SECTION 1– DESIGN FOR SIMPLE STRESSES

SECTION 1– DESIGN FOR SIMPLE STRESSES
2
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 bh 5.1= . 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)
ksisu 96=
ksisy 59=
psiE 6
1030×=
A
F
sd =
where
lbF 8000=
bhA =
but
bh 5.1=
therefore 2
5.1 bA =
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
A
F
N
s
s u
d ==
2
5.1
8000
6
000,96
b
=
inb 577.0= say in
8
5
.

3
inbh
16
15
5.1 ==
(b) Based on yield strength
A
F
N
s
s u
d ==
2
5.1
8000
3
000,59
b
=
inb 521.0= say in
16
9
.
inbh
32
27
5.1 ==
(c) Elongation =
AE
FL
=δ
where,
in005.0=δ
lbF 8000=
psiE 6
1030×=
inL 15=
2
5.1 bA =
then,
AE
FL
=δ
( )( )
( )( )62
10305.1
158000
005.0
×
=
b
inb 730.0= say in
4
3
.
inbh
8
1
15.1 ==
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)
ksisu 55=
ksisy 5.36=
psiE 6
1025×=

4
A
F
sd =
where
lbF 8000=
bhA =
but
bh 5.1=
therefore 2
5.1 bA =
A
F
N
s
s u
d ==
2
5.1
8000
6
000,55
b
=
inb 763.0= say in
8
7
.
inbh
16
5
15.1 ==
A
F
N
s
s u
d ==
2
5.1
8000
3
500,36
b
=
inb 622.0= say in
16
11
.
inbh
32
1
15.1 ==
(c) Elongation =
AE
FL
=δ
where,
in005.0=δ
lbF 8000=
psiE 6
1025×=
inL 15=
2
5.1 bA =
then,

5
AE
FL
=δ
( )( )
( )( )62
10255.1
158000
005.0
×
=
b
inb 8.0= say in
8
7
.
inbh
16
5
15.1 ==
3. The same as 1 except that the material is gray iron, ASTM 30.
Solution:
For ASTM 30 (Table AT 6)
ksisu 30= , no ys
psiE 6
105.14 ×=
Note: since there is no ys for brittle materials. Solve only for (a) and (c)
A
F
sd =
where
lbF 8000=
bhA =
but
bh 5.1=
therefore 2
5.1 bA =
N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1)
A
F
N
s
s u
d ==
2
5.1
8000
5.7
000,30
b
=
inb 1547.1= say in
16
3
1 .
inbh
32
25
15.1 ==
(c) Elongation =
AE
FL
=δ
where,
in005.0=δ
lbF 8000=
psiE 6
105.14 ×=

6
inL 15=
2
5.1 bA =
then,
AE
FL
=δ
( )( )
( )( )62
105.145.1
158000
005.0
×
=
b
inb 050.1= say in
16
1
1 .
inbh
32
19
15.1 ==
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
ksisu 5.152=
ksisy 5.132=
( ) ( ) kipslbforceF 27.39270,3912520
4
2
====
π
From Table 1.1, page 20
8=uN
4=yN
u
u
s
FN
A =
( )( )
5.152
27.398
4
2
=d
π
ind 62.1= say in
8
5
1
y
y
s
FN
A =
( )( )
5.132
27.394
4
2
=d
π

7
ind 23.1= say in
4
1
1
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 io DD 2= .
Solution:
ksisu 65=
8=uN
kipsF 1500=
( ) ( ) 4
3
4
44
2
2222 i
iiio
D
DDDDA
πππ
=−=−=
( )( )
65
15008
4
3 2
===
u
ui
s
FND
A
π
inDi 85.8= say in
8
7
8
inDD io
4
3
17
8
7
822 =





==
6. A short compression member with io DD 2= 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
ksisu 127=
ksisy 114=
From Table 1.1 page 20, for dead load
4~3=uN , say 4
2~5.1=yN , say 2
Area, ( ) ( ) 4
3
4
44
2
2222 i
iiio
D
DDDDA
πππ
=−=−=
kipstonsF 5025 ==
(a) Based on yield strength
( )( )
114
502
4
3 2
===
y
yi
s
FND
A
π

8
inDi 61.0= say in
8
5
inDD io
4
1
1
8
5
22 =





==
(b) Based on ultimate strength
( )( )
127
504
4
3 2
===
u
ui
s
FND
A
π
inDi 82.0= say in
8
7
inDD io
4
3
1
8
7
22 =





==
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
=δ or
E
FL
A
δ
=
where,
lbF 7000=
inL 55=
in030.0=δ
psiE 6
1030×=
( )( )
( )( )6
2
1030030.0
557000
4 ×
== dA
π
ind 74.0= say in
4
3
(b) For gradually applied and repeated (not reversed) load
3=yN
( )( )
( )
psi
A
FN
s
y
y 534,47
75.0
4
70003
2
===
π
ksisy 48≈
say C1015 normalized condition ( ksisy 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

9
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
ksisus 48=
r
g
W
F 2
ω=
where
lbW 332.0=
2
2.32 fpsg =
( ) sec1047
60
000,102
60
2
rad
n
===
ππ
ω
inr 12=
( ) ( ) kipslbr
g
W
F 3.11300,1111047
2.32
332.0 22
==== ω
From Table 1.1, page 20
4~3=N , say 4
u
u
s
FN
A =
( )( )
48
3.114
4
2 2
=





d
π
for double shear
ind 774.0= say in
32
25
CHECK PROBLEMS
9. The link shown is made of AISIC1020 annealed steel, with inb
4
3
= and
inh
2
1
1= . (a) What force will cause breakage? (b) For a design factor of 4 based
on the ultimate strength, what is the maximum allowable load? (c) If 5.2=N
based on the yield strength, what is the allowable load?
Problem 9.

10
Solution:
For AISI C1020 annealed steel, from Table AT 7
ksisu 57=
ksisy 42=
(a) AsF u=
2
125.1
2
1
1
4
3
inbhA =











==
( )( ) kipsF 64125.157 ==
(b)
u
u
N
As
F =
4=uN
2
125.1
2
1
1
4
3
inbhA =











==
( )( ) kipsF 16
4
125.157
==
(c)
y
y
N
As
F =
5.2=yN
2
125.1
2
1
1
4
3
inbhA =











==
( )( ) kipsF 9.18
2
125.142
==
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
sL
=δ
from Table AT 7 for cold-finished B1113
ksisy 72=
then, ( ) ksiss y 6.57728.080.0 ===
ksipsiE 000,301030 6
=×=
( )( ) in
E
sL
0096.0
000,30
56.57
===δ

11
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, psiksisus 000,6262 ==
( ) 2
2
2209.0
8
3
4
1
2 inA =











= π
Asr
g
W
F us== 2
ω
( ) ( )( )2209.0000,6214
2.32
4 2
=ω
sec74.88 rad=ω
74.88
60
2
==
nπ
ω
rpmn 847=
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
ksisu 80=
( ) ksiksiss uus 608075.075.0 ===
2
4418.0
16
3
4
3
intdA =











== ππ
kipstonsF 12060 ==
n = number of holes
( )( )
5
604415.0
120
===
usAs
F
n holes
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:
WpDL =
where
psip 200=
inD 4=

12
lbW 6400=
( )( ) 64004200 =L
inL 8=
BENDING STRESSES
DESIGN PROBLEMS
14. A lever keyed to a shaft is inL 15= long and has a rectangular cross section of
th 3= . 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 ina 13= ? (b) at inb 4= ? (c) What
should be the size where the load is applied?
Problem 14.
Solution:
For AISI C1020, as rolled, Table AT 7
ksisu 65=
ksisy 49=
Design factors for gradually applied and reversed load
8=uN
4=yN
12
3
th
I = , moment of inertial
but th 3=
36
4
h
I =
Moment Diagram (Load Upward)

13
Based on ultimate strength
u
u
N
s
s =
(a)
I
Fac
I
Mc
s ==
2
h
c =
kipslbsF 22000 ==
( )( )












==
36
2
132
8
65
4
h
h
s
inh 86.3=
in
h
t 29.1
3
86.3
3
===
say
ininh
2
1
45.4 ==
inint
2
1
15.1 ==
(b)
I
Fbc
I
Mc
s ==
2
h
c =
kipslbsF 22000 ==
( )( )












==
36
2
42
8
65
4
h
h
s
inh 61.2=
in
h
t 87.0
3
61.2
3
===
say
inh 3=
int 1=
(c)

14
413
35.4
4
3
−
−
=
− h
inh 33.2=
413
15.1
4
1
−
−
=
− t
int 78.0=
say
inh 625.2= or inh
8
5
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 bh 3≈ ). (a) Determine the
dimensions for 3=N 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)
ksisy 40=
psiE 6
1030×=

15
From Table AT 2
Max.
( )( ) inkips
FL
M −=== 54
4
544
4
12
3
bh
I =
but bh 3=
36
4
h
I =
(a)
I
Mc
N
s
s
y
y
==
2
h
c =
( )












=
36
2
54
3
40
4
h
h
inh 18.4=
in
h
b 39.1
3
18.4
3
===
say inh
2
1
4= , inin
h
b
2
1
15.1
3
5.4
3
====
(b)
( )( )
( ) ( )( )
in
EI
FL
0384.0
12
5.45.1
103048
544000
48 3
6
33
=






×
==δ
(c)






=
36
48
4
3
h
E
FL
δ
( )( ) ( )
( )( )46
3
103048
36544000
03.0
h×
=
inh 79.4=
in
h
b 60.1
3
79.4
3
===
say ininh
4
1
525.5 == , inin
h
b
4
3
175.1
3
25.5
3
====

16
16. The same as 15, except that the beam is to have a circular cross section.
Solution:
(a)
I
Mc
N
s
s
y
y
==
64
4
d
I
π
=
2
d
c =
34
32
64
2
d
M
d
d
M
s
ππ
=












=
( )
3
5432
3
40
dπ
=
ind 46.3=
say ind
2
1
3=
(b)
EI
FL
48
3
=δ
64
4
d
I
π
=
( )
( )( )
( )( )( )
in
dE
FL
0594.0
5.3103048
54400064
48
64
46
3
4
3
=
×
==
ππ
δ
(c)
( )4
3
48
64
dE
FL
π
δ =
( )( )
( )( ) 46
3
103048
54400064
03.0
dπ×
=
ind 15.4=
say ind
4
1
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 bh 2= . (b)
Determine the dimensions based on yield strength. (c) Determine the dimensions
using the principle of “limit design.”

17
Solution:
From Table AT 7 and Table 1.1
ksisu 65=
ksisy 48=
4~3=uN , say 4
2~5.1=yN , say 2
( )( ) kipsin
FL
M −=== 72
4
486
4
I
Mc
s =
2
h
c =
12
3
bh
I =
but
2
h
b =
24
4
h
I =
34
12
24
2
h
M
h
h
M
s =






=
3
12
h
M
N
s
s
u
u
==
( )
3
7212
4
65
h
=
inh 76.3=

18
in
h
b 88.1
2
76.3
2
===
say ininh
4
3
375.3 == , inin
h
b
8
7
1875.1
2
75.3
2
====
3
12
h
M
N
s
s
y
y
==
( )
3
7212
2
48
h
=
inh 30.3=
in
h
b 65.1
2
30.3
2
===
say ininh
2
1
35.3 == , inin
h
b
4
3
175.1
2
5.3
2
====
(c) Limit design (Eq. 1.6)
4
2
bh
sM y=
( )
4
2
4872
2
h
h






=
inh 29.2=
in
h
b 145.1
2
29.2
2
===
say ininh
2
1
25.2 == , inin
h
b
4
1
125.1
2
5.2
2
====
18. The bar shown is subjected to two vertical loads, 1F and 2F , of 3000 lb. each, that
are inL 10= apart and 3 in. ( a ,d ) from the ends of the bar. The design factor is 4
based on the ultimate strength; bh 3= . 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.

19
Problems18, 19.
Solution:
lbRRFF 30002121 ====
Moment Diagram
( )( ) inkipsinlbsaRM −=−=== 99000330001
N = factor of safety = 4 based on us
12
3
bh
I =
2
h
c =
3612
3 4
3
h
h
h
I =






=
(a) For gray cast iron, SAE 111
ksisu 30= , Table AT 6
34
18
36
2
h
M
h
h
M
I
Mc
N
s
s u
=












===
( )
3
918
4
30
h
s ==
inh 78.2=
in
h
b 93.0
3
78.2
3
===
say inh 5.3= , inb 1=
(b) For malleable cast iron, ASTM A47-52, grade 35 018

20
34
18
36
2
h
M
h
h
M
I
Mc
N
s
s u
=












===
( )
3
918
4
55
h
s ==
inh 28.2=
in
h
b 76.0
3
28.2
3
===
say inh
4
1
2= , inb
4
3
=
(c) For AISI C1040, as rolled
ksisu 90= , Fig. AF 1
34
18
36
2
h
M
h
h
M
I
Mc
N
s
s u
=












===
( )
3
918
4
90
h
s ==
inh 93.1=
in
h
b 64.0
3
93.1
3
===
say inh
8
7
1= , inb
8
5
=
19. The same as 18, except that 1F acts up ( 2F acts down).
Solution:
[ ]∑ = 0AM lbRR 187521 ==

21
Shear Diagram
Moment Diagram
=M maximum moment = 5625 lb-in = 5.625 kips-in
(a) For gray cast iron
3
18
h
M
N
s
s u
==
( )
3
625.518
4
30
h
=
inh 38.2=
in
h
b 79.0
3
38.2
3
===
say inh
4
1
2= , inb
4
3
=
(b) For malleable cast iron
3
18
h
M
N
s
s u
==
( )
3
625.518
4
55
h
=
inh 95.1=
in
h
b 65.0
3
95.1
3
===
say inh
8
7
1= , inb
8
5
=

22
3
18
h
M
N
s
s u
==
( )
3
625.518
4
90
h
=
inh 65.1=
in
h
b 55.0
3
65.1
3
===
say inh
2
1
1= , inb
2
1
=
20. The bar shown, supported at A and B , is subjected to a static load F of 2500 lb.
at 0=θ . Let ind 3= , inL 10= and bh 3= . 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:

23
[ ]∑ = 0AM ( )2500133 =BR
lbRB 833,10=
[ ]∑ = 0BM ( )2500103 =AR
lbRA 8333=
Shear Diagram
Moment Diagram
( )( ) inkipsinlbM −=−== 25000,25102500
bh 3=
12
3
bh
I =
36
4
h
I =
2
h
c =
34
18
36
2
h
M
h
h
M
I
Mc
s =












==
6~5=N , say 6 for cast iron, dead load
3
18
h
M
N
s
s u
==
( )
3
2518
6
20
h
=

24
inh 13.5=
in
h
b 71.1
3
==
say inh
4
1
5= , inb
4
3
1=
(b) For malleable cast iron, ASTM A47-32 grade 32510
ksisu 52= , ksisy 34=
4~3=N , say 4 for ductile, dead load
3
18
h
M
N
s
s u
==
( )
3
2518
4
52
h
=
inh 26.3=
in
h
b 09.1
3
==
say inh
4
3
3= , inb
4
1
1=
4=N , based on ultimate strength
3
18
h
M
N
s
s u
==
( )
3
2518
4
85
h
=
inh 77.2=
in
h
b 92.0
3
==
say inh 3= , inb 1=
(d) For AISI C1035, as rolled
ksissu 64=
4=N , kipsRB 833.10=
A
R
N
s
s Bsu
s ==
22
24
2 DDA
ππ
=





=
2
2
833.10
4
64
D
ss
π
==
inD 657.0=

25
say inD
16
11
=
(e) Limit Design
4
2
bh
sM y=
For AISI C1035 steel, ksisy 55=
3
h
b =
( )
4
3
5525
2
h
h
M






==
inh 76.1=
in
h
b 59.0
3
==
say ininh
8
7
1875.1 == , inb
8
5
=
21. The same as 20, except that o
30=θ . Pin B takes all the horizontal thrust.
Solution:
θcosFFV =
[ ]∑ = 0AM VB FR 133 =
( ) 30cos2500133 =BR
lbRB 9382=
[ ]∑ = 0BM VA FR 103 =
( ) 30cos2500103 =AR
lbRA 7217=
Shear Diagram

26
Moment Diagram
( )( ) inkipsinlbM −=−== 65.21650,21102165
3
18
h
M
s =
6~5=N , say 6 for cast iron, dead load
3
18
h
M
N
s
s u
==
( )
3
65.2118
6
20
h
=
inh 89.4=
in
h
b 63.1
3
==
say inh
4
1
5= , inb
4
3
1=
(b) For malleable cast iron, ASTM A47-32 grade 32510
4~3=N , say 4 for ductile, dead load
3
18
h
M
N
s
s u
==
( )
3
65.2118
4
52
h
=
inh 11.3=
in
h
b 04.1
3
==
say inh 3= , inb 1=

27
3
18
h
M
N
s
s u
==
( )
3
65.2118
4
85
h
=
inh 64.2=
in
h
b 88.0
3
==
say inh
8
5
2= , inb
8
7
=
(d) For AISI C1035, as rolled
ksissu 64=
4=N , lbRBV 9382=
lbFFR HBH 125030sin2500sin ==== θ
( ) ( )22222
12509382 +=+= BHBVB RRR
lbRB 9465=
A
R
N
s
s Bsu
s ==
22
24
2 DDA
ππ
=





=
2
2
465.9
4
64
D
ss
π
==
inD 614.0=
say inD
8
5
=
(e) Limit Design
4
2
bh
sM y=
For AISI C1035 steel, ksisy 55=
3
h
b =
( )
4
3
5565.21
2
h
h
M






==
inh 68.1=
in
h
b 56.0
3
==
say ininh
8
7
1875.1 == , inb
8
5
=

28
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 ine 10= from the free end,
and one of 1000 lb at the free end. (a) Determine the dimensions of the cross
section if acb 3≈= . (b) The same as (a) except that the top of the tee is below.
Problem 22.
Solution:
For cast iron, ASTM 50
ksisu 50= , ksisuc 164=
For gradually applied, repeated load
8~7=N , say 8
( )edFdFM ++= 21
where:
lbF 20001 =
lbF 10002 =
ind 201030 =−=
ined 30=+
( )( ) ( )( ) inkipsinlbM −=−=+= 70000,70301000202000
I
Mc
s =
Solving for I , moment of inertia
( )( ) ( )( ) ( )( ) ( )( )[ ]yaaaa
a
aa
a
aa 33
2
5
3
2
3 +=





+





2
3a
y =

29
( )( ) ( )( )( ) ( )( ) ( )( )( ) 2
17
3
12
3
3
12
3 4
2
3
2
3
a
aaa
aa
aaa
aa
I =+++=
(a)
2
3a
ct =
2
5a
cc =
Based on tension
I
Mc
N
s
s tu
t ==
( )












=
2
17
2
3
70
8
50
4
a
a
ina 255.1=
Based on compression
I
Mc
N
s
s cuc
c ==
( )












=
2
17
2
5
70
8
164
4
a
a
ina 001.1=
Therefore ina 255.1=
Or say ina
4
1
1=
And ( ) inacb 75.325.133 ====

30
Or incb
4
3
3==
(b) If the top of the tee is below
2
5a
ct =
2
3a
cc =
2
17 4
a
I =
inkipsM −= 70
Based on tension
I
Mc
N
s
s tu
t ==
( )












=
2
17
2
5
70
8
50
4
a
a
ina 488.1=
Based on compression
I
Mc
N
s
s cuc
c ==
( )












=
2
17
2
3
70
8
164
4
a
a
ina 845.0=
Therefore ina 488.1=
Or say ina
2
1
1=
And inacb
2
1
43 ===
CHECK PROBLEMS

31
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; 1F and 12 2FF = ; 1F is 5 in. from one end and
2F is 5 in. from the other ends. The beam is 25 in. long; flange width is
inb 509.2= ; 4
9.2 inIx = . 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:
[ ]∑ = 0AM ( ) BRFF 252205 11 =+
18.1 FRB =
[ ]∑ = 0VF BA RRFF +=+ 11 2
111 2.18.13 FFFRA =−=
Shear Diagram
Moment Diagram

32
19FM = = maximum moment
For AISI C1020, as rolled
ksisy 48=
(a)
I
Mc
sy =
where in
d
c 5.1
2
3
2
===
( )( )
9.2
5.19
48 1F
sy ==
kipsF 31.101 =
kipsFF 62.202 12 ==
(b)
I
Mc
N
s
s y
==
( )( )
9.2
5.19
3
48 1F
s ==
kipsF 44.31 =
kipsFF 88.62 12 ==
(c) 1596.9
509.2
25
<==
b
L
(page 34)
ksisc 20= ( page 34, i1.24)
I
Mc
sc =
( )( )
9.2
5.19
20 1F
=
kipsF 30.41 =
kipsFF 60.82 12 ==
(d) For maximum deflection,
by method of superposition, Table AT 2
( ) 2
3
max
33 


 ′+′
=
bLa
EIL
bF
y , ba ′>
or
( ) 2
3
max
33 


 +′
=
aLb
EIL
Fa
y , ab >′

33
maxy caused by 1F
( ) 2
3
1111
max
331 


 +′
=
aLb
EIL
aF
y , 11 ab >′
where ksiE 000,30=
ina 51 =
inb 201 =′
inL 25=
4
9.2 inI =
( )
( )( )( )
( )
1
2
3
1
max 0022.0
3
52520
259.2000,303
5
1
F
F
y =


 +
=
maxy caused by 2F
( ) 2
3
2222
max
332 


 ′+′
=
bLa
EIL
bF
y , 22 ba ′>
where inb 52 =′
ina 202 =
( )
( )( )( )
( )
1
2
3
1
max 0043.0
3
52520
259.2000,303
52
2
F
F
y =


 +
=
Total deflection = δ
111maxmax 0065.00043.0022.021
FFFyy =+=+=δ
Deflection caused by the safe loads in (a)
( ) ina 067.031.100065.0 ==δ
Deflection caused by the safe loads in (b)
( ) inb 022.044.30065.0 ==δ
Deflection caused by the safe loads in (c)
( ) inc 028.030.40065.0 ==δ
24. The same as 23, except that the material is aluminum alloy, 2024-T4, heat treated.
Solution:
For aluminum alloy, 2024-T4, heat treated
ksisy 47=
(a)
I
Mc
sy =

34
( )( )
9.2
5.19
47 1F
sy ==
kipsF 10.101 =
kipsFF 20.202 12 ==
(b)
I
Mc
N
s
s y
==
( )( )
9.2
5.19
3
47 1F
s ==
kipsF 36.31 =
kipsFF 72.62 12 ==
(c) 1596.9
509.2
25
<==
b
L
(page 34)
ksisc 20= ( page 34, i1.24)
I
Mc
sc =
( )( )
9.2
5.19
20 1F
=
kipsF 30.41 =
kipsFF 60.82 12 ==
(d) Total deflection = δ
111maxmax 0065.00043.0022.021
FFFyy =+=+=δ
Deflection caused by the safe loads in (a)
( ) ina 066.010.100065.0 ==δ
Deflection caused by the safe loads in (b)
( ) inb 022.036.30065.0 ==δ
Deflection caused by the safe loads in (c)
( ) inc 028.030.40065.0 ==δ
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 ind 4= , a flange width of inb 66.2= ,
and 4
0.6 inIx = (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.

35
Solution:
(a) For C1020, as rolled, ksisu 65=
Consider flange buckling
30
66.2
80
==
b
L
since 4015 <<
b
L
( )
ksi
b
L
sc 15
1800
30
1
5.22
18001
5.22
22
=
+
=






+
=
I
Mc
s =
in
d
c 2
2
4
2
===
From Table AT 2
( ) F
FFL
M 20
4
80
4
===
I
Mc
ss c ==
( )( )
6
220
15
F
=
kipsF 25.2= , safe load
(b) Considering stress owing to the weight of the beam
add’l
8
2
wL
M = (Table AT 2)
where ftlbw 7.7=

36
add’l
( ) inkipsinlb
wL
M −=−=





== 513.0513
8
80
12
7.7
8
22
513.020 += FM = total moment
I
Mc
ss c ==
( )( )
6
2513.020
15
+
=
F
kipsF 224.2=
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 01325.00265.0
2
===
inr 76325.1301325.075.13 =+=
Using Eq. (1.4) page 11 (Text)
r
Ec
s =
where psiE 6
1030×=
( )( ) psis 881,28
76325.13
01325.01030 6
=
×
=
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:
ksisu 65= (Table AT 7)
Designing based on ultimate strength,
6=N , for repeated, minor shock load

37
ksi
N
s
s u
8.10
6
65
===
Loading Diagram
x
h 1
30
14 −
=
−
110.0 += xh
12
3
wh
I =
2
h
c =
FxM =
( )
( )2223
110.0
33
2
6
12
2
+
===












==
x
Fx
h
Fx
h
Fx
wh
h
Fx
I
Mc
s
Differentiating with respect to x then equate to zero to solve for x giving maximum
stress.
( ) ( ) ( )( )( )
( )
0
110.0
10.0110.021110.0
3 4
2
=





+
+−+
=
x
xxx
F
dx
ds
( ) 010.02110.0 =−+ xx
inx 10=
( ) inh 211010.0 =+=
2
3
h
Fx
N
s
s u
==
( )
( )2
2
103
8.10
F
=
kipsF 44.1=
TORSIONAL STRESSES
DESIGN PROBLEMS

38
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,
ksisy 59=
ksisus 72=
Designing based on ultimate strength
N
s
s us
= , 6=N (Table 1.1)
ksis 12
6
72
==
Torque,
( )
( )
kipsinlbinlbft
n
hp
T −=−=−=== 540.054045
17502
15000,33
2
000,33
ππ
For diameter,
3
16
d
T
s
π
=
( )
3
540.016
12
dπ
=
ind 612.0=
say ind
8
5
=
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, io DD 2= , 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:
( )
( )
kipsinlbft
n
hp
T −=−=== 276036,23
5702
2500000,33
2
000,33
ππ
For AISI 1137, annealed
ksisy 50= (Table AT 8)
ksiss yys 306.0 ==
Designing based on yield strength
3=N for medium shock, one direction

39
Design stress
ksi
N
s
s ys
10
3
30
===
(a) Let D = shaft diameter
J
Tc
s =
32
4
D
J
π
=
2
D
c =
3
16
D
T
s
π
=
( )
3
27616
10
Dπ
=
inD 20.5=
say inD
4
1
5=
(b)
( ) ( )[ ]
32
15
32
2
32
44444
iiiio DDDDD
J
πππ
=
−
=
−
=
i
io
D
DD
c ===
2
2
2
34
15
32
32
15 ii
i
D
T
D
TD
s
ππ
=






=
( )
3
15
27632
10
iDπ
=
inDi 66.2=
inDD io 32.52 ==
say
inDi
8
5
2=
inDo
4
1
5=
(c) Density, 3
284.0 inlb=ρ (Table AT 7)

40
For solid shaft
=w weight per foot of length
( )( ) ftlbDDw 8.7325.5284.033
4
12
222
===





= ππρ
π
ρ
For hollow shaft
( ) ( ) ( )( ) ( )[ ] ftlbDDDDw ioio 3.55625.225.5284.033
4
12
222222
=−=−=−





= ππρ
π
ρ
Therefore hollow shaft is lighter
Percentage lightness = ( ) %5.33%100
3.55
3.558.73
=
−
(d) Torsional Deflection
JG
TL
=θ
where
inftL 12010 ==
ksiG 3
105.11 ×=
For solid shaft,
32
4
D
J
π
=
( )( )
( ) ( )
( ) o
2.2
180
039.0039.0
105.1125.5
32
120276
34
=





==
×





=
ππ
θ rad
For hollow shaft,
( )
32
44
io DD
J
−
=
π
( )( )
( ) ( )[ ]( )
( ) o
4.2
180
041.0041.0
105.11625.225.5
32
120276
344
=





==
×−





=
ππ
θ rad
Therefore, solid shaft is more rigid, oo
4.22.2 <
30. The same as 29, except that the material is AISI 4340, OQT 1200 F.
Solution:
( )
( )
kipsinlbft
n
hp
T −=−=== 276036,23
5702
2500000,33
2
000,33
ππ
For AISI 4340, OQT 1200 F
ksisy 130=
( ) ksiss yys 781306.06.0 ===
Designing based on yield strength

41
3=N for mild shock
Design stress
ksi
N
s
s ys
26
3
78
===
(a) Let D = shaft diameter
J
Tc
s =
32
4
D
J
π
=
2
D
c =
3
16
D
T
s
π
=
( )
3
27616
26
Dπ
=
inD 78.3=
say inD
4
3
3=
(b)
( ) ( )[ ]
32
15
32
2
32
44444
iiiio DDDDD
J
πππ
=
−
=
−
=
i
io
D
DD
c ===
2
2
2
34
15
32
32
15 ii
i
D
T
D
TD
s
ππ
=






=
( )
3
15
27632
26
iDπ
=
inDi 93.1=
inDD io 86.32 ==
say
inDi 2=
inDo 4=
(c) Density, 3
284.0 inlb=ρ (Table AT 7)

42
For solid shaft
=w weight per foot of length
( )( ) ftlbDDw 6.3775.3284.033
4
12
222
===





= ππρ
π
ρ
For hollow shaft
( ) ( ) ( )( ) ( )[ ] ftlbDDDDw ioio 1.3224284.033
4
12
222222
=−=−=−





= ππρ
π
ρ
Therefore hollow shaft is lighter
Percentage lightness = ( ) %1.17%100
1.32
1.326.37
=
−
(d) Torsional Deflection
JG
TL
=θ
where
inftL 12010 ==
ksiG 3
105.11 ×=
For solid shaft,
32
4
D
J
π
=
( )( )
( ) ( )
( ) o
48.8
180
148.0148.0
105.1175.3
32
120276
34
=





==
×





=
ππ
θ rad
For hollow shaft,
( )
32
44
io DD
J
−
=
π
( )( )
( ) ( )[ ]( )
( ) o
99.6
180
122.0122.0
105.1124
32
120276
344
=





==
×−





=
ππ
θ rad
Therefore, hollow shaft is more rigid, oo
48.899.6 < .
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 D20 ? (b) If deflection is
primary what kind of steel would be satisfactory?
Solution:
(a)
( )
( )
kipsinlbft
n
hp
T −=−=== 04.5420
5002
40000,33
2
000,33
ππ
ksiG 3
105.11 ×=
DL 20=

43
rad
180
1
π
θ == o
JG
TL
=θ
( )( )
( )3
4
105.11
32
2004.5
180
×





=
D
D
π
π
inD 72.1=
say inD
4
3
1=
(b)
( )
( )
ksi
D
T
s 8.4
75.1
04.51616
33
===
ππ
Based on yield strength
3=N
( )( ) ksiNssys 4.148.43 ===
ksi
s
s ys
y 24
6.0
4.14
6.0
===
Use C1117 normalized steel ksisy 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
ksisy 75=
ksiss yys 456.0 ==
3=N for medium shock
Z
T
N
s
s ys
′
==
where, bh =
9
2
9
2 32
bhb
Z ==′ (Table AT 1)
kipsinlbinT −=−= 2.11200
( )
3
2
92.1
3
45
b
s ==
inhb 71.0==
say inhb
4
3
==

44
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:
ksisus 49=
( )DtsF us π=
where inD
16
15
=
int
2
1
=
( ) kipsF 2.72
2
1
16
15
49 =











= π
FrT =
where inr
4
3
=
( ) kipsinT −=





= 2.54
4
3
2.72
(a) 3
16
d
T
s
π
=
where ind 5.3=
( )
( )
ksis 44.6
5.3
2.5416
3
==
π
(b) For AISI 1035 steel, ksisus 64=
for shock loading, traditional factor of safety, 15~10=N
Design factor , 94.9
44.6
64
===
s
s
N us
, the design is safe ( 10≈N )
34. The same as 33, except that the shaft diameter is 2 ¾ in.
Solution:

45
ind 75.2=
(a) 3
16
d
T
s
π
=
( )
( )
ksis 3.13
75.2
2.5416
3
==
π
(b) For AISI 1035 steel, ksisus 64=
for shock loading, traditional factor of safety, 15~10=N
Design factor , 8.4
3.13
64
===
s
s
N us
, the design is not safe ( 10<N )
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,
ksisus 98= (Table AT 3)
4~3=N , for dead load, based on ultimate strength
(a)
J
Tc
s =
( ) ( ) ( )[ ] 4
4444
3086
32
5.65.13
32
in
DD
J io
=
−
=
−
=
ππ
in
D
c o
75.6
2
5.13
2
===
( )
( )
kipsinlbft
n
hp
T −=−=== 3152606,262
2002
000,10000,33
2
000,33
ππ
( )( ) ksis 9.6
3086
75.63152
==
(b) Factor of safety,
2.14
9.6
98
===
s
s
N us
, not risky

46
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 dD 2.11 = . (c) Choose standard fractional dimensions
which you think would be satisfactory.
Problems 36 – 38.
Solution:
s = axial stress
ss = shear stress
(a)
2
2
4
4
1 d
F
d
F
s
ππ
==
Equation (1)
s
F
d
π
4
=

47
( ) ( ) ( )[ ] ( )22222
1
2
2
1
2 44.1
4
2.1
44
4
1 dD
F
dD
F
DD
F
DD
F
s
−
=
−
=
−
=
−
=
ππππ
Equation (2) 2
44.1
4
d
s
F
D +=
π
dh
F
hD
F
ss
ππ 2.11
==
Equation (3)
sds
F
h
π2.1
=
Dt
F
ss
π
=
Equation (4)
sDs
F
t
π
=
(b) Designing based on ultimate strength,
Table AT 7, AISI C1020, as rolled
ksisu 65=
ksisus 49=
4~3=N say 4, design factor for static load
ksi
N
s
s u
16
4
65
===
ksi
N
s
s us
s 12
4
49
===
kipslbF 12000,12 ==
From Equation (1)
( )
( )
in
s
F
d 98.0
16
1244
===
ππ
From Equation (2)
( )
( )
( ) ind
s
F
D 53.198.044.1
16
124
44.1
4 22
=+=+=
ππ
From Equation (3)
( )( )
in
ds
F
h
s
27.0
1298.02.1
12
2.1
===
ππ
From Equation (4)
( )( )
in
Ds
F
t
s
21.0
1253.1
12
===
ππ

48
(c) Standard fractional dimensions
ind 1=
inD
2
1
1=
inh
4
1
=
int
4
1
=
37. The same as 36, except that a shock load of 4000 lb. is repeatedly applied.
Solution:
(a) Same as 36.
(b) 15~10=N for shock load, based on ultimate strength
say 15=N , others the same.
ksi
N
s
s u
4
15
65
===
ksi
N
s
s us
s 3
15
49
===
kipslbF 44000 ==
From Equation (1)
( )
( )
in
s
F
d 13.1
4
444
===
ππ
From Equation (2)
( )
( )
( ) ind
s
F
D 76.113.144.1
4
44
44.1
4 22
=+=+=
ππ
From Equation (3)
( )( )
in
ds
F
h
s
31.0
313.12.1
4
2.1
===
ππ
From Equation (4)
( )( )
in
Ds
F
t
s
24.0
376.1
4
===
ππ

49
(c) Standard fractional dimensions
ind
8
1
1=
inD
4
3
1=
inh
8
3
=
int
4
1
=
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 dD 2.11 = , uus ss 75.0= ,
and that bending in the plate is negligible.
Solution:
2
4
1
d
F
s
π
=
sdF 2
4
1
π=
(1) 





=
N
s
dF u2
4
1
π

50
( ) ( )222
1
2
44.1
4
1
4
1
dD
F
DD
F
s
−
=
−
=
ππ
( )sdDF 22
44.1
4
1
−= π
(2) ( ) 





−=
N
s
dDF u22
44.1
4
1
π
dh
F
hD
F
ss
ππ 2.11
==
sdhsF π2.1=






=





=
N
s
dh
N
s
dhF uus 75.0
2.12.1 ππ
(3) 





=
N
s
dhF u5
9.0 π
Dt
F
ss
π
=
sDtsF π=






=





=
N
s
Dt
N
s
DtF uus 75.0
ππ
(4) 





=
N
s
DtF u
π75.0
Equate (2) and (1)
( ) 





=





−=
N
s
d
N
s
dDF uu 222
4
1
44.1
4
1
ππ
22
44.2 dD =
dD 562.1=
Equate (3) and (1)






=





=
N
s
d
N
s
dhF uu 2
4
1
9.0 ππ
( )
d
d
h 278.0
9.04
==
Equate (4) and (1)






=





=
N
s
d
N
s
DtF uu 2
4
1
75.0 ππ
( )( ) 





=





=
N
s
d
N
s
tdF uu 2
4
1
562.175.0 ππ
( )( )
d
d
t 214.0
562.175.04
==

51
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)






=
2
4
1
5 D
F
ss
π
Equation (1)
ss
F
D
π5
4
=
( )Dbt
F
s
2−
=
Equation (2) D
ts
F
b 2+=
Dt
F
s
5
=
Equation (3)
Ds
F
t
5
=
(b) For AISI C1020, as rolled
ksiss yys 286.0 ==
4=N for repeated and reversed load (mild shock) based on yield strength
ksis 12
4
48
==
ksiss 7
4
28
==
From Equation (1)

52
ss
F
D
π5
4
=
where
kipslbF 5.22500 ==
( )
( )
in
s
F
D
s
30.0
75
5.24
5
4
===
ππ
say in
16
5
From Equation (3)
( )
in
Ds
F
t 13.0
12
16
5
5
5.2
5
=






== say in
32
5
From Equation (2)
( )
inD
ts
F
b 96.1
16
5
2
12
32
5
5.2
2 =





+






=+= say in2
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
ksiss yys 2555.0 ==
4=N for repeated and reversed load (mild shock) based on yield strength
ksis 12
4
47
==
ksiss 6
4
25
==
From Equation (1)
ss
F
D
π5
4
=
where
kipslbF 5.22500 ==
( )
( )
in
s
F
D
s
33.0
65
5.24
5
4
===
ππ
say in
8
3
From Equation (3)

53
( )
in
Ds
F
t 11.0
12
8
3
5
5.2
5
=






== say in
8
1
From Equation (2)
( )
inD
ts
F
b 42.2
8
3
2
12
8
1
5.2
2 =





+






=+= say in
2
1
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)
( )Dbt
F
sP
−
= or
( )Dbt
F
sP
2
4
3
−
= Equation (1)
( )2
4
1
4 2






=
D
F
ssR
π
Equation (2)

54
Dt
F
sR
4
= Equation (3)
ksisuR 61= , ksiss uRusR 4575.0 ==
ksisuP 65=
ksi
N
s
s uP
P 8.10
6
65
===
ksi
N
s
s uR
R 1.10
6
61
===
ksi
N
s
s usR
sR 5.7
6
45
===
kipslbF 88000 ==
Solving for D
2
2 D
F
ssR
π
=
( )
in
s
F
D
sR
412.0
5.72
8
2
===
ππ
say in
16
7
Solving for t
Dt
F
sR
4
=
( )
in
Ds
F
t
R
453.0
1.10
16
7
4
8
4
=






== say in
2
1
Solving for b
Using
( )Dbt
F
sP
−
=
( )
inD
ts
F
b
P
92.1
16
7
8.10
2
1
8
=+






=+= say in2
Using
( )Dbt
F
sP
2
4
3
−
=

55
( )
( )
inD
ts
F
b
P
99.1
16
7
2
8.10
2
1
4
83
2
4
3
=





+






=+= say in2
Therefore
inb 2=
inD
16
7
=
int
2
1
=
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
2
12
1
4
4
1 D
F
D
F
s
ππ
== Equation (1)
( )eDL
F
s
2−
= Equation (2)

56
eD
F
s
2
= Equation (3)
( ) ( )2
2
2
2
2
2
4
4
1 Da
F
Da
F
s
−
=
−
=
ππ
Equation (4)
eDD
F
eDD
F
s
2
2
2
2
2
2
4
4
4
1 −
=
−
=
ππ
Equation (5)
Shear Stresses
eb
F
ss
2
= Equation (6)
( )teDL
F
ss
+−
=
22
Equation (7)

57
at
F
ss
π
= Equation (8)
mD
F
ss
1π
= Equation (9)
hD
F
ss
22
= Equation (10)
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
kipsF 10= , 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)
inint 875.0
8
7
== , ysy ss 6.0=
For steel rod, AISI C1035, as rolled
ksisy 551
=
ksissy 331
=
For plate and cotter, AISI C1020, as rolled
ksisy 482
=
ksissy 282
=
7~5=N based on yield strength
say 7=N
From Equation (1) (Prob. 42)
2
1
41
D
F
N
s
s
y
π
==
( )
2
1
104
7
55
Dπ
=
inD 27.11 =
say inD
4
1
11 =

58
From Equation (9)
mD
F
N
s
s
sy
s
1
1
π
==
m





=
4
1
1
10
7
33
π
inm 54.0=
say inm
16
9
=
From Equation (3)
eD
F
N
s
s
y
2
1
==
eD
s
2
10
7
55
==
273.12 =eD
From Equation (5)
eDD
F
N
s
s
y
2
2
2 4
41
−
==
π
( )
( )273.14
104
7
55
2
2 −
=
Dπ
inD 80.12 =
say inD
4
3
12 =
and 273.12 =eD
273.1
4
3
1 =





e
ine 73.0=
say ine
4
3
=
By further adjustment
Say inD 22 = , ine
8
5
=
From Equation (8)
at
F
N
s
s
sy
s
π
== 2
( )875.0
10
7
28
aπ
=
ina 91.0=
say ina 1=

59
From Equation (4)
( )2
2
2
42
Da
F
N
s
s
y
−
==
π
( )
( )22
2
104
7
48
−
=
aπ
ina 42.2=
say ina
2
1
2=
use ina
2
1
2=
From Equation (7)
( )teDL
F
N
s
s
sy
s
+−
==
22
2
( )875.0
8
5
22
10
7
28






+−
=
L
inL 80.2=
say inL 3=
From Equation (6)
eb
F
N
s
s
sy
s
2
2
==
b





=
8
5
2
10
7
28
inb 2=
From Equation (10)
hD
F
N
s
s
sy
s
22
2
==
( )h22
10
7
28
=
ininh
8
5
625.0 ==
Summary of Dimensions
inL 3=
inh
8
5
=
inb 2=
int
8
7
=

60
inm
16
9
=
ina
2
1
2=
inD
4
1
11 =
inD 22 =
ine
8
5
=
(b) Tolerances and allowances, No fit, tolerance = in010.0±
inL 010.03±=
inh 010.0625.0 ±=
int 010.0875.0 ±=
inm 010.05625.0 ±=
ina 010.0500.2 ±=
inD 010.025.11 ±=
For Free Running Fits (RC 7) Table 3.1
Female Male
inb
0000.0
0030.0
0.2
−
+
= inb
0058.0
0040.0
0.2
−
−
=
allowance = 0.0040 in
inD
0000.0
0030.0
0.22
−
+
= inD
0058.0
0040.0
0.22
−
−
=
ine
0000.0
0016.0
625.0
−
+
= ine
0030.0
0020.0
625.0
−
−
=
44. A 1-in. ( 1D ) 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
ksisu 851
=
ksisus 641
=

61
ksisu 652
=
ksisus 482
=
Ultimate strength
Use Equation (1)
( ) ( ) kipsDsF uu 8.661
4
1
85
4
1 22
11
=





=





= ππ
Equation (9)
mDsF usu 11
π=
( )( )( )m1648.66 π=
inm 33.0=
say inm
8
3
=
From Equation (3)
eDsF uu 21
=
( ) eD2858.66 =
7859.02 =eD
From Equation (5)






−= eDDsF uu 2
2
2
4
1
1
π
( ) 





−= 7859.0
4
1
858.66 2
2Dπ
inD 42.12 =
say inD
8
3
12 =
7859.0
8
3
12 =





= eeD
ine 57.0=
say ine
16
9
=
From Equation (4)
( )



−= 2
2
2
4
1
2
DasF uu π
( )














−





=
2
2
8
3
1
4
1
658.66 aπ
ina 79.1=
say ina
4
3
1=
From Equation (8)

62
atsF usu π2
=
( )( )( )( )1488.66 aπ=
ina 44.0=
say ina
2
1
=
use ina
4
3
1=
From Equation (2)
( )eDLsF uu 22
−=
( ) 











−=
16
9
8
3
1658.66 L
inL 20.3=
say inL
4
1
3=
From Equation (7)
( )teDLsF usu −−= 22
2
( ) ( )1
16
9
8
3
14828.66 





−−= L
inL 51.1=
say inL
2
1
1=
use inL
4
1
3=
From Equation (6)
ebsF usu 1
2=
( ) b





=
16
9
6428.66
inb 93.0=
say inb 1=
From Equation (10)
hDsF usu 21
2=
( ) h





=
8
3
16428.66
inh 38.0=
say inh
8
3
=
Dimensions
inL
4
1
3=

63
inh
8
3
=
inb 1=
int 1=
inm
8
3
=
ina
4
3
1=
inD 11 =
inD
8
3
12 =
ine
16
9
=
(b) Tolerances and allowances, No fit, tolerance = in010.0±
inL 010.025.3 ±=
inh 010.0375.0 ±=
int 010.0000.1 ±=
inm 010.0375.0 ±=
ina 010.075.1 ±=
inD 010.0000.11 ±=
For Loose Running Fits (RC 8) Table 3.1
Female Male
inb
0000.0
0035.0
0.1
−
+
= inb
0065.0
0045.0
0.1
−
−
=
inD
0000.0
0040.0
375.12
−
+
= inD
0075.0
0050.0
375.12
−
−
=
ine
0000.0
0028.0
5625.0
−
+
= ine
0051.0
0035.0
5625.0
−
−
=
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 uus ss 75.0= .

64
Problems 45 – 47.
Solution:
(a) Neglecting bending
Equation (1): 





= 2
4
1
DsF π
Equation (2): 











= 2
4
1
2 csF s π
Equation (3): ( )bcsF 2=
Equation (4): ( )acsF =
Equation (5): ( )[ ]bcdsF −= 2
Equation (6): ( )mbsF s 4=
Equation (7): ( )nbsF s 2=
Equation (8): ( )acdsF −=
(b)
N
s
s u
= and
N
s
s us
s =
Therefore
sss 75.0=
Equate (2) and (1)






=











= 22
4
1
4
1
2 DscsF s ππ






=




 22
4
1
2
1
75.0 Dscs
Dc 8165.0=
Equate (3) and (1)
( ) 





== 2
4
1
2 DsbcsF π
( ) 2
4
1
8165.02 DDb π=
Db 4810.0=

65
Equate (4) and (1)






== 2
4
1
DssacF π
( ) 2
4
1
8165.0 DDa π=
Da 9619.0=
Equate (5) and (1)
( )[ ] 





=−= 2
4
1
2 DsbcdsF π
( )( ) 2
4
1
4810.08165.02 DDd π=−
Dd 6329.1=
Equate (6) and (1)
( ) 





== 2
4
1
4 DsmbsF s π
( )( ) 2
4
1
4810.0475.0 DDm π=
Dm 5443.0=
Equate (7) and (1)
( ) 





== 2
4
1
2 DsnbsF s π
( )( ) 2
4
1
4810.0275.0 DDn π=
Dn 0886.1=
Equate (8) and (1)
( ) 





=−= 2
4
1
DsacdsF π
( ) 2
4
1
8165.06329.1 DaDD π=−−
Da 9620.0=
Summary
Da 9620.0=
Db 4810.0=
Dc 8165.0=
Dd 6329.1=
Dm 5443.0=
Dn 0886.1=

66
46. The same as 45, except that the calculations are to be based on yield strengths. Let
ysy ss 6.0= .
Solution: (Refer to Prob. 45)
(a) Neglecting bending
Equation (1): 





= 2
4
1
DsF π
Equation (2): 











= 2
4
1
2 csF s π
Equation (3): ( )bcsF 2=
Equation (4): ( )acsF =
Equation (5): ( )[ ]bcdsF −= 2
Equation (6): ( )mbsF s 4=
Equation (7): ( )nbsF s 2=
Equation (8): ( )acdsF −=
(b)
N
s
s y
= and
N
s
s sy
s =
Therefore
sss 6.0=
Equate (2) and (1)






=











= 22
4
1
4
1
2 DscsF s ππ






=




 22
4
1
2
1
6.0 Dscs
Dc 9129.0=
Equate (3) and (1)
( ) 





== 2
4
1
2 DsbcsF π
( ) 2
4
1
9129.02 DDb π=
Db 4302.0=
Equate (4) and (1)






== 2
4
1
DssacF π
( ) 2
4
1
9129.0 DDa π=
Da 8603.0=

67
Equate (5) and (1)
( )[ ] 





=−= 2
4
1
2 DsbcdsF π
( )( ) 2
4
1
4302.09129.02 DDd π=−
Dd 8257.1=
Equate (6) and (1)
( ) 





== 2
4
1
4 DsmbsF s π
( )( ) 2
4
1
4302.046.0 DDm π=
Dm 7607.0=
Equate (7) and (1)
( ) 





== 2
4
1
2 DsnbsF s π
( )( ) 2
4
1
4302.026.0 DDn π=
Dn 5214.1=
Equate (8) and (1)
( ) 





=−= 2
4
1
DsacdsF π
( ) 2
4
1
9129.08257.1 DaDD π=−−
Da 8604.0=
Summary
Da 8604.0=
Db 4302.0=
Dc 9129.0=
Dd 8257.1=
Dm 7607.0=
Dn 5214.1=
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)

68
4=N based on yield strength for gradually applied and reversed load.
For AISI C1040, annealed
ksisy 47= (Fig. AF 7)
ksiss ysy 286.0 ==
ksi
N
s
s y
75.11
4
47
===






= 2
4
1
DsF π






= 2
4
1
75.1112 Dπ
inD 14.1=
say inD
8
1
1=
inDa 97.0
8
1
18604.08604.0 =





==
but Da >
say ina
4
1
1=
inb 48.0
8
1
14302.0 =





=
say inb
2
1
=
inc 030.1
8
1
19129.0 =





=
say inc 1=
ind 05.2
8
1
18257.1 =





=
say ind 2=
inm 86.0
8
1
17607.0 =





=
say inm
8
7
=
inn 71.1
8
1
15214.1 =





=
say inn
4
3
1=
Dimension:
ina
4
1
1=

69
inb
2
1
=
inc 1=
ind 2=
inm
8
7
=
inn
4
3
1=
inD
8
1
1=
48. Give all the strength equations for the union of rods shown.
Problems 48 – 68.
Solution:






= 2
4
1
dsF π Equation (1)
( )adsF s π= Equation (2)
( )tcsF s 2= Equation (3)

70
( )[ ]beDsF s −= 2 Equation (4)
setF = Equation (5)
( )teDsF −= Equation (6)
( )



−= 22
4
1
eksF π Equation (7)
( ) ( ) 



−−−= tememsF 22
4
1
π Equation (8)
( )efsF s 2= Equation (9)

71






−= etesF 2
4
1
π 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 3000 1020
50 3500 1030
51 4000 1117
52 4500 1020
52 5000 1015
54 5500 1035
55 6000 1040
56 6500 1020
57 7000 1015
58 7500 1118
59 8000 1022
60 8500 1035
61 9000 1040
62 9500 1117
63 10,000 1035
64 10,500 1022
65 11,000 1137
66 11,500 1035
67 12,000 1045
68 12,500 1030

72
Solution: (For Prob. 49 only)
(a) For AISI 1020, as rolled
ksisy 48=
( ) ksiss yys 8.28486.06.0 ===
For reversing load, 4=N based on yield strength
ksi
N
s
s y
12
4
48
===
ksi
N
s
s ys
s 2.7
4
8.28
===
kipslbF 33000 ==
Equation (1)






= 2
4
1
dsF π






= 2
4
1
123 dπ
ind 5642.0=
say ind
16
9
=
Equation (2)
( )adsF s π=
( ) 





=
16
9
2.73 aπ
ina 236.0=
say ina
4
1
=
Equation (5)
setF =
et123 =
25.0=et
Equation (10)






−= etesF 2
4
1
π






−= 25.0
4
1
123 2
eπ
ine 798.0=
say ine
16
13
=
25.0=et

73
25.0
16
13
=





t
int 308..0=
say int
16
5
=
Equation (6)
( )teDsF −=












−=
16
5
16
13
123 D
inD 6125.1=
say inD
8
5
1=
Equation (4)
( )[ ]beDsF s −= 2












−= b
16
13
8
5
122.73
inb 256.0=
say inb
4
1
=
Equation (7)
( )



−= 22
4
1
eksF π














−





=
2
2
16
13
4
1
123 kπ
ink 989.0=
say ink 1=
Equation (9)
( )efsF s 2=
( ) f





=
16
13
22.73
inf 256.0=
say inf
4
1
=
Equation (8)
( ) ( ) 



−−−= tememsF 22
4
1
π




















−−














−=
16
5
16
13
16
13
4
1
123
2
2
mmπ

74
2539.03125.05185.07854.025.0 2
+−−= mm
05146.03125.07854.0 2
=−− mm
06552.03979.02
=−− mm
inm 032.1=
say inm 1=
Equation (3)
( )tcsF s 2=
( ) c





=
16
5
22.73
inc 667.0=
say inc
16
11
=
DIMENSIONS:
ind
16
9
=
ina
4
1
=
inb
4
1
=
inc
16
11
=
inf
4
1
=
ine
16
13
=
int
16
5
=
ink 1=
inD
8
5
1=
inm 1=
(b) Modified dimensions
ind
16
9
=
ina
4
1
=
inb
4
3
=
inc
16
11
=

75
inf
2
1
=
ine
16
13
=
int
16
5
=
ink 1=
inD
8
5
1=
inm
4
1
1=
(c) Tolerances and allowances
No fit, in010.0±
ind 010.05625.0 ±=
ina 010.0250.0 ±=
inf 010.0500.0 ±=
inD 010.0625.1 ±=
ink 010.0000.1 ±=
inm 010.0250.1 ±=
Fits, Table 3.1, loose-running fits, say RC 8
Female Male
inb
0000.0
0035.0
750.0
−
+
= inb
0065.0
0045.0
750.0
−
−
=
inc
0000.0
0028.0
6875.0
−
+
= inc
0051.0
0035.0
6875.0
−
−
=
ine
0000.0
0035.0
8125.0
−
+
= ine
0065.0
0045.0
8125.0
−
−
=
allowance = 0.0045
int
0000.0
0022.0
3125.0
−
+
= int
0040.0
0030.0
3125.0
−
−
=

76
(d)
ROD
COTTER

77
SOCKET
CHECK PROBLEMS
69. The connection shown has the following dimensions: ind
4
1
1= , inD
2
1
2= ,
inD
2
1
11 = , inh
8
5
= , int
2
1
= ; it supports a load of 15 kips. Compute the tensile,
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)?

78
Problem 69.
Solution:
Tensile Stresses
(1) ksi
d
F
s 22.12
4
1
1
4
1
15
4
1 2
2
1 =






==
ππ
(2) ksi
D
F
s 4.8
2
1
1
4
1
15
4
1 2
2
1
2 =






==
ππ
Compressive Stress
(3)
( )
ksi
DD
F
s 78.4
2
1
1
2
1
2
4
1
15
4
1 22
2
1
2
3 =














−





=
−
=
ππ
Shear Stresses
(4) ksi
Dt
F
ss 82.3
2
1
2
1
2
15
4
=












==
π
π
(5) ksi
hD
F
ss 09.5
8
5
2
1
1
15
1
5
=












==
π
π
For AISI C1045, as rolled (rod and nut)
ksisy 591
=
( ) ksiss yys 4.35596.06.01
===
For structural steel plate (1020)
ksisy 482
=
( ) ksiss yys 8.28486.06.01
===
Solving for design factor

79
(1) 83.4
22.12
59
1
1
1
===
s
s
N
y
(2) 95.6
49.8
59
2
2
1
===
s
s
N
y
(3) 04.10
78.4
48
3
3
2
===
s
s
N
y
(4) 54.7
82.3
8.28
4
2
4 ===
s
ys
s
s
N
(5) 96.6
09.5
4.35
5
1
5 ===
s
ys
s
s
N
The corresponding design factor is 83.4=N
70. In the figure, let inD
4
3
= , int
16
7
= , inb
4
3
3= , and let the load, which is applied
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?
Problem 70.
Solution:
(a) Tensile stresses
( )
ksi
Dbt
F
s 43.11
4
3
4
3
3
16
7
15
1 =






−





=
−
=
( )
( )
ksi
Dbt
F
s 43.11
4
3
2
4
3
3
16
7
15
4
3
2
4
3
2 =












−











=
−
=
Compressive bearing stress
ksi
Dt
F
s 43.11
16
7
4
3
4
15
4
3 =












==

80
Shearing stress
( ) ( )
ksi
D
F
ss 24.4
2
4
3
15
2
4
1
4
2
2
4 =






=






=
ππ
ksisy 48=
ksiss yys 8.286.0 ==
s
s
N y
= or
s
ys
s
s
N =
Using
s
s
N y
=
2.4
43.11
48
===
s
s
N y
Using
s
ys
s
s
N =
8.6
24.4
8.28
===
s
ys
s
s
N
Therefore the design factor is 2.4=N
71. For the connection shown, let ina
16
15
= , inb
16
9
= , inc
4
3
= , ind
2
1
1= ,
inD
4
3
= , innm
16
15
== . The material is AISI C1040, annealed (see Fig. AF 1).
(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.
Problem 71.
Solution:
(a) Tensile stresses

81
ksi
D
F
s 98.16
4
3
4
1
5.7
4
1 2
2
1 =






==
ππ
( )
ksi
cdb
F
s 89.8
4
3
2
1
1
16
9
2
5.7
2
2 =






−





=
−
=
( )
ksi
cda
F
s 67.10
4
3
2
1
1
16
15
5.7
3 =






−





=
−
=
Compressive Stresses (Bearing)
ksi
bc
F
s 89.8
4
3
16
9
2
5.7
2
4 =












==
ksi
ac
F
s 67.10
4
3
16
15
5.7
5 =












==
Shearing Stresses
ksi
mb
F
ss 56.3
16
9
16
15
4
5.7
46
=












==
ksi
nb
F
ss 11.7
16
9
16
15
2
5.7
27
=












==
For AISI C1040, annealed,Fig. AF 1
ksisy 47=
ksisu 79=
ksiss yys 286.0 ==
ksiss uus 44760 .. ==
(b) Based on ultimate strength
65.4
98.16
79
1
===
s
s
N u
(c) Based on yield strength
77.2
98.16
47
1
===
s
s
N
y
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

82
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:
lbsF 000,60=
(a)
( )
( )
ksipsisc 75.11753,11
3
2
1
3
4
1
2000,60
2
2
==








−





=
π
(b)
( )
( )
ksipsisb 19.10186,10
2
1
34
4
1
2000,60
2
2
==














−
=
π
(c)
( )
( )( )
ksipsisc 18.3183,3
13
2000,60
===
π
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:
lbsF 000,22=
inD 2=
inftL 12010 ==
in03.0=δ

83
(a)
EA
FL
=δ
( )( )
( )( )( )
psi
D
FL
A
FL
E 6
22
1028
203.0
120000,2244
×====
πδπδ
(b) Use both stainless steel, Table AT 4, psiE 6
1028×= and wrought iron , Table AT 7,
psiE 6
1028×= .
(c) Energy absorbed = ( )( ) inlbF −== 33003.0000,22
2
1
2
1
δ
(d) For Aluminum alloy, 3003-H14
psiE 6
1010×=
ksisy 21=
( )( )
( )( )( )
in
DE
FL
EA
FL
084.0
21010
120000,2244
262
=
×
===
ππ
δ
( )
( )( ) ysksipsi
D
F
s <==== 0.77003
2
000,2244
22
ππ
, within the elastic limit.
74. The same as 73, except that kipsF 88= and total in112.0=δ . Is the
computation for part (d) valid? Explain.
Solution:
(a) kipsF 88=
in112.0=δ
( )( )
( )( )( )
psi
D
FL
A
FL
E 6
22
1030
2112.0
120000,8844
×====
πδπδ
(b) Use wrought steel, Table AT 4, psiE 6
1030×=
(c) Energy absorbed = ( )( ) inlbF −== 4928112.0000,88
2
1
2
1
δ
(d) For Aluminum alloy, 3003-H14
psiE 6
1010×=
ksisy 21=
( )( )
( )( )( )
in
DE
FL
EA
FL
336.0
21010
120000,8844
262
=
×
===
ππ
δ

84
( )
( )( ) ysksipsi
D
F
s >==== 0.28011,28
2
000,8844
22
ππ
, not within the elastic limit, therefore
not valid.
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:
inftL 484 ==
(a) For SAE 1020, as rolled
ksisu 65= , ksiE 000,30=
A
F
N
s
s u
==
2
40
5
65
x
=
inx 754.1=
EA
FL
=δ
( )( )
( ) 2
000,30
4840
024.0
x
=
inx 633.1=
Therefore say inx
4
3
1=
(b) For SAE 1045, as rolled
ksisu 96= , ksiE 000,30=
A
F
N
s
s u
==
2
40
5
96
x
=
inx 443.1=
EA
FL
=δ
( )( )
( ) 2
000,30
4840
024.0
x
=
inx 633.1=

85
Therefore say inx
8
5
1=
(c) For SAE 4640, as rolled
ksisu 152= , ksiE 000,30=
A
F
N
s
s u
==
2
40
5
152
x
=
inx 15.1=
EA
FL
=δ
( )( )
( ) 2
000,30
4840
024.0
x
=
inx 633.1=
Therefore say inx
8
5
1=
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:
L
tL
LE
s ∆
==
αδ
For steel Finin −= 000007.0α
psiE 6
1030×=
( )( )( )6
103040120000007.0 ×−=∆= tEs α
psis 800,16=
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?

86
Solution:
tEs ∆= α
For steel Finin −= 000007.0α
psiE 6
1030×=
( )( )( ) ksis 30.111000,3070600000007.0 =−=
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., inL 30= ., 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)
Finin −= 000007.01α
ksipsiE 000,301030 6
1 =×=
For aluminum alloy, 3003-H14 (Table AT 3)
Finin −= 0000129.02α
ksipsiE 000,101010 6
2 =×=
(a) CA PP =
FPPP CBA =++
(1) kipsFPP BA 242 ==+
( ) 2
2 25.0
8
1
2 inA =





=
( ) 2
1 1
2
1
2 inA =





=

87
BA δδ =
1122 EA
LP
EA
LP BA
=
( )( ) ( )( )000,301000,1025.0
BA PP
=
(2) AB PP 12=
(1) kipsPP AA 24122 =+
kipsPA 714.1=
( ) kipsPB 568.20714.112 ==
Stresses:
Aluminum plate
ksi
A
P
ss A
CA 856.6
25.0
714.1
2
====
Chromium steel plate
ksi
A
P
s B
B 568.20
1
568.20
1
===
(b)
( )( )
( )( )
in
EA
LPA
021.0
000,1025.0
30714.1
22
===δ
(c) Energy absorbed by steel plate ( )( ) inkipsPB −=== 216.0021.0568.20
2
1
2
1
δ
(d) Energy absorbed by aluminum plate ( )( ) inkipsPA −=== 018.0021.0714.1
2
1
2
1
δ
(e) kipsFPP BA 242 ==+
BTAT BA
δδδδ +=+
tLAT ∆= 2αδ
tLBT ∆= 1αδ
( )( )( ) inAT 0387.0100300000129.0 ==δ
( )( )( ) inBT 021.010030000007.0 ==δ
( )
( )( ) A
AA
A P
P
EA
LP
012.0
000,1025.0
30
22
===δ
( )
( )( ) B
BB
B P
P
EA
LP
001.0
000,301
30
11
===δ
Then
BA PP 001.0021.0012.00387.0 +=+
( )AA PP 224001.0012.00177.0 −=+
AA PP 002.0024.0012.00177.0 −=+

88
kipsPA 45.0=
( ) kipsPB 1.2345.0224 =−=
Stresses:
Aluminum plate
ksi
A
P
s A
A 8.1
25.0
45.0
2
===
ksi
A
P
s B
B 1.23
1
1.23
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)
ksiE 000,152 =
Finin −= 0000092.02α
(a)
(1) kipsFPP BA 242 ==+
BA δδ =
1122 EA
LP
EA
LP BA
=
( )( ) ( )( )000,301000,1525.0
BA PP
=
(2) AB PP 8=
kipsPP AA 2482 =+
kipsPA 4.2=
( ) kipsPB 2.194.28 ==
Stresses:
Aluminum plate
ksi
A
P
ss A
CA 6.9
25.0
4.2
2
====
ksi
A
P
s B
B 2.19
1
2.19
1
===

89
(b)
( )( )
( )( )
in
EA
LPA
019.0
000,1525.0
304.2
22
===δ
(c) Energy absorbed by steel plate ( )( ) inkipsPB −=== 182.0019.02.19
2
1
2
1
δ
(d) Energy absorbed by aluminum plate ( )( ) inkipsPA −=== 023.0019.04.2
2
1
2
1
δ
(e) kipsFPP BA 242 ==+
BTAT BA
δδδδ +=+
tLAT ∆= 2αδ
tLBT ∆= 1αδ
( )( )( ) inAT 0276.0100300000092.0 ==δ
( )( )( ) inBT 021.010030000007.0 ==δ
( )
( )( ) A
AA
A P
P
EA
LP
008.0
000,1525.0
30
22
===δ
( )
( )( ) B
BB
B P
P
EA
LP
001.0
000,301
30
11
===δ
Then
BA PP 001.0021.0008.00276.0 +=+
( )AA PP 224001.0008.00066.0 −=+
AA PP 002.0024.0008.00066.0 −=+
kipsPA 74.1=
( ) kipsPB 52.2074.1224 =−=
Stresses:
Aluminum plate
ksi
A
P
s A
A 96.6
25.0
74.1
2
===
ksi
A
P
s B
B 52.20
1
52.20
1
===
80. A machine part shown is made of AISI C1040, annealed steel; inL 151 = .,
inL 62 = ., inD
4
3
1 = ., and inD
2
1
2 = . Determine (a) the elongation due to a force
lbF 6000= ., (b) the energy absorbed by each section of the part if the load is
gradually applied.

90
Problems 80, 81
Solution:
For AISI C1040, annealed steel
psiE 6
1030×=
(a) 21 δδδ +=
( )( )
( )
in
EA
FL
0068.0
1030
4
3
4
156000
6
2
1
1
1 =
×











==
π
δ
( )( )
( )
in
EA
FL
0061.0
1030
2
1
4
66000
6
2
2
2
2 =
×











==
π
δ
in0129.00061.00068.021 =+=+= δδδ
(b) Energy absorbed
( )( ) inlbFU ==== 4.200068.06000
2
1
2
1
11 δ
( )( ) inlbFU ==== 3.180061.06000
2
1
2
1
22 δ
81. A rod as shown is made of AISI 2340 steel, OQT 1000 F, and has the following
dimensions: inL 201 = ., inL 122 = ., inD
8
7
1 = ., and inD
4
3
2 = . The unit strain at
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 ksiE 30000=
(a)
EA
F
L 22
2
== ε
δ

91
EAF 2ε=
1
2
1
2
1
1
2
1
12
1 L
D
D
L
A
A
EA
ELA






=





== εε
ε
δ
( ) inLL
D
D
T 067.01220
87
43
0025.0
2
21
2
1
2
21 =








+





=








+





=+= εδδδ
(b) ( ) kipsEAF 13.33000,30
4
3
4
0025.0
2
2 =











==
π
ε
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, 2=N based on the yield strengths?
Problem 82.
Solution:
For aluminum 2024-T4 (Table AT 3)
ksisy 471
= , ksiE 600101 ,=
For steel AISI 1045, as rolled (Table AT 7)
ksisy 592
= , ksiE 000,302 =
[ ]0=∑ GM ( ) ( ) ( )20241224 21 =+ RR
402 21 =+ RR Equation (1)

92
1224
21 δδ
=
21 2δδ =
11
11
1
AE
LR
=δ
22
22
2
AE
LR
=δ
inftL 9681 ==
inftL 144122 ==
2
1 5.0 inA =
21 2δδ =
22
22
11
11 2
AE
LR
AE
LR
=
( )
( )( )
( )
( ) 2
21
00030
1442
5060010
96
A
RR
,.,
=
2
2
1
530
A
R
R
.
=
But
N
s
A
R
s
y2
2
2
2 ==
5.29
2
59
2
2
==
A
R
( ) kipsR 64155295301 ... ==
N
s
A
R
s
y1
1
1
1 ==
2
47
5.0
1
=
R
kipsR 75.111 =
use kipsR 75.111 =
( ) kipsR 5.1675.112402 =−=
22
2 56.0
5.29
5.16
5.29
in
R
A ===

93
83. The bar shown supports a static load kipsF 5.2= with 0=θ ; ind 3= .,
inL 10= ., inh
4
3
2= . inb 1= . It is made of AISI 1035, as rolled. (a) How far
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?
Problem 83.
Solution:
[ ]∑ = 0AM ( )FLddRB +=
( )( )5.21033 +=BR
kipsRB 83.10=
[ ]∑ = 0BM LFdRA =
( )5.2103 =AR
kipsRA 33.8=

94
3−−= xRxRM BA
383.1033.82
2
−−== xx
dy
yd
EIM
1
22
3415.5165.4 Cxx
dy
dy
EI +−−=
21
33
3805.1388.1 CxCxxEIy ++−−=
When 0=x , 0=y
( ) ( ) ( ) 21
33
00805.10388.10 CCEI ++−=
02 =C
When 3=x , 0=y
( ) ( ) ( ) 030805.13388.10 1
33
++−= CEI
492.121 −=C
xxxEIy 492.123805.1388.1
33
−−−=
When inLdx 13=+=
( ) ( ) 108213492.1210805.113388.1
33
=−−=EIy
For AISI 1035, as rolled , ksiE 000,30=
( )( ) 4
33
7331.1
12
75.21
12
in
bh
I ===
1082=EIy
( )( ) 10827331.1000,30 =y
iny 021.0= , upward.

95
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 us . 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
ksisu 57=
ksi
N
s
s u
25.14
4
57
===
Solving for plate thickness
ηt
pD
s
2
=
ksipsip 2.0200 ==
inD 36=
( )( )
( )9.02
362.5
25.14
t
s ==
int 281.0=
say int
16
5
=
Stress on diametral section
( )( )
( )
ksi
t
pD
s 40.6
9.0
16
5
4
362.0
4
=






==
η
Stress on longitudinal section
( )( )
( )
ksi
t
pD
s 80.12
9.0
16
5
2
362.0
2
=






==
η
Stress on diametral section < stress on longitudinal section
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?

96
Solution:
(a) For 301, ¼ hard, stainless steel
ksi
N
s
s y
50
5.1
75
===
ksipsip 33000 ==
ηt
pD
s
4
=
( )( )
( )90.04
73
50
t
=
int 117.0=
3
286.0 inlb=γ
( ) ( )( ) lbtDtrW 2.5286.0117.074
222
==== πγπγπ
(b) For annealed titanium B265, gr. 5
ksi
N
s
s y
67.86
5.1
130
===
ksipsip 33000 ==
ηt
pD
s
4
=
( )( )
( )90.04
73
67.86
t
=
int 061.0=
3
160.0 inlb=γ
( ) ( )( ) lbtDtrW 5.1160.0061.074
222
==== πγπγπ
(c) For hardened titanium
ksi
N
s
s y
105
5.1
158
===
ksipsip 33000 ==
ηt
pD
s
4
=
( )( )
( )90.04
73
105
t
=

97
int 056.0=
3
160.0 inlb=γ
( ) ( )( ) lbtDtrW 38.1160.0056.074
222
==== πγπγπ
Savings in weight = ( ) %8%100
50.1
38.150.1
=
−
Circumstances: less in weight and small thickness.
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, 30-
ft.-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, ksisu 100= , 4=N (Table 1.1)
ksi
N
s
s u
25
4
100
===
ηt
pD
s
2
=
ksipsip 75.22750 ==
inD 12=
%87=η
( )( )
( )87.02
1275.2
25
t
s ==
int 759.0=
say int
4
3
=
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:

98
(a) inr 75.02 1 = , inr 375.01 =
inr 333.32 =
kipsF 5.2=
inb 5.0=
2
1
21
21
max
11
11
35.0




















+






+
=
EE
b
rr
F
sc
ksiE 000,30=
( )
ksisc 279
000,30
2
5.0
333.3
1
375.0
1
5.235.0
2
1
max =
























+
=
(b)
( )
ksisc 126
000,30
2
5.0
333.3
1
333.3
1
5.235.0
2
1
max =
























+
=
(c)
( )
ksisc 249
000,30
2
5.0
333.3
1
375.0
1
5.235.0
2
1
max =
























−
=
(d) Maximum shear stress
( ) ksiss cs 842793.03.0 maxmax ===
Location:
( ) ( )( )
in
rr
EE
s
w
c
023.0
333.3
1
375.0
1
000,30
2
3.012794
11
11
14 2
21
21
2
max
=






+






−
=






+






+−
=
µ
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?

99
Solution:
inr 03.11 =
inr 42.32 =
inb 3=
ksisc 200max =
2
1
21
21
max
11
11
35.0




















+






+
=
EE
b
rr
F
sc
ksiE 000,30=
2
1
max
000,30
2
3
42.3
1
03.1
1
35.0
200
























+
==
F
sc
kipsF 18=
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:
inD 75.0=
For pin
inD
003.0
000.0
749.0
−
+
=
For hole
inD
000.0
003.0
750.0
−
+
=
Sketch

100
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: inD 8=
(a) For close-sliding fit, RC 1
Hole, in Shaft, in
+0.0008 - 0.0006
-0.0000 -0.0012
Allowance = 0.0006 in
With tolerances,
Hole inD
0000.0
0008.0
0000.8
−
+
=
Shaft inD
0006.0
0000.0
9994.7
−
+
=
Limit dimension,
Hole intoD 0008.80000.8=
Shaft intoD 9988.79994.7=
Sketch

101
(b) For a precision-running fit, RC 3
Hole, in Shaft, in
+0.0012 -0.0020
-0.0000 -0.0032
With tolerances,
Hole inD
0000.0
0012.0
0000.8
−
+
=
Shaft inD
0012.0
0000.0
9980.7
−
+
=
Limit dimension,
Hole intoD 0012.80000.8=
Sketch
(c) For medium-running fit, RC5, RC 6. Say RC 5
Hole, in Shaft, in
+0.0018 -0.0040
-0.0000 -0.0058
With tolerances,
Hole inD
0000.0
0018.0
0000.8
−
+
=
Shaft inD
0018.0
0000.0
9960.7
−
+
=
Limit dimension,
Hole intoD 0018.80000.8=
Sketch

102
(d) For loose-running fit, RC 8, RC 9. Say RC 8
Hole, in Shaft, in
+0.0070 -0.0100
-0.0000 -0.0145
With tolerances,
Hole inD
0000.0
0070.0
0000.8
−
+
=
Shaft inD
0045.0
0000.0
9900.7
−
+
=
Limit dimension,
Hole intoD 0070.80000.8=
Sketch
91. The same as 90, except that the nominal diameter is 4 in.
Solution:
inD 4=
(a) For close-sliding fit, RC 1

103
Hole, in Shaft, in
+0.0006 -0.0005
-0.0000 -0.0009
With tolerances,
Hole inD
0000.0
0006.0
0000.4
−
+
=
Shaft inD
0004.0
0000.0
9995.3
−
+
=
Limit dimension,
Hole intoD 0006.40000.4=
Sketch
(b) For a precision-running fit, RC 3
Hole, in Shaft, in
+0.0009 -0.0014
-0.0000 -0.0023
With tolerances,
Hole inD
0000.0
0009.0
0000.4
−
+
=
Shaft inD
0009.0
0000.0
9986.3
−
+
=
Limit dimension,
Hole intoD 0009.40000.4=
Sketch

104
(c) For medium-running fit, RC5, RC 6. Say RC 6
Hole, in Shaft, in
+0.0022 -0.0030
-0.0000 -0.0052
With tolerances,
Hole inD
0000.0
0022.0
0000.4
−
+
=
Shaft inD
0022.0
0000.0
9970.3
−
+
=
Limit dimension,
Hole intoD 0022.40000.4=
Sketch
(d) For loose-running fit, RC 8, RC 9. Say RC 9
Hole, in Shaft, in
+0.0090 -0.0100
-0.0000 -0.0150

105
With tolerances,
Hole inD
0000.0
0090.0
0000.4
−
+
=
Shaft inD
0050.0
0000.0
9900.3
−
+
=
Limit dimension,
Hole intoD 0090.40000.4=
Sketch
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: inD 3=
(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)

106
(c) For maximum interference
D
Ei
s =
ksiE 000,23=
( )( ) ksis 6.14
3
0019.0000,23
==
For minimum interference
( )( ) ksis 4.5
3
0007.0000,23
==
93. The same as 92, except that the gear hub is C1035 steel and class of fit is FN 3.
Solution: inD 3=
(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)

107
(c) For C1035 steel, ksiE 000,30=
For maximum interference
D
Ei
s =
( )( ) ksis 37
3
0037.0000,30
==
For minimum interference
( )( ) ksis 18
3
0018.0000,30
==
94. 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
Therefore, inD
0000.0
0005.0
3775.1
−
+
=

108
Shaft, 1.3784 – 1.3775 = 0.0009 in
1.3779 – 1.3775 = 0.0004 in
inD
0005.0
0000.0
37845.1
−
+
=
Tolerances
Hole, in Shaft, in
+0.0005 +0.0009
-0.0000 +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 Shaft, in
+0.0005 +0.0003
-0.0000 +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

109
Tolerances
Hole, in Shaft, in
+0.0006 +0.0015
-0.0000 +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)
( ) inNST 0036.00006.066 ==== σ
(b)
inA 0228.0
2
0456.0
==

110
From Table 3.3
0.2=
σ
z
σ2=z
σ42 == zT
( ) inT 0024.00006.04 ==
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
( ) in
NS
pin 00025.0
6
0015.0
6
1
1 ===σ
( ) in
NS
hole 00033.0
6
0020.0
6
2
2 ===σ
( ) ( )222
2
2
1
2
00033.000025.0 +=+= σσσD
inD 00041.0=σ
Natural Spread of Difference = ( ) inD 00246.000041.066 ==σ

111
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:
( ) in
NS
rod 00025.0
6
0015.0
6
1
1 ===σ
( ) in
NS
pin 00033.0
6
0020.0
6
2
2 ===σ
( ) ( )222
2
2
1
2
00033.000025.0 +=+= σσσD
inD 00041.0=σ

112
( ) inNS D 00246.000041.066 === σ
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) DNS σ6=
in
NS
D 00033.0
6
002.0
6
===σ
(b) σσ 2=D
inD
000233.0
2
00033.0
2
===
σ
σ

113
( ) inNS 0014.0000233.066 === σ
Tolerance = 0.0014 in
Corresponing allowance = 0.0016 in
(c) From Fig. 3.4, NST >
Tolerance = ( ) inNST 0018.00014.03.13.1 ===
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.

114
Solution:
From Table RC 6, inD 4=
Hole Shaft
+0.0022 -0.0030
-0.0000 -0.0052
or
( ) inholeD
0000.0
0022.0
000.4
−
+
=
( ) inshaftD
0022.0
0000.0
997.3
−
+
=
σ6=NS
in00028.0
6
00165.0
==σ
( ) inD 0004.000028.022 === σσ
( ) inNS DD 0024.00004.066 === σ

115
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, inD 4=
Hole Shaft
+0.0009 -0.0014
-0.0000 -0.0023
or
( ) inholeD
0000.0
0009.0
000.4
−
+
=
( ) inshaftD
0009.0
0000.0
9986.3
−
+
=

116
σ6=NS
in0001125.0
6
000675.0
==σ
( ) inD 00016.00001125.022 === σσ
( ) inNS DD 00096.000016.066 === σ
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 ( 21 σσ = ).
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 σ3± limits
so that virtually no scrap is manufactured.
Solution:

117
inD 00067.0=σ
22
2
2
1 Dσσσ =+
( )22
1 00067.02 =σ
in00047.02 =σ
170.0 NSNS =
( )( ) inNS 00197.000047.0670.0 ==
( )( ) inNST 00282.000047.061 ===

118
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?

119
Problem 103.
Solution:
inT 004.0000.4004.41 =−=
inT 008.0000.5008.52 =−=
inT 007.0700.6707.63 =−=
( ) ( ) ( )2222
3
2
2
2
1
2
007.0008.0004.0 ++=++= TTTT
inT 0114.0=
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

120
Processes:
1. Contour sawing
2. Rough grinding
4. Drilling
6. Finish turning
7. Broaching
8. Boring
9. Reaming
10. Commercial grinding
11. Barrel finishing
(c) 0.005 in
Processes:
2. Drilling
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
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

121
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
8. Superfinishing
(f) 0.00005 in
Surface finishes = 0 to 2 µin.
Processes:
1. Honing
3. Superfinishing
105. If it cost $100 to finish a certain surface to 500 microinches rms, what would be
the approximate cost to finish it to the following roughness: (a) 125, (b) 32, (c) 8,
(d) 2 µin. rms?
Solution:
From Fig. 3.9
Relative cost of 500 µin rms = 1.75
(a) Relative cost of 125 µin rms = 3
Cost = 171$
75.1
3
100$ =






122
(b) Relative cost of 32 µin rms = 5
Cost = 286$
75.1
5
100$ =





(c) Relative cost of 8 µin rms = 7.75
Cost = 443$
75.1
75.7
100$ =





(d) Relative cost of 2 µin rms = 11.5
Cost = 657$
75.1
5.11
100$ =





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 ( 1221 FFLL = ), the force 1F , the length 1L , 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 1F 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 1F 1L AISI No. As
Rolled
Mech,
Advantage
106 700 12 C1020 1.5
107 650 14 C1020 2
108 600 15 C1022 2.5
109 550 18 C1035 3
110 500 20 C1040 4
111 800 12 C1020 1.5
112 750 14 C1020 2
113 750 14 C1020 2.5
114 650 18 C1035 3
115 600 20 C1040 4
116 900 12 C1020 1.5
117 850 14 C1020 2

123
118 800 15 C1022 2.5
119 750 18 C1035 3
120 700 20 C1040 4
121 1000 12 C1020 1.5
122 950 14 C1020 2
123 900 15 C1022 2.5
124 850 18 C1035 3
125 800 20 C1040 4
Problems 106 to 125.
Solution:
kiplbF 7.07001 ==
inL 121 =
5.1=MA
For AISI C1020 as rolled (Table AT 7)
ksisu 65=
ksisus 49=
Designing based on ultimate strength
6=N (Table 1.1) mild shock, one direction
ksi
N
s
s u
10
6
65
===
ksi
N
s
s us
s 8
6
49
===
Consider yoke connection A

124
2
1
1
2
1
1
2
1
4
1
2 d
F
d
F
ss
ππ
=






=
2
1
2
1
7.0
8
dπ
=
ind 24.01 =
say ind
4
1
1 =
11
1
da
F
s =






=
4
1
7.0
10
1a
ina 28.01 =
say ina
16
5
1 =
since 11 at <
say int
4
1
1 =
11
1
tD
F
s =






=
4
1
7.0
10
1D
inD 28.01 =
say inD
16
5
1 =
Consider yoke connection B.
5.1
2
1
=
L
L
inL 8
5.1
12
2 ==
5.1
1
2
=
F
F
( ) kipF 05.17.05.12 ==
2
2
2
2
2
2
2
1
4
1
2 d
F
d
F
ss
ππ
=






=

125
2
2
2
1
05.1
8
dπ
=
ind 29.02 =
say ind
16
5
2 =
22
2
da
F
s =






=
16
5
05.1
10
2a
ina 34.02 =
say ina
8
3
2 =
since 22 at <
say int
16
5
2 =
11
1
tD
F
s =






=
16
5
05.1
10
2D
inD 34.02 =
say inD
8
3
2 =
For shaft diameter
Assume torque, ( )( ) kipsinLFT −=== 4.81270.0111
3
116
s
s
d
T
s
π
=
( )
3
4.816
8
sdπ
=
ininds
4
3
175.1 ==
Tolerances and allowances, consider RC 4 (Table 3.1)
Hole Shaft
ind
0000.0
0006.0
2500.01
−
+
= ind
0006.0
0000.0
2495.01
−
+
=

126
ind
0000.0
0006.0
3125.02
−
+
= ind
0006.0
0000.0
3120.02
−
+
=
inds
0000.0
0010.0
7500.1
−
+
= inds
0010.0
0000.0
7490.1
−
+
=
Female Male
ina
0000.0
0006.0
3125.01
−
+
= ina
0006.0
0000.0
3120.01
−
+
=
ina
0000.0
0006.0
3750.02
−
+
= ina
0006.0
0000.0
3745.02
−
+
=
(b) For good proporion use the following dimension
inDD 121 ==
indd
4
3
21 ==
intt
4
3
21 ==
inaa 121 ==
inds 2=
Tolerances and allowances, consider RC 4 (Table 3.1)
Hole Shaft
ind
0000.0
0008.0
7500.01
−
+
= ind
0008.0
0000.0
7492.01
−
+
=
ind
0000.0
0008.0
7500.02
−
+
= ind
0008.0
0000.0
7492.02
−
+
=
inds
0000.0
0012.0
0000.2
−
+
= inds
0012.0
0000.0
9988.1
−
+
=
Female Male
ina
0000.0
0008.0
0000.11
−
+
= ina
0008.0
0000.0
9992.01
−
+
=
ina
0000.0
0008.0
0000.12
−
+
= ina
0008.0
0000.0
9992.02
−
+
=

127
(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:
5.2=N
(a) For AISI C1020, as rolled, Table AT 7
ksisy 48=
N
s
I
Mc
s y
==

128
ksis 19
5.2
48
==
( )( ) inlb
FL
M −=== 360
4
14410
4
Z
M
s =
Z = section modulus
3
95.18
19
360
in
s
M
Z ===
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 = 3
2.29 inZ =
4
8.145 inI = , moment of inertia
Size (Depth) = 10.0 in
Weight of beam = (35)(12) = 420 lb
(b)
EI
FL
48
3
=δ
ksiE 000,30=
( )( )
( )( )
in142.0
8.145000,3048
14410
3
==δ
(c)
EI
FL
48
3
=δ
( )( )
( )I000,3048
14410
25.0
3
=
4
9.82 inI =
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
4
8.145 inI = , moment of inertia
Size (Depth) = 10.0 in
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:

129
( )( )
inlbw 7.30
1218
6306000
=
+
=
Table B.3, From Strength of Materials, F.L. Singer, 2nd
Edition
For 10-in., 35-lb. I-beam
4
8.145 inI =
3
2.29 inZ =
Z
M
s max
max =
( )( ) inkipsinlb
wL
M −=−=== 179042,179
8
2167.30
8
22
max
ksis 13.6
2.29
179
max ==
EI
FL
384
5 3
max =δ
ksiE 000,30=
( )( ) kipslbswLF 631.666312167.30 ====
( )( )
( )( )
in20.0
8.145000,30384
216631.65
3
max ==δ
128. The same as 127, except the beam is a cantilever.
Solution:
( )( ) inkipsinlb
wL
M −=−=== 17.716170,716
2
2167.30
8
22
max
ksi
Z
M
s 53.24
2.29
17.716max
max ===

130
( )( )
inlbw 7.30
1218
6306000
=
+
=
( )( )
( )( )
in
EI
wL
91.1
8.145000,308
2167.30
8
44
max ===δ
129. Two equal angles, placed back to back as shown, act as a simple beam and are to
support a load of lbF 000,2= .; inL 40= .; ina 15= . 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
Z
M
s =
L
Fab
M =
ina 15=
inL 40=
inaLb 251540 =−=−=
( )( )( ) inkipsM −== 75.18
40
25152
3
9375.0
20
75.18
in
s
M
Z ===
For each angles, ( ) 3
46875.09375.0
2
1
inZ ==
From Strength of Materials, F.L. Singer, 2nd
Edition
Table B-5
Say size 3” x 3”, thickness = ¼ in
3
58.0 in
c
I
Z ==
130. The same as 129, except that a rolled T-section is to be used.
Solution:
From Table AT 1, No. 6

SECTION 1– DESIGN FOR SIMPLE STRESSES

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SECTION 1– DESIGN FOR SIMPLE STRESSES