Download - Problems in Mechanical Design
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 1 of 62
VARYING STRESSES – NO CONCENTRATION
DESIGN PROBLEMS
141. The maximum pressure of air in a 20-in. cylinder (double-acting air compressor)
is 125 psig. What should be the diameter of the piston rod if it is made of AISI
3140, OQT at 1000 F, and if there are no stress raisers and no column action? Let
75.1=N ; indefinite life desired. How does your answer compare with that
obtained for 4?
Solution:
For AISI 3140, OQT 1000 F
ksisu 153=
ksisy 134=
( ) ksiss un 5.761535.05.0 ===
For axial loading, with size factor
( )( )( ) ksiss un 525.7685.08.05.0 ===
Soderberg line
n
a
y
m
s
s
s
s
N+=
1
For double-acting
( ) ( ) kipslbpAFF 27.39270,39204
1252
max ==
===
π
kipsFF 27.39min −=−=
0=ms
( )222
5027.3944
ddd
Fsa ===
ππ
52
50
075.1
11 2
+==d
N
ind 2972.1=
say ind16
51=
comparative to Problem 4.
142. A link as shown is to be made of AISI 2330, WQT 1000 F. The load kipsF 5=
is repeated and reversed. For the time being, ignore stress concentrations. (a) If
its surface is machined, what should be its diameter for 40.1=N . (b) The same
as (a), except that the surface is mirror polished. What would be the percentage
saving in weight? (c) The same as (a), except that the surface is as forged.
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 2 of 62
Prob. 142 – 144
Solution:
For AISI 2330, WQT 1000 F
ksisu 105=
ksisy 85=
( ) ksiss un 5.521055.05.0 ===
0=ms
( )222
20544
ddd
Fsa
πππ===
Soderberg line
n
a
y
m
s
s
s
s
N+=
1
n
a
s
s
N+= 0
1
N
ss n
a =
Size factor = 0.85
Factor for axial loading = 0.80
(a) Machined surface
Surface factor = 0.85 (Fig. AF 5)
( )( )( )( ) ksiksiss un 345.305.5285.085.080.05.0 ===
4.1
345.30202
==D
saπ
inD 542.0=
say inD16
9=
(b) Mirror polished surface
Surface factor = 1.00 (Fig. AF 5)
( )( )( )( ) ksiksiss un 7.355.5200.185.080.05.0 ===
4.1
7.35202
==D
saπ
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 3 of 62
inD 5.0=
Savings in weight = ( ) %21%100
16
9
2
1
16
9
2
22
=
−
(c) As forged surface
Surface factor = 0.40 (Fig. AF 5)
( )( )( )( ) ksiksiss un 28.145.5240.085.080.05.0 ===
4.1
28.14202
==D
saπ
inD 79.0=
say inD4
3=
143. The same as 142, except that, because of a corrosive environment, the link is
made from cold-drawn silicon bronze B and the number of reversals of the load
is expected to be less than 3 x 107.
Solution:
For cold-drawn silicon bronze, Type B.
ksisn 30= at 3 x 108
ksisy 69=
ksisu 75.93=
ns at 3 x 107 ( ) ksi5.36
103
10330
085.0
7
8
=
×
×=
( )( )( ) ksisn 82.245.3685.080.0 ==
4.1
82.24202
==D
saπ
inD 60.0=
say inD8
5=
144. The same as 142, except that the link is made of aluminum alloy 2024-T4 with a
minimum life of 107 cycles.
Solution:
For AA 2024-T4
ksisy 47=
ksisu 68=
ksisn 20= at 5 x108
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 4 of 62
ns at 107 ( ) ksi9.27
10
10520
085.0
7
8
=
×
( )( )( ) ksisn 199.2785.080.0 ==
4.1
19202
==D
saπ
inD 685.0=
say inD16
11=
145. A shaft supported as a simple beam, 18 in. long, is made of carburized AISI 3120
steel (Table AT 10). With the shaft rotating, a steady load of 2000 lb. is appliled
midway between the bearings. The surfaces are ground. Indefinite life is desired
with 6.1=N based on endurance strength. What should be its diameter if there
are no surface discontinuities?
Solution:
For AISI 3120 steel, carburized
ksisn 90=
ksisy 100=
ksisu 141=
Size Factor = 0.85
Surface factor (ground) = 0.88
( )( )( ) ksisn 32.679088.085.0 ==
0=ms
3
32
D
Msa
π=
( )( )kipsinlbin
FLM −=−=== 0.99000
4
182000
4
Soderberg line
n
a
y
m
s
s
s
s
N+=
1
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 5 of 62
n
a
s
s
N+= 0
1
N
ss n
a =
( )6.1
32.679323
=Dπ
inD 2964.1=
say inD4
11=
146. (a) A lever as shown with a rectangular section is to be designed for indefinite
life and a reversed load of lbF 900= . Find the dimensions of a section without
discontinuity where tb 8.2= and inL 14= . for a design factor of 2=N . The
material is AISI C1020, as rolled, with an as-forged surface. (b) compute the
dimensions at a section where ine 4= .
Problems 146, 147
Solution:
For AISI C1020, as rolled
ksisu 65=
ksisy 48=
ksiss un 5.325.0 ==
Surface factor (as forged) = 0.55
(a) 0=ms
I
Mcsa =
( ) 4
33
8293.112
8.2
12t
tttbI ===
ttb
c 4.12
8.2
2===
( )( ) kipsinlbinFLM −=−=== 6.12600,1214900
( )( )34
643.9
8293.1
4.16.12
tt
tsa ==
( )( )( ) ksisn 20.155.3255.085.0 ==
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 6 of 62
Soderberg line
n
a
y
m
s
s
s
s
N+=
1
n
a
s
s
N+= 0
1
N
ss n
a =
2
20.15643.93
=t
int 08.1=
( ) intb 0.308.18.28.2 ===
say int16
11= , inb 0.3=
(b) ( )( ) kipsinlbinFeM −=−=== 6.3600,34900
( )( )34
755.2
18293
4.16.3
tt
tsa ==
2
20.15755.23
=t
int 713.0=
( ) intb 996.1713.08.28.2 ===
say int32
23= , inb 2=
147. The same as 146, except that the reversal of the load are not expected to exceed
105 (Table AT 10).
Solution:
ksisn 5.32=
ns at 105 ( ) ksi5.39
10
105.32
085.0
5
6
=
=
( )( )( ) ksisn 5.185.3955.085.0 ==
(a) N
ss n
a =
2
5.18643.93
=t
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 7 of 62
int 014.1=
( ) intb 839.2014.18.28.2 ===
say int 1= , inb16
132=
(b) N
ss n
a =
2
5.18755.23
=t
int 6678.0=
( ) intb 870.16678.08.28.2 ===
say int16
11= , inb
8
71=
148. A shaft is to be subjected to a maximum reversed torque of 15,000 in-lb. It is
machined from AISI 3140 steel, OQT 1000 F (Fig. AF 2). What should be its
diameter for 75.1=N ?
Solution:
For AISI 3140 steel, OQT 1000 F
ksisu 152=
ksisy 134=
ksiss un 765.0 ==
For machined surface,
Surface factor = 0.78
Size factor = 0.85
( )( )( )( ) ksisns 3.5313478.085.06.0 ==
( ) ksiss yys 4.801346.06.0 ===
ns
as
ys
ms
s
s
s
s
N+=
1
0=mss
3
16
D
Tsas
π=
kipsinT −= 15
( )33
2401516
DDsas
ππ==
ns
as
s
s
N+= 0
1
N
ss ns
as =
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 8 of 62
75.1
3.532403
=Dπ
inD 3587.1=
say inD8
31=
149. The same as 148, except that the shaft is hollow with the outside diameter twice
the inside diameter.
Solution:
io DD 2=
( )( )( )
( )[ ] 34444
32
2
2151616
iii
i
io
oas
DDD
D
DD
TDs
πππ=
−=
−=
N
ss ns
as =
75.1
3.53323
=iDπ
inDi 694.0=
say inDi16
11= , inDo
8
31=
150. The link shown is machined from AISI 1035 steel, as rolled, and subjected to a
repeated tensile load that varies from zero to 10 kips; bh 2= . (a) Determine these
dimensions for 40.1=N (Soderberg) at a section without stress concentration.
(b) How much would these dimensions be decreased if the surfaces of the link
were mirror polished?
Problems 150, 151, 158.
Solution:
For AISI 1035, steel as rolled
ksisu 85=
ksisy 55=
ksiss un 5.425.0 ==
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 9 of 62
( ) kipsFm 50102
1=+=
( ) kipsFa 50102
1=−=
22 3
10
5.1
5
bbbh
Fs m
m ===
22 3
10
5.1
5
bbbh
Fs a
a ===
(a) Soderberg line
n
a
y
m
s
s
s
s
N+=
1
For machined surface,
Factor = 0.88
Size factor = 0.85
( )( )( )( ) ksisn 4.255.4288.085.080.0 ==
( ) ( )4.253
10
553
10
40.1
122
bb+=
inb 5182.0=
say inb16
9=
inbh32
275.1 ==
(b) Mirror polished,
Factor = 1.00
Size factor = 0.85
( )( )( )( ) ksisn 9.285.4200.185.080.0 ==
( ) ( )9.283
10
553
10
40.1
122
bb+=
inb 4963.0=
say inb2
1=
inbh4
35.1 ==
151. The same as 150, except that the link operates in brine solution. (Note: The
corroding effect of the solution takes precedence over surface finish.)
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 10 of 62
Solution:
Table AT 10, in brine, AISI 1035,
ksisn 6.24=
ksisy 58=
( )( )( ) ksisn 73.166.2485.080.0 ==
( ) ( )73.163
10
553
10
40.1
122
bb+=
inb 60.0=
say inb8
5=
inbh16
155.1 ==
152. The simple beam shown, 30-in. long ( dLa ++= ), is made of AISI C1022 steel,
as rolled, left a forged. At ina 10= , .30001 lbF = is a dead load. At
ind 10= , .24002 lbF = is repeated, reversed load. For 5.1=N , indefinite life,
and bh 3= , determine b and h . (Ignore stress concentration).
Problem 152, 153
Solution:
For AISI C1022, as rolled
ksisu 72=
ksisy 52=
ksiss un 365.0 ==
For as forged surface
Figure AF 5, factor = 0.52
Size factor = 0.85
( )( )( ) ksisn 163652.085.0 ==
Loading:
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 11 of 62
∑ = 0AM
( ) ( ) 230240020300010 R=+
lbR 26002 =
∑ = 0VF
2121 FFRR +=+
2400300026001 +=+R
lbR 28001 =
Shear Diagram
( )( ) kipsinlbinMC −=−== 28000,28102800
1
( )( ) kipsinlbinM D −=−== 26000,261026001
Then
Loading
∑ = 0AM
( ) ( )24002030300010 2 =+ R
lbR 6002 =
∑ = 0VF
2121 RFFR +=+
600300024001 +=+R
lbR 12001 =
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 12 of 62
Shear Diagram
( )( ) kipsinlbinMC −=−== 12000,121012002
( )( ) kipsinlbinM D −=−== 6000,6106002
Then using
kipsinMM C −== 281max
kipsinMM C −== 122min
( ) ( ) kipsinMMMm −=+=+= 2012282
1
2
1minmax
( ) ( ) kipsinMMM a −=−=−= 812282
1
2
1minmax
I
cMs m
m = , I
cMs a
a =
( ) 4
33
25.212
3
12b
bbbhI ===
bh
c 5.12
==
35.1 b
Ms m
m = ,35.1 b
Ms a
a =
n
a
y
m
s
s
s
s
N+=
1
16
5.1
8
52
5.1
20
5.1
1 33
+
=bb
inb 96.0=
say inb 1=
inbh 33 ==
153. The same as 152, except that the cycles of 2F will not exceed 100,000 and all
surfaces are machined.
Solution:
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 13 of 62
ns at 105 cycles ( ) ksi8.43
10
1036
085.0
5
6
=
=
ksisu 72=
Machined surface, factor = 0.90
( )( )( ) ksisn 5.338.4390.085.0 ==
5.33
5.1
8
52
5.1
20
5.1
1 33
+
=bb
inb 8543.0=
say inb8
7=
inbh8
523 ==
154. A round shaft, made of cold-finished AISI 1020 steel, is subjected to a variable
torque whose maximum value is 6283 in-lb. For 5.1=N on the Soderberg
criterion, determine the diameter if (a) the torque is reversed, (b) the torque varies
from zero to a maximum, (c) the torque varies from 3141 in-lb to maximum.
Solution:
For AISI 1020, cold-finished
ksisu 78=
ksisy 66=
ksiss un 395.0 ==
size factor = 0.85
( )( )( ) ksisns 203985.06.0 ==
( ) ksiss yys 40666.06.0 ===
ns
as
ys
ms
s
s
s
s
N+=
1
(a) Reversed torque
0=mss
3
16
D
Tsas
π=
lbinT −= 6283
( )ksi
Dpsi
DDsas 333
32000,32628316===
π
ns
as
s
s
N+= 0
1
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 14 of 62
20
32
05.1
1 3
+=D
inD 34.1=
say inD8
31=
(b) 0min =T , lbinT −= 6283max
( ) lbinTm −== 314162832
1
( ) lbinTa −== 314162832
1
( )ksi
Dpsi
DDsms 333
16000,16314116===
π
( )ksi
Dpsi
DDsas 333
16000,16314116===
π
20
16
40
16
5.1
1 33
+
=DD
inD 22.1=
say inD4
11=
(c) lbinT −= 3141min , lbinT −= 6283max
( ) lbinTm −=+= 4712314162832
1
( ) lbinTa −=−= 1571314162832
1
( )ksi
Dpsi
DDsms 333
24000,24471216===
π
( )ksi
Dpsi
DDsas 333
8000,8157116===
π
20
8
40
24
5.1
1 33
+
=DD
inD 145.1=
say inD32
51=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 15 of 62
CHECK PROBLEMS
155. A simple beam 2 ft. long is made of AISI C1045 steel, as rolled. The dimensions
of the beam, which is set on edge, are 1 in. x 3 in. At the midpoint is a repeated,
reversed load of 4000 lb. What is the factor of safety?
Solution:
For AISI C1045, as rolled
ksisu 96=
ksisy 59=
( ) ksiss un 48965.05.0 ===
size factor = 0.85
( )( ) ksisn 8.404885.0 ==
n
a
y
m
s
s
s
s
N+=
1
0=ms
2
6
bh
Msa =
inh 3=
inb 1=
( )( )kipsinlbin
FLM −=−=== 24000,24
4
244000
4
( )( )( )
ksisa 1631
2462
==
8.40
160
1+=
N
55.2=N
156. The same as 155, except that the material is normalized and tempered cast steel,
SAE 080.
Solution:
Table AT 6
ksisn 35=′
ksisy 40=
( )( ) ksisn 75.293585.0 ==
75.29
160
1+=
N
86.1=N
157. A 1 ½-in. shaft is made of AISI 1045 steel, as rolled. For 2=N , what repeated
and reversed torque can the shaft sustain indefinitely?
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 16 of 62
Solution:
For AISI 1045, as rolled
ksisu 96=
ksisy 59=
( ) ksiss un 48965.05.0 ===′
( )( )( ) ksisns 48.244885.06.0 ==
( )( ) ksiss yys 4.35596.06.0 ===
ns
as
ys
ms
s
s
s
s
N+=
1
0=mss
48.240
2
1 ass+=
ksisas 24.12=
24.1216
3==
D
Tsas
π
kipsinT −= 8
VARIABLE STRESSES WITH STRESS CONCENTRATIONS
DESIGN PROBLEMS
158. The load on the link shown (150) is a maximum of 10 kips, repeated and
reversed. The link is forged from AISI C020, as rolled, and it has a ¼ in-hole
drilled on the center line of the wide side. Let bh 2= and 5.1=N . Determine b
and h at the hole (no column action) (a) for indefinite life, (b) for 50,000
repetitions (no reversal) of the maximum load, (c) for indefinite life but with a
ground and polished surface. In this case, compute the maximum stress.
Solution:
For AISI C1020, as rolled
ksisu 65=
ksisy 48=
( ) ksiss un 5.32655.05.0 ===
For as forged surface
Surface factor = 0.55
Size factor = 0.85
( )( )( )( ) ksisn 2.125.3255.085.080.0 ==
n
af
y
m
s
sK
s
s
N+=
1
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 17 of 62
Fig. AF 8, 1>hb
Assume 5.3=tK
Figure AF 7, inind
r 125.08
1
2===
ina 01.0=
926.0
125.0
01.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 3.3115.3926.011 =+−=+−= tf KqK
0=ms
( ) ( )25.02
10
−=
−=
bbdhb
Fsa
(a) n
af
s
sK
N+= 0
1
( )( )( )( )2.1225.02
103.30
5.1
1
−+=
bb
06.425.02 2 =− bb
003.2125.02 =−− bb
inb 489.1=
say inb2
11= , inbh 32 ==
(b) For 50,000 repetitions or 50,000 cycles
( ) ksisn 74.15105
102.12
085.0
4
6
=
×=
( ) ( )( )
0.210
105
103.3log
33.3log4
log
3log
=×
==f
f
K
K
fl
nK
n
afl
s
sK
N=
1
( )( )( )( )74.1525.02
100.2
5.1
1
−=
bb
906.125.02 2 =− bb
0953.0125.02 =−− bb
inb 04.1=
say inb16
11= , inbh
8
122 ==
(c) For indefinite life, ground and polished surface
Surface factor = 0.90
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 18 of 62
( )( )( )( ) ksisn 205.3290.085.080.0 ==
n
af
s
sK
N=
1
( )( )( )( )2025.02
103.3
5.1
1
−=
bb
02375.1125.02 =−− bb
inb 18.1=
say inb16
31= , inbh
8
322 ==
Maximum stress = ( )dhb
FK f
−
1>hb , 105.0375.225.0 ==hd
Figure AF 8
5.3=tK
( ) ( ) 315.3115.3926.011 =+−=+−= tf KqK
( )( )( )
ksis 14.1325.0375.21875.1
10315.3max =
−=
159. A connecting link as shown, except that there is a 1/8-in. radial hole drilled
through it at the center section. It is machined from AISI 2330, WQT 1000 F, and
it is subjected to a repeated, reversed axial load whose maximum value is 5 kips.
For 5.1=N , determine the diameter of the link at the hole (a) for indefinite life;
(b) for a life of 105 repetitions (no column action). (c) In the link found in (a)
what is the maximum tensile stress?
Problem 159
Solution:
For AISI 2330, WQT 1000 F
ksisu 135=
ksisy 126=
( ) ksiss un 5.671355.05.0 ===
For machined surface, Fig. AF 7, surface factor = 0.80
Size factor = 0.85
( )( )( )( ) ksisn 72.365.6780.085.080.0 ==
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 19 of 62
n
af
y
m
s
sK
s
s
N+=
1
Fig. AF 8, 1>hb
Assume 5.2=tK
Figure AF 7, inind
r 0625.016
1
2===
ina 0025.0=
96.0
0625.0
0025.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 44.2115.296.011 =+−=+−= tf KqK
(a) Indefinite life, 44.2=fK
ksisn 72.36=
0=ms
( )DD
DDDdD
F
DdD
Fsa
5.0
20
8
14
54
4
4
4
22
22−
=
−
=−
=
−
=π
πππ
n
af
s
sK
N+= 0
1
( )( )( )DD 5.072.36
2044.2
5.1
12 −
=π
00.25.02 =− DDπ
inD 88.0=
say inD8
7=
(b) For a life of 105 repetitions or cycles
( ) ksisn 66.4410
1072.36
085.0
5
6
=
=
( ) ( )( )
81.110
10
1044.2log
34.2log5
log
3log
===f
f
K
K
fl
nK
n
afl
s
sK
N=
1
( )( )( )DD 5.066.44
2081.1
5.1
12 −
=π
216.15.02 =− DDπ
inD 71.0=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 20 of 62
say inD4
3=
(c) DD
FKs
f
5.0
42max
−=
π
inD8
7= , 14.0
875.0
125.0==
D
d
Figure AF 8
6.2=tK
( ) ( ) 54.2116.296.011 =+−=+−= tf KqK
( )( )ksis 82.25
8
75.0
8
7
554.242max =
−
=
π
160. A machine part of uniform thickness 5.2bt = is shaped as shown and machined
all over from AISI C1020, as rolled. The design is for indefinite life for a load
repeated from 1750 lb to 3500 lb. Let bd = . (a) For a design factor of 1.8
(Soderberg), what should be the dimensions of the part? (b) What is the
maximum tensile stress in the part designed?
Problems 160, 161
Solution:
For AISI C1020, as rolled
ksisu 65=
ksisy 48=
( ) ksiss un 5.32655.05.0 ===′
For machined surface
Surface factor = 0.90
Size factor = 0.85
( )( )( )( ) ksisn 205.3290.085.080.0 ==
n
af
y
m
s
sK
s
s
N+=
1
(a) For flat plate with fillets
Figure AF 9
33
dbr ==
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 21 of 62
333.03
1==
d
r
22
==b
b
d
h
65.1=tK
ina 01.0=
0.1
1
1≈
+
=
r
aq
65.1=≈ tf KK
bt
Fs m
m =
bt
Fs a
a =
5.2
bt =
( ) lbFm 2625175035002
1=+=
( ) lbFa 875175035002
1=−=
2
5.6562
5.2
2625
bbb
sm =
=
2
5.2187
5.2
875
bbb
sa =
=
( )( )22 000,20
5.218765.1
000,48
5.6562
8.1
1
bb+=
inb 7556.0=
or inb 75.0=
inb
t 3.05.2
75.0
5.2===
For flat plate with central hole
Fig. AF 8, 1>hb , 212 == bbhd
Assume 9.2=≈ tf KK
( ) ( ) bt
F
tbb
F
tdh
Fs mmm
m =−
=−
=2
( ) ( ) bt
F
tbb
F
tdh
Fs aaa
a =−
=−
=2
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 22 of 62
2
5.6562
5.2
2625
bbb
sm =
=
2
5.2187
5.2
875
bbb
sa =
=
( )( )22 000,20
5.21879.2
000,48
5.6562
8.1
1
bb+=
inb 904.0=
or ininb16
159375.0 ==
inb
t8
3
5.2==
inbd16
15==
use inb16
15= , int
8
3= , ind
16
15=
(b) afm sKss +=max
ind
r32
15
2==
98.0
32
15
01.01
1=
+
=q
9.2=tK
( ) ( ) 86.2119.298.011 =+−=+−= tf KqK
psibbt
Fs m
m 7467
16
15
5.65625.656222
=
===
psibbt
Fs a
a 2489
16
15
5.21875.218722
=
===
( )( ) psisKss afm 586,14248986.27467max =+=+=
162. The beam shown has a circular cross section and supports a load F that
varies from 1000 lb to 3000 lb; it is machined from AISI C1020 steel, as
rolled. Determine the diameter D if Dr 2.0= and 2=N ; indefinite life.
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 23 of 62
Problems 162 – 164.
Solution:
For AISI C1020, as rolled
ksisu 65=
ksisy 48=
( ) ksiss un 5.32655.05.0 ===′
For machined surface
Surface factor = 0.90
Size factor = 0.85
( )( )( ) ksisn 86.245.3290.085.0 ==
∑ = 0AM
BF 2412 =
BF 2=
2
FB =
2
FBA ==
At discontinuity
FF
M 32
6==
( ) kipsinlbinlbinM −=−=−= 9900030003max
( ) kipsinlbinlbinM −=−=−= 3300010003min
( ) kipsinM m −=+= 6392
1
( ) kipsinM a −=−= 3392
1
3
32
D
Ms
π=
Figure AF 12
5.15.1 == dddD
2.02.0 == dddr
42.1=tK
assume 42.1=≈ tf KK
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 24 of 62
n
af
y
m
s
sK
s
s
N+=
1
( )( ) ( )( )( )33 86.24
33242.1
48
632
2
1
DD ππ+=
inD 821.1=
say inD16
131=
At maximum moment
FF
M 62
12==
( ) kipsinlbinlbinM −=−=−= 181800030006max
( ) kipsinlbinlbinM −=−=−= 6600010006min
( ) kipsinM m −=+= 126182
1
( ) kipsinM a −=−= 66182
1
3
32
D
Ms
π=
00.1=fK
n
af
y
m
s
sK
s
s
N+=
1
( )( ) ( )( )( )33 86.24
6320.1
48
1232
2
1
DD ππ+=
inD 4368.1=
Therefore use inD16
131=
164. The shaft shown is machined from C1040, OQT 1000 F (Fig. AF 1). It is
subjected to a torque that varies from zero to 10,000 in-lb. ( 0=F ). Let Dr 2.0=
and 2=N . Compute D . What is the maximum torsional stress in the shaft?
Solution:
For C1040, OQT 1000 F
ksisu 104=
ksisy 72=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 25 of 62
( ) ksiss un 521045.05.0 ===′
For machined surface
Surface factor = 0.85
Size factor = 0.85
( )( )( )( ) ksisns 5.225285.085.060.0 ==
( ) kipsinlbinTT ma −=−=== 55000000,102
1
( ) ksiss yys 2.43726.06.0 ===
3
16
D
Tss asms
π==
ns
asfs
ys
ms
s
sK
s
s
N+=
1
Figure AF 12
5.15.1 == dddD
2.02.0 == dddr
2.1=tsK
assume 2.1=≈ tsfs KK
( )( ) ( )( )( )33 5.22
5162.1
2.43
516
2
1
DD ππ+=
inD 5734.1=
say inD16
91=
afm sKss +=max
( )( ) ( )( )( )ksis 686.14
16
91
5162.1
16
91
51633max =
+
=
ππ
165. An axle (nonrotating) is to be machined from AISI 1144, OQT 1000 F, to the
proportions shown, with a fillet radius Dr 25.0≈ ; F varies from 400 lb to 1200
lb.; the supports are to the left of BB not shown. Let 2=N (Soderberg line). (a)
At the fillet, compute D and the maximum tensile stress. (b) Compute D at
section BB. (c) Specify suitable dimensions keeping the given proportions, would
a smaller diameter be permissible if the fillet were shot-peened?
Problems 165 – 167
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 26 of 62
Solution:
For AISI 1144, OQT 1000 F
ksisu 118=
ksisy 83=
ksiss un 595.0 ==′
For machined surface
Surface factor = 0.83
Size factor = 0.85
( )( )( ) ksisn 62.415983.085.0 ==
(a) At the fillet
5.15.1 == dddD
25.025.0 == dddr
35.1=tK
assume 35.1=≈ tf KK
FM 6=
( ) kipsinlbinlbinM −=−=−= 2.7720012006max
( ) kipsinlbinlbinM −=−=−= 4.224004006min
( ) kipsinM m −=+= 8.44.22.72
1
( ) kipsinM a −=−= 4.24.22.72
1
3
32
D
Ms
π=
n
af
y
m
s
sK
s
s
N+=
1
( )( ) ( )( )( )33 62.41
4.23235.1
83
8.432
2
1
DD ππ+=
inD 4034.1=
say inD16
71=
(b) At section BB,
FM 30=
( ) kipsinlbinlbinM −=−=−= 3636000120030max
( ) kipsinlbinlbinM −=−=−= 121200040030min
( ) kipsinM m −=+= 8.44.22.72
1
( ) kipsinM a −=−= 4.24.22.72
1
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 27 of 62
3
32
D
Ms
π=
0.1=fK
n
af
y
m
s
sK
s
s
N+=
1
( )( )( )
( )( )( )( )33
5.162.41
12320.1
5.183
3632
2
1
DD ππ+=
inD 6335.1=
say inD16
111=
(c) Specified dimension:
inD 2= , inD 35.1 =
A smaller diameter is permissible if the fillet were shot-peened because of increased
fatigue strength.
166. A pure torque varying from 5 in-kips to 15 in-kips is applied at section C.
( 0=F ) of the machined shaft shown. The fillet radius 8Dr = and the torque
passes through the profile keyway at C. The material is AISI 1050, OQT 1100 F,
and 6.1=N . (a) What should be the diameter? (b) If the fillet radius were
increased to 4D would it be reasonable to use a smaller D ?
Solution:
kipsinT −= 15max
kipsinT −= 5min
( ) kipsinTm −=+= 105152
1
( ) kipsinTa −=−= 55152
1
For AISI 1050, OQT 1100 F
ksisu 101=
ksisy 5.58=
( ) ksiss un 5.501015.05.0 ===
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 28 of 62
For machined surface
Surface factor = 0.85
Size factor = 0.85
( )( )( )( ) ksisns 9.215.5085.085.060.0 ==
(a) At the fillet
81=== Drdr
5.1=dD
3.1=tsK
assume 3.1=≈ tsfs KK
At the key profile
6.1=fsK
use 6.1=fsK
( ) ksiss yys 1.355.586.06.0 ===
ns
asfs
ys
ms
s
sK
s
s
N+=
1
( )( ) ( )( )( )33 9.21
5166.1
1.35
1016
6.1
1
DD ππ+=
inD 7433.1=
say inD4
31=
(b) 4Dr =
25.0=Dr
5.1=dD
Figure AF 12
18.1=tsK
6.118.1 <=≈ tsfs KK
Therefore, smaller D is not reasonable.
170. The beam shown is made of AISI C1020 steel, as rolled; ine 8= . The load F is
repeated from zero to a maximum of 1400 lb. Assume that the stress
concentration at the point of application of F is not decisive. Determine the
depth h and width t if th 4≈ ; 1.05.1 ±=N for Soderberg line. Iteration is
necessary because fK depends on the dimensions. Start by assuming a logical
fK for a logical h (Fig. AF 11), with a final check of fK . Considerable
estimation inevitable.
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 29 of 62
Problem 170
Solution:
FBA2
1==
At the hole
( ) FF
eBM 42
8 =
==
FM 4max =
0min =M
( ) ( ) kipsinFFM m −==== 8.24.12242
1
( ) ( ) kipsinFFM a −==== 8.24.12242
1
I
Mcs =
( )12
23tdh
I−
=
inind 5.02
1==
inc 75.12
1
2
1
2
11 =
+=
For AISI C1020, as rolled
ksisu 65=
ksisy 48=
( ) ksiss un 5.32655.05.0 ===
Size factor = 0.85
( )( ) ksisn 62.275.3285.0 ==
Fig. AF 7, 5.05.35.075.1 >==dc
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 30 of 62
Assume 5.3=tK
inr 25.02
1
2
1=
=
ina 010.0=
962.0
25.0
010.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 4.3115.3962.011 =+−=+−= tf KqK
n
af
y
m
s
sK
s
s
N+=
1
( )( )( )
( )( )( )( )( ) tdhtdh
33262.27
75.18.2124.3
248
75.18.212
5.1
1
−+
−=
( ) 70.1223
=− tdh
( )[ ] 70.1250.023
=− th
( ) 70.12143
=− tt
int 8627.0=
say int8
7=
inth 5.34 ==
inh2
1
2
11
2
11 ++>
inh 5.3>
Figure AF 11, 10>dh
( ) indh 550.01010 ===
5.0
2
11
2
52
1
=
−
=b
d
Therefore 5.3=tK , 4.3=fK
Use inh 5= , int4
11=
171. Design a crank similar to that shown with a design factor of 16.06.1 ± based on
the modified Goodman line. The crank is to be forged with certain surfaces
milled as shown and two ¼-in. holes. It is estimated that the material must be of
the order of AISI 8630, WQT 1100 F. The length .17 inL = , .5 ina = , and the
load varies form + 15 kips to –9 kips. (a) Compute the dimensions at section AB
with bh 3= . Check the safety of the edges (forged surfaces). (Iteration involves;
one could first make calculations for forged surfaces and then check safety at
holes.) (b) Without redesigning but otherwise considering relevant factors ,
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 31 of 62
quantitatively discuss actions that might be taken to reduce the size; holes must
remain as located.
Problems 171-174.
Solution:
(a) AISI 8630, WQT 1100 F
ksisu 96=
( ) ksiss un 48965.05.0 ===
Size factor = 0.85
As-forged surface (Fig. AF I)
Surface factor = 0.4
( )( )( ) ksisn 174842.085.0 ==
Milled surface (Machined)
Surface factor = 0.85
( )( )( ) ksisn 68.344885.085.0 ==
At AB, machined
n
af
u
m
s
sK
s
s
N+=
1
Figure AF 11
ininb 5.02
1==
inind 25.04
1==
5.05.0
25.0== in
b
d
Assume 50.3=fK
998.0=q
( ) ( ) 495.3115.3998.011 =+−=+−= tf KqK
I
Mcs =
( )12
23bdh
I−
=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 32 of 62
( ) ( )348
1144
8
1
4
11
2
1
4
1
2
1
2
1
2−=+−=
+−=
+−= hhh
hc
bh 3=
( )
12
4
12
348
1
3
bh
hM
s
−
−
=
( )
( ) bb
bM
s3
5.03
3122
3
−
−=
( )( ) bb
bMs
35.03
145.4
−
−=
( )aLFM −=
( )( ) kipsinM −=−= 18051715max
( )( ) kipsinM −=−−= 1085179min
( ) kipsinMm −=−
= 36108180
2
1
( ) kipsinM a −=+
= 144108180
2
1
n
af
u
m
s
sK
s
s
N+=
1
( )( )( )
( )( )( )( )( ) bb
b
bb
b33
5.0368.34
141445.4495.3
5.0396
14365.4
6.1
1
−
−+
−
−=
( )( ) 2.107
1
5.03
143
=−
−
bb
b
( )( )
2.10714
5.033
=−
−
b
bb
inb 6.2=
say inb8
52=
inbh8
773 ==
Checking at the edges (as forged)
( )( ) kipsinM −== 2551715max
( )( ) kipsinM −−=−= 153179min
( ) kipsinMm −=−
= 51153255
2
1
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 33 of 62
( ) kipsinM a −=+
= 204153255
2
1
332 3
2
9
66
b
M
b
M
bh
Ms ===
0.1≈fK
n
af
u
m
s
sK
s
s
N+=
1
( )( )
( )( )( )( )173
20420.1
963
512
6.1
133
bb+=
inb 373.2=
say inb8
32=
since ininb8
32
8
52 >= , ∴ safe.
(c) Action: reduce number of repetitions of load.
CHECK PROBLEMS
173. For the crank shown, inL 15= , ina 3= , ind 5.4= , inb 5.1= . It is as forged
from AISI 8630, WQT 1100 F, except for machined areas indicated. The load F
varies from +5 kips to –3 kips. The crank has been designed without detailed
attention to factors that affect its endurance strength. In section AB only,
compute the factor of safety by the Soderberg criterion. Suppose it were desired
to improve the margin of safety, with significant changes of dimensions
prohibited, what various steps could be taken? What are your particular
recommendations?
Solution:
For as forged surface
ksisn 17=
For machined surface
ksisn 68.34=
ksisn 72=
In section AB, machined
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 34 of 62
( )aLFM −=
( )( ) kipsinM −=−+= 603155max
( )( ) kipsinM −−=−−= 363153min
( ) kipsinMm −=−
= 123660
2
1
( ) kipsinM a −=+
= 483660
2
1
inhd 5.4== , inb 5.1=
3=b
h
( )( ) bb
bMs
35.03
145.4
−
−=
( ) ( )[ ]( )[ ] ( )
ksism 8125.25.15.05.13
15.14125.43
=−
−=
( ) ( )[ ]( )[ ] ( )
ksisa 25.115.15.05.13
15.14485.43
=−
−=
n
af
y
m
s
sK
s
s
N+=
1
495.3=fK from Problem 171.
( )( )68.34
25.11495.3
72
8125.21+=
N
185.0 <=N , unsafe
To increase the margin of safety
1. reduce the number of repetitions of loads
2. shot-peening
3. good surface roughness
Recommendation:
No. 1, reducing the number of repetitions of loads.
175. The link shown is made of AISI C1020, as rolled, machined all over. It is loaded
in tension by pins in the inD8
3= holes in the ends; ina
16
9= , int
16
5= ,
inh8
11= . Considering sections at A, B, and C, determine the maximum safe
axial load for 2=N and indefinite life (a) if it is repeated and reversed; (b) if it
is repeated varying from zero to maximum; (c) if it is repeatedly varies or
WF −= to WF 3= . (d) Using the results from (a) and (b), determine the ratio of
the endurance strength for a repeated load to that for a reversed load (Soderberg
line).
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 35 of 62
Problems 175 - 178
Solution:
For AISI C1020, as rolled
ksisu 65=
ksisy 48=
( ) ksiss un 5.32655.05.0 ===
Size factor = 0.85
For machined all over
Surface factor = 0.90
( )( )( )( ) ksisn 205.3280.090.085.0 ==
n
af
y
m
s
sK
s
s
N+=
1
at A, Figure AF 8
inb16
9=
inh8
11=
inDd8
3==
int16
5=
33.0
8
11
8
3
==h
d
5.0
8
11
16
9
==h
b
6.3=tAK
ind
r16
3
2==
ina 01.0=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 36 of 62
95.0
16
3
01.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 47.3116.395.011 =+−=+−= tAfA kqk
( ) 15
64
16
5
8
3
8
11
FF
tdh
Fs =
−
=−
=
( )( )( )2015
6447.3
4815
64
2
1 am FF+=
am FF 48.145
81 += at A
At B Figure AF 9
inad16
9==
inh8
11=
inr16
3=
int16
5=
33.0
16
916
3
==d
r
2
16
98
11
==d
h
63.1=tBK
ina 01.0=
95.0
16
3
01.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 6.11163.195.011 =+−=+−= tBfB kqk
45
256
16
5
16
9
FF
dt
Fs =
==
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 37 of 62
( )( )
( )2045
2566.1
4845
256
2
1 am FF+=
am FF 455.0135
321 += at B
at C, Figure AF 8, 1>h
b
inD8
1=
inah16
9==
22.0
16
98
1
==h
d
5.3=tCK
ind
r16
1
2==
ina 01.0=
862.0
16
1
01.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 2.3115.3862.011 =+−=+−= tCfC kqk
( ) 35
256
16
5
8
1
16
9
FF
tdh
Fs =
−
=−
=
( )( )
( )2035
2562.3
4835
256
2
1 am FF+=
am FF 17.1105
321 += at C
Equations
At A, am FF 48.145
81 +=
At B, am FF 455.0135
321 +=
At C, am FF 17.1105
321 +=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 38 of 62
(a) Repeated and reversed load
0=mF
FFa =
use at A
am FF 48.145
81 +=
( ) aF48.1045
81 +=
kipF 676.0=
(b) FFF am ==
at A, FF 48.145
81 +=
kipF 603.0=
at B, FF 455.0135
321 +=
kipsF 480.1=
at C, FF 17.1105
321 +=
kipF 678.0=
use kipF 603.0=
(c) WF −=min , WF 3max =
( ) WWWFm =−= 32
1
( ) WWWFa 232
1=+=
at A, ( )WW 248.145
81 +=
kipW 319.0=
at B, ( )WW 2455.0135
321 +=
kipW 884.0=
at C, ( )WW 217.1105
321 +=
kipW 378.0=
use kipW 319.0=
kipF 957.0max =
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 39 of 62
(d) ( )( )
892.0676.0
603.0===
aF
bFRatio
179. A steel rod shown, AISI 2320, hot rolled, has been machined to the following
dimensions: .1 inD = , .4
3inc = , .
8
1ine = A semicircular groove at the
midsection has .8
1inr = ; for radial hole, .
4
1ina = An axial load of 5 kips is
repeated and reversed ( 0=M ). Compute the factor of safety (Soderberg) and
make a judgement on its suitability (consider statistical variations of endurance
strength – i4.4). What steps may be taken to improve the design factor?
Problems 179-183
Solution:
AISI 2320 hot-rolled (Table AT 10)
ksisu 96=
ksisy 51=
ksisn 48=
Size factor = 0.85
Surface factor = 0.85 (machined)
( )( )( )( ) ksisn 74.274885.085.080.0 ==
n
af
y
m
s
sK
s
s
N+=
1
0=ms , reversed
ssa =
n
af
s
sK
N=
1
f
na
NK
ss =
at the fillet, Figure AF 12
iner8
1==
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 40 of 62
incd4
3==
inD 1=
17.0
4
38
1
==d
r
3.1
4
3
1==
d
D
55.1=tK
ina 010.0=
926.0
8
1
010.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 51.11155.1926.011 =+−=+−= tf KqK
( )ksissa 32.11
4
3
542
=
==
π
( )( )62.1
51.132.11
74.27===
fa
n
Ks
sN
At the groove, Figure AF 14
inininrDbd4
3
8
1212 =
−=−==
inD 1=
inr8
1=
17.0
4
38
1
==d
r
3.1
4
3
1==
d
D
75.1=tK
ina 010.0=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 41 of 62
926.0
8
1
010.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 7.11175.1926.011 =+−=+−= tf KqK
( )ksi
d
Fssa 32.11
4
3
54422
=
===
ππ
( )( )44.1
7.132.11
74.27===
fa
n
Ks
sN
At the hole, Figure AF8
inhD 1==
inad4
1==
25.014
1
==h
D
44.2=tK
ina 010.0=
926.0
8
1
010.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 33.21144.2926.011 =+−=+−= tf KqK
( ) ( )ksi
DdD
Fssa 34.9
4
11
4
1
5
4
22=
−
=
−
==ππ
( )( )27.1
33.234.9
74.27===
fa
n
Ks
sN
Factor of safety is 1.27
From i4.4
nss 76.0=
27.1min32.176.0
>==n
n
s
sN
Therefore, dimensions are not suitable.
Steps to be taken:
1. Reduce number of cycle to failure
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 42 of 62
2. Good surface condition
3. Presetting
186. A stock stud that supports a roller follower on a needle bearing for a cam is
made as shown, where ina8
5= , inb
16
7= , inc
4
3= . The nature of the junction
of the diameters at B is not defined. Assume that the inside corner is sharp. The
material of the stud is AISI 2317, OQT 1000 F. Estimate the safe, repeated load
F for 2=N . The radial capacity of the needle bearing is given as 1170 lb. at
2000 rpm for a 2500-hr life. See Fig. 20.9, p. 532, Text.
Problem 186
Solution:
AISI 2317, OQT 1000 F
ksisu 106=
ksisy 71=
ksiss un 535.0 ==
Size factor = 0.85
( )( ) ksisn 455385.0 ==
Figure AF 12
inad8
5==
incD4
3==
0≈dr , sharp corner
2.1
8
54
3
==d
D
Assume 7.2=tK
7.2=≈ tf KK
3
32
a
Ms
π=
FFFbM 4375.016
7=
==
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 43 of 62
inina 625.08
5==
( )( )
FF
s 25.18625.0
4375.0323
==π
Fsss am 25.18===
n
af
y
m
s
sK
s
s
N+=
1
( )( )45
25.187.2
71
25.18
2
1 FF+=
lbkipF 370370.0 == < less than radial capacity of the needle bearing. Ok.
187. The link shown is made of AISI C1035 steel, as rolled, with the following
dimensions .8
3ina = , .
8
7inb = , .1 inc = , .
2
1ind = , .12 inL = , .
16
1inr = The
axial load F varies from 3000 lb to 5000 lb and is applied by pins in the holes.
(a) What are the factors of safety at points A, B, and C if the link is machined all
over? What are the maximum stresses at these points?
Problems 187, 188
Solution:
AISI C1035, as rolled
ksisu 85=
ksisy 55=
ksiss un 5.425.0 ==
size factor = 0.85
( )( )( ) ksisn 68.215.4285.06.0 ==
n
af
y
m
s
sK
s
s
N+=
1
( ) kipsFm 4352
1=+=
( ) kipFa 1352
1=−=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 44 of 62
(a) at A, Figure AF 9
inr16
1=
inad8
3==
inbh8
7==
17.0
8
316
1
==d
r
33.2
8
38
7
==d
h
9.1=tK
ina 010.0=
862.0
16
1
010.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 78.1119.1862.011 =+−=+−= tf KqK
ac
Fs =
( )ksism 67.10
18
3
4=
=
( )ksisa 67.2
18
3
1=
=
( )( )68.21
67.278.1
55
67.101+=
N
42.2=N
At B, same as A, 78.1=fK
( )cab
Fs
−=
( )ksism 8
18
3
8
7
4=
−
=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 45 of 62
( )ksisa 2
18
3
8
7
1=
−
=
( )( )68.21
278.1
55
81+=
N
23.3=N
At C, Figure AF 8
ind2
1=
inch 1==
1>hb
5.012
1
==h
d
2.2=tK
ina 010.0=
inind
r 25.04
1
2===
962.0
25.0
010.01
1
1
1=
+
=
+
=
r
aq
( ) ( ) 15.2112.2962.011 =+−=+−= tf KqK
( )( )dcab
Fs
−−=
ksism 16
2
11
8
3
8
7
4=
−
−
=
ksism 4
2
11
8
3
8
7
1=
−
−
=
( )( )68.21
415.2
55
161+=
N
45.1=N
(b) Maximum stresses
at A
( ) ksisKss afmA 42.1567.278.167.10 =+=+=
at B
( ) ksisKss afmB 56.11278.18 =+=+=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 46 of 62
at C
( ) ksisKss afmC 6.24415.216 =+=+=
IMPACT PROBLEMS
189. A wrought-iron bar is 1in. in diameter and 5 ft. long. (a) What will be the stress
and elongation if the bar supports a static load of 5000 lb? Compute the stress
and elongation if a 5000 lb. weight falls freely 0.05 in. and strikes a stop at the
end of the bar. (b) The same as (a), except that the bar is aluminum alloy 3003-
H14.
Solution:
.1 inD = , ftL 5=
For wrought iron,
psiE 61028×=
(a) elongation
lbF 5000=
( )( )( )
( ) ( )in
AE
FL01364.0
102814
1255000
62
=
×
==π
δ
Stress and elongation
inh 05.0=
lbW 5000=
inftL 605 ==
2
1
21
++=
LW
hEA
A
W
A
Ws
( ) ( )
( )( ) ( )
( )( )psis 741,24
500060
14
102805.02
1
14
5000
14
5000
2
1
26
22
=
×
++=
π
ππ
( )( )in
E
sL053.0
1028
60741,246
=×
==δ
(b) Aluminum alloy 3003-H14
psiE 61010×=
lbF 5000=
( )( )( )
( ) ( )in
AE
FL038.0
101014
1255000
62
=
×
==π
δ
Stress and elongation
inh 05.0=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 47 of 62
lbW 5000=
inftL 605 ==
2
1
21
++=
LW
hEA
A
W
A
Ws
( ) ( )
( )( ) ( )
( )( )psis 475,18
500060
14
101005.02
1
14
5000
14
5000
2
1
26
22
=
×
++=
π
ππ
( )( )in
E
sL111.0
1010
60475,186
=×
==δ
190. What should be the diameter of a rod 5 ft. long, made of an aluminum alloy
2024-T4, if it is to resist the impact of a weight of lbW 500= dropped through a
distance of 2 in.? The maximum computed stress is to be 20 ksi.
Solution:
For aluminum alloy, 2024-T4
psiE 6106.10 ×=
lbW 500=
inh 2=
inftL 605 ==
psiksis 000,2020 ==
2
1
21
++=
LW
hEA
A
W
A
Ws
( )( )( )( )
2
16
50060
106.10221
50005000000,20
×++=
A
AA
( )2
1
14131140 AA ++=
9332.04
2
==D
Aπ
inD 09.1= , say inD16
11=
191. A rock drill has the heads of the cylinder bolted on by 7/8-in. bolts somewhat as
shown. The grip of the bolt is 4 in. (a) If the shank of the bolt is turned down to
the minor diameter of the coarse-thread screw, 0.7387 in., what energy may each
bolt absorb if the stress is not to exceed 25 ksi? (b) Short bolts used as described
above sometimes fail under repeated shock loads. It was found in one instance
that if long bolts, running from head to head, were used, service failures were
eliminated. How much more energy will the bolt 21 in. long absorb for a stress of
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 48 of 62
25 ksi. That the bolt 4 in. long? As before, let the bolt be turned down to the
minor diameter. The effect of the threads on the strength is to be neglected.
Problem 191
Solution:
( )E
ALsAL
E
sU
22
22
==
(a) 4
2D
Aπ
=
inL 4=
inD 7387.0=
psiE 61030×=
psiksis 000,2525 ==
( ) ( ) ( )
( )lbinU −=
×
= 86.1710302
47387.04
000,25
6
22 π
(b) inL 21=
( ) ( ) ( )
( )lbinU −=
×
= 75.9310302
217387.04
000,25
6
22 π
lbinU −=−=∆ 89.7586.1775.93
192. As seen in the figure, an 8.05-lb body A moving down with a constant
acceleration of 12 fps2, having started from rest at point C. If A is attached to a
steel wire, W & M gage 8 (0.162 in. diameter) and if for some reason the sheave
D is instantly stopped, what stress is induced in the wire?
Problems 192, 193
Solution:
E
ALsU
2
2
=
( ) maLmahahmmvU ==== 22
1
2
1 2
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 49 of 62
maLE
ALs=
2
2
gA
WaE
A
maEs
222 ==
lbW 05.8= 212 fpsa = 232 fpsg =
212 fpsb =
psiE 61030×=
4
2D
Aπ
=
( )( )( )( ) ( )32162.0
10301205.8882
6
2
2
ππ
×==
gD
WaEs
psis 741,93=
193. The hoist A shown, weighing 5000 lb. and moving at a constant fpsv 4= is
attached to a 2 in. wire rope that has a metal area of 1.6 sq. in. and a modulus
psiE 61012×= . When fth 100= , the sheave D is instantly stopped by a brake
(since this is impossible, it represents the worst conceivable condition).
Assuming that the stretching is elastic, compute the maximum stress in the rope.
Solution:
E
ALsU
2
2
=
22
22
1v
g
WmvU ==
22
22v
g
W
E
ALs=
gAL
EWvs
22 =
lbW 5000=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 50 of 62
fpsv 4=
psiE 61012×= 26.1 inA =
fthL 100== 232 fpsg =
( )( ) ( )( )( )( )1006.132
101245000 62
2 ×=s
psis 693,13=
194. A coarse-thread steel bolt, ¾ in. in diameter, with 2 in. of threaded and 3 in. of
unthreaded shank, receives an impact caused by a falling 500-lb weight. The area
at the root of the thread is 0.334 sq. in. and the effects of threads are to be
neglected. (a) What amount of energy in in-lb. could be absorbed if the maximum
calculated stress is 10 ksi? (b) From what distance h could the weight be
dropped for this maximum stress? (c) How much energy could be absorbed at the
same maximum stress if the unthreaded shank were turned down to the root
diameter.
Solution:
E
ALsU
2
2
=
(a) 21 UUU +=
E
LAsU
2
11
2
11 =
E
LAsU
2
22
2
22 =
2
1 334.0 inA =
( ) 2
2 442.075.04
inA ==π
psis 000,101 =
( )( )psi
A
Ass 7556
442.0
334.0000,10
2
112 ===
inL 21 =
inL 32 =
psiE 61030 ×=
( ) ( )( )( )
lbinU −=×
= 113.110302
2334.0000,106
2
1
( ) ( )( )( )
lbinU −=×
= 262.110302
3442.075566
2
2
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 51 of 62
lbinUUU −=+=+= 375.2262.1113.121
(b)
++=
2
1
211
LW
hEA
A
Ws
+
++=
2
1
2
2
1
11
211
A
L
A
LW
hE
A
Ws
( )
+++=
2
1
2112
21
1
211
LALAW
AhEA
A
Ws
lbW 500= 2
1 334.0 inA = 2
2 442.0 inA =
inL 21 =
inL 32 =
psiE 61030×=
psis 000,10=
( )( )( )( )( ) ( )( )[ ]
2
16
3334.02442.0500
442.0334.01030211
334.0
500000,10
+
×++=
h
inh 0033.0=
(c) E
ALsU
2
2
=
2334.0 inA =
inL 5=
psiE 61030×=
psis 000,10=
( ) ( )( )( )
lbinU −=×
= 783.210302
5334.0000,106
2
196. A part of a machine that weighs 1000 lb. raised and lowered by 1 ½-in. steel rod
that has Acme threads on one end (see i8.18 Text, for minor diameter). The
length of the rod is 10 ft. and the upper 4 ft are threaded. As the part being
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 52 of 62
lowered it sticks, then falls freely a distance of 1/8 in. (a) Compute the maximum
stress in the rod. (b) What would be the maximum stress in the rod if the lower
end had been turned down to the root diameter?
Solution:
++=
2
1
211
LW
hEA
A
Ws
+
++=
2
1
2
2
1
11
211
A
L
A
LW
hE
A
Ws
( )
+++=
2
1
2112
21
1
211
LALAW
AhEA
A
Ws
see i8.18 , inD2
112 = , inD 25.11 =
( ) 2
2
1 227.14
25.1inA ==
π
( ) 2
2
2 767.14
5.1inA ==
π
inL 41 =
inL 62 =
ininh 125.08
1==
lbW 1000=
psiE 61030×=
( )( )( )( )( )( ) ( )( )[ ]
psis 186,286227.14767.11000
767.1227.11030125.0211
227.1
1000 2
16
=
+
×++=
(b)
++=
2
1
211
LW
hEA
A
Ws
2
1 227.1 inAA ==
inLLL 1021 =+=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 53 of 62
( )( )( )( )
psis 552,25100010
227.11030125.0211
227.1
1000 2
16
=
×
++=
197. A weight W of 50 lb is moving on a smooth horizontal surface with a velocity of
2 fps when it strikes head-on the end of a ¾-in. round steel rod, 6 ft. long.
Compute the maximum stress in the rod. What design factor based on yield
strength is indicated for AISI 1010, cold drawn?
Solution:
2
1
2
1
+
=
W
WALg
EWvs
eo
3
be
WW =
ALWb ρ=
3284.0 inlb=ρ
2
2
442.04
3
4inA =
=
π
inftL 726 ==
( )( )( ) lbWb 038.972442.0284.0 ==
lbWe 013.33
038.9==
lbW 50=
fpsv 2= 232 fpsgo =
psiE 61030×=
ftL 6=
( )( ) ( )
( )( )( )psis 8166
50
013.316442.032
1030250
2
1
62
=
+
×=
For AISI 1010, cold drawn
psiksisy 000,5555 ==
74.68166
000,55===
s
sN
y
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 54 of 62
199. A rigid weight of 100 lb is dropped a distance of 25 in. upon the center of a 12
in., 50-lb. I-beam ( 46.301 inI x = ) that is simply supported on supports 10 ft
apart. Compute the maximum stress in the I-beam both with and without
allowing for the beam’s weight.
Solution:
Without beams weight
st
sty
yss =
EI
FLy
48
3
=
3
48
L
EI
y
Fk ==
++==
2
1
211
W
hk
k
Wy δ
psiE 61030×=
inftL 12010 == 46.301 inI =
( )( )( )
inlbk 333,251120
6.3011030483
6
=×
=
lbW 100=
inh 25=
( )( )iny 1415.0
100
333,25125211
333,251
100 2
1
=
++
=
( )( )( )( )
inEI
WLyst 0004.0
6.301103048
120100
48 6
33
=×
==
I
Mcsst =
( )( )lbin
WLM −=== 3000
4
120100
4
inh
c 62
12
2===
( )( )psisst 68.59
6.301
63000==
( ) psis 112,210004.0
1415.068.59 =
=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 55 of 62
with mass of beam
2
1
21
++=
st
ststy
hyyy
h - correction factor =
W
We+1
1
35
17 be
WW =
( )( ) lbftftlbWb 5001050 ==
( )lbWe 243
35
50017==
h - correction factor = 292.0
100
2431
1=
+
( )( )iny 0764.0
0004.0
292.0252110004.0
2
1
=
++=
( ) psiy
yss
st
st 400,110004.0
0764.068.59 =
==
201. A 3000 lb. automobile (here considered rigid) strikes the midpoint of a guard rail
that is an 8-in. 23-lb. I-beam, 40 ft. long; 42.64 inI = . Made of AISI C1020, as
rolled, the I-beam is simply supported on rigid posts at its ends. (a) What level
velocity of the automobile results in stressing the I-beam to the tensile yield
strength? Compare results observed by including and neglecting the beam’s
mass.
Solution:
For AISI C1020, as rolled
psiksisy 000,4848 ==
og
WvF
22
2
=δ
3
48
L
EIFk ==
δ
I
FLc
I
Mcs
4==
Lc
IsF
4=
( ) 2
2
22
32232
696
16
962 Ec
ILs
EIcL
LsI
EI
LFF===
δ
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 56 of 62
neglecting mass of beam
og
Wv
Ec
ILsF
262
2
2
2
==δ
ILg
EcWvs
o2
3 222 =
lbW 3000= 232 fpsgo =
inh
c 42
8
2===
psiE 61030×= 42.64 inI =
ftL 40=
psiksiss y 000,4848 ===
( ) ( ) ( )( )( )( )402.6432
4103030003
2
3000,48
2622222 ×
===v
ILg
EcWvs
o
fpsv 62.6=
Including mass of beam
+
=
W
WILg
EcWvs
eo 1
1
2
3 222
35
17 be
WW =
( )( ) lbftftlbWb 9204023 ==
( )lbWe 447
35
92017==
( ) ( ) ( )( )( )( )
+
×===
3000
4471
1
402.6432
4103030003
2
3000,48
2622222 v
ILg
EcWvs
o
fpsv 10.7=
DATA LACKING – DESIGNER’S DECISIONS
202. A simple beam is struck midway between supports by a 32.2-lb. weight that has
fallen 20 in. The length of the beam is 12 ft. If the stress is not to exceed 20 ksi,
what size I-beam should be used?
Solution:
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 57 of 62
2
1
21
++=
st
ststy
hyyy
st
sty
yss =
inh 20=
psis 000,20=
EI
WLyst
48
3
=
2
1
3
9611
++=
WL
EIh
y
y
st
with correction factor
2
1
3
1
19611
+
++=
W
WWL
EIh
y
y
est
I
WLd
I
Mcsst
8==
35
17wLWe =
+
++=
2
1
3
35
171
19611
8
W
wLWL
EIh
I
WLds
lbW 2.32=
inh 20=
inftL 14412 ==
psiE 61030×=
( )( ) ( )( )( )( )( ) ( )( )
( )
+
×++=
2
1
3
6
2.3235
12171
1
1442.32
2010309611
8
1442.32
w
I
I
ds
+++=
2
1
181.01
159911
6.579
wI
I
ds
From The Engineer’s Manual
By Ralph G. Hudson, S.B.
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 58 of 62
Use 3”, 5.7 lb, 45.2 inI =
( ) ( )( )
psipsis 000,20600,197.5181.01
15.259911
5.2
36.579 2
1
<=
+++=
Therefore use 3-in depth, 5.7-lb I-beam ( 45.2 inI = )
204. A 10-in., 25.4-lb.., I-bean, AISI 1020, as rolled, is 10 ft. long and is simply
supported at the ends shown. There is a static load of kipsF 101 = , 4 ft from the
left end, and a repeated reversed load of kipsF 102 = , 3 ft from the right end. It is
desired to make two attachments to the beam through holes as shown. No
significant load is supported by these attachments, but the holes cause stress
concentration. Will it be safe to make these attachments as planned? Determine
the factor of safety at the point of maximum moment and at points of stress
concentration.
Problem 204
Solution:
Mass of beam negligible
For AISI C1020, as rolled
ksisy 48=
ksisu 65=
( )∑ = 0AM
( ) BFF 103104 21 =−+
( )21 7410
1FFB +=
( )∑ = 0BM
( ) AFF 104103 12 =−+
( )21 3610
1FFA +=
kipsF 101 =
kipstoF 10102 −=
( ) ( )[ ] kipsB 310710410
1min −=−+=
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 59 of 62
( ) ( )[ ] kipsB 1110710410
1max =+=
( ) ( )[ ] kipsA 310310610
1min =−+=
( ) ( )[ ] kipsA 930710610
1max =+=
Figure AF 11,
ine2
11= ,
ind4
1=
inc 625.14
12
2
11 =
+=
inh 10=
ineh
b 5.32
11
2
10
2=−=−=
07.05.3
25.0==
b
d
5.0625.0
50.1>==
d
e
Use 0.3=tK
926.0
8
1
010.01
1=
+
=q
( ) ( ) 85.2113926.011 =+−=+−= tf KqK
( ) ksiss un 5.32655.05.0 ===
size factor = 0.85
( ) ksisn 6.275.3285.0 ==
left hole, ( )AM 2=
( ) kipsftM −== 1892max
( ) kipsftM −== 632min
I
Mcs =
( ) kipsinkipsftMm −=−=+= 144126182
1
( ) kipsinkipsftM a −=−=−= 7266182
1
inc 625.1= 41.122 inI = (Tables)
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 60 of 62
( )( )ksism 92.1
1.122
625.1144==
( )( )ksisa 96.0
1.122
625.172==
n
af
y
m
s
sK
s
s
N+=
1
( )( )6.27
96.085.2
48
92.11+=
N
2.7=N
right hole , ( )BM 5.1=
( ) kipsftM −== 5.16115.1max
( ) kipsftM −−=−= 5.435.1min
I
Mcs =
( ) kipsinkipsftMm −=−=−= 7265.45.162
1
( ) kipsinkipsftM a −=−=+= 1265.105.45.162
1
inc 625.1= 41.122 inI = (Tables)
( )( )ksism 96.0
1.122
625.172==
( )( )ksisa 68.1
1.122
625.1126==
n
af
y
m
s
sK
s
s
N+=
1
( )( )6.27
68.185.2
48
96.01+=
N
67.5=N
at maximum moment, or at , 2F
( ) kipsftM −== 33113max
( ) kipsftM −−=−= 933min
I
Mcs =
( ) kipsinkipsftMm −=−=−= 144129332
1
( ) kipsinkipsftM a −=−=+= 252219332
1
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 61 of 62
inc 52
10==
41.122 inI = (Tables)
( )( )ksism 90.5
1.122
5144==
( )( )ksisa 32.10
1.122
5252==
0.1=fK
n
af
y
m
s
sK
s
s
N+=
1
( )( )6.27
32.100.1
48
90.51+=
N
2=N
Since the design factor at the holes is much larger than at the point of maximum moment,
it is safe to make these attachment as planned.
205. The runway of a crane consists of .20 ftL = lengths of 15-in., 42.9-lb. I-beams,
as shown, each section being supported at its ends; AISI C1020, as rolled. The
wheels of the crane are 9 ft apart, and the maximum load expected is
lbF 000,10= on each wheel. Neglecting the weight of the beam, find the design
factor (a) based on variable stresses for 105 cycles, (b) based on the ultimate
strength. (Hint. Since the maximum moment will occur under the wheel, assume
the wheels at some distance x from the point of support, and determine the
reaction, 1R as a function of x ; 0=dx
dM gives position for a maximum bending
moment.)
Problem 205.
Solution:
( )∑ = 02RM
( ) ( ) 1LRFaxLFxL =−−+−
( )L
FaxLR
−−=
221
( )FaxLL
xxRM −−== 221
SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS
Page 62 of 62
( ) ( )[ ] 0222 =−+−−= xaxLL
F
dx
dM
0222 =−−− xaxL
−=
22
1 aLx
L
Fa
L
Faa
LLL
aL
M2
2
222
2
max
−
=
−
−−
−=
inftL 24020 ==
infta 1089 ==
kipslbF 10000,10 ==
( )
( )kipsinM −=
−
= 75.7202402
102
108240
2
max
For 15-in., 42.9 lb, I-beam 48.441 inI =
inc 5.72
15==
( )( )ksi
I
Mcs 24.12
8.441
5.775.720max ===
For AISI C1029, as rolled
ksisu 65=
ksiss un 5.325.0 ==
size factor = 0.85
( ) ksisn 6.275.3285.0 ==
(a) at 105
cycles
ksisn 3410
106.27
085.0
5
6
=
=
724.12
34===
s
sN n
(b) 31.524.12
65===
s
sN u
- end -