

e = tan λ 
 cos φn tan λ + f 
Page 17 of 19
SECTION 14 – WORM GEARS
tan λ =
N t Pc 4(1.0 )
=
πDw π (2 )
λ = 32.5o
 cos 25 − f tan 32.5 

0.9035 = tan 32.5
 cos 25 tan 32.5 + f 
0.5217 + 0.9035 f = 0.5774 − 0.4059 f
f = 0.0425
FORCE ANALYSIS
804.
1
The input to a 4-threaded worm is 21 hp at 1750 rpm; e = 90% , Dw = 2 in ,
4
o
Dg = 14 in , N g = 44 , φn = 25 . (a) From the horsepowers in and out, compute
the tangential forces on the worm Wt and the gear Ft . (b) Using this value of Ft ,
compute Wt from equation (k), i16.8, Text. (Check?) (c) Compute the separating
force. (d) What is the end thrust on the worm shaft? On the gear shaft?
Solution:
hpi = 21 hp
hpo = (hpi )(e ) = (21)(0.90) = 18.9 hp
N
Dg
mw = g =
N t Dw tan λ
44
14
=
4  1
 2  tan λ
 4
λ = 29.5o
 cos φn − f tan λ 

e = tan λ 
 cos φn tan λ + f 
 cos 25 − f tan 29.5 

0.90 = tan 29.5
 cos 25 tan 29.5 + f 
0.4615 + 0.90 f = 0.5128 − 0.32 f
f = 0.0420
πDg ng
vmg =
12
ng N t
=
nw N g
Page 18 of 19
SECTION 14 – WORM GEARS
ng
4
=
1750 44
ng = 159 rpm
vmg =
π (14)(159)
(a) Ft =
12
= 583 fpm
33,000hpo 33,000(18.9)
=
= 1070 lb
vmg
583
33,000hpi
vw
πD n π (2.25)(1750)
vw = w w =
= 1031 fpm
12
12
33,000(21)
Wt =
= 672 lb
1031
Wt =
 cos φn sin λ + f cos λ 
cos 25 sin 29.5 + 0.0420 cos 29.5 
 = 1070
(b) Wt = Ft 
 = 672 lb
 cos 25 cos 29.5 − 0.0420 sin 29.5 
 cos φn cos λ − f sin λ 
Ft sin φn
1070 sin 25
(c) S =
=
= 589 lb
cos φn cos λ − f sin λ cos 25 cos 29.5 − 0.0420 sin 29.5
(d) End thrust
Worm shaft = Ft = 1070 lb
Gear shaft = Wt = 672 lb
- end -
Page 19 of 19
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
LEATHER BELTS
DESIGN PROBLEMS
841.
A belt drive is to be designed for F1 F2 = 3 , while transmitting 60 hp at 2700
rpm of the driver D1 ; mw ≈ 1.85 ; use a medium double belt, cemented joint, a
squirrel-cage, compensator-motor drive with mildly jerking loads; center distance
is expected to be about twice the diameter of larger pulley. (a) Choose suitable
iron-pulley sizes and determine the belt width for a maximum permissible
s = 300 psi . (b) How does this width compare with that obtained by the ALBA
procedure? (c) Compute the maximum stress in the straight port of the ALBA
belt. (d) If the belt in (a) stretches until the tight tension F1 = 525 lb ., what is
F1 F2 ?
Solution:
(a) Table 17.1, Medium Double Ply,
Select D1 = 7 in . min.
20
t=
in
64
πD n π (7 )(2700 )
vm = 1 1 =
= 4948 fpm
12
12
4000 fpm < 4948 fpm < 6000 fpm
(F − F2 )vm
hp = 1
33,000
F1 − F2 )(4948)
(
60 =
33,000
F1 − F2 = 400 lb
F1 = 3F2
3F2 − F2 = 400 lb
F2 = 200 lb
F1 = 3F2 = 3(200 ) = 600 lb
F1 = sbt
sd = 300η
For cemented joint, η = 1.0
sd = 300 psi
 20 
F1 = 600 = (300)(b ) 
 64 
b = 6.4 in
say b = 6.5 in
757
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
(b) ALBA Procedure
hp = (hp in., Table 17.1)(bC mC p )(C f1 C f 2 L)
Table 17.1, vm = 4948 fpm
Medium Double Ply
hp in = 12.448
Table 17.2
Squirrel cage, compensator, starting
Cm = 0.67
Pulley Size, D1 = 7 in
C p = 0.6
Jerky loads, C f = 0.83
hp = 60 = (12.448)(b )(0.67 )(0.6)(0.83)
b = 14.5 in
say b = 15 in
(c) s =
F1
=
ηbt
1
1
600
= 128 psi
 20 
(1)(15) 
 64 
1
1
1
(d) 2 Fo2 = F12 + F22 = (600) 2 + (200 )2
Fo = 373.2 lb
F1 = 525 lb
1
2
1
2
1
2
2
2(373.2 ) = (525) + F
F2 = 247 lb
F1 525
=
= 2.1255
F2 247
842.
A 20-hp, 1750 rpm, slip-ring motor is to drive a ventilating fan at 330 rpm. The
horizontal center distance must be about 8 to 9 ft. for clearance, and operation is
continuous, 24 hr./day. (a) What driving-pulley size is needed for a speed
recommended as about optimum in the Text? (b) Decide upon a pulley size (iron
or steel) and belt thickness, and determine the belt width by the ALBA tables. (c)
Compute the stress from the general belt equation assuming that the applicable
coefficient of friction is that suggested by the Text. (d) Suppose the belt is
installed with an initial tension Fo = 70 lb in . (§17.10), compute F1 F2 and the
stress on the tight side if the approximate relationship of the operating tensions
1
1
1
and the initial tensions is F12 + F22 = 2 Fo2 .
758
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Solution:
vm = 4000 to 4500 fpm
assume vm = 4250 fpm
πD n
vm = 1 1
12
πD (1750 )
4250 = 1
12
D1 = 9.26 in
say D1 = 10 in
23
in
64
Minimum pulley diameter for vm ≈ 4250 fpm , D1 = 10 in
(b) Using Heavy Double Ply Belt, t =
Use D1 = 10 in
πD n π (10 )(1750)
vm = 1 1 =
= 4581 fpm
12
12
ALBA Tables
hp = (hp in., Table 17.1)(bC mC p )(C f1 C f 2 L)
hp in = 13.8
Slip ring motor, Cm = 0.4
Pulley Size, D1 = 10 in
C p = 0.7
Table 17.7, 24 hr/day, continuous
N sf = 1.8
Assume C f = 0.74
hp = (1.8)(20 ) = (13.8)(b )(0.4)(0.7 )(0.74 )
b = 12.59 in
use b = 13 in
(c) General belt equation
 12 ρvs2  e fθ − 1 


F1 − F2 = bt  s −
32.2  e fθ 

4581
= 76.35 fps
60
ρ = 0.035 lb cu. in. for leather
23
t=
in
64
b = 13 in
vs =
759
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
33,000(1.8)(20 )
= 260 lb
4581
f = 0.3 on iron or steel
D − D1
θ ≈π ± 2
C
C = 8 ~ 9 ft use 8.5 ft
 1750 
D2 = 
(10 ) = 53 in
 330 
53 − 10
θ =π −
= 2.72 rad
8.5(12 )
fθ = (0.3)(2.72 ) = 0.816
F1 − F2 =
e fθ − 1 e 0.816 − 1
= 0.816 = 0.5578
e fθ
e
2
 23   12(0.035)(76.35) 
F1 − F2 = 260 = (13)   s −
 (0.5578)
32.2
 64  

s = 176 psi
1
1
1
(d) F12 + F22 = 2 Fo2
Fo = (70 lb in )(13 in ) = 910 lb
F1 − F2 = 260 lb
F2 = F1 − 260 lb
1
2
1
1
F1 + ( F1 − 260 )2 = 2(910)2 = 60.33
F1 = 1045 lb
F2 = 1045 − 260 = 785 lb
1045
= 224 psi
 23 
(13) 
 64 
F1 1045
=
= 1.331
F2 785
s=
F1
=
bt
843.
A 100-hp squirrel-cage, line-starting electric motor is used to drive a Freon
reciprocating compressor and turns at 1140 rpm; for the cast-iron motor pulley,
D1 = 16 in ; D2 = 53 in , a flywheel; cemented joints;l C = 8 ft . (a) Choose an
appropriate belt thickness and determine the belt width by the ALBA tables. (b)
Using the design stress of §17.6, compute the coefficient of friction that would be
needed. Is this value satisfactory? (c) Suppose that in the beginning, the initial
tension was set so that the operating F1 F2 = 2 . Compute the maximum stress in
a straight part. (d) The approximate relation of the operating tensions and the
760
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
1
2
1
2
2
1
2
o
initial tension Fo is F1 + F = 2 F . For the condition in (c), compute Fo . Is it
reasonable compared to Taylor’s recommendation?
Solution:
(a) Table 17.1
πD n π (16)(1140)
vm = 1 1 =
= 4775 fpm
12
12
Use heavy double-ply belt
23
t=
in
64
hp in = 14.1
hp = (hp in., Table 17.1)(bC mC p ) C f1 C f 2 L
(
)
line starting electric motor , C m = 0.5
Table 17.7, squirrel-cage, electric motor, line starting, reciprocating compressor
N sf = 1.4
D1 = 16 in , C p = 0.8
assume, C f = 0.74
hp = (1.4)(100 ) = 140 hp
hp = 140 = (14.1)(b )(0.5)(0.8)(0.74)
b = 33.5 in
use b = 34 in
(b) §17.6, s d = 400η
η = 1.00 for cemented joint.
s d = 400 psi
 12 ρvs2  e fθ − 1 


F1 − F2 = bt  s −
32.2  e fθ 

4775
vs =
= 79.6 fps
60
ρ = 0.035 lb cu. in. for leather
23
t=
in
64
b = 34 in
33,000(1.4 )(100)
F1 − F2 =
= 968 lb
4775
2
12(0.035)(79.6 )  e fθ − 1 
 23  
F1 − F2 = 968 = (34)  400 −
 fθ 
32.2
 64  

 e
761
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
e fθ − 1
= 0.2496
e fθ
fθ = 0.28715
D − D1
θ ≈π ± 2
C
C = 8 ft
53 − 16
θ =π −
= 2.7562 rad
8(12 )
f (2.7562) = 0.28715
f = 0.1042 < 0.3
Therefore satisfactory.
(c) F1 − F2 = 968 lb
F1 = 2F2
2 F2 − F2 = 968 lb
F1 = 2 F2 = 2(968) = 1936 lb
F
1936
s= 1 =
= 159 psi
bt
 23 
(34) 
 64 
(d) F1 = 1936 lb , F2 = 968 lb
1
1
1
2 Fo2 = F12 + F22
1
2
o
1
1
2 F = (1936) 2 + (968)2
Fo = 1411 lb
Fo =
844.
1411
= 41.5 lb in of width is less than Taylor’s recommendation and is reasonable.
34
A 50-hp compensator-started motor running at 865 rpm drives a reciprocating
compressor for a 40-ton refrigerating plant, flat leather belt, cemented joints. The
diameter of the fiber driving pulley is 13 in., D2 = 70 in ., a cast-iron flywheel;
C = 6 ft.11 in. Because of space limitations, the belt is nearly vertical; the
surroundings are quite moist. (a) Choose a belt thickness and determine the width
by the ALBA tables. (b) Using recommendations in the Text, compute s from
the general belt equation. (c) With this value of s , compute F1 and F1 F2 . (d)
1
1
1
Approximately, F12 + F22 = 2 Fo2 , where Fo is the initial tension. For the
condition in (c), what should be the initial tension? Compare with Taylor, §17.10.
(e) Compute the belt length. (f) The data are from an actual drive. Do you have
any recommendations for redesign on a more economical basis?
762
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Solution:
(a) vm =
πD1n1
=
π (13)(865)
= 2944 fpm
12
12
Table 17.1, use Heavy Double Ply,
Dmin = 9 in for vm = 2944 fpm
belts less than 8 in wide
23
t=
in
64
(
)
hp = (hp in., Table 17.1)(bC mC p ) C f1 C f 2 L
hp in = 9.86
Table 17.2
Cm = 0.67
C p = 0.8
C f = (0.74 )(0.83) = 0.6142
Table 17.7, electric motor, compensator-started (squirrel cage) and reciprocating
compressor
N sf = 1.4
hp = (1.4 )(50) = 70 hp
hp = 70 = (9.86)(b)(0.67 )(0.8)(0.6142)
b = 21.6 in
use b = 25 in
(b) General Belt Equation
 12 ρvs2  e fθ − 1 


F1 − F2 = bt  s −
32.2  e fθ 

b = 25 in
23
t=
in
64
ρ = 0.035 lb cu. in. for leather
2944
vs =
= 49.1 fps
60
Leather on iron, f = 0.3
D − D1
θ =π − 2
C
70 − 13
= 2.455 rad
6(12) + 11
fθ = (0.3)(2.455) = 0.7365
θ =π −
763
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
e fθ − 1 e 0.7365 − 1
= 0.7365 = 0.5212
e fθ
e
F1 − F2 =
33,000(1.4 )(50)
= 785 lb
2944
2
 23   12(0.035)(49.1) 
F1 − F2 = 785 = (25)   s −
 (0.5212 )
32.2
 64  

s = 199 psi
Cemented joint, η = 1.0
s = 199 psi
(c) F1 = sbt = (199)(25)
23 
 = 1788 lb
 64 
F2 = 1788 − 785 = 1003 lb
F1 1788
=
= 1.783
F2 1003
1
2
o
1
2
1
2
2
(d) 2 F = F1 + F
1
1
1
2Fo2 = (1788) 2 + (1003) 2
Fo = 1367 lb
Fo =
1367
= 54.7 lb in
25
Approximately less than Taylor’s recommendation ( = 70 lb/in.)
(e)
2
(
D2 − D1 )
L ≈ 2C + 1.57(D2 + D1 ) +
L = 2[(6)(12 ) + 11] + 1.57(70 + 13) +
4C
(70 − 13)2
= 306 in
4[(6)(12 ) + 11]
(f) More economical basis
πD n
vm = 1 1
12
πD (865)
4500 = 1
12
D1 = 19.87 in
use D1 = 20 in
764
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
CHECK PROBLEMS
846.
An exhaust fan in a wood shop is driven by a belt from a squirrel-cage motor that
runs at 880 rpm, compensator started. A medium double leather belt, 10 in. wide
is used; C = 54 in .; D1 = 14 in . (motor), D2 = 54 in ., both iron. (a) What
horsepower, by ALBA tables, may this belt transmit? (b) For this power,
compute the stress from the general belt equation. (c) For this stress, what is
F1 F2 ? (d) If the belt has stretched until s = 200 psi on the tight side, what is
F1 F2 ? (e) Compute the belt length.
Solution:
(a) For medium double leather belt
20
t=
in
64
hp = (hp in )(b )C mC pC f
Table 17.1 and 17.2
Cm = 0.67
C p = 0.8
C f = 0.74
b = 10 in
πD n π (14)(880)
vm = 1 1 =
= 3225 fpm
12
12
hp in = 6.6625
hp = (6.6625)(10 )(0.67 )(0.8)(0.74) = 26.43 hp
 12 ρvs2  e fθ − 1 


(b) F1 − F2 = bt  s −
32.2  e fθ 

b = 10 in
20
t=
in
64
ρ = 0.035 lb cu. in.
3225
vs =
= 53.75 fps
60
D − D1
θ =π − 2
C
54 − 14
θ =π −
= 2.4 rad
54
Leather on iron f = 0.3
fθ = (0.3)(2.4 ) = 0.72
765
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
e fθ − 1 e 0.72 − 1
= 0.72 = 0.51325
e fθ
e
33,000(26.43)
F1 − F2 =
= 270 lb
3225
2
 20   12(0.035)(53.75) 
F1 − F2 = 270 = (10 )   s −
 (0.51325)
32.2
 64  

s = 206 psi
 20 
(c) F1 = sbt = (206 )(10)  = 644 lb
 64 
F2 = 644 − 270 = 374 lb
F1 644
=
= 1.72
F2 374
(d) s = 200 psi
 20 
F1 = sbt = (200 )(10)  = 625 lb
 64 
F2 = 625 − 270 = 355 lb
F1 625
=
= 1.76
F2 355
(e)
2
(
D2 − D1 )
L ≈ 2C + 1.57(D2 + D1 ) +
4C
(54 − 14)2 = 222 in
L = 2(54) + 1.57(54 + 14) +
4(54 )
847.
A motor is driving a centrifugal compressor through a 6-in. heavy, single-ply
leather belt in a dusty location. The 8-in motor pulley turns 1750 rpm;
D2 = 12 in . (compressor shaft); C = 5 ft . The belt has been designed for a net
belt pull of F1 − F2 = 40 lb in of width and F1 F2 = 3 . Compute (a) the
horsepower, (b) the stress in tight side. (c) For this stress, what needed value of
f is indicated by the general belt equation? (d) Considering the original
data,what horsepower is obtained from the ALBA tables? Any remarks?
Solution:
(a) vm =
πD1n1
12
=
π (8)(1750)
12
= 3665 fpm
b = 6 in
F1 − F2 = (40)(6) = 240 lb
766
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
hp =
(F1 − F2 )vm = (240)(3665) = 26.65 hp
33,000
33,000
(b) F1 = 3F2
3F2 − F2 = 240 lb
F2 = 120 lb
F1 = 360 lb
F
s= 1
bt
For heavy single-ply leather belt
13
t=
in
64
360
s=
= 295 psi
 13 
(6) 
 64 
 12 ρvs2  e fθ − 1 


(c) F1 − F2 = bt  s −
32.2  e fθ 

ρ = 0.035 lb cu. in.
3665
vs =
= 61.1 fps
60
F1 − F2 = 240 lb
2
12(0.035)(61.1)  e fθ − 1 
 13  
F1 − F2 = 240 = (6)   295 −
 fθ 
32.2
 64  

 e
e fθ − 1
= 0.7995
e fθ
D − D1
θ =π − 2
C
12 − 8
θ =π −
= 3.075 rad
5(12)
e fθ = 4.9875
fθ = 1.607
f (3.075) = 1.607
f = 0.5226
(d) ALBA Tables (Table 17.1 and 17.2)
hp = (hp in )(b )C mC pC f
vm = 3665 fpm
hp in = 6.965
767
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
b = 10 in
Cm = 1.0 (assumed)
C p = 0.6
C f = 0.74
hp = (6.965)(6 )(1.0 )(0.6 )(0.74) = 18.6 hp < 26.65 hp
848.
A 10-in. medium double leather belt, cemented joints, transmits 60 hp from a 9in. paper pulley to a 15-in. pulley on a mine fab; dusty conditions. The
compensator-started motor turns 1750 rpm; C = 42 in . This is an actual
installation. (a) Determine the horsepower from the ALBA tables. (b) Using the
general equation, determine the horsepower for this belt. (c) Estimate the service
factor from Table 17.7 and apply it to the answer in (b). Does this result in better
or worse agreement of (a) and (b)? What is your opinion as to the life of the belt?
Solution:
πD n π (9)(1750 )
vm = 1 1 =
= 4123 fpm
12
12
(a) hp = (hp in )(b )C mC pC f
Table 17.1 and 17.2
Medium double leather belt
20
t=
in
64
vm = 4123 fpm
hp in = 11.15
Cm = 0.67
C p = 0.7
C f = 0.74
b = 10 in
hp = (11.15)(10)(0.67 )(0.7 )(0.74) = 38.7 hp
 12 ρvs2  e fθ − 1 


(b) F1 − F2 = bt  s −
32.2  e fθ 

b = 10 in
ρ = 0.035 lb cu. in.
s = 400η
η = 1.0 cemented joint
s = 400 psi
D − D1
θ =π − 2
C
768
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
15 − 9
= 2.9987 rad
42
Leather on paper pulleys, f = 0.5
fθ = (0.5)(2.9987 ) = 1.5
θ =π −
e fθ − 1
= 0.77687
e fθ
4123
= 68.72 fps
vs =
60
2
12(0.035)(68.72) 
 20  
F1 − F2 = (10 )  400 −
 (0.77687 ) = 822 lb
32.2
 64  

hp =
(F1 − F2 )vm = (822)(4123) = 102.7 hp
33,000
33,000
(c) Table 17.7
N sf = 1.6
102.7
= 64.2 hp < 102.7 hp
1. 6
Therefore, better agreement
hp =
Life of belt, not continuous, 60 hp > 38.7 hp .
MISCELLANEOUS
849.
Let the coefficient of friction be constant. Find the speed at which a leather belt
may transmit maximum power if the stress in the belt is (a) 400 psi, (b) 320 psi.
(c) How do these speeds compare with those mentioned in §17.9, Text? (d)
Would the corresponding speeds for a rubber belt be larger or smaller? (HINT:
Try the first derivative of the power with respect to velocity.)
Solution:
 12 ρvs2  e fθ − 1 


F1 − F2 = bt  s −
32.2  e fθ 

(F − F2 )vm
hp = 1
33,000
60(F1 − F2 )vs
hp =
33,000
hp =
60vs bt  12 ρvs2  e fθ − 1 
s −


33,000 
32.2  e fθ 
60bt  e fθ − 1  12 ρvs2 

 s −
vs
hp =
33,000  e fθ 
32.2 
769
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
d (hp )
60bt  e fθ − 1   12 ρvs2  24 ρvs2 

  s −
−
=
=0
d (vs ) 33,000  e fθ  
32.2  32.2 
36 ρvs2
s=
32.2
ρ = 0.035 lb cu. in.
(a) s = 400 psi
36(0.035)vs2
32.2
vs = 101.105 fps
vm = 6066 fpm
400 =
(b) s = 320 psi
36(0.035)vs2
32.2
vs = 90.431 fps
vm = 5426 fpm
320 =
(c) Larger than those mentioned in §17.9 (4000 – 4500 fpm)
(d) Rubber belt, ρ = 0.045 lb cu. in.
(a) s = 400 psi
36(0.045)vs2
32.2
vs = 89.166 fps
vm = 5350 fpm < 6066 fpm
400 =
Therefore, speeds for a rubber belt is smaller.
850.
A 40-in. pulley transmits power to a 20-in. pulley by means of a medium double
leather belt, 20 in. wide; C = 14 ft , let f = 0.3 . (a) What is the speed of the 40-in
pulley in order to stress the belt to 300 psi at zero power? (b) What maximum
horsepower can be transmitted if the indicated stress in the belt is 300 psi? What
is the speed of the belt when this power is transmitted? (See HINT in 849).
Solution:
 12 ρvs2  e fθ − 1 
 fθ 
F1 − F2 = bt  s −
32
.
2
 e


770
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
hp =
60(F1 − F2 )vs
33,000
hp =
60bt  e fθ − 1  12 ρvs2 

 s −
vs
33,000  e fθ 
32.2 
d (hp )
60bt  e fθ − 1   12 ρvs2  24 ρvs2 

  s −
−
=
=0
d (vs ) 33,000  e fθ  
32.2  32.2 
36 ρvs2
for maximum power
32.2
(a) At zero power:
s=
12 ρvs2
s=
32.2
s = 300 psi
ρ = 0.035 lb cu. in.
12(0.035)vs2
32.2
vs = 151.6575 fps
vm = 9100 fpm
300 =
Speed, 40 in pulley, n2 =
12vm 12(9100)
=
= 869 rpm
πD2
π (40 )
(b) Maximum power
36 ρvs2
s=
32.2
36(0.035)vs2
300 =
32.2
vs = 87.5595 fps
vm = 5254 fpm
60bt  e fθ − 1  12 ρvs2 

 s −
vs
hp =
33,000  e fθ 
32.2 
20
in
64
b = 20 in
D − D1
θ =π − 2
C
40 − 20
θ =π −
= 3.0225 rad
14(12)
f = 0.3
t=
771
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
fθ = (0.3)(3.0225) = 0.90675
e fθ − 1
= 0.5962
e fθ
 20 
60(20 ) 
2
 64  (0.5962 )300 − 12(0.035)(87.5595)  (87.5595) = 118.64
hp =


33,000
32.2


vm = 5254 fpm
AUTOMATIC TENSION DEVICES
851.
An ammonia compressor is driven by a 100-hp synchronous motor that turns
1200 rpm; 12-in. paper motor pulley; 78-in. compressor pulley, cast-iron;
C = 84 in . A tension pulley is placed so that the angle of contact on the motor
pulley is 193o and on the compressor pulley, 240o. A 12-in. medium double
leather belt with a cemented joint is used. (a) What will be the tension in the
tight side of the belt if the stress is 375 psi? (b) What will be the tension in the
slack side? (c) What coefficient of friction is required on each pulley as indicated
by the general equation? (d) What force must be exerted on the tension pulley to
hold the belt tight, and what size do you recommend?
Solution:
(a) F1 = sbt
b = 12 in
20
t=
in
64
 20 
F1 = (375)(12 ) 
 64 
33,000hp
vm
πD n π (12)(1200)
vm = 1 1 =
= 3770 fpm
12
12
Table 17.7, N sf = 1.2
(b) F1 − F2 =
33,000(1.2 )(100 )
= 1050 lb
3770
F2 = F1 − 1050 = 1406 − 1050 = 356 lb
F1 − F2 =
 12 ρvs2  e fθ − 1 
 fθ 
(c) F1 − F2 = bt  s −
32
.
2

 e

3770
vs =
= 62.83 fps
60
772
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
ρ = 0.035 lb cu. in.
12(0.035)(62.83)  e fθ − 1 
 20  
1050 = (12 )  375 −
 e fθ 
32.2
 64  


e fθ − 1
= 0.8655
e fθ
fθ = 2.006
Motor pulley
 π 
θ = 193o = 193
 = 3.3685 rad
 180 
f (3.3685) = 2.006
f = 0.5955
Compressor Pulley
 π 
θ = 2403o = 240
 = 4.1888 rad
 180 
f (4.1888) = 2.006
f = 0.4789
(d) Force:
Without tension pulley
D − D1
78 − 12
θ1 = π − 2
=π −
= 2.356 rad
C
84
D − D1
78 − 12
θ2 = π + 2
=π +
= 3.9273 rad
C
84
773
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
α1 = θ1′ − θ1 −
2
= 3.3685 − 2.356 −
π − 2.356
2
= 0.6197 rad = 35.5o
3.9273 − π
+ 4.1888 − 3.9273 = 0.6544 rad = 37.5o
2
2
Q = F1 (sin α1 + sin α 2 ) = 1406(sin 35.5 + sin 37.5) = 1672 lb of force exerted
Size of pulley; For medium double leather belt,
vm = 3770 fpm , width = 12 in > 8 in
D = 6 + 2 = 8 in
α2 =
θ2 − π
π − θ1
+ θ 2′ − θ 2 =
774
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
852.
A 40-hp motor, weighing 1915 lb., runs at 685 rpm and is mounted on a pivoted
3
base. In Fig. 17.11, Text, e = 10 in ., h = 19 in . The center of the 11 ½-in.
16
motor pulley is 11 ½ in. lower than the center of the 60-in. driven pulley;
C = 48 in . (a) With the aid of a graphical layout, find the tensions in the belt for
maximum output of the motor if it is compensator started. What should be the
width of the medium double leather belt if s = 300 psi ? (c) What coefficient of
friction is indicated by the general belt equation? (Data courtesy of Rockwood
Mfg. Co.)
Solution:
(a)
R = 1915 lb
Graphically
b ≈ 26 in
a ≈ 9 in
[∑ M
B
=0
]
eR = F1a + F2b
(10)(1915) = (F1 )(9) + (F2 )(26)
9 F1 + 26 F2 = 19,150
For compensator started
hp = 1.4(rated hp ) = 1.4(40 ) = 56 hp
33,000hp
F1 − F2 =
vm
775
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
vm =
πD1n1
π (11.5)(685)
= 2062 fpm
12
33,000(56)
F1 − F2 =
= 896 lb
2062
F2 = F1 − 896
Substituting
9 F1 + 26(F1 − 896) = 19,150
F1 = 1213 lb
F2 = 1213 − 896 = 317 lb
12
=
For medium leather belt, t =
20
in
64
F1 = sbt
 20 
1213 = (300)(b ) 
 64 
b = 13 in
 12 ρvs2  e fθ − 1 


(c) F1 − F2 = bt  s −
32.2  e fθ 

2062
vs =
= 34.37 fps
60
ρ = 0.035 lb cu. in.
12(0.035)(34.37 )   e fθ − 1 
 20  
896 = (13)  300 −
  e fθ 
32.2
 64  


e fθ − 1
= 0.775
e fθ
fθ = 1.492
D − D1
60 − 11.5
θ =π − 2
=π −
= 2.1312 rad
C
48
f (2.1312) = 1.492
f = 0.70
853.
A 50-hp motor, weighing 1900 lb., is mounted on a pivoted base, turns 1140 rpm,
3
and drives a reciprocating compressor; in Fig. 17.11, Text, e = 8 in .,
4
5
h = 17 in . The center of the 12-in. motor pulley is on the same level as the
16
center of the 54-in. compressor pulley; C = 40 in . (a) With the aid of a graphical
layout, find the tensions in the belt for maximum output of the motor if it is
compensator started. (b) What will be the stress in the belt if it is a heavy double
776
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
leather belt, 11 in. wide? (c) What coefficient of friction is indicated by the
general belt equation? (Data courtesy of Rockwood Mfg. Co.)
Solution:
(a) For compensator-started
hp = 1.4(50 ) = 70 hp
33,000hp
F1 − F2 =
vm
πD n π (12 )(1140)
vm = 1 1 =
= 3581 fpm
12
12
33,000(70)
F1 − F2 =
= 645 lb
2062
b ≈ 25 in
a ≈ 5 in
R = 1900 lb
eR = F 1a + F2b
(8.75)(1900 ) = F 1 (5) + F2 (25)
F 1+5F2 = 3325 lb
645 + F 2+5 F2 = 3325 lb
F2 = 447 lb
F1 = 645 + F2 = 645 + 447 = 1092 lb
(b) For heavy double leather belt
23
t=
in
64
b = 11 in
777
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
1092
= 276 psi
 20 
(11) 
 64 
 12 ρvs2  e fθ − 1 


(c) F1 − F2 = bt  s −
32.2  e fθ 

3581
vs =
= 59.68 fps
60
ρ = 0.035 lb cu. in.
s=
F1
=
bt
12(0.035)(59.68)  e fθ − 1 
 23  
645 = (11)  276 −
 e fθ 
32.2
 64  


fθ = 1.241
D − D1
54 − 12
θ =π − 2
=π −
= 2.092 rad
C
40
f (2.092) = 1.492
f = 0.60
RUBBER BELTS
854.
A 5-ply rubber belt transmits 20 horsepower to drive a mine fan. An 8-in., motor
pulley turns 1150 rpm; D2 = 36 in ., fan pulley; C = 23 ft . (a) Design a rubber
belt to suit these conditions, using a net belt pull as recommended in §17.15,
Text. (b) Actually, a 9-in., 5-ply Goodrich high-flex rubber belt was used. What
are the indications for a good life?
Solution:
(a) θ = π −
36 − 8
D2 − D1
=π −
= 3.040 rad = 174o
23(12 )
C
Kθ = 0.976
bv N K
hp = m p θ
2400
Kθ = 0.976
πD n π (8)(1150)
vm = 1 1 =
= 2409 fpm
12
12
Np = 5
hp = 20 =
b(2409 )(5)(0.976)
2400
b = 4.1 in
min. b = 5 in
778
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
(b) With b = 9 in is safe for good life.
855.
A 20-in., 10-ply rubber belt transmits power from a 300-hp motor, running at 650
rpm, to an ore crusher. The center distance between the 33-in. motor pulley and
the 108-in. driven pulley is 18 ft. The motor and crusher are so located that the
belt must operate at an angle 75o with the horizontal. What is the overload
capacity of this belt if the rated capacity is as defined in §17.15, Text?
Solution:
bv N
hp = m p
2400
b = 20 in
πD n π (33)(650 )
vm = 1 1 =
= 5616 fpm
12
12
N p = 10
hp =
(20 )(5616)(10 ) = 468 hp
2400
Overlaod Capacity =
468 − 300
(100%) = 56%
300
V-BELTS
NOTE: If manufacturer’s catalogs are available, solve these problems from catalogs as
well as from data in the Text.
856.
A centrifugal pump, running at 340 rpm, consuming 105 hp in 24-hr service, is to
be driven by a 125-hp, 1180-rpm, compensator-started motor; C = 43 to 49 in .
Determine the details of a multiple V-belt drive for this installation. The B.F.
Goodrich Company recommended six C195 V-belts with 14.4-in. and 50-in.
sheaves; C ≈ 45.2 in .
Solution:
Table 17.7
N sf = 1.2 + 0.2 = 1.4 (24 hr/day)
Design hp = N sf (transmitted hp) = (1.4 )(125) = 175 hp
Fig. 17.4, 175 hp, 1180 rpm
Dmin = 13 in , D-section
D2 1180 50
=
=
D1 340 14.4
use D1 = 14.4 in > 13 in
D2 = 50 in
779
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
vm =
πD1n1
12
=
π (14.4 )(1180)
12
= 4449 fpm
  103  0.09
v2  v
c
 −
Rated hp =  a
− e m6  m3
K d D1
10  10
  vm 

Table 17.3, D-section
a = 18.788 , c = 137.7 , e = 0.0848
D
Table 17.4, 2 = 3.47
D1
K d = 1.14
0.09

 103 
(4449 )2  4449 = 28.294 hp
137.7
 −
Rated hp = 18.788
− (0.0848)
(1.14)(14.4)
106  103

 4449 

Back to Fig. 17.14, C-section must be used.
a = 8.792 , c = 38.819 , e = 0.0416
  103  0.09
v2  v
c
 −
Rated hp =  a
− e m6  m3
K d D1
10  10
  vm 

0
.
09

 103 
(4449 )2  4449 = 20.0 hp
38.819
 −
Rated hp = 8.792
− (0.0416 )
(1.14 )(14.4 )
10 6  103

 4449 

Adjusted rated hp = Kθ K L (rated hp )
Table 17.5,
D2 − D1 50 − 14.4
=
= 0.77
C
46
Kθ = 0.88
Table 17.6
(D − D1 )2
L ≈ 2C + 1.57(D2 + D1 ) + 2
4C
2
(
50 − 14.4)
L = 2(46) + 1.57(50 + 14.4 ) +
= 200 in
4(46 )
use C195, L = 197.9 in
K L = 1.07
Adjusted rated hp = (0.88)(1.07 )(20 ) = 18.83 hp
Design hp
175
No. of belts =
=
= 9.3 belts use 9 belts
Adjusted rated hp 18.83
Use 9 , C195 V-belts with 14.4 in and 50 in sheaves
780
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
B + B 2 − 32(D2 − D1 )
C=
16
B = 4 L − 6.28(D2 + D1 ) = 4(197.9) − 6.28(50 + 14.4 ) = 387.2 in
2
C=
857.
387.2 +
(387.2 )2 − 32(50 − 14.4)2
16
= 44.9 in
A 50-hp, 1160-rpm, AC split-phase motor is to be used to drive a reciprocating
pump at a speed of 330 rpm. The pump is for 12-hr. service and normally
requires 44 hp, but it is subjected to peak loads of 175 % of full load; C ≈ 50 in .
Determine the details of a multiple V-belt drive for this application. The Dodge
Manufacturing Corporation recommended a Dyna-V Drive consisting of six
5V1800 belts with 10.9-in. and 37.5-in. sheaves; C ≈ 50.2 in .
Solution:
Table 17.7, (12 hr/day)
N sf = 1.4 − 0.2 = 1.2
Design hp = (1.2 )(1.75)(50 ) = 105 hp
Fig. 17.4, 105 hp, 1160 rpm
Dmin = 13 in , D-section
D2 1160 46.4
=
≈
D1 330 13.2
use D1 = 13.2 in > 13 in
D2 = 46.4 in
πD n π (13.2 )(1160)
vm = 1 1 =
= 4009 fpm
12
12
  103  0.09
vm2  vm
c


Rated hp =  a
 − K D − e 10 6  103
  vm 

d 1
Table 17.3, D-section
a = 18.788 , c = 137.7 , e = 0.0848
D
46.4
Table 17.4, 2 =
= 3.5
D1 13.2
K d = 1.14
0.09
2

 103 
(
137.7
4009 )  4009
 −

Rated hp = 18.788
− (0.0848)
= 24.32 hp
(1.14)(13.2)
106  103

 4009 

Back to Fig. 17.14, C-section must be used.
a = 8.792 , c = 38.819 , e = 0.0416
Dmin = 9 in
781
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
D2 1160 32
=
≈
D1 330 9.1
use D1 = 9.1 in
πD n π (9.1)(1160)
= 2764 fpm
vm = 1 1 =
12
12
0.09

 103 
(2764 )2  2764 = 10.96 hp
38.819
 −
Rated hp = 8.792
− (0.0416 )
(1.14 )(9.1)
106  103

 2764 

Adjusted rated hp = Kθ K L (rated hp )
Table 17.5,
D2 − D1 32 − 9.1
=
= 0.458
C
50
Kθ = 0.935
Table 17.6
2
(
D2 − D1 )
L ≈ 2C + 1.57(D2 + D1 ) +
4C
(32 − 9.1)2 = 167 in
L = 2(50 ) + 1.57(32 + 9.1) +
4(50 )
use C158, L = 160.9 in
K L = 1.02
Adjusted rated hp = (0.935)(1.02 )(10.96 ) = 10.45 hp
Design hp
105
No. of belts =
=
= 10 belts
Adjusted rated hp 10.43
B + B 2 − 32(D2 − D1 )
16
B = 4 L − 6.28(D2 + D1 ) = 4(160.9) − 6.28(32 + 9.1) = 385.5 in
2
C=
C=
385.5 +
(385.5)2 − 32(32 − 9.1)2
16
Use 10-C158 belts, D1 = 9.1 in
D2 = 32 in , C = 46.8 in
858.
= 46.8 in
A 200-hp, 600-rpm induction motor is to drive a jaw crusher at 125 rpm; starting
load is heavy; operating with shock; intermittent service; C = 113 to 123 in .
Recommend a multiple V-flat drive for this installation. The B.F. Goodrich
Company recommended eight D480 V-belts with a 26-in. sheave and a 120.175in. pulley; C ≈ 116.3 in .
Solution:
782
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Table 17.7
N sf = 1.6 − 0.2 = 1.4
hp = (1.4 )(200 ) = 280 hp
Fig. 17.14, 280 hp, 600 rpm
Use Section E
But in Table 17.3, section E is not available, use section D
Dmin = 13
D2 600
=
= 4.8
D1 125
For D1max :
D + D2
min C = 1
+ D1
2
D + 4.8 D1
113 = 1
+ D1
2
D1 = 28 in
min C = D2
D2 = 113 in
113
= 23.5 in
D1 =
4. 8
1
use D1 ≈ (13 + 23.5) = 18 in
2
D2 = (4.8)(18) = 86.4 in
L ≈ 2C + 1.57(D2 + D1 ) +
(D2 − D1 )2
4C
2
(
86.4 − 18)
(
)
(
)
L = 2 118 + 1.57 86.4 + 18 +
= 410 in
4(118)
using D1 = 19 in , D2 = 91.2 in , C = 118 in
2
(
91.2 − 19)
L = 2(118) + 1.57(91.2 + 19) +
4(118)
= 420 in
Therefore use D420 sections
D1 = 19 in , D2 = 91.2 in
πD n π (19)(600 )
vm = 1 1 =
= 2985 fpm
12
12
  103  0.09
v2  v
c
 −
Rated hp =  a
− e m6  m3
K d D1
10  10
  vm 

Table 17.3, D-section
a = 18.788 , c = 137.7 , e = 0.0848
D
Table 17.4, 2 = 4.8
D1
783
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
K d = 1.14
0.09

 103 
(2985)2  2985 = 29.6 hp
137.7
 −
Rated hp = 18.788
− (0.0848)
(1.14 )(19)
106  103

 2985 

Therefore, Fig. 17.14, section D is used.
Adjusted rated hp = Kθ K L (rated hp )
Table 17.5,
D2 − D1 91.2 − 19
=
= 0.612
C
118
Kθ = 0.83 (V-flat)
Table 17.6, D420
L = 420.8 in
K L = 1.12
Adjusted rated hp = (0.83)(1.12 )(29.6 ) = 27.52 hp
Design hp
280
No. of belts =
=
= 10 belts
Adjusted rated hp 27.52
Use10 , D420, D1 = 19 in , D2 = 91.2 in , C = 118 in
859.
A 150-hp, 700-rpm, slip-ring induction motor is to drive a ball mill at 195 rpm;
heavy starting load; intermittent seasonal service; outdoors. Determine all details
for a V-flat drive. The B.F. Goodrich Company recommended eight D270 Vbelts, 17.24-in sheave, 61-in. pully, C ≈ 69.7 in .
Solution:
Table 17.7,
N sf = 1.6 − 0.2 = 1.4
Design hp = (1.4 )(150 ) = 210 hp
Fig. 17.4, 210 hp, 700 rpm
Dmin = 13 in , D-section
  103  0.09
v2  v
c
 −
Rated hp =  a
− e m6  m3
K d D1
10  10
  vm 

d (hp )
For Max. Rated hp,
=0
 vm 
d 3 
 10 
v 
Rated hp = a m3 
 10 
v
Let X = m3
10
0.91
c  vm   vm 
−

 − e

K d D1  103   103 
3
784
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
c
X − eX 3
K d D1
v
πD1n1 πD1 (700)
X = m3 =
=
10 12 × 103 12 × 103
12 × 103 X
D1 =
700π
700πc
hp = aX 0.91 −
− eX 3
3
12 × 10 K d
d (hp )
= 0.91aX − 0.09 − 3eX 2 = 0
d (X )
0.91a
X 2.09 =
3e
Table 17.3, D-section
a = 18.788 , c = 137.7 , e = 0.0848
hp = aX 0.91 −
2.09
0.91(18.788)
v 
X 2.09 =  m3  =
3(0.0848)
 10 
vm = 7488 fpm
πD n
vm = 1 1 = 7488
12
πD (700)
vm = 1
= 7488
12
D1 = 40.86 in
max D1 = 40.86 in
1
ave. D1 = (13 + 40.86 ) = 26.93 in
2
use D1 = 22 in
D2 700 79
=
≈
D1 195 22
D1 = 22 in , D2 = 79 in
22 + 79
D + D2
Min. C = 1
+ D1 =
+ 22 = 72.5 in
2
2
Or Min. C = D2 = 79 in
2
(
D2 − D1 )
L ≈ 2C + 1.57(D2 + D1 ) +
4C
(79 − 22 )2 = 327 in
L = 2(79) + 1.57(79 + 22 ) +
4(79 )
use D330, L = 330.8 in
B + B 2 − 32(D2 − D1 )
C=
16
2
785
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
B = 4 L − 6.28(D2 + D1 ) = 4(330.8) − 6.28(79 + 22 ) = 689 in
C=
689 +
(689)2 − 32(79 − 22 )2
= 81.12 in
16
πD n π (22 )(700 )
= 4032 fpm
vm = 1 1 =
12
12
K d = 1.14
0.09

 103 
(4032)2  4032 = 39.124 hp
137.7
 −
Rated hp = 18.788
− (0.0848)
(1.14)(22)
106  103

 4032 

Adjusted rated hp = Kθ K L (rated hp )
Table 17.5,
D2 − D1 79 − 22
=
= 0.70
C
81.12
Kθ = 0.84 (V-flat)
Table 17.6
D330
K L = 1.07
Adjusted rated hp = (0.84 )(1.07 )(39.124) = 35.165 hp
Design hp
210
No. of belts =
=
= 5.97 belts use 6 belts
Adjusted rated hp 35.165
Use 6 , D330 V-belts , D1 = 22 in , D2 = 79 in , C ≈ 81.1 in
860.
A 30-hp, 1160-rpm, squirrel-cage motor is to be used to drive a fan. During the
summer, the load is 29.3 hp at a fan speed of 280 rpm; during the winter, it is 24
hp at 238 rpm; 44 < C < 50 in .; 20 hr./day operation with no overload. Decide
upon the size and number of V-belts, sheave sizes, and belt length. (Data
courtesy of The Worthington Corporation.)
Solution:
Table 17.7
N sf = 1.6 + 0.2 = 1.8
Design hp = (1.8)(30 ) = 54 hp
Speed of fan at 30 hp
30 − 24
n2 =
(280 − 238) + 238 = 286 rpm
29.3 − 24
at 54 hp, 1160 rpm. Fig. 17.4
use either section C or section D
Minimum center distance:
C = D2
786
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
D1 + D2
+ D1
2
D2 1160
=
= 4.056
D1 286
use C = 4.056 D1
44 in < C < 50 in , use C = 47 in
47
D1max =
= 11.6 in
4.056
use C-section, Dmin = 9 in
Let D1 = 10.1 in , D2 = 41 in
or C =
2
(
D2 − D1 )
L ≈ 2C + 1.57(D2 + D1 ) +
4C
(41 − 10.1)2 = 179.3 in
L = 2(47 ) + 1.57(41 + 10.1) +
4(47 )
use C137, L = 175.9 in
B + B 2 − 32(D2 − D1 )
C=
16
B = 4 L − 6.28(D2 + D1 ) = 4(175.9) − 6.28(41 + 10.1) = 328.7 in
2
C=
382.7 +
(382.7 )2 − 32(41 − 10.1)2
16
C173, satisfies 44 in < C < 50 in
0.91
= 45.2 in ≈ 44 in
3
c  vm   vm 
v 
Rated hp = a m3  −

 − e

K d D1  103   103 
 10 
πD n π (10.1)(1160 )
vm = 1 1 =
= 3067 fpm
12
12
Table 17.4
D2
= 4.056 , K d = 1.14
D1
Table 17.3, C-section
a = 8.792 , c = 38.819 , e = 0.0416
0.09
2

 103 
(
38.819
3067 )  3067
 −

Rated hp = 8.792
− (0.0416 )
= 12.838 hp
(1.14)(10.1)
106  103

 3067 

Adjusted rated hp = Kθ K L (rated hp )
Table 17.5,
D2 − D1 41 − 10.1
=
= 0.68
C
45.2
Kθ = 0.90
787
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Table 17.6
L = 175.9 , C173
K L = 1.04
Adjusted rated hp = (0.90 )(1.04 )(12.838) = 12.02 hp
Design hp
54
No. of belts =
=
= 4.5 belts use 5 belts
Adjusted rated hp 12.02
Use 5 , C173 V-belts , D1 = 10.1 in , D2 = 41 in
POWER CHAINS
NOTE: If manufacturer’s catalogs are available, solve these problems from catalogs as
well as from data in the Text.
861.
A roller chain is to be used on a paving machine to transmit 30 hp from the 4cylinder Diesel engine to a counter-shaft; engine speed 1000 rpm, counter-shaft
speed 500 rpm. The center distance is fixed at 24 in. The cain will be subjected to
intermittent overloads of 100 %. (a) Determine the pitch and the number of
chains required to transmit this power. (b) What is the length of the chain
required? How much slack must be allowed in order to have a whole number of
pitches? A chain drive with significant slack and subjected to impulsive loading
should have an idler sprocket against the slack strand. If it were possible to
change the speed ratio slightly, it might be possible to have a chain with no
appreciable slack. (c) How much is the bearing pressure between the roller and
pin?
Solution:
(a) design hp = 2(30 ) = 60 hp intermittent
D2 n1 1000
≈
=
=2
D1 n2 500
D2 = 2D1
D
C = D2 + 1 = 24 in
2
D
2 D1 + 1 = 24
2
D1max = 9.6 in
D2 max = 2D1max = 2(9.6) = 19.2 in
vm =
πD1n1
=
π (9.6 )(1000)
= 2513 fpm
12
12
Table 17.8, use Chain No. 35,
Limiting Speed = 2800 fpm
788
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Minimum number of teeth
Assume N1 = 21
N 2 = 2 N1 = 42
[Roller-Bushing Impact]
1.5
 100 N ts 
0.8
hp = K r 
 P
 n 
Chain No. 35
3
P = in
8
N ts = 21
n = 1000 rpm
K r = 29
1.5
100(21)   3 
hp = 29 
  
 1000   8 
0.8
= 40.3 hp
[Link Plate Fatigue]
hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P
3
3− 0.07  
8
3
hp = 0.004(21) (1000 )  
= 2.91 hp
8
design hp
60
No. of strands =
=
= 21
rated hp 2.91
3
Use Chain No. 35, P = in , 21 strands
8
Check for diameter and velocity
1.08
D1 =
P
=
0.9
0.375
= 2.516 in
 180 
sin

 21 
 180 

sin
 Nt 
πD n π (2.516)(1000 )
vm = 1 1 =
= 659 fpm
12
12
Therefore, we can use higher size,
Max. pitch
 180 
180 
 = 9.6 sin
P = D1 sin
 = 1.43 in

 21 
 Nt 
say Chain no. 80
P = 1 in
hp = 0.004(21)1.08 (1000)0.9 (1)3−0.07 (1) = 53.7 hp
A single-strand is underdesign, two strands will give almost twice over design,
Try Chain no. 60
789
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
P=
3
in
4
hp = 0.004(21)
1.08
0.9 
(1000)
3
 
 4
 3
3− 0.07  
 4
= 23 hp
design hp 60
=
= 2.61 or 3 strands
rated hp 23
P
0.75
D1 =
=
= 5.0 in
 180 
 180 
 sin

sin
 21 
 Nt 
vm =
πD1n1
=
12
π (5.0 )(1000)
12
= 1309 fpm
The answer is Chain No. 60, with P = ¾ in and 3 chains, limiting velocity is 1800 fpm
2
N + N 2 ( N 2 − N1 )
(b) L ≈ 2C + 1
+
pitches
2
40C
C=
24
= 32
3
 
4
N1 = 21
N 2 = 42
L = 2(32) +
21 + 42 (42 − 21)2
+
= 95.845 pitches ≈ 96 pitches
2
40(32 )
Amount of slack
(
h = 0.433 S 2 − L2
L = C = 24 in
1
2
)
(96 − 95.845) 3 in
S = 24 in +
[
2
 4  = 24.058in
h = 0.433 (24.058)2 − (24)2
1
2
]
= 0.7229 in
(c) pb = bearing pressure
Table 17.8, Chain No. 60
C = 0.234 in
1
E = in
2
J = 0.094 in
1

A = C (E + 2 J ) = 0.234 + 2(0.094) = 0.160992 in 2
2

FV
= 60 hp
33,000
790
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
F (1309)
= 60 hp
33,000
F = 1512.6 lb
1512.6
F=
= 504.2 lb strand
3
504.2
pb =
= 3131 psi
0.160992
862.
A conveyor is driven by a 2-hp high-starting-torque electric motor through a
flexible coupling to a worm-gear speed reducer, whose mw ≈ 35 , and then via a
roller chain to the conveyor shaft that is to turn about 12 rpm; motor rpm is 1750.
Operation is smooth, 8 hr./day. (a) Decide upon suitable sprocket sizes, center
distance, and chain pitch. Compute (b) the length of chain, (c) the bearing
pressure between the roller and pin. The Morse Chain Company recommended
15- and 60-tooth sprockets, 1-in. pitch, C = 24 in ., L = 88 pitches .
Solution:
Table 17.7
N sf = 1.2 − 0.2 = 1.0 (8 hr/day)
design hp = 1.0(2 ) = 2.0 hp
1750
n1 =
= 50 rpm
35
n2 = 12 rpm
Minimum number of teeth = 12
Use N1 = 12
[Link Plate Fatigue]
hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P
hp
2.0
P 3−0.07 P ≈ P 3 =
=
= 1.0
1.08
0.9
1.08 0.9
0.004 N ts n
0.004(12 ) (50)
Use Chain No. 80, P = 1.0 in
To check for roller-bushing fatigue
1.5
 100 N ts 
0.8
hp = K r 
 P
 n 
K r = 29
100(12 ) 
hp = 17 

 1000 
1.5
(1)0.8 = 2747 hp > 2 hp
(a) N1 = 12
n 
 50 
N 2 =  1  N1 =  (12) = 50 teeth
 12 
 n2 
791
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
C = D2 +
D1 ≈
D2 ≈
PN1
π
PN1
π
D1
2
=
(1.0 )(12) = 3.82 in
=
(1.0 )(50) = 15.92 in
π
π
3.82
= 17.83 in
2
use C = 18 in
C = 18 pitches
chain pitch = 1.0 in, Chain No. 80
C ≈ 15.92 +
N + N 2 ( N 2 − N1 )
+
(b) L ≈ 2C + 1
2
40C
2
12 + 59 (50 − 12 )
L ≈ 2(18) +
+
= 69 pitches
2
40(18)
use L = 70 pitches
2
(c) pb = bearing pressure
Table 17.8, Chain No. 80
C = 0.312 in
5
E = in
8
J = 0.125 in
PN ts n1 (1)(12 )(50)
vm =
=
= 50 fpm
12
12
3

A = C (E + 2 J ) = 0.141 + 2(0.05) = 0.04054 in 2
16

FV
= 60 hp
33,000
33,000(2 )
F=
= 1320 lb
50
F
1320
pb =
=
= 4835 psi
C (E + 2 J )
5

0.312  + 2(0.125)
8

863.
A roller chain is to transmit 5 hp from a gearmotor to a wood-working machine,
with moderate shock. The 1-in output shaft of the gearmotor turns n = 500 rpm .
The 1 ¼-in. driven shaft turns 250 rpm; C ≈ 16 in . (a) Determine the size of
sprockets and pitch of chain that may be used. If a catalog is available, be sure
maximum bore of sprocket is sufficient to fit the shafts. (b) Compute the center
792
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
distance and length of chain. (c) What method should be used to supply oil to the
chain? (d) If a catalog is available, design also for an inverted tooth chain.
Solution:
Table 17.7
N sf = 1.2
design hp = 1.2(5) = 6 hp
D2 500
=
=2
D1 250
D
C = D2 + 1
2
D
16 = 2 D1 + 1
2
D1max = 6.4 in
D2 max = 2D1max = 2(6.4 ) = 12.8 in
vm =
πD1n1
π (6.4 )(500)
=
12
12
(a) Link Plate Fatigue
hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P
= 838 fpm
Nts = N1 = 21
hp = 0.004(21)1.08 (500)0.9 P 3−0.07 P
Max. pitch
 180 
180 
 = 6.4 sin
P = D1 sin
 = 0.95 in

 21 
 Nt 
Try chain no. 60
5
P = in
8
hp = 0.004(21)
1.08
0.9  5 
(500)
 
8
5
3− 0.07  
8
= 7.17 hp
5
8
P
0.625
D1 =
=
= 4.194 in
 180 
 180 
sin



sin
 21 
 N1 
use P = in , Chain No. 60
n 
 500 
N2 = N1  1  = 21
 = 42
n
 250 
 2
P
0.625
D2 =
=
= 8.364 in
 180 
 180 
sin



sin
 42 
 N2 
793
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
5
8
Size of sprocket, N1 = 21 , N2 = 42 , P = in , D1 = 4.194 in , D2 = 8.364 in
(b) C = 16 in
C=
16 in
= 25.6 pitches
5
in
8
N1 + N 2 ( N 2 − N1 )
+
2
40C
2
L ≈ 2C +
21 + 42 (42 − 21)2
+
= 83.13 use 84 pitches
2
40(25.6)
use L = 84 pitches
L ≈ 2(25.6) +
(c) Method:
πD1n1 π (4.194)(500)
vm =
12
=
12
= 549 fpm .
Use Type II Lubrication ( vmax = 1300 fpm ) – oil is supplied from a drip lubricator to link
plate edges.
864.
A roller chain is to transmit 20 hp from a split-phase motor, turning 570 rpm, to a
reciprocating pump, turning at 200 rpm; 24 hr./day service. (a) Decide upon the
tooth numbers for the sprockets, the pitch and width of chain, and center
distance. Consider both single and multiple strands. Compute (b) the chain
length, (c) the bearing pressure between the roller and pin, (d) the factor of safety
against fatigue failure (Table 17.8), with the chain pull as the force on the chain.
(e) If a catalog is available, design also an inverted-tooth chain drive.
Solution:
Table 17.7
N sf = 1.4 + 0.2 (24 hr/day)
design hp = 1.6(20) = 32 hp
n 570
(a) 1 =
= 2.85
n2 200
D2 n1
≈
= 2.85
D1 n2
Considering single strand
hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P
min N ts = 17
hp = 32 = 0.004(17 )
1.08
(570)0.9 P 3−0.07 P
P 3−0.07 P = 1.24
P = 1.07 in
use P = 1.0 in
794
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
3− 0.07 (1)
hp = 32 = 0.004( N1 ) (570 ) (1)
N1 = 21
 570 
N2 = 
(21) = 60
 200 
5
Roller width = in
8
D
C = D2 + 1
2
PN1 (1)(21)
D1 ≈
=
= 6.685 in
1.08
π
D2 ≈
π
PN 2
π
(1)(60 ) = 19.10 in
=
π
C = 19.10 +
C=
0.9
6.685
= 22.44 in
2
22.44
= 22.44 pitches
1
Considering multiple strands
5
8
Assume, P = in
hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P
hp = 0.004(21)1.08 (570)0.9 (0.625)3−0.07 (0.625 ) = 8.07 hp
32 hp
No. of strands =
= 4.0
8.07 hp
Use 4 strands
Roller width = 0.4 in
D1 ≈
D2 ≈
PN1
π
PN2
π
5
 (21)
8
= 
= 4.178 in
π
5
 (60)
8
= 
= 11.937 in
C = 11.937 +
C=
π
4.178
= 14.026 in
2
14.026
= 22.44 pitches
5
8
(b) Chain Length
For single strands
795
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
N1 + N 2 ( N 2 − N1 )
+
2
40C
2
L ≈ 2C +
21 + 60 (60 − 21)2
+
= 87.08 pitches
2
40(22.44)
use L = 88 pitches
L ≈ 2(22.44) +
For multiple strands
2
N1 + N 2 ( N 2 − N1 )
L ≈ 2C +
+
2
40C
21 + 60 (60 − 21)2
+
= 87.08 pitches
2
40(22.44)
use L = 88 pitches
L ≈ 2(22.44) +
(c) pb = bearing pressure
Table 17.8, P = 1 in
5
E = in
8
J = 0.125 in
C = 0.312 in
33,000hp
F=
vm
πD n π (6.685)(570 )
vm = 1 1 =
= 998 fpm
12
12
33,000(32 )
F=
= 1058 lb
998
F
1058
pb =
=
= 3876 psi
C (E + 2 J )
5

0.312  + 2(0.125)
8

5
8
Table 17.8, P = in
3
E = in
8
J = 0.080 in
C = 0.200 in
33,000hp
vm
πD n π (6.685)(570 )
vm = 1 1 =
= 998 fpm
12
12
F=
796
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
F=
33,000(32 )
= 1058 lb
998
1058
= 264.5 lb
4
F
264.5
pb =
=
= 2472 psi
C (E + 2 J )
3

0.200  + 2(0.080)
8

F=
(d) Factor of Safety =
Fu
, based on fatigue
4F
Fu = 14,500 lb , Table 17.8
F
14,500
= 3.43
Factor of Safety = u =
4 F 4(1058)
4-strand, chain no.50
Fu = 6100 lb , Table 17.8
Factor of Safety =
865.
Fu
6100
=
= 5.77
4F 4(264.5)
A 5/8-in. roller chain is used on a hoist to lift a 500-lb. load through 14 ft. in 24
sec. at constant velocity. If the load on the chain is doubled during the speed-up
period, compute the factor of safety (a) based on the chain’s ultimate strength, (b)
based on its fatigue strength. (c) At the given speed, what is the chain’s rated
capacity ( N s = 20 teeth ) in hp? Compare with the power needed at the constant
speed. Does it look as though the drive will have a “long” life?
Solution:
Table 17.8
5
P = in
8
Fu = 6100 lb
Fu
F
F = (500 )(2 ) = 1000 lb
6100
Factor of Safety =
= 6. 1
1000
F
(b) Factor of Safety = u (fatigue)
4F
(a) Factor of Safety =
797
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Factor of Safety =
(c) vm =
6100
= 1.5
4(1000)
14 ft  60 sec 

 = 35 fpm
24 sec  1 min 
N s = 20
5
P = in
8
Rated hp = 0.004 N ts1.08 n 0.9 P 3− 0.07 P [Link Plate Fatigue]
5
 (20 )n
PN s n  8 
vm =
=
= 35 fpm
12
12
n = 33.6 rpm
Rated hp = 0.004(20 )
1.08
(33.6 )
0.9
5
 
8
5
3− 0.07  
8
= 0.6 hp
Hp needed at constant speed
Fvm
(500)(35) = 0.53 hp < 0.6 hp
hp =
=
33,000
33,000
Therefore safe for “long” life.
WIRE ROPES
866.
In a coal-mine hoist, the weight of the cage and load is 20 kips; the shaft is 400
ft. deep. The cage is accelerated from rest to 1600 fpm in 6 sec. A single 6 x 19,
IPS, 1 ¾-in. rope is used, wound on an 8-ft. drum. (a) Include the inertia force
but take the static view and compute the factor of safety with and without
allowances for the bending load. (b) If N = 1.35 , based on fatigue, what is the
expected life? (c) Let the cage be at the bottom of the shaft and ignore the effect
of the rope’s weight. A load of 14 kips is gradually applied on the 6-kip cage.
How much is the deflection of the cable due to the load and the additional energy
absorbed? (d) For educational purposes and for a load of 0.2 Fu , compute the
energy that this 400-ft rope can absorb and compare it with that for a 400-ft., 1
¾-in., as-rolled-1045 steel rod. Omit the weights of the rope and rod. What is the
energy per pound of material in each case?
Solution:
(a)
798
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
(1600
 1 min 

fpm)
 60 sec  = 4.445 fps 2
6 sec
v2 − v1
=
t
Wh = 20 kips
For 6 x 19 IPS,
w ≈ 1.6 Dr2 lb ft
 400 
2
wL = 1.6 Dr2 
 kips = 0.64 Dr kips
 1000 
Ft − wL − Wh = ma
a=
m=
20 + 0.64 Dr2
32.2
 20 + 0.64 Dr2 
(4.445)
Ft − 0.64 Dr2 − 20 = 
32
.
2


2
Ft = 22.76 + 0.73Dr
3
Dr = 1 in
4
2
 3
Ft = 22.76 + 0.731  = 25 kips
 4
F − Fb
N= u
Ft
Table AT 28, IPS
Fu ≈ 42 Dr2 tons
Fu = 42(1.75) = 129 tons = 258 kips
with bending load
Fb = sb Am
EDw
sb =
Ds
EAm Dw
Fb =
Ds
Table At 28, 6 x 19 Wire Rope
2
799
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Dw = 0.067 Dr = 0.067(1.75) = 0.11725 in
Ds = 8 ft = 96 in
E = 30,000 ksi
Am ≈ 0.4 Dr2
Am = 0.4(1.75) = 1.225 sq in
(30,000)(1.225)(0.11725) = 45 kips
Fb =
(96)
F − Fb 258 − 45
N= u
=
= 8.52
Ft
25
without bending load
258
F
N= u =
= 10.32
Ft
25
2
(b) N = 1.35 on fatigue
IPS, su ≈ 260 ksi
2 NFt
Dr Ds =
( p su )su
(1.75)(96) = 2(1.35)(25)
( p su )(260)
p su = 0.0015
Fig. 17.30, 6 x 19 IPS
Number of bends to failure = 7 x 105
FL
Am Er
Am = 1.225 sq in
Er ≈ 12,000 ksi (6 x 19 IPS)
F = 14 kips
L = 400 ft = 4800 in
(14)(4800) = 4.57 in
δ=
(1.225)(12,000)
1
1
U = Fδ = (14 )(4.57 ) = 32 in − kips
2
2
(c) δ =
(d) F = 0.2 Fu = 0.2(258) = 51.6 kips
FL
δ=
Am Er
(51.6)(4800) = 16.85 in
δ=
(1.225)(12,000)
800
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
1
1
Fδ = (51.6)(16.85) = 434 in − kips
2
2
For 1 ¾ in, as-rolled 1045 steel rod
su = 96 ksi
U=
π 
2
Fu = su A = (96) (1.75) = 230.9 kips
4
F = 0.2 Fu = 0.2(230.9) = 46.2 kips
FL
δ=
AE
(46.2 )(4800 ) = 3.073 in
δ=
π 
2
 (1.75) (30,000)
4
1
1
U = Fδ = (46.2 )(3.073) = 71 in − kips < U of wire rope.
2
2
868.
A hoist in a copper mine lifts ore a maximum of 2000 ft. The weight of car, cage,
and ore per trip is 10 kips, accelerated in 6 sec. to 2000 fpm; drum diameter is 6
ft. Use a 6 x 19 plow-steel rope. Determine the size (a) for a life of 200,000
cycles and N = 1.3 on the basis of fatigue, (b) for N = 5 by equation (v), §17.25,
Text. (c) What is the expected life of the rope found in (b) for N = 1.3 on the
basis of fatigue? (d) If a loaded car weighing 7 kips can be moved gradually onto
the freely hanging cage, how much would the rope stretch? (e) What total energy
is stored in the rope with full load at the bottom of te shaft? Neglect the rope’s
weight for this calculation. (f) Compute the pressure of the rope on the cast-iron
drum. Is it reasonable?
Solution:
(2000
 1 min 

fpm)
 60 sec  = 5.56 fps 2
6 sec
v2 − v1
=
t
For 6 x 19 IPS,
w ≈ 1.6 Dr2 lb ft
 2000 
2
wL = 1.6 Dr2 
 kips = 3.2 Dr kips
 1000 
a=
801
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Wh = 10 kips
 wL + Wh 
Ft − wL − Wh = 
a
 32.2 
 a

 5.56 
Ft = 
+ 1(wL + Wh ) = 
+ 1 3.2 Dr2 + 10 = 1.17267 3.2 Dr2 + 10
 32.2 
 32.2 
2 NFt
(a) Dr Ds =
( p su )su
Fig. 17.30, 200,000 cycles, 6 x 19
p su = 0.0028
PS: su ≈ 225 ksi
Ds = 6 ft = 72 in
N = 1.3
2(1.3)(1.17267 ) 3.2 Dr2 + 10
Dr (72) =
(0.0028)(225)
45.36 Dr = 9.7566 Dr2 + 30.49
Dr2 − 4.64916 Dr + 3.1251 = 0
Dr = 0.815 in
7
say Dr = in
8
(
(
)
)
(b) by N = 5 , Equation (v)
F − Fb
N= u
Ft
EDw
sb =
Ds
Dw = 0.067 Dr
(30,000 )(0.067 Dr ) = 27.92 D
sb =
r
72
Fb = sb Am
Am = 0.4 Dr2
(
)
Fb = (27.92 Dr ) 0.4 Dr2 = 11.17 Dr3
Fu = 36 Dr2 tons for PS
Fu = 72 Dr2 kips
Fu − Fb = NFt
72 Dr2 − 11.17 Dr3 = (5)(1.17267 )(3.2 Dr2 + 10 )
72 Dr2 − 11.17 Dr3 = (5.8634 )(3.2 Dr2 + 10)
Dr = 1.216 in
802
(
)
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
1
use Dr = 1 in
4
2 NFt
(c) Dr Ds =
( p su )su
(1.25)(72) = 2(1.3)(1.17267 )[3.2(1.25)
( p su )(225)
2
+ 10
]
p su = 0.00226
Fig. 17.20
Expected Life = 3 x 105 cycles
(d) F = 7 kips
Er = 12,000 ksi
L = 2000 ft = 24,000 in
7
For (a) Dr = in
8
FL
δ=
Am Er
2
7
Am ≈ 0.4 D = 0.4  = 0.30625 sq in
8
(7)(24,000) = 45.7 in
δ=
(0.30625)(12,000)
1
For (b) Dr = 1 in
4
FL
δ=
Am Er
3
r
2
 1
Am ≈ 0.4 D = 0.41  = 0.625 sq in
 4
(7)(24,000) = 22.4 in
δ=
(0.625)(12,000)
3
r
1
1
Fδ = (7 )(45.7 ) = 160 in − kips
2
2
1
1
For (b) U = Fδ = (7 )(22.4 ) = 78.4 in − kips
2
2
(f) Limiting pressure, cast-iron sheaves, 6 x19, p = 500 psi .
(e) For (a) U =
For (a) p su = 0.0028
p = 0.0028(225) = 0.630 kips = 630 psi > 500 psi , not reasonable.
For (b) p su = 0.00226
803
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
p = 0.00226(225) = 0.5085 kips = 508.5 psi ≈ 500 psi , reasonable.
869.
For a mine hoist, the cage weighs 5900 lb., the cars 2100 lb., and the load of coal
in the car 2800 lb.; one car loaded loaded at a time on the hoist. The drum
diameter is 5 ft., the maximum depth is 1500 ft. It takes 6 sec. to accelerate the
loaded cage to 3285 fpm. Decide on a grade of wire and the kind and size of rope
on the basis of (a) a life of 2× 105 cycles and N = 1.3 against fatigue failure, (b)
static consideration (but not omitting inertia effect) and N = 5 . (c) Make a final
recommendation. (d) If the loaded car can be moved gradually onto the freely
hanging cage, how much would the rope stretch? (e) What total energy has the
rope absorbed, fully loaded at the bottom of the shaft? Neglect the rope’s weight
for this calculation. (f) Compute the pressure of the rope on the cast-iron drum. Is
it all right?
Solution:
Wh = 5900 + 2100 + 2800 = 10,800 lb = 10.8 kips
 1 min 

fpm)
60 sec 
v2 − v1

a=
=
= 9.125 fps 2
t
6 sec
 wL + Wh 
Ft − wL − Wh = 
a
 32.2 
Assume 6 x 19 IPS,
w ≈ 1.6 Dr2 lb ft
 1500 
2
wL = 1.6 Dr2 
 kips = 2.4 Dr kips
 1000 
 a

 9.125 
Ft = 
+ 1(wL + Wh ) = 
+ 1 2.4 Dr2 + 10 = 3.08 Dr2 + 13.86
 32.2 
 32.2

5
(a) Fig. 17.30, 2 x 10 cycles
p su = 0.0028
2 NFt
Dr Ds =
( p su )su
Ds = 5 ft = 60 in
(3285
(
)
804
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Ds ≈ 45Dr
60
Dr max =
= 1.33 in
45
1
use Dr = 1 in
4
2
 1
Ft = 3.081  + 13.86 = 18.67 kips
 4
2(1.3)(18.67 )
su =
= 231 ksi
 1
(0.0028)1 (60 )
 4
1
Use Plow Steel, 6 x 19 Wire Rope, Dr = 1 in .
4
(b) N =
sb =
Fu − Fb
Ft
EDw
Ds
 1
Dw = 0.067 Dr = 0.0671  = 0.08375 in
 4
Ds = 60 in
E = 30,000 ksi
(30,000 )(0.08375) = 41.875 ksi
sb =
60
2
 1
2
Am = 0.4 Dr = 0.41  = 0.625 in 2
 4
Fb = sb Am = (41.875)(0.625) = 26.17 kips
N =5
Fu = NFt + Fb = (5)(18.67 ) + 26.17 = 119.52 kips = 59.76 tons
Fu
59.76
=
= 38.25
2
Dr  1  2
1 
 4
Table AT 28,
Use IPS, 6 x 19,
Fu
= 42 > 38.25
Dr2
(c) Recommendation:
1
6 x 19, improved plow steel, Dr = 1 in
4
805
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
FL
Am Er
F = 2100 + 2800 = 4900 lb
Er ≈ 12×10 6 psi
L = 1500 ft = 18,000 in
(4900)(18,000) = 11.76 in
δ=
(0.625) 12 ×106
(d) δ =
(
)
1
1
Fδ = (4900)(11.76) = 28,800 in − lb
2
2
(f) p su = 0.0028
(e) U =
su = 231 ksi
p = 0.0028(231,000 ) = 646.8 psi
For cast-iron sheave, limiting pressure is 500 psi
p = 646.8 psi > 500 psi , not al right.
870.
The wire rope of a hoist with a short lift handles a total maximum load of 14 kips
each trip. It is estimated that the maximum number of trips per week will be
1000. The rope is 6 x 37, IPS, 1 3/8 in. in diameter, with steel core. (a) On the
basis of N = 1 for fatigue, what size drum should be used for a 6-yr. life? (n)
Because of space limitations, the actual size used was a 2.5-ft. drum. What is the
factor of safety on a static basis? What life can be expected ( N = 1 )?
Solution:
(a)
 365 days  1 wk  1000 trips 


 = 312,857 cycles ≈ 3 × 105 cycles
No. of cycles = (6 yr )
 1 yr  7 days  1 wk 
Figure 17.30, 6 x 37, IPS
p su = 0.00225
2 NFt
Dr Ds =
( p su )su
For IPS, su ≈ 260 ksi
Ft = 14 kips
N = 1.0
Dr = 1.375 in
2 NFt
Dr Ds =
( p su )su
(1.375)Ds = 2(1.0)(14)
(0.00225)(260)
806
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Ds = 34.8 in
(b) Ds = 2 ft = 30 in
Static Basis
F − Fb
N= u
Ft
Table AT 28, 6 x 37
Dw ≈ 0.048 Dr = 0.048(1.375) = 0.066 in
Am ≈ 0.4 Dr2 = 0.4(1.375) = 0.75625 in 2
2
Fu = su Am = (260)(0.75625) = 196.6 kips
EDw Am (30,000 )(0.066)(0.75625)
Fb = sb Am =
=
= 49.9 kips
Ds
30
F − Fb 196.6 − 49.9
N= u
=
= 10.5
Ft
10.5
Life: N = 1.0 (fatigue)
2 NFt
Dr Ds =
( p su )su
(1.375)(30) = 2(1.0)(14)
( p su )(260)
p su = 0.0026
Figure 17.30, Life ≈ 2.5 × 105 cycles , 6 x 37.
871.
A wire rope passes about a driving sheave making an angle of contact of 540o, as
shown. A counterweight of 3220 lb. is suspended from one side and the
acceleration is 4 fps2. (a) If f = 0.1 , what load may be noised without slipping on
the rope? (b) If the sheave is rubber lined and the rope is dry, what load may be
raised without slipping? (c) Neglecting the stress caused by bending about the
sheave, find the size of 6 x 19 MPS rope required for N = 6 and for the load
found in (a). (d) Compute the diameter of the sheave for indefinite life with say
N = 1.1 on fatigue. What changes could be made in the solution to allow the use
of a smaller sheave?
807
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Problems 871 – 874.
Solution:

4 fps 2 
 = 2820 lb
F2 = (3220 lb )1 −
2 
 32.2 fps 
(a) F1 = F2 e fθ
θ = 540o = 3π
f = 0.10
F1 = (2820 )e (0.10 )(3π ) = 7237 lb
(b) For rubber lined, dry rope
f = 0.495
F1 = (2820 )e (0.495 )(3π ) = 249,466 lb
(c) Ft = F1 = 7237 lb
F − (Fb ≈ 0 ) Fu
N= u
=
Ft
Ft
Fu ≈ 32 Dr2 tons for MPS
Fu ≈ 64 Dr2 kips
Fu = 64,000 Dr2 lb
Fu = NFt
64,000 Dr2 = (6)(7237 )
Dr = 0.824 in
use Dr = 0.875 in
808
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
2 NFt
( p su )su
Indefinite life, p su = 0.0015
MPS: su ≈ 195 ksi = 195,000 psi
(0.875)Ds = 2(1.1)(7237)
(0.0015)(195,000)
Ds = 62.2 in
To reduce the size of sheave, increase the size of rope.
(d) Dr Ds =
A traction elevator with a total weight of 8 kips has an acceleration of 3 fps2; the
6 cables pass over the upper sheave twice, the lower one once, as shown..
Compute the minimum weight of counterweight to prevent slipping on the
driving sheave if it is (a) iron with a greasy rope, (b) iron with a dry rope, (c)
rubber lined with a greasy rope. (d) Using MPS and the combination in (a),
decide upon a rope and sheave size that will have indefinite life ( N = 1 will do).
(e) Compute the factor of safety defined in the Text. (f) If it were decided that
5× 105 bending cycles would be enough life, would there be a significant
difference in the results?
872.
Solution:

3 fps 2 
 = 8.745 kips
F1 = (8 kips )1 +
2 
 32.2 fps 
θ = 3 180o = 3π
F
F2 = f1θ
e
Wc = weight of counterweight
F2
Wc =
= 1.10274 F2
3
1−
32.2
1.10274 F1
Wc =
e fθ
(a) Iron sheave, greasy rope, f = 0.07
1.10274(8.745)
Wc =
= 4.986 kips
e (0.07 )(3π )
(b) Iron sheave, dry rope, f = 0.12
1.10274(8.745)
Wc =
= 3.112 kips
e (0.12 )(3π )
(c) Rubber lined with a greasy rope, f = 0.205
1.10274(8.745)
Wc =
= 1.397 kips
e (0.205 )(3π )
(
)
809
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
2 NFt
( p su )su
Indefinite life, p su = 0.0015
Ft = F1 = 8.745 kips total
8.745
Ft =
= 1.458 kips each rope
6
Ft = 1458 lbs
N =1
Table AT 28, 6 x 19
Ds ≈ 45Dr
2(1)(1458)
Dr (45Dr ) =
(0.0015)(195,000)
Dr = 0.47 in
1
Use Dr = in = 0.5 in
2
(d) Dr Ds =
Fu − Fb
Ft
Table AT 28, MPS
2
Fu = 32 Dr2 tons = 64,000 Dr2 lb = 64,000(0.5) lb = 16,000 lb
(e) N =
Fb =
EDw Am
Ds
E = 30× 106 psi
6 x 19, Dw = 0.067 Dr
Ds ≈ 45Dr
Am = 0.4 Dr2 = 0.4(0.5) = 0.1 sq. in.
2
(30 ×10 )(0.067)(0.1) = 4467 lb
6
Fb =
45
16,000 − 4467
N=
= 7.91
1458
(f) 5 x 105 cycles
Fig. 17.30, 6 x 19.
p su = 0.0017
2 NFt
Dr Ds =
( p su )su
810
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
Dr (45Dr ) =
2(1)(1458)
(0.0017)(195,000)
Dr = 0.44 in
since Dr = 0.44 in ≈ 0.47 in as in (d), therefore, no significant difference will result.
873.
A 5000-lb. elevator with a traction drive is supported by a 6 wire ropes, each
passing over the driving sheave twice, the idler once, as shown. Maximum values
are 4500-lb load, 4 fps2 acceleration during stopping. The brake is applied to a
drum on the motor shaft, so that the entire decelerating force comes on the
cables, whose maximum length will be 120 ft. (a) Using the desirable Ds in
terms of Dr , decide on the diameter and type of wire rope. (b) For this rope and
N = 1.05 , compute the sheave diameter that would be needed for indefinite life.
(c) Compute the factor of safety defined in the Text for the result in (b). (d)
Determine the minimum counterweight to prevent slipping with a dry rope on an
iron sheave. (e) Compute the probable life of the rope on the sheave found in (a)
and recommend a final choice.
Solution:
(a)
Ft = 4500 lb
Wh = 5000 lb
 W + wL 
Wh + wL − Ft =  h
a
 32.2 
assume 6 x 19
w = 1.6 Dr2 lb ft
wL = (1.6 Dr2 )(120 ) = 192 Dr2 per rope
wL = 6(192 Dr2 ) = 1152 Dr2
 5000 + 1152 Dr2 
(4 )
5000 + 1152 D − 4500 = 
32
.
2


2
2
1152 Dr + 500 = 621.12 + 143.11Dr
Dr = 0.3465 in
3
say Dr = 0.375 in = in
8
2
r
811
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
7
 3
Ds ≈ 45Dr = 45  = 16 in
8
8
3
Six – 6 x 19 rope, Dr = in
8
(a) Dr = 0.375 in =
3
in
8
4500
= 750 lb
6
N = 1.05
2 NFt
Dr Ds =
( p su )su
assume IPS, su = 260 ksi = 260,000 psi
Ft =
Indefinite life, p su = 0.0015
(0.375)Ds = 2(1.05)(750)
(0.0015)(260,000)
Ds = 10.77 in
Fu − Fb
Ft
Ft = 750 lb
IPS
(c) N =
2
3
Fu ≈ 42 D tons = 84,000 D lb = 84,000  lb = 11,813 lb
8
2
r
Fb =
2
r
EDw Am
Ds
6 x 19,
Ds = 10.77 in as in (b)
 3
Dw = 0.067 Dr = 0.067  = 0.025 in
8
2
3
Am = 0.4 D = 0.4  = 0.05625 sq. in.
8
6
E = 30× 10 psi
2
r
(30 ×10 )(0.025)(0.05625) = 3917 lb
6
Fb =
10.77
812
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
N=
11,813 − 3917
= 10.53
750
(c) F1 = Ft = 4500 lb
F1 = F2 e fθ
For iron sheave, dry rope, f = 0.12
θ = 540o = 3π
F2 =
F1
4500
= (0.12 )(3π ) = 1452 lb
fθ
e
e
a 

CW 1 +
 = F2
 32.2 
4 

CW 1 +
 = 1452
 32.2 
CW = 1291 lb
874.
A traction elevator has a maximum deceleration of 8.05 fps2 when being braked
on the downward motion with a total load of 10 kips. There are 5 cables that pass
twice over the driving sheave. The counterweight weighs 8 kips. (a) Compute the
minimum coefficient of friction needed between ropes and sheaves for no
slipping. Is a special sheave surface needed? (b) What size 6 x 19 mild-plowsteel rope should be used for N = 4 , including the bending effect? (Static
approach.) (c) What is the estimated life of these ropes ( N = 1 )?
Solution:
a = 8.05 fps 2
(a) F1 = 10 kips
 8.05 
F2 = (8 kips )1 −
 = 6 kips
 32.2 
θ = 3π
F1
= e fθ
F2
813
SECTION 15 – FLEXIBLE POWER-TRANSMITTING ELEMENTS
10
= e f (3π )
6
f = 0.0542
Special sheave surface is needed for this coefficient of friction, §17.21.
F − Fb
(b) N = u
Ft
10
Ft =
= 2 kips
5
EDw Am
Fb =
Ds
Table AT 28, 6 x 19, MPS
Dw = 0.067 Dr
Ds ≈ 45Dr
Am ≈ 0.4 Dr2
E = 30× 106 psi
(30 ×10 )(0.067 D )(0.4D ) = 17.87 D
6
Fb =
r
45Dr
2
r
2
r
kips
Fu ≈ 32 Dr2 tons = 64 Dr2 kips
64 Dr2 − 17.87 Dr2
2
Dr = 0.4164 in
7
use Dr = in
16
7
(c) Ds ≈ 45Dr = 45  = 20 in
 16 
2 NFt
Dr Ds =
( p su )su
Ft = 2 kips each rope
MPS, su = 195 ksi
N = 1.0
2(1.0 )(2 )
7
 (20) =
( p su )(195)
 16 
p su = 0.0023
Expected life, Figure 17.30, 3 x 105 bending cycles.
- end N =4=
814
SECTION 16 – BRAKES AND CLUTCHES
ENERGY TO BRAKES
881.
A motor operates a hoist through a pair of spur gears, with a velocity ratio of 4.
The drum on which the cable wraps is on the same shaft as the gear, and the
torque cause by the weight of the load and hoist is 12,000 ft-lb. The pinion is on
the motor shaft. Consider first on which shaft to mount the brake drum; in the
process make trial calculations, and try to think of pros and cons. Make a
decision and determine the size of a drum that will not have a temperature rise
greater than ∆t = 150o F when a 4000-lb. load moves down 200 ft. at a constant
speed. Include a calculation for the frp/sq. in. of the drum’s surface.
Solution:
Consider that brake drum is mounted on motor shaft that has lesser torque.
12,000 ft − lb
T f=
= 3000 ft − lb = 36,000 in − lb
4
From Table AT 29,
Assume f = 0.35 , p = 75 psi , max. vm = 5000 fpm
Tf =
FD
2
F = fN =
N=
2T f
D
2T f
fD
N
p=
A
A = π Db
p=
2T f
N
2(36,000 )
=
=
= 75
2
π Db π D bf π D 2b(0.35)
D 2b = 873
use D 2b = 873
873
b= 2
D
Then,
U ft − lb
∆t o F = f
Wm c
Assume a cast-iron,
ρ = 0.253 lb in3
c = 101
Wm = ρ V
Page 1 of 97
SECTION 16 – BRAKES AND CLUTCHES

D2 

D 2t = π t  Db +
4
4 

U f = (4000 )(200 ) = 800,000 ft − lb
V = π Dbt +
π
∆t = 150o F
Wm = ρ V =
Uf
c∆t
800,000
0.253V =
(150)(101)
V = 208.7 in 3
But

D2 

V = π t  Db +
4 

873
b= 2
D
 873 D 2 

V = π t 
+
4 
 D
For minimum V :
dV
 − 873 D 
= π t 2 +  = 0
dD
2
 D
3
D = 2(873)
D = 12 in
For t :
 873 (12 )2 
V = 208.7 = π t 
+

4 
 12
t = 0.611 in
5
say t = in
8
873
1
= 6.0625 in = 6 in
2
16
(12 )
5
1
Therefore use D = 12 in , t = in , b = 6 in
8
16
b=
For fhp sq. in. =
fhp =
Fvm
33,000
Page 2 of 97
fhp
A
SECTION 16 – BRAKES AND CLUTCHES
2(36,000)
= 6000 lb
D
12
vm = 5000 fpm (max.)
(6000)(5000) = 909 hp
fhp =
33,000
 1
A = π Db = π (12 ) 6  in 2
 16 
fhp
909
fhp sq. in. =
=
= 3.98 (peak value)
A
228.55
F=
882.
2T f
=
A 3500-lb. automobile moving on level ground at 60 mph, is to be stopped in a
distance of 260 ft. Tire diameter is 30 in.; all frictional energy except for the
brake is to be neglected. (a) What total averaging braking torque must be
applied? (b) What must be the minimum coefficient of friction between the tires
and the road in order for the wheels not to skid if it is assumed that weight is
equally distributed among the four wheels (not true)? (c) If the frictional energy
is momentarily stored in 50 lb. of cast iron brake drums, what is the average
temperature rise of the drums?
Solution:
(a) Solving for the total braking torque.
W 2
U f = − ∆KE =
vs − vs22
2g 1
W = 3500 lb
vs1 = 60 mph = 88 fps
(
)
vs2 = 0 mph = 0 fps
g = 32.2 fps 2
3500
Uf =
(882 − 02 ) = 421,000 ft − lb
2(32.2)
(T ft − lb )ωm = (T f in − lb )n
fhp = f
33,000
63,000
2
2
2
vs − vs1 0 − (88)
a= 2
=
= −14.892 fps 2
2s
2(260)
vs − vs1
0 − 88
t= 2
=
= 5.91 sec
a
− 14.892
U
− ∆KE
421,000
fhp =
= f =
= 130 hp
(t )(550) 550t 550(5.91)
Page 3 of 97
SECTION 16 – BRAKES AND CLUTCHES
1
(88 fps )(60 sec min )
vm
2
n=
=
= 336 rpm
πD
 30 
π  ft 
 12 
T n
fhp = f
63,000
63,000(130 )
Tf =
= 24,375 in − lb
336
(b) f =
F
N
for each wheel, N =
3500
= 875 lb
4
24,375
= 6094 in − lb
4
2T
2(6094)
F= f =
= 406 in − lb
D
30
F 406
f = =
= 0.464
N 875
Tf =
(c) ∆t =
Uf
Wm c
U f = 421,000 ft − lb
Wm = 50 lb
c = 101 ft − lb lb − F for cast-iron
421,000
∆t =
= 83.4o F
(50)(101)
884.
An overhead traveling crane weighs 160,000 lb. with its load and runs 253 fpm.
It is driven by a 25-hp motor operating at 1750 rpm.The speed reduction from the
motor to the 18-in. wheels is 32 to 1. Frictional energy other than at the brake is
negligible. (a) How much energy must be absorbed by the brake to stop this crane
in a distance of 18 ft.? (b) Determine the constant average braking torque that
must be exerted on the motor shaft. (c) If all the energy is absorbed by the rim of
the cast-iron brake drum, which is 8 in. in diameter, 1 ½ in. thick, with a 3 ¼-in.
face, what will be its temperature rise? (d) Compute the average rate at which the
energy is absorbed during the first second (fhp). Is it reasonable?
Solution:
Page 4 of 97
SECTION 16 – BRAKES AND CLUTCHES
W 2
(
vs1 − vs22 )
2g
W = 160,000 lb
g = 32.2 fps 2
vs1 = 253 fpm = 4.22 fps
U f = −∆KE =
vs2 = 0 fps
Uf =
[
]
160,000
(4.22)2 − 02 = 44,245 ft − lb
2(32.2)
fhp (63,000 )
n
2
2
vs − vs1 0 − (4.22)2
a= 2
=
= −0.495 fps 2
2s
2(18)
vs − vs1 0 − 4.22
t= 2
=
= 8.53 sec
a
− 0.495
U
44,245
fhp = f =
= 9.43 hp
550t 550(8.53)
fhp (63,000 ) (9.43)(63,000 )
Tf =
=
= 68 in − lb on the motor shaft.
1
n
(1750)
2
(b) T f =
(c) ∆t =
Uf
Wm c
V = π Dbt (rim only) on the motor shaft
D = 8 in
b = 3.25 in
t = 0.5 in
V = π (8)(3.25)(0.5) = 40.84 in 3
Wm = ρ V
ρ = 0.253 lb in3 for cast iron
c = 101 ft − lb lb − F for cast-iron
Wm = (0.253)(40.84) = 10.33 lb
44,245
∆t =
= 42.4o F
(10.33)(101)
(d) First second:
Page 5 of 97
SECTION 16 – BRAKES AND CLUTCHES
vs1 = 4.22 fps
a = −0.495 fps 2
vs2 = vs1 + at = 4.22 − 0.495(1) = 3.73 fps
U f = −∆KE =
fhp =
885.
Uf
550t
=
[
]
160,000
(4.22)2 − (3.73)2 = 9680 ft − lb
2(32.2)
9680
= 17.6 hp < 25 hp , therefore reasonable.
550(1)
The diagrammatic hoist shown with its load weighs 6000 lb. The drum weighs
8000 lb., has a radius of gyration k = 1.8 ft ; D = 4 ft . A brake on the drum shaft
brings the hoist to rest in 10 ft. from vs = 8 fps (down). Only the brake frictional
energy is significant, and it can be reasonably assumed that the acceleration is
constant. (a) From the frictional energy, compute the average braking torque. (b)
If the average fhp/sq. in. is limited to 0.15 during the first second, what brake
contact area is needed?
Problems 885, 886
Solution:
63,000 fhp
Tf =
n
U f = −∆KE1 − ∆KE2 =
(
I1 2
W
(
ω1 − ω22 ) + 2 vs21 − vs21
2
2g
vs1 = 8 fps , vs2 = 0 fps
ω1 =
2vs1
=
2(8)
= 4 rad s , ω2 = 0 rad s
4
D
W k2
I1 = 1
g
W1 = 8000 lb
k = 1.8 ft
W2 = 6000 lb
Page 6 of 97
)
SECTION 16 – BRAKES AND CLUTCHES
g = 32.2 fps 2
I1 2
W
80000(1.8)
(
ω1 − ω22 ) + 2 vs21 − vs21 =
(4)2 + 60000 (8)2 = 12,400 ft − lb
2
2g
2(32.2)
2(32.2)
2
2
vs2 − vs1
(
Uf =
a=
2
)
2s
s = 10 ft
0 2 − 82
= −3.2 fps 2
2(10 )
vs − vs1 0 − 8
t= 2
= 2.5 sec
=
a
− 3.2
U
12,400
fhp = f =
= 9 hp
550t 550(2.5)
60ω
n=
rpm
2π
1
ω = (4 rad s − 0) = 2 rad s − 0
2
60(2 )
n=
= 19.1 rpm
2π
63,000 fhp 63,000(9 )
Tf =
=
= 29,700 in − lb
n
19.1
a=
(b) fhp sq. in. = 0.15 (first second)
vs2 = vs1 + at = 8 − 3.2(1) = 4.8 fps
2(4.8)
= 2.4 rad sec
D
4
2
80000(1.8)
Uf =
(4)2 − (2.4)2 + 60000 (8)2 − (4.8)2 = 6106 ft − lb
2(32.2 )
2(32.2)
U
6106
fhp = f =
= 11.10 hp
550t 550(1)
fhp
11.10
A=
=
= 74 in 2
fhp sq.in. 0.15
ω2 =
2vs2
=
[
887.
]
[
]
The same as 885, except that a traction drive, arranged as shown, is used; the
counterweight weighs 4000 lb. The ropes pass twice about the driving sheave; the
brake drum is on this same shaft.
Page 7 of 97
SECTION 16 – BRAKES AND CLUTCHES
Problem 887.
Solution:
(
)
WT 2
vs − vs22
2g 1
WT = 4000 lb + 6000 lb = 10,000 lb
− ∆KE of pulley is negligible
vs1 = 8 fps , vs2 = 0 fps
(a) U f = −∆KE =
10,000 2
(8) = 9,940 ft − lb
2(32.2)
vs22 − vs21 0 2 − 82
a=
=
= −3.2 fps 2
2s
2(10)
vs − vs1 0 − 8
t= 2
=
= 2.5 sec
a
− 3.2
U
9940
fhp = f =
= 7.23 hp
550t 550(2.5)
D = 4 ft
2v
2(8)
ω1 = s1 =
= 4 rad sec
D
4
2v
2(0)
ω 2 = s2 =
= 0 rad sec
D
4
1
1
ω = (ω1 + ω2 ) = (4 + 0) = 2 rad sec
2
2
60ω 60(2 )
n=
=
19.1 rpm
2π
2π
Uf =
Page 8 of 97
SECTION 16 – BRAKES AND CLUTCHES
Braking torque, T f =
63,000 fhp 63,000(7.23)
=
= 23,850 in − lb
n
19.1
(b) fhp sq. in. = 0.15 (first second)
vs1 = 8 fps
vs2 − vs1 = at
vs2 − 8 = −3.2(1)
vs2 = 4.8 fps
[
]
10,000
(8)2 − (4.8)2 = 6360 ft − lb
2(32.2)
U
6360
fhp = f =
= 11.56 hp
550t 550(1)
fhp
11.56
=
= 77.1 in 2
Contact area = A =
fhp sq.in. 0.15
Uf =
SINGLE-SHOE BRAKES
888.
For the single-shoe, short-block brake shown (solid lines) derive the expressions
for brake torque for (a) clockwise rotation, (b) counterclockwise rotation. (c) In
which direction of rotation does the brake have self-actuating properties? If
f = 0.25 , for what proportions of e and c would the brake be self-actuating?
Problems 888 – 891, 893.
Solution:
(a) Clockwise rotation (as shown)
Page 9 of 97
SECTION 16 – BRAKES AND CLUTCHES
FD
2
F = fN
Tf =
[∑ M
]
=0
fN e + Wa = N c
N c − fN e = Wa
Wa
N=
c − fe
fWa
F=
c − fe
fWaD
Tf =
2(c − fe )
H
(b) Counter Clockwise Rotation
FD
2
F = fN
Tf =
[∑ M
H
=0
]
Page 10 of 97
SECTION 16 – BRAKES AND CLUTCHES
Wa = fN e + N c
Wa
N=
c + fe
fWa
F=
c + fe
fWaD
Tf =
2(c + fe )
(c) Clockwise rotation is self-actuating
c > fe
with f = 0.25
c > 0.25e
The same as 888, except that the wheel and brake shoe are grooved, 2θ degrees
between the sides of the grooves (as in a sheave, Fig. 17.38, Text).
889.
Solution:
[∑ F
V
=0
]
2 N1 sin θ = N
F = 2 f N1
Page 11 of 97
SECTION 16 – BRAKES AND CLUTCHES
 N  fN
F =2f
=
 2 sin θ  sin θ
(a) Clockwise rotation
Wa
N=
c − fe
fWa
F=
(c − fe )sin θ
fWaD
Tf =
2(c − fe )sin θ
(b) Counter clockwise rotation
Wa
c + fe
fWa
F=
(c + fe)sin θ
fWaD
Tf =
2(c + fe ) sin θ
N=
(c) Clockwise rotation is self-actuating
c > fe
with f = 0.25
c > 0.25e
890.
Consider the single-shoe, short-block brake shown (solid lines) with the drum
rotating clockwise; let e be positive measured downward and D = 1.6c . (a) Plot
the mechanical advantage MA (ordinate) against f values of 0.1, 0.2, 0.3, 0.4,
0.5 (abscissa) when e c has values 2, 0.5, 0, -0.5, -1. (b) If f may vary from 0.3
to 0.4, which proportions give the more nearly constant brake response? Are
proportions good? (c) What proportions are best if braking is needed for both
directions of rotation?
Solution:
Page 12 of 97
SECTION 16 – BRAKES AND CLUTCHES
(a) MA =
Tf
, Clockwise rotation
Wa
fD
MA =
2(c − fe )
D = 1 .6 c
1.6 fc
MA =
2(c − fe )
0 .8 f
MA =
fe 

1 − 
c 

Tabulation:
f
0.1
0.2
0.3
0.4
0.5
Page 13 of 97
2
0.100
0.267
0.600
1.600
∞
Values of MA
ec
0.5
0
0.084
0.08
0.178
0.16
0.284
0.24
0.400
0.32
0.533
0.40
-0.5
0.076
0.145
0.209
0.267
0.320
-1
0.073
0.133
0.185
0.229
0.267
SECTION 16 – BRAKES AND CLUTCHES
Plot:
(b) f = 0.3 to 0.4 , e c = −1 , with MA ≈ constant .
They are good because 1 >
fe
except e c = 2 .
c
(c) e c = 0 is the best if braking is needed for both directions of rotation with MA the
same.
891.
A single-block brake has the dimensions: cast-iron wheel of D = 15 in .,
1
3
11
a = 32 in ., c = 9 in ., e = 4 in ., width of contact surface = 2 in. The brake
2
8
16
block lined with molded asbestos, subtends 80o, symmetrical about the center
line; it is permitted to absorb energy at the rate of 0.4 hp/in.2; n = 200 rpm .
Assume that p is constant, that F and N act at K , and compute (a) pvm and
the approximate braking torque, (b) the force W to produce this torque, (c) the
mechanical advantage, (d) the temperature rise of the 3/8-in.-thick rim, if it
absorbs all the energy with operation as specified, in 1 min. (e) How long could
this brake be so applied for ∆t = 400 o F ? See 893.
Solution:
Page 14 of 97
SECTION 16 – BRAKES AND CLUTCHES
D = 15 in
a = 32.5 in
c = 9.375 in
e = 4.6875 in
b = 2 in
(a) Solving for pvm
Fvm = fpAvm ft − lb min
Fvm
= 0.4 hp in 2
A
Fvm (0.4 hp )(33,000 ft − lb hp − min ) 13,200 ft − lb min
=
=
A
in 2
in 2
Fvm
= fpvm
A
f = 0.35 from Table AT 29, molded asbestos on cast iron
Fvm
= 13,200 = 0.35 pvm
A
pvm = 37,700 ft − lb sq in − min
Solving for braking torque
Fvm
= 13,200 ft − lb sq. in. − min
A
 15 
vm = π Dn = π  (200 ) = 785 fpm
 12 
θDb
A=
2
 π 
θ = (80)
 = 1.3963 rad
 180 
Page 15 of 97
SECTION 16 – BRAKES AND CLUTCHES
A=
θDb (1.3963)(15)(2)
=
= 21 sq. in.
2
2
F (785)
= 13,200
21
F = 353 lb
FD (353)(15)
Tf =
=
= 2650 in − lb
2
2
(b) Solving for W
f Wa
c − fe
F = 353 lb
f = 0.35
a = 32.5 in
e = 4.6875 in
c = 9.375 in
F (c − fe ) (353)[9.375 − (0.35)(4.6875)]
W=
=
= 240 lb
(0.35)(32.5)
fa
F=
(c) Solving for MA
MA =
(0.35)(15)
fD
=
= 0.34
2(c − fe ) 2[9.375 − (0.35)(4.6875)]
(d) Solving for ∆t
∆t o F =
U f , ft − lb
Wm c
Wm = ρπ Dbt
D = 15 in
b = 2 in
3
t = in = 0.375 in
8
ρ = 0.253 lb in3 for cast iron
Wm = (0.253)(π )(15)(2)(0.375) = 8.942 lb
c = 101 ft − lb lb − F for cast iron
Page 16 of 97
SECTION 16 – BRAKES AND CLUTCHES
U f = 550t ′( fhp )
t ′ = 1 min = 60 sec
U f = 550(60 )( fhp ) = 33,000 fhp
fhp =
Tf n
=
(2650)(200) = 8.4127 hp
63,000
63,000
U f = 33,000(8.4127 ) = 277,619 ft − lb
∆t =
Uf
Wm c
=
277,619
= 310o F
(8.942)(101)
(e) Solving for t ′ , (time) with ∆t = 400o F
U f = Wm c∆t
U f = (8.942 )(101)(400 ) = 361,260 ft − lb
550( fhp )(t ′) = U f
550(8.4127 )(t ′) = 361,260
t ′ = 78 sec = 1.3 min
892.
1
For a single-block brake, as shown, a = 26 in ., c = 7 in ., e = 3.75 in .,
2
1
D = 15 in ., drum contact width b = 3 in . The molded asbestos lining subtends
2
o
θ = 60 , symmetrical about the vertical axis; force W = 300 lb .; n = 600 rpm .
Assume that p is constant, that F and N act at K , and compute (a) pvm and
the braking torque, (b) the energy rate in fhp/in.2 of contact surface. (c) the
mechanical advantage, (d) the temperature of the 3/8-in.-thick rim, if it absorbs
all the energy with the operation as specified in 1 min. (e) How long could this
brake be so applied for ∆t rim = 400o F ? See 894.
Page 17 of 97
SECTION 16 – BRAKES AND CLUTCHES
Problems 892, 894.
Solution:
For greater braking torque, T f , use counterclockwise rotation
[∑ M
]
=0
aW + efN = cN
Wa
N=
c − ef
f Wa
F=
c − ef
From Table AT 29, f = 0.35 for molded asbestos
W = 300 lb
a = 26 in
c = 7.5 in
e = 3.75
(0.35)(300)(26) = 442 lb
F=
7.5 − (3.75)(0.35)
A
(a) Solving for pvm
Fvm = fpAvm
π Dn π (15)(600)
vm =
=
= 2536 fpm
12
12
θDb
A=
2
 π 
θ = 60
 = 1.047 rad
 180 
(1.047 )(15)(3.5) = 27.5 in 2
A=
2
Fvm = (442)(2536) = (0.35)(27.5) pvm
pvm = 116,500 ft − lb sq. in. − min
Page 18 of 97
SECTION 16 – BRAKES AND CLUTCHES
Solving for the braking torque,
FD (442 )(15)
Tf =
=
= 3315 in − lb
2
2
(b) Energy rate, fhp.in2.
fhp =
Tf n
=
(3315)(600) = 31.6 hp
63,000
63,000
2
A = 27.5 in
31.6 hp
fhp in 2 =
= 1.15 hp in 2
27.5 in 2
T
3315
(c) MA = f =
= 0.425
Wa (300)(26)
U , ft − lb
(d) ∆t o F = f
Wm c
Wm = ρπ Dbt
3
t = in = 0.375 in
8
D = 15 in
b = 3.5 in
ρ = 0.253 lb in3 for cast iron
c = 101 ft − lb lb − F for cast iron
Wm = (0.253)(π )(15)(3.5)(0.375) = 15.648 lb
For 1 min
U f = 33,000(1)( fhp ) = 33,000(1)(31.6 ) = 1,042,800 ft − lb
∆t =
1,042,800
= 660o F
(15.648)(101)
(e) ∆t rim = 400o F
U f = (400 )(15.648)(101) = 632,179 ft − lb
′ =
t min
Uf
33,000 fhp
Page 19 of 97
=
632,179
= 0.61 min
33,000(31.6)
SECTION 16 – BRAKES AND CLUTCHES
LONG-SHOE BRAKES
FIXED SHOES
893.
The brake is as described in 891 and is to absorb energy at the same rate but the
pressure varies as p = P sin θ . Derive the equations needed and compute (a) the
maximum pressure, (b) the moment M F H of F about H , (c) the moment M N H
of N about H , (d) the force W , (e) the braking torque, (f) the x and y
components of the force at H .
Solution:
p = P sin θ = P sin φ
D
r=
2
dN = pbrdφ
dF = fpbrdφ
Page 20 of 97
SECTION 16 – BRAKES AND CLUTCHES
T f = ∫ rdF
T f = ∫ fpbr 2 dφ
T f = fbr 2 P ∫ sin φdφ
T f = fbr 2 P(cos φ1 − cos φ2 )
(a) Solving for P
Tf
P=
2
fbr (cos φ1 − cos φ2 )
D
r=
2
c
tan α =
r −e
c = 9.375 in
15
r = = 7.5 in
2
e = 4.6875 in
9.375
tan α =
7.5 − 4.6875
α = 73.3o
θ = 80o
θ
80
φ1 = α − = 73.3 − = 33.3o
2
2
θ
80
φ2 = α + = 73.3 + = 113.3o
2
2
f = 0.35
b = 2 in
r = 7.5 in
Tf
P=
fbr 2 (cos φ1 − cos φ2 )
Tf
Tf
P=
=
psi
(0.35)(2)(7.5)2 (cos 33.3 − cos113.3) 48.5
63,000 fhp
Tf =
n
fhp = fhp in 2 ( A)
θDb
A=
2
 π 
θ = 80
 = 1.396 rad
 180 
(
Page 21 of 97
)
SECTION 16 – BRAKES AND CLUTCHES
A=
(1.396)(15)(2) = 21 in 2
2
fhp in 2 = 0.4
fhp = (0.4)(21) = 8.4 hp
n = 200 rpm
63,000(8.4 )
Tf =
= 2646 in − lb
200
T
2646
P= f =
= 55 psi = max .P (φ2 > 90o )
48.5 48.5
(b) M F H = ∫ (r − R cos φ )dF
φ2
MF H = ∫
φ1
(r − R cos φ ) fbrP sin φdφ
φ2
M F H = fbrP ∫
φ1
(r sin φ − R sin φ cos φ )dφ
φ2
MF H
R


= fbrP − r cos φ − sin 2 φ 
2

 φ1
R


M F H = fbrP r (cos φ1 − cos φ2 ) − sin 2 φ2 − sin 2 φ1 
2


(
R = c 2 + (r − e ) =
2
(9.375)2 + (7.5 − 4.6875)2
)
= 9.788 in
9.788


sin 2 113.3 − sin 2 33.3 
M F H = (0.35)(2 )(7.5)(55)7.5(cos 33.3 − cos113.3) −
2


M F H = 1900 in − lb
(
(c) M N
H
= ∫ R sin φdN
φ2
MN
H
= ∫ RP sin 2 φbrdφ
MN
H
= brRP ∫ sin 2 φdφ
MN
H
=
φ1
φ2
φ1
(
)
brRP φ2
1 − cos 2 φ dφ
2 ∫φ1
φ
MN
H
2
brRP  1

=
φ
−
sin
2
φ

2 
2
φ1
brRP
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]
4
φ2 − φ1 = θ = 1.396 rad
2φ2 = 2(113.3) = 226.6o
MN
H
=
2φ1 = 2(33.3) = 66.6o
Page 22 of 97
)
SECTION 16 – BRAKES AND CLUTCHES
MN
H
MN
H
(d)
(2)(7.5)(9.788)(55) [2(1.396) − (sin 226.6 − sin 66.6)]
=
4
= 8956 in − lb
∑M
H
=0
Wa + M F H − M N
H
=0
a = 32.5 in
W (32.5) + 1900 − 8956 = 0
W = 217 lb
(e) T f = 2646 in − lb
(f)
∑F
x
=0
− H x − W cos α + ∫ dN sin φ + ∫ dF cos φ = 0
φ2
φ2
φ1
φ1
− H x = W cos α − Pbr ∫ sin 2 φdφ − fPbr ∫ sin φ cos φdφ
(
brP
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] − fbrP sin 2 φ2 − sin 2 φ1
4
2
(
2 )(7.5)(55)
− H x = 217 cos 73.3 −
[2(1.396) − (sin 226.6 − sin 66.6)]
4
(0.35)(2)(7.5)(55) sin 2 113.3 − sin 2 113.3
−
2
− H x = −931 lb
− H x = W cos α −
(
)
H x = 931 lb
∑F
y
=0
− H y + W sin α − ∫ dN cos φ + ∫ dF sin φ = 0
φ2
φ2
φ1
φ1
− H y = ∫ brP sin φ cos φdφ − ∫ fbrP sin 2 φdφ − W sin α
(
)
brP
fbrP
sin 2 φ2 − sin 2 φ1 −
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] − W sin α
2
4
(2)(7.5)(55) sin 2 113.3 − sin 2 33.3
− Hy =
2
(0.35)(2)(7.5)(55) [2(1.396) − (sin 226.6 − sin 66.6)] − 217 sin 73.3
−
4
− H y = −305 lb
− Hy =
(
H y = 305 lb
Page 23 of 97
)
)
SECTION 16 – BRAKES AND CLUTCHES
894.
The brake is as described in 892, but the pressure varies as p = P sin φ . Assume
the direction of rotation for which a given W produces the greater T f , derive the
equations needed, and compute (a) the maximum pressure, (b) the moment of F
about A , (c) The moment of N about A , (d) the braking torque, (e) the x and y
components of the force at A .
Solution:
p = P sin φ
dN = pbrdφ
dN = Pbr sin φdφ
dF = fdN = fPbr sin φdφ
Solving for φ1 and φ2
Page 24 of 97
SECTION 16 – BRAKES AND CLUTCHES
tan α =
c
r+e
D
= 7.5 in
2
7 .5
tan α =
7.5 + 3.75
α = 33.69o
r=
θ
60
= 3.69o
2
2
θ
60
φ1 = α + = 33.69 + = 63.69o
2
2
M F A = ∫ (R cos φ − r )dF
φ1 = α −
φ2
MF A = ∫
φ1
= 33.69 −
(R cos φ − r ) fPbr sin φdφ
φ2
M F A = fPbr ∫
φ1
(R sin φ cos φ − r sin φ )dφ
R

M F A = fPbr  sin 2 φ2 − sin 2 φ1 + r (cos φ2 − cos φ1 )
2

(
)
R = c 2 + (e + r ) =
2
(7.5)2 + (3.75 + 7.5)2
= 13.52 in
 13.52 

2
2
M F A = (0.35)P(3.5)(7.5)
(sin 63.69 − sin 3.69) + 7.5(cos 63.69 − cos 3.69)
 2 

M F A = 11.43P
MN
A
= ∫ R sin φdN
MN
A
= ∫ RPbr sin 2 φdφ
φ2
φ1
(
)
brPR φ2
1 − cos 2 φ dφ
2 ∫φ1
brPR
MN A =
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]
4
φ2 − φ1 = θ = 1.047 rad
2φ2 = 2(63.69) = 127.38o
MN
A
=
2φ1 = 2(3.69) = 7.38o
(3.5)(7.5)P(13.52) [2(1.047 ) − (sin 127.38 − sin 7.38)]
MN A =
4
M N A = 126.68 P
(a)
∑M
A
=0
Wa + M F A − M N
W = 300 lb
Page 25 of 97
A
=0
SECTION 16 – BRAKES AND CLUTCHES
a = 26 in
(300)(26) + 11.43P − 126.68P = 0
P = 67.68 psi
max. p = P sin φ2 = 67.68 sin 63.69 = 60.67 psi
(b) M F A = 11.43(67.68) = 774 in − lb
(c) M N A = 126.68(67.68) = 8575 in − lb
(d) T f = ∫ rdF
φ2
T f = ∫ fPbr 2 sin φdφ
φ1
T f = fPbr 2 (cos φ1 − cos φ2 )
T f = (0.35)(60.68)(3.5)(7.5) (cos 3.69 − cos 63.69)
2
T f = 2587 in − lb
(e)
[∑ F
x
=0
]
− H x − W cos α + ∫ dN sin φ − ∫ dF cos φ = 0
φ2
φ2
φ1
φ1
− H x = W cos α − Pbr ∫ sin 2 φdφ + fPbr ∫ sin φ cos φdφ
(
Pbr
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] + fPbr sin 2 φ2 − sin 2 φ1
4
2
(67.68)(3.5)(7.5) [2(1.047 ) − (sin 127.38 − sin 7.38)]
− H x = 300 cos 33.69 −
4
(0.35)(67.68)(3.5)(7.5) sin 2 63.69 − sin 2 3.69
+
2
− H x = −136 lb
− H x = W cos α −
(
)
H x = 136 lb
[∑ F
y
=0
]
H y + W sin α − ∫ dN cos φ − ∫ dF sin φ = 0
φ2
φ2
φ1
φ1
H y = Pbr ∫ sin φ cos φdφ + fPbr ∫ sin 2 φdφ − W sin α
(
)
Pbr
fPbr
sin 2 φ2 − sin 2 φ1 +
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )] − W sin α
2
4
(67.68)(3.5)(7.5) sin 2 63.69 − sin 2 3.69
Hy =
2
(
0.35)(67.68)(3.5)(7.5)
+
[2(1.047) − (sin 127.38 − sin 7.38)] − 300 sin 33.69
4
Hy =
(
Page 26 of 97
)
)
SECTION 16 – BRAKES AND CLUTCHES
H y = 766 lb
895.
(a) For the brake shown, assume p = P cos α and the direction of rotation for
which a given force W results in the greater braking torque, and derive equations
for T f in terms of W , f , and the dimensions of the brake. (b) Under what
circumstances will the brake be self-acting? (c) Determine the magnitude and
location of the resultant forces N and F .
Solution:
(a) Clockwise rotation has greatest braking torque.
p = P cos α
dN = pbrdα = Pbr cos αdα
dF = fdN = fpbrdα = fPbr cos αdα
θ2
MF H = ∫
−θ1
θ2
MF H = ∫
−θ1
(r + c sin α )dF
(r + c sin α ) fPbr cos αdα
Page 27 of 97
SECTION 16 – BRAKES AND CLUTCHES
θ2
MF H = ∫
−θ1
fPbr (r cos α + c sin α cos α )dα
θ
2
1


M F H = fPbr r sin α + c sin 2 α 
2

 −θ1
[
]
1


M F H = fPbr r [sin (θ 2 ) − sin (− θ1 )] + c sin 2 (θ 2 ) − sin 2 (− θ1 ) 
2


1


M F H = fPbr r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1 
2


(
)
θ2
MN
H
= ∫ cos αdN
MN
H
= ∫ cPbr cos 2 αdα
−θ1
θ2
−θ1
cPbr θ 2
(1 + cos 2α )dα
2 ∫−θ1
cPbr
MN H =
[2 + sin 2α ]θ−2θ1
4
cPbr
MN H =
[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )]
4
∑MH = 0
MN
=
H
[
]
Wa + M F H − M N
H
=0
1

 cPbr
[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )]
Wa + fPbr r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1  =
2
4


Wa
P=
cbr
[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − fbr 2r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1
4
2
4Wa
P=
br c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − 2 f 2r (sin θ 2 + sin θ1 ) + c sin 2 θ 2 − sin 2 θ1
(
)
[
[
{
)]
(
)]}
(
T f = ∫ rdF
θ2
Tf = ∫
−θ1
fPbr 2 cos αdα
T f = fPbr 2 [sin α ]−2θ1
θ
T f = fPbr 2 (sin θ 2 + sin θ1 )
4 fWabr 2 (sin θ 2 + sin θ1 )
br {c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − 2 f 2r (sin θ 2 + sin θ1 ) + c(sin 2 θ 2 − sin 2 θ1 ) }
4 fWar (sin θ 2 + sin θ1 )
Tf =
c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] − 2 f 2r (sin θ 2 + sin θ1 ) + c (sin 2 θ 2 − sin 2 θ1 )
D
where r = e =
2
Tf =
[
[
Page 28 of 97
]
]
SECTION 16 – BRAKES AND CLUTCHES
[
]
(b) c[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )] > 2 f 2r (sin θ 2 + sin θ1 ) + c (sin 2 θ 2 − sin 2 θ1 )
4 fr (sin θ 2 + sin θ1 )
c>
2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 ) − 2 f (sin 2 θ 2 − sin 2 θ1 )
(c) N = ∫ dN
θ2
N = ∫ Pbr cos αdα
−θ1
N = Pbr [sin α ]−2θ1
θ
N = Pbr (sin θ 2 + sin θ1 )
F = fN
F = fPbr (sin θ 2 + sin θ1 )
Solving for the location of F and N .
Let A = vertical distance from O .
θ2
( A − r cos α )dF
∑M
F Loc .
=∫
∑M
F Loc .
= ∫ P A cos α − r cos 2 α fbrdα
∑M
F Loc .
= Pfbr ∫
∑M
F Loc .
∑M
−θ1
θ2
−θ1
(
θ2
−θ1
)
(A cos α − r cos α )dα
2
θ2 
1

= Pfbr ∫  A cos α − r (1 + cos 2α ) dα
−θ1
2


θ2
F Loc .

1 
1

= Pfbr  A sin α − r  α + sin 2α 
2 
2
 −θ1


1 
1

= Pfbr [ A(sin θ 2 + sin θ1 )] − r (θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 ) 
2 
2


Then ∑ M F Loc . = 0
∑M
F Loc .
[A(sin θ 2 + sin θ1 )] − 1 r (θ 2 + θ1 ) + 1 (sin 2θ 2 + sin 2θ1 ) = 0
2 
2

1 
1

A(sin θ 2 + sin θ1 ) = r (θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )
2 
2

1 
1

r (θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )
2 
2

A=
(sin θ 2 + sin θ1 )
r [2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )]
A=
4(sin θ 2 + sin θ1 )
Page 29 of 97
SECTION 16 – BRAKES AND CLUTCHES
896.
For the brake shown with θ1 ≠ θ 2 , assume that the direction of rotation is such
that a given W results in the greater braking torque and that p = P sin φ . (a)
Derive equations in terms of θ1 and θ 2 for the braking torque, for the moment
M F H and for M N H . (b) Reduce the foregoing equations for the condition
θ1 = θ 2 . (c) Now suppose that θ , taken as θ = θ1 + θ 2 , is small enough that
θ
sin θ ≈ θ , cos θ ≈ 1 , θ1 = θ 2 = . What are the resulting equations?
2
Solution:
(a) Use clockwise rotation
p = P sin φ
dN = Pbr sin φdφ
dF = fdN = fPbr sin φdφ
φ1 = 90 − θ1
φ2 = 90 + θ 2
T f = ∫ rdF
Page 30 of 97
SECTION 16 – BRAKES AND CLUTCHES
φ2
T f = fPbr 2 ∫ sin φdφ
φ1
T f = fPbr (cos φ1 − cos φ2 )
2
T f = fPbr 2 [cos(90 − θ1 ) − cos(90 + θ 2 )]
T f = fPbr 2 (sin θ1 + sin θ 2 )
M F H = ∫ (r − c cos φ )dF
φ2
M F H = ∫ fbPr (r − c cos φ )sin φdφ
φ1
φ2
M F H = fPbr ∫
φ1
(r sin φ − c sin φ cos φ )dφ
φ
MF H
2
1


= fPbr − r cos φ − c sin 2 φ 
2

 φ1
1


M F H = fPbr r (cos φ1 − cos φ2 ) − c sin 2 φ2 − sin 2 φ1 
2


1


M F H = fPbr r [cos(90 − θ1 ) − cos(90 + θ 2 )] − c sin 2 (90 + θ 2 ) − sin 2 (90 − θ1 ) 
2


1


M F H = fPbr r (sin θ1 + sin θ 2 ) − c cos 2 θ 2 − cos 2 θ1 
2


1


M F H = fPbr r (sin θ1 + sin θ 2 ) − c 1 − sin 2 θ 2 − 1 − sin 2 θ1 
2


1


M F H = fPbr r (sin θ1 + sin θ 2 ) + c sin 2 θ 2 − sin 2 θ1 
2


(
)
[
(
[(
(
MN
H
= ∫ r sinφdN
MN
H
= bPr 2 ∫ sin 2 φdφ
]
)
)]
) (
)
φ2
φ1
2
MN
H
=
MN
H
=
MN
H
=
MN
H
=
MN
H
=
Pbr φ2
(1 − cos 2φ )dφ
2 ∫φ1
Pbr 2
[2 − sin 2φ ]φφ12
4
Pbr 2
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]
4
Pbr 2
{2[(90 + θ 2 ) − (90 − θ1 )] − [sin 2(90 + θ 2 ) − sin 2(90 − θ1 )]}
4
Pbr 2
[2(θ 2 + θ1 ) − (− sin 2θ 2 − sin 2θ1 )]
4
Page 31 of 97
SECTION 16 – BRAKES AND CLUTCHES
MN
H
=
Pbr 2
[2(θ 2 + θ1 ) + (sin 2θ 2 + sin 2θ1 )]
4
(b) θ1 = θ 2
T f = fbPr 2 (sin θ1 + sin θ 2 )
T f = 2 fbPr 2 sin θ1
1


M F H = fPbr r (sin θ1 + sin θ 2 ) + c sin 2 θ 2 − sin 2 θ1 
2


2
M F H = 2 fPbr sin θ1
(
MN
H
MN
H
MN
H
MN
H
Pbr 2
=
[2(θ 2 + θ1 ) − (− sin 2θ 2 − sin 2θ1 )]
4
bPr 2
=
(4θ1 + 2 sin 2θ1 )
4
bPr 2
=
(4θ1 + 4 sin θ1 cos θ1 )
4
= bPr 2 (θ1 + sin θ1 cos θ1 )
(c) θ = θ1 + θ 2
sin θ ≈ θ
cos θ ≈ 1
θ1 = θ 2 =
θ
2
T f = 2 fbPr 2 sin θ1
θ 
θ 
T f = 2 fbPr 2 sin   = 2 fbPr 2   = fbPr 2θ
2
2
MF
H
= 2 fPbr 2 sin θ1
θ 
θ 
M F H = 2 fbPr 2 sin   = 2 fbPr 2   = fbPr 2θ
2
2
MN
H
MN
H
= bPr 2 (θ1 + sin θ1 cos θ1 )
θ θ 
= bPr 2  + (1) = bPr 2θ
2 2 
Page 32 of 97
)
SECTION 16 – BRAKES AND CLUTCHES
897.
The brake shown is lined with woven asbestos; the cast-iron wheel is turning at
60 rpm CC; width of contact surface is 4 in. A force W = 1300 lb . is applied via
linkage systemnot shown; θ = 90o . Let p = P sin φ . (a) With the brake lever as a
free body, take moments about the pivot J and determine the maximum pressure
and compare with permissible values. Compute (b) the braking torque, (c) the
frictional energy in fhp. (d) Compute the normal force N , the average pressure
on the projected area, and decide if the brake application can safely be
continuous.
Solution:
(a)
dF = fdN
p = P sin φ
dN = pbrdφ = Pbr sin φdφ
dF = fPbr sin φdφ
M F J = ∫ (R cos φ − r )dF
φ2
M F J = fPbr ∫
φ1
Page 33 of 97
(R cos φ − r )sin φdφ
SECTION 16 – BRAKES AND CLUTCHES
φ2
M F J = fPbr ∫
φ1
(R sin φ cos φ − r sin φ )dφ
φ
2
1

M F J = fPbr  R sin 2 φ + r cos φ 
2
φ1
1

M F J = fPbr  R sin 2 φ2 − sin 2 φ1 + r (cos φ2 − cos φ1 )
2

12.5
tan β =
10
β = 51.34o
(
φ1 = β −
θ = 90
)
θ
2
o
90
= 6.34o
2
θ
90
φ1 = β + = 51.34 + = 96.34o
2
2
b = 4 in
r = 10 in
for woven asbestos f = 0.4 (Table At 29)
φ1 = 51.34 −
(12.5)2 + (10)2
R=
= 16 in
1

M F J = fPbr  R sin 2 φ2 − sin 2 φ1 + r (cos φ2 − cos φ1 )
2

16

M F J = (0.4 )P(4 )(10 ) sin 2 96.34 − sin 2 6.34 + 10(cos 96.34 − cos 6.34 )
2

M F J = −51.81P
(
)
(
)
M N J = ∫ R sin φdN
φ2
M N J = PbrR ∫ sin 2 φdφ
φ1
PbrR
[1 − cos 2φ ]φφ12
2
PbrR
=
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]
4

P(4)(10)(16) 
 π 
=
2(96.34 − 6.34)
 − (sin 2(96.34) − sin 2(6.34))

4
 180 


= 572.9 P
MN J =
MN
J
MN
J
MN
J
∑M
J
= Wa +M F J − M N J = 0
Page 34 of 97
SECTION 16 – BRAKES AND CLUTCHES
(1300)(25) + (− 51.81P ) − 572.9P = 0
P = 52 psi
max. p = P = 52 psi , φ2 > 90
From Table AT 29, permissible p = 50 psi
Therefore pmax ≈ p permissible
(b) T f = ∫ rdF
φ2
T f = fPbr ∫ sin φdφ
φ1
T f = fPbr (cos φ1 − cos φ2 )
T f = (0.4 )(52 )(4 )(10 )(cos 6.34 − cos 96.34 ) = 9188 in − lb
Tf n
(c) fhp =
, n = 60 rpm
63,000
(9188)(60) = 8.75 hp
fhp =
63,000
(d) N = ∫ dN
φ2
N = Pbr ∫ sin φdφ
φ1
N = Pbr (cos φ1 − cos φ2 )
N = (52)(4)(10)(cos 6.34 − cos 96.34) = 2297 lb
ave. p =
N
2br sin
θ = 90
θ
2
o
ave. p =
2297
2(4 )(10 )sin
90
2
= 40.6 psi
 π Dn 
π 
pvm = p
 = (40.6 ) (20 )(60 ) = 12,755 ft − lb sq. in. − min
 12 
 12 
since pvm < 28,000 ft − lb sq. in. − min (§18.4)
Application is continuous.
Page 35 of 97
SECTION 16 – BRAKES AND CLUTCHES
PIVOTED-SHOE BRAKES
898.
In the brake shown, the shoe is lined with flexible woven asbestos, and pivoted at
point K in the lever; face width is 4 in.; θ = 90o . The cast-iron wheel turns 60
rpm CL; let the maximum pressure be the value recommended in Table At 29.
On the assumption that K will be closely at the center of pressure, as planned,
compute (a) the brake torque, (b) the magnitude of force W , (c) the rate at which
frictional energy grows, (d) the time of an application if it is assumed that all this
energy is stored in the 1-in. thick rim with ∆t rim = 350 F , (e) the average pressure
on projected area. May this brake be applied for a “long time” without damage?
(f) What would change for CC rotation?
Problem 898.
Solution:
a = 27 in , b = 4 in , n = 60 rpm CL
2 D sin
c=
θ
2
θ + sin θ
D = 20 in , r = 10 in
θ = 90o = 1.571 rad
90
2 = 11.0 in
c=
1.571 + sin 90
2(20)sin
(a) T f = 2 fPbr 2 sin
θ
2
For woven asbestos, Table AT 29, f = 0.4
P = 50 psi
90
2
T f = 2(0.4 )(50 )(4 )(10 ) sin
= 11,314 in − lb
2
Page 36 of 97
SECTION 16 – BRAKES AND CLUTCHES
(b)
 θ + sin θ
N = Pbr 
2

∑MJ = 0
Wa = 12 N
W (15) = 12(2571)
W = 2057 lb
[

 1.571 + sin 90 
 = (50 )(4 )(10 )
 = 2571 lb
2



]
(c) fhp =
Tf n
=
(11,314)(60) = 10.78 hp
63,000
63,000
rate of frictional energy = 33,000 fhp = 33,000(10.78) = 355,740 ft − lb min
(d) Time (min) =
∆t o F =
Uf
33,000 fhp
U f ft − lb
Wm c
Wm = ρπ Dbt
For cast iron
ρ = 0.253 lb in3
c = 101 ft − lb lb − F
t = 1 in
Wm = (0.253)π (20)(4)(1) = 63.6 lb
U f ft − lb
∆t o F = 350 =
(63.6 lb )(101 ft − lb lb − F )
U f = 2,248,260 ft − lb
Time (min) =
2,248,260
= 6.32 min
33,000(10.78)
N
(e) Ave. p =
2br sin
Page 37 of 97
θ
2
=
2571
90
2(4 )(10 ) sin
2
= 45.45 psi
SECTION 16 – BRAKES AND CLUTCHES
pπ Dn (45.45)(π )(20 )(60 )
=
= 14,280 ft − lb sq. in. − F
12
12
since pvm < 28,000 , this brake may be applied for a long time.
pvm =
(f) Since the moment arn of F is zero, no change or CC rotation.
The pivoted-shoe brake shown is rated at 450 ft-lb. of torque; θ = 90o ; contact
width is 6.25 in.; cast-iron wheel turns at 600 rpm; assume a symmetric
sinusoidal distribution of pressure. (a) Locate the center of pressure and compute
with the location of K. Compute (b) the maximum pressure and compare with
allowable value, (c) the value of force W , (d) the reaction at the pin H , (e) the
average pressure and pvm , and decide whether or not the application could be
continuous at the rated torque. (f) Compute the frictional work from Tω and
estimate the time it will take for the rim temperature to reach 450 F (ambient, 100
F).
899.
Problem 899.
Solution:
2 D sin
(a) c =
θ
2
θ + sin θ
D = 18 in
θ = 90o = 1.571 rad
90
2 = 9.9011 in
c=
1.571 + sin 90
but location of K = 9.8125 in
then, c ≈ location K
2(18)sin
(b) T f = 2 fPbr 2 sin
θ
2
T f = 450 ft − lb = 5400 in − lb
Page 38 of 97
SECTION 16 – BRAKES AND CLUTCHES
b = 6.25 in
r = 9 in
use f = 0.4 (on cast-iron)
T f = 2 fPbr 2 sin
θ
2
90
2
P = 18.86 psi < allowable (Table AT 9)
5400 = 2(0.4 )P(6.25)(9 ) sin
2
(c) W (20.375) = N (10.375)
 θ + sin θ 
 1.571 + sin 90 
N = Pbr 
 = (18.86 )(6.25)(9 )
 = 1364 lb
2
2




(
1364 )(10.375)
W=
= 695 lb
20.375
(d) H = N − W = 1364 − 695 = 669 lb ↓
N
(e) Ave. p =
2br sin
θ
2
=
1364
90
2(6.25)(9 ) sin
2
= 17.15 psi
n = 600 rpm
pπ Dn (17.15)(π )(18)(600 )
pvm =
=
= 48,490 ft − lb sq. in. − F
12
12
since pvm > 28,000 , not continuous
 2π (600 rpm ) 
(f) Frictional work = Tω = (450 ft − lb )
 = 28,275 ft − lb per sec
 60 sec min 
U ft − lb
∆t o F = f
Wm c
Wm = ρπ Dbt
For cast iron
ρ = 0.253 lb in3
c = 101 ft − lb lb − F
π (18)
Wm = (0.253)π (18)(6.25)t +

2

t  = 154 t

4

∆t = 450 − 100 = 350 F
U f = ∆tWm c = (350 )(154 t )(101) = 5,443,900 t ft − lb
Page 39 of 97
SECTION 16 – BRAKES AND CLUTCHES
U f = 2,248,260 ft − lb
5,443,900 t
= 192.5 t sec
28,275
1
Assume t = in
2
Time = 96 sec
Time =
TWO-SHOE BRAKES
PIVOTED SHOES
900.
The double-block brake shown is to be used on a crane; the force W is applied
by a spring, and the brake is released by a magnet (not shown); θ = 90o ; contact
width = 2.5 in. Assume that the shoes are pivoted at the center of pressure. The
maximum pressure is the permissible value of Table AT 29. Compute (a) the
braking torque, (b) the force W , (c) the rate of growth of frictional energy at 870
rpm, (d) the time it would take to raise the temperature of the 0.5-in.-thick rim by
∆t = 300 F (usual assumption of energy storage), (e) pvm . (f) Where should the
pivot center be for the calculations to apply strictly?
Problem 900.
Solution:
θ
90
2 =
2 = 5.5 in
c=
π
θ + sin θ
+ sin 90
2
2 D sin
Page 40 of 97
2(10)sin
SECTION 16 – BRAKES AND CLUTCHES
[∑ M
]
R1
=0
R2
=0
F1 (5.5 − 0.875) + 12.75W = 6.75 N1
fN1 (4.625) + 12.75W = 6.75 N1
12.75W
N1 =
6.25 − 4.625 f
[∑ M
]
12.75W = F2 (5.5 − 0.875) + 6.75 N 2
12.75W = 4.625 fN 2 + 6.75 N 2
12.75W
N2 =
6.25 + 4.625 f
Assume flexible woven asbestos,
f = 0.40 , p = 50 psi
12.75W
= 2.898W
6.25 − 4.625(0.40)
F1 = fN1 = (0.4)(2.898W ) = 1.16W
N1 =
Page 41 of 97
SECTION 16 – BRAKES AND CLUTCHES
12.75W
= 1.574W
6.25 + 4.625(0.40)
F2 = fN 2 = (0.4)(1.574W ) = 0.63W
N2 =
max . T f = T f1
2 fPbr 2 sin
θ
2
= F1c
2
90
 10 
2(0.40 )(50 )(2.5)  sin
= (1.16W )(5.5)
2
2
W = 277 lb
(a) Braking torque = T f1 + T f 2 = (F1 + F2 )c = (1.16 + 0.63)(277 )(5.5) = 2727 in − lb
(b) W = 277 lb
Tf n
(2727 )(870) = 37.66 hp
=
(c) fhp =
63,000
63,000
(d) Solving for tine:
U ft − lb
∆t o F = f
Wm c
∆t o F = 300o F
c = 101 , ρ = 0.253 for cast iron
Wm = ρV
V = π Dbt +
π D 2t
= π (10 )(2.5)(0.5) +
4
Wm = (0.253)(78.54) = 19.87 lb
π (10)2 (0.5)
4
= 78.54 in 3
U f = (300 )(19.87 )(101) = 602,061 ft − lb
Time =
Uf
33,000 fhp
=
602,061
= 0.4844 min = 29 sec
33,000(37.66)
(e) pvm :
π Dn π (10)(870)
vm =
=
= 2278 fpm
12
12
pvm = (50)(2278) = 113,900
(f) c = 5.5 in
901.
A pivoted-shoe brake, rated at 900 ft-lb. torque, is shown. There are 180 sq. in. of
braking surface; woven asbestos lining; 600 rpm of the wheel; 90o arc of brake
contact on each shoe. The effect of spring A is negligible. (a) Is the pin for the
Page 42 of 97
SECTION 16 – BRAKES AND CLUTCHES
shoe located at the center of pressure? (b) How does the maximum pressure
compare with that in Table AT 29? (c) What load W produces the rated torque?
(d) At what rate is energy absorbed? Express in horsepower. Is it likely that this
brake can operate continuously without overheating? (e) Does the direction of
rotation affect the effectiveness of this brake?
Problem 901.
Solution:
θ
90
2 =
2 = 9.9 in
(a) c =
π
θ + sin θ
+ sin 90
2
13
19
and 16 ≈ 9.9 in , therefore the pin located at the center of pressure
2
2 D sin
(b)
4
13
19
16
o
α = 11.4
tan α =
[∑ M
Q
=0
]
4 FA cos α = 8.5W
Page 43 of 97
2(18) sin
SECTION 16 – BRAKES AND CLUTCHES
4 FA cos 11.4 = 8.5W
FA = 2.168W
∑ FV = 0 and ∑ FH = 0
[
]
[
]
Qv = FA sin α + W = (2.168W )sin 11.4 + W = 1.429W
Qh = FA cos α = (2.168W ) cos 11.4 = 2.125W
[∑ M
R1
=0
]
N1 (10.375) = 20.375Qh
N1 (10.375) = 20.375(2.125W )
N1 = 4.173W
F1 = f N1
For woven asbestos lining, f = 0.40 , p = 50 psi
F1 = (0.40)(4.173W ) = 1.67W (either direction)
Page 44 of 97
SECTION 16 – BRAKES AND CLUTCHES
[∑ M
=0
R2
]
10.375 N 2 = 20.375FA cos α
20.375
N2 =
(2.168W ) cos11.4 = 4.174W
10.375
F2 = (0.40)(4.174W ) = 1.67W (either direction)
T f = (F1 + F2 )c
(900)(12) = (1.67 + 1.67 )(W )(9.9)
W = 326.6 lb
T f1 = T f 2 = Fc = 2 fPbr 2 sin
but A = θ br
Ar
br 2 =
θ
2
θ
(1.67)(326.6)(9.9) =
2(0.4)(P )(180)(9)sin
π
90
2
2
P = 9.26 psi < 50 psi
(c) W = 326.6 lb
(d) fhp =
Tf n
=
(900)(12)(600) = 103 hp
63,000
63,000
π Dn π (18)(600)
vm =
=
= 2827 fpm
12
12
pvm = (9.26)(2827 ) = 26,178 ft − lb sq.in. − F
since pvm < 28,000 , it is likely to operate continuously.
(e) Since the value of F is independent of rotation, the direction doesn’t affect the
effectiveness of this brake.
902.
Refer to the diagrammatic representation of the brake of Fig. 18.2, Text, and let
9
9
the dimensions be: a = b = m = t = 4 , c = 14 , D = 15 , h = 9 in ., and the
16
16
o
contact width is 4 in.; arc of contact = 90 ; lining is asbestos in resin binder,
wheel rotation of 100 rpm CC; applied load W = 2000 lb . (a) Locate the center of
pressure for a symmetrical sinusoidal pressure distribution and compare with the
actual pin centers. Assume that this relationship is close enough for approximate
Page 45 of 97
SECTION 16 – BRAKES AND CLUTCHES
results and compute (b) the dimensions k and e if the braking force on each
shoe is to be the same, (c) the normal force and the maximum pressure, (d) the
braking torque, (e) pvm . Would more-or-less continuous application be
reasonable?
Figure 18.2
Solution:
θ
90
2 =
2 = 8.25 in
(a) c =
π
θ + sin θ
+ sin 90
2
2 D sin
2(15) sin
On Centers:
9
9
+ 4 = 9.125 in > c
16
16
9
9
B : a + b = 4 + 4 = 9.125 in > c
16
16
K :t + m = 4
[∑ M
RC
=0
]
Page 46 of 97
SECTION 16 – BRAKES AND CLUTCHES
eRF = (e + c )W
e+c
RF = 
W
 e 
RC = RF − W
cW
e+c
RC = 
W − W =
e
 e 
[∑ M
RH
=0
]
N1h − F1b = RF a
N1h − fN1b = RF a
R a
N1 = F
h − fb
fRF a
F1 =
h − fb
fa(e + c )W
F1 =
e(h − fb )
[∑ M
RE
=0
]
N 2 h + F2t = RC k
Page 47 of 97
SECTION 16 – BRAKES AND CLUTCHES
N 2 h + fN 2t = RC k
R k
N2 = C
h + ft
fR k
F2 = C
h − ft
fkcW
F2 =
e(h + ft )
(b) T f 1 = T f 2
F1c = F2c
F1 = F2
fa(e + c )W
fkcW
=
e(h − fb )
e(h + ft )
a(e + c )
kc
=
h − fb h + ft
For asbestos in resin binder,
f = 0.35 , Table AT 29
9
a = 4 in = 4.5625 in
16
9
b = 4 in = 4.5625 in
16
9
m = 4 in = 4.5625 in
16
9
t = 4 in = 4.5625 in
16
c = 14 in
9
h = 9 in = 9.5625 in
16
4.5625(e + 14)
k (14)
=
9.5625 − 0.35(4.5625) 9.5625 + 0.35(4.5625)
e + 14 = 2.1903k
but k + m = e
or e = k + 4.5625
then k + 4.5625 + 14 = 2.1903k
k = 15.6 in
e = 15.6 + 4.5625 = 20.1625 in
(c) N = N1 = N 2 =
Page 48 of 97
kcW
(15.6)(14)(2000)
=
= 2720 lb
e(h + ft ) (20.1625)[9.5625 − 0.35(4.5625)]
SECTION 16 – BRAKES AND CLUTCHES
(d) T f = T f 1 + T f 2 = f ( N1 + N 2 )c = 0.35(2)(2720)(8.25) = 15,708 in − lb
(e) vm =
π Dn
=
π (15)(100)
= 393 fpm
12
12
pvm = (64.11)(393) = 25,195 ft − lb sq.in. − F
since pvm < 28,000 , continuous application is reasonable.
FIXED SHOES
903.
A double-block brake has certain dimensions as shown. Shoes are lined with
woven asbestos; cast-iron wheel turns 60 rpm; applied force W = 70 lb . For each
direction of rotation, compute (a) the braking torque, (b) the rate of generating
frictional energy (fhp). (c) If the maximum pressure is to be P = 50 psi (Table
AT 29), what contact width should be used? (d) With this width, compute pvm
and decide whether or not the applications must be intermittent.
Problems 903, 904.
Solution:
[∑ M
=0
4Q = 26W
B
]
Page 49 of 97
SECTION 16 – BRAKES AND CLUTCHES
Q = 6.5W
[∑ M
R
=0
]
2.25S = 6Q = 6(6.5W )
S = 17.33W
RH = S = 17.33W
RV = Q = 6.5W
e = 10 in
R = 12.5 in
a = 2.25 + 9 + 12.5 = 23.75 in
∑M
∑M
M F1
M N1
H
= Sa − M F1
H
−M N1
H
= 0 (CC)
H
= Sa + M F1
H
−M N1
H
= 0 (CL)
R


= fbrP r (cos φ1 − cos φ2 ) − sin 2 φ2 − sin 2 φ1 
2


brRP
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]
H =
4
(
H
T f1 = 2 fPbr 2 sin
T f1
Pbr =
2 fr sin
r = 10 in
Page 50 of 97
θ
2
θ
2
)
SECTION 16 – BRAKES AND CLUTCHES
2r sin
θ
2
2(10 ) sin
= 11 in
θ
= 11 in
2
θ = 66.43o = 1.165 rad
f = 0.4 for woven asbestos
M F1
H
M F1
H
R


fT f1 r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 )
2


=
θ
2 fr sin
2
R


T f1 r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 )
2

= 
2r sin
θ
2
66.73
φ1 = 90 − = 90 −
= 56.64o = 0.9886 rad
2
2
2φ1 = 113.28o
θ
66.73
φ2 = 90 + = 90 +
= 123.36o = 2.1530 rad
2
2
o
2φ2 = 246.72
φ2 − φ1 = θ
12.5


T f1 10(cos 56.64 − cos 123.36) −
sin 2 123.36 − sin 2 56.64 
2
 =T
M F1 H = 
f1
66.73
2(10)sin
2
RT f1 [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]
M N1 H =
θ

4 2 fr sin 
2

12.5T f1 [2(1.165) − (sin 246.72 − sin 113.28)]
M N1 H =
= 2.96T f1
66.73
8(0.4)(10) sin
2
θ
(
CC:
∑ M H = Sa − M F1
H
−M N1
(17.33)(70)(23.75) − T f
1
H
=0
− 2.960T f1 = 0
T f1 = 7276 in − lb
CL:
∑ M H = Sa + M F1
Page 51 of 97
H
−M N1
H
=0
)
SECTION 16 – BRAKES AND CLUTCHES
(17.33)(70)(23.75) + T f
1
− 2.960T f1 = 0
T f1 = 14,700 in − lb
e = 10 in
d = 12.5 in
CC:
[∑ M
H
=0
]
RH a′ − RV d + M F2
CL:
[∑ M
H
=0
]
RH a′ − RV d − M F2
H
− M N2
H
=0
H
− M N2
H
=0
 T f2 

 = Tf
=
M
H
F1 H
2
 Tf 
 1
 Tf 
M N 2 H = M N1 H  2  = 2.960T f 2
 Tf 
 1
CC:
RH a′ − RV d + M F2 H − M N 2 H = 0
M F2
[(17.33)(21.5) − (6.5)(12.5)](70) + T f
T f 2 = 10,405 in − lb
CL:
RH a′ − RV d − M F2 H − M N 2
H
− 2.960T f 2 = 0
2
− 2.960T f 2 = 0
=0
[(17.33)(21.5) − (6.5)(12.5)](70) − T f
T f 2 = 5150 in − lb
(a) Braking Torque = T f1 + T f 2
Page 52 of 97
2
SECTION 16 – BRAKES AND CLUTCHES
CC:
T f = T f1 + T f 2 = 7276 + 10,405 = 17,681 in − lb
CL:
T f = T f1 + T f 2 = 14,700 + 5150 = 19,850 in − lb
(b) Rate of generating frictional energy
T n
fhp = f
63,000
(17,681)(60) = 16.84 hp
CC: fhp =
63,000
(19,850)(60) = 18.90 hp
CL: fhp =
63,000
(c) p = 50 psi
T f1 or T f 2 = 2 fPbr 2 sin
CC: b =
T f1
θ
2
10,405
= 4.73 in
66.73
2 fPr sin
2(0.4)(50)(10) sin
2
2
T f2
14,700
CL: b =
=
= 6.68 in
θ
66.73
2
2
2 fPr sin
2(0.4)(50)(10) sin
2
2
2
θ
=
2
(d) pvm
πDn π (20)(60)
vm =
=
= 314 fpm
12
12
pvm = (50)(314) = 15,700 < 55,000
pvm = (50)(314) = 15,700 < 28,000
application can be continuous or intermittent.
904.
If the brake shown has a torque rating of 7000 lb-in. for counter-clockwise
rotation, what braking torque would it exert for clockwise rotation, force W the
same?
Solution:
CC:
Sa − M F1 H − M N1
M F1
H
= T f1
M N1
H
= 2.960T f1
Page 53 of 97
H
=0
SECTION 16 – BRAKES AND CLUTCHES
S = 17.33W
a = 23.75 in
(17.33W )(23.75) − T f
1
− 2.96T f1 = 0
T f1 = 103.9W
RH a′ − RV d + M F2
H
− M N2
H
=0
RH = 17.33W
RV = 6.53W
a′ = 21.5 in
d = 12.5 in
M F2
H
= T f2
M N2
H
= 2.960T f 2
(17.33W )(21.5) − (6.5W )(12.5) + T f
2
− 2.960T f 2 = 0
T f 2 = 148.65W
T f = T f1 + T f 2
7000 = 103.9W + 148.65W
W = 27.7 lb
CL:
Sa + M F1
H
− M N1
H
=0
(17.33)(27.7 )(23.75) − T f
1
− 2.96T f1 = 0
T f1 = 5817 in − lb
RH a′ − RV d − M F2
H
− M N2
H
=0
[(17.33)(21.5) − (6.5)(12.5)](27.7) + T f
2
− 2.960T f 2 = 0
T f 2 = 2038 in − lb
T f = T f1 + T f 2 = 5817 + 2038 = 7855 in − lb (CL)
905.
A double-block brake is shown for which θ = 90o , b = 5 in ., n = 300 rpm , rim
thickness = ¾ in., and W = 400 lb . The shoes are lined with asbestos in resin
binder. Determine the frictional torque for (a) clockwise rotation, (b)
counterclockwise rotation. (c) How much energy is absorbed by the brake?
Express in horsepower. (d) Will the brake operate continuously without danger of
overheating? How long for a ∆t rim = 300 F ? How does pvm compare with Text
values?
Page 54 of 97
SECTION 16 – BRAKES AND CLUTCHES
Problem 905
Solution:
4
4+4
α = 26.565o
tan α =
[∑ M
R
=0
]
(Q cos α )(4) = 16W
(Q cos 26.565)(4) = 16(400)
Q = 1789 lb
RH = Q cos α = 1789 cos 26.565 = 1600 lb
RV = Q sin α + W = 1789 sin 26.565 + 400 = 1200 lb
Page 55 of 97
SECTION 16 – BRAKES AND CLUTCHES
R


M F H = fbrP r (cos φ1 − cos φ2 ) − sin 2 φ2 − sin 2 φ1 
2


brRP
MN H =
[2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]
4
(
T f = 2 fPbr 2 sin
MF H
θ
2
R


T f r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 )
2

= 
2r sin
MN
H
)
=
θ
2
RT f [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]
8 fr sin
θ
2
20
= 10 in
2
4
tan β =
12
β = 18.435o
r=
θ = 90o = 1.571 rad
θ
90
φ1 = 90 − − β = 90 − − 18.435 = 26.565o = 0.464 rad
2
2
2φ1 = 2(26.565) = 53.13o
θ
90
φ2 = 90 + − β = 90 + − 18.435 = 116.565o = 2.034 rad
2
2
2φ2 = 2(116.565) = 233.13o
R = 4 2 + 12 2 = 12.65 in
Asbestos in resin binder f = 0.35
Page 56 of 97
SECTION 16 – BRAKES AND CLUTCHES
MF H
MN
H
12.65


T f 10(cos 26.565 − cos 116.5656) −
(
sin 2 116.565 − sin 2 26.565)
2
 = 0.6803T
= 
f
90
2(10) sin
2
(12.65)T f [2(2.034 − 0.464) − (sin 233.13 − sin 53.13)]
=
= 3.03T f
90
8(0.35)(10) sin
2
(a) Clockwise
[∑ M
H1
=0
]
(Q sin α )(2.5) + (Q cos α )(24) + M F H − M N H = 0
(1789 sin 26.565)(2.5) + (1789 cos 26.565)(24) + 0.6803T f
1
1
1
1
1
T f1 = 17,195 in − lb
[∑ M
H2
=0
]
2.5 RV − 24 RH + M N 2
H2
+ M F2
H2
=0
2.5(1200 ) − 24(1600 ) + 3.03T f 2 + 0.6803T f 2 = 0
T f 2 = 9541 in − lb
Page 57 of 97
− 3.03T f1 = 0
SECTION 16 – BRAKES AND CLUTCHES
T f = T f1 + T f 2 = 17,195 + 9541 = 26,736 in − lb
(b) Counterclockwise
[∑ M
H1
=0
]
24Q cos α + 2.5Q sin α − M F1
H1
− M N1
H1
=0
(24)(1789 cos 26.565) + (2.5)(1789 sin 26.565) − 0.6803T f
1
T f1 = 10,890 in − lb
[∑ M
H2
=0
]
2.5 RV − 24 RH − M F2
H2
+ M N2
H2
=0
2.5(1200 ) − 24(1600 ) − 0.6803T f 2 + 3.03T f 2 = 0
T f 2 = 15,066 in − lb
T f = T f1 + T f 2 = 10,890 + 15,066 = 25,956 in − lb
Page 58 of 97
− 3.03T f1 = 0
SECTION 16 – BRAKES AND CLUTCHES
(c) CL:
fhp =
Tf n
63,000
=
(26,736)(300) = 127.3 hp
=
(25,956)(300) = 123.6 hp
CC:
fhp =
Tf n
63,000
(d) vm =
πDn
12
=
63,000
63,000
π (20)(300)
12
= 1571 fpm
For p :
T f = 2 fPbr 2 sin
θ
2
= T f1 (CL)
17,195 = 2(0.35)(P )(5)(10 ) sin
2
90
2
P = 69.48 psi
pvm = (69.48)(1571) = 109,153 > 28,000
the brake operate continuously with danger of overheating.
For time:
U ft − lb
∆t o F = f
Wm c
c = 101 , ρ = 0.253
Wm = ρV
V = πDbt +
πD 2t
4
2
 3  π (20)  3 
3
V = π (20 )(5)  +
  = 471.24 in
4 4
4
Wm = ρV = (0.253)(471.24) = 119.22 lb
U f = Wm c∆t = (119.22 )(101)(300 ) = 3,612,366 ft − lb
Time =
Uf
33,000 fhp
Uf
3,612,366
CL: Time =
=
= 0.886 min = 53 sec
33,000 fhp 33,000(123.6)
Uf
3,612,366
CC: Time =
=
= 0.860 min = 52 sec
33,000 fhp 33,000(127.3)
pvm > 28,000 , not good for continuous application.
Page 59 of 97
SECTION 16 – BRAKES AND CLUTCHES
906.
The double-block brake for a crane has the dimensions: a = 14.3 , b = 2.37 ,
D = 10 , e = 11.05 , g = 7.1 , h = 12 , j = 6.6 , k = 10.55 , m = 3.5 in ., the width of
shoes is 4 in., and the subtended angle is θ = 90o ; wocen asbestos lining. Its
rated braking torque is 200 ft-lb. The shoes contact the arms in such a manner
that they are virtually fixed to the arms. What force W must be exerted by a
hydraulic cylinder to develop the rated torque for (a) counterclockwise rotation,
(b) clockwise rotation? Is the torque materially affected by the direction of
rotation? (c) Compute the maximum pressure and compare with that in Table AT
29. (Data courtesy of Wagner Electric Corporation.)
Problem 906.
Solution:
b
2.37
=
e − c 11.05 − 0.83
α = 13.056o
tan α =
Page 60 of 97
SECTION 16 – BRAKES AND CLUTCHES
[∑ M
]
=0
bQ cos α + cQ sin α = eW
(2.37)(Q cos13.056) + (0.83)(Q sin 13.056) = 14.3W
Q = 5.7286W
R
RH = Q cos α = 5.7286W cos 13.056 = 5.58W
RV = Q sin α − W = 5.7286W sin 13.056 − W = 0.294W
k 5.275
=
2j
6.6
o
β = 38.63
tan β =
θ
90
φ1 = 90 − − β = 90 − − 38.63 = 6.37o = 0.1112 rad
2
2φ1 = 12.74o
φ2 = 90 +
θ
− β = 90 +
2
2φ2 = 192.74o
2
2
90
− 38.63 = 96.37 o = 1.6820 rad
2
2
k 
 10.55 
2
R =   + j2 = 
 + (6.6 ) = 8.449 in
2
 2 
D 10
r= =
= 5 in
2
2
Page 61 of 97
SECTION 16 – BRAKES AND CLUTCHES
MF H
R


T f r (cos φ1 − cos φ2 ) − (sin 2 φ2 − sin 2 φ1 )
2

= 
2r sin
MN
H
=
θ
2
RT f [2(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )]
8 fr sin
θ
2
For woven asbestos lining, f = 0.40
8.449


T f 5(cos 6.37 − cos 96.37 ) −
(
sin 2 96.37 − sin 2 6.37 )
2
 = 0.1985T
MF H = 
f
90
2(5) sin
2
8.449T f [2(1.682 − 0.1112) − (sin 192.74 − sin 12.74 )]
MN H =
= 2.6755T f
90
8(0.4 )(5) sin
2
(a) CC:
[∑ M
H1
=0
]
RH (12 ) − RV (0.25) − M F1
H1
− M N1
H1
=0
(5.58W )(12) − (0.294W )(0.25) − 0.1985T f
1
T f1 = 23.3W
Page 62 of 97
− 2.6755T f1 = 0
SECTION 16 – BRAKES AND CLUTCHES
[∑ M
H2
=0
]
12Q cos α + 0.25Q sin α + M F2
H2
− M N2
H2
− 3.5W = 0
12(14.3W ) cos 13.056 + 0.25(14.3W )sin 13.056 − 3.5W = 2.6755T f 2 − 0.1985T f 2
T f 2 = 66.4W
T f = T f1 + T f 2
T f = 200 ft − lb = 2400 in − lb
2400 = 23.3W + 66.4W
W = 26.8 lb
(b) CL:
Page 63 of 97
SECTION 16 – BRAKES AND CLUTCHES
[∑ M
H1
=0
]
RH (12 ) − RV (0.25) + M F1
H1
− M N1
H1
=0
(5.58W )(12) − (0.294W )(0.25) + 0.1985T f
1
− 2.6755T f1 = 0
T f1 = 27.0W
[∑ M
H2
=0
]
12Q cos α + 0.25Q sin α − M F2
H2
− M N2
H2
− 3.5W = 0
12(14.3W ) cos 13.056 + 0.25(14.3W )sin 13.056 − 3.5W = 2.6755T f 2 + 0.1985T f 2
T f 2 = 57.2W
T f = T f1 + T f 2
2400 = 27.0W + 57.2W
W = 28.5 lb
Since W has different values, torque is materially affected by the direction of rotation.
(c) T f = 2 fPbr 2 sin
θ
2
For woven asbestos lining, f = 0.40
Use T f = 66.4W = 66.4(26.8) = 1780 in − lb
b = 4 in
r = 5 in
θ = 90o
T f = 1780 = 2(0.4 )P(4 )(5) sin
2
P = 31.47 psi
Page 64 of 97
90
2
SECTION 16 – BRAKES AND CLUTCHES
From Table AT 29, pmax = 50 psi
31.47 psi < 50 psi
INTERNAL-SHOE BRAKES
908.
Assuming that the distribution of pressure on the internal shoe shown is given by
p = P sin φ , show that the moments M N B , M F B , and TF O of N with respect to
B and of F with respect to B and to O are ( b = face width)
MN
B
= (Pbar 2 )[θ − (sin 2φ2 − sin 2φ1 ) 2] ,
[
(
) ]
M F B = fPbr r (cos φ1 − cos φ2 ) − a sin 2 φ2 − sin 2 φ1 2 ,
TF O = fPbr 2 (cos φ1 − cos φ2 ) .
Problems 908 – 910.
Solution:
p = P sin φ
d (M N B ) = kdN
dN = P sin φ (brdφ ) = Pbr sin φdφ
k = a cos(φ − 90) = a sin φ
d (M N
MN
B
) = (a sin φ )(Pbr sin φ )dφ = Pabr sin
φ2
B
= Pabr ∫ sin 2 φdφ =
φ1
B
φdφ
Pabr φ2
(1 − cos 2φ )dφ
2 ∫φ1
2
Pabr  1
Pabr 

(φ2 − φ1 ) − (sin 2φ2 − sin 2φ1 )
φ
φ
=
−
sin
2
=



2 
2
2 
2
 φ1

φ
MN
2
but φ2 − φ1 = θ
Page 65 of 97
SECTION 16 – BRAKES AND CLUTCHES
MN
B
=
(sin 2φ2 − sin 2φ1 )
Pabr 
θ−


2
2 

d (M F B ) = edF
dF = fdN = fPbr sin φdφ
e = r + a sin (φ − 90) = r − a cos φ
d (M F B ) = (r − a cos φ )( fPbr sin φdφ ) = fPbr (r sin φ − a sin φ cos φ )dφ
[
]
φ2
M F B = fPbr − r cos φ − a sin 2 φ φ1
(
)

a sin 2 φ2 − sin 2 φ1 
M F B = fPbr r (cos φ1 − cos φ2 ) −

2


d (TF O ) = rdF = fPbr 2 sin φdφ
TF O = fPbr 2 [− cos φ ]φ12
φ
TF O = fPbr 2 (cos φ1 − cos φ2 )
The same as 908, except that a pressure distribution of p = P cos α is assumed.
909.
[
(
) ]
M N B = Pbr h(2θ + sin 2α 2 + sin 2α1 ) 4 + c sin 2 α 2 − sin 2 α1 2 ,
[
(
)
M F B = fPbr r (sin α 2 + sin α1 ) + h sin 2 α 2 − sin 2 α1 2 − c(2θ + sin 2α 2 + sin 2α1 ) 4
M F O = fPbr 2 (sin α 2 + sin α1 ) .
Solution:
k = h cos α + c sin α
e = r + h sin α − c cos α
dN = pbrdα = Pbr cos αdα
dF = fdN = fPbr cos αdα
dM N
B
= kdN = (h cos α + c sin α )(Pbr cos αdα )
(
)
dM N B = Pbr h cos 2 α + c sin α cos α dα
MN
α2
B
= Pbr ∫
−α1
(h cos
2
α + c sin α cos α )dα
α2
MN B
 h(2α + sin 2α ) c sin 2 α 
= Pbr 
+

4
2  −α

1
but θ = α1 + α 2
[
]
 h[2(α 2 + α1 ) + sin 2α 2 − sin (− 2α1 )] c sin 2 α 2 − sin 2 (− α1 ) 
M N B = Pbr 
+

4
2


Page 66 of 97
]
SECTION 16 – BRAKES AND CLUTCHES
(
)
 h(2θ + sin 2α 2 + sin 2α1 ) c sin 2 α 2 − sin 2 α1 
M N B = Pbr 
+

4
2


dM F B = edF = (r + h sin α − c cos α )( fPbr cos αdα )
(
)
dM F B = fPbr r cos α + h sin α cos α − c cos 2 α dα
α2
MF B

h sin 2 α c(2α + sin 2α ) 
= fPbr r sin α +
−

2
4

 −α1
[
]

h sin 2 α 2 − sin 2 (− α1 ) c[2(α 2 + α1 ) + sin 2α 2 − sin (− 2α1 )]
M F B = fPbr r [sin α 2 − sin (− α1 )] +
−

2
4


(
)

h sin 2 α 2 − sin 2 α1 c(2θ + sin 2α 2 + sin 2α1 ) 
M F B = fPbr r (sin α 2 + sin α1 ) +
−

2
4


dM F O = rdF = r ( fPbr cos αdα ) = fPbr 2 cos αdα
M F O = fPbr 2 [− sin α ]−α2 1 = fPbr 2 [sin α 2 − sin (− α1 )]
α
M F O = fPbr 2 (sin α 2 + sin α1 )
The same as 909, except that the α is to be measured from OG , a perpendicular
to OB ; limits from − α1 to + α 2 .
910.
Solution:
k = a cos α
e = r + a sin α
dM N
B
= kdN = a cos α (Pbr cos αdα ) = Pbar cos 2 αdα =
Pbar
(1 + cos 2α )dα
2
α
MN
B
MN
B
Pbar  2α + sin 2α  2
Pbar
[2(α 2 + α1 ) + sin 2α 2 − sin 2(− α1 )]
=
=


2 
2
4
 −α1
=
Pbar
(2θ + sin 2α 2 − sin 2α1 )
4
dM F B = edF = (r + a sin α )( fPbr cos αdα ) = fPbr (r cos α + a sin α cos α )dα
[
α2
MF B
]


a sin 2 α 
a sin 2 α 2 − sin 2 (− α1 ) 
= fPbr r sin α +
=
fPbr
r
[
sin
−
sin
(
−
)
]
+
α
α


2
1

2  −α
2



1
(
)

a sin 2 α 2 − sin 2 α1 
M F B = fPbr r (sin α 2 + sin α1 ) +

2


2
dM F O = rdF = r ( fPbr cos αdα ) = fPbr cos αdα
Page 67 of 97
SECTION 16 – BRAKES AND CLUTCHES
M F O = fPbr 2 [− sin α ]−α2 1 = fPbr 2 [sin α 2 − sin (− α1 )]
α
M F O = fPbr 2 (sin α 2 + sin α1 )
911.
The following dimensions apply to a two-shoe truck brake somewhat as shown:
face b = 5 , r = 8 , h = 5.1 , c = 2.6 , w = u = 6.4 in ., θ = 110 o , φ1 = 15o . Lining is
asbestos in rubber compound. For a maximum pressure on each shoe of 100 psi,
determine the force Q , and the braking torque for (a) clockwise rotation, (b)
counterclockwise rotation. See 908. (Data courtesy of Wagner Electric
Corporation.)
Problems 911, 912.
Solution: See 908.
(sin 2φ2 − sin 2φ1 )
Pbar 
θ−
MN B =


2 
2

(
)

a sin 2 φ2 − sin 2 φ1 
M F B = fPbr r (cos φ1 − cos φ2 ) −
,
2


2
TF O = fPbr (cos φ1 − cos φ2 )
φ1 = 15o , 2φ1 = 30o , φ2 = φ1 + θ = 15 + 110 = 125o , 2φ1 = 250o
p = 100 psi
b = 5 in
r = 8 in
a = h2 + c 2 =
(5.1)2 + (2.6)2
= 5.7245 in
θ = 110o = 1.92 rad
For asbestos in rubber compound
f = 0.35
Page 68 of 97
SECTION 16 – BRAKES AND CLUTCHES
(a) Both sides (clockwise rotation)
Q(h + w) + M F B − M N
MN
=
B
=0
(100)(5)(5.7245)(8) 1.92 − (sin 250 − sin 30) = 30,224 in − lb



5.7245 sin 2 125 − sin 2 30
M F B = (0.35)(100)(5)(8)8(cos 30 − cos 125) −
2

h = 5.1 in , w = 6.4 in
B


2
2
(
Q(5.1 + 6.4) + 14,436 − 30,224 = 0
Q = 1373 lb
TF O = (0.35)(100)(5)(8) (cos 15 − cos 125) = 17,242 in − lb
2
T f = 2TF O = 2(17,242 ) = 34,484 in − lb
(b) Counterclockwise rotation
Q(h + w) − M N B − M F B = 0
Page 69 of 97
) = 14,436 in − lb


SECTION 16 – BRAKES AND CLUTCHES
Q(5.1 + 6.4) − 30,224 − 14,436 = 0
Q = 3883 lb
T f = 2TF O = 2(17,242 ) = 34,484 in − lb
913.
The data are the same as 911, but the shoe arrangement is as shown for this
problem. For a maximum pressure on the shoes of 100 psim determine the force
Q and TF O for (a) Cl rotation, (b) CC rotation, See 908.
Problem 913.
Solution:
TF O = T f = fPbr 2 (cos φ1 − cos φ2 )
Pbr =
Tf
fr (cos φ1 − cos φ2 )
MN B =
aT f
(sin 2φ2 − sin 2φ1 ) =
(sin 2φ2 − sin 2φ1 )
Pbar 

θ
−
−
θ
 2 fr (cos φ − cos φ ) 

2 
2
2
1
2
(
)

a sin 2 φ2 − sin 2 φ1 
M F B = fPbr r (cos φ1 − cos φ2 ) −

2


Tf

a sin 2 φ2 − sin 2 φ1 
(
)
φ
φ
MF B =
r
cos
−
cos
−
1
2


2r (cos φ1 − cos φ2 ) 
2

From 911:
φ1 = 15o , 2φ1 = 30o , φ2 = φ1 + θ = 15 + 110 = 125o , 2φ1 = 250o
θ = 110o = 1.92 rad
(
a = h2 + c 2 =
f = 0.35
Page 70 of 97
(5.1)2 + (2.6)2
= 5.7245 in
)
SECTION 16 – BRAKES AND CLUTCHES
(sin 250 − sin 30) = 1.753T

1.92 −
f


2(0.35)(8)(cos 15 − cos 125) 
2
Tf

5.7245 sin 2 125 − sin 2 15
(
)
=
8
cos
15
−
cos
125
−

28(cos 15 − cos 125) 
2
5.7245T f
MN B =
MF B
(
(a) CL rotation:
Left Side
[∑ M
B
=0
]
Q(h + w) − M N1 B − M F1
B
=0
Q(5.1 + 6.4 ) − 1.753T f1 − 0.43T f1 = 0
T f1 = 5.268Q
Right Side:
[∑ M
B
=0
]
Page 71 of 97
) = 0.43T


f
SECTION 16 – BRAKES AND CLUTCHES
Q(h + w) + M F2 B − M N 2
B
=0
Q(5.1 + 6.4 ) + 0.43T f 2 − 1.753T f 2 = 0
T f 2 = 8.6924Q
T f max = T f 2 = 8.6924Q
T f max = fPbr 2 (cos φ1 − cos φ2 )
8.6924Q = (0.35)(100 )(5)(8) (cos 15 − cos 125)
Q = 1984 lb
2
T f1 = 5.268Q = 5.268(1984 ) = 10,452 in − lb
T f 2 = 8.6924Q = 8.6924(1984 ) = 17,246 in − lb
Total TF O = T f1 + T f 2 = 10,452 + 17,246 = 27,698 in − lb
(b) CC rotation
Left Side
[∑ M
B
=0
]
Q(h + w) + M F1 B − M N1
B
=0
Q(5.1 + 6.4 ) + 0.43T f1 − 1.753T f1 = 0
T f1 = 8.6924Q
Right Side:
Page 72 of 97
SECTION 16 – BRAKES AND CLUTCHES
[∑ M
B
=0
]
Q(h + w) − M F2
B
− M N2
B
=0
Q(5.1 + 6.4 ) − 0.43T f 2 − 1.753T f 2 = 0
T f 2 = 5.268Q
Since values are just interchanged
Q = 1984 lb
Total TF O = 27,698 in − lb as in (a)
914.
A double-shoe internal brake is actuated by an involute cam as shown, where QR
is the force on the right shoe at a radius wR and QL is the force on the left shoe at
a radius wL . The pressure of each shoe is proportional to the rotation of the shoe
about B which is inversely proportional to w ; therefore, the ratio of the
maximum pressures is PL PR = wR wL . The dimensions are: face width b = 4 ,
9
1
5
5
r = 6 , h = 4 , c = 1 , wL = 9 , wR = 8 in .: for each shoe, θ = 120 o ,
16
8
16
16
o
φ1 = 30 . The lining is asbestos in rubber compound, Determine the braking
torque and forces QR and QL for the maximum permissible pressure for (a)
clockwise rotation, (b) counterclockwise rotation.
Page 73 of 97
SECTION 16 – BRAKES AND CLUTCHES
Problem 914.
Solution:
(sin 2φ2 − sin 2φ1 )
Pbar 
θ−
MN B =


2 
2

(
)

a sin 2 φ2 − sin 2 φ1 
M F B = fPbr r (cos φ1 − cos φ2 ) −
,
2


2
TF O = fPbr (cos φ1 − cos φ2 )
2
2
 9   1
a = h + c =  4  + 1  = 4.70 in
 16   8 
5
8
pL wR
=
= 16 = 0.8926
pR wL 9 5
16
For asbestos in rubber compound, f = 0.35 , p = 75 psi
pR = 75 psi
pL = 0.8926(75) = 67 psi
2
2
(a) Clockwise rotation
Left Side:
Page 74 of 97
SECTION 16 – BRAKES AND CLUTCHES
[∑ M
BL
=0
QL wL − M FL
M FL
BL
]
BL
− M NL
BL
=0
(
)

a sin 2 φ2 − sin 2 φ1 
= fPLbr r (cos φ1 − cos φ2 ) −

2


φ1 = 30o
2φ1 = 60o
φ2 = θ + φ1 = 120 + 30 = 150o
2φ2 = 300o
θ = 120o = 2.094 rad
(

4.70 sin 2 150 − sin 2 30
(
)(
)(
)(
)
(
)
=
0
.
35
67
4
6
6
cos
30
−
cos
150
−
BL

2

(sin 2φ2 − sin 2φ1 )
P bar 
θ−
M NL BL = L


2 
2

(67 )(4)(4.7 )(6) 2.094 − (sin 300 − sin 60) = 11,185 in − lb
M NL BL =


2
2


M FL
 5
QL  9  − 5849 − 11,185 = 0
 16 
QL = 1829 lb
T( F O )L = fPL br 2 (cos φ1 − cos φ2 )
T( F O )L = (0.35)(67 )(4 )(6 ) (cos 30 − cos 150 ) = 5849 in − lb
2
Right side:
Page 75 of 97
) = 5849 in − lb


SECTION 16 – BRAKES AND CLUTCHES
[∑ M
BR
=0
]
QR wR + M FR
M FR
BR
M NR
BR
=
=
− M NR
BR
M FL
BL
PR
PL
M NL
BL
PL
PR
=
BR
=0
(5849)(75) = 6547 in − lb
=
67
(11,185)(75) = 12,520 in − lb
67
 5
QR  8  + 6547 − 12,520 = 0
 16 
QR = 719 lb
T( F O )L PR (5849)(75)
T( F O )R =
=
= 6547 in − lb
PL
67
T( F O ) = T( F O )R + T( F O )L = 6547 + 5849 = 12,396 in − lb
(b) Counterclockwise rotation
Left side:
Page 76 of 97
SECTION 16 – BRAKES AND CLUTCHES
[∑ M
BL
=0
]
QL wL + M FL
BL
− M NL
BL
=0
 5
QL  9  + 5849 − 11,185 = 0
 16 
QL = 573 lb
T( F O )L = 5849 in − lb
Right Side:
[∑ M
BR
=0
]
QR wR − M FR
BR
− M NR
BR
=0
 5
QR  8  − 6547 − 12,520 = 0
 16 
QR = 2294 lb
T( F O )R = 6547 in − lb
Page 77 of 97
SECTION 16 – BRAKES AND CLUTCHES
T( F O ) = T( F O )R + T( F O )L = 6547 + 5849 = 12,396 in − lb
BAND BRAKES
915.
The steel band for the brake shown is lined with flexible asbestos and it is
expected tha the permissible pressure of Table AT 29 is satisfactory; θ = 245o ,
1
a = 20 in ., m = 3 in ., D = 18 in ., and face width b = 4 in .; rotation CL. The
2
cast-iron wheel turns 200 rpm. Set up suitable equations, use the average f
given and compute (a) the force in each end of the band, (b) the brake torque and
fhp. (c) Determine the mechanical advantage for the limit values of f in Table
AT 29 and its percentage variation fron that for the average f . (d) Investigate
the overheating problem using relevant information given in the Text.
Problem 915.
Solution:
(1)
F1
= e fθ
F2
[∑ M
Fixed point
Wa = F2 m
Wa
(2) F2 =
m
Page 78 of 97
=0
]
SECTION 16 – BRAKES AND CLUTCHES
(3) F = F1 − F2 = fpA
FD
(4) T f =
2
From Table AT 29, flexible asbestos
Ave. f = 0.40 , p = 50 psi
(a) For F1 and F2 :
θDb
A=
2
θ = 245o = 4.276 rad
(4.276)(18)(4) = 154 in 2
A=
2
F = F1 − F2 = fpA = (0.40)(50 )(154) = 3080 lb
F1
= e fθ = e (0.40 )(4.276 ) = 5.5312
F2
F1 = 5.5312 F2
F = 5.5312 F2 − F2 = 3080
F2 = 680 lb
F1 = 5.5312(680) = 3760 lb
(b) T f and fhp
FD (3080 )(18)
=
= 27,720 in − lb
2
2
Tf n
(27,720)(200) = 88 hp
fhp =
=
63,000
63,000
(c) For MA
T
T
MA = f = f
Wa F2 m
FD
Tf =
2
F
F2 = fθ
e −1
 FD 


D e fθ − 1
2 

MA =
=
2m
 Fm 
 fθ

 e −1 
D = 18 in
m = 3.5 in
θ = 4.276 rad
Tf =
(
Page 79 of 97
)
SECTION 16 – BRAKES AND CLUTCHES
Limit values (Table AT 29) f = 0.35 to 0.45 .
f = 0.35
[
]
[
]
18 e (0.35 )(4.276 ) − 1
= 8.914
2(3.5)
f = 0.45
MA =
18 e (0.45 )(4.276 ) − 1
MA =
= 15.042
2(3.5)
with f = 0.40 (average)
[
]
18 e (0.40 )(4.276 ) − 1
= 11.652
2(3.5)
Percentage variation from f = 0.40 .
f = 0.35
11.652 − 8.914
(100% ) = 23.5%
% var =
11.652
f = 0.45
15.042 − 11.652
% var =
(100% ) = 29.1%
11.652
MA =
(d) Overheating problem
fhp 88
=
= 0.57 fhp in 2
A 154
Therefore, a problem of overheating is expected as Rasmussen recommends 0.2 to 0.3
fhp per square inch of brake contact area.
916.
(a) For the band brake shown, derive the expressions for the braking torque in
terms of W , etc., for CL rotation and for CC rotation, and specify the ratio c b
for equal effectiveness in both directions of rotation. Are there any proportions of
b and c as shown that would result in the brae being self locking? (b) When
θ = 270o , a = 16 in ., b = c = 3 in ., and D = 12 in ., it was found that a
force W = 50 lb . Produced a frictional torque of 1000 in-lb. Compute the
coefficient of friction.
Page 80 of 97
SECTION 16 – BRAKES AND CLUTCHES
Problem 916.
Solution:
(a)
CL:
[∑ M
O
=0
]
aW = F1b + F2c
F1 = F2 e fθ
aW = F2e fθ b + F2 c
aW
F2 = fθ
be + c
aWe fθ
F1 = F2 e fθ = fθ
be + c
aWe fθ − aW aW (e fθ − 1)
F = F1 − F2 =
=
be fθ + c
be fθ + c
Page 81 of 97
SECTION 16 – BRAKES AND CLUTCHES
Tf =
FD WaD  e fθ − 1 


=
2
2  be fθ + c 
CC:
[∑ M
O
=0
]
aW = F2b + F1c
WaD  e fθ − 1 


Tf =
2  ce fθ + b 
No proportions of b and c as shown that would result in the brake being self-locking.
(b) W = 50 lb
T f = 1000 in − lb
D = 12 in
a = 16 in
b = c = 3 in
θ = 270o = 4.7124 rad
T f = 1000 =
(50)(16)(12) 
2
e fθ − 1 
 3e fθ + 3 


e fθ − 1
= 0.625
e fθ + 1
e fθ = e 4.7124 f = 4.333
f = 0.311
917.
(a) For the brake shown, assume the proper direction of rotation of the cast-iron
wheel for differential acion and derive expressions for the braking torque. (b) Let
3
D = 14 in ., n = 1 in ., m = 4 in ., θ = 235o , and assume the band to be lined with
4
woven asbestos. Is there a chance that this brake will be self-acting? If true, will
Page 82 of 97
SECTION 16 – BRAKES AND CLUTCHES
it always be for the range of values of f given in Table AT 29? (c) The ratio
n m should exceed what value in order for the brake to be self-locking? (d) If the
direction of rotation of the wheel is opposite to that taken in (a), what is the
braking torque with a force W = 10 lb . at a = 8 in .? (e) Suppose the brake is
used as a stop to prevent reverse motion on a hoist. What is the frictional
horsepower for the forward motion if the wheel turns 63 rpm?
Problems 917, 918.
Solution:
(a) Assume CL
[∑ M
O
=0
]
Wa + F1n = F2 m
F1 = F2 e fθ
Wa = F2 m − F2 ne fθ = F2 (m − ne fθ )
Wa
F2 =
(m − ne fθ )
Page 83 of 97
SECTION 16 – BRAKES AND CLUTCHES
F1 =
Wae fθ
m − ne fθ
(
)
Wa (e fθ − 1)
F = F1 − F2 =
, Braking force.
m − ne fθ
FD WaD  e fθ − 1 

 , Braking torque.
Tf =
=
2
2  m − ne fθ 
(b) D = 14 in
3
n = 1 in
4
m = 4 in
θ = 235o = 4.10 rad
Table AT 29, woven asbestos
f = 0.35 to 0.45
There is a chance of self-acting if
ne fθ > m
m = 4 in
use f = 0.40
ne fθ = 1.75e(0.40 )(4.10 ) = 9.0 > m
use f = 0.35
ne fθ = 1.75e(0.35 )(4.10 ) = 7.35 > m
use f = 0.45
ne fθ = 1.75e (0.45 )(4.10 ) = 11.07 > m
Therefore true for the range of values of f .
(c) ne fθ > m , f = 0.40 (average)
n
1
> fθ
m e
n
1
> (0.4 )(4.10 )
m e
n
> 0 .2
m
(d) For CC:
Page 84 of 97
SECTION 16 – BRAKES AND CLUTCHES
[∑ M
O
=0
]
Wa + F2 n = F1m
Wa = F1m − F2 n
F1 = F2 e fθ
Wa = F2 e fθ m − F2 n = F2 (me fθ − n )
Wa
F2 =
me fθ − n
Wae fθ
F1 =
me fθ − n
Wa (e fθ − 1)
F = F1 − F2 =
me fθ − n
FD WaD  e fθ − 1 


Tf =
=
2
2  me fθ − n 
Tf =
(10)(8)(14) 
2
(e) fhp =
918.
e (0.40 )(4.10 ) − 1 
 4e (0.40 )(4.10 ) − 1.75  = 123.3 in − lb


Tf n
63,000
=
(123.3)(63) = 0.1233 hp
63,000
A differential band brake similar to that shown and lined with woven asbestos,
has the dimensions: D = 18 in ., n = 2 in ., m = 12 in ., θ = 195o . (a) Is there a
chance that this brake will be self-acting? (b) If W = 30 lb . and a = 26 in . ,
compute the maximum braking torque and the corresponding mechanical
advantage. (c) What is the ratio of the braking torque for CL rotation to the
braking torque for CC rotation? (d) A 1/16-in.-thick steel band, SAE 1020 as
rolled, carries the asbestos lining. What should be its width for a factor of safety
of 8, based on the ultimate stress? What should be the face width if the average
pressure is 50 psi?
Solution:
Page 85 of 97
SECTION 16 – BRAKES AND CLUTCHES
(a) For CL:
ne fθ > m
θ = 195o = 3.4 rad
m = 12 in
n = 2 in
f = 0 .4
2e 0.4(3.4 ) = 7.8 < m , not self-acting
For CC:
n > me fθ
me fθ < n
12e 0.4(3.4 ) = 46.8 > n , not self-acting
Therefore, there is no change that this brake will be self-acting.
(b) T f max = T f (CL)
fθ
0.4 (3.4 )
−1 
 Wad  e − 1   (30)(26)(18)   e


Tf = 
=


fθ 
0
.
 12 − 2e 4(3.4 )  = 4832 in − lb
2
 2  m − ne  


Tf
4832
MA =
=
= 6.2
Wa (30)(26)
(c) T f (CL ) = 4832 in − lb
fθ
0.4 (3.4 )
−1 
 Wad  e − 1   (30)(26)(18)   e

T f (CC ) = 
=

 0.4(3.4 )
 = 454 in − lb
fθ



− 2
2
 2  me − n  
 12e
T f (CL ) 4832
Ratio =
=
= 10.64
T f (CC ) 454
(d) For SAE 1020, as rolled.
su = 65 ksi
s
65
s= u =
= 8.125 ksi = 8125 psi
N
8
F
s= 1
bt
1
t = in = 0.0625 in
16
Wae fθ
max. F1 =
(CL)
m − ne fθ
(30)(26)e 0.4(3.4 ) = 722.3 lb
F1 =
12 − 2e 0.4(3.4 )
Page 86 of 97
SECTION 16 – BRAKES AND CLUTCHES
722.3
b(0.0625)
b = 1.422 in
With p = 50 psi
F = fpA
θbD
A=
2
Wa (e fθ − 1) (30)(26)(e 0.4(3.4 ) − 1)
max. F =
=
= 536.9 lb
m − ne fθ
12 − 2e 0.4(3.4 )
fpθbD
F=
2
(0.4)(50)(3.4)(b )(18)
536.9 =
2
b = 0.88 in
s = 8125 =
919.
A differential band brake is to be design to absorb 10 fhp at 250 rpm. (a)
Compute the maximum and minimum diameters from both equations (z) and (a),
p. 495, Text. Decide on a size. (b) The band is to be lined with woven asbestos.
The Rasmussen recommendation (§18.4) will help in deciding on the face width.
Also check the permissible pressure in Table AT 29. Choose dimensions of the
lever, its location and shape and the corresponding θ . Be sure the brake is not
self locking. What is the percentage variation of the mechanical advantage from
the minimum value ( f min ) for the f limits in Table AT 29?
Solution:
63,000hp 63,000(10 )
Tf =
=
= 2520 in − lb
n
250
(a) Eq. (z)
1
1
1
1
 Tf
= 
 5
 3  2520  3
 = 
 = 7.96 in
  5 
 Tf
Dmax = 
 4
Eq. (a)
 3  2520  3
 = 
 = 8.57 in
  4 
Dmin
1
1
1
1
Dmin = (60 fhp )3 = [60(10)]3 = 8.44 in
Dmin = (80 fhp )3 = [80(10)]3 = 9.28 in
use D = 8.5 in
(b) By Rasmussen
Energy absorption capacity = 0.2 to 0.3 fhp per sq. in. of brake contact area.
Page 87 of 97
SECTION 16 – BRAKES AND CLUTCHES
Say 0.25 fhp per sq. in.
θ bD
A=
2
fhp = 10 hp
D = 8.5 in
assume θ = 180o = π rad
fhp
fhp in 2 =
A
10
0.25 =
 π b(8.5) 
 2 
b = 3 in
From Table AT 29, f = 0.40 , p per . = 50 psi
F
fA
π (3)(8.5)
A=
= 40 in 2
2
2T
2(2520)
F= f =
= 593 lb
D
8.5
593
p=
= 37.1 psi < 50 psi (OK)
0.4(40)
p=
For MA :
Tf
D e fθ − 1
MA =
=
WA 2 c − be fθ
Not self-locking
c > be fθ
c
> e fθ
b
c
> e 0.4π
b
c
> 3 .5
b
c
say = 4 or c = 4b
b
(
(
)
)
For f = 0.40
T
D e fθ − 1
8.5 e 0.4π − 1
21.96
MA = f =
=
=
fθ
0.4π
WA 2 c − be
2 4b − be
b
For f = 0.35 = f min
(
Page 88 of 97
(
)
(
) (
)
)
SECTION 16 – BRAKES AND CLUTCHES
(
)
(
) (
)
)
D e fθ − 1
8.5 e 0.35π − 1
8.54
=
=
fθ
0.35π
WA 2 c − be
2 4b − be
b
21.96 − 8.54
% variation =
(100%) = 157%
8.54
MA =
Tf
=
(
DISK CLUTCHES
920.
An automobile engine develops its maximum brake torque at 2800 rpm when the
bhp = 200. A design value of f = 0.25 is expected to be reasonable for the
asbestos facing and it is desired that the mean diameter not exceed 8.5 in.;
permissible pressure is 35 psi. Designing for a single plate clutch, Fig. 18.10,
Text, determine the outer and inner diameters of the disk.
Solution:
1
Dm = (Do + Di ) = 8.5 in
2
ro + ri = 8.5 in
63,000hp 63,000(200 )
Tf =
=
= 4500 in − lb
n
2800
p = 35 psi
f N (ro + ri )
Tf =
2
(0.25)(N )(8.5)
4500 =
2
N = 4235 lb
N
ave. p =
2
π (ro − ri 2 )
4235
35 =
π (ro2 − ri 2 )
ro2 − ri 2 = 38.5
ro = 8.5 − ri
(8.5 − ri )2 − ri2 = 38.5
72.25 − 17 ri + ri 2 − ri 2 = 38.5
ri = 1.985 in
say ri = 2.0 in
ro = 8.5 − 2.0 = 6.5 in
Do = 2ro = 2(6.5) = 13 in
Di = 2ri = 2(2.0) = 4 in
Page 89 of 97
SECTION 16 – BRAKES AND CLUTCHES
921.
An automobile engine can develop a maximum brake torque of 2448 in-lb.
Which of the following plate clutches, which make up a manufacturer’s standard
7
1
“line,” should be chosen for this car? Facing sizes: (a) Do = 8 , Di = 6 in ., (b)
8
8
1
1
1
Do = 10 , Di = 6 in ., (c) Do = 11 , Di = 6 in . In each case, assume f = 0.3 .
8
16
8
The unit pressures are (a) 34 psi, (b) 30 psi, and (c) 26.2 psi.
Solution:
f N (ro + ri )
2
f N (Do + Di )
Tf =
4
π
 
N = pave   Do2 − Di2
4
π pave Do2 − Di2 (Do + Di )
Tf =
16
Tf =
(
)
(
)
(a) Do = 8.875 in ; Di = 6.125 in , p = 34 psi , f = 0.3
Tf =
π pave (Do2 − Di2 )(Do + Di )
16
π (0.3)(34)(8.875)2 − (6.125)2 (8.875 + 6.125)
Tf =
= 1239 in − lb
16
[
]
(b) Do = 10 in ; Di = 6.125 in , p = 30 psi , f = 0.3
Tf =
π pave (Do2 − Di2 )(Do + Di )
16
π (0.3)(30)(10)2 − (6.125)2 (10 + 6.125)
Tf =
= 1780 in − lb
16
[
]
(c) Do = 11.0625 in ; Di = 6.125 in , p = 26.2 psi , f = 0.3
Tf =
π pave (Do2 − Di2 )(Do + Di )
16
π (0.3)(26.2)(11.0625)2 − (6.125)2 (11.0625 + 6.125)
Tf =
= 2251 in − lb
16
[
use (c)
Page 90 of 97
]
SECTION 16 – BRAKES AND CLUTCHES
922.
A single-disk clutch for an industrial application, similar to that in Fig. 18.11,
Text, except that there are two disks attached to one shaft and one attached to the
other. The clutch is rated at 50 hp at 500 rpm. The asbestos-in-resin-binder facing
1
3
has a Do = 8 in . and Di = 4 in . What must be the axial force and average
2
4
pressure? How does this pressure compare with that recommended by Table AT
29?
Solution:
n = 2 pairs in contact
f = 0.35 (Table AT 29)
p = 75 psi
63,000hp 63,000(50 )
Tf =
=
= 6300 in − lb
nm
500
Do = 8.5 in
Di = 4.75 in
nf N (ro + ri )
Tf =
2
nf N (Do + Di )
Tf =
4
(2)(0.35)(N )(8.5 + 4.75)
6300 =
4
N = 2717 lb
4N
4(2717 )
ave. p =
=
= 69.6 psi < 75 psi
2
2
π (Do − Di ) π (8.5)2 − (4.75)2
[
923.
]
A multiple-disk clutch similar to Fig. 18.11, Text, is rated at 22 hp at 100 rpm.
The outside and inside diameters of the disks are 14 and 7 ½ in., respectively. If
f = 0.25 , find (a) the axial force required to transmit the rated load, and (b) the
unit pressure between the disks.
Solution:
(a)
Fig. 18-11, n = 4 pairs in contact
63,000hp 63,000(22 )
Tf =
=
= 13,860 in − lb
nm
100
nf N (Do + Di )
Tf =
4
Do = 14 in
Di = 7.5 in
4(0.25)( N )(14 + 7.5)
13,860 =
4
Page 91 of 97
SECTION 16 – BRAKES AND CLUTCHES
N = 2579 lb
(b) p =
924.
4N
4(2579 )
=
= 23.5 psi
2
2
π Do − Di
π (14)2 − (7.5)2
(
)
[
]
A multiple-disk clutch for a machine tool operation has 4 phosphor-bronze
driving disks and 5 hardened-steel driven disks. This clutch is rated at 5.8 hp at
100 rpm when operated dry. The outside and inside diameters of the disks are 5
½ and 4 3/16 in., respectively. (a) If the pressure between the disks is that
recommended for metal on metal in Table AT 29, what coefficient of friction is
required to transmit the rated power? (b) What power may be transmitted for f
and p as recommended in Table AT 29?
Solution:
Do = 5.5 in
Di = 4.1875 in
63,000hp 63,000(5.8)
Tf =
=
= 3654 in − lb
nm
100
n = 8 pairs in contact
(a) Table AT 29, p = 150 psi , metal to metal
nf N (Do + Di )
Tf =
4
π 
π 
2
2
N = p (Do2 − Di2 ) = 150  (5.5) − (4.1875) = 1498 lb
4
4
(8)( f )(1498)(5.5 + 4.1875)
T f = 3654 =
4
f = 0.126
[
]
(b) from Table AT 29, p = 150 psi , f = 0.2
(8)(0.2)(1498)(5.5 + 4.1875) = 5805 in − lb
Tf =
4
T n
(5805)(100) = 9.2 hp
hp = f m =
63,000
63,000
925.
A multiple-disk clutch with three disks on one shaft and two on the other, similar
to that in Fig. 18.11, Text, is rated at 53 hp at 500 rpm. (a) What is the largest
value of Di if f and p are given by Table AT 29 for asbestos in resin binder
and Do = 10.5 in . (b) For the diameter used of Di = 7 in .,what is the required
axial force and the average pressure?
Solution:
Page 92 of 97
SECTION 16 – BRAKES AND CLUTCHES
Table AT 29, asbestos in resin binder, f = 0.3 , p = 75 psi
nf N (ro + ri )
Tf =
2
nf N (Do + Di )
Tf =
4
4T f
N=
nf (Do + Di )
but
π 
N = p (Do2 − Di2 )
4
π 
4T f = nfp  Do2 − Di2 (Do + Di )
4
63,000hp 63,000(53)
Tf =
=
= 6678 in − lb
nm
500
n = 4 pairs in contact
π 
4T f = nfp  Do2 − Di2 (Do + Di )
4
π 
2
4(6678) = (4 )(0.3)(75)  (10.5) − Di2 (10.5 + Di )
4
Di = 9.5607 in
(
)
(
)
[
]
(b) Di = 7 in
nf N (Do + Di )
Tf =
4
(
4 )(0.3)( N )(10.5 + 7 )
6678 =
4
N = 1272 lb
4N
4(1272 )
ave. p =
=
= 26.44 psi
2
2
π (Do − Di ) π (10.5)2 − (7 )2
[
]
MISCELLANEOUS CLUTCHES AND BRAKES
926.
For the cone brake shown, find an expression for the braking torque for a given
applied force W on the bell crank. Consider the force F ′ , Fig. 18.12, Text, in
obtaining the expression.
Page 93 of 97
SECTION 16 – BRAKES AND CLUTCHES
Problems 926-928.
Solution:
927.
For the cone brake similar to that shown, certain dimensions are: Dm = 15 in .,
1
c = 2 in ., α = 12o , b = 9 in ., and a = 20 in . The contact surfaces are metal and
2
asbestos. (a) For an applied force W = 80 lb ., what braking torque may be
expected of this brake? Consider the resistance F ′ , Fig. 18.12, Text. (b) If the
rotating shaft comes to rest from 300 rpm during 100 revolutions, what frictional
work has been done? (c) What must be the diameter of the steel pin P , SAE
1020 as rolled, for a factor fo safety of 6 against being sheard off? The diameter
1
of the hub d = 4 in . (d) What is the unit pressure on the face of the brake?
2
Solution:
f Dm R
f Dm aW
=
2(sin α + f cos α ) 2b(sin α + f cos α )
Table AT 29, asbestos on metal, f = 0.40
(0.40)(15)(20)(80) = 890 in − lb
Tf =
2(9)(sin 12 + 0.4 cos 12)
(a)
Tf =
2π (300 )
= 31.42 rad sec
60
ω2 = 0 rad sec
θ = 100(2π ) = 628.3 rad
1
θ = (ω1 + ω2 )t
2
1
628.3 = (31.42 + 0 )t
2
(b) ω1 =
Page 94 of 97
SECTION 16 – BRAKES AND CLUTCHES
t = 40 sec
1
nm = (300 + 0 ) = 150 rpm
2
T n
(890)(150) = 2.119 hp
fhp = f m =
63,000
63,000
U f = 550( fhp )(t ) = 550(2.119 )(40 ) = 46,618 ft − lb
(c) For SAE 1020, as rolled,
ssu = 49 ksi
s
49
ss = su =
= 8.17 ksi
N
6
4R
ss =
πd2
aW (20 )(80 )
R=
=
= 177.8 lb
b
9
4(177.8)
ss = 8170 =
πd2
d = 0.1665 in
3
say d = in
16
N
π Dm c
R
177.8
N=
=
= 297 lb
sin α + f cos α sin 12 + 0.4 cos 12
297
p=
= 2.52 psi
π (15)(2.5)
(d) p =
928.
A cone clutch for industrial use is to transmit 15 hp at 400 rpm. The mean
diameter of the clutch is 10 in. and the face angle α = 10o ; let f = 0.3 for the
cast-iron cup and the asbestos lined cone; permissible p = 35 psi . Compute (a)
the needed axial force, (b) the face width, (c) the minimum axial force to achieve
engagement under load.
Solution:
63,000hp 63,000(15)
Tf =
=
= 2362.5 in − lb
n
400
f Dm R
(a) T f =
2(sin α + f cos α )
Page 95 of 97
SECTION 16 – BRAKES AND CLUTCHES
2362.5 =
(0.3)(10)R
2(sin 10 + 0.3 cos 10)
R = 739 lb
N
π Dm c
R
739
N=
=
= 1575 lb
sin α + f cos α sin 10 + 0.3 cos 10
1575
35 =
π (15)c
c = 1.44 in
(b) p =
(c) max. f = 0.4 (Table AT 29)
f Dm R
Tf =
2(sin α + f cos α )
(0.4)(10)R
2362.5 =
2(sin 10 + 0.4 cos 10)
R = 670 lb , minimum.
929.
An “Airflex” clutch, Fig. 18.15, Text, has a 16-in drum with a 5-in. face. This
clutch is rated at 110 hp at 100 rpm with an air pressure of 75 psi. What must be
the coefficient of friction if the effect of centrifugal force is neglected? (Data
courtesy of Federal Fawick Corporation.)
Solution:
D = 16 in
b = 5 in
hp = 110 hp
rpm = 100 rpm
p = 75 psi
63,000hp 63,000(110 )
Tf =
=
= 69,300 in − lb
n
100
FD
Tf =
2
F (16 )
69,300 =
2
F = 8662.5 lb
N = p(π Db ) = (75)(π )(16)(5) = 18,850 lb
F 8662.5
f = =
= 0.46
N 18,850
Page 96 of 97
SECTION 16 – BRAKES AND CLUTCHES
930.
The same as 929 except that the diameter is 6 in., the face width is 2 in., and the
rated horsepower is 3.
Solution:
63,000hp 63,000(3)
Tf =
=
= 1890 in − lb
n
100
FD
Tf =
2
F (6 )
69,300 =
2
F = 630 lb
N = p(π Db ) = (75)(π )(6)(2 ) = 2827 lb
F
630
f = =
= 0.22
N 2827
- end -
Page 97 of 97
SECTION 17– WELDING
DESIGN PROBLEMS
941.
A joint welded with a coated rod is to support a steady load of 10 kips; the design
is to be for a 3/8-in weld. Compute the length of weld needed for (a) a reinforced
butt joint, Fig. 19-2, Text, (b) a lap joint, Fig. 19-3(a), (c) a T-joint, Fig. 19-3(b),
(d) fillet welds, parallel loaded, Fig. 19-3(d).
Solution:
(a) Reinforced butt joint, Fig. 19-2.
F = ηs t tL
From Table AT 30, using Jennings recommendation, st = 16 ksi
F = 10 kips
η = 0.90
3
t = in = 0.375 in
8
3
F = 10 = (0.90 )(16 )  L
8
L = 1.85 in
(b) Lap joint
F = 2ss Lb cos 45
Table AT 30, normal or transverse loading, ss = 16 ksi (Lincoln-Electric)
F = 10 kips
3
b = in = 0.375 in
8
3
F = 10 = 2(16 )L  cos 45
8
L = 1.18 in
(c) T-joint
F = 2ss Lb cos 45
Table AT 30, all loading, ss = 14 ksi (Jennings recommendation)
F = 10 kips
3
b = in = 0.375 in
8
3
F = 10 = 2(14 )L  cos 45
8
Page 1 of 41
SECTION 17– WELDING
L = 1.35 in
(d) Fillet welds, parallel load
F = 2ss Lb cos 45
Table AT 30, ss = 13.6 ksi (AWS Code)
F = 10 kips
3
b = in = 0.375 in
8
3
F = 10 = 2(13.6 )L  cos 45
8
L = 1.39 in
943.
The load F varies from 5 to 10 kips and the arrangement is such that the
location is given by m = 3 in and n = 7 in . It is desired that the weld lengths L1
and L2 be such that the line of action of F passes through the centroid G of the
weld metal, thereby avoiding eccentric loading. Determine (a) the ratio of L1 L2 ,
(b) the lengths L1 and L2 of 3/8 in. fillet welds made with E6010 rod, for
indefinite life. (c) The same as (b) except that the life expectancy is 105 cycles.
Problem 943
Solution:
F = 2ss Lb cos 45
F1 = 2 ss L1b cos 45
F2 = 2s s L2b cos 45
(a) Solving for the ratio of L1 L2
[∑ M
G
=0
]
Page 2 of 41
SECTION 17– WELDING
F1m = F2 n
F1 n
=
F2 m
F1 L1 7
=
=
F2 L2 3
L1 : L2 = 7 : 3
(b) Solving for the lengths
F1 + F2 = F
3F1 + 3F2 = 3F
7 F2 + 3F2 = 3F
F2 = 0.3F
F1 = 0.7 F
For indefinite life use nc = 2× 10 6
Table AT 30
7 .2
sd =
1 − 0 .5 R
5 kips
R=
= 0.50
10 kips
7.2
sd =
= 9.6 ksi
1 − 0.5(0.5)
F = 10 kips
F1 = 0.7(10 ) = 7 kips
F2 = 0.3(10) = 3 kips
F1 = 7 = 2(9.6)L1 (0.375) cos 45
L1 = 1.375 in
F2 = 3 = 2(9.6)L2 (0.375) cos 45
L2 = 0.589 in
(c) Solving for the lengths, 105 cycles
For indefinite life use nc = 105
Table AT 30
12.2
sd =
1 − 0 .5 R
12.5
sd =
= 16.67 ksi
1 − 0.5(0.5)
Page 3 of 41
SECTION 17– WELDING
F1 = 7 = 2(16.67 )L1 (0.375) cos 45
L1 = 0.792 in
F2 = 3 = 2(16.67 )L2 (0.375) cos 45
L2 = 0.339 in
944.
A bracket somewhat as shown is made of structural steel and supports a repeated
( R = 0 , nc = 2× 10 6 ) load of 2000 lb at a distance a = 10 in from the wall. What
should be the length L of a 3/8-in. fillet weld that resists the entire load? Adapt
the design shear stress from Table AT 30 (fillet weld).
Problem 944, 958
Solution:
Fa (L 2 )
I
3
2tL
I=
12
FaL
3Fa
st =
= 2
3
 2tL  tL

2
12


t = b cos 45
st =
1
1
2 2

 F  2  3Fa  2  2
s  
τ =  ss2 +  t   = 
 + 2  
 2  
 2tL   2tL  

From Table AT 30, R = 0 , nc = 2× 10 6
7.2
7.2
τ=
=
= 7.2 ksi
1 − 0.5R 1 − 0.5(0)
F = 2000 lb = 2 kips
a = 10 in
Page 4 of 41
SECTION 17– WELDING
b=
3
in
8
2
 F   3Fa 
τ =
 + 2 
 2tL   2tL 
2
2
2

2
1
(7.2) = 12 
+ 4

L  2(0.375) cos 45  L
14.22 12,800
51.84 = 2 +
L
L4
51.84 L4 = 14.22 L2 + 12,800
51.84 L4 − 14.22 L2 − 12,800 = 0
L = 3.98 in
say L = 4 in
2
945.
 3(2)(10) 
 2(0.375) cos 45 


2
A bracket is to be fabricated from flat plates by bending and welding with a
shielded rod, E6010. A steady load F = 5000 lb , L = 18 in , h = 4 in , and
a = 6 in . (a) Take the design shear stress for a design factor N = 3.75 on the
ultimate shear strength, which may be estimated at 80% of su of rod, and find
the size of fillet weld ABCDA. Compare the design stress used with values in
Table AT 30. (b) Compute the thickness of the SAE 1020, rolled-steel plates of
all are the same (cantilever part).
Problem 945
Solution:
h
FL 
2
st =
I
Page 5 of 41
SECTION 17– WELDING
I=
2th 3 2at 3
h
+
+ 2(at ) 
12
12
2
2
th 3 at 3 ath 2
+
+
6
6
2
h = 4 in , a = 6 in , L = 18 in , F = 5 kips
I=
t (4 ) 6t 3 (6 )t (4 )
I=
+
+
= 58.67t + t 3
6
6
2
(5)(18) 4 
180
2 =
st =
3
58.67t + t
58.67t + t 3
F
5
0.25
ss =
=
=
2at + 2th 2(6)t + 2t (4)
t
3
s 
τ = s + t 
2
s
(a) τ d = us
N
sus = 0.8su
2
2
2
2
s
su = 60 ksi for E6010
N = 3.75
(0.8)(60) = 12.8 ksi
τd =
3.75
2
180
 0.25  
 + 
3
 t   2 58.67t + t
0.0625
8100
163.84 =
+
2
2
t
58.67t + t 3
τ 2 = (12.8)2 = 
(
(
)



2
)
By trial and error method
t = 0.121399 in
t = b cos 45
0.121399 = b cos 45
b = 0.1717 in
3
use b =
= 0.1875 in
16
Comparing the design stress, from Table AT 30
τ = 13.6 ksi (AISC Building Code)
τ = 13.6 ksi > 12.8 ksi
(b) Solving for the thickness of the SAE 1020, rolled-steel plates
Table 1.1, use N = 2 based on yield strength
Page 6 of 41
SECTION 17– WELDING
Table AT 7, s y = 48 ksi
sd =
sy
=
48
= 24 ksi 6391
2
N
Mc
sd =
I
h 4
c = = = 2 in
2 2
I = 58.67t + t 3
M = FL = (5)(18) = 90 in − kips
Mc
sd =
I
(90 in − kips )(2 in )
24 ksi =
58.67t + t 3 in 4
58.67t + t 3 = 7.5
By trial and error method
t = 0.128 in
but t ≥ b
3
use t = in
16
946.
A bracket is made with two 3/8-in. steel plates A welded with a coated electrode
to a vertical I-beam with fillet welds on both sides of the plate, as indicated. It
supports a steady vertical load F = 12 kips in a center position; a = 14 in ;
h = 8 in . What size and length of weld should be used? Is the stress at G the
maximum one? Justify your answer.
Problem 946 – 948
Page 7 of 41
SECTION 17– WELDING
Solution:
a = 14 in
h = 8 in
F = 12 kips (steady load)
use weld size = 3/8 in = plate thickness = 0.375 in
τ B > τ G , therefore the stress at G is not the maximum one.
L
 
2
cos θ =   =
1
ρ′
2
By cosine law
L
 
2
=
L2 + h 2
1
τ = (ss2 + ss2 + 2ss ss cos θ )2
1
2
F ′
 eρ
2
ss1 =  
Jc
L
e=a+
2
Page 8 of 41
1
2
L
L2 + h 2
SECTION 17– WELDING
2
3
2tL3
 h  tL 1
+ 2tL  =
+ tLh 2
12
6 2
2
F
 
2
s s2 =  
2tL
Jc =
Substituting values
t = b cos 45 = (0.375) cos 45 = 0.2652 in
1 2
1 2
1 2
2
L + h2 =
L + (8) =
L + 64
ρ′ =
2
2
2
L
cos θ =
L2 + 64
e = 14 + 0.5 L
0.2652 L3 1
2
Jc =
+ (0.2652)L(8) = 0.0442 L3 + 8.1984 L
6
2
 12 
1 2
 (14 + 0.5L )  L + 64
33.94(28 + L ) L2 + 64
2
2
ss1 =  
=
0.0442 L3 + 8.1984 L
L3 + 185.484 L
F
 12 
 
 
11.43
2

2
s s2 =
=
=
2tL 2(0.2652 )L
L
2
2
2
τ = ss1 + ss2 + 2ss1 ss2 cos θ
From Table AT 30,
Use τ d = 14 ksi as recommended by Jennings
Solving this by trial and error method
τd >τ
)
(
(
L, in
8
7
5
4
3.5
3.75
3.875
)
)
cos θ
0.7071
0.6585
0.5300
0.4472
0.4008
0.4244
0.4359
Use L = 3.875 in , τ = 13.92 ksi ≈ 14ksi
7
Or L = 3 in each weld
8
Page 9 of 41
(
ss1 , ksi
ss2 , ksi
6.9260
7.6930
10.040
12.053
13.489
12.723
12.377
1.429
1.633
2.286
2.858
3.266
3.048
2.950
τ , ksi
8.00
8.85
11.42
13.57
15.10
14.28
13.92
SECTION 17– WELDING
947.
The same as 946, except that the material is aluminum alloy, welded with
shielded 1100 wire, and the load is 5 kips. Let the design factor N = 3.4 for the
information in Table At 30; but consider other approaches, as available, to a
design stress.
Solution:
With F = 5 kips
5
1 2
 (14 + 0.5L )  L + 64
14.14(28 + L ) L2 + 64
2
2
ss1 =  
=
0.0442 L3 + 8.1984 L
L3 + 185.484 L
F
5
 
 
4.714
2
2
s s2 =   =
=
2tL 2(0.2652 )L
L
L
cos θ =
2
L + 64
2
2
τ = ss1 + ss22 + 2ss1 ss2 cos θ
From Table AT 30,
ultimate strength of aluminum alloy welded with 1100 wire, = 12.7 ksi.
12.7
τd =
= 3.735 ksi
3 .4
Solving this by trial and error method
τd >τ
)
(
(
(
)
)
L, in
8
7
6.5
6.75
6.875
cos θ
0.7071
0.6585
0.6306
0.6449
0.6518
ss1 , ksi
ss2 , ksi
2.886
3.205
3.397
3.298
3.251
0.589
0.673
0.725
0.698
0.686
τ , ksi
3.326
3.683
3.895
3.786
3.735
Use L = 6.875 in , τ = 3.735 ksi = τ d
7
Or L = 6 in for each weld
8
948.
(a) Two ¾-in plates A, arranged as shown, are to be welded with coated
electrodes, E6020; a = 12 in ; h = 4 in , and F repeats from 0 to 10 kips. Choose
a design stress from Table AT 30 for 2x106 cycles and specify the size and length
of weld. (b) The same as (a), except that the design is for 105 cycles. (c)
Demonstrate that the stress at G is or is not the maximum.
Page 10 of 41
SECTION 17– WELDING
Problem 946 – 948
Solution:
This problem is the same as 946 except that
a = 12 in
h = 4 in
F = 0 to 10 kips
3
b = in = same as plate thickness
4
t = b cos 45 = (0.75) cos 45 = 0.53 in
L
L
L
cos θ =
=
=
2
2
2
2
2
L +h
L +4
L + 16
F ′
 eρ
2
ss1 =  
Jc
L
e = a + = 12 + 0.5 L
2
1 2
1 2
1 2
2
L + h2 =
L + (4 ) =
L + 16
ρ′ =
2
2
2
0.53L3 1
2
Jc =
+ (0.53)L(4) = 0.088L3 + 4.24 L
6
2
F = 10 kips
 10 
1 2
 (12 + 0.5L )  L + 16
14.205(24 + L ) L2 + 16
2
2
ss1 =  
=
0.088L3 + 4.24 L
L3 + 48.18L
F
 10 
 
 
4.714
2
2

s s2 =
=   =
2tL 2(0.53)L
L
2
2
2
τ = ss1 + ss2 + 2ss1 ss2 cos θ
)
(
(
)
(a) From Table AT 30 , nc = 2× 10 6
Page 11 of 41
(
)
SECTION 17– WELDING
7 .2
ksi
1 − 0 .5 R
0
R=
=0
10
7.2
τd =
= 7.2 ksi
1 − 0.5(0)
Solving for length by trial and error method
τd >τ
τd =
L, in
8
7
6
5.5
5.625
cos θ
0.8944
0.8682
0.8321
0.8087
0.8150
ss1 , ksi
ss2 , ksi
4.530
5.219
6.084
6.607
6.469
0.590
0.674
0.786
0.858
0.839
τ , ksi
5.06
5.81
6.87
7.32
7.17
Use L = 5.625 in , τ = 7.17 ksi ≈ τ d
5
Or L = 5 in for each weld
8
(b) From Table AT 30, nc = 105
12.5
τd =
ksi
1 − 0 .5 R
0
R=
=0
10
12.5
τd =
= 12.5 ksi
1 − 0.5(0)
Solving for length by trial and error method
τd >τ
L, in
4
3
cos θ
0.7071
0.6000
ss1 , ksi
ss2 , ksi
8.764
11.179
1.179
1.573
Use L = 3 in , τ = 12.188 ksi ≈ τ d
Or L = 3 in for each weld
(c) Demonstrating that the stress at G is or is not the maximum
Page 12 of 41
τ , ksi
9.634
12.188
SECTION 17– WELDING
From the figure τ G is not the maximum.
949.
One arm of a bracket that is to support a steady load of F = 18 kips without
twisting is welded with an E6010 rod, as shown. The plate is 10 in. ( ≈ L2 ) deep.
Assume a value of L1 (not less than 5 in.) and compute the size of fillet weld. By
sketching vectors (only), compare the stress at C with that of B.
Problems 949-951
Solution:
Use L1 = 5 in , L2 = 10 in
F = 18 kips
Solving for the center G.
(L1 + L1 + L2 )x = 2 L1  L1 
  2 
Page 13 of 41
SECTION 17– WELDING
(5) = 1.25 in
L12
=
2 L1 + L2 2(5) + 10
2
x=
e = 11.25 in − 1.25 in = 10 in
at B, where τ is maximum
Feρ1
Jc
For J c :
ss1 =
r12 = (2.5 − 1.25) + (5)
r1 = 5.154 in
r2 = 1.25 in
2
2
2tL13 tL32
2t (5) t (10 )
2
2
Jc =
+
+ tL2 r22 + 2tL1r12 =
+
+ t (10 )(1.25) + 2t (5)(5.154 ) = 385.43t
12 12
12
12
2
2
2
ρ1 = (5) + (5 − 1.25)
ρ1 = 6.25 in
(18)(10)(6.25) = 2.92
ss1 =
385.43t
t
3
Page 14 of 41
3
SECTION 17– WELDING
F
18
0 .9
=
=
2tL1 + tL2 2t (5) + t (10 )
t
L − x 5 − 1.25
cos θ1 = 1
=
= 0.60
ρ1
6.25
s s2 =
τ B2 = τ 2 = ss2 + ss2 + 2ss ss cos θ1
1
2
1
2
2
2
 2.92   0.90 
 2.92  0.90 
 +
 + 2

(0.6 )
 t   t 
 t  t 
3.534
τ=
t
From Table AT 30, steady load, use code
τ = 13.6 ksi
τ2 =
τ = 13.6 =
3.534
t
t = 0.26 in
t = b cos 45
0.26 = b cos 45
b = 0.368 in
Therefore use size of fillet weld = b = 0.375 in =
3
in
8
Comparing stress at C with that of B.
Stress at B
Stress at C
τ B > τ C , ρ1 > ρ 2 , ss′ > ss
1
950.
1
The same as 949, except that F makes a 30o angle with the vertical as indicated
by the dotted line in the figure. Consider all computed stress to be shear.
Solution:
Page 15 of 41
SECTION 17– WELDING
L1 = 5 in
L2 = 10 in
F = 18 kips
T = (F cos 30)e − (F sin 30)(3)
e = 10 in as in 949.
T = (18 cos 30)(10) − (18 sin 30)(3) = 128.89 in − kips
As in 949,
Tρ
ss1 = 1
Jc
ρ1 = 6.25 in
(128.89)(6.25) = 2.09
ss1 =
385.43t
t
0 .9
s s2 =
(from 949)
t
at B
cos θ1 = 0.60 from 949.
θ1 = 53.13
α = 90 − 30 − 53.13 = 6.87 o
τ B2 = τ 2 = ss21 + ss22 + 2ss1 ss2 cos α
2
2
 2.09   0.90 
 2.09  0.90 
τ =
 +
 + 2

 cos 6.87
 t   t 
 t  t 
2.986
τ=
t
but τ = 13.6 ksi as in 949
2
Page 16 of 41
SECTION 17– WELDING
τ = 13.6 =
2.986
t
t = 0.22 in
t = b cos 45
0.22 = b cos 45
b = 0.311 in
Therefore use size of fillet weld = b = 0.3125 in =
5
in
16
Comparing stress at C with that of B.
Stress at B
Stress at C
τ B >τC
951.
The same as 949, except that the load varies from 4 to 18 kips; expected life, 2 x
106 cycles. Solve (a) by using a design stress from Table AT 30 for the given
value of R , and (b) by using a design stress for R = −1 , and the Soderberg
criterion.
Solution: same as 949, but
F = 4 to 18 kips
F = 18 kips (is the same as in 949)
3.534
then τ =
t
(a) Table AT 30, nc = 2 × 10 6 cycles
4
R = = 0.222
18
7.2
7.2
τd =
=
= 8.1 ksi
1 − 0.5 R 1 − 0.5(0.222 )
Page 17 of 41
SECTION 17– WELDING
3.534
t
3.534
8 .1 =
t
t = 0.4363 in
t = b cos 45
0.4363 = b cos 45
b = 0.617 in
5
say in fillet weld.
8
τ =τd =
(b) Using a design stress for R = −1 and the Soderberg criterion, nc = 2 × 10 6 cycles
1 sm K f sa
=
+
N sy
sn
For R = −1
sm = 0
sa = τ d
s
τd = n
KfN
Table 19.1, sn = 11.7 ksi ,
i4.20,
use N = 1.4
assume K f = 1.67
11.7
= 5.0 ksi
(1.4)(1.67 )
3.534
τ =τd = 5 =
t
t = 0.7068 in
t = b cos 45
0.7068 = b cos 45
b = 0.9996 in
say b = 1 − in fillet weld.
τd =
952.
A steel plate, welded to a column as shown with E6010 rod, is to support a steady
load of F = 5 kips , applied so as to produce no twisting of the plate; m = 24 in ,
n = 18 in ; the initial design is for a 3/8-in. fillet weld. Compute L . Demonstrate
by sketches which stress ss A or ssB is the larger.
Page 18 of 41
SECTION 17– WELDING
Problems 952-954, 959
Solution:
F = 5 kips
m = 24 in
n = 18 in
3
b = in = 0.375 in
8
t = b cos 45 = (0.375) cos 45 = 0.2652 in
maximum τ = ssA > ssD
ρ=
1
2
(m − n )2 + L2
cos θ =
L
(m − n )2 + L2
τ 2 = ss2 + ss2 + 2ss ss cos θ
1
ρ=
1
2
2
1
2
(24 − 18)2 + L2
Page 19 of 41
=
1 2
L + 36
2
SECTION 17– WELDING
cos θ =
L
L2 + 36
Feρ
Jc
F = 5 kips (steady load)
1
1
1
e = n + (m − n ) = (m + n ) = (24 + 18) = 21 in
2
2
2
ss1 =
2tL3
m−n
Jc =
+ 2tL

12
 2 
2
2(0.2652 )L3
 24 − 18 
+ 2(0.2652 )L

12
 2 
J c = 0.0884 L3 + 4.774 L
2
Jc =
(5)(21) 1 
L2 + 36
594 L2 + 36
2

ss1 =
=
0.0884 L3 + 4.774 L
L3 + 54 L
F
5
9.427
=
=
s s2 =
2tL 2(0.2652 )L
L
From Table AT 30, use AISC Building Code
τ d = 13.6 ksi
Solving for L by trial and error method.
τd ≥τ
L, in
6
5
4.875
4.75
cos θ
0.7071
0.6402
0.6306
0.6207
ss1 , ksi
ss2 , ksi
τ , ksi
9.334
11.745
12.113
12.500
1.571
1.885
1.934
1.985
10.50
13.03
13.42
13.82
7
∴ use L = 4 in = 4.875 in , τ d ≥ τ
8
953.
A steel plate, welded as shown with E6010 rod, is to support a load that varies
from –5 to 5 kips, without twisting; m = 14 in , n = 8 in ; the initial design is for a
3/8-in. fillet weld; indefinite life. Compute L .
Solution:
The same as 952, except
m = 14 in
n = 8 in
F = −5 to 5 kips
Page 20 of 41
SECTION 17– WELDING
3
in fillet weld
8
t = b cos 45
t = (0.375) cos 45 = 0.2652 in
F = 5 kips
Then;
1
ρ=
(m − n )2 + L2
2
1
ρ=
(14 − 8)2 + L2 = 1 L2 + 36
2
2
L
cos θ =
(m − n )2 + L2
L
cos θ =
L2 + 36
1
1
1
e = n + (m − n ) = (m + n ) = (14 + 8) = 11 in
2
2
2
b=
Jc =
2tL3
m−n
+ 2tL

12
 2 
2
2(0.2652 )L3
 14 − 8 
Jc =
+ 2(0.2652 )L

12
 2 
J c = 0.0884 L3 + 4.774 L
Feρ
ss1 =
Jc
(5)(11) 1 
2
L2 + 36
311 L2 + 36
2
=
0.0884 L3 + 4.774 L
L3 + 54 L
F
5
9.427
s s2 =
=
=
2tL 2(0.2652 )L
L
2
2
2
τ = ss1 + ss2 + 2ss1 ss2 cos θ
ss1 =
From Table AT 30, indefinite life ( nc = 2 × 10 6 )
7 .2
τd =
1 − 0 .5 R
−5
R=
= −1
5
7.2
τd =
= 4.8 ksi
1 − 0.5(− 1)
Solving for L by trial and error method.
τd ≥τ
Page 21 of 41
SECTION 17– WELDING
L, in
6
7
8
7.5
7.75
7.25
7.375
cos θ
0.7071
0.7593
0.8000
0.7809
0.7907
0.7704
0.7757
ss1 , ksi
ss2 , ksi
τ , ksi
4.887
3.977
3.295
3.613
3.448
3.788
3.699
1.571
1.347
1.178
1.257
1.216
1.300
1.278
6.10
5.01
4.30
4.66
4.47
4.86
4.76
3
∴ use L = 7 in = 7.375 in , τ = 4.76 ≈ τ d
8
3
or L = 7 in for each weld.
8
The same as 953, except that F varies from 0 to 5 kips with a life expectancy of
105 cycles.
954.
Solution:
Same as 953, but
F = 0 to 5 kips
nc = 105
From table AT 30, nc = 105
12.5
τd =
1 − 0 .5 R
0
R= =0
5
12.5
τd =
= 12.5 ksi
1 − 0.5(0)
L
cos θ =
L2 + 36
311 L2 + 36
ss1 =
L3 + 54 L
9.427
s s2 =
L
2
2
τ = ss1 + ss22 + 2ss1 ss2 cos θ
Solving by trial and error method.
τd ≥τ
L, in
3
3.5
3.125
Page 22 of 41
cos θ
0.4472
0.5039
0.4619
ss1 , ksi
ss2 , ksi
τ , ksi
11.038
9.317
10.558
3.142
2.694
3.017
12.76
10.93
12.25
SECTION 17– WELDING
∴ use L = 3.125 in , τ = 12.25 ≈ τ d
1
or L = 3 in for each weld.
8
955.
An arm for a machine is to be fabricated by welding, coated welding rod. See
figure. The left end is a hollow cylinder with Do = 3 in , and it is keyed to a 2-in.
shaft; L = 14 in , steady load F = 600 lb . The arm material is SAE 1020, rolledsteel plate, ½-in. thick. Compute (a) the depth h of the arm at the hub, and (b) the
size of the weld.
Problem 955
Solution:
(a) Solving for the length or depth h
For SAE 1020, rolled-steel plate,
s y = 48 ksi (Table AT 7)
N = 2 (Table 1.1)
48
sd =
= 24 ksi
2
h
Fe  
2
s=  
I
1
e ≈ L − (Do + Ds )
2
Do = 3 in
Ds = 2 in
L = 14 in
1
e = 14 − (3 + 2 ) = 11.5 in
2
3
th
I=
12
1
t = in = 0.5 in
2
Page 23 of 41
SECTION 17– WELDING
0.5h 3
12
F = 600 lb steady
(600)(11.5) h 
 2  = 82,800
s=
3
h2
 0.5h 


 12 
h = 1.857 in
7
say h = 1 in
8
I=
(b) Solving for the size of weld
t = b cos 45
h
Fe 
2
st =  
I
3
2th
I=
12
h
Fe 
3Fe
2
st =  3  = 2
 2th  th


 12 
3(0.6 )(11.5) 5.888
st =
=
2
t
t (1.875)
F
0.6
0.16
st =
=
=
2th 2t (1.875)
t
1
2 2

s  
τ =  ss2 +  t  
 2  

From Table AT 30, use Jennings recommendations, τ = 14 ksi
1
 0.16  2  5.888  2  2
τ = 14 = 
 +
 
 t   2t  
t = 0.2106 in
0.2106 = b cos 45
Page 24 of 41
SECTION 17– WELDING
b = 0.2978 in
5
say b = in fillet weld.
16
956.
A pair of gusset plates, 3/8-in. thick, are to be welded with E6010 electrodes, as
shown. The load F on the plates varies from 0 to 10 kips (no twisting of plates).
For the first approximation, assume that BC = AD = L = 5 in and compute the
size of weld. With free hand sketches, compare the resultant stress at each corner
A, B, C, and D.
Problems 956, 957
Solution:
ρ1 =
(2)2 + (6)2
Page 25 of 41
= 6.325 in
SECTION 17– WELDING
ρ2 =
(2)2 + (9)2
= 9.220 in
F
2tL
F = 0 to 10kips
F = 10 kips
L = 5 in
10
1
s s2 =
=
2t (5) t
τC =τD ; τ A =τB
Feρ
ss1 =
Jc
e = 6 + 2 = 8 in
s s2 =
2
2
2tL3
3
2t (5)
3


Jc =
+ 2tL 6 +  =
+ 2t (5) 6 +  = 583.3t
12
2
12
2


at D
2
= 0.3162
ρ1 6.325
Feρ (10 )(8)(6.325) 0.8675
=
=
=
Jc
583.3t
t
cos θ1 =
ss1D
2
3
=
1
τ D = (ss2 + ss2 + 2ss ss cos θ1 )2
1D
2
1D
2
1
 0.8675  2  1  2  0.8675  1 
 2 1.5170
τ D = 
 +   + 2
 (0.3162) =
t
 t  t 
 t   t 

at A
cos θ 2 =
2
ρ2
=
2
9.220
θ 2 = 77.47 o
α = 180 − 77.47 = 102.53o
Page 26 of 41
SECTION 17– WELDING
ss1 A =
Feρ (10 )(8)(9.220 ) 1.2645
=
=
Jc
583.3t
t
1
τ A = (ss2 + ss2 + 2ss ss cos α )2
1A
2
1A
2
1
 1.2645  2  1  2  1.2645  1 
 2 1.4319
τ A = 
 +   + 2
  cos 102.53 =
t
 t  t 
 t   t 

1.5170
t
From Table AT 30, assume nc = 2 × 10 6 .
7 .2
τd =
1 − 0 .5 R
0
R=
=0
10
7.2
τd =
= 7.2 ksi
1 − 0.5(0)
1.5170
τ = 7 .2 =
t
t = 0.2107 in
t = b cos 45
0.2107 = b cos 45
b = 0.2980 in
5
say b = in for each weld.
16
τ max = τ D =
CHECK PROBLEMS
957.
A 3/8-in. gusset plate is welded with an E6010 electrode; ¼-in. fillet weld, as
shown. The loading does not twist the plate and the force varies from 0.2 F to
F . For a life expectancy of 105 cycles, what is a safe F ? Make clear how you
decide upon the point of maximum stress.
Solution:
Same as 956, except that F is unknown and varies from 0.2 F to F , nc = 105 cycles.
Stress vector is the same as shown in 956, τ D = τ C is maximum.
At D (prob. 956)
ρ1 = 6.325 in
F
s s2 =
2tL
Page 27 of 41
SECTION 17– WELDING
t = b cos 45
b = 0.25 in
t = (0.25) cos 45 = 0.1768 in
L = 5 in
F
s s2 =
= 0.5656 F
2(0.1768)(5)
Feρ1
ss1 =
Jc
e = 6 + 2 = 8 in
ρ1 = 6.325 in
2
2
2tL3
3
2(0.1768)(5)
3


Jc =
+ 2tL 6 +  =
+ 2(0.1768)(5) 6 +  = 23.573 in 4
12
2
12
2


Feρ1 F (8)(6.325)
ss1 =
=
= 2.1465 F
Jc
23.573
3
τ 2 = ss2 + ss2 + 2ss ss cos θ
cos θ = 0.3162 (prob. 956)
1
2
1
2
From Table AT 30, nc = 105 cycles.
12.5
τd =
1 − 0 .5 R
0 .2 F
R=
= 0 .2
F
12.5
τd =
= 13.89 ksi
1 − 0.5(0.2)
τ 2 = ss21 + ss22 + 2ss1 ss2 cos θ
(13.89)2 = (2.1465 F )2 + (0.5656 F )2 + 2(2.1465F )(0.5656 F )(0.3162)
F = 5.82 kips
958.
A bracket of the type shown (944) is to support a load of F = 6 kips ; a = 8 in ,
L = 5 in , weld size is 3/8-in. (a) Determine the stress in the weld. Is this safe
value for a steady load? (b) If the welding is shielded and the load varies with
R = 0 , is the weld safe for 2 × 10 6 cycles? For 105 cycles? For Q & T alloy and
2 × 10 6 cycles?
Solution:
Page 28 of 41
SECTION 17– WELDING
F = 6 kips
a = 8 in
L = 5 in
3
b = in = 0.375 in
8
t = b cos 45 = (0.375) cos 45 = 0.2652 in
F
6
ss =
=
= 2.3 ksi
2tL 2(0.2652)(5)
L
Fa 
3Fa
3(6)(8)
2
st =  3  = 2 =
= 21.7 ksi
2
 2tL  tL
(
0.2652)(5)


12


1
 2  st  2  2
τ =  ss +   
 2  

1
2

 21.7   2
2
τ = (2.3) + 
  = 11.1 ksi
 2  

(a) Steady load, Table AT 30, AISC Building Code
τ d = 13.6 ksi
Since τ d > 11.1 ksi ∴ a safe value
(b) Table AT 30, nc = 2 × 10 6 , R = 0
7.2
7.2
τd =
=
= 7.2 ksi
1 − 0.5R 1 − 0.5(0)
Page 29 of 41
SECTION 17– WELDING
Since τ d < 11.1 ksi ∴ not a safe value
Table AT 30, nc = 105 , R = 0
12.5
12.5
τd =
=
= 12.5 ksi
1 − 0.5 R 1 − 0.5(0 )
Since τ d > 11.1 ksi ∴ a safe value
Table AT 30, nc = 2 × 10 6 ,Q & T alloy, R = 0
9
9
τd =
=
= 9 ksi
1 − 0.5R 1 − 0.5(0)
Since τ d < 11.1 ksi ∴ not a safe value
959.
The 1-in. plate shown (952) is attached with ½-in fillet welds, laid with E6010
rods; L = 4 in , m = 15 in , n = 9 in . What maximum load may be carried if it is
(a) static, (b) varies from 0.5 F to F for 2 × 10 6 cycles and for 105 cycles, (c)
R = 0 , indefinite life. (d) Considering strengths given in Table 19.1, Text,
determine the design factor for parts (b) and (c).
Solution:
Same as 952, except that m = 15 in , n = 9 in , b =
1
in , L = 4 in
2
t = b cos 45 = (0.5) cos 45 = 0.3536 in
τ 2 = ss21 + ss22 + 2ss1 ss2 cos θ
ss1 =
Feρ
Jc
1
(m + n ) = 1 (15 + 9) = 12 in
2
2
1
ρ=
(m − n )2 + L2 = 1 (15 − 9)2 + (4)2 = 3.606 in
2
2
e=
2
2
2tL3
2(0.3536 )(4 )
m−n
 15 − 9 
4
Jc =
+ 2tL
+ 2(0.3536 )(4 )
 =
 = 29.23 in
12
12
 2 
 2 
ss1 =
Feρ F (12 )(3.606 )
=
= 1.4804 F
Jc
29.23
Page 30 of 41
3
SECTION 17– WELDING
F
F
=
= 0.3535F
2tL 2(0.3536)(4)
L
4
cos θ =
=
= 0.5547
2
2
(m − n ) + L
(15 − 9)2 + 42
s s2 =
1
τ = (ss2 + ss2 + 2ss ss cos θ )2
1
2
1
2
[
1
]
τ = (1.4804 F )2 + (0.3535F )2 + 2(1.4804 F )(0.3535 F )(0.5547 ) 2 = 1.7021F
(a) Table AT 30, static, AISC Building Code
τ d = 13.6 ksi
τ = 13.6 = 1.7021F
F = 7.99 kips
(b) Table AT 30, static, nc = 2 × 10 6 cycles
7 .2
τd =
1 − 0 .5 R
0 .5 F
= 0 .5
R=
F
7.2
τd =
= 9.6 ksi
1 − 0.5(0.5)
τ = 9.6 = 1.7021F
F = 5.64 kips
Table AT 30, static, nc = 105 cycles
12.5
τd =
1 − 0 .5 R
0 .5 F
R=
= 0 .5
F
12.5
τd =
= 16.67 ksi
1 − 0.5(0.5)
τ = 16.67 = 1.7021F
F = 9.79 kips
(c) Table AT 30, R = 0 , indefinite life, nc = 2 × 10 6 cycles
7 .2
τd =
1 − 0 .5 R
7.2
τd =
= 7.2 ksi
1 − 0.5(0)
Page 31 of 41
SECTION 17– WELDING
τ = 7.2 = 1.7021F
F = 4.23 kips
(d) Design factors
nc = 2 × 10 6 cycles
R = 0 .5
τ = 9.6 ksi
Table 19-1, sn = 40.1 ksi
s
40.1
N= n =
= 4 .2
τ
9 .6
nc = 105 cycles
R = 0 .5
τ = 16.67 ksi
Table 19-1, sn = 46.1 ksi
s
46.1
N= n =
= 2 .8
τ 16.67
Indefinite life ( nc = 2 × 10 6 ), R = 0
τ = 7.2 ksi
Table 19-1, sn = 18.1 ksi
s 18.1
N= n =
= 2 .5
τ
7 .2
960.
The plate for a bracket, as shown must be welded to a member in the manner
shown; 5/16-in. welds with shielded arc. Compute the safe load for this plate (no
twisting) (a) for static loading, (b) for R = 0 and indefinite life, (c) for R = −1
and indefinite life, (d) for R = 0.2 indefinite life.
Problem 960
Page 32 of 41
SECTION 17– WELDING
Solution:
For x
(3 + 6)x = (3) 3  + (6) 6 
2
2
x = 2.5 in
5
in = 0.3125 in
16
t = b cos 45 = (0.3125) cos 45 = 0.2210 in
F
F
s s2 =
=
= 0.503F
t (3 + 6) (0.2210)(9)
Feρ
ss1 =
Jc
e = 10 − 2.5 = 7.5 in
b=
2
6
2
Solving for J c ;
ρ =   + (6 − 2.5)2 = 4.61 in
2
2
6 3

r1 =   +  − 2.5  = 3.1623 in
2 2

Page 33 of 41
SECTION 17– WELDING
2
2
6 6

r2 =   +  − 2.5  = 3.0414 in
2 2

t (3)
t (6 )
+ 3tr12 +
+ 6tr22
12
12
3
3
(
0.221)(3)
(
0.221)(6 )
2
2
Jc =
+ 3(0.221)(3.1623) +
+ 6(0.221)(3.0414 ) = 23.37 in 4
12
12
F (7.5)(4.61)
= 1.4795 F
ss1 =
23.37
ss2 = 0.503F
3
3
Jc =
cos θ =
6 − 2.5
ρ
[
=
6 − 2.5
= 0.7592
4.61
1
2
]
τ = (1.4795F ) + (0.503F ) + 2(1.4795F )(0.503F )(0.7592) = 1.890 F
2
2
(a) For static loading, Table AT 30, AISC Building Code
τ d = 13.6 ksi
τ = 13.6 = 1.890 F
F = 7.20 kips
(b) For R = 0 and indefinite life, ( ≈ nc = 2× 10 6 )
7.2
7.2
=
= 7.2 ksi
1 − 0.5R 1 − 0.5(0)
τ = 7.2 = 1.890 F
F = 3.8 kips
τd =
(c) For R = −1 and indefinite life
7.2
7.2
=
= 4.8 ksi
1 − 0.5R 1 − 0.5(− 1)
τ = 4.8 = 1.890 F
F = 2.5 kips
τd =
(d) For R = 0.2 and indefinite life
7.2
7.2
=
= 8.0 ksi
1 − 0.5R 1 − 0.5(0.2)
τ = 8.0 = 1.890 F
F = 4.2 kips
τd =
Page 34 of 41
SECTION 17– WELDING
961.
A 2-in. round bar is welded to a vertical wall by a 3/8-in. fillet weld as shown in
Fig. 19.8, Text. The bar supports a vertical load of 800 lb at a distance of 10 in.
from the wall. What is the maximum computed stress in the weld? Would this
result be safe for a varying load with R = 0 , shielded weld?
Solution:
Fig. 19.8, Text, with additional
3
in = 0.375 in
8
D = 2 in
F = 800 lb
e = 10 in
5.66 M 5.66 Fe 5.66(800 )(10 )
st =
=
=
= 9609 psi = 9.609 ksi
π bD 2
π bD 2
π (0.375)(2)2
F
F
800
ss =
=
=
= 480 psi = 0.48 ksi
π Dt π Db cos 45 π (2)(0.375) cos 45
b=
1
2 2

s  
τ =  ss2 +  t  
 2  

1
2 2

 9.609  
τ = (0.48)2 + 
  = 4.83 ksi
2

 

Table AT 30, R = 0
Assume nc = 2× 10 6 cycles
7 .2
τd =
= 7.2 > 4.83 ksi
1 − 0 .5 R
∴ safe for a varying load, nc = 2× 10 6 cycles.
962.
The 14-in. structural-steel disk is welded to the plate by a 7/16-in. fillet weld,
360o, shielded arc. The force F acts on a pin attached to the disk. The pin is
short enough that the moment arm to the disk is negligible. Determine a safe
force F for (a) static loading, (b) reversed loading, indefinite life, (c) a varying
load from 0.3F to F indefinite life.
Page 35 of 41
SECTION 17– WELDING
Problem 962
Solution:
b=
7
in
16
7
t = b cos 45 =   cos 45 = 0.3094 in
 16 
D = 14 in
Feρ
ss1 =
Jc
e = 6 sin 60
ρ = 14 in
1
J c = 2π r 3t = π D 3t
4
F (6 sin 60 )(14 )
ss1 =
= 0.1091F
1
π (14)3 (0.3094)
4
F
F
sss =
=
= 0.0735 F
π Dt π (14)(0.3094)
τ = ss1 + s s2 = 0.1091F + 0.0735F = 0.1826 F
(a) Static loading, Table AT 30, AISC Building Code
τ = 13.6 ksi
τ = 13.6 = 0.1826 F
F = 74.48 kips
(b) Reversed loading, indefinite life, Table AT 30
τ = 5 ksi (Jenning recommendation)
τ = 5 = 0.1826 F
F = 27.38 kips
(c) Varying load, R = 0.3 , indefinite life, Table AT30, nc ≈ 2× 10 6 cycles.
Page 36 of 41
SECTION 17– WELDING
7.2
7.2
=
= 8.47 ksi
1 − 0.5 R 1 − 0.5(0.3)
τ = 8.47 = 0.1826 F
F = 46.38 kips
τd =
963.
A bracket is fabricated from ½-in., AISI-1020 rolled plate with 3/8-in. fillet
welds on both sides of the plates G, H, and J as shown. The welds are made with
E6016 welding rod; a central load at L = 30 in ; h = 11 in . Determine the repeated
load that the welding can support.
Problem 963.
Solution:
3
b = in = 0.375 in
8
t = b cos 45 = 0.375 cos 45 = 0.2652 in
L = 30 in
h = 11 in
ss =
F
1

2t (2 ) + 2t (11 − 1) + 2t  2 − 
2

h
FL 
2
st =
I
Page 37 of 41
=
F
F
=
= 0.13966 F
27t 27(0.2652 )
SECTION 17– WELDING
t = 0.2652 in
1

3
2 2 − (0.2652)
2
2(2)(0.2652 )
1
2
 11 

 11 − 1 
I=
+ 2(2)(0.2652)  + 
+ 2 2 − (0.2652)

12
12
2
2

 2 
2
3
2(0.2652)(11 − 1)
+
= 96.2 in 4
12
 11 
F (30) 
 2  = 1.715F
st =
96.2
From Table AT 30, using Jennings recommendation, τ = 14 ksi and assume a strength
reduction factor = 1.4
14
τ=
= 10 ksi
1 .4
3
1
2 2

s  
τ =  ss2 +  t  
 2  

1
2

 1.715F   2
2
10 = (0.13966 F ) + 
 
 2  

F = 11.51 kips
964.
The bracket shown is made of ½-in. AISI-C1020 rolled plates. The 3/8-in. fillet
welds are on both sides of each plate A and B, E7010 welding rod. The entire
bracket is normalized after welding; L = 12 in , a = 8 in , and h = 8 in . What is
the safe maximum load if it is (a) static, (b) varies from 0.5 F to F for 2× 10 6
cycles and for 105 cycles, (c) R = 0 , indefinite life. (d) Considering the strengths
given in Table 19.1, Text, determine the design factor for part (b) and (c).
Problem 964
Solution:
3
b = in = 0.375 in
8
t = b cos 45 = 0.375 cos 45 = 0.2652 in
L = 12 in
Page 38 of 41
SECTION 17– WELDING
h = 8 in
a = 8 in
 1 

8 − 


2  1
 1 
1  1 

(
)
8
+
8
−
1
+
4
8
−
y
=
8
+
4
8
−
+ 







2 
2  2
2

2 



y = 2.922 in
IG
2
3
3
(
(
0.2652 ) (8)
0.2652 ) (8 − 1)
1

2
=
+ (0.2652 )(8)(2.922 ) +
+ (0.2652 )(8 − 1) 2.922 −
12
3
 1

1 

4(0.2652 ) 8 − 
8 − 


1
2

 + 4(0.2652 ) 8 −  (8 − 2.922 ) − 
2
+



12
2 
 2 






4
I = 80.3524 in
 1
A = 8(0.2652 ) + (8 − 1)(0.2652 ) + 4 8 − (0.2652 ) = 11.934 in 2
2

F
F
=
= 0.0838 F
A 11.934
FLc
st =
I
c = 8 − 2.922 = 5.078 in
ss =
Page 39 of 41


12
2

2
SECTION 17– WELDING
st =
FLc F (12 )(5.078)
=
= 0.7584 F
I
80.3524
1
 2  st  2  2
τ =  ss +   
 2  

1
2 2

 0.7584 F  
2
τ = (0.0838F ) + 
  = 0.3884 F
2

 

(a) Static loading, Table AT 30, AISC Building Code
τ = 13.6 ksi
τ = 13.6 = 0.3884 F
F = 35 kips
(b) Variable, R = 0.5
For nc = 2× 10 6 cycles (Table AT 30)
7.2
7.2
τ=
=
= 9.6 ksi
1 − 0.5R 1 − 0.5(0.5)
τ = 9.6 = 0.3884 F
F = 24.7 kips
For nc = 105 cycles (Table AT 30)
12.5
12.5
τ=
=
= 16.67 ksi
1 − 0.5R 1 − 0.5(0.5)
τ = 16.67 = 0.3884 F
F = 42.9 kips
(c) R = 0 , indefinite life (Table AT 30)
7 .2
= 7.2 ksi
1 − 0 .5 R
τ = 7.2 = 0.3884 F
F = 18.5 kips
τ=
(d) From Table 19.1
sn = 40.1 ksi , R = 0.5 , nc = 2× 10 6
sn = 46.1 ksi , R = 0.5 , nc = 105
sn = 18.1 ksi , R = 0 , nc ≈ 2× 10 6 (indefinite life)
Page 40 of 41
SECTION 17– WELDING
Design Factor, N
R = 0.5 , nc = 2× 10 6
40.1
N=
= 4 .2
9 .6
R = 0.5 , nc = 105
46.1
N=
= 2 .8
16.67
R = 0 , nc ≈ 2× 10 6 (indefinite life)
18.1
N=
= 2 .5
7 .2
- end -
Page 41 of 41
SECTION 18 – MISCELLANEOUS PROBLEMS
THIN SHELLS, EXTERNAL PRESSURE
981.
A closed cylindrical tank is used for a steam heater. The inner shell, 200 in.
outside diameter and 50 ft. long, is subjected to an external pressure of 40 psi.
The material is equivalent to SA 30 (ASME Pressure-Vessel Code: min.
su = 55 ksi ); assume an elastic limit of s y = su 2 ; let N = 5 . (a) What thickness
of shell is needed from a stress standpoint? (b) For this thickness, what must be
the maximum length of unsupported section to insure against collapse? (c)
Choose a spacing L to give a symmetric arrangement and determine the moment
of inertia of the steel stiffening rings. (d) For a similar problem, the Code
recommends that t ≥ 0.76 in , L = 50 in , and I = 96 in 4 . How do these values
check with those obtained above? (e) Without stiffening rings, what thickness
would be needed?
Solution:
(a) Solving for the thickness of shell,
t=
pc D
2sy
pc = 5 p = 5(40) = 200 psi
s
55,000
sy = u =
= 27,500 psi
2
2
D = 200 in
p D (200)(200 )
t= c =
= 0.73 in
2sy
2(27,500 )
say t = 0.75 in
(b) Solving for the maximum length of unsupported section, use Eq. (20-1)
5
 t 2
2.60 E  
 D
pc =
psi
1
2
L
t 
− 0.45 
D
D
5
 t 2
1
2.60 E  
2
L
t
D


  + 0.45
=
 
D
pc
D
E = 30× 106 psi
pc = 200 psi
t = 0.75 in
Page 1 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
D = 200 in
5
 0.75  2
1
2.60 30 ×10 6 

L
0
.
75
2
 200  + 0.45
=


200
200
 200 
L = 72.68 in
(
)
(c) Solving for the moment of inertia of the steel stiffening rings.
Choosing L = 60 in for 50 ft long shell
0.035D 3 Lpc 0.035(200) (60 )(200 )
=
= 112 in 4
E
30 ×106
3
I=
(d) For t ≥ 0.76 in - above minimum
L = 50 in - below maximum
I = 96 in 4 - lighter than above.
(e) L = 50 ft = 600 in
Solving for thickness without stiffening rings
By Saunders and Windenburg, Eq. 20-1
5
pc =
 t 2
2.60 E  
 D
L
t 
− 0.45 
D
D
1
2
psi
1


2
L
t



p
−
0
.
45
  
c
5

D
 D 
 t 2


=
 
2.60 E
 D
1


2
600
t



200
− 0.45
 
5

200
200

 
 t 2



 =
6
200
2
.
60
30
×
10


(
5
2
137.886t = 600 − 6.364t
5
2
1
2
)
1
2
137.886t + 6.364t = 600
t = 1.791 in
Page 2 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
say t = 1
982.
13
in
16
The same as 981, except that p = 175 psi , D = 4 ft , and the length of the tank is
18 ft.
(a) Solving for the thickness of shell,
t=
pc D
2sy
pc = 5 p = 5(175) = 875 psi
s
55,000
sy = u =
= 27,500 psi
2
2
D = 4 ft = 48 in
p D (875)(48)
t= c =
= 1.53 in
2 s y 2(27,500 )
say t = 1
9
= 1.5625 in
16
(b) Solving for the maximum length of unsupported section, use Eq. (20-1)
5
 t 2
2.60 E  
 D
pc =
psi
1
L
 t 2
− 0.45 
D
D
5
 t 2
1
2.60 E  
2
L
t
D


  + 0.45
=
 
D
pc
D
E = 30×106 psi
pc = 875 psi
t = 1.5625 in
D = 48 in
5
 1.5625  2
1
2.60 30 ×10 

L
48 
 1.5625  2

=
+ 0.45

48
875
 48 
L = 822 in
(
6
)
(c) Since length of shell = 18 ft = 216 in < 822 in, there is no need for stiffeners.
Page 3 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
(d) L = 216 in
Solving for thickness without stiffening rings
By Saunders and Windenburg, Eq. 20-1
5
 t 2
2.60 E  
 D
pc =
psi
1
2
L
t 
− 0.45 
D
D
1


2
L
t



p
−
0
.
45
  
c
5

D
 D 
 t 2


=
 
2.60 E
 D
1


2
216
t



875
− 0.45  
5

48
 48  
 t 2


  =
6
2.60 30 ×10
 48 
(
)
5
2
4886.428t = 3937.5 − 56.833t
5
2
1
2
1
2
4886.428t + 56.833t = 3937.5
t = 0.9122 in
15
say t = in
16
19
but minimum t = 1 in
16
19
use t = 1 in
16
Approximate ratio of weight of this shell to the weight of the shell found in (a) =
thickness of shell without stiffening rings / thickness of shell with stiffening rings
= 0.34375 / 0.09375 = 3.6667
STEEL TUBES, EXTERNAL PRESSURE
983.
A closed cylindrical tank, 6 ft in diameter, 10 ft long, is subjected to an internal
pressure of 1 psi absolute. The atmospheric pressure on the outside is 14.7 psi.
The material is equivalent to SA 30 (ASME Pressure Vessel Code:
min su = 55 ksi ); assume an elastic limit of s y = su 2 ; let pc p = 5 . (a) What
thickness of shell is needed for the specified design stress? (b) For this thickness,
what must be the maximum length of unsupported section to insure against
collapse? (c) Choose a symmetric spacing L of stiffening rings, and compute
Page 4 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
their moment of inertia and the cross-sectional dimensions h and b if they are
rectangular with h = 2b . (d) Suppose that the tank had no stiffening rings. What
thickness of shell would be needed? What is the approximate ratio of the weight
of the shell found in (a)? Material costs are roughly proportional to the weight.
Solution:
(a) Solving for the thickness of shell
pD
t= c
2sy
pc = 5 p = 5(14.7 − 1) = 68.5 psi
s
55,000
sy = u =
= 27,500 psi
2
2
D = 6 ft = 72 in
p D (68.5)(72 )
t= c =
= 0.08967 in
2 s y 2(27,500 )
say t =
3
= 0.09375 in
32
(b) Solving for the maximum length of unsupported section, use Eq. (20-1)
5
 t 2
2.60 E  
 D
pc =
psi
1
2
L
t 
− 0.45 
D
D
5
 t 2
1
2.60 E  
2
L
t
D


  + 0.45
=
 
D
pc
D
E = 30×106 psi
pc = 68.5 psi
t = 0.09375 in
D = 72 in
5
 0.09375  2
1
2.60 30 ×10 

L
72 
 0.09375  2

=
+ 0.45

72
68.5
 72 
L = 6.2 in
(
6
)
(c) For the length of shell = 10 ft = 120 in.
use L = 6 .0 in
Page 5 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
Moment of inertia of stiffening rings.
0.035D 3 Lpc 0.035(72) (6 )(68.5)
I=
=
= 0.1790 in 4
E
30 ×106
Solving for cross-sectional dimension
h = 2b
bh 3 8b 4 2b 4
I=
=
=
= 0.1790 in 4
12
12
3
b = 0.72 in
3
say b = in
4
1
3
h = 2  = 1 in
2
4
3
(d) L = 10 ft = 120 in
Solving for thickness without stiffening rings
By Saunders and Windenburg, Eq. 20-1
5
 t 2
2.60 E  
D
pc =
psi
1
L
 t 2
− 0.45 
D
D
1


2
L
t



pc
− 0.45  
5

D
 D 
 t 2


  =
2.60 E
 D
1


120
t
 2 

68
.
5
−
0
.
45


5
 72
 72  
 t 2


  =
6
2.60(30 ×10 )
 72 
5
1
25.8865t 2 = 1.667 − 0.053t 2
5
1
25.8865t 2 + 0.053t 2 = 1.667
t = 0.3314 in
11
say t =
in = 0.34375 in
32
Page 6 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
984.
A long lap-welded steel tube, 8-in. OD, is to withstand an external pressure of
120 psi. with N = 5 . (a) What should be the thickness of the wall of the tube? (b)
What is the ratio D t ? Is it within the range of the Stewart equation? (c)
Assuming the internal pressure to be negligible relative to the external pressure,
calculate the maximum principal stress from equation (8.13), p. 255, text. What
design factor is given by this stress compared with s y for AISI C1015 annealed?
(d) Compute the stress from the thin shell formula.
Solution:
(a) Solving for the thickness of the wall
Stewart’s formula
3
t 
pc = 50,200,000 
 D
pc = Np = (5)(120) = 600 psi
3
t 
600 = 50, 200,000 
8
t = 0.1829 in
3
say t = in = 0.1875 in
16
D
8
=
= 42.67
t 0.1875
or D = 42.67t
outside the range of the Steward equation ( D < 40t )
(b) ratio
(c) Using eq. (8.13) (Lame’s formula)
pi ri 2 − po ro2 + ri 2 ro2 ( pi − po ) r 2
ro2 − ri 2
po = 120 psi
pi ≈ 0
OD 8
ro =
= = 4 in
2
2
ri = 4 − 0.1875 = 3.8125 in
r = ri
σt =
− po ro2 + ro2 (0 − po ) − 2 po ro2
− 2(120)(4 )
= 2 2 = 2
= −2621 psi
2
2
ro − ri
ro − ri
(4 ) − (3.8125)2
s y of AISI C1015, annealed = 42 ksi
2
σt =
Page 7 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
Design factor, N y =
sy
σt
=
42,000
= 16
2621
(d) Solving for the stress from the thin shell formula
s=
pD
(120)(8) = 2560 psi
=
2t 2(0.1875)
985.
A long lap-welded steel tube, 3 –in. OD, is to withstand an external pressure of
150 psi with N = 5 . Parts (a) – (c) are the same as in 984.
Solution:
(a) Solving for the thickness of the wall
Stewart’s equation
3
t 
pc = 50,200,000 
 D
pc = Np = (5)(150) = 750 psi
3
t 
750 = 50, 200,000 
3
t = 0.074 in
5
say t =
in = 0.078125 in
64
D
3
=
= 38.4
t 0.078125
or D = 38.4t
within the range of the Steward equation ( D < 40t )
(b) ratio
(c) Using eq. (8.13) (Lame’s formula)
po = 150 psi
OD 3
ro =
= = 1.5 in
2
2
ri = 1.5 − 0.078125 = 1.421875 in
− 2 po ro2
− 2(150 )(1.5)
=
= −2957 psi
2
2
ro − ri
(1.5)2 − (1.421875)2
s y of AISI C1015, annealed = 42 ksi
2
σt =
Design factor, N y =
Page 8 of 25
sy
σt
=
42,000
= 14.2
2927
SECTION 18 – MISCELLANEOUS PROBLEMS
986.
A long lap-welded tube, 3-in. OD, is made of SAE 1015, annealed. Let the shell
thickness t = D 40 and N = 5 . (a) What is the corresponding safe external
pressure? (b) Compute the maximum principal stress (p. 255, Text), assuming a
negligible internal pressure. What design factor is given by this stress compared
with s y ? (c) Compare with stress computed from the thin-shell formula.
Solution:
(a) Solving for safe external pressure,
t 
pc = 50,200,000 
 D
t
1
=
D 40
3
3
 1 
pc = 50,200,000  = 784 psi
 40 
p
784
p= c =
= 157 psi
N
5
(b) Solving for maximum principal stress, neglecting internal pressure
po = 157 psi
OD 3
ro =
= = 1.5 in
2
2
D
3
t=
=
= 0.075 in
40 40
ri = 1.5 − 0.075 = 1.425 in
− 2 po ro2
− 2(157 )(1.5)
=
= −3221 psi
2
2
ro − ri
(1.5)2 − (1.425)2
s y of SAE 1015 annealed = 42 ksi
2
σt =
Design factor, N y =
sy
σt
=
42,000
= 13.0
3221
(c) Solving for stress from the thin-shell formula
s=
pD (157 )(3)
=
= 3140 psi
2t 2(0.075)
FLAT PLATES
987.
A circular plate 24 in. in diameter and supported but not fixed at the edges, is
subjected to a uniformly distributed load of 125 psi. The material is SAE 1020, as
Page 9 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
rolled, and N = 2.5 based on the yield strength. Determine the thickness of the
plate.
Solution:
Solving for the thickness of the plate
2
r
s = p  psi
t
for SAE 1020,a s rolled, s y = 48 ksi
sy
48,000
= 19,200 psi
N
2.5
p = 125 psi
24
r=
= 12 in
2
2
 12 
s = 19,200 = 125 
 t 
t = 0.968 in
say t = 1 in
s=
988.
=
The cylinder head of a compressor is a circular cast-iron plate (ASTM class 20),
mounted on a 12-in. cylinder in which the pressure is 250 psi. Assuming the head
to be supported but not fixed at the edges, compute its thickness for N = 6 based
on ultimate strength.
Solution:
Solving for the thickness of the head
2
r
s = p  psi
t
for cast-iron (ASTM class 20), su = 20 ksi
s
20,000
s= u =
= 3333 psi
N
6
p = 250 psi
12
r=
= 6 in
2
2
6
s = 3333 = 250 
t
t = 1.6432 in
21
say t = 1 in = 1.65625 in
32
Page 10 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
989.
A 10x15-in. rectangular opening in the head of a pressure vessel, whose internal
pressure is 175 psi, is covered with a flat plate of SAE 1015, annealed. Assuming
the plate to be supported at the edges, compute its thickness for N = 6 based on
ultimate strength.
Solution:
Solving for the thickness of the head
a 2b 2 p
s= 2 2
psi
2t (a + b 2 )
for SAE 1015, annealed, su = 56 ksi
s
56,000
s= u =
= 9333 psi
N
6
a = 10 in
b = 15 in
p = 175 psi
s = 9333 =
(10 )2 (15)2 (175) psi
2
2
2t 2 [(10 ) + (15) ]
t = 0.8056 in
13
say t = in = 0.8125 in
16
CAMS
990.
The force between a 5/8-in. hardened steel roller and a cast-iron (140 BHN) cam
is 100 lb.; radius of cam curvature at this point is 1 ¼ in. Compute the contact
width.
Solution:
Solving for the contact width
Kcb
P=
1 1
N  + 
 r1 r2 
From Table 20-2, hardened steel and a cast-iron, BHN = 400
Use K c2 = 900
P = 100 lb
15
r1 =   = 0.3125 in
28
1
r2 = 1 = 1.25 in
4
N = 1.15
Page 11 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
900b
1 
 1
1.15
+

 0.3125 1.25 
b = 0.511 in
100 =
A radial cam is to lift a roller follower 3 in. with harmonic motion during a 150o
turn of the cam; 1 1.2-in. roller of hardened steel. The reciprocating parts weigh
10 lb., the spring force is 175 lb., the external force during the lift is 250 lb. The
cast-iron (225 BHN) cam turns 175 rpm. The cam curvature at the point of
maximum acceleration is 1 ½-in. radius. Compute the contact width.
991.
Solution:
Neglecting frictional forces
Q + Fg + Fs + Fre = P cos φ
Q = external force during lift = 250 lb
Fg = weight of reciprocating parts = 10 lb
Fs = spring force = 175 lb
Fre = reversed effective force = − qma
Fre = −2ma for harmonic motion
2
π 
L  πω 
 cos θ 
a = &x& = 
2 β 
β 
L = 3 in
2π (175)
ω=
= 18.326 rad sec
60
150
β = 150o =
π = 2.618 rad
180
at maximum acceleration
2
3  π (18.326) 
&x& = 
= 726 in sec 2

2  2.618 
P cos φ = Q + Fg + Fs + Fre = 250 + 10 + 175 − 2m&x&
10 lb
10 lb − sec 2
=
g (32.2 ft sec 2 )(12 in ft ) 386.4
in
assume cosφ ≈ 1
 10 
P = 435 − 2
(726 ) = 397 lb
 386.4 
m=
Fg
=
Page 12 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
Solving for the contact width
Kcb
P=
1 1
N  + 
 r1 r2 
From Table 20-2, hardened steel and a cast-iron, BHN = 225
Use K c2 = 2100
1
r1 = 1 in = 1.5 in
2
1
r2 = 1 in = 1.5 in
2
N = 1.15
2100b
397 =
1 
 1
1.15
+

 1. 5 1. 5 
b = 0.29 in
992.
The same as 991, except that the motion of the follower is cycloidal.
Solution: Fre = −1.1ma for cycloidal motion
As a continuation of 991, but
 2π 
2π Lω 2
a = &x& =
sin  θ  for cycloidal
2
β
 β 
L = 3 in
2π (175)
ω=
= 18.326 rad sec
60
150
β = 150o =
π = 2.618 rad
180
at maximum acceleration
2
2π Lω 2 2π (3)(18.326)
&x& =
=
= 924 in sec 2
2
2
β
(2.618)
 10 
P = 435 − 1.1
(924 ) = 409 lb
 386.4 
Solving for the contact width
Kcb
P=
1 1
N  + 
 r1 r2 
From Table 20-2, hardened steel and a cast-iron, BHN = 225
Use K c2 = 2100
Page 13 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
1
r1 = 1 in = 1.5 in
2
1
r2 = 1 in = 1.5 in
2
N = 1.15
2100b
409 =
1 
 1
1.15
+

 1.5 1.5 
b = 0.30 in
993.
The same as 991, except that the motion of the follower is parabolic.
Solution: Fre = −3ma for parabolic motion
As a continuation of 991, but
2
 2ω 
 for parabolic
a = &x& = L
 β 
L = 3 in
2π (175)
ω=
= 18.326 rad sec
60
150
β = 150o =
π = 2.618 rad
180
at maximum acceleration
2
 2(18.326 ) 
&x& = (3)
= 588 in sec 2

 2.618 
 10 
P = 435 − 3
(588) = 389 lb
 386.4 
Solving for the contact width
Kcb
P=
1 1
N  + 
 r1 r2 
From Table 20-2, hardened steel and a cast-iron, BHN = 225
Use K c2 = 2100
1
r1 = 1 in = 1.5 in
2
1
r2 = 1 in = 1.5 in
2
N = 1.15
Page 14 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
2100b
1 
 1
1.15
+

 1. 5 1. 5 
b = 0.284 in
389 =
FLYWHEELS AND DISK
994.
A cast-iron flywheel with a mean diameter of 36 in. changes speed from 400 rpm
to 380 rpm while it gives up 8000 ft-lb of energy. What is the coefficient of
fluctuation, the weight, and the approximate sectional area of the rim?
Solution:
Solving for coefficient of fluctuation
n −n
Cf = 1 2
n
n1 + n2
n=
2
2(n1 − n2 )
Cf =
n1 + n2
n1 = 400 rpm
n2 = 380 rpm
2(400 − 380 )
Cf =
= 0.0513
400 + 380
Solving for the weight
32.2∆KE
w=
C f vs2
∆KE = 8000 ft − lb
C f = 0.0513
vs =
π Dn
12(60 )
D = 36 in
400 + 380
n=
= 390 rpm
2
π (36 )(390)
vs =
= 61.26 fps
12(60 )
32.2(8000 )
w=
= 1338 lb
2
0.0513(61.26)
Page 15 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
Solving for the approximate sectional area of the rim
w = ρV
assume ρ = 0.254 lb in 3 for cast iron
V = π DA
D = 36 in
w = ρπ DA
1338 = 0.254π (36 ) A
A = 36 in 2
995.
The energy required to shear a 1-in. round bar is approximately 1000 ft-lb. In use,
the shearing machine is expected to make a maximum of 40 cutting strokes a
minute. The frictional losses should not exceed 15 % of the motor output. The
shaft carrying the flywheel is to average 150 rpm. (a) What motor horsepower is
required? (b) Assuming a size of flywheel and choosing appropriate C f , find the
mass and sectional dimensions of the rim of a cast-iron flywheel. The width of
the rim is to equal the depth and is not to exceed 3 ½ in. It would be safe to
assume that all the work of shearing is supplied by the kinetic energy given up by
the flywheel.
Solution:
(a) Solving for the horsepower required
(Energy required )(Strokes per min )
(1 − Frictional losses )(33,000 ft − lb hp − min )
(1000)40
hp =
= 1.426 hp
(1 − 0.15)(33,000 ft − lb hp − min )
hp =
(b) Solving for the mass of the rim and size of section
32.2∆KE
C f vs2
w = ρV
assume ρ = 0.254 lb in 3 for cast iron
w=
V = π DA
w = ρπ DA =
32.2∆KE
C f vs2
assume C f = 0.06 (Table 20-3)
∆KE = 1000 ft − lb
Page 16 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
vs =
π Dn
12(60 )
n = 150 rpm
w = ρπ DA =
D3 =
32.2∆KE
 5π D 
Cf 

 24 
32.2∆KE
2
2
 5π 
Cf 
 ρπ A
 24 
using width = depth = 3 ½ in
A = (3.5)(3.5) = 12.25 in 2
32.2(1000 )
D3 =
2
 5π 
0.06
 (0.254)(π )(12.25)
 24 
D = 50.42 in
assume D = 51 in
5π ( 51)
vs =
= 33.38 fps
24
32.2(1000)
w=
= 482 lb
2
0.06(33.38)
w 482
m= =
= 15 slugs
g 32.2
w
482
A = (depth )(width ) =
=
= 11.84 in 2
ρπ D 0.254π (51)
depth = width = 11.84 in 2 = 3.44 in
1
say depth = width = 3 in
2
996.
The same as 995, except that the capacity of the machine is such as to cut 1 ½-in.
round brass rod, for which the energy required is about 400 ft-lb./sq. in. of
section.
Solution:
(a) Solving for the horsepower required
hp =
(Energy required )(Strokes per min )
(1 − Frictional losses )(33,000 ft − lb hp − min )
Page 17 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
π 
2
Energy required = 400 ft − lb in 2  (1.5 in ) = 707 ft − lb
4
(707)(40)
hp =
= 1.01 hp
(1 − 0.15)(33,000 ft − lb hp − min )
(
)
(b) Solving for the mass of the rim and size of section
32.2∆KE
D3 =
2
 5π 
Cf 
 ρπ A
 24 
∆KE = 707 ft − lb
C f = 0.06
ρ = 0.254 lb in 3
A = (3.5)(3.5) = 12.25 in 2
32.2(707 )
D3 =
2
 5π 
0.06
 (0.254)(π )(12.25)
 24 
D = 45 in
use D = 45 in
32.2∆KE
w=
C f vs2
5π D 5π (45)
=
= 29.45 fps
24
24
32.2(707 )
w=
= 438 lb
2
0.06(29.45)
w 438
m= =
= 13.6 slugs
g 32.2
w
438
A = (depth )(width ) =
=
= 12.20 in 2
ρπ D 0.254π (45)
vs =
depth = width = 12.20 in 2 = 3.49 in
1
say depth = width = 3 in
2
997.
A 75-hp Diesel engine, running at 517 rpm, has a maximum variation of output
of energy of 3730 ft-lb. The engine has three 8 x 10 ½ in. cylinders and is
directly connected to an a-c generator. (a) What should be the weight and
sectional area of the flywheel rim if it has an outside diameter of 48-in.? (b) The
actual flywheel and generator have Wk 2 = 6787 lb − ft 2 . Compute the
corresponding coefficient of fluctuation and compare.
Page 18 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
Solution:
(a) Solving for the weight ad sectional areas
w=
32.2∆KE
lb
C f vs2
assume C f = 0.0035 , Table 20-3
∆KE = 3730 ft − lb
π Dm n
vs =
12(60)
n = 570 rpm
Dm = Do − t = 48 − t
π (48 − t )(517 )
vs =
= 2.25584(48 − t )
12(60)
32.2(3730)
6,743,418
w=
=
(0.0035)[2.25584(48 − t )]2 (48 − t )2
w = ρπ DA
assume ρ = 0.254 lb in 3 (cast iron), b = 2t
w
A=
, A = bt = 2t 2
ρπ D
0.254π (48 − t )(2t )(t ) =
6,743,418
(48 − t )2
t 2 (48 − t ) = 4,225,387
t = 8.18 in
Then
w=
6,743,418
= 4253 lb
(48 − 8.18)2
A = 2t 2 = 2(8.18)2 = 133.8 in 2
(b) Solving for coefficient of fluctuation
Ig = Wk 2 = 6787 lb − ft 2
(
)
I ω12 − ω22
∆KE =
ft − lb
2
Wk 2 (ω12 − ω22 )
∆KE =
ft − lb
g
2
Wk 2
(ω1 − ω2 ) (ω1 + ω2 ) ft − lb
∆KE =
g
2
(ω1 + ω2 ) = ω
2
Page 19 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
2π (517 )
= 54.14 rad sec
60
∆KE = 3730 ft − lb
6787
∆KE = 3730 =
(ω1 − ω2 )54.14
32.2
ω1 − ω 2 = 0.327
ω − ω 2 0.327
Cf = 1
=
= 0.006 > 0.0035
ω
54.14
ω=
998.
A 4-ft flywheel, with an rim 4 in. thick and 3 in. wide, rotates at 400 rpm. If there
are 6 arms, what is the approximate stress in the rim? Is this a safe stress? At
what maximum speed should this flywheel rotate if it is made of cast iron, class
30?
Solution:
Solving for the approximate stress,
v2 ρ
s= s
psi
144 g o
π Dn π (4 )(400 )
vs =
=
= 83.78 fps
60
60
ρ = 0.254 lb in 3 (class 30, cast iron)
ρ = 0.254 lb in 3 = (0.254)(1728) lb ft 3 = 439 lb ft 3
g o = 32.2 ft s 2
(83.78)2 (439) = 665 psi
144(32.2 )
since v = (83.78)(60) fpm = 5027
s=
fpm < 6000 fpm (cast iron)
this is a safe stress
Solving for maximum speed, max. v = 6000 fpm
v = π Dn = π (4 )n = 6000 fpm
maximum, n = 477 rpm
999.
A hollow steel shaft with Do = 6 in and Di = 3 in rotates at 10,000 rpm. (a) What
is the maximum stress in the shaft due to rotation? Will this stress materially
affect the strength of the shaft? (b) The same as (a), except that the shaft is solid.
Solution:
(a) Solving for maximum stress
Page 20 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
st =
ρω 2
[(3 + µ )r
2
+ (1 − µ )ri
2
]
psi
4 go
where
D
6
ro = o = = 3 in
2
2
D 3
ri = i = = 1.5 in
2 2
for steel, ρ = 0.284 lb in 3 , µ = 0.30
o
g o = 386 in sec
 10,000 
 = 1047 rad sec
 60 
ω = 2π 
st =
ρω 2
4 go
[(3 + µ )r
2
o
+ (1 − µ )ri
2
] psi
2
(
0.284)(1047 )
[
(3 + 0.3)(3)2 + (1 − 0.3)(1.5)2 ] = 6306 psi
st =
4(386 )
(does not affect the strength of the shaft)
(b) Solving for the maximum stress for solid
st =
ρω 2 (3 + µ )ro 2
2
2
(
0.284 )(1047 ) (3 + 0.3)(3)
=
4(386 )
4 go
1000.
A circular steel disk has an outside diameter Do = 10 in and an inside diameter
Di = 2 in . Compute the maximum stress for a speed of (a) 10,000 rpm, (b)
20,000 rpm. (c) What will be the maximum speed without danger of permanent
deformation if the material is AISI 3150, OQT at 1000 F?
Solution:
st =
= 2994 psi
ρω 2
[(3 + µ )r
o
2
4 go
where
ρ = 0.284 lb in 3
µ = 0.30
D 10
ro = o =
= 5 in
2
2
D 2
ri = i = = 1 in
2 2
g o = 386 in sec
Page 21 of 25
+ (1 − µ )ri
2
] psi
SECTION 18 – MISCELLANEOUS PROBLEMS
(a) Solving for maximum stress for a speed of 10,000 rpm
 10,000 
ω = 2π 
 = 1047 rad sec
 60 
2
(
0.284)(1047 )
(3 + 0.3)(5)2 + (1 − 0.3)(1)2 = 16,776 psi
st =
4(386 )
[
]
(b) Solving for maximum stress for a speed of 10,000 rpm
 20,000 
ω = 2π 
 = 2094 rad sec
 60 
2
(
0.284)(2094)
(3 + 0.3)(5)2 + (1 − 0.3)(1)2 = 67,104 psi
st =
4(386)
[
]
(c) Solving for maximum speed, ω .
For AISI 3150, OQT at 1000 F, s y = 130 ksi
st = s y =
ρω 2
4go
[(3 + µ )r
o
2
+ (1 − µ )ri
2
] psi
(
0.284 )ω 2
[
(3 + 0.3)(5)2 + (1 − 0.3)(1)2 ]
130,000 =
4(386)
ω = 2914.57 rad sec
60ω 60(2914.57 )
RPM =
=
= 27,832 rpm
2π
2π
1001.
The same as 1000, except that Di = 1 in .
Solution:
Di = 1 in
D 1
ri = i = in = 0.5 in
2 2
use other data as in 1000.
(a) Solving for maximum stress for a speed of 10,000 rpm
 10,000 
ω = 2π 
 = 1047 rad sec
 60 
2
(
0.284)(1047 )
(3 + 0.3)(5)2 + (1 − 0.3)(0.5)2 = 16,670 psi
st =
4(386 )
[
]
(b) Solving for maximum stress for a speed of 10,000 rpm
Page 22 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
 20,000 
 = 2094 rad sec
 60 
2
(
0.284)(2094)
(3 + 0.3)(5)2 + (1 − 0.3)(0.5)2 = 66,680 psi
st =
4(386)
ω = 2π 
[
]
(c) Solving for maximum speed, ω .
For AISI 3150, OQT at 1000 F, s y = 130 ksi
st = s y =
ρω 2
4go
130,000 =
[(3 + µ )r
o
2
+ (1 − µ )ri
2
] psi
(0.284)ω 2 [(3 + 0.3)(5)2 + (1 − 0.3)(0.5)2 ]
4(386 )
ω = 2923.8 rad sec
60ω 60(2923.8)
RPM =
=
= 27,920 rpm
2π
2π
1002.
A circular steel disk, with Do = 8 in and Di = 2 in , is shrunk onto a solid steel
shaft with an interference of metal i = 0.002 in . (a) At what speed will the
pressure in the fit become zero as a result of the rotation? Assume that the shaft is
unaffected by centrifugal action. (This effect is relatively small.) (b) Compute the
maximum stress in the disk and the pressure at the interface when the speed is
10,000 rpm. Note: The maximum stress in the disk is obtained by adding
equations (8.15) of i8.26, Text, and (n of i20.9. The resulting equation together
with equation (s) of i8.27 can then be used to obtain pi and σ th ; where
σ ts = − pi for a solid shaft.
Solution:
(a) Solving for speed, pi = 0
From equation 8.15, i8.26, Text.
σ + µ h pi σ ts + µ s pi 
i = Di  th
−

Eh
Es


Di = 2 in
i = 0.002 in
Eh = Es = 30 × 10 6 psi
µ h = µ s = 0.30
σ ts = − pi = 0
Page 23 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
σ + (0.3)(0 ) 0 + (0.3)(0) 
i = 0.002 = 2  th
−
6
30 ×10 6 
 30 ×10
σ th = 30,000 psi
σ th = σ ti + st
From Equation 8-15
pi (ro2 + ri 2 ) − 2 po ro2
σ ti =
ro2 − ri 2
but po = 0
pi (ro2 + ri 2 )
ro2 − ri 2
D
8
ro = o = = 4 in
2
2
Di 2
ri =
= = 1 in
2 2
pi = 0
σ ti = 0
σ ti =
From Equation (n) i20.9
st =
ρω 2
4 go
[(3 + µ )r
o
2
+ (1 − µ )ri
2
] psi
ρ = 0.284 lb in 3
µ = 0.30
ro = 4 in
ri = 1 in
g o = 386 in sec
σ th = σ ti + st
30,000 = 0 +
(0.284)ω 2 [(3 + 0.3)(4)2 + (1 − 0.3)(1)2 ]
4(386)
ω = 1746 rad sec
60ω 60(1746 )
RPM =
=
= 16,673 rpm
2π
2π
(b) Solving for the maximum stress in the disk and the pressure within the interface.
σ + µ h pi σ ts + µ s pi 
i = Di  th
−

Eh
Es


σ + 0.3 pi (− pi + 0.3 pi ) 
0.002 = 2  th
−
6
30 ×10 
 30 ×10
30,000 = σ th + pi
Page 24 of 25
SECTION 18 – MISCELLANEOUS PROBLEMS
σ th = 30,000 − pi
σ th = σ ti + st
σ ti =
st =
 4 2 + 12  17
pi (ro2 + ri 2 )

=
=
p
pi
i 2
2 
ro2 − ri 2
 4 − 1  15
ρω 2
[(3 + µ )r
2
+ (1 − µ )ri
2
]
psi
4 go
2π (10,000)
ω=
= 1047 rad sec
60
(0.284)(1047 )2 (3 + 0.3)(4)2 + (1 − 0.3)(1)2 = 10,788 psi
st =
4(386 )
σ th = σ ti + st
17
σ th = 30,000 − pi = pi + 10,788
15
pi = 9000 psi (interface pressure)
σ th = 30,000 − pi = 30,000 − 9000 = 21,000 psi (maximum stress)
o
[
]
- end -
Page 25 of 25