Machine Design by Faires (Section 1)

SECTION 1– DESIGN FOR SIMPLE STRESSES

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TENSION, COMPRESSION, SHEAR

DESIGN PROBLEMS

1. The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load

of 8000 lb. Let bh 5.1= . If the load is repeated but not reversed, determine the

dimensions of the section with the design based on (a) ultimate strength, (b) yield

strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005

in., what should be the dimensions of the cross section?

Problems 1 – 3.

Solution:

For AISI C1045 steel, as rolled (Table AT 7)

ksisu

96=

ksisy 59=

psiE 61030×=

A

Fs

d=

where

lbF 8000=

bhA =

but

bh 5.1=

therefore 25.1 bA =

(a) Based on ultimate strength

N = factor of safety = 6 for repeated but not reversed load (Table 1.1)

A

F

N

ss u

d==

25.1

8000

6

000,96

b=

inb 577.0= say in8

5.


of 64

inbh16

155.1 ==

(b) Based on yield strength


A

F

N

ss u

d==

25.1

8000

3

000,59

b=

inb 521.0= say in16

9.

inbh32

275.1 ==

(c) Elongation = AE

FL=δ

where,

in005.0=δ

lbF 8000=

psiE 61030×=

inL 15= 25.1 bA =

then,

AE

FL=δ

( )( )( )( )62

10305.1

158000005.0

×=

b

inb 730.0= say in4

3.

inbh8

115.1 ==

2. The same as 1 except that the material is malleable iron, ASTM A47-52, grade 35

018.

Solution:

For malleable iron, ASTM A47-52, grade 35 018(Table AT 6)

ksisu

55=

ksisy 5.36=

psiE 61025×=


of 64

A

Fs

d=

where

lbF 8000=

bhA =

but

bh 5.1=

therefore 25.1 bA =



A

F

N

ss u

d==

25.1

8000

6

000,55

b=

inb 763.0= say in8

7.

inbh16

515.1 ==



A

F

N

ss u

d==

25.1

8000

3

500,36

b=

inb 622.0= say in16

11.

inbh32

115.1 ==

(c) Elongation = AE

FL=δ

where,

in005.0=δ

lbF 8000=

psiE 61025×=

inL 15= 25.1 bA =

then,


of 64

AE

FL=δ

( )( )( )( )62

10255.1

158000005.0

×=

b

inb 8.0= say in8

7.

inbh16

515.1 ==

3. The same as 1 except that the material is gray iron, ASTM 30.

Solution:

For ASTM 30 (Table AT 6)

ksisu

30= , no ys

psiE 6105.14 ×=

Note: since there is no ys for brittle materials. Solve only for (a) and (c)

A

Fs

d=

where

lbF 8000=

bhA =

but

bh 5.1=

therefore 25.1 bA =


N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1)

A

F

N

ss u

d==

25.1

8000

5.7

000,30

b=

inb 1547.1= say in16

31 .

inbh32

2515.1 ==

(c) Elongation = AE

FL=δ

where,

in005.0=δ

lbF 8000=

psiE 6105.14 ×=


of 64

inL 15= 25.1 bA =

then,

AE

FL=δ

( )( )( )( )62

105.145.1

158000005.0

×=

b

inb 050.1= say in16

11 .

inbh32

1915.1 ==

4. A piston rod, made of AISI 3140 steel, OQT 1000 F (Fig. AF 2), is subjected to a

repeated, reversed load. The rod is for a 20-in. air compressor, where the

maximum pressure is 125 psig. Compute the diameter of the rod using a design

factor based on (a) ultimate strength, (b) yield strength.

Solution:

From Fig. AF 2 for AISI 3140, OQT 1000 F

ksisu

5.152=

ksisy 5.132=

( ) ( ) kipslbforceF 27.39270,39125204

2====

π

From Table 1.1, page 20

8=u

N

4=yN


u

u

s

FNA =

( )( )5.152

27.398

4

2 =dπ

ind 62.1= say in8

51


y

y

s

FNA =

( )( )5.132

27.394

4

2 =dπ


of 64

ind 23.1= say in4

11

5. A hollow, short compression member, of normalized cast steel (ASTM A27-58,

65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the

ultimate strength. Determine the outside and inside diameters if io

DD 2= .

Solution:

ksisu

65=

8=u

N

kipsF 1500=

( ) ( )4

34

44

22222 iiiio

DDDDDA

πππ=−=−=

( )( )65

15008

4

32

===u

ui

s

FNDA

π

inDi 85.8= say in8

78

inDD io4

317

8

7822 =

==

6. A short compression member with io DD 2= is to support a dead load of 25 tons.

The material is to be 4130 steel, WQT 1100 F. Calculate the outside and inside

diameters on the basis of (a) yield strength, (b) ultimate strength.

Solution:

From Table AT 7 for 4130, WQT 1100 F

ksisu 127=

ksisy 114=

From Table 1.1 page 20, for dead load

4~3=uN , say 4

2~5.1=yN , say 2

Area, ( ) ( )4

34

44

22222 iiiio

DDDDDA

πππ=−=−=

kipstonsF 5025 ==

(a) Based on yield strength

( )( )114

502

4

32

===y

yi

s

FNDA

π


of 64

inDi 61.0= say in8

5

inDD io4

11

8

522 =

==

(b) Based on ultimate strength

( )( )127

504

4

32

===u

ui

s

FNDA

π

inDi 82.0= say in8

7

inDD io4

31

8

722 =

==

7. A round, steel tension member, 55 in. long, is subjected to a maximum load of

7000 lb. (a) What should be its diameter if the total elongation is not to exceed

0.030 in? (b) Choose a steel that would be suitable on the basis of yield strength if

the load is gradually applied and repeated (not reversed).

Solution:

(a) AE

FL=δ or

E

FLA

δ=

where,

lbF 7000=

inL 55=

in030.0=δ

psiE6

1030×=

( )( )( )( )6

2

1030030.0

557000

4 ×== dA

π

ind 74.0= say in4

3

(b) For gradually applied and repeated (not reversed) load

3=yN

( )( )

( )psi

A

FNs

y

y 534,47

75.04

70003

2

===π

ksisy 48≈

say C1015 normalized condition ( ksisy 48= )

8. A centrifuge has a small bucket, weighing 0.332 lb. with contents, suspended on a

manganese bronze pin (B138-A, ½ hard) at the end of a horizontal arm. If the pin

is in double shear under the action of the centrifugal force, determine the diameter


of 64

needed for 10,000 rpm of the arm. The center of gravity of the bucket is 12 in.

from the axis of rotation.

Solution:

From Table AT 3, for B138-A, ½ hard

ksisus 48=

rg

WF 2ω=

where

lbW 332.0= 22.32 fpsg =

( )sec1047

60

000,102

60

2rad

n===

ππω

inr 12=

( ) ( ) kipslbrg

WF 3.11300,1111047

2.32

332.0 22 ==== ω

From Table 1.1, page 20

4~3=N , say 4

u

u

s

FNA =

( )( )48

3.114

42

2 =

d

π for double shear

ind 774.0= say in32

25

CHECK PROBLEMS

9. The link shown is made of AISIC1020 annealed steel, with inb4

3= and

inh2

11= . (a) What force will cause breakage? (b) For a design factor of 4 based

on the ultimate strength, what is the maximum allowable load? (c) If 5.2=N

based on the yield strength, what is the allowable load?

Problem 9.


of 64

Solution:

For AISI C1020 annealed steel, from Table AT 7

ksisu 57=

ksisy 42=

(a) AsF u=

2125.1

2

11

4

3inbhA =

==

( )( ) kipsF 64125.157 ==

(b) u

u

N

AsF =

4=uN

2125.1

2

11

4

3inbhA =

==

( )( )kipsF 16

4

125.157==

(c) y

y

N

AsF =

5.2=yN

2125.1

2

11

4

3inbhA =

==

( )( )kipsF 9.18

2

125.142==

10. A ¾-in.bolt, made of cold-finished B1113, has an effective stress area of 0.334 sq.

in. and an effective grip length of 5 in. The bolt is to be loaded by tightening until

the tensile stress is 80 % of the yield strength, as determined by measuring the

total elongation. What should be the total elongation?

Solution:

E

sL=δ

from Table AT 7 for cold-finished B1113

ksisy 72=

then, ( ) ksiss y 6.57728.080.0 ===

ksipsiE 000,301030 6 =×=

( )( )in

E

sL0096.0

000,30

56.57===δ


of 64

11. A 4-lb. weight is attached by a 3/8-in. bolt to a rotating arm 14-in. from the center

of rotation. The axis of the bolts is normal to the plane in which the centrifugal

force acts and the bolt is in double shear. At what speed will the bolt shear in two

if it is made of AISI B1113, cold finish?

Solution:

From Table AT 7, psiksisus 000,6262 ==

( ) 2

2

2209.08

3

4

12 inA =

= π

Asrg

WF us== 2ω

( ) ( )( )2209.0000,62142.32

4 2 =ω

sec74.88 rad=ω

74.8860

2==

nπω

rpmn 847=

12. How many ¾-in. holes could be punched in one stroke in annealed steel plate of

AISI C1040, 3/16-in. thick, by a force of 60 tons?

Solution:

For AISI C1040, from Figure AF 1

ksisu 80=

( ) ksiksiss uus 608075.075.0 ===

24418.0

16

3

4

3intdA =

== ππ

kipstonsF 12060 ==

n = number of holes

( )( )5

604415.0

120===

usAs

Fn holes

13. What is the length of a bearing for a 4-in. shaft if the load on the bearing is 6400

lb. and the allowable bearing pressure is 200 psi of the projected area?

Solution:

WpDL =

where

psip 200=

inD 4=


of 64

lbW 6400=

( )( ) 64004200 =L

inL 8=

BENDING STRESSES

DESIGN PROBLEMS

14. A lever keyed to a shaft is inL 15= long and has a rectangular cross section of

th 3= . A 2000-lb load is gradually applied and reversed at the end as shown; the

material is AISI C1020, as rolled. Design for both ultimate and yield strengths. (a)

What should be the dimensions of a section at ina 13= ? (b) at inb 4= ? (c) What

should be the size where the load is applied?

Problem 14.

Solution:

For AISI C1020, as rolled, Table AT 7

ksisu 65=

ksisy 49=

Design factors for gradually applied and reversed load

8=uN

4=yN

12

3thI = , moment of inertial

but th 3=

36

4hI =

Moment Diagram (Load Upward)


of 64

Based on ultimate strength

u

u

N

ss =

(a) I

Fac

I

Mcs ==

2

hc =

kipslbsF 22000 ==

( )( )

==

36

2132

8

654h

h

s

inh 86.3=

inh

t 29.13

86.3

3===

say

ininh2

145.4 ==

inint2

115.1 ==

(b) I

Fbc

I

Mcs ==

2

hc =

kipslbsF 22000 ==

( )( )

==

36

242

8

654h

h

s

inh 61.2=

inh

t 87.03

61.2

3===

say

inh 3=

int 1=

(c)


of 64

413

35.4

4

3

−

−=

− h

inh 33.2=

413

15.1

4

1

−

−=

− t

int 78.0=

say

inh 625.2= or inh8

52=

15. A simple beam 54 in. long with a load of 4 kips at the center is made of cast steel,

SAE 080. The cross section is rectangular (let bh 3≈ ). (a) Determine the

dimensions for 3=N based on the yield strength. (b) Compute the maximum

deflection for these dimensions. (c) What size may be used if the maximum

deflection is not to exceed 0.03 in.?

Solution:

For cast steel, SAE 080 (Table AT 6)

ksisy 40=

psiE 61030×=


of 64

From Table AT 2

Max. ( )( )

inkipsFL

M −=== 544

544

4

12

3bhI =

but bh 3=

36

4hI =

(a) I

Mc

N

ss

y

y==

2

hc =

( )

=

36

254

3

404h

h

inh 18.4=

inh

b 39.13

18.4

3===

say inh2

14= , inin

hb

2

115.1

3

5.4

3====

(b) ( )( )

( ) ( )( )in

EI

FL0384.0

12

5.45.1103048

544000

48 3

6

33

=

×

==δ

(c)

=

3648

4

3

hE

FLδ

( )( ) ( )( )( )46

3

103048

3654400003.0

h×=

inh 79.4=

inh

b 60.13

79.4

3===

say ininh4

1525.5 == , inin

hb

4

3175.1

3

25.5

3====


of 64

16. The same as 15, except that the beam is to have a circular cross section.

Solution:

(a) I

Mc

N

ss

y

y==

64

4dI

π=

2

dc =

34

32

64

2

d

M

d

dM

sππ

=

=

( )3

5432

3

40

dπ=

ind 46.3=

say ind2

13=

(b) EI

FL

48

3

=δ

64

4dI

π=

( )( )( )

( )( )( )in

dE

FL0594.0

5.3103048

54400064

48

6446

3

4

3

=×

==ππ

δ

(c) ( )4

3

48

64

dE

FL

πδ =

( )( )( )( ) 46

3

103048

5440006403.0

dπ×=

ind 15.4=

say ind4

14=

17. A simple beam, 48 in. long, with a static load of 6000 lb. at the center, is made of

C1020 structural steel. (a) Basing your calculations on the ultimate strength,

determine the dimensions of the rectangular cross section for bh 2= . (b)

Determine the dimensions based on yield strength. (c) Determine the dimensions

using the principle of “limit design.”


of 64

Solution:

From Table AT 7 and Table 1.1

ksisu 65=

ksisy 48=

4~3=uN , say 4

2~5.1=yN , say 2

( )( )kipsin

FLM −=== 72

4

486

4

I

Mcs =

2

hc =

12

3bhI =

but 2

hb =

24

4hI =

34

12

24

2

h

M

h

hM

s =

=


3

12

h

M

N

ss

u

u ==

( )3

7212

4

65

h=

inh 76.3=


of 64

inh

b 88.12

76.3

2===

say ininh4

3375.3 == , inin

hb

8

71875.1

2

75.3

2====


3

12

h

M

N

ss

y

y==

( )3

7212

2

48

h=

inh 30.3=

inh

b 65.12

30.3

2===

say ininh2

135.3 == , inin

hb

4

3175.1

2

5.3

2====

(c) Limit design (Eq. 1.6)

4

2bhsM y=

( )4

24872

2hh

=

inh 29.2=

inh

b 145.12

29.2

2===

say ininh2

125.2 == , inin

hb

4

1125.1

2

5.2

2====

18. The bar shown is subjected to two vertical loads, 1F and 2F , of 3000 lb. each, that

are inL 10= apart and 3 in. ( a , d ) from the ends of the bar. The design factor is 4

based on the ultimate strength; bh 3= . Determine the dimensions h and b if the

bar is made of (a) gray cast iron, SAE 111; (b) malleable cast iron, ASTM A47-

52, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and

moment diagrams approximately to scale.


of 64

Problems18, 19.

Solution:

lbRRFF 30002121 ====

Moment Diagram

( )( ) inkipsinlbsaRM −=−=== 99000330001

N = factor of safety = 4 based on us

12

3bhI =

2

hc =

3612

34

3

hh

h

I =

=

(a) For gray cast iron, SAE 111

ksisu 30= , Table AT 6

34

18

36

2

h

M

h

hM

I

Mc

N

ss u =

===

( )3

918

4

30

hs ==

inh 78.2=

inh

b 93.03

78.2

3===

say inh 5.3= , inb 1=

(b) For malleable cast iron, ASTM A47-52, grade 35 018


of 64

ksisu 55= , Table AT 6

34

18

36

2

h

M

h

hM

I

Mc

N

ss u =

===

( )3

918

4

55

hs ==

inh 28.2=

inh

b 76.03

28.2

3===

say inh4

12= , inb

4

3=

(c) For AISI C1040, as rolled

ksisu 90= , Fig. AF 1

34

18

36

2

h

M

h

hM

I

Mc

N

ss u =

===

( )3

918

4

90

hs ==

inh 93.1=

inh

b 64.03

93.1

3===

say inh8

71= , inb

8

5=

19. The same as 18, except that 1F acts up ( 2F acts down).

Solution:

[ ]∑ = 0AM lbRR 187521 ==


of 64

Shear Diagram

Moment Diagram

=M maximum moment = 5625 lb-in = 5.625 kips-in

(a) For gray cast iron

3

18

h

M

N

ss u ==

( )3

625.518

4

30

h=

inh 38.2=

inh

b 79.03

38.2

3===

say inh4

12= , inb

4

3=

(b) For malleable cast iron

3

18

h

M

N

ss u ==

( )3

625.518

4

55

h=

inh 95.1=

inh

b 65.03

95.1

3===

say inh8

71= , inb

8

5=


of 64


3

18

h

M

N

ss u ==

( )3

625.518

4

90

h=

inh 65.1=

inh

b 55.03

65.1

3===

say inh2

11= , inb

2

1=

20. The bar shown, supported at A and B , is subjected to a static load F of 2500 lb.

at 0=θ . Let ind 3= , inL 10= and bh 3= . Determine the dimensions of the

section if the bar is made of (a) gray iron, SAE 110; (b) malleable cast iron,

ASTM A47-52, grade 32 510; (c) AISI C1035 steel, as rolled. (d) For economic

reasons, the pins at A, B, and C are to be the same size. What should be their

diameter if the material is AISI C1035, as rolled, and the mounting is such that

each is in double shear? Use the basic dimensions from (c) as needed. (e) What

sectional dimensions would be used for the C1035 steel if the principle of “limit

design” governs in (c)?

Problems 20, 21.

Solution:


of 64

[ ]∑ = 0AM ( )2500133 =B

R

lbRB

833,10=

[ ]∑ = 0BM ( )2500103 =A

R

lbRA

8333=

Shear Diagram

Moment Diagram

( )( ) inkipsinlbM −=−== 25000,25102500

bh 3=

12

3bhI =

36

4hI =

2

hc =

34

18

36

2

h

M

h

hM

I

Mcs =

==


ksisu

20= , Table AT 6

6~5=N , say 6 for cast iron, dead load

3

18

h

M

N

ss u ==

( )3

2518

6

20

h=


of 64

inh 13.5=

inh

b 71.13

==

say inh4

15= , inb

4

31=

(b) For malleable cast iron, ASTM A47-32 grade 32510

ksisu

52= , ksisy 34=

4~3=N , say 4 for ductile, dead load

3

18

h

M

N

ss u ==

( )3

2518

4

52

h=

inh 26.3=

inh

b 09.13

==

say inh4

33= , inb

4

11=


ksisu

85= , ksisy 55=

4=N , based on ultimate strength

3

18

h

M

N

ss u ==

( )3

2518

4

85

h=

inh 77.2=

inh

b 92.03

==

say inh 3= , inb 1=

(d) For AISI C1035, as rolled

ksissu

64=

4=N , kipsRB

833.10=

A

R

N

ss Bsu

s==

22

242 DDA

ππ=

=

2

2

833.10

4

64

D

ss π

==

inD 657.0=


of 64

say inD16

11=

(e) Limit Design

4

2bhsM y=

For AISI C1035 steel, ksisy 55=

3

hb =

( )4

35525

2hh

M

==

inh 76.1=

inh

b 59.03

==

say ininh8

71875.1 == , inb

8

5=

21. The same as 20, except that o30=θ . Pin B takes all the horizontal thrust.

Solution:

θcosFF

V=

[ ]∑ = 0AM VB

FR 133 =

( ) 30cos2500133 =B

R

lbRB

9382=

[ ]∑ = 0BM VA

FR 103 =

( ) 30cos2500103 =A

R

lbRA

7217=

Shear Diagram


of 64

Moment Diagram

( )( ) inkipsinlbM −=−== 65.21650,21102165

3

18

h

Ms =


ksisu

20= , Table AT 6

6~5=N , say 6 for cast iron, dead load

3

18

h

M

N

ss u ==

( )3

65.2118

6

20

h=

inh 89.4=

inh

b 63.13

==

say inh4

15= , inb

4

31=

(b) For malleable cast iron, ASTM A47-32 grade 32510

ksisu

52= , ksisy 34=

4~3=N , say 4 for ductile, dead load

3

18

h

M

N

ss u ==

( )3

65.2118

4

52

h=

inh 11.3=

inh

b 04.13

==

say inh 3= , inb 1=


ksisu

85= , ksisy 55=



of 64

3

18

h

M

N

ss u ==

( )3

65.2118

4

85

h=

inh 64.2=

inh

b 88.03

==

say inh8

52= , inb

8

7=

(d) For AISI C1035, as rolled

ksissu

64=

4=N , lbRBV

9382=

lbFFRHBH

125030sin2500sin ==== θ

( ) ( )22222 12509382 +=+=BHBVB

RRR

lbRB

9465=

A

R

N

ss Bsu

s==

22

242 DDA

ππ=

=

2

2

465.9

4

64

D

ss π

==

inD 614.0=

say inD8

5=

(e) Limit Design

4

2bhsM y=

For AISI C1035 steel, ksisy 55=

3

hb =

( )4

35565.21

2hh

M

==

inh 68.1=

inh

b 56.03

==

say ininh8

71875.1 == , inb

8

5=


of 64

22. A cast-iron beam, ASTM 50, as shown, is 30 in. long and supports two gradually

applied, repeated loads (in phase), one of 2000 lb. at ine 10= from the free end,

and one of 1000 lb at the free end. (a) Determine the dimensions of the cross

section if acb 3≈= . (b) The same as (a) except that the top of the tee is below.

Problem 22.

Solution:

For cast iron, ASTM 50

ksisu

50= , ksisuc

164=

For gradually applied, repeated load

8~7=N , say 8

( )edFdFM ++= 21

where:

lbF 20001 =

lbF 10002 =

ind 201030 =−=

ined 30=+

( )( ) ( )( ) inkipsinlbM −=−=+= 70000,70301000202000

I

Mcs =

Solving for I , moment of inertia

( )( ) ( )( ) ( )( ) ( )( )[ ]yaaaaa

aaa

aa 332

53

23 +=

+

2

3ay =


of 64

( )( )( )( )( ) ( )( )

( )( )( )2

173

12

33

12

3 42

3

2

3a

aaaaa

aaaaa

I =+++=

(a)

2

3ac

t=

2

5ac

c=

Based on tension

I

Mc

N

ss tu

t==

( )

=

2

17

2

370

8

504a

a

ina 255.1=

Based on compression

I

Mc

N

ss cuc

c==

( )

=

2

17

2

570

8

1644a

a

ina 001.1=

Therefore ina 255.1=

Or say ina4

11=

And ( ) inacb 75.325.133 ====


of 64

Or incb4

33==

(b) If the top of the tee is below

2

5ac

t=

2

3ac

c=

2

17 4aI =

inkipsM −= 70

Based on tension

I

Mc

N

ss tu

t==

( )

=

2

17

2

570

8

504a

a

ina 488.1=

Based on compression

I

Mc

N

ss cuc

c==

( )

=

2

17

2

370

8

1644a

a

ina 845.0=

Therefore ina 488.1=

Or say ina2

11=

And inacb2

143 ===

CHECK PROBLEMS


of 64

23. An I-beam is made of structural steel, AISI C1020, as rolled. It has a depth of 3

in. and is subjected to two loads; 1F and 12 2FF = ; 1F is 5 in. from one end and

2F is 5 in. from the other ends. The beam is 25 in. long; flange width is

inb 509.2= ; 49.2 inIx

= . Determine (a) the approximate values of the load to

cause elastic failure, (b) the safe loads for a factor of safety of 3 based on the yield

strength, (c) the safe load allowing for flange buckling (i1.24), (f) the maximum

deflection caused by the safe loads.

Problems 23 – 25.

Solution:

[ ]∑ = 0AM ( )

BRFF 252205 11 =+

18.1 FRB

=

[ ]∑ = 0VF BA

RRFF +=+ 11 2

111 2.18.13 FFFRA

=−=

Shear Diagram

Moment Diagram


of 64

19FM = = maximum moment

For AISI C1020, as rolled

ksisy 48=

(a) I

Mcs

y=

where ind

c 5.12

3

2===

( )( )9.2

5.1948 1F

sy

==

kipsF 31.101 =

kipsFF 62.202 12 ==

(b) I

Mc

N

ss

y==

( )( )9.2

5.19

3

48 1Fs ==

kipsF 44.31 =

kipsFF 88.62 12 ==

(c) 1596.9509.2

25<==

b

L (page 34)

ksisc

20= ( page 34, i1.24)

I

Mcs

c=

( )( )9.2

5.1920 1F

=

kipsF 30.41 =

kipsFF 60.82 12 ==

(d) For maximum deflection,

by method of superposition, Table AT 2

( ) 2

3

max33

′+′=

bLa

EIL

bFy , ba ′>

or

( ) 2

3

max33

+′=

aLb

EIL

Fay , ab >′


of 64

maxy caused by 1F

( ) 2

3

1111max

331

+′=

aLb

EIL

aFy , 11 ab >′

where ksiE 000,30=

ina 51 =

inb 201 =′

inL 25=

49.2 inI =

( )( )( )( )

( )1

2

3

1max 0022.0

3

52520

259.2000,303

51

FF

y =

+=

maxy caused by 2F

( ) 2

3

2222max

332

′+′=

bLa

EIL

bFy , 22 ba ′>

where inb 52 =′

ina 202 =

( )( )( )( )

( )1

2

3

1max 0043.0

3

52520

259.2000,303

522

FF

y =

+=

Total deflection = δ

111maxmax 0065.00043.0022.021

FFFyy =+=+=δ

Deflection caused by the safe loads in (a)

( ) ina

067.031.100065.0 ==δ

Deflection caused by the safe loads in (b)

( ) inb

022.044.30065.0 ==δ

Deflection caused by the safe loads in (c)

( ) inc

028.030.40065.0 ==δ

24. The same as 23, except that the material is aluminum alloy, 2024-T4, heat treated.

Solution:

For aluminum alloy, 2024-T4, heat treated

ksisy 47=

(a) I

Mcs

y=


of 64

( )( )9.2

5.1947 1F

sy

==

kipsF 10.101 =

kipsFF 20.202 12 ==

(b) I

Mc

N

ss

y==

( )( )9.2

5.19

3

47 1Fs ==

kipsF 36.31 =

kipsFF 72.62 12 ==

(c) 1596.9509.2

25<==

b

L (page 34)

ksisc

20= ( page 34, i1.24)

I

Mcs

c=

( )( )9.2

5.1920 1F

=

kipsF 30.41 =

kipsFF 60.82 12 ==

(d) Total deflection = δ

111maxmax 0065.00043.0022.021

FFFyy =+=+=δ

Deflection caused by the safe loads in (a)

( ) ina

066.010.100065.0 ==δ

Deflection caused by the safe loads in (b)

( ) inb

022.036.30065.0 ==δ

Deflection caused by the safe loads in (c)

( ) inc

028.030.40065.0 ==δ

25. A light I-beam is 80 in. long, simply supported, and carries a static load at the

midpoint. The cross section has a depth of ind 4= , a flange width of inb 66.2= ,

and 40.6 inIx

= (see figure). (a) What load will the beam support if it is made of

C1020, as-rolled steel, and flange buckling (i1.24) is considered? (b) Consider the

stress owing to the weight of the beam, which is 7.7 lb/ft, and decide whether or

not the safe load should be less.


of 64

Solution:

(a) For C1020, as rolled, ksisu

65=

Consider flange buckling

3066.2

80==

b

L

since 4015 <<b

L

( )ksi

b

Ls

c15

1800

301

5.22

18001

5.2222

=

+

=

+

=

I

Mcs =

ind

c 22

4

2===

From Table AT 2

( )F

FFLM 20

4

80

4===

I

Mcss

c==

( )( )6

22015

F=

kipsF 25.2= , safe load

(b) Considering stress owing to the weight of the beam

add’l 8

2wLM = (Table AT 2)

where ftlbw 7.7=


of 64

add’l ( )

inkipsinlbwL

M −=−=

== 513.0513

8

80

12

7.7

8

22

513.020 += FM = total moment

I

Mcss

c==

( )( )6

2513.02015

+=

F

kipsF 224.2=

Therefore, the safe load should be less.

26. What is the stress in a band-saw blade due to being bent around a 13 ¾-in. pulley?

The blade thickness is 0.0265 in. (Additional stresses arise from the initial tension

and forces of sawing.)

Solution:

int

c 01325.00265.02

===

inr 76325.1301325.075.13 =+=

Using Eq. (1.4) page 11 (Text)

r

Ecs =

where psiE 61030×=

( )( )psis 881,28

76325.13

01325.01030 6

=×

=

27. A cantilever beam of rectangular cross section is tapered so that the depth varies

uniformly from 4 in. at the fixed end to 1 in. at the free end. The width is 2 in. and

the length 30 in. What safe load, acting repeated with minor shock, may be

applied to the free end? The material is AISI C1020, as rolled.

Solution:


ksisu

65= (Table AT 7)

Designing based on ultimate strength,

6=N , for repeated, minor shock load


of 64

ksiN

ss u 8.10

6

65===

Loading Diagram

x

h 1

30

14 −=

−

110.0 += xh

12

3whI =

2

hc =

FxM =

( )

( )2223 110.0

33

2

6

12

2

+===

==x

Fx

h

Fx

h

Fx

wh

hFx

I

Mcs

Differentiating with respect to x then equate to zero to solve for x giving maximum

stress.

( ) ( ) ( )( )( )( )

0110.0

10.0110.021110.03

4

2

=

+

+−+=

x

xxxF

dx

ds

( ) 010.02110.0 =−+ xx

inx 10=

( ) inh 211010.0 =+=

2

3

h

Fx

N

ss u ==

( )( )22

1038.10

F=

kipsF 44.1=

TORSIONAL STRESSES

DESIGN PROBLEMS


of 64

28. A centrifugal pump is to be driven by a 15-hp electric motor at 1750 rpm. What

should be the diameter of the pump shaft if it is made of AISI C1045 as rolled?

Consider the load as gradually repeated.

Solution:

For C1045 as rolled,

ksisy 59=

ksisus

72=

Designing based on ultimate strength

N

ss us= , 6=N (Table 1.1)

ksis 126

72==

Torque, ( )

( )kipsinlbinlbft

n

hpT −=−=−=== 540.054045

17502

15000,33

2

000,33

ππ

For diameter,

3

16

d

Ts

π=

( )3

540.01612

dπ=

ind 612.0=

say ind8

5=

29. A shaft in torsion only is to transmit 2500 hp at 570 rpm with medium shocks. Its

material is AISI 1137 steel, annealed. (a) What should be the diameter of a solid

shaft? (b) If the shaft is hollow, io

DD 2= , what size is required? (c) What is the

weight per foot of length of each of these shafts? Which is the lighter? By what

percentage? (d) Which shaft is the more rigid? Compute the torsional deflection

of each for a length of 10 ft.

Solution:

( )( )

kipsinlbftn

hpT −=−=== 276036,23

5702

2500000,33

2

000,33

ππ

For AISI 1137, annealed

ksisy 50= (Table AT 8)

ksiss yys 306.0 ==

Designing based on yield strength

3=N for medium shock, one direction


of 64

Design stress

ksiN

ss

ys10

3

30===

(a) Let D = shaft diameter

J

Tcs =

32

4DJ

π=

2

Dc =

3

16

D

Ts

π=

( )3

2761610

Dπ=

inD 20.5=

say inD4

15=

(b) ( ) ( )[ ]

32

15

32

2

32

44444

iiiioDDDDD

Jπππ

=−

=−

=

iio D

DDc ===

2

2

2

34 15

32

32

15 ii

i

D

T

D

TDs

ππ=

=

( )3

15

2763210

iDπ=

inDi

66.2=

inDDio

32.52 ==

say

inDi

8

52=

inDo

4

15=

(c) Density, 3284.0 inlb=ρ (Table AT 7)


of 64

For solid shaft

=w weight per foot of length

( )( ) ftlbDDw 8.7325.5284.0334

12222 ===

= ππρ

πρ

For hollow shaft

( ) ( ) ( ) ( ) ( )[ ] ftlbDDDDw ioio 3.55625.225.5284.0334

12222222 =−=−=−

= ππρ

πρ

Therefore hollow shaft is lighter

Percentage lightness = ( ) %5.33%1003.55

3.558.73=

−

(d) Torsional Deflection

JG

TL=θ

where

inftL 12010 ==

ksiG 3105.11 ×=

For solid shaft, 32

4DJ

π=

( )( )

( ) ( )( ) o2.2

180039.0039.0

105.1125.532

120276

34

=

==

×

=

ππθ rad

For hollow shaft, ( )

32

44

ioDD

J−

=π

( )( )

( ) ( )[ ]( )( ) o4.2

180041.0041.0

105.11625.225.532

120276

344

=

==

×−

=

ππθ rad

Therefore, solid shaft is more rigid, oo 4.22.2 <

30. The same as 29, except that the material is AISI 4340, OQT 1200 F.

Solution:

( )( )

kipsinlbftn

hpT −=−=== 276036,23

5702

2500000,33

2

000,33

ππ

For AISI 4340, OQT 1200 F

ksisy 130=

( ) ksiss yys 781306.06.0 ===

Designing based on yield strength


of 64

3=N for mild shock

Design stress

ksiN

ss

ys26

3

78===

(a) Let D = shaft diameter

J

Tcs =

32

4DJ

π=

2

Dc =

3

16

D

Ts

π=

( )3

2761626

Dπ=

inD 78.3=

say inD4

33=

(b) ( ) ( )[ ]

32

15

32

2

32

44444

iiiioDDDDD

Jπππ

=−

=−

=

iio D

DDc ===

2

2

2

34 15

32

32

15 ii

i

D

T

D

TDs

ππ=

=

( )3

15

2763226

iDπ=

inDi

93.1=

inDDio

86.32 ==

say

inDi

2=

inDo

4=

(c) Density, 3284.0 inlb=ρ (Table AT 7)


of 64

For solid shaft

=w weight per foot of length

( )( ) ftlbDDw 6.3775.3284.0334

12222 ===

= ππρ

πρ

For hollow shaft

( ) ( ) ( ) ( ) ( )[ ] ftlbDDDDw ioio 1.3224284.0334

12222222 =−=−=−

= ππρ

πρ

Therefore hollow shaft is lighter

Percentage lightness = ( ) %1.17%1001.32

1.326.37=

−

(d) Torsional Deflection

JG

TL=θ

where

inftL 12010 ==

ksiG 3105.11 ×=

For solid shaft, 32

4DJ

π=

( )( )

( ) ( )( ) o48.8

180148.0148.0

105.1175.332

120276

34

=

==

×

=

ππθ rad

For hollow shaft, ( )

32

44

ioDD

J−

=π

( )( )

( ) ( )[ ]( )( ) o99.6

180122.0122.0

105.112432

120276

344

=

==

×−

=

ππθ rad

Therefore, hollow shaft is more rigid, oo 48.899.6 < .

31. A steel shaft is transmitting 40 hp at 500 rpm with minor shock. (a) What should

be its diameter if the deflection is not to exceed 1o in D20 ? (b) If deflection is

primary what kind of steel would be satisfactory?

Solution:

(a) ( )

( )kipsinlbft

n

hpT −=−=== 04.5420

5002

40000,33

2

000,33

ππ

ksiG 3105.11 ×=

DL 20=


of 64

rad180

1π

θ == o

JG

TL=θ

( )( )

( )34

105.1132

2004.5

180×

=

D

D

π

π

inD 72.1=

say inD4

31=

(b) ( )

( )ksi

D

Ts 8.4

75.1

04.5161633

===ππ

Based on yield strength

3=N

( )( ) ksiNssys 4.148.43 ===

ksis

sys

y 246.0

4.14

6.0===

Use C1117 normalized steel ksisy 35=

32. A square shaft of cold-finish AISI 1118 transmits a torsional moment of 1200 in-

lb. For medium shock, what should be its size?

Solution:

For AISI 1118 cold-finish

ksisy 75=

ksiss yys 456.0 ==

3=N for medium shock

Z

T

N

ss

ys

′==

where, bh =

9

2

9

2 32 bhbZ ==′ (Table AT 1)

kipsinlbinT −=−= 2.11200

( )32

92.1

3

45

bs ==

inhb 71.0==

say inhb4

3==


of 64

CHECK PROBLEMS

33. A punch press is designed to exert a force sufficient to shear a 15/16-in. hole in a

½-in. steel plate, AISI C1020, as rolled. This force is exerted on the shaft at a

radius of ¾-in. (a) Compute the torsional stress in the 3.5-in. shaft (bending

neglected). (b) What will be the corresponding design factor if the shaft is made

of cold-rolled AISI 1035 steel (Table AT 10)? Considering the shock loading that

is characteristics of this machine, do you thick the design is safe enough?

Solution:


ksisus

49=

( )DtsFus

π=

where inD16

15=

int2

1=

( ) kipsF 2.722

1

16

1549 =

= π

FrT =

where inr4

3=

( ) kipsinT −=

= 2.54

4

32.72

(a) 3

16

d

Ts

π=

where ind 5.3=

( )( )

ksis 44.65.3

2.54163

==π

(b) For AISI 1035 steel, ksisus

64=

for shock loading, traditional factor of safety, 15~10=N

Design factor , 94.944.6

64===

s

sN us , the design is safe ( 10≈N )

34. The same as 33, except that the shaft diameter is 2 ¾ in.

Solution:


of 64

ind 75.2=

(a) 3

16

d

Ts

π=

( )( )

ksis 3.1375.2

2.54163

==π

(b) For AISI 1035 steel, ksisus

64=

for shock loading, traditional factor of safety, 15~10=N

Design factor , 8.43.13

64===

s

sN us , the design is not safe ( 10<N )

35. A hollow annealed Monel propeller shaft has an external diameter of 13 ½ in. and

an internal diameter of 6 ½ in.; it transmits 10,000 hp at 200 rpm. (a) Compute the

torsional stress in the shaft (stress from bending and propeller thrust are not

considered). (b) Compute the factor of safety. Does it look risky?

Solution:

For Monel shaft,

ksisus

98= (Table AT 3)

4~3=N , for dead load, based on ultimate strength

(a) J

Tcs =

( ) ( ) ( )[ ] 4

4444

308632

5.65.13

32in

DDJ io =

−=

−=

ππ

inD

c o 75.62

5.13

2===

( )( )

kipsinlbftn

hpT −=−=== 3152606,262

2002

000,10000,33

2

000,33

ππ

( )( )ksis 9.6

3086

75.63152==

(b) Factor of safety,

2.149.6

98===

s

sN us , not risky


of 64

STRESS ANALYSIS

DESIGN PROBLEMS

36. A hook is attached to a plate as shown and supports a static load of 12,000 lb. The

material is to be AISI C1020, as rolled. (a) Set up strength equations for

dimensions d , D , h , and t . Assume that the bending in the plate is negligible.

(b) Determine the minimum permissible value of these dimensions. In estimating

the strength of the nut, let dD 2.11 = . (c) Choose standard fractional dimensions

which you think would be satisfactory.

Problems 36 – 38.

Solution:

s = axial stress

ss = shear stress

(a)

22

4

4

1 d

F

d

Fs

ππ==

Equation (1) s

Fd

π

4=


of 64

( ) ( ) ( )[ ] ( )22222

1

22

1

2 44.1

4

2.1

44

4

1 dD

F

dD

F

DD

F

DD

Fs

−=

−=

−=

−

=ππππ

Equation (2) 244.1

4d

s

FD +=

π

dh

F

hD

Fs

sππ 2.11

==

Equation (3) sds

Fh

π2.1=

Dt

Fs

sπ

=

Equation (4) sDs

Ft

π=

(b) Designing based on ultimate strength,

Table AT 7, AISI C1020, as rolled

ksisu

65=

ksisus

49=

4~3=N say 4, design factor for static load

ksiN

ss u 16

4

65===

ksiN

ss us

s12

4

49===

kipslbF 12000,12 ==

From Equation (1)

( )( )

ins

Fd 98.0

16

1244===

ππ

From Equation (2)

( )( )

( ) inds

FD 53.198.044.1

16

12444.1

4 22 =+=+=ππ

From Equation (3)

( )( )in

ds

Fh

s

27.01298.02.1

12

2.1===

ππ

From Equation (4)

( )( )in

Ds

Ft

s

21.01253.1

12===

ππ


of 64

(c) Standard fractional dimensions

ind 1=

inD2

11=

inh4

1=

int4

1=

37. The same as 36, except that a shock load of 4000 lb. is repeatedly applied.

Solution:

(a) Same as 36.

(b) 15~10=N for shock load, based on ultimate strength

say 15=N , others the same.

ksiN

ss u 4

15

65===

ksiN

ss us

s3

15

49===

kipslbF 44000 ==

From Equation (1)

( )( )

ins

Fd 13.1

4

444===

ππ

From Equation (2)

( )( )

( ) inds

FD 76.113.144.1

4

4444.1

4 22 =+=+=ππ

From Equation (3)

( )( )in

ds

Fh

s

31.0313.12.1

4

2.1===

ππ

From Equation (4)

( )( )in

Ds

Ft

s

24.0376.1

4===

ππ


of 64

(c) Standard fractional dimensions

ind8

11=

inD4

31=

inh8

3=

int4

1=

38. The connection between the plate and hook, as shown, is to support a load F .

Determine the value of dimensions D , h , and t in terms of d if the connection

is to be as strong as the rod of diameter d . Assume that dD 2.11 = , uus

ss 75.0= ,

and that bending in the plate is negligible.

Solution:

2

4

1d

Fs

π=

sdF 2

4

1π=

(1)

=

N

sdF u2

4

1π


of 64

( ) ( )222

1

2 44.14

1

4

1dD

F

DD

Fs

−

=

−

=

ππ

( )sdDF 22 44.14

1−= π

(2) ( )

−=

N

sdDF u22

44.14

1π

dh

F

hD

Fs

sππ 2.11

==

sdhsF π2.1=

=

=

N

sdh

N

sdhF uus 75.0

2.12.1 ππ

(3)

=

N

sdhF u5

9.0 π

Dt

Fs

sπ

=

sDtsF π=

=

=

N

sDt

N

sDtF uus 75.0

ππ

(4)

=

N

sDtF uπ75.0

Equate (2) and (1)

( )

=

−=

N

sd

N

sdDF uu 222

4

144.1

4

1ππ

22 44.2 dD =

dD 562.1=

Equate (3) and (1)

=

=

N

sd

N

sdhF uu 2

4

19.0 ππ

( )d

dh 278.0

9.04==

Equate (4) and (1)

=

=

N

sd

N

sDtF uu 2

4

175.0 ππ

( )( )

=

=

N

sd

N

stdF uu 2

4

1562.175.0 ππ

( )( )d

dt 214.0

562.175.04==


of 64

39. (a) For the connection shown, set up strength equations representing the various

methods by which it might fail. Neglect bending effects. (b) Design this

connection for a load of 2500 lb. Both plates and rivets are of AISI C1020, as

rolled. The load is repeated and reversed with mild shock. Make the connection

equally strong on the basis of yield strengths in tension, shear, and compression.

Problems 39, 40

Solution:

(a)

=

2

4

15 D

Fs

s

π

Equation (1) s

s

FD

π5

4=

( )Dbt

Fs

2−=

Equation (2) Dts

Fb 2+=

Dt

Fs

5=

Equation (3) Ds

Ft

5=

(b) For AISI C1020, as rolled


ksiss yys 286.0 ==

4=N for repeated and reversed load (mild shock) based on yield strength

ksis 124

48==

ksiss

74

28==

From Equation (1)


of 64

ss

FD

π5

4=

where

kipslbF 5.22500 ==

( )( )

ins

FD

s

30.075

5.24

5

4===

ππ say in

16

5

From Equation (3)

( )in

Ds

Ft 13.0

1216

55

5.2

5=

== say in

32

5

From Equation (2)

( )inD

ts

Fb 96.1

16

52

1232

5

5.22 =

+

=+= say in2

40. The same as 39, except that the material is 2024-T4, aluminum alloy.

Solution:

(a) Same as 39.

(b) ) For 2024-T4, aluminum alloy


ksiss yys 2555.0 ==

4=N for repeated and reversed load (mild shock) based on yield strength

ksis 124

47==

ksiss

64

25==

From Equation (1)

ss

FD

π5

4=

where

kipslbF 5.22500 ==

( )( )

ins

FD

s

33.065

5.24

5

4===

ππ say in

8

3

From Equation (3)


of 64

( )in

Ds

Ft 11.0

128

35

5.2

5=

== say in

8

1

From Equation (2)

( )inD

ts

Fb 42.2

8

32

128

1

5.22 =

+

=+= say in

2

12

41. (a) For the connection shown, set up strength equations representing the various

methods by which it might fail. (b) Design this connection for a load of 8000 lb.

Use AISI C1015, as rolled, for the rivets, and AISI C1020, as rolled, for the

plates. Let the load be repeatedly applied with minor shock in one direction and

make the connection equally strong on the basis of ultimate strengths in tension,

shear, and compression.

Problem 41.

Solution:

(a)

( )Dbt

Fs

P−

= or ( )Dbt

F

sP2

4

3

−= Equation (1)

( )24

14 2

=

D

Fs

sR

π

Equation (2)


of 64

Dt

Fs

R4

= Equation (3)

(b) For AISI C1015, as rolled

ksisuR

61= , ksissuRusR

4575.0 ==


ksisuP

65=


ksiN

ss uP

P8.10

6

65===

ksiN

ss uR

R1.10

6

61===

ksiN

ss usR

sR5.7

6

45===

kipslbF 88000 ==

Solving for D

22 D

Fs

sRπ

=

( )in

s

FD

sR

412.05.72

8

2===

ππ say in

16

7

Solving for t

Dt

Fs

R4

=

( )in

Ds

Ft

R

453.0

1.1016

74

8

4=

== say in

2

1

Solving for b

Using ( )Dbt

Fs

P−

=

( )inD

ts

Fb

P

92.116

7

8.102

1

8=+

=+= say in2

Using ( )Dbt

F

sP2

4

3

−=


of 64

( )

( )inD

ts

Fb

P

99.116

72

8.102

14

832

4

3=

+

=+= say in2

Therefore

inb 2=

inD16

7=

int2

1=

42. Give the strength equations for the connection shown, including that for the shear

of the plate by the cotter.

Problems 42 – 44.

Solution:

Axial Stresses

2

12

1

4

4

1 D

F

D

Fs

ππ== Equation (1)

( )eDL

Fs

2−= Equation (2)


of 64

eD

Fs

2

= Equation (3)

( ) ( )2

2

22

2

2

4

4

1 Da

F

Da

Fs

−=

−

=ππ

Equation (4)

eDD

F

eDD

Fs

2

2

22

2

2

4

4

4

1 −=

−

=ππ

Equation (5)

Shear Stresses

eb

Fs

s2

= Equation (6)

( )teDL

Fs

s+−

=22

Equation (7)


of 64

at

Fs

sπ

= Equation (8)

mD

Fs

s

1π= Equation (9)

hD

Fs

s

22= Equation (10)

43. A steel rod, as-rolled AISI C1035, is fastened to a 7/8-in., as-rolled C1020 plate

by means of a cotter that is made of as-rolled C1020, in the manner shown. (a)

Determine all dimensions of this joint if it is to withstand a reversed shock load

kipsF 10= , basing the design on yield strengths. (b) If all fits are free-running

fits, decide upon tolerances and allowances.

Solution: (See figure of Prob. 42)

inint 875.08

7== , ysy ss 6.0=

For steel rod, AISI C1035, as rolled

ksisy 551

=

ksissy 331

=

For plate and cotter, AISI C1020, as rolled

ksisy 482

=

ksissy 282

=

7~5=N based on yield strength

say 7=N

From Equation (1) (Prob. 42)

2

1

41

D

F

N

ss

y

π==

( )2

1

104

7

55

Dπ=

inD 27.11 =

say inD4

111 =


of 64

From Equation (9)

mD

F

N

ss

sy

s

1

1

π==

m

=

4

11

10

7

33

π

inm 54.0=

say inm16

9=

From Equation (3)

eD

F

N

ss

y

2

1 ==

eDs

2

10

7

55==

273.12 =eD

From Equation (5)

eDD

F

N

ss

y

2

2

2 4

41

−==

π

( )( )273.14

104

7

552

2 −=

Dπ

inD 80.12 =

say inD4

312 =

and 273.12 =eD

273.14

31 =

e

ine 73.0=

say ine4

3=

By further adjustment

Say inD 22 = , ine8

5=

From Equation (8)

at

F

N

ss

sy

sπ

== 2

( )875.0

10

7

28

aπ=

ina 91.0=

say ina 1=


of 64

From Equation (4)

( )2

2

2

42

Da

F

N

ss

y

−==

π

( )( )22

2

104

7

48

−=

aπ

ina 42.2=

say ina2

12=

use ina2

12=

From Equation (7)

( )teDL

F

N

ss

sy

s+−

==22

2

( )875.08

522

10

7

28

+−

=

L

inL 80.2=

say inL 3=

From Equation (6)

eb

F

N

ss

sy

s2

2 ==

b

=

8

52

10

7

28

inb 2=

From Equation (10)

hD

F

N

ss

sy

s

22

2 ==

( )h22

10

7

28=

ininh8

5625.0 ==

Summary of Dimensions

inL 3=

inh8

5=

inb 2=

int8

7=


of 64

inm16

9=

ina2

12=

inD4

111 =

inD 22 =

ine8

5=

(b) Tolerances and allowances, No fit, tolerance = in010.0±

inL 010.03 ±=

inh 010.0625.0 ±=

int 010.0875.0 ±=

inm 010.05625.0 ±=

ina 010.0500.2 ±=

inD 010.025.11 ±=

For Free Running Fits (RC 7) Table 3.1

Female Male

inb0000.0

0030.00.2

−

+= inb

0058.0

0040.00.2

−

−=

allowance = 0.0040 in

inD0000.0

0030.00.22

−

+= inD

0058.0

0040.00.22

−

−=


ine0000.0

0016.0625.0

−

+= ine

0030.0

0020.0625.0

−

−=


44. A 1-in. ( 1D ) steel rod (as-rolled AISI C1035) is to be anchored to a 1-in. steel

plate (as-rolled C1020) by means of a cotter (as rolled C1035) as shown. (a)

Determine all the dimensions for this connection so that all parts have the same

ultimate strength as the rod. The load F reverses direction. (b) Decide upon

tolerances and allowances for loose-running fits.

Solution: (Refer to Prob. 42)

(a) For AISI C1035, as rolled

ksisu 851

=

ksisus 641

=



of 64

ksisu 652

=

ksisus 482

=

Ultimate strength

Use Equation (1)

( ) ( ) kipsDsF uu 8.6614

185

4

1 22

11=

=

= ππ

Equation (9)

mDsF usu 11π=

( )( )( )m1648.66 π=

inm 33.0=

say inm8

3=

From Equation (3)

eDsF uu 21=

( ) eD2858.66 =

7859.02 =eD

From Equation (5)

−= eDDsF uu 2

2

24

11

π

( )

−= 7859.0

4

1858.66

2

2Dπ

inD 42.12 =

say inD8

312 =

7859.08

312 =

= eeD

ine 57.0=

say ine16

9=

From Equation (4)

( )

−= 2

2

2

4

12

DasF uu π

( )

−

=

2

2

8

31

4

1658.66 aπ

ina 79.1=

say ina4

31=

From Equation (8)


of 64

atsF usu π2

=

( )( )( )( )1488.66 aπ=

ina 44.0=

say ina2

1=

use ina4

31=

From Equation (2)

( )eDLsF uu 22−=

( )

−=

16

9

8

31658.66 L

inL 20.3=

say inL4

13=

From Equation (7)

( )teDLsF usu −−= 222

( ) ( )116

9

8

314828.66

−−= L

inL 51.1=

say inL2

11=

use inL4

13=

From Equation (6)

ebsF usu 12=

( ) b

=

16

96428.66

inb 93.0=

say inb 1=

From Equation (10)

hDsF usu 212=

( ) h

=

8

316428.66

inh 38.0=

say inh8

3=

Dimensions

inL4

13=


of 64

inh8

3=

inb 1=

int 1=

inm8

3=

ina4

31=

inD 11 =

inD8

312 =

ine16

9=

(b) Tolerances and allowances, No fit, tolerance = in010.0±

inL 010.025.3 ±=

inh 010.0375.0 ±=

int 010.0000.1 ±=

inm 010.0375.0 ±=

ina 010.075.1 ±=

inD 010.0000.11 ±=

For Loose Running Fits (RC 8) Table 3.1

Female Male

inb0000.0

0035.00.1

−

+= inb

0065.0

0045.00.1

−

−=


inD0000.0

0040.0375.12

−

+= inD

0075.0

0050.0375.12

−

−=


ine0000.0

0028.05625.0

−

+= ine

0051.0

0035.05625.0

−

−=


45. Give all the simple strength equations for the connection shown. (b) Determine

the ratio of the dimensions a , b , c , d , m , and n to the dimension D so that the

connection will be equally strong in tension, shear, and compression. Base the

calculations on ultimate strengths and assume uus

ss 75.0= .


of 64

Problems 45 – 47.

Solution:

(a) Neglecting bending

Equation (1):

= 2

4

1DsF π

Equation (2):

= 2

4

12 csF s π

Equation (3): ( )bcsF 2=

Equation (4): ( )acsF =

Equation (5): ( )[ ]bcdsF −= 2

Equation (6): ( )mbsFs

4=

Equation (7): ( )nbsFs

2=

Equation (8): ( )acdsF −=

(b) N

ss u= and

N

ss us

s=

Therefore

sss

75.0=

Equate (2) and (1)

=

= 22

4

1

4

12 DscsF s ππ

=

22

4

1

2

175.0 Dscs

Dc 8165.0=

Equate (3) and (1)

( )

== 2

4

12 DsbcsF π

( ) 2

4

18165.02 DDb π=

Db 4810.0=


of 64

Equate (4) and (1)

== 2

4

1DssacF π

( ) 2

4

18165.0 DDa π=

Da 9619.0=

Equate (5) and (1)

( )[ ]

=−= 2

4

12 DsbcdsF π

( )( ) 2

4

14810.08165.02 DDd π=−

Dd 6329.1=

Equate (6) and (1)

( )

== 2

4

14 DsmbsF s π

( )( ) 2

4

14810.0475.0 DDm π=

Dm 5443.0=

Equate (7) and (1)

( )

== 2

4

12 DsnbsF s π

( )( ) 2

4

14810.0275.0 DDn π=

Dn 0886.1=

Equate (8) and (1)

( )

=−= 2

4

1DsacdsF π

( ) 2

4

18165.06329.1 DaDD π=−−

Da 9620.0=

Summary

Da 9620.0=

Db 4810.0=

Dc 8165.0=

Dd 6329.1=

Dm 5443.0=

Dn 0886.1=

Machine Design by Faires (Section 1)

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Transcript of Machine Design by Faires (Section 1)