Solucionario faires

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?
F
sd =
where
F = 8000 lb
A = bh
but
h =1.5b
therefore 2 A =1.5b
F
s
s u
d = =
8000
96,000
Page 1 of 131
Problems 1 – 3.
Solution:
For AISI C1045 steel, as rolled (Table AT 7)
s ksi u = 96
s ksi y = 59
E psi 6 = 30×10
A
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
A
N
2 1.5
6
b
=
5
b = 0.577 in say in
8
.

15
=1.5 =
h b in
16
(b) Based on yield strength
F
A
s
s u
d = =
N
8000
2 1.5
59,000
3
b
=
9
b = 0.521in say in
27
FL
1
Page 2 of 131
16
.
h b in
32
=1.5 =
(c) Elongation =
FL
AE
d =
where,
d = 0.005 in
F = 8000 lb
E psi 6 = 30×10
L =15 in
2 A =1.5b
then,
AE
d =
( 8000 )( 15
)
( 1.5 2 )( 30 10
6 ) 0.005
×
=
b
3
b = 0.730 in say in
4
.
h b in
8
=1.5 =1
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)
s ksi u = 55
s ksi y = 36.5
E psi 6 = 25×10

F
s=
d A
where
F = 8000 lb
A = bh
but
h =1.5b
therefore A =1.5b
2 (a) Based on ultimate strength
F
A
s
s u
d = =
N
8000
2 1.5
55,000
6
b
=
7
b = 0.763 in say in
5
F
s
s u
d = =
8000
36,500
1
Page 3 of 131
8
.
h b in
16
=1.5 =1
A
N
2 1.5
3
b
=
11
b = 0.622 in say in
16
.
h b in
32
=1.5 =1
(c) Elongation =
FL
AE
d =
where,
d = 0.005 in
F = 8000 lb
E psi 6 = 25×10
L =15 in
2 A =1.5b
then,

FL
AE
d =
( 8000 )( 15
)
( 1.5 2 )( 25 10
6 ) 0.005
5
F
sd =
where
F = 8000 lb
A = bh
but
h =1.5b
therefore 2 A =1.5b
F
s
s u
d = =
8000
30,000
25
Page 4 of 131
×
=
b
7
b = 0.8 in say in
8
.
h b in
16
=1.5 =1
3. The same as 1 except that the material is gray iron, ASTM 30.
Solution:
For ASTM 30 (Table AT 6)
s ksi u = 30 , no y s
E psi 6 =14.5×10
Note: since there is no y s for brittle materials. Solve only for (a) and (c)
A
N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1)
A
N
2 1.5
7.5
b
=
3
1 .
b =1.1547 in say in
16
h b in
32
=1.5 =1
(c) Elongation =
FL
AE
d =
where,
d = 0.005 in
F = 8000 lb
E psi 6 =14.5×10

L =15 in
2 A =1.5b
then,
FL
AE
d =
( 8000 )( 15
)
( 1.5 2 )( 14.5 10
6 ) 0.005
19
p
2 = = = =
N F
u
s
N F
y
s
Page 5 of 131
×
=
b
1
1 .
b =1.050 in say in
16
h b in
32
=1.5 =1
4. 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
s ksi u =152.5
s ksi y =132.5
F force (20) (125) 39,270 lb 39.27 kips
4
From Table 1.1, page 20
= 8 u N
= 4 y N
u
A =
( 8 )( 39.27
)
152.5
4
2 d =
p
5
1
d =1.62 in say in
8
y
A =
( 4 )( 39.27
)
132.5
4
2 d =
p

1
1
d =1.23 in say in
p p p
= − = − =
3 D 2
N F
= i = u
=

7
8 2 2 =
p p p
= − = − =
3 D 2
N F
= i = y
=
Page 6 of 131
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 o i D = 2D .
Solution:
s ksi u = 65
= 8 u N
F =1500 kips
( ) ( )
D
4
3
4
4 4
2
2 2 2 2 i
o i i i
A D D D D
( 8 )( 1500
)
65
4
s
u
A
p
7
8
D in i = 8.85 say in
8
3
17
D D in o i 8
4

= =
6. A short compression member with o i D = 2D 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
s ksi u =127
s ksi y =114
From Table 1.1 page 20, for dead load
= 3 ~ 4 u N , say 4
=1.5 ~ 2 y N , say 2
Area, ( ) ( )
D
4
3
4
4 4
2
2 2 2 2 i
o i i i
A D D D D
F = 25 tons = 50 kips
(a) Based on yield strength
( 2 )( 50
)
114
4
y
s
A
p

5

2 2 =
3 D 2
N F
= i = u
=

2 2 =
FL
d = or
p
4 ×
N F
3 7000
s = y
= =
47,534
y Page 7 of 131
8
1
1
5
D D in o i 8
4

= =
(b) Based on ultimate strength
( 4 )( 50
)
127
4
s
u
A
p
7
8
3
1
7
D D in o i 8
4

= =
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)
AE
FL
E
A
d
=
where,
F = 7000 lb
L = 55 in
d = 0.030 in
E psi 6 = 25×10
( )( )
( )( 6 )
2
7000 55
0.030 30 10
A = d =
3
d = 0.74 in say in
4
(b) For gradually applied and repeated (not reversed) load
= 3 y N
( )( )
( )
psi
A
0.75
4
2
p
s ksi y » 48
say C1015 normalized condition ( s ksi y = 48 )
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

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
s ksi us = 48
r
W
F 2 = w
g
where
W = 0.332 lb
2 g = 32.2 fps
p p
2 10,000
2
W
0.332 = w 2 = 2 = =
F 1047 1 11,300 11.3
N F
u
s
4 11.3

2 2 =
1
=1 . (a) What force will cause breakage? (b) For a design factor of 4 based
Page 8 of 131
( )
1047 sec
60
60
rad
n
= = =
w
r =12 in
r ( ) ( ) lb kips
g
32.2
From Table 1.1, page 20
N = 3 ~ 4 , say 4
u
A =
( )( )
48
4

d
p
for double shear
25
d = 0.774 in say in
32
CHECK PROBLEMS
3
= and
9. The link shown is made of AISIC1020 annealed steel, with b in
4
h in
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.

Solution:
For AISI C1020 annealed steel, from Table AT 7
s ksi u = 57
s ksi y = 42
(a) F s A u =
3

in bh A =

s A
3

in bh A =

57 1.125
= =
s A
3

in bh A =

42 1.125
= =
sL
sL
57.6 5
d = = =
Page 9 of 131
2 1.125
1
1
2
4

= =
F = (57)(1.125) = 64 kips
(b)
u
u
N
F =
= 4 u N
2 1.125
1
1
2
4

= =
( )( )
F 16 kips
4
(c)
y
y
N
F =
= 2.5 y N
2 1.125
1
1
2
4

= =
( )( )
F 18.9 kips
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:
E
d =
from Table AT 7 for cold-finished B1113
s ksi y = 72
then, s s ( ) ksi y = 0.80 = 0.8 72 = 57.6
E 30 10 psi 30,000 ksi 6 = × =
( )( )
in
E
0.0096
30,000

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, s ksi psi us = 62 = 62,000
3

( ) 2
1

2 in A =
W
F us = = 2 w
4 2 w =
p n
2
= =
F
= = =
n holes
Page 10 of 131
2
0.2209
8
4

= p
r s A
g
(14) (62,000)(0.2209)
32.2
w = 88.74 rad sec
88.74
60
w
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
s ksi u = 80
s s ( ) ksi ksi us u = 0.75 = 0.75 80 = 60
A =p dt
F = 60 tons =120 kips
n = number of holes
120
( )( )
9
0.2209 60
us As
13. 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
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?
3 th
I = , moment of inertial
4 h
Page 11 of 131
Problem 14.
Solution:
For AISI C1020, as rolled, Table AT 7
s ksi u = 65
s ksi y = 49
Design factors for gradually applied and reversed load
= 8 u N
= 4 y N
12
but h = 3t
36
I =
Moment Diagram (Load Upward)

Based on ultimate strength
s
u
N
u
s =
(a)
Fac
Mc
s = =
h

65
3.86
= = =
1
1
Fbc
Mc
s = =
h

65
2.61
= = =
Page 12 of 131
I
I
2
c =
F = 2000 lbs = 2 kips

( )( )

= =
36
2
2 13
8
4 h
h
s
h = 3.86 in
in
h
t 1.29
3
3
say
h in in
2
= 4.5 = 4
t in in
2
=1.5 =1
(b)
I
I
2
c =
F = 2000 lbs = 2 kips
( )( )

= =
36
2
2 4
8
4 h
h
s
h = 2.61in
in
h
t 0.87
3
3
say
h = 3 in
t =1in
(c)

−
4.5 3
−
13 4
− h
4
3
=
h = 2.33 in
−
1.5 1
−
13 4
− t
4
1
=
t = 0.78 in
say
5
h = 2.625 in or h in
Page 13 of 131
8
= 2
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 ksi y = 40
E psi 6 = 30×10

From Table AT 2
Max.
3 bh
4 h
Mc
s
y = =
h

40
4.18
= = =
1
1
4.5
= 4 , in in
3 3
FL

48 3
FL

4000 54 36
4.79
= = =
3
5.25
= 5.25 = 5 , in in
Page 14 of 131
( )( )
kips in
FL
4 54
M = = = 54 −
4
4
12
I =
but h = 3b
36
I =
(a)
I
N
s
y
2
c =
( )

=
36
2
54
3
4 h
h
h = 4.18 in
in
h
b 1.39
3
3
say h in
2
h
b
2
= = = =
1.5 1
3
3
(b)
( )( )
( ) ( )( )
in
EI
0.0384
4000 54
1.5 4.5
12
48 30 10
6
=

×
d = =
(c)

=
36
48
4
3
h
E
d
( )( ) 3
( )
48 ( 30 10
6 )( 4 )
0.03
× h
=
h = 4.79 in
in
h
b 1.60
3
3
1
say h in in
4
h
b
4
= = = =
1.75 1
3
3

16. The same as 15, except that the beam is to have a circular cross section.
Solution:
(a)
Mc
I
s
= y =
N
s
y
4 d
p
=
64
I
d
2
c =
32
d
2
M
=

4 3

64

32 54
40
1
3
FL
48
4 d
FL
64
3
64
E d
64 4000 54
1
Page 15 of 131
d
d
M
s
p p

=
( )
3
3
p d
=
d = 3.46 in
say d in
2
= 3
(b)
EI
d =
64
I
p
=
( )
( )( )
( )( )( )
in
E d
0.0594
64 4000 54
3
48 30 10 3.5
48
6 4
4
3
=
×
= =
p p
d
(c) 48
( 4 )
FL
p
d =
( )( )
3
( 6 )( ) 4
48 30 10
0.03
× p d
=
d = 4.15 in
say d in
4
= 4
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.”

Solution:
From Table AT 7 and Table 1.1
s ksi u = 65
s ksi y = 48
= 3 ~ 4 u N , say 4
=1.5 ~ 2 y N , say 2
( 6 )( 48
)
Mc
h
3 bh
h
4 h
12
2

s
= u =
65
Page 16 of 131
in kips
FL
M = = = 72 −
4
4
I
s =
2
c =
12
I =
but
2
b =
24
I =
M
4 3
24
h
h
h
M
s =
=
3
12
h
M
N
s
u
( )
3
12 72
4
h
=
h = 3.76 in

3.76
= = =
7
3.75
= 3.75 = 3 , in in
s
y = =
48
3.30
= = =
3
3.5
= 3.5 = 3 , in in
2 bh
2

2.29
= = =
1
2.5
= 2.5 = 2 , in in
Page 17 of 131
in
h
b 1.88
2
2
3
say h in in
4
h
b
8
= = = =
1.875 1
2
2
3
12
h
M
N
s
y
( )
3
12 72
2
h
=
h = 3.30 in
in
h
b 1.65
2
2
1
say h in in
2
h
b
4
= = = =
1.75 1
2
2
(c) Limit design (Eq. 1.6)
4
M sy =
( )
4
72 48
2 h
h

=
h = 2.29 in
in
h
b 1.145
2
2
1
say h in in
2
h
b
4
= = = =
1.25 1
2
2
18. The bar shown is subjected to two vertical loads, 1 F and 2 F , 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 A47-
52, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and
moment diagrams approximately to scale.

3 bh
h
3
h
h

3 h
4

I =
Mc
s

s u =
30
2.78
= = =
Page 18 of 131
Problems18, 19.
Solution:
F F R R 3000 lb 1 2 1 2 = = = =
Moment Diagram
M = R a = (3000)(3) = 9000 lbs − in = 9 kips − in 1
N = factor of safety = 4 based on u s
12
I =
2
c =
12 36

=
(a) For gray cast iron, SAE 111
s ksi u = 30 , Table AT 6
18
2
M

4 3
36
h
h
h
M
I
N

= = =
( )
3
18 9
4
h
s = =
h = 2.78 in
in
h
b 0.93
3
3
say h = 3.5 in , b =1in
(b) For malleable cast iron, ASTM A47-52, grade 35 018

Mc
s

s u =
55
2.28
= = =
1
3
=
= 2 , b in
Mc
s

s u =
90
1.93
= = =
7
5
=
=1 , b in
Page 19 of 131
18
2
M

4 3
36
h
h
h
M
I
N

= = =
( )
3
18 9
4
h
s = =
h = 2.28 in
in
h
b 0.76
3
3
say h in
4
4
(c) For AISI C1040, as rolled
s ksi u = 90 , Fig. AF 1
18
2
M

4 3
36
h
h
h
M
I
N

= = =
( )
3
18 9
4
h
s = =
h =1.93 in
in
h
b 0.64
3
3
say h in
8
8
19. The same as 18, except that 1 F acts up ( 2 F acts down).
Solution:
[
= 0] A M R R 1875 lb 1 2 = =

Shear Diagram
Moment Diagram
M =maximum moment = 5625 lb-in = 5.625 kips-in
(a) For gray cast iron
M
3
18
h
s
s = u =
N
( )
18 5.625
3
30
4
h
=
h = 2.38 in
2.38
= = =
1
3
=
= 2 , b in
s
s = u =
18 5.625
55
1.95
= = =
7
5
=
=1 , b in
Page 20 of 131
in
h
b 0.79
3
3
say h in
4
4
(b) For malleable cast iron
3
18
h
M
N
( )
3
4
h
=
h =1.95 in
in
h
b 0.65
3
3
say h in
8
8

M
3
18
h
s
s = u =
N
( )
18 5.625
3
90
4
h
=
h =1.65 in
1.65
= = =
1
1
=
=1 , b in
Page 21 of 131
in
h
b 0.55
3
3
say h in
2
2
20. The bar shown, supported at A and B , is subjected to a static load F of 2500 lb.
at q = 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:

[
= 0] A M 3 =13(2500) B R
3 bh
4 h
h
2
Mc

s =

s
s = u =
20
Page 22 of 131
R lb B =10,833
[
= 0] B M 3 =10(2500) A R
R lb A = 8333
Shear Diagram
Moment Diagram
M = (2500)(10) = 25,000 lb − in = 25 kips − in
h = 3b
12
I =
36
I =
2
c =
18
M

4 3
36
h
h
h
M
I

= =
N = 5 ~ 6 , say 6 for cast iron, dead load
3
18
h
M
N
( )
3
18 25
6
h
=

h = 5.13 in
in
h
= =
b 1.71
3
1
3
= 5 , b in
say h in
4
s
s = u =
52
= =
3
1
= 3 , b in
s
s = u =
85
= =
R
s
s su B
s = =

p p
=

A 2 D D
10.833
64
= =
Page 23 of 131
4
=1
(b) For malleable cast iron, ASTM A47-32 grade 32510
s ksi u = 52 , s ksi y = 34
N = 3 ~ 4 , say 4 for ductile, dead load
3
18
h
M
N
( )
3
18 25
4
h
=
h = 3.26 in
in
h
b 1.09
3
say h in
4
4
=1
s ksi u = 85 , s ksi y = 55
N = 4 , based on ultimate strength
3
18
h
M
N
( )
3
18 25
4
h
=
h = 2.77 in
in
h
b 0.92
3
say h = 3 in , b =1in
(d) For AISI C1035, as rolled
s ksi su = 64
N = 4 , R kips B =10.833
A
N
2 2
4 2

=
2
2
4
D
ss p
D = 0.657 in

11
=
say D in
16
(e) Limit Design
2 bh
4
M sy =
For AISI C1035 steel, s ksi y = 55
h
3
b =
( )

= =
5
=
=1.875 =1 , b in
Page 24 of 131
3

4
25 55
2 h
h
M

= =
h =1.76 in
in
h
b 0.59
3
7
say h in in
8
8
21. The same as 20, except that o q = 30 . Pin B takes all the horizontal thrust.
Solution:
F F cosq V =
[
= 0] A M B V 3R =13F
3 =13(2500)cos30 B R
R lb B = 9382
[
= 0] B M A V 3R =10F
3 =10(2500)cos30 A R
R lb A = 7217
Shear Diagram

Moment Diagram
M = (2165)(10) = 21,650 lb − in = 21.65 kips − in
M
3
18
h
s =
N = 5 ~ 6 , say 6 for cast iron, dead load
M
3
18
h
s
s = u =
N
( )
18 21.65
3
20
6
h
=
h = 4.89 in
in
h
= =
b 1.63
3
1
3
= 5 , b in
say h in
4
s
s = u =
18 21.65
52
= =
Page 25 of 131
4
=1
(b) For malleable cast iron, ASTM A47-32 grade 32510
s ksi u = 52 , s ksi y = 34
N = 3 ~ 4 , say 4 for ductile, dead load
3
18
h
M
N
( )
3
4
h
=
h = 3.11in
in
h
b 1.04
3
say h = 3 in , b =1in
s ksi u = 85 , s ksi y = 55
N = 4 , based on ultimate strength

M
3
18
h
s
s = u =
N
( )
18 21.65
3
85
4
h
=
h = 2.64 in
in
h
= =
b 0.88
3
5
7
=
= 2 , b in
say h in
8
R
s
s su B
s = =

p p
=

A 2 D D
9.465
64
= =
5
=
2 bh
h
= =
5
=
=1.875 =1 , b in
Page 26 of 131
8
(d) For AISI C1035, as rolled
s ksi su = 64
N = 4 , R lb BV = 9382
R F F lb BH H = = sinq = 2500sin 30 =1250
( ) ( )2 2 2 2 2 = + = 9382 + 1250 B BV BH R R R
R lb B = 9465
A
N
2 2
4 2

=
2
2
4
D
ss p
D = 0.614 in
say D in
8
(e) Limit Design
4
M sy =
For AISI C1035 steel, s ksi y = 55
3
b =
( )
3

4
21.65 55
2 h
h
M

= =
h =1.68 in
in
h
b 0.56
3
7
say h in in
8
8

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.
Mc

3 + =
3a
Page 27 of 131
Problem 22.
Solution:
For cast iron, ASTM 50
s ksi u = 50 , s ksi uc =164
For gradually applied, repeated load
N = 7 ~ 8 , say 8
M = F d + F (d + e) 1 2
where:
F 2000 lb 1 =
F 1000 lb 2 =
d = 30 −10 = 20 in
d + e = 30 in
M = (2000)(20)+ (1000)(30) = 70,000 lb − in = 70 kips −in
I
s =
Solving for I , moment of inertia
a
5
( )( ) ( a )( a
) [( a)(a) ( a)(a)]y
a
a a 3 3
2
3
2

+

2
y =

( )( ) ( )( )( ) ( )( ) ( )( )( )
3 a
3 4
3a
5a
cc =
Based on tension
Mc
s
s u t
t = =

a

17
3
50
Mc
s
s uc c
c = =

a

17
5
164
1
Page 28 of 131
17
2
3
3
12
3
12
2
3
2
a a a
a a
a a a
a a
I = + + + =
(a)
2
ct =
2
I
N
( )

=
2
2
70
8
4 a
a =1.255 in
Based on compression
I
N
( )

=
2
2
70
8
4 a
a =1.001in
Therefore a =1.255 in
Or say a in
4
=1
And b = c = 3a = 3(1.25) = 3.75 in

3
= = 3
Or b c in
4
(b) If the top of the tee is below
5a
2
ct =
3a
2
cc =
17 4 a
2
I =
M = 70 kips − in
Based on tension
Mc
s u t
t = =
I
s
N

( )

a

70

=
17
4 a
2
5
2
50
8
a =1.488 in
Based on compression
Mc
s uc c
c = =
I
s
N

( )

a

17
3
164
1
Page 29 of 131

=
2
2
70
8
4 a
a = 0.845 in
Therefore a =1.488 in
Or say a in
2
=1
1
And b c a in
2
= = 3 = 4
CHECK PROBLEMS

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 2 1 F = 2F ; 1 F is 5 in. from one end and
2 F is 5 in. from the other ends. The beam is 25 in. long; flange width is
b = 2.509 in ; 4 I 2.9 in x = . 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.
Page 30 of 131
Problems 23 – 25.
Solution:
[
= 0] A M ( ) B 5F 20 2F 25R 1 1 + =
1 R 1.8F B =
[
= 0] V F A B F + F = R + R 1 1 2
1 1 1 R 3F 1.8F 1.2F A = − =
Shear Diagram
Moment Diagram

M = 9F1 = maximum moment
For AISI C1020, as rolled
s ksi y = 48
(a)
Mc
I
sy =
d
3
= = =
( )( )
where c 1.5
in
9 1.5
48 1 F
Mc
s
s y = =
48 1 F
9 1.5
s = =
25
= =
b
Mc
9 1.5
20 1 F
¢

¢ +
Fa
y , b¢ a
Page 31 of 131
2
2
2.9
sy = =
F 10.31 kips 1 =
F 2F 20.62 kips 2 1 = =
(b)
I
N
( )( )
2.9
3
F 3.44 kips 1 =
F 2F 6.88 kips 2 1 = =
(c) 9.96 15
2.509
L
(page 34)
s ksi c = 20 ( page 34, i1.24)
I
sc =
( )( )
2.9
=
F 4.30 kips 1 =
F 2F 8.60 kips 2 1 = =
(d) For maximum deflection,
by method of superposition, Table AT 2
3
( ) 2
max 3 3

=
a L b
EIL
or
3
( ) 2
max 3 3

=
b L a
EIL

max y caused by 1 F

+ ¢
F b
y , 2 2 a b¢
2 5
d = y + y = F + F = F
Mc
Page 32 of 131
3
( ) 2
1 1 1 1

max 3 3 1

=
b L a
EIL
where E = 30,000 ksi
a 5 in 1 =
b 20 in 1 ¢ =
L = 25 in
4 I = 2.9 in
( 5
)
( )( )( )
( )
1
3
2
1

+
20 25 5

max 0.0022
3
3 30,000 2.9 25
1
F
F
y =

=
max y caused by 2 F
3
( ) 2
2 2 2 2

max 3 3 2

=
a L b
EIL
where b 5 in 2 ¢ =
a 20 in 2 =
( )
( )( )( )
( )
1
3
2
1

+
20 25 5

max 0.0043
3
3 30,000 2.9 25
2
F
F
y =

=
Total deflection = d
max max 1 1 1 0.022 0.0043 0.0065
1 2
Deflection caused by the safe loads in (a)
( ) in a d = 0.0065 10.31 = 0.067
Deflection caused by the safe loads in (b)
( ) in b d = 0.0065 3.44 = 0.022
Deflection caused by the safe loads in (c)
( ) in c d = 0.0065 4.30 = 0.028
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 ksi y = 47
(a)
I
sy =

( 9 )( 1.5
)
47 1 F
2.9
sy = =
F 10.10 kips 1 =
F 2F 20.20 kips 2 1 = =
(b)
Mc
I
s
s y = =
N
( )( )
47 1 F
9 1.5
2.9
s = =
3
F 3.36 kips 1 =
F 2F 6.72 kips 2 1 = =
25
L
= =
b
(c) 9.96 15
2.509
Mc
9 1.5
20 1 F
d = y + y = F + F = F
Page 33 of 131
(page 34)
s ksi c = 20 ( page 34, i1.24)
I
sc =
( )( )
2.9
=
F 4.30 kips 1 =
F 2F 8.60 kips 2 1 = =
(d) Total deflection = d
max max 1 1 1 0.022 0.0043 0.0065
1 2
Deflection caused by the safe loads in (a)
( ) in a d = 0.0065 10.10 = 0.066
Deflection caused by the safe loads in (b)
( ) in b d = 0.0065 3.36 = 0.022
Deflection caused by the safe loads in (c)
( ) in c d = 0.0065 4.30 = 0.028
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 4 I 6.0 in x = (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.

Solution:
(a) For C1020, as rolled, s ksi u = 65
Consider flange buckling
30
80
L
= =
b
2.66
L
since 15 40
b
22.5
2 2 =
sc 15
L

+
Mc
4
= = =
80
= = =
Mc
20 2
2 wL
M = (Table AT 2)
Page 34 of 131
( )
ksi
b
30
1800
1
22.5
1 1800
+
=

=
I
s =
in
d
c 2
2
2
From Table AT 2
( )
F
FL F
M 20
4
4
I
s sc = =
( )( )
6
15
F
=
F = 2.25 kips , safe load
(b) Considering stress owing to the weight of the beam
add’l
8
where w = 7.7 lb ft

add’l
Mc
20 0.513 2
= = =
Ec
30 10 0.01325 6
Page 35 of 131
( )
lb in kips in
wL

80
7.7

M = = = 513 − = 0.513
− 8
12
8
2 2
M = 20F + 0.513 = total moment
I
s sc = =
( )( )
6
15
+
=
F
F = 2.224 kips
Therefore, the safe load should be less.
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:
in
t
c 0.0265 0.01325
2
r =13.75 + 0.01325 =13.76325 in
Using Eq. (1.4) page 11 (Text)
r
s =
where E psi 6 = 30×10
( )( )
s 28,881 psi
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:
s ksi u = 65 (Table AT 7)
Designing based on ultimate strength,
N = 6, for repeated, minor shock load

s
65
s u 10.8
= = =
4 1 −
h 1
3 wh
h

Mc
+ − +
0.10 1 1 2 0.10 1 0.10

x x x
3 4
ds
s
s = u =
Page 36 of 131
ksi
N
6
Loading Diagram
x
30
=
−
h = 0.10x +1
12
I =
2
c =
M = Fx
( )
3 3
6

Fx
Fx
Fx
3 2
2 2 ( 0.10 1
)2 12
2
+
= = =

= =
x
h
h
wh
h
Fx
I
s
Differentiating with respect to x then equate to zero to solve for x giving maximum
stress.
( ) ( ) ( )( )( )
( )
0
0.10 1
2
=

+
=
x
F
dx
0.10x +1− 2(0.10x) = 0
x =10 in
h = 0.10(10)+1 = 2 in
2
3
h
Fx
N
( )
( )2 2
3 10
10.8
F
=
F =1.44 kips
TORSIONAL STRESSES
DESIGN PROBLEMS

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 ksi y = 59
s ksi us = 72
Designing based on ultimate strength
s
s = us , N = 6 (Table 1.1)
N
72
= =
s 12 ksi
6
Torque,
33,000
16 0.540
5
=
33,000
p p
Page 37 of 131
( )
( )
ft lb in lb in kips
p n
p
hp
33,000 15
T = = = 45 − = 540 − = 0.540 −
2 1750
2
For diameter,
3
16
d
T
s
p
=
( )
3
12
p d
=
d = 0.612 in
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, o i D = 2D , 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:
( )
( )
ft lb in kips
n
hp
33,000 2500
T = = = 23,036 − = 276 −
2 570
2
For AISI 1137, annealed
s ksi y = 50 (Table AT 8)
s s ksi ys y = 0.6 = 30
Designing based on yield strength
N = 3 for medium shock, one direction

Design stress
s
30
s ys 10
= = =
Tc
4 D
D
16 276
1
4 4 4 4 4
o i i i i D D D D D
o i D
TD
i
T
=

15 i i

32 276
5
= 2
1
= 5
Page 38 of 131
ksi
N
3
(a) Let D = shaft diameter
J
s =
32
J
p
=
2
c =
3
16
D
T
s
p
=
( )
3
10
p D
=
D = 5.20 in
say D in
4
= 5
(b)
( ) [( ) ]
15
32
32
2
32
J
p p p
=
−
=
−
=
i
D D
c = = =
2
2
2
32
4 3 15
32
D
D
s
p p

=
( )
3 15
10
i p D
=
D in i = 2.66
D D in o i = 2 = 5.32
say
D in i 8
D in o 4
(c) Density, 3 r = 0.284 lb in (Table AT 7)

For solid shaft
w = weight per foot of length

p
w D 3 D 3 (0.284)(5.25) 73.8 lb ft
12 2 = 2 = 2 = 4

p
12 2 2 2 2 2 2 = − = − = −
TL
q rad

q rad

33,000
p p
Page 39 of 131

= r
pr p
For hollow shaft
w (D D ) (D D ) ( )[( ) ( ) ] lb ft o i o i 3 3 0.284 5.25 2.625 55.3
4

= r
pr p
Therefore hollow shaft is lighter
73.8 55.3
Percentage lightness = (100%) 33.5%
55.3
=
−
(d) Torsional Deflection
JG
q =
where
L =10 ft =120 in
G ksi 3 =11.5×10
For solid shaft,
4 D
32
J
p
=
( )( )
( ) ( )
180

( ) o 2.2
0.039 0.039
5.25 11.5 10
32
276 120
4 3
=

= =
×

=
p p
For hollow shaft,
( )
4 4
o i D D
32
J
−
=
p
( )( )
[( ) ( ) ]( )
180

( ) o 2.4
0.041 0.041
5.25 2.625 11.5 10
32
276 120
4 4 3
=

= =
× −

=
p p
Therefore, solid shaft is more rigid, o o 2.2 2.4
30. The same as 29, except that the material is AISI 4340, OQT 1200 F.
Solution:
( )
( )
ft lb in kips
n
hp
33,000 2500
T = = = 23,036 − = 276 −
2 570
2
For AISI 4340, OQT 1200 F
s ksi y =130
s s ( ) ksi ys y = 0.6 = 0.6 130 = 78
Designing based on yield strength

N = 3 for mild shock
Design stress
s
78
s ys 26
= = =
Tc
4 D
D
16 276
3
4 4 4 4 4
o i i i i D D D D D
o i D
TD
i
T
=

15 i i

32 276
Page 40 of 131
ksi
N
3
(a) Let D = shaft diameter
J
s =
32
J
p
=
2
c =
3
16
D
T
s
p
=
( )
3
26
p D
=
D = 3.78 in
say D in
4
= 3
(b)
( ) [( ) ]
15
32
32
2
32
J
p p p
=
−
=
−
=
i
D D
c = = =
2
2
2
32
4 3 15
32
D
D
s
p p

=
( )
3 15
26
i p D
=
D in i =1.93
D D in o i = 2 = 3.86
say
D in i = 2
D in o = 4
(c) Density, 3 r = 0.284 lb in (Table AT 7)

For solid shaft
w = weight per foot of length

p
w D 3 D 3 (0.284)(3.75) 37.6 lb ft
12 2 = 2 = 2 = 4

p
12 2 2 2 2 2 2 = − = − = −
TL
q rad

q rad

33,000
p p
Page 41 of 131

= r
pr p
For hollow shaft
w (D D ) (D D ) ( )[( ) ( ) ] lb ft o i o i 3 3 0.284 4 2 32.1
4

= r
pr p
Therefore hollow shaft is lighter
37.6 32.1
Percentage lightness = (100%) 17.1%
32.1
=
−
(d) Torsional Deflection
JG
q =
where
L =10 ft =120 in
G ksi 3 =11.5×10
For solid shaft,
4 D
32
J
p
=
( )( )
( ) ( )
180

( ) o 8.48
0.148 0.148
3.75 11.5 10
32
276 120
4 3
=

= =
×

=
p p
For hollow shaft,
( )
4 4
o i D D
32
J
−
=
p
( )( )
[( ) ( ) ]( )
180

( ) o 6.99
0.122 0.122
4 2 11.5 10
32
276 120
4 4 3
=

= =
× −

=
p p
Therefore, hollow shaft is more rigid, o o 6.99 8.48 .
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 20D? (b) If deflection is
primary what kind of steel would be satisfactory?
Solution:
(a)
( )
( )
ft lb in kips
n
hp
33,000 40
T = = = 420 − = 5.04 −
2 500
2
G ksi 3 =11.5×10
L = 20D

rad
p
180
q = 1
o
= TL
JG
q =
( 5.04 )( 20
)
( 3 )
4

×

11.5 10
D
32
180

=
3
16 T
16 5.04
3 3 = = =
p p
s 4.8
14.4
= = =
T
s
s ys
2 b 2 h 2
b
3 Z¢ = = (Table AT 1)
45
3
Page 42 of 131
D
p
p
D =1.72 in
say D in
4
=1
(b)
( )
( )
ksi
D
1.75
Based on yield strength
N = 3
s Ns ( )( ) ksi ys = = 3 4.8 =14.4
ksi
s
s ys
y 24
0.6
0.6
Use C1117 normalized steel s ksi y = 35
32. A square shaft of cold-finish AISI 1118 transmits a torsional moment of 1200 in-lb.
For medium shock, what should be its size?
Solution:
For AISI 1118 cold-finish
s ksi y = 75
s s ksi ys y = 0.6 = 45
N = 3 for medium shock
Z
N
¢
= =
where, h = b
9
9
T =1200 in −lb =1.2 in − kips
( )
3 2
1.2 9
3
b
s = =
b = h = 0.71in
say b h in
4
= =

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:
s ksi us = 49
F s ( Dt ) us = p
15
=
where D in
16
1
=
t in
2
1
15

F ( ) 72.2 kips
49 =

3
=

3
16
d
T
16 54.2
3 = =
p
Page 43 of 131
2
16

= p
T = Fr
where r in
4
( ) kips in T − =

= 54.2
4
72.2
(a) s
3
p
=
where d = 3.5 in
( )
( )
s 6.44 ksi
3.5
(b) For AISI 1035 steel, s ksi us = 64
for shock loading, traditional factor of safety, N =10 ~ 15
64
s
N = us = =
, the design is safe ( N »10 )
Design factor , 9.94
6.44
s
34. The same as 33, except that the shaft diameter is 2 ¾ in.
Solution:

d = 2.75 in
16
d
T
=
( )
( )
(a) s
3
p
16 54.2
3 = =
p
s 13.3 ksi
2.75
(b) For AISI 1035 steel, s ksi us = 64
for shock loading, traditional factor of safety, N =10 ~ 15
64
s
N = us = =
, the design is not safe ( N 10 )
Design factor , 4.8
Tc
s =
( ) [( ) ( ) ] 4
13.5
= = =
33,000
p p
3152 6.75
= =
98
s
N us , not risky
= = =
Page 44 of 131
13.3
s
35. 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?
Solution:
For Monel shaft,
s ksi us = 98 (Table AT 3)
N = 3 ~ 4 , for dead load, based on ultimate strength
(a)
J
4 4 4 4
3086
−
13.5 6.5
32
32
in
D D
J o i =
=
−
=
p p
in
D
c o 6.75
2
2
( )
( )
ft lb in kips
n
hp
33,000 10,000
T = = = 262,606 − = 3152 −
2 200
2
( )( )
s 6.9 ksi
3086
(b) Factor of safety,
14.2
6.9
s

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 D 1.2d 1 = . (c) Choose standard fractional dimensions
which you think would be satisfactory.
F
= =
1 d
Page 45 of 131
Problems 36 – 38.
Solution:
s = axial stress
s s = shear stress
(a)
2
2
4
4
F
d
s
p
p
Equation (1)
F
s
d
4
p
=

F
F
F
( ) ( 2 ) [ 2 ( )2 ] ( 2 2 )
1 D d
2 1.44
D = +
F
= =
ss p 1.2p 1
F
ss p
s
65
s u 16
= = =
s
49
s = us
= =
12
s 4 F
4 12
= = =
d 0.98
4 F
4 12
= + 2 = + 2 =
D 1.44 0.98 1.53
= = =
p p
F
= = =
p p
Page 46 of 131
1
2
2
1
4
1.2
4 4
4
F
D d
D D
D D
s
−
=
−
=
−
=
−
=
p p p
p
4
F
Equation (2) 1.44
d
2 p
s
dh
F
D h
Equation (3)
s ds
F
h
1.2p
=
Dt
=
Equation (4)
F
s Ds
t
p
=
(b) Designing based on ultimate strength,
Table AT 7, AISI C1020, as rolled
s ksi u = 65
s ksi us = 49
N = 3 ~ 4 say 4, design factor for static load
ksi
N
4
ksi
N
4
F =12,000 lb =12 kips
From Equation (1)
( )
( )
in
s
16
p p
From Equation (2)
( )
( )
d ( ) in
s
16
1.44
p p
From Equation (3)
12
( )( )
in
ds
F
h
s
0.27
1.2 0.98 12
1.2
From Equation (4)
12
( )( )
in
Ds
t
s
0.21
1.53 12

(c) Standard fractional dimensions
d =1in
1
=1
D 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
= = =
s
49
s = us
= =
3
s 4 F
4 4
= = =
d 1.13
4 F
4 4
= + 2 = + 2 =
D 1.44 1.13 1.76
F
= = =
p p
F
= = =
p p
Page 47 of 131
ksi
N
15
ksi
N
15
F = 4000 lb = 4 kips
From Equation (1)
( )
( )
in
s
4
p p
From Equation (2)
( )
( )
d ( ) in
s
4
1.44
p p
From Equation (3)
4
( )( )
in
ds
h
s
0.31
1.2 1.13 3
1.2
From Equation (4)
4
( )( )
in
Ds
t
s
0.24
1.76 3

(c) Standard fractional dimensions
1
=1
d in
8
3
=1
D 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 D 1.2d 1 = , us u s = 0.75s ,
and that bending in the plate is negligible.
Solution:
2
1
4
F
p
d
s
=
1
= p
F d s 2
4

s
1
p
F d 2 u
(1)
Page 48 of 131

=
N
4

1
F
F
( 2 2 ) ( 2 1.44
2 )
1
1
= p −
1
p
F D d 2 2 u 1.44
F
= =
ss p 1.2p 1

s
0.75
F dh us u 1.2p 1.2p
F dh u 5
F
ss p

s
0.75
F Dt us u p p
F Dt u 0.75p
1
1
p p
F D d 2 2 u 2 u

s
1
F dh u 2 u
0.9p p
= =

s
1
F Dt u 2 u
0.75p p
1
F d t u 2 u
0.75p 1.562 p
= =
Page 49 of 131
1
4
4
D d
D D
s
−
=
−
=
p p
F (D d )s 2 2 1.44
4

s
(2) ( )

= −
N
4
dh
F
D h
s F =1.2p dhs

=

=
N
s
dh
N

s
(3)

=
N
0.9p
Dt
=
s F =p Dts

=

=
N
s
Dt
N

s
(4)

=
N
Equate (2) and (1)

s
s
( )

=

= −
N
d
N
4
1.44
4
2 2 D = 2.44d
D =1.562d
Equate (3) and (1)

=

=
s
N
d
N
4
( )
d
d
h 0.278
4 0.9
Equate (4) and (1)

=

=
s
N
d
N
4

s
s
( )( )

=

=
N
d
N
4
( )( )
d
d
t 0.214
4 0.75 1.562

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.

1
5 D
F
b = + 2
F
48
= =
28
= =
Page 50 of 131
Problems 39, 40
Solution:
(a)

=
2
4
F
ss
p
Equation (1)
F
s s
D
4
5p
=
t(b D)
s
− 2
=
F
Equation (2) D
ts
Dt
s
5
=
Equation (3)
F
Ds
t
5
=
(b) For AISI C1020, as rolled
s s ksi ys y = 0.6 = 28
N = 4 for repeated and reversed load (mild shock) based on yield strength
s 12 ksi
4
s ksi s 7
4
From Equation (1)

F
4
5p
s s
D
=
where
F = 2500 lb = 2.5 kips
F
4
= = =
p p
F
5
= = say in
t 0.13
5

5
2.5

2 =
= + = say 2 in
b 1.96
D in
F

47
= =
25
= =
F
4
F
4
= = =
p p
Page 51 of 131
( )
( )
in
s
D
s
0.30
4 2.5
5 7
5
5
say in
16
From Equation (3)
( )
in
Ds
12
16
5
2.5
5
=

32
From Equation (2)
( )
ts
16
2
12
5
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 s ksi ys y = 0.55 = 25
N = 4 for repeated and reversed load (mild shock) based on yield strength
s 12 ksi
4
s ksi s 6
4
From Equation (1)
s s
D
5p
=
where
F = 2500 lb = 2.5 kips
( )
( )
in
s
D
s
0.33
4 2.5
5 6
5
3
say in
8
From Equation (3)

F
1
= = say in
t 0.11
3

3
2.5

2 =
1
2
= + = say in
b 2.42
D in
F

F
= or
sP −
Page 52 of 131
( )
in
Ds
12
8
5
2.5
5
=

8
From Equation (2)
( )
ts
8
2
12
1
8

+

2
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)
t(b D)
F
3
4
= Equation (1)
sP 2
t(b −
D)
(2)
1

F
4 2
4

=
D
ssR
p
Equation (2)

s
65
s = uP
= =
10.8
P s
61
s = uR
= =
10.1
R s
45
s = usR
= =
7.5
sR F
ssR p
F
= = =
p p
F
=
F
1
= = say in

F
sP −
7
= + D = in
say 2 in
F

3
4
sP 2
Page 53 of 131
F
= Equation (3)
Dt
sR 4
(b) For AISI C1015, as rolled
s ksi uR = 61 , s s ksi usR uR = 0.75 = 45
s ksi uP = 65
N = 6, based on ultimate strength
ksi
N
6
ksi
N
6
ksi
N
6
F = 8000 lb = 8 kips
Solving for D
2 2 D
=
8
( )
in
s
D
sR
0.412
2 7.5
2
7
say in
16
Solving for t
Dt
sR 4

( )
in
Ds
t
R
0.453
10.1
7
16
4
8
4
=

2
Solving for b
Using
t(b D)
=
( )
ts
b
P
1.92
16
10.8
1
2
8
+ =

Using
F
t(b −
D)
=

F
3
= + = say 2 in
7
1
=
Page 54 of 131
( )
( )
7

D in
ts
b
P
1.99
16
2
10.8
1
2

4
3 8
2
4
=

+

Therefore
b = 2 in
D in
16
=
t in
2
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
= = Equation (1)
2
4
F
1 D
2 1
1
4
F
D
s
p
p
F
= Equation (2)
(L D )e
s
2 −

Page 55 of 131
F
= Equation (3)
D e
s
2
F
( ) ( 2 )
1 a D
2
2
2
2
2
4
4
F
a D
s
−
=
−
=
p
p
Equation (4)
F
1 −
D D e
F
D D e
s
2
2
2
2
2
2
4
4
4
=
−
=
p
p
Equation (5)
Shear Stresses
F
= Equation (6)
eb
ss 2
F
ss − +
(L D e)t
=
2 2
Equation (7)

7
= = , sy y s = 0.6s
For steel rod, AISI C1035, as rolled
s ksi y 55
F
s
s y
4 10
55
1
1 1 =
Page 56 of 131
F
= Equation (8)
ss p
at
F
= Equation (9)
D m
ss
1 p
F
= Equation (10)
D h
ss
2 2
43. 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)
t in 0.875 in
8
1
=
s ksi sy 33
1
=
For plate and cotter, AISI C1020, as rolled
s ksi y 48
2
=
s ksi sy 28
2
=
N = 5 ~ 7 based on yield strength
say N = 7
From Equation (1) (Prob. 42)
2
1
4 1
D
N
p
= =
( )
2
1
7
p D
=
D 1.27 in 1 =
say D in
4

From Equation (9)
F
p
D m
s
s sy
s
= 1
=
N
1

m

=
10
1
1
4
33
7
p
m = 0.54 in
9
=
say m in
16
From Equation (3)
F
D e
s
s y
1 = =
N
2
10
D e
s
2
55
= =
7
1.273 2 D e =
From Equation (5)
4 1
D D e
s
s y
4 10
55
=
p D
3
1 2 =
3
1 =

3
=
F
s
s = sy
2
= s p
10
28
Page 57 of 131
F
N
2
2
−
4
2 = =
p
( )
4(1.273)
7
2
2 −
D 1.80 in 2 =
say D in
4
and 1.273 2 D e =
1.273
4

e
e = 0.73 in
say e in
4
By further adjustment
5
=
Say D 2 in 2 = , e in
8
From Equation (8)
at
N
(0.875)
7
p a
=
a = 0.91in
say a =1in

From Equation (4)
4 2
F
( 2 )
s
s y
48
1
1
s
s = sy
=
s − +
28

F
s
s = sy
2 =
s 2

5
28
F
s
s sy
s
10
28
5
5
=
7
=
Page 58 of 131
2
2
a D
N
−
= =
p
4 ( 10
)
( 2 2
2 ) 7
−
=
p a
a = 2.42 in
say a in
2
= 2
use a in
2
= 2
From Equation (7)
F
(L D e)t
N
2 2
2

(0.875)
5
8
2 2
10
7

− +
=
L
L = 2.80 in
say L = 3 in
From Equation (6)
eb
N
b

=
8
2
10
7
b = 2 in
From Equation (10)
D h
N
2 2
2 = =
2(2)h
7
=
h in in
8
= 0.625 =
Summary of Dimensions
L = 3 in
h in
8
b = 2 in
t in
8

9
=
m in
16
1
= 2
a in
2
1
1 1 =
D in
4
D 2 in 2 =
5
=
e in
8
(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
D 1.25 0.010 in 1 = ±
For Free Running Fits (RC 7) Table 3.1
Female Male
0.0030
0.0040
−
= 2.0
b in
+
b in
0.0000
−
0.0030
+
0.0040
−
= D in
0.0016
+
0.0020
−
= e in
Page 59 of 131
0.0058
2.0
−
=
allowance = 0.0040 in
D in
0.0000
2.0 2 −
0.0058
=
2.0 2 −
e in
0.0000
0.625
−
0.0030
0.625
−
=
44. A 1-in. ( 1 D ) 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
s ksi u 85
1
=
s ksi us 64
1
=

=
s ksi u 65
2
=
s ksi us 48
2
Ultimate strength
Use Equation (1)

1 1
2 2

F s D ( ) ( ) kips u u 1 66.8
1 1
3
=

1
2
2 4

1
66.8 85 2
3
1 2 =

3
1 2 =
9

1
= − 2
F s a D u u p

1
66.8 65 p a
3
Page 60 of 131
4
85
4
=

=

= p p
Equation (9)
F = s p
D m u us1 1
66.8 = (64)(p )(1)m
m = 0.33 in
say m in
8
From Equation (3)
F =
s D e u u1 2
( )D e 2 66.8 = 85
0.7859 2 D e =
From Equation (5)

F = s D − D e u u 2
1
p

( )

= − 0.7859
4
2 p D
D 1.42 in 2 =
say D in
8
0.7859
8

D e = e
e = 0.57 in
say e in
16
=
From Equation (4)

( )

2
2
4
2
( )

−

=
2
2
3
1
8
4
a =1.79 in
say a in
4
=1
From Equation (8)

=
F s at u us p
2
66.8 = (48)(p )(a)(1)
a = 0.44 in
1
=
say a in
2
3
=1
use a in
4
From Equation (2)
F = s (L −
D )e u u2 2

9
3
( )
1

3
1 48 2 8 . 66
1
1

9

3
3
=
1
Page 61 of 131

= −
16
8
66.8 65 L 1
L = 3.20 in
say L in
4
= 3
From Equation (7)
F s (L D e)t u us = − − 2 2
2
( ) 9
(1)
16
8

= L − −
L =1.51in
say L in
2
=1
use L in
4
= 3
From Equation (6)
F = 2
s eb u us1

( ) b

=
16
66.8 2 64
b = 0.93 in
say b =1in
From Equation (10)
F = 2
s D h u us1 2

( ) h

=
8
66.8 2 64 1
h = 0.38 in
say h in
8
Dimensions
L in
4
= 3

3
=
h in
8
b =1in
t =1in
3
=
m in
8
3
=1
a in
4
D 1in 1 =
3
1 2 =
D in
8
9
=
e in
16
(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
D 1.000 0.010 in 1 = ±
For Loose Running Fits (RC 8) Table 3.1
Female Male
0.0035
0.0045
−
= 1.0
b in
+
b in
0.0000
−
0.0040
+
0.0050
−
= D in
0.0028
+
0.0035
−
= e in
Page 62 of 131
0.0065
1.0
−
=
D in
0.0000
1.375 2 −
0.0075
=
1.375 2 −
e in
0.0000
0.5625
−
0.0051
0.5625
−
=
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 us u s = 0.75s .

F s p D
F s 2 c s p
s
s = u and

1
1
= 2 2
F s 2 c s D s p p
1

2 2
0.75s c s D

F s 2bc s p D
2b 0.8165D = p D
Page 63 of 131
Problems 45 – 47.
Solution:
(a) Neglecting bending

1
= 2

Equation (1): 4

1
= 2
Equation (2):

4
Equation (3): F = s(2bc)
Equation (4): F = s(ac)
Equation (5): F = s[2(d − c)b]
Equation (6): F s ( mb) s = 4
Equation (7): F s ( nb) s = 2
Equation (8): F = s(d − c)a
(b)
N
s
s us
s =
N
Therefore
s s s = 0.75
Equate (2) and (1)

=

4
4

=

1
4
2
c = 0.8165D
Equate (3) and (1)

1
( )
= = 2
4
1
( ) 2
4
b = 0.4810D

Equate (4) and (1)

1
= = 2
F sac s p D
1
a 0.8165D = p D
1
= − = 2
F s 2 d c b s p D
2 d − 0.8165D 0.4810 = p D
1
= = 2
F s 4mb s D s p
0.75 4m 0.4810D = p D
1
= = 2
F s 2nb s D s p
0.75 2n 0.4810D = p D
1
= − = 2
F s d c a s p D
1.6329 − D − 0.8165D a = p D
Page 64 of 131

4
( ) 2
4
a = 0.9619D
Equate (5) and (1)

( ) [ ]

4
1
( )( ) 2
4
d =1.6329D
Equate (6) and (1)

( )

4
1
( )( ) 2
4
m = 0.5443D
Equate (7) and (1)

( )

4
1
( )( ) 2
4
n =1.0886D
Equate (8) and (1)

( )

4
1
( ) 2
4
a = 0.9620D
Summary
a = 0.9620D
b = 0.4810D
c = 0.8165D
d =1.6329D
m = 0.5443D
n =1.0886D
46. The same as 45, except that the calculations are to be based on yield strengths. Let
sy y s = 0.6s .

Solution: (Refer to Prob. 45)
(a) Neglecting bending

1
= 2
Equation (1):
F s p D
F s 2 c s p
s
s y = and

1
1
= 2 2
F s 2 c s D s p p
1

2 2
0.6s c s D

1
= = 2
F s 2bc s p D
2b 0.9129D = p D

1
= = 2
F sac s p D
1
a 0.9129D = p D
1
= − = 2
F s 2 d c b s p D
Page 65 of 131

4

1
= 2
Equation (2):

4
Equation (3): F = s(2bc)
Equation (4): F = s(ac)
Equation (5): F = s[2(d − c)b]
Equation (6): F s ( mb) s = 4
Equation (7): F s ( nb) s = 2
Equation (8): F = s(d − c)a
(b)
N
s
s sy
s =
N
Therefore
s s s = 0.6
Equate (2) and (1)

=

4
4

=

1
4
2
c = 0.9129D
Equate (3) and (1)

( )

4
1
( ) 2
4
b = 0.4302D
Equate (4) and (1)

4
( ) 2
4
a = 0.8603D
Equate (5) and (1)

( ) [ ]

4

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:
s ksi u =153
s ksi y =134
s s ( ) ksi n u = 0.5 = 0.5 153 = 76.5
For axial loading, with size factor
s s ( )( )( ) ksi n u = 0.5 = 0.8 0.85 76.5 = 52
Soderberg line
a
n
s
m
= +
y
s
s
s
1
N
For double-acting

F F pA ( ) (20) 39,270 lb 39.27 kips
max = =
4 F
4 39.27 50
d d d
sa = = =

1 1 2
5
Page 1 of 62
125 2
4

= = =
p
F F 39.27 kips min = − = −
= 0 m s
( )
2 2 2
p p
50
52
0
1.75

= = +
d
N
d =1.2972 in
say d in
16
=1
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.

4 F
4 5 20
d d d
= = =
sa p p p
a
s
m
s
a
s
s
s n
a =
Size factor = 0.85
Factor for axial loading = 0.80
20 30.345
sa p
D = 0.542 in
9
20 35.7
sa p
Page 2 of 62
Prob. 142 – 144
Solution:
For AISI 2330, WQT 1000 F
s ksi u =105
s ksi y = 85
s s ( ) ksi n u = 0.5 = 0.5 105 = 52.5
= 0 m s
( )
2 2 2
Soderberg line
n
y
s
s
N
= +
1
n
s
N
= 0 +
1
N
(a) Machined surface
Surface factor = 0.85 (Fig. AF 5)
s s ( )( )( )( ) ksi ksi n u = 0.5 = 0.80 0.85 0.85 52.5 = 30.345
1.4
2 = =
D
say D in
16
=
(b) Mirror polished surface
s s ( )( )( )( ) ksi ksi n u = 0.5 = 0.80 0.85 1.00 52.5 = 35.7
1.4
2 = =
D

D = 0.5 in
2 2
1
9

−

Savings in weight = (100%) 21%
20 14.28
sa p
D = 0.79 in
3
=
20 24.82
sa p
D = 0.60 in
5
=
Page 3 of 62
9
16
2
16
2
=

(c) As forged surface
s s ( )( )( )( ) ksi ksi n u = 0.5 = 0.80 0.85 0.40 52.5 =14.28
1.4
2 = =
D
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.
s ksi n = 30 at 3 x 108
s ksi y = 69
s ksi u = 93.75
×
3 10
8

n s at 3 x 107 ( ) 36.5 ksi
3 10
30
0.085
7
=

×
=
s ( )( )( ) ksi n = 0.80 0.85 36.5 = 24.82
1.4
2 = =
D
say D in
8
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 ksi y = 47
s ksi u = 68
s ksi n = 20 at 5 x108

×
5 10

sn at 107 ( ) 27.9 ksi
20 19
sa p
D = 0.685 in
11
=
=
2000 18
a
s
m
s
Page 4 of 62
10
20
0.085
7
8
=

s ( )( )( ) ksi n = 0.80 0.85 27.9 =19
1.4
2 = =
D
say D in
16
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
s ksi n = 90
s ksi y =100
s ksi u =141
Size Factor = 0.85
Surface factor (ground) = 0.88
s ( )( )( ) ksi n = 0.85 0.88 90 = 67.32
= 0 m s
3
32
D
M
sa p
( )( )
in lb in kips
FL
M = = = 9000 − = 9.0 −
4
4
Soderberg line
n
y
s
s
N
= +
1

s
a
s
n
1
N
= 0 +
s
s n
a =
( )
N
32 9 67.32
1.6
3 =
pD
D =1.2964 in
1
=1
say D 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 .
Mc
2.8
2.8
= = =
12.6 1.4
sa = =
Page 5 of 62
Problems 146, 147
Solution:
s ksi u = 65
s ksi y = 48
s s ksi n u = 0.5 = 32.5
Surface factor (as forged) = 0.55
(a) = 0 m s
I
sa =
( ) 4
3 3
1.8293
12
12
t
tb t t
I = = =
t
b t
c 1.4
2
2
M = FL = (900)(14) =12,600 in − lb =12.6 in − kips
( )( )
9.643
t
4 3
1.8293
t t
s ( )( )( ) ksi n = 0.85 0.55 32.5 =15.20

Soderberg line
a
n
s
m
= +
y
s
s
s
1
N
s
a
s
n
1
N
= 0 +
s
s n
a =
N
9.643 15.20
2
3 =
t
t =1.08 in
b = 2.8t = 2.8(1.08) = 3.0 in
1
=1 , b = 3.0 in
say t in
16
(b) M = Fe = (900)(4) = 3,600 in −lb = 3.6 in − kips
( 3.6 )( 1.4
)
sa = =
2.755 15.20
23
= , b = 2 in
s
s n
a =
9.643 18.5
Page 6 of 62
2.755
t
4 3
18293
t t
2
3 =
t
t = 0.713 in
b = 2.8t = 2.8(0.713) =1.996 in
say t in
32
147. The same as 146, except that the reversal of the load are not expected to exceed
105 (Table AT 10).
Solution:
s ksi n = 32.5
10
6

n s at 105 ( ) 39.5 ksi
10
32.5
0.085
5
=

=
s ( )( )( ) ksi n = 0.85 0.55 39.5 =18.5
(a)
N
2
3 =
t

t =1.014 in
b = 2.8t = 2.8(1.014) = 2.839 in
13
say t =1in , b in
s
s n
a =
2.755 18.5
11
= , b in
as
s
ms
s
=
16 15 240
= =
sas p p
as
s
s
s ns
as =
Page 7 of 62
16
= 2
(b)
N
2
3 =
t
t = 0.6678 in
b = 2.8t = 2.8(0.6678) =1.870 in
say t in
16
7
8
=1
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
s ksi u =152
s ksi y =134
s s ksi n u = 0.5 = 76
For machined surface,
Surface factor = 0.78
Size factor = 0.85
s ( )( )( )( ) ksi ns = 0.6 0.85 0.78 134 = 53.3
s s ( ) ksi ys y = 0.6 = 0.6 134 = 80.4
ns
ys
s
s
N
= +
1
= 0 ms s
3
16
D
T
sas p
T =15 in − kips
( )
3 3
D D
ns
s
N
= 0 +
1
N

240 53.3
1.75
3 =
pD
D =1.3587 in
3
=1
say D in
8
149. The same as 148, except that the shaft is hollow with the outside diameter twice
the inside diameter.
Solution:
o i D = 2D
16 TD
16 15 2
o
( )
=
as D D D
s
s ns
as =
32 53.3
11
= , D in o 8
Page 8 of 62
( )( )
32
i
4 4 [( 2
)4 4 ] 3
i i i
o i
D
D D
s
p p p
−
=
−
=
N
1.75
3 =
i pD
D in i = 0.694
say D in i 16
3
=1
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
s ksi u = 85
s ksi y = 55
s s ksi n u = 0.5 = 42.5

1
= + =
F ( ) kips m 10 0 5
2
1
= − =
F ( ) kips a 10 0 5
2
F
s m
m = = =
5
F
s a
a = = =
(a) Soderberg line
a
s
m
s
10
1
= +
9
27
10
1
= +
1
=
3
Page 9 of 62
10
2 2 3
5
1.5
bh b b
10
2 2 3
1.5
bh b b
n
y
s
s
N
= +
1
For machined surface,
Factor = 0.88
Size factor = 0.85
s ( )( )( )( ) ksi n = 0.80 0.85 0.88 42.5 = 25.4
10
( ) 3 (25.4)
3 55
1.40
2 2 b b
b = 0.5182 in
sayb in
16
=
h b in
32
=1.5 =
(b) Mirror polished,
Factor = 1.00
Size factor = 0.85
s ( )( )( )( ) ksi n = 0.80 0.85 1.00 42.5 = 28.9
10
( ) 3 (28.9)
3 55
1.40
2 2 b b
b = 0.4963 in
sayb in
2
h b in
4
=1.5 =
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.)

Solution:
Table AT 10, in brine, AISI 1035,
s ksi n = 24.6
s ksi y = 58
s ( )( )( ) ksi n = 0.80 0.85 24.6 =16.73
10
= +
( ) 3 (16.73)
1
5
=
15
Page 10 of 62
10
3 55
1.40
2 2 b b
b = 0.60 in
sayb in
8
h b in
16
=1.5 =
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 , 3000 . 1 F = lb is a dead load. At
d =10 in , 2400 . 2 F = 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:
s ksi u = 72
s ksi y = 52
s s ksi n u = 0.5 = 36
For as forged surface
Figure AF 5, factor = 0.52
Size factor = 0.85
s ( )( )( ) ksi n = 0.85 0.52 36 =16
Loading:

= 0 A M
10(3000)+ 20(2400) = 30R2
R 2600 lb 2 =
= 0 V F
1 2 1 2 R + R = F + F
2600 3000 2400 1 R + = +
R 2800 lb 1 =
Shear Diagram
M ( )( ) in lb in kips C = 2800 10 = 28,000 − = 28 −
1
M ( )( ) in lb in kips D = 2600 10 = 26,000 − = 26 −
1
Then
Loading
= 0 A M
10(3000) 30 20(2400) 2 + R =
R 600 lb 2 =
= 0 V F
1 2 1 2 R + F = F + R
2400 3000 600 1 R + = +
R 1200 lb 1 =
Page 11 of 62

Shear Diagram
M ( )( ) in lb in kips C = 1200 10 =12,000 − =12 −
2
M ( )( ) in lb in kips D = 600 10 = 6,000 − = 6 −
2
Then using
M M in kips C = = 28 −
max 1
M M in kips C = =12 −
min 2
1
1
M (M M ) ( ) in kips m = + = 28 +12 = 20 −
1
M c
s m
m = ,
s =
a
a ( 3
) 4
= =
M
s m
m = , 3 1.5b
a
s
m
s
20

1 3 3
1.5
Page 12 of 62
2
2
max min
1
M (M M ) ( ) in kips a = − = 28 −12 = 8 −
2
2
max min
I
M c
I
3 3
2.25
12
12
b
bh b b
I = = =
b
h
c 1.5
2
3 1.5b
M
s a
a =
n
y
s
s
N
= +
1
8
1.5
16
52
1.5

+

=
b b
b = 0.96 in
sayb =1in
h = 3b = 3 in
153. The same as 152, except that the cycles of 2 F will not exceed 100,000 and all
surfaces are machined.
Solution:

10
6

n s at 105 cycles ( ) 43.8 ksi
20

1 3 3
1.5
7
=
5
as
s
ms
s
=
16 6283 32,000 32
= = =
sas 3 3 3
p
as
s
Page 13 of 62
10
36
0.085
5
=

=
s ksi u = 72
Machined surface, factor = 0.90
s ( )( )( ) ksi n = 0.85 0.90 43.8 = 33.5
8
1.5
33.5
52
1.5

+

=
b b
b = 0.8543 in
sayb in
8
h b in
8
= 3 = 2
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
s ksi u = 78
s ksi y = 66
s s ksi n u = 0.5 = 39
size factor = 0.85
s ( )( )( ) ksi ns = 0.6 0.85 39 = 20
s s ( ) ksi ys y = 0.6 = 0.6 66 = 40
ns
ys
s
s
N
= +
1
(a) Reversed torque
= 0 ms s
3
16
D
T
sas p
T = 6283 in − lb
( )
ksi
D
psi
D D
ns
s
N
= 0 +
1

32

1 3
20
0
1.5
3
1
1
16 3141 16,000 16
= = =
sms 3 3 3
p
( )
16 3141 16,000 16
= = =
sas 3 3 3
p
16

1 3 3
1
1
1
16 4712 24,000 24
= = =
sms 3 3 3
p
( )
16 1571 8,000 8
= = =
sas 3 3 3
p

24

1 3 3 5
Page 14 of 62

= +
D
D =1.34 in
say D in
8
=1
(b) 0 min T = , T = 6283 in − lb max
T ( ) in lb m = 6283 = 3141 −
2
T ( ) in lb a = 6283 = 3141 −
2
( )
ksi
D
psi
D D
ksi
D
psi
D D
16
20
40
1.5

+

=
D D
D =1.22 in
say D in
4
=1
(c) T = 3141in −lb min , T = 6283 in − lb max
T ( ) in lb m = 6283+ 3141 = 4712 −
2
T ( ) in lb a = 6283−3141 =1571 −
2
( )
ksi
D
psi
D D
ksi
D
psi
D D
8
20
40
1.5

+

=
D D
D =1.145 in
say D in
32
=1

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:
s ksi u = 96
s ksi y = 59
s s ( ) ksi n u = 0.5 = 0.5 96 = 48
size factor = 0.85
s ( )( ) ksi n = 0.85 48 = 40.8
a
n
s
m
= +
y
s
s
s
1
N
= 0 m s
M
2
6
bh
sa =
h = 3 in
b =1in
( 4000 )( 24
)
6 24
2 = =
16
16
Page 15 of 62
in lb in kips
FL
M = = = 24,000 − = 24 −
4
4
( )
( )( )
s ksi a 16
1 3
40.8
0
1
= +
N
N = 2.55
156. The same as 155, except that the material is normalized and tempered cast steel,
SAE 080.
Solution:
Table AT 6
s ksi n = 35 ¢
s ksi y = 40
s ( )( ) ksi n = 0.85 35 = 29.75
29.75
0
1
= +
N
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?

Solution:
For AISI 1045, as rolled
s ksi u = 96
s ksi y = 59
s s ( ) ksi n u = 0.5 = 0.5 96 = 48 ¢
s ( )( )( ) ksi ns = 0.6 0.85 48 = 24.48
s s ( )( ) ksi ys y = 0.6 = 0.6 59 = 35.4
as
ns
s
ms
= +
ys
s
s
s
1
N
= 0 ms s
1 as s
= +
24.48
0
2
s ksi as =12.24
12.24
16
T
3 = =
sas p
D
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:
s ksi u = 65
s ksi y = 48
s s ( ) ksi n u = 0.5 = 0.5 65 = 32.5
Size factor = 0.85
s ( )( )( )( ) ksi n = 0.80 0.85 0.55 32.5 =12.2
K s
f a
n
s
m
= +
y
s
s
1
N
Page 16 of 62

Fig. AF 8, b h 1
Assume = 3.5 t K
1
= = =
Figure AF 7, in in
a
F
=
=
b h d b b
K s
1
1
=1 , h = 2b = 3 in
10

4 log3.3 3
log 3
10 log3.3
K s
fl a
s
2.0 10
1
1
1
=1 , h b in
Page 17 of 62
d
r 0.125
8
2
a = 0.01in
0.926
0.01
0.125
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.926(3.5 −1)+1 = 3.3 f t K q K
= 0 m s
10
−
( −
) (2 0.25)
sa
(a)
f a
s
n
N
= 0 +
1
( 3.3 )( 10
)
(2 0.25)(12.2)
0
1.5
−
= +
b b
2 0.25 4.06 2 b − b =
0.125 2.03 0 2 b − b − =
b =1.489 in
say b in
2
(b) For 50,000 repetitions or 50,000 cycles

s ( ) ksi n 15.74
5 10
12.2
0.085
4
6
=

×
=
( ) ( )( )
2.0
5 10
10
log
=
×
= =
f
f
K
K
fl
n
K
n
N
=
1
( )( )
(2 0.25)(15.74)
1.5
−
=
b b
2 0.25 1.906 2 b − b =
0.125 0.953 0 2 b − b − =
b =1.04 in
say b in
16
8
= 2 = 2
(c) For indefinite life, ground and polished surface

s ( )( )( )( ) ksi n = 0.80 0.85 0.90 32.5 = 20
K s
f a
s
n
1
N
=
( 3.3 )( 10
)
(2 0.25)(20)
1
=
1.5
b b
−
b 2 − 0.125 b − 1.2375 =
0 b =1.18 in
3
3
=1 , h b in
say b in
16
3.315 10
max =
Page 18 of 62
8
= 2 = 2
Maximum stress =
K F f
−
b(h d )
b h 1, d h = 0.25 2.375 = 0.105
Figure AF 8
= 3.5 t K
= ( −1)+1 = 0.926(3.5 −1)+1 = 3.315 f t K q K
( )( )
( )
s 13.14 ksi
−
1.1875 2.375 0.25
=
159. 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
s ksi u =135
s ksi y =126
s s ( ) ksi n u = 0.5 = 0.5 135 = 67.5
For machined surface, Fig. AF 7, surface factor = 0.80
Size factor = 0.85
s ( )( )( )( ) ksi n = 0.80 0.85 0.80 67.5 = 36.72

K s
f a
n
s
m
= +
y
s
s
1
N
Fig. AF 8, b h 1
Assume = 2.5 t K
1
= = =
Figure AF 7, in in
a

2 2 −
K s
f a
s
2.44 20
1
7
=
10

5 log 2.4 3
log 3
= = =
10 log 2.44
K s
fl a
s
1.81 20
1
Page 19 of 62
d
r 0.0625
16
2
a = 0.0025 in
0.96
0.0025
0.0625
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.96(2.5 −1)+1 = 2.44 f t K q K
(a) Indefinite life, = 2.44 f K
s ksi n = 36.72
= 0 m s
( )
20
D D
4 5
D D
F
D Dd
Dd
D
F
sa 1
0.5
8
4
4
4
4
2
2
=

−
=
−
=
−
=
p
p
p p
n
N
= 0 +
1
( )( )
36.72( D 0.5D)
1.5
2 −
=
p
0.5 2.00 2 pD − D =
D = 0.88 in
say D in
8
(b) For a life of 105 repetitions or cycles
6

s ( ) ksi n 44.66
10
36.72
0.085
5
=

=
( ) ( )( )
1.81
10
10
log
f
f
K
K
fl
n
K
n
N
=
1
( )( )
44.66( D 0.5D)
1.5
2 −
=
p
0.5 1.216 2 pD − D =
D = 0.71in

3
=
say D in
4
(c)
=
p
D D
7
d
= , 0.14
4 2.54 5
max 2 =
7

=
p

K s
f a
s
m
b d
r = =
Page 20 of 62
K F
s f
0.5
4
max 2 −
D in
8
0.125
= =
0.875
D
Figure AF 8
= 2.6 t K
= ( −1)+1 = 0.96(2.6 −1)+1 = 2.54 f t K q K
( )( )
s 25.82 ksi
7
8

0.5
8

−

160. 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:
s ksi u = 65
s ksi y = 48
s s ( ) ksi n u = 0.5 = 0.5 65 = 32.5 ¢
For machined surface
Size factor = 0.85
s ( )( )( )( ) ksi n = 0.80 0.85 0.90 32.5 = 20
n
y
s
s
N
= +
1
(a) For flat plate with fillets
Figure AF 9
3 3

0.333
1
= =
3
r
d
2
2
b
= =
b
h
d
=1.65 t K
a = 0.01in
1.0
1
1
»
+
=
a
r
q
» =1.65 f t K K
F
s m
m =
bt
F
s a
a =
bt
b
2.5
t =
1
= + =
F ( ) lb m 3500 1750 2625
2
1
= − =
F ( ) lb a 3500 1750 875
2
2625
sm =

b b

875
b b
b
sa =

6562.5
1
= +
0.75
= = =
F
s m m m
m =
F
s a a a
a =
Page 21 of 62
6562.5
2
2.5
b

=
2187.5
2
2.5

=
( 1.65 )( 2187.5
)
2 2 20,000
48,000
1.8
b b
b = 0.7556 in
or b = 0.75 in
in
b
t 0.3
2.5
2.5
For flat plate with central hole
Fig. AF 8, b h 1, d h = b 2b =1 2
Assume » = 2.9 f t K K
F
F
( h d ) t
( b −
b ) t
bt
=
−
=
2
F
F
( h d ) t
( b −
b ) t
bt
=
−
=
2

2625
sm =

b b

875
b b
b
sa =

6562.5
1
= +
3
= =
15
15
= , t in
15
= =
0.01
15

F
6562.5 6562.5
s m
m 7467
F
2187.5 2187.5
s a
a 2489
Page 22 of 62
6562.5
2
2.5
b

=
2187.5
2
2.5

=
( 2.9 )( 2187.5
)
2 2 20,000
48,000
1.8
b b
b = 0.904 in
15
or b in in
16
= 0.9375 =
in
b
t
8
2.5
d b in
16
= =
use b in
16
3
= , d in
8
15
=
16
(b) m f a s = s + K s max
in
d
r
32
2
0.98
32
1
1
=

+
q =
= 2.9 t K
= ( −1)+1 = 0.98(2.9 −1)+1 = 2.86 f t K q K
psi
bt b
2 2 =
15
16

= = =
psi
bt b
2 2 =
15
16

= = =
s s K s ( )( ) psi m f a 7467 2.86 2489 14,586 max = + = + =
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 = 0.2D and N = 2 ; indefinite life.

F
F
6
= =
1
1
Page 23 of 62
Problems 162 – 164.
Solution:
s ksi u = 65
s ksi y = 48
s s ( ) ksi n u = 0.5 = 0.5 65 = 32.5 ¢
Size factor = 0.85
s ( )( )( ) ksi n = 0.85 0.90 32.5 = 24.86
= 0 A M
12F = 24B
F = 2B
2
B =
2
A = B =
At discontinuity
F
F
M 3
2
M = 3(3000) in − lb = 9000 in − lb = 9 in − kips max
M = 3(1000) in − lb = 3000 in −lb = 3 in − kips min
M ( ) in kips m = 9 + 3 = 6 −
2
M ( ) in kips a = 9 −3 = 3 −
2
3
32
D
M
s
p
=
Figure AF 12
D d =1.5d d =1.5
r d =0.2d d =0.2
=1.42 t K
assume » =1.42 f t K K

K s
f a
n
s
m
= +
y
s
s
1
N
( 32 )( 6
) ( )( )( )
= +
3 3 24.86
1
13
12
= =
1
1
K s
f a
s
m
32 12
1
= +
Page 24 of 62
1.42 32 3
48
2
pD pD
D =1.821in
say D in
16
=1
At maximum moment
F
F
M 6
2
M = 6(3000) in − lb =18000 in − lb =18 in − kips max
M = 6(1000) in − lb = 6000 in −lb = 6 in − kips min
M ( ) in kips m = 18 + 6 =12 −
2
M ( ) in kips a = 18 − 6 = 6 −
2
3
32
D
M
s
p
=
=1.00 f K
n
y
s
s
N
= +
1
( )( ) ( 1.0 )( 32 )( 6
)
3 3 24.86
48
2
pD pD
D =1.4368 in
13
Therefore use D in
16
=1
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 = 0.2D
and N = 2 . Compute D . What is the maximum torsional stress in the shaft?
Solution:
For C1040, OQT 1000 F
s ksi u =104
s ksi y = 72

s s ( ) ksi n u = 0.5 = 0.5 104 = 52 ¢
Size factor = 0.85
s ( )( )( )( ) ksi ns = 0.60 0.85 0.85 52 = 22.5
1
T T ( ) in lb in kips a m = = 10,000 = 5000 − = 5 −
2
s s ( ) ksi ys y = 0.6 = 0.6 72 = 43.2
= =
K s
fs as
s
ms
16 5
1
= +
9
16 5
=
p p
max 3 3 =
9
1

Page 25 of 62
3
16
D
T
s sms as p
ns
ys
s
s
N
= +
1
Figure AF 12
D d =1.5d d =1.5
r d =0.2d d =0.2
=1.2 ts K
assume » =1.2 fs ts K K
( )( ) ( 1.2 )( 16 )( 5
)
3 3 22.5
43.2
2
pD pD
D =1.5734 in
say D in
16
=1
m f a s = s + K s max
( )( ) ( 1.2 )( 16 )( 5
)
s 14.686 ksi
9
1
16
16

+

165. An axle (nonrotating) is to be machined from AISI 1144, OQT 1000 F, to the
proportions shown, with a fillet radius r » 0.25D; 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

Solution:
s ksi u =118
s ksi y = 83
s s ksi n u = 0.5 = 59 ¢
Size factor = 0.85
s ( )( )( ) ksi n = 0.85 0.83 59 = 41.62
(a) At the fillet
D d =1.5d d =1.5
r d =0.25d d =0.25
=1.35 t K
assume » =1.35 f t K K
M = 6F
M = 6(1200) in − lb = 7200 in −lb = 7.2 in − kips max
M = 6(400) in − lb = 2400 in − lb = 2.4 in − kips min
1
M ( ) in kips m = 7.2 + 2.4 = 4.8 −
2
1
M ( ) in kips a = 7.2 − 2.4 = 2.4 −
2
M
3
32
D
s
p
=
K s
f a
n
s
m
= +
y
s
s
1
N
( 32 )( 4.8
) ( )( )( )
= +
3 3 41.62
1
7
1
1
Page 26 of 62
1.35 32 2.4
83
2
pD pD
D =1.4034 in
say D in
16
=1
(b) At section BB,
M = 30F
M = 30(1200) in − lb = 36000 in −lb = 36 in − kips max
M = 30(400) in − lb =12000 in − lb =12 in − kips min
M ( ) in kips m = 7.2 + 2.4 = 4.8 −
2
M ( ) in kips a = 7.2 − 2.4 = 2.4 −
2

M
3
32
D
s
p
=
=1.0 f K
K s
f a
n
s
m
= +
y
s
s
1
N
( 32 )( 36
)
( )
1
= +
11
1
1
Page 27 of 62
( 1.0 )( 32 )( 12
)
( )3 3 41.62 1.5
83 1.5
2
p D p D
D =1.6335 in
say D in
16
=1
(c) Specified dimension:
D = 2 in , 1.5D = 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:
T =15 in − kips max
T = 5 in − kips min
T ( ) in kips m = 15 + 5 =10 −
2
T ( ) in kips a = 15 − 5 = 5 −
2
s ksi u =101
s ksi y = 58.5
s s ( ) ksi n u = 0.5 = 0.5 101 = 50.5

Size factor = 0.85
s ( )( )( )( ) ksi ns = 0.60 0.85 0.85 50.5 = 21.9
(a) At the fillet
r d =r D = =1 8
D d =1.5
=1.3 ts K
assume » =1.3 fs ts K K
At the key profile
=1.6 fs K
use =1.6 fs K
s s ( ) ksi ys y = 0.6 = 0.6 58.5 = 35.1
K s
fs as
ns
s
ms
= +
ys
s
s
1
N
( )( ) ( )( )( )
= +
3 3 21.9
16 10
1
3
Page 28 of 62
1.6 16 5
35.1
1.6
pD pD
D =1.7433 in
say D in
4
=1
(b) r = D 4
r D =0.25
D d =1.5
Figure AF 12
=1.18 ts K
» =1.18 1.6 fs ts K K
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 f K depends on the dimensions. Start by assuming a logical
f K for a logical h (Fig. AF 11), with a final check of f K . Considerable
estimation inevitable.

1

8 =
1
1
Mc
2 3h d t
1
= =

1
1
1
1
= Page 29 of 62
Problem 170
Solution:
A B F
2
= =
At the hole
F
( ) F
M eB 4
2

= =
M 4F max =
0 min M =
M ( F) F ( ) in kips m = 4 = 2 = 2 1.4 = 2.8 −
2
M ( F) F ( ) in kips a = 4 = 2 = 2 1.4 = 2.8 −
2
I
s =
( )
12
I
−
=
d in 0.5 in
2
c 1.75 in
2
2
2

= +
s ksi u = 65
s ksi y = 48
s s ( ) ksi n u = 0.5 = 0.5 65 = 32.5
Size factor = 0.85
s ( )( ) ksi n = 0.85 32.5 = 27.62
Fig. AF 7, c d =1.75 0.5 = 3.5 0.5

Assume = 3.5 t K
1
1

=

=

r 0.25 in
2
2
a = 0.010 in
a
K s
f a
s
m
12 2.8 1.75
1
7
=
1
1
1
1 + +
1
1
1
5
d
Page 30 of 62
0.962
0.010
0.25
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.962(3.5 −1)+1 = 3.4 f t K q K
n
y
s
s
N
= +
1
( )( )
( )
( 3.4 )( 12 )( 2.8 )( 1.75
)
h d t (h d ) t 3 3 27.62 2
48 2
1.5
−
+
−
=
( 2 ) 12.70 3 h − d t =
[ 2(0.50)] 12.70 3 h − t =
(4 1) 12.70 3 t − t =
t = 0.8627 in
say t in
8
h = 4t = 3.5 in
1
h in
2
2
2
h 3.5 in
Figure AF 11, h d 10
h =10d =10(0.50) = 5 in
0.5
2
2
2
=
−
=
b
Therefore = 3.5 t K , = 3.4 f K
1
Use h = 5 in , t in
4
=1
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 ,

quantitatively discuss actions that might be taken to reduce the size; holes must
remain as located.
K s
f a
s
m
1
= =
1
= =
0.25
= in =
b
Mc
2 3h d b
Page 31 of 62
Problems 171-174.
Solution:
(a) AISI 8630, WQT 1100 F
s ksi u = 96
s s ( ) ksi n u = 0.5 = 0.5 96 = 48
Size factor = 0.85
As-forged surface (Fig. AF I)
s ( )( )( ) ksi n = 0.85 0.42 48 =17
Milled surface (Machined)
s ( )( )( ) ksi n = 0.85 0.85 48 = 34.68
At AB, machined
n
u
s
s
N
= +
1
Figure AF 11
b in 0.5 in
2
d in 0.25 in
4
0.5
0.5
d
Assume = 3.50 f K
q = 0.998
= ( −1)+1 = 0.998(3.5 −1)+1 = 3.495 f t K q K
I
s =
( )
12
I
−
=

1
1

h
4 3

M h
1
4
1

2

12 3
M b
3
2
s 3 3 0.5
4.5 M 4 b
1
s 3 3 0.5

1

= 180 108 36

1

= 180 108 144
K s
f a
s
m
4.5 36 4 1
1
4 b
1
3 0.5 3
5
7

1

= 255 153 51
Page 32 of 62
1
( ) (4 3)
8
4 4 1
1
8
1
4
= − 1
+
1
2
1
4
2
2
2

− = + − =

= − + h h h
c
h = 3b
( )
12
8
3

h b
s

−
−

=
( )
( b −
) b
−
=
( )
( b −
) b
−
=
M = F(L − a)
M = (15)(17 − 5) =180 in − kips max
M = (− 9)(17 −5) =108 in − kips min
M ( ) in kips m − = −

2
M ( ) in kips a − = +

2
n
u
s
s
N
= +
1
( )( )
( )
( )( )( )( b
)
( b ) b
b
3 3 34.68 3 0.5
b b
3.495 4.5 144 4 1
96 3 0.5
1.6
−
−
+
−
−
=
( −
)
1
=
( 3 b −
0.5
) 3 b
107.2
( )
( )
107.2
4 1
=
−
−
b
b b
b = 2.6 in
say b in
8
= 2
h b in
8
= 3 = 7
Checking at the edges (as forged)
M = (15)(17) = 255 in − kips max
M = (− 9)(17) = −153 in − kips min
M ( ) in kips m − = −

2


1

= 255 153 204
M ( ) in kips a − = +

2
6 6
M
M
M
s = = =
2 3 3 3
K s
f a
s
m
2 51
1
= +
3
5
= 2 , safe.
Page 33 of 62
2
9
b
b
bh
»1.0 f K
n
u
s
s
N
= +
1
( )
( )
( 1.0 )( 2 )( 204
)
3 (17)
3 96
1.6
3 3 b b
b = 2.373 in
say b in
8
= 2
3
2
since b in in
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:
s ksi n =17
s ksi n = 34.68
s ksi n = 72
In section AB, machined

M = F(L − a)
M = (+ 5)(15 − 3) = 60 in − kips max
M = (− 3)(15 − 3) = −36 in − kips min

1

= 60 36 12
M ( ) in kips m − = −

2

1

= 60 36 48
M ( ) in kips a − = +

2
d = h = 4.5 in , b =1.5 in
h
= 3
b
( −
)
4.5 M 4 b
1
=
s 3 3 0.5
( b −
) b
( )[ ( ) ]
4.5 12 4 1.5 1
s ksi m 2.8125
[ 3 ( 1.5 ) 0.5 ] ( 1.5
)
−
4.5 48 4 1.5 1
3 =
s ksi a 11.25
−
K s
f a
s
m
1 2.8125
= +
1
=1 . Considering sections at A, B, and C, determine the maximum safe
Page 34 of 62
3 =
−
−
=
( )[ ( ) ]
[ 3 ( 1.5 ) 0.5 ] ( 1.5
)
=
n
y
s
s
N
= +
1
= 3.495 f K from Problem 171.
( 3.495 )( 11.25
)
34.68
72
N
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
= holes in the ends; a in
in tension by pins in the D in
8
9
5
= , t in
16
= ,
16
h in
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).

K s
f a
s
m
9
1
3
5
3
8
= =
h
1
1
9
16
= =
h
1
1
3
= =
Page 35 of 62
Problems 175 - 178
Solution:
s ksi u = 65
s ksi y = 48
s s ( ) ksi n u = 0.5 = 0.5 65 = 32.5
Size factor = 0.85
For machined all over
s ( )( )( )( ) ksi n = 0.85 0.90 0.80 32.5 = 20
n
y
s
s
N
= +
1
at A, Figure AF 8
b in
16
=
h in
8
=1
d D in
8
= =
t in
16
=
0.33
8
d
0.5
8
b
= 3.6 tA K
in
d
r
16
2
a = 0.01in

a
F
s =
1 64
F F
= m +
a 8
1 = + at A
9
1
3
5
3
16
= =
9
r
1
1
= 8
=
9
h
a
s =
9
F

Page 36 of 62
0.95
0.01
3
16
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.95(3.6 −1)+1 = 3.47 fA tA k q k
64
( ) 5
15
16
3
8
1
1
8
F F
h d t

−
=
−
=
( )
( )
15(20)
3.47 64
15 48
2
m a F 1.48F
45
At B Figure AF 9
d a in
16
= =
h in
8
=1
r in
16
=
t in
16
=
0.33
16
d
2
16
d
=1.63 tB K
a = 0.01in
0.95
0.01
3
16
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.95(1.63−1)+1 =1.6 fB tB k q k
256
45
5
16
16
F F
dt

= =

1 256
F F
= m +
a ( )
32
1 = + at B
1
=
9
1
8
= =
h
9
1
= =
a
F
s =
1 256
F F
= m +
a 32
1 = + at C
8
1 = +
32
1 = +
32
1 = +
Page 37 of 62
( )
45(20)
1.6 256
45 48
2
m a F 0.455F
135
b
at C, Figure AF 8, 1
h
D in
8
h a in
16
= =
0.22
16
d
= 3.5 tC K
in
d
r
16
2
a = 0.01in
0.862
0.01
1
16
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.862(3.5 −1)+1 = 3.2 fC tC k q k
256
( ) 5
35
16
1
8
9
16
F F
h d t

−
=
−
=
( )
( )
35(20)
3.2 256
35 48
2
m a F 1.17F
105
Equations
At A, m a F 1.48F
45
At B, m a F 0.455F
135
At C, m a F 1.17F
105

(a) Repeated and reversed load
= 0 m F
F F a =
use at A
1 = +
m a F 1.48F
8
45
( ) a 0 1.48F
8
1 = +
45
F = 0.676 kip
(b) F F F m a = =
8
1 = +
at A, F 1.48F
45
F = 0.603 kip
32
1 = +
at B, F 0.455F
135
F =1.480 kips
32
1 = +
at C, F 1.17F
105
F = 0.678 kip
use F = 0.603 kip
(c) F = −W min , F 3W max =
1
F ( W W) W m = 3 − =
2
1
= + =
F ( W W) W a 3 2
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
F 0.957 kip max =
Page 38 of 62

(d)
= = =
F a
K s
f a
s
m
K s
f a
s
s
s =
1
Page 39 of 62
( )
( )
0.892
0.603
0.676
F b
Ratio
179. A steel rod shown, AISI 2320, hot rolled, has been machined to the following
3
dimensions: D =1in. , .
1
c = in , .
4
e = in A semicircular groove at the
8
1
1
r = in ; for radial hole, .
midsection has .
8
a = in An axial load of 5 kips is
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)
s ksi u = 96
s ksi y = 51
s ksi n = 48
Size factor = 0.85
Surface factor = 0.85 (machined)
s ( )( )( )( ) ksi n = 0.80 0.85 0.85 48 = 27.74
n
y
s
s
N
= +
1
= 0 m s , reversed
s s a =
n
N
=
1
f
n
a NK
at the fillet, Figure AF 12
r e in
8
= =

3
= =
d c in
4
D =1in
0.17
1
= 8
=
3
4
r
d
1.3
1
= =
3
4
D
d
=1.55 t K
a = 0.010 in
a
4 5
3

n
s
= = =
s K

2 1 2 =
1
=
1
= 8
=
3
r
1
= =
3
D
Page 40 of 62
0.926
0.010
1
8
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.926(1.55 −1)+1 =1.51 f t K q K
( )
s s ksi a 11.32
4
2 =

= =
p
27.74
( )( )
1.62
11.32 1.51
a f
N
At the groove, Figure AF 14
3
1
d b D r in in in
4
8

= = − = −
D =1in
r in
8
0.17
4
d
1.3
4
d
=1.75 t K
a = 0.010 in

a
4 F
4 5
2 2 =
s s a 11.32
s
n
= = =
s K
1
1
= 4
=
D
a
n
s
= = =
s K
= =
Page 41 of 62
0.926
0.010
1
8
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.926(1.75 −1)+1 =1.7 f t K q K
( )
ksi
d
3
4

= = =
p
p
27.74
( )( )
1.44
11.32 1.7
a f
N
At the hole, Figure AF8
D = h =1in
d a in
4
= =
0.25
1
h
= 2.44 t K
a = 0.010 in
0.926
0.010
1
8
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.926(2.44 −1)+1 = 2.33 f t K q K
( ) ( )
ksi
Dd
D
F
s s a 9.34
1
4
1
1
4
5
4
2 2 =

−
=
−
= =
p p
27.74
( )( )
1.27
9.34 2.33
a f
N
Factor of safety is 1.27
From i4.4
n s = 0.76s
1.32 min 1.27
0.76
n
n
s
s
N
Therefore, dimensions are not suitable.
Steps to be taken:
1. Reduce number of cycle to failure

2. Good surface condition
3. Presetting
186. A stock stud that supports a roller follower on a needle bearing for a cam is
5
= , b in
made as shown, where a in
5
3
3
= 4
=
5
D
M

Page 42 of 62
8
7
3
= . The nature of the junction
= , c in
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
s ksi u =106
s ksi y = 71
s s ksi n u = 0.5 = 53
Size factor = 0.85
s ( )( ) ksi n = 0.85 53 = 45
Figure AF 12
d a in
8
= =
D c in
4
= =
r d » 0 , sharp corner
1.2
8
d
Assume = 2.7 t K
» = 2.7 f t K K
3
32
a
s
p
=
7

M Fb F 0.4375F
16
=

= =

5
= =
a in 0.625 in
8
( )
( )
32 0.4375
3 = =
K s
f a
s
m
1 18.25
F F
= +
K s
f a
s
m
1
= + =
1
= − =
Page 43 of 62
F
F
s 18.25
0.625
p
s s s F m a = = =18.25
n
y
s
s
N
= +
1
( 2.7 )( 18.25
)
45
71
2
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
a = in , .
dimensions .
8
1
b = in , c =1in. , .
8
1
d = in , L =12 in. , .
2
r = in The
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
s ksi u = 85
s ksi y = 55
s s ksi n u = 0.5 = 42.5
size factor = 0.85
s ( )( )( ) ksi n = 0.6 0.85 42.5 = 21.68
n
y
s
s
N
= +
1
F ( ) kips m 5 3 4
2
F ( ) kip a 5 3 1
2

(a) at A, Figure AF 9
1
=
r in
16
3
= =
d a in
8
7
= =
h b in
8
0.17
1
16
= =
3
8
r
d
2.33
7
= 8
=
3
8
h
d
=1.9 t K
a = 0.010 in
a
F
4
s ksi m 10.67
3

1
s ksi a 2.67
3

1 10.67
= +
F
s ksi m 8
3
7

Page 44 of 62
0.862
0.010
1
16
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.862(1.9 −1)+1 =1.78 f t K q K
ac
s =
( 1
)
8
=

=
( 1
)
8
=

=
( 1.78 )( 2.67
)
21.68
55
N
N = 2.42
At B, same as A, =1.78 f K
(b a)c
s
−
=
( 1
)
8
8
4
=

−
=

1
s ksi a 2
( 1
)
3
8
7
8

1 8
= +
1
=
1
= 2
=
h
1
= = =
a
F
3
7

−

3
7

−

1 16
= +
Page 45 of 62
=

−
=
( 1.78 )( 2
)
21.68
55
N
N = 3.23
At C, Figure AF 8
d in
2
h = c =1in
b h 1
0.5
1
d
= 2.2 t K
a = 0.010 in
in in
d
r 0.25
4
2
0.962
0.010
0.25
1
1
1
1
=
+
=
+
=
r
q
= ( −1)+1 = 0.962(2.2 −1)+1 = 2.15 f t K q K
(b a)(c d )
s
− −
=
s ksi m 16
1
2
1
8
8
4
=

−
=
s ksi m 4
1
2
1
8
8
1
=

−
=
( 2.15 )( 4
)
21.68
55
N
N =1.45
(b) Maximum stresses
at A
s s K s ( ) ksi A m f a = + =10.67 +1.78 2.67 =15.42
at B
s s K s ( ) ksi B m f a = + = 8 +1.78 2 =11.56

at C
s s K s ( ) ksi C m f a = + =16 + 2.15 4 = 24.6
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 3003-
H14.
Solution:
D =1in. , L = 5 ft
For wrought iron,
E psi 6 = 28×10
(a) elongation
F = 5000 lb
( )( )( )
( ) ( )
FL
5000 5 12

W
W
5000
5000
1

s 24,741 psi
sL
24,741 60
FL
5000 5 12
Page 46 of 62
in
AE
0.01364
1 28 10
4
2 6
=
×
= =
p
d
Stress and elongation
h = 0.05 in
W = 5000 lb
L = 5 ft = 60 in
1
2
2

hEA
1

= + +
LW
A
A
s
( ) ( )
( )( ) ( )
( 60 )( 5000
)
1
4
2 0.05 28 10
1
1
4
1
4
2
6 2
2 2
=

×
= + +
p
p p
( )( )
in
E
0.053
28 10
6 =
×
d = =
(b) Aluminum alloy 3003-H14
E psi 6 =10×10
F = 5000 lb
( )( )( )
( ) ( )
in
AE
0.038
1 10 10
4
2 6
=
×
= =
p
d
Stress and elongation
h = 0.05 in

W = 5000 lb
L = 5 ft = 60 in

W
W
5000
5000
1

s 18,475 psi
sL
18,475 60

W
W

20 , 000 = + +
D
Page 47 of 62
1
2
2

hEA
1

= + +
LW
A
A
s
( ) ( )
( )( ) ( )
( 60 )( 5000
)
1
4
2 0.05 10 10
1
1
4
1
4
2
6 2
2 2
=

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