machine design by faires (section 1)
DESCRIPTION
Solution Manual Section 01TRANSCRIPT
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 1 of 64
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 61030×=
A
Fs
d=
where
lbF 8000=
bhA =
but
bh 5.1=
therefore 25.1 bA =
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
A
F
N
ss u
d==
25.1
8000
6
000,96
b=
inb 577.0= say in8
5.
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 2 of 64
inbh16
155.1 ==
(b) Based on yield strength
N = factor of safety = 3 for repeated but not reversed load (Table 1.1)
A
F
N
ss u
d==
25.1
8000
3
000,59
b=
inb 521.0= say in16
9.
inbh32
275.1 ==
(c) Elongation = AE
FL=δ
where,
in005.0=δ
lbF 8000=
psiE 61030×=
inL 15= 25.1 bA =
then,
AE
FL=δ
( )( )( )( )62
10305.1
158000005.0
×=
b
inb 730.0= say in4
3.
inbh8
115.1 ==
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 61025×=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 3 of 64
A
Fs
d=
where
lbF 8000=
bhA =
but
bh 5.1=
therefore 25.1 bA =
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
A
F
N
ss u
d==
25.1
8000
6
000,55
b=
inb 763.0= say in8
7.
inbh16
515.1 ==
(b) Based on yield strength
N = factor of safety = 3 for repeated but not reversed load (Table 1.1)
A
F
N
ss u
d==
25.1
8000
3
500,36
b=
inb 622.0= say in16
11.
inbh32
115.1 ==
(c) Elongation = AE
FL=δ
where,
in005.0=δ
lbF 8000=
psiE 61025×=
inL 15= 25.1 bA =
then,
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 4 of 64
AE
FL=δ
( )( )( )( )62
10255.1
158000005.0
×=
b
inb 8.0= say in8
7.
inbh16
515.1 ==
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 6105.14 ×=
Note: since there is no ys for brittle materials. Solve only for (a) and (c)
A
Fs
d=
where
lbF 8000=
bhA =
but
bh 5.1=
therefore 25.1 bA =
(a) Based on ultimate strength
N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1)
A
F
N
ss u
d==
25.1
8000
5.7
000,30
b=
inb 1547.1= say in16
31 .
inbh32
2515.1 ==
(c) Elongation = AE
FL=δ
where,
in005.0=δ
lbF 8000=
psiE 6105.14 ×=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 5 of 64
inL 15= 25.1 bA =
then,
AE
FL=δ
( )( )( )( )62
105.145.1
158000005.0
×=
b
inb 050.1= say in16
11 .
inbh32
1915.1 ==
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 27.39270,39125204
2====
π
From Table 1.1, page 20
8=u
N
4=yN
(a) Based on ultimate strength
u
u
s
FNA =
( )( )5.152
27.398
4
2 =dπ
ind 62.1= say in8
51
(b) Based on yield strength
y
y
s
FNA =
( )( )5.132
27.394
4
2 =dπ
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 6 of 64
ind 23.1= say in4
11
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=u
N
kipsF 1500=
( ) ( )4
34
44
22222 iiiio
DDDDDA
πππ=−=−=
( )( )65
15008
4
32
===u
ui
s
FNDA
π
inDi 85.8= say in8
78
inDD io4
317
8
7822 =
==
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
34
44
22222 iiiio
DDDDDA
πππ=−=−=
kipstonsF 5025 ==
(a) Based on yield strength
( )( )114
502
4
32
===y
yi
s
FNDA
π
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 7 of 64
inDi 61.0= say in8
5
inDD io4
11
8
522 =
==
(b) Based on ultimate strength
( )( )127
504
4
32
===u
ui
s
FNDA
π
inDi 82.0= say in8
7
inDD io4
31
8
722 =
==
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
FLA
δ=
where,
lbF 7000=
inL 55=
in030.0=δ
psiE6
1030×=
( )( )( )( )6
2
1030030.0
557000
4 ×== dA
π
ind 74.0= say in4
3
(b) For gradually applied and repeated (not reversed) load
3=yN
( )( )
( )psi
A
FNs
y
y 534,47
75.04
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
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 8 of 64
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=
rg
WF 2ω=
where
lbW 332.0= 22.32 fpsg =
( )sec1047
60
000,102
60
2rad
n===
ππω
inr 12=
( ) ( ) kipslbrg
WF 3.11300,1111047
2.32
332.0 22 ==== ω
From Table 1.1, page 20
4~3=N , say 4
u
u
s
FNA =
( )( )48
3.114
42
2 =
d
π for double shear
ind 774.0= say in32
25
CHECK PROBLEMS
9. The link shown is made of AISIC1020 annealed steel, with inb4
3= and
inh2
11= . (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.
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 9 of 64
Solution:
For AISI C1020 annealed steel, from Table AT 7
ksisu 57=
ksisy 42=
(a) AsF u=
2125.1
2
11
4
3inbhA =
==
( )( ) kipsF 64125.157 ==
(b) u
u
N
AsF =
4=uN
2125.1
2
11
4
3inbhA =
==
( )( )kipsF 16
4
125.157==
(c) y
y
N
AsF =
5.2=yN
2125.1
2
11
4
3inbhA =
==
( )( )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
sL0096.0
000,30
56.57===δ
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 10 of 64
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
2209.08
3
4
12 inA =
= π
Asrg
WF us== 2ω
( ) ( )( )2209.0000,62142.32
4 2 =ω
sec74.88 rad=ω
74.8860
2==
nπω
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 ===
24418.0
16
3
4
3intdA =
== ππ
kipstonsF 12060 ==
n = number of holes
( )( )5
604415.0
120===
usAs
Fn 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=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 11 of 64
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
3thI = , moment of inertial
but th 3=
36
4hI =
Moment Diagram (Load Upward)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 12 of 64
Based on ultimate strength
u
u
N
ss =
(a) I
Fac
I
Mcs ==
2
hc =
kipslbsF 22000 ==
( )( )
==
36
2132
8
654h
h
s
inh 86.3=
inh
t 29.13
86.3
3===
say
ininh2
145.4 ==
inint2
115.1 ==
(b) I
Fbc
I
Mcs ==
2
hc =
kipslbsF 22000 ==
( )( )
==
36
242
8
654h
h
s
inh 61.2=
inh
t 87.03
61.2
3===
say
inh 3=
int 1=
(c)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 13 of 64
413
35.4
4
3
−
−=
− h
inh 33.2=
413
15.1
4
1
−
−=
− t
int 78.0=
say
inh 625.2= or inh8
52=
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 61030×=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 14 of 64
From Table AT 2
Max. ( )( )
inkipsFL
M −=== 544
544
4
12
3bhI =
but bh 3=
36
4hI =
(a) I
Mc
N
ss
y
y==
2
hc =
( )
=
36
254
3
404h
h
inh 18.4=
inh
b 39.13
18.4
3===
say inh2
14= , inin
hb
2
115.1
3
5.4
3====
(b) ( )( )
( ) ( )( )in
EI
FL0384.0
12
5.45.1103048
544000
48 3
6
33
=
×
==δ
(c)
=
3648
4
3
hE
FLδ
( )( ) ( )( )( )46
3
103048
3654400003.0
h×=
inh 79.4=
inh
b 60.13
79.4
3===
say ininh4
1525.5 == , inin
hb
4
3175.1
3
25.5
3====
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 15 of 64
16. The same as 15, except that the beam is to have a circular cross section.
Solution:
(a) I
Mc
N
ss
y
y==
64
4dI
π=
2
dc =
34
32
64
2
d
M
d
dM
sππ
=
=
( )3
5432
3
40
dπ=
ind 46.3=
say ind2
13=
(b) EI
FL
48
3
=δ
64
4dI
π=
( )( )( )
( )( )( )in
dE
FL0594.0
5.3103048
54400064
48
6446
3
4
3
=×
==ππ
δ
(c) ( )4
3
48
64
dE
FL
πδ =
( )( )( )( ) 46
3
103048
5440006403.0
dπ×=
ind 15.4=
say ind4
14=
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.”
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 16 of 64
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
FLM −=== 72
4
486
4
I
Mcs =
2
hc =
12
3bhI =
but 2
hb =
24
4hI =
34
12
24
2
h
M
h
hM
s =
=
(a) Based on ultimate strength
3
12
h
M
N
ss
u
u ==
( )3
7212
4
65
h=
inh 76.3=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 17 of 64
inh
b 88.12
76.3
2===
say ininh4
3375.3 == , inin
hb
8
71875.1
2
75.3
2====
(b) Based on yield strength
3
12
h
M
N
ss
y
y==
( )3
7212
2
48
h=
inh 30.3=
inh
b 65.12
30.3
2===
say ininh2
135.3 == , inin
hb
4
3175.1
2
5.3
2====
(c) Limit design (Eq. 1.6)
4
2bhsM y=
( )4
24872
2hh
=
inh 29.2=
inh
b 145.12
29.2
2===
say ininh2
125.2 == , inin
hb
4
1125.1
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.
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 18 of 64
Problems18, 19.
Solution:
lbRRFF 30002121 ====
Moment Diagram
( )( ) inkipsinlbsaRM −=−=== 99000330001
N = factor of safety = 4 based on us
12
3bhI =
2
hc =
3612
34
3
hh
h
I =
=
(a) For gray cast iron, SAE 111
ksisu 30= , Table AT 6
34
18
36
2
h
M
h
hM
I
Mc
N
ss u =
===
( )3
918
4
30
hs ==
inh 78.2=
inh
b 93.03
78.2
3===
say inh 5.3= , inb 1=
(b) For malleable cast iron, ASTM A47-52, grade 35 018
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 19 of 64
ksisu 55= , Table AT 6
34
18
36
2
h
M
h
hM
I
Mc
N
ss u =
===
( )3
918
4
55
hs ==
inh 28.2=
inh
b 76.03
28.2
3===
say inh4
12= , inb
4
3=
(c) For AISI C1040, as rolled
ksisu 90= , Fig. AF 1
34
18
36
2
h
M
h
hM
I
Mc
N
ss u =
===
( )3
918
4
90
hs ==
inh 93.1=
inh
b 64.03
93.1
3===
say inh8
71= , inb
8
5=
19. The same as 18, except that 1F acts up ( 2F acts down).
Solution:
[ ]∑ = 0AM lbRR 187521 ==
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 20 of 64
Shear Diagram
Moment Diagram
=M maximum moment = 5625 lb-in = 5.625 kips-in
(a) For gray cast iron
3
18
h
M
N
ss u ==
( )3
625.518
4
30
h=
inh 38.2=
inh
b 79.03
38.2
3===
say inh4
12= , inb
4
3=
(b) For malleable cast iron
3
18
h
M
N
ss u ==
( )3
625.518
4
55
h=
inh 95.1=
inh
b 65.03
95.1
3===
say inh8
71= , inb
8
5=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 21 of 64
(c) For AISI C1040, as rolled
3
18
h
M
N
ss u ==
( )3
625.518
4
90
h=
inh 65.1=
inh
b 55.03
65.1
3===
say inh2
11= , 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:
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 22 of 64
[ ]∑ = 0AM ( )2500133 =B
R
lbRB
833,10=
[ ]∑ = 0BM ( )2500103 =A
R
lbRA
8333=
Shear Diagram
Moment Diagram
( )( ) inkipsinlbM −=−== 25000,25102500
bh 3=
12
3bhI =
36
4hI =
2
hc =
34
18
36
2
h
M
h
hM
I
Mcs =
==
(a) For gray cast iron, SAE 110
ksisu
20= , Table AT 6
6~5=N , say 6 for cast iron, dead load
3
18
h
M
N
ss u ==
( )3
2518
6
20
h=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 23 of 64
inh 13.5=
inh
b 71.13
==
say inh4
15= , inb
4
31=
(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
ss u ==
( )3
2518
4
52
h=
inh 26.3=
inh
b 09.13
==
say inh4
33= , inb
4
11=
(c) For AISI C1035, as rolled
ksisu
85= , ksisy 55=
4=N , based on ultimate strength
3
18
h
M
N
ss u ==
( )3
2518
4
85
h=
inh 77.2=
inh
b 92.03
==
say inh 3= , inb 1=
(d) For AISI C1035, as rolled
ksissu
64=
4=N , kipsRB
833.10=
A
R
N
ss Bsu
s==
22
242 DDA
ππ=
=
2
2
833.10
4
64
D
ss π
==
inD 657.0=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 24 of 64
say inD16
11=
(e) Limit Design
4
2bhsM y=
For AISI C1035 steel, ksisy 55=
3
hb =
( )4
35525
2hh
M
==
inh 76.1=
inh
b 59.03
==
say ininh8
71875.1 == , inb
8
5=
21. The same as 20, except that o30=θ . Pin B takes all the horizontal thrust.
Solution:
θcosFF
V=
[ ]∑ = 0AM VB
FR 133 =
( ) 30cos2500133 =B
R
lbRB
9382=
[ ]∑ = 0BM VA
FR 103 =
( ) 30cos2500103 =A
R
lbRA
7217=
Shear Diagram
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 25 of 64
Moment Diagram
( )( ) inkipsinlbM −=−== 65.21650,21102165
3
18
h
Ms =
(a) For gray cast iron, SAE 110
ksisu
20= , Table AT 6
6~5=N , say 6 for cast iron, dead load
3
18
h
M
N
ss u ==
( )3
65.2118
6
20
h=
inh 89.4=
inh
b 63.13
==
say inh4
15= , inb
4
31=
(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
ss u ==
( )3
65.2118
4
52
h=
inh 11.3=
inh
b 04.13
==
say inh 3= , inb 1=
(c) For AISI C1035, as rolled
ksisu
85= , ksisy 55=
4=N , based on ultimate strength
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 26 of 64
3
18
h
M
N
ss u ==
( )3
65.2118
4
85
h=
inh 64.2=
inh
b 88.03
==
say inh8
52= , inb
8
7=
(d) For AISI C1035, as rolled
ksissu
64=
4=N , lbRBV
9382=
lbFFRHBH
125030sin2500sin ==== θ
( ) ( )22222 12509382 +=+=BHBVB
RRR
lbRB
9465=
A
R
N
ss Bsu
s==
22
242 DDA
ππ=
=
2
2
465.9
4
64
D
ss π
==
inD 614.0=
say inD8
5=
(e) Limit Design
4
2bhsM y=
For AISI C1035 steel, ksisy 55=
3
hb =
( )4
35565.21
2hh
M
==
inh 68.1=
inh
b 56.03
==
say ininh8
71875.1 == , inb
8
5=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 27 of 64
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
Mcs =
Solving for I , moment of inertia
( )( ) ( )( ) ( )( ) ( )( )[ ]yaaaaa
aaa
aa 332
53
23 +=
+
2
3ay =
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 28 of 64
( )( )( )( )( ) ( )( )
( )( )( )2
173
12
33
12
3 42
3
2
3a
aaaaa
aaaaa
I =+++=
(a)
2
3ac
t=
2
5ac
c=
Based on tension
I
Mc
N
ss tu
t==
( )
=
2
17
2
370
8
504a
a
ina 255.1=
Based on compression
I
Mc
N
ss cuc
c==
( )
=
2
17
2
570
8
1644a
a
ina 001.1=
Therefore ina 255.1=
Or say ina4
11=
And ( ) inacb 75.325.133 ====
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 29 of 64
Or incb4
33==
(b) If the top of the tee is below
2
5ac
t=
2
3ac
c=
2
17 4aI =
inkipsM −= 70
Based on tension
I
Mc
N
ss tu
t==
( )
=
2
17
2
570
8
504a
a
ina 488.1=
Based on compression
I
Mc
N
ss cuc
c==
( )
=
2
17
2
370
8
1644a
a
ina 845.0=
Therefore ina 488.1=
Or say ina2
11=
And inacb2
143 ===
CHECK PROBLEMS
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 30 of 64
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= ; 49.2 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
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 31 of 64
19FM = = maximum moment
For AISI C1020, as rolled
ksisy 48=
(a) I
Mcs
y=
where ind
c 5.12
3
2===
( )( )9.2
5.1948 1F
sy
==
kipsF 31.101 =
kipsFF 62.202 12 ==
(b) I
Mc
N
ss
y==
( )( )9.2
5.19
3
48 1Fs ==
kipsF 44.31 =
kipsFF 88.62 12 ==
(c) 1596.9509.2
25<==
b
L (page 34)
ksisc
20= ( page 34, i1.24)
I
Mcs
c=
( )( )9.2
5.1920 1F
=
kipsF 30.41 =
kipsFF 60.82 12 ==
(d) For maximum deflection,
by method of superposition, Table AT 2
( ) 2
3
max33
′+′=
bLa
EIL
bFy , ba ′>
or
( ) 2
3
max33
+′=
aLb
EIL
Fay , ab >′
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 32 of 64
maxy caused by 1F
( ) 2
3
1111max
331
+′=
aLb
EIL
aFy , 11 ab >′
where ksiE 000,30=
ina 51 =
inb 201 =′
inL 25=
49.2 inI =
( )( )( )( )
( )1
2
3
1max 0022.0
3
52520
259.2000,303
51
FF
y =
+=
maxy caused by 2F
( ) 2
3
2222max
332
′+′=
bLa
EIL
bFy , 22 ba ′>
where inb 52 =′
ina 202 =
( )( )( )( )
( )1
2
3
1max 0043.0
3
52520
259.2000,303
522
FF
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
Mcs
y=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 33 of 64
( )( )9.2
5.1947 1F
sy
==
kipsF 10.101 =
kipsFF 20.202 12 ==
(b) I
Mc
N
ss
y==
( )( )9.2
5.19
3
47 1Fs ==
kipsF 36.31 =
kipsFF 72.62 12 ==
(c) 1596.9509.2
25<==
b
L (page 34)
ksisc
20= ( page 34, i1.24)
I
Mcs
c=
( )( )9.2
5.1920 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 40.6 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.
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 34 of 64
Solution:
(a) For C1020, as rolled, ksisu
65=
Consider flange buckling
3066.2
80==
b
L
since 4015 <<b
L
( )ksi
b
Ls
c15
1800
301
5.22
18001
5.2222
=
+
=
+
=
I
Mcs =
ind
c 22
4
2===
From Table AT 2
( )F
FFLM 20
4
80
4===
I
Mcss
c==
( )( )6
22015
F=
kipsF 25.2= , safe load
(b) Considering stress owing to the weight of the beam
add’l 8
2wLM = (Table AT 2)
where ftlbw 7.7=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 35 of 64
add’l ( )
inkipsinlbwL
M −=−=
== 513.0513
8
80
12
7.7
8
22
513.020 += FM = total moment
I
Mcss
c==
( )( )6
2513.02015
+=
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:
int
c 01325.00265.02
===
inr 76325.1301325.075.13 =+=
Using Eq. (1.4) page 11 (Text)
r
Ecs =
where psiE 61030×=
( )( )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:
For AISI C1020, as rolled
ksisu
65= (Table AT 7)
Designing based on ultimate strength,
6=N , for repeated, minor shock load
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 36 of 64
ksiN
ss u 8.10
6
65===
Loading Diagram
x
h 1
30
14 −=
−
110.0 += xh
12
3whI =
2
hc =
FxM =
( )
( )2223 110.0
33
2
6
12
2
+===
==x
Fx
h
Fx
h
Fx
wh
hFx
I
Mcs
Differentiating with respect to x then equate to zero to solve for x giving maximum
stress.
( ) ( ) ( )( )( )( )
0110.0
10.0110.021110.03
4
2
=
+
+−+=
x
xxxF
dx
ds
( ) 010.02110.0 =−+ xx
inx 10=
( ) inh 211010.0 =+=
2
3
h
Fx
N
ss u ==
( )( )22
1038.10
F=
kipsF 44.1=
TORSIONAL STRESSES
DESIGN PROBLEMS
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 37 of 64
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
ss us= , 6=N (Table 1.1)
ksis 126
72==
Torque, ( )
( )kipsinlbinlbft
n
hpT −=−=−=== 540.054045
17502
15000,33
2
000,33
ππ
For diameter,
3
16
d
Ts
π=
( )3
540.01612
dπ=
ind 612.0=
say ind8
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:
( )( )
kipsinlbftn
hpT −=−=== 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
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 38 of 64
Design stress
ksiN
ss
ys10
3
30===
(a) Let D = shaft diameter
J
Tcs =
32
4DJ
π=
2
Dc =
3
16
D
Ts
π=
( )3
2761610
Dπ=
inD 20.5=
say inD4
15=
(b) ( ) ( )[ ]
32
15
32
2
32
44444
iiiioDDDDD
Jπππ
=−
=−
=
iio D
DDc ===
2
2
2
34 15
32
32
15 ii
i
D
T
D
TDs
ππ=
=
( )3
15
2763210
iDπ=
inDi
66.2=
inDDio
32.52 ==
say
inDi
8
52=
inDo
4
15=
(c) Density, 3284.0 inlb=ρ (Table AT 7)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 39 of 64
For solid shaft
=w weight per foot of length
( )( ) ftlbDDw 8.7325.5284.0334
12222 ===
= ππρ
πρ
For hollow shaft
( ) ( ) ( ) ( ) ( )[ ] ftlbDDDDw ioio 3.55625.225.5284.0334
12222222 =−=−=−
= ππρ
πρ
Therefore hollow shaft is lighter
Percentage lightness = ( ) %5.33%1003.55
3.558.73=
−
(d) Torsional Deflection
JG
TL=θ
where
inftL 12010 ==
ksiG 3105.11 ×=
For solid shaft, 32
4DJ
π=
( )( )
( ) ( )( ) o2.2
180039.0039.0
105.1125.532
120276
34
=
==
×
=
ππθ rad
For hollow shaft, ( )
32
44
ioDD
J−
=π
( )( )
( ) ( )[ ]( )( ) o4.2
180041.0041.0
105.11625.225.532
120276
344
=
==
×−
=
ππθ 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:
( )( )
kipsinlbftn
hpT −=−=== 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
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 40 of 64
3=N for mild shock
Design stress
ksiN
ss
ys26
3
78===
(a) Let D = shaft diameter
J
Tcs =
32
4DJ
π=
2
Dc =
3
16
D
Ts
π=
( )3
2761626
Dπ=
inD 78.3=
say inD4
33=
(b) ( ) ( )[ ]
32
15
32
2
32
44444
iiiioDDDDD
Jπππ
=−
=−
=
iio D
DDc ===
2
2
2
34 15
32
32
15 ii
i
D
T
D
TDs
ππ=
=
( )3
15
2763226
iDπ=
inDi
93.1=
inDDio
86.32 ==
say
inDi
2=
inDo
4=
(c) Density, 3284.0 inlb=ρ (Table AT 7)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 41 of 64
For solid shaft
=w weight per foot of length
( )( ) ftlbDDw 6.3775.3284.0334
12222 ===
= ππρ
πρ
For hollow shaft
( ) ( ) ( ) ( ) ( )[ ] ftlbDDDDw ioio 1.3224284.0334
12222222 =−=−=−
= ππρ
πρ
Therefore hollow shaft is lighter
Percentage lightness = ( ) %1.17%1001.32
1.326.37=
−
(d) Torsional Deflection
JG
TL=θ
where
inftL 12010 ==
ksiG 3105.11 ×=
For solid shaft, 32
4DJ
π=
( )( )
( ) ( )( ) o48.8
180148.0148.0
105.1175.332
120276
34
=
==
×
=
ππθ rad
For hollow shaft, ( )
32
44
ioDD
J−
=π
( )( )
( ) ( )[ ]( )( ) o99.6
180122.0122.0
105.112432
120276
344
=
==
×−
=
ππθ 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
hpT −=−=== 04.5420
5002
40000,33
2
000,33
ππ
ksiG 3105.11 ×=
DL 20=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 42 of 64
rad180
1π
θ == o
JG
TL=θ
( )( )
( )34
105.1132
2004.5
180×
=
D
D
π
π
inD 72.1=
say inD4
31=
(b) ( )
( )ksi
D
Ts 8.4
75.1
04.5161633
===ππ
Based on yield strength
3=N
( )( ) ksiNssys 4.148.43 ===
ksis
sys
y 246.0
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
ss
ys
′==
where, bh =
9
2
9
2 32 bhbZ ==′ (Table AT 1)
kipsinlbinT −=−= 2.11200
( )32
92.1
3
45
bs ==
inhb 71.0==
say inhb4
3==
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 43 of 64
CHECK PROBLEMS
33. A punch press is designed to exert a force sufficient to shear a 15/16-in. hole in a
½-in. steel plate, AISI C1020, as rolled. This force is exerted on the shaft at a
radius of ¾-in. (a) Compute the torsional stress in the 3.5-in. shaft (bending
neglected). (b) What will be the corresponding design factor if the shaft is made
of cold-rolled AISI 1035 steel (Table AT 10)? Considering the shock loading that
is characteristics of this machine, do you thick the design is safe enough?
Solution:
For AISI C1020, as rolled
ksisus
49=
( )DtsFus
π=
where inD16
15=
int2
1=
( ) kipsF 2.722
1
16
1549 =
= π
FrT =
where inr4
3=
( ) kipsinT −=
= 2.54
4
32.72
(a) 3
16
d
Ts
π=
where ind 5.3=
( )( )
ksis 44.65.3
2.54163
==π
(b) For AISI 1035 steel, ksisus
64=
for shock loading, traditional factor of safety, 15~10=N
Design factor , 94.944.6
64===
s
sN us , the design is safe ( 10≈N )
34. The same as 33, except that the shaft diameter is 2 ¾ in.
Solution:
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 44 of 64
ind 75.2=
(a) 3
16
d
Ts
π=
( )( )
ksis 3.1375.2
2.54163
==π
(b) For AISI 1035 steel, ksisus
64=
for shock loading, traditional factor of safety, 15~10=N
Design factor , 8.43.13
64===
s
sN 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
Tcs =
( ) ( ) ( )[ ] 4
4444
308632
5.65.13
32in
DDJ io =
−=
−=
ππ
inD
c o 75.62
5.13
2===
( )( )
kipsinlbftn
hpT −=−=== 3152606,262
2002
000,10000,33
2
000,33
ππ
( )( )ksis 9.6
3086
75.63152==
(b) Factor of safety,
2.149.6
98===
s
sN us , not risky
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 45 of 64
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)
22
4
4
1 d
F
d
Fs
ππ==
Equation (1) s
Fd
π
4=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 46 of 64
( ) ( ) ( )[ ] ( )22222
1
22
1
2 44.1
4
2.1
44
4
1 dD
F
dD
F
DD
F
DD
Fs
−=
−=
−=
−
=ππππ
Equation (2) 244.1
4d
s
FD +=
π
dh
F
hD
Fs
sππ 2.11
==
Equation (3) sds
Fh
π2.1=
Dt
Fs
sπ
=
Equation (4) sDs
Ft
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
ksiN
ss u 16
4
65===
ksiN
ss us
s12
4
49===
kipslbF 12000,12 ==
From Equation (1)
( )( )
ins
Fd 98.0
16
1244===
ππ
From Equation (2)
( )( )
( ) inds
FD 53.198.044.1
16
12444.1
4 22 =+=+=ππ
From Equation (3)
( )( )in
ds
Fh
s
27.01298.02.1
12
2.1===
ππ
From Equation (4)
( )( )in
Ds
Ft
s
21.01253.1
12===
ππ
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 47 of 64
(c) Standard fractional dimensions
ind 1=
inD2
11=
inh4
1=
int4
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.
ksiN
ss u 4
15
65===
ksiN
ss us
s3
15
49===
kipslbF 44000 ==
From Equation (1)
( )( )
ins
Fd 13.1
4
444===
ππ
From Equation (2)
( )( )
( ) inds
FD 76.113.144.1
4
4444.1
4 22 =+=+=ππ
From Equation (3)
( )( )in
ds
Fh
s
31.0313.12.1
4
2.1===
ππ
From Equation (4)
( )( )in
Ds
Ft
s
24.0376.1
4===
ππ
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 48 of 64
(c) Standard fractional dimensions
ind8
11=
inD4
31=
inh8
3=
int4
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
1d
Fs
π=
sdF 2
4
1π=
(1)
=
N
sdF u2
4
1π
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 49 of 64
( ) ( )222
1
2 44.14
1
4
1dD
F
DD
Fs
−
=
−
=
ππ
( )sdDF 22 44.14
1−= π
(2) ( )
−=
N
sdDF u22
44.14
1π
dh
F
hD
Fs
sππ 2.11
==
sdhsF π2.1=
=
=
N
sdh
N
sdhF uus 75.0
2.12.1 ππ
(3)
=
N
sdhF u5
9.0 π
Dt
Fs
sπ
=
sDtsF π=
=
=
N
sDt
N
sDtF uus 75.0
ππ
(4)
=
N
sDtF uπ75.0
Equate (2) and (1)
( )
=
−=
N
sd
N
sdDF uu 222
4
144.1
4
1ππ
22 44.2 dD =
dD 562.1=
Equate (3) and (1)
=
=
N
sd
N
sdhF uu 2
4
19.0 ππ
( )d
dh 278.0
9.04==
Equate (4) and (1)
=
=
N
sd
N
sDtF uu 2
4
175.0 ππ
( )( )
=
=
N
sd
N
stdF uu 2
4
1562.175.0 ππ
( )( )d
dt 214.0
562.175.04==
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 50 of 64
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
15 D
Fs
s
π
Equation (1) s
s
FD
π5
4=
( )Dbt
Fs
2−=
Equation (2) Dts
Fb 2+=
Dt
Fs
5=
Equation (3) Ds
Ft
5=
(b) For AISI C1020, as rolled
ksisy 48= (Table AT 7)
ksiss yys 286.0 ==
4=N for repeated and reversed load (mild shock) based on yield strength
ksis 124
48==
ksiss
74
28==
From Equation (1)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 51 of 64
ss
FD
π5
4=
where
kipslbF 5.22500 ==
( )( )
ins
FD
s
30.075
5.24
5
4===
ππ say in
16
5
From Equation (3)
( )in
Ds
Ft 13.0
1216
55
5.2
5=
== say in
32
5
From Equation (2)
( )inD
ts
Fb 96.1
16
52
1232
5
5.22 =
+
=+= 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
ksisy 47= (Table AT 3)
ksiss yys 2555.0 ==
4=N for repeated and reversed load (mild shock) based on yield strength
ksis 124
47==
ksiss
64
25==
From Equation (1)
ss
FD
π5
4=
where
kipslbF 5.22500 ==
( )( )
ins
FD
s
33.065
5.24
5
4===
ππ say in
8
3
From Equation (3)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 52 of 64
( )in
Ds
Ft 11.0
128
35
5.2
5=
== say in
8
1
From Equation (2)
( )inD
ts
Fb 42.2
8
32
128
1
5.22 =
+
=+= say in
2
12
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
Fs
P−
= or ( )Dbt
F
sP2
4
3
−= Equation (1)
( )24
14 2
=
D
Fs
sR
π
Equation (2)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 53 of 64
Dt
Fs
R4
= Equation (3)
(b) For AISI C1015, as rolled
ksisuR
61= , ksissuRusR
4575.0 ==
For AISI C1020, as rolled
ksisuP
65=
6=N , based on ultimate strength
ksiN
ss uP
P8.10
6
65===
ksiN
ss uR
R1.10
6
61===
ksiN
ss usR
sR5.7
6
45===
kipslbF 88000 ==
Solving for D
22 D
Fs
sRπ
=
( )in
s
FD
sR
412.05.72
8
2===
ππ say in
16
7
Solving for t
Dt
Fs
R4
=
( )in
Ds
Ft
R
453.0
1.1016
74
8
4=
== say in
2
1
Solving for b
Using ( )Dbt
Fs
P−
=
( )inD
ts
Fb
P
92.116
7
8.102
1
8=+
=+= say in2
Using ( )Dbt
F
sP2
4
3
−=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 54 of 64
( )
( )inD
ts
Fb
P
99.116
72
8.102
14
832
4
3=
+
=+= say in2
Therefore
inb 2=
inD16
7=
int2
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
Fs
ππ== Equation (1)
( )eDL
Fs
2−= Equation (2)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 55 of 64
eD
Fs
2
= Equation (3)
( ) ( )2
2
22
2
2
4
4
1 Da
F
Da
Fs
−=
−
=ππ
Equation (4)
eDD
F
eDD
Fs
2
2
22
2
2
4
4
4
1 −=
−
=ππ
Equation (5)
Shear Stresses
eb
Fs
s2
= Equation (6)
( )teDL
Fs
s+−
=22
Equation (7)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 56 of 64
at
Fs
sπ
= Equation (8)
mD
Fs
s
1π= Equation (9)
hD
Fs
s
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 875.08
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
ss
y
π==
( )2
1
104
7
55
Dπ=
inD 27.11 =
say inD4
111 =
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 57 of 64
From Equation (9)
mD
F
N
ss
sy
s
1
1
π==
m
=
4
11
10
7
33
π
inm 54.0=
say inm16
9=
From Equation (3)
eD
F
N
ss
y
2
1 ==
eDs
2
10
7
55==
273.12 =eD
From Equation (5)
eDD
F
N
ss
y
2
2
2 4
41
−==
π
( )( )273.14
104
7
552
2 −=
Dπ
inD 80.12 =
say inD4
312 =
and 273.12 =eD
273.14
31 =
e
ine 73.0=
say ine4
3=
By further adjustment
Say inD 22 = , ine8
5=
From Equation (8)
at
F
N
ss
sy
sπ
== 2
( )875.0
10
7
28
aπ=
ina 91.0=
say ina 1=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 58 of 64
From Equation (4)
( )2
2
2
42
Da
F
N
ss
y
−==
π
( )( )22
2
104
7
48
−=
aπ
ina 42.2=
say ina2
12=
use ina2
12=
From Equation (7)
( )teDL
F
N
ss
sy
s+−
==22
2
( )875.08
522
10
7
28
+−
=
L
inL 80.2=
say inL 3=
From Equation (6)
eb
F
N
ss
sy
s2
2 ==
b
=
8
52
10
7
28
inb 2=
From Equation (10)
hD
F
N
ss
sy
s
22
2 ==
( )h22
10
7
28=
ininh8
5625.0 ==
Summary of Dimensions
inL 3=
inh8
5=
inb 2=
int8
7=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 59 of 64
inm16
9=
ina2
12=
inD4
111 =
inD 22 =
ine8
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
inb0000.0
0030.00.2
−
+= inb
0058.0
0040.00.2
−
−=
allowance = 0.0040 in
inD0000.0
0030.00.22
−
+= inD
0058.0
0040.00.22
−
−=
allowance = 0.0040 in
ine0000.0
0016.0625.0
−
+= ine
0030.0
0020.0625.0
−
−=
allowance = 0.0020 in
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
=
For AISI C1020, as rolled
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 60 of 64
ksisu 652
=
ksisus 482
=
Ultimate strength
Use Equation (1)
( ) ( ) kipsDsF uu 8.6614
185
4
1 22
11=
=
= ππ
Equation (9)
mDsF usu 11π=
( )( )( )m1648.66 π=
inm 33.0=
say inm8
3=
From Equation (3)
eDsF uu 21=
( ) eD2858.66 =
7859.02 =eD
From Equation (5)
−= eDDsF uu 2
2
24
11
π
( )
−= 7859.0
4
1858.66
2
2Dπ
inD 42.12 =
say inD8
312 =
7859.08
312 =
= eeD
ine 57.0=
say ine16
9=
From Equation (4)
( )
−= 2
2
2
4
12
DasF uu π
( )
−
=
2
2
8
31
4
1658.66 aπ
ina 79.1=
say ina4
31=
From Equation (8)
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 61 of 64
atsF usu π2
=
( )( )( )( )1488.66 aπ=
ina 44.0=
say ina2
1=
use ina4
31=
From Equation (2)
( )eDLsF uu 22−=
( )
−=
16
9
8
31658.66 L
inL 20.3=
say inL4
13=
From Equation (7)
( )teDLsF usu −−= 222
( ) ( )116
9
8
314828.66
−−= L
inL 51.1=
say inL2
11=
use inL4
13=
From Equation (6)
ebsF usu 12=
( ) b
=
16
96428.66
inb 93.0=
say inb 1=
From Equation (10)
hDsF usu 212=
( ) h
=
8
316428.66
inh 38.0=
say inh8
3=
Dimensions
inL4
13=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 62 of 64
inh8
3=
inb 1=
int 1=
inm8
3=
ina4
31=
inD 11 =
inD8
312 =
ine16
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
inb0000.0
0035.00.1
−
+= inb
0065.0
0045.00.1
−
−=
allowance = 0.0045 in
inD0000.0
0040.0375.12
−
+= inD
0075.0
0050.0375.12
−
−=
allowance = 0.0050 in
ine0000.0
0028.05625.0
−
+= ine
0051.0
0035.05625.0
−
−=
allowance = 0.0035 in
45. Give all the simple strength equations for the connection shown. (b) Determine
the ratio of the dimensions a , b , c , d , m , and n to the dimension D so that the
connection will be equally strong in tension, shear, and compression. Base the
calculations on ultimate strengths and assume uus
ss 75.0= .
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 63 of 64
Problems 45 – 47.
Solution:
(a) Neglecting bending
Equation (1):
= 2
4
1DsF π
Equation (2):
= 2
4
12 csF s π
Equation (3): ( )bcsF 2=
Equation (4): ( )acsF =
Equation (5): ( )[ ]bcdsF −= 2
Equation (6): ( )mbsFs
4=
Equation (7): ( )nbsFs
2=
Equation (8): ( )acdsF −=
(b) N
ss u= and
N
ss us
s=
Therefore
sss
75.0=
Equate (2) and (1)
=
= 22
4
1
4
12 DscsF s ππ
=
22
4
1
2
175.0 Dscs
Dc 8165.0=
Equate (3) and (1)
( )
== 2
4
12 DsbcsF π
( ) 2
4
18165.02 DDb π=
Db 4810.0=
SECTION 1– DESIGN FOR SIMPLE STRESSES
Page 64 of 64
Equate (4) and (1)
== 2
4
1DssacF π
( ) 2
4
18165.0 DDa π=
Da 9619.0=
Equate (5) and (1)
( )[ ]
=−= 2
4
12 DsbcdsF π
( )( ) 2
4
14810.08165.02 DDd π=−
Dd 6329.1=
Equate (6) and (1)
( )
== 2
4
14 DsmbsF s π
( )( ) 2
4
14810.0475.0 DDm π=
Dm 5443.0=
Equate (7) and (1)
( )
== 2
4
12 DsnbsF s π
( )( ) 2
4
14810.0275.0 DDn π=
Dn 0886.1=
Equate (8) and (1)
( )
=−= 2
4
1DsacdsF π
( ) 2
4
18165.06329.1 DaDD π=−−
Da 9620.0=
Summary
Da 9620.0=
Db 4810.0=
Dc 8165.0=
Dd 6329.1=
Dm 5443.0=
Dn 0886.1=