problems in mechanical design

SECTION 2 – VARYING LOADS AND STRESS CONCENTRATIONS

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.


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


( )( )( )( ) ksiksiss un 7.355.5200.185.080.05.0 ===

4.1

7.35202

==D

saπ


of 62

inD 5.0=

Savings in weight = ( ) %21%100

16

9

2

1

16

9

2

22

=

−

(c) As forged surface


( )( )( )( ) 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


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


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 ==


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


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 =


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 ==


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.)


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:


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:


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 =


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:


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


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=


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:


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?


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:


ksisu 65=

ksisy 48=

( ) ksiss un 5.32655.05.0 ===



Size factor = 0.85

( )( )( )( ) ksisn 2.125.3255.085.080.0 ==

n

af

y

m

s

sK

s

s

N+=

1


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



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 ==


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=


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:


ksisu 65=

ksisy 48=

( ) ksiss un 5.32655.05.0 ===′

For machined surface


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 ==


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


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.


of 62

Problems 162 – 164.

Solution:


ksisu 65=

ksisy 48=

( ) ksiss un 5.32655.05.0 ===′



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


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==



( ) 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=


of 62

( ) ksiss un 521045.05.0 ===′



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


of 62

Solution:


ksisu 118=

ksisy 83=

ksiss un 595.0 ==′



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=



( ) kipsinM m −=+= 8.44.22.72

1

( ) kipsinM a −=−= 4.24.22.72

1


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


ksisu 101=

ksisy 5.58=

( ) ksiss un 5.501015.05.0 ===


of 62



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.


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 =

+=


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


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 ,


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)


( )( )( ) ksisn 174842.085.0 ==

Milled surface (Machined)


( )( )( ) 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−

=


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


= 51153255

2

1


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:


ksisn 17=


ksisn 68.34=

ksisn 72=

In section AB, machined


of 62

( )aLFM −=

( )( ) kipsinM −=−+= 603155max

( )( ) kipsinM −−=−−= 363153min


= 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).


of 62

Problems 175 - 178

Solution:


ksisu 65=

ksisy 48=

( ) ksiss un 5.32655.05.0 ===

Size factor = 0.85

For machined all over


( )( )( )( ) 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=


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 =

==


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 +=


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 =


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==


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=


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


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=

==


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=−=


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=

−

=


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 =+=+=


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=


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


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


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=


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


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


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 =+=


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


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 =

=


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:


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


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:


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.


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


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 −=−+=


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)


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


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


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 ===


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 -

problems in mechanical design

Documents

lwheaawaws

ank ssn

ma

maximum moment

bb ms35

maximum stress

maximum stresses

sy ys4