Design for Simple Stresses

SECTION 1 DESIGN FOR SIMPLE STRESSES

2

TENSION, COMPRESSION, SHEAR

DESIGN PROBLEMS

1. The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load of 8000 lb. Let bh 5.1= . If the load is repeated but not reversed, determine the dimensions of the section with the design based on (a) ultimate strength, (b) yield strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005 in., what should be the dimensions of the cross section?

Problems 1 3. Solution: For AISI C1045 steel, as rolled (Table AT 7)

ksisu 96= ksisy 59=

psiE 61030=

AF

sd =

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)

AF

Ns

s ud ==

25.18000

6000,96

b=

inb 577.0= say in85

.


3

inbh16155.1 ==

(b) Based on yield strength


AF

Ns

s ud ==

25.18000

3000,59

b=

inb 521.0= say in169

.

inbh32275.1 ==

(c) Elongation = AEFL

= where,

in005.0= lbF 8000=

psiE 61030= inL 15=

25.1 bA = then,

AEFL

= ( )( )

( )( )62 10305.1158000005.0

=

b

inb 730.0= say in43

.

inbh8115.1 ==

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=


4

AF

sd =

where lbF 8000=

bhA = but




AF

Ns

s ud ==

25.18000

6000,55

b=

inb 763.0= say in87

.

inbh16515.1 ==



AF

Ns

s ud ==

25.18000

3500,36

b=

inb 622.0= say in1611

.

inbh32115.1 ==


= where,

in005.0= lbF 8000=

psiE 61025= inL 15=

25.1 bA = then,


5

AEFL

= ( )( )

( )( )62 10255.1158000005.0

=

b

inb 8.0= say in87

.

inbh16515.1 ==

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)

AF

sd =

where lbF 8000=

bhA = but



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

AF

Ns

s ud ==

25.18000

5.7000,30

b=

inb 1547.1= say in1631 .

inbh322515.1 ==


= where,

in005.0= lbF 8000=

psiE 6105.14 =


6

inL 15= 25.1 bA =

then,

AEFL

= ( )( )

( )( )62 105.145.1158000005.0

=

b

inb 050.1= say in1611 .

inbh321915.1 ==

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====

pi

From Table 1.1, page 20 8=uN 4=yN


u

u

s

FNA =

( )( )5.15227.398

42

=dpi

ind 62.1= say in851


y

y

s

FNA =

( )( )5.13227.394

42

=dpi


7

ind 23.1= say in411

5. A hollow, short compression member, of normalized cast steel (ASTM A27-58, 65 ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the ultimate strength. Determine the outside and inside diameters if io DD 2= .

Solution: ksisu 65=

8=uN kipsF 1500=

( ) ( )4

3444

22222 iiiio

DDDDDA pipipi ===

( )( )6515008

43 2

===

u

ui

s

FNDA pi

inDi 85.8= say in878

inDD io 4317

87822 =

==

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

3444

22222 iiiio

DDDDDA pipipi ===

kipstonsF 5025 ==

(a) Based on yield strength ( )( )

114502

43 2

===

y

yi

s

FNDA pi


8

inDi 61.0= say in85

inDD io 411

8522 =

==

(b) Based on ultimate strength ( )( )

127504

43 2

===

u

ui

s

FNDA pi

inDi 82.0= say in87

inDD io 431

8722 =

==

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) AEFL

= or E

FLA = where,

lbF 7000= inL 55=

in030.0= psiE

61030=

( )( )( )( )62 1030030.0

5570004

== dA pi

ind 74.0= say in43

(b) For gradually applied and repeated (not reversed) load 3=yN

( )( )( )

psiAFN

sy

y 534,4775.0

4

700032

===

pi

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, hard) at the end of a horizontal arm. If the pin is in double shear under the action of the centrifugal force, determine the diameter


9

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

Solution: From Table AT 3, for B138-A, hard

ksisus 48=

rg

WF 2=

where lbW 332.0=

22.32 fpsg = ( )

sec104760

000,10260

2radn === pipi

inr 12=

( ) ( ) kipslbrg

WF 3.11300,11110472.32

332.0 22====

From Table 1.1, page 20 4~3=N , say 4

u

u

s

FNA =

( )( )48

3.1144

2 2 =

dpi for double shear

ind 774.0= say in3225

CHECK PROBLEMS

9. The link shown is made of AISIC1020 annealed steel, with inb43

= and

inh211= . (a) What force will cause breakage? (b) For a design factor of 4 based

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

Problem 9.


10

Solution: For AISI C1020 annealed steel, from Table AT 7

ksisu 57= ksisy 42=

(a) AsF u= 2125.1

211

43 inbhA =

==

( )( ) kipsF 64125.157 == (b)

u

u

NAsF =

4=uN

2125.1211

43 inbhA =

==

( )( ) kipsF 164

125.157==

(c) y

y

NAs

F =

5.2=yN

2125.1211

43 inbhA =

==

( )( ) kipsF 9.182

125.142==

10. A -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:

EsL

= from Table AT 7 for cold-finished B1113

ksisy 72= then, ( ) ksiss y 6.57728.080.0 ===

ksipsiE 000,301030 6 == ( )( ) in

EsL 0096.0

000,3056.57

===


11

11. A 4-lb. weight is attached by a 3/8-in. bolt to a rotating arm 14-in. from the center of rotation. The axis of the bolts is normal to the plane in which the centrifugal force acts and the bolt is in double shear. At what speed will the bolt shear in two if it is made of AISI B1113, cold finish?

Solution: From Table AT 7, psiksisus 000,6262 ==

( ) 22

2209.083

412 inA =

= pi

Asrg

WF us==2

( ) ( )( )2209.0000,62142.32

4 2=

sec74.88 rad=

74.8860

2==

npi

rpmn 847=

12. How many -in. holes could be punched in one stroke in annealed steel plate of AISI C1040, 3/16-in. thick, by a force of 60 tons?

Solution:

For AISI C1040, from Figure AF 1 ksisu 80=

( ) ksiksiss uus 608075.075.0 === 2

4418.016

3

4

3intdA =

== pipi

kipstonsF 12060 == n = number of holes

( )( ) 5604415.0120

===

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=


12

lbW 6400= ( )( ) 64004200 =L

inL 8=

BENDING STRESSES

DESIGN PROBLEMS

14. A lever keyed to a shaft is inL 15= long and has a rectangular cross section of th 3= . A 2000-lb load is gradually applied and reversed at the end as shown; the

material is AISI C1020, as rolled. Design for both ultimate and yield strengths. (a) What should be the dimensions of a section at ina 13= ? (b) at inb 4= ? (c) What should be the size where the load is applied?

Problem 14. Solution: For AISI C1020, as rolled, Table AT 7

ksisu 65= ksisy 49=

Design factors for gradually applied and reversed load 8=uN 4=yN

12

3thI = , moment of inertial

but th 3=

36

4hI =

Moment Diagram (Load Upward)


13

Based on ultimate strength

u

u

Ns

s =

(a) I

FacI

Mcs ==

2h

c =

kipslbsF 22000 ==

( )( )

==

36

2132

865

4h

h

s

inh 86.3=

inht 29.1386.3

3===

say

ininh2145.4 ==

inint2115.1 ==

(b) I

FbcI

Mcs ==

2h

c =

kipslbsF 22000 ==

( )( )

==

36

242

865

4h

h

s

inh 61.2=

inht 87.0361.2

3===

say inh 3=

int 1=

(c)


14

41335.4

43

= h

inh 33.2=

41315.1

41

= t

int 78.0= say

inh 625.2= or inh852=

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=


15

From Table AT 2

Max. ( )( ) inkipsFLM === 544544

4

12

3bhI =

but bh 3=

36

4hI =

(a) I

McNs

sy

y==

2h

c =

( )

=

36

254

340

4h

h

inh 18.4=

inhb 39.1318.4

3===

say inh214= , ininhb

2115.1

35.4

3====

(b) ( )( )( ) ( )( )

inEI

FL 0384.0

125.45.1103048

54400048 36

33

=

==

(c)

=

3648

4

3

hE

FL

( )( ) ( )( )( )46

3

1030483654400003.0h

=

inh 79.4=

inhb 60.1379.4

3===

say ininh41525.5 == , ininhb

43175.1

325.5

3====


16

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

Solution:

(a) I

McNs

sy

y==

64

4dI pi=

2d

c =

34

32

64

2dM

d

dMs

pipi=

=

( )3

5432340

dpi=

ind 46.3=

say ind213=

(b) EI

FL48

3

=

64

4dI pi=

( )( )( )

( )( )( ) indEFL 0594.0

5.310304854400064

4864

46

3

4

3

=

==

pipi

(c) ( )43

4864

dEFLpi

=

( )( )( )( ) 46

3

1030485440006403.0

dpi=

ind 15.4=

say ind414=

17. A simple beam, 48 in. long, with a static load of 6000 lb. at the center, is made of C1020 structural steel. (a) Basing your calculations on the ultimate strength, determine the dimensions of the rectangular cross section for bh 2= . (b) Determine the dimensions based on yield strength. (c) Determine the dimensions using the principle of limit design.


17

Solution:

From Table AT 7 and Table 1.1 ksisu 65= ksisy 48=

4~3=uN , say 4 2~5.1=yN , say 2

( )( ) kipsinFLM === 724486

4

IMc

s =

2h

c =

12

3bhI =

but 2hb =

24

4hI =

3412

24

2hM

h

hMs =

=


312

hM

Ns

su

u==

( )37212

465

h=

inh 76.3=


18

inhb 88.1276.3

2===


871875.1

275.3

2====


312

hM

Ns

sy

y==

( )37212

248

h=

inh 30.3=

inhb 65.1230.3

2===


43175.1

25.3

2====

(c) Limit design (Eq. 1.6)

4

2bhsM y=

( )4

24872

2hh

=

inh 29.2=

inhb 145.1229.2

2===


41125.1

25.2

2====

18. The bar shown is subjected to two vertical loads, 1F and 2F , of 3000 lb. each, that are inL 10= apart and 3 in. ( a , d ) from the ends of the bar. The design factor is 4 based on the ultimate strength; bh 3= . Determine the dimensions h and b if the bar is made of (a) gray cast iron, SAE 111; (b) malleable cast iron, ASTM A47-52, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and moment diagrams approximately to scale.


19

Problems18, 19. Solution:

lbRRFF 30002121 ====

Moment Diagram

( )( ) inkipsinlbsaRM ==== 99000330001 N = factor of safety = 4 based on us

12

3bhI =

2h

c =

36123 4

3

hhh

I =

=

(a) For gray cast iron, SAE 111 ksisu 30= , Table AT 6

34

18

36

2hM

h

hM

IMc

Ns

s u =

===

( )3918

430

hs ==

inh 78.2=

inhb 93.0378.2

3===

say inh 5.3= , inb 1=

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


20

ksisu 55= , Table AT 6

34

18

36

2hM

h

hM

IMc

Ns

s u =

===

( )3918

455

hs ==

inh 28.2=

inhb 76.0328.2

3===

say inh412= , inb

43

=

(c) For AISI C1040, as rolled ksisu 90= , Fig. AF 1

34

18

36

2hM

h

hM

IMc

Ns

s u =

===

( )3918

490

hs ==

inh 93.1=

inhb 64.0393.1

3===

say inh871= , inb

85

=

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

Solution:

[ ] = 0AM lbRR 187521 ==


21

Shear Diagram

Moment Diagram

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

(a) For gray cast iron

318

hM

Ns

s u ==

( )3625.518

430

h=

inh 38.2=

inhb 79.0338.2

3===

say inh412= , inb

43

=

(b) For malleable cast iron

318

hM

Ns

s u ==

( )3625.518

455

h=

inh 95.1=

inhb 65.0395.1

3===

say inh871= , inb

85

=


22

(c) For AISI C1040, as rolled

318

hM

Ns

s u ==

( )3625.518

490

h=

inh 65.1=

inhb 55.0365.1

3===

say inh211= , inb

21

=

20. The bar shown, supported at A and B , is subjected to a static load F of 2500 lb. at 0= . Let ind 3= , inL 10= and bh 3= . Determine the dimensions of the section if the bar is made of (a) gray iron, SAE 110; (b) malleable cast iron, ASTM A47-52, grade 32 510; (c) AISI C1035 steel, as rolled. (d) For economic reasons, the pins at A, B, and C are to be the same size. What should be their diameter if the material is AISI C1035, as rolled, and the mounting is such that each is in double shear? Use the basic dimensions from (c) as needed. (e) What sectional dimensions would be used for the C1035 steel if the principle of limit design governs in (c)?

Problems 20, 21. Solution:


23

[ ] = 0AM ( )2500133 =BR lbRB 833,10= [ ] = 0BM ( )2500103 =AR lbRA 8333= Shear Diagram

Moment Diagram

( )( ) inkipsinlbM === 25000,25102500

bh 3=

12

3bhI =

36

4hI =

2h

c =

34

18

36

2hM

h

hM

IMc

s =

==

(a) For gray cast iron, SAE 110 ksisu 20= , Table AT 6 6~5=N , say 6 for cast iron, dead load

318

hM

Ns

s u ==

( )32518

620

h=


24

inh 13.5=

inhb 71.13

==

say inh415= , inb

431=

(b) For malleable cast iron, ASTM A47-32 grade 32510 ksisu 52= , ksisy 34= 4~3=N , say 4 for ductile, dead load

318

hM

Ns

s u ==

( )32518

452

h=

inh 26.3=

inhb 09.13

==

say inh433= , inb

411=

(c) For AISI C1035, as rolled ksisu 85= , ksisy 55=

4=N , based on ultimate strength

318

hM

Ns

s u ==

( )32518

485

h=

inh 77.2=

inhb 92.03

==

say inh 3= , inb 1=

(d) For AISI C1035, as rolled ksissu 64=

4=N , kipsRB 833.10=

AR

Ns

s Bsus ==

22

242 DDA pipi =

=

2

2

833.104

64

Dss pi

==

inD 657.0=


25

say inD1611

=

(e) Limit Design

4

2bhsM y=

For AISI C1035 steel, ksisy 55=

3hb =

( )4

35525

2hh

M

==

inh 76.1=

inhb 59.03

==

say ininh871875.1 == , inb

85

=

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

Solution:

cosFFV = [ ] = 0AM VB FR 133 = ( ) 30cos2500133 =BR

lbRB 9382= [ ] = 0BM VA FR 103 = ( ) 30cos2500103 =AR

lbRA 7217= Shear Diagram


26

Moment Diagram

( )( ) inkipsinlbM === 65.21650,21102165 3

18hM

s =

(a) For gray cast iron, SAE 110 ksisu 20= , Table AT 6 6~5=N , say 6 for cast iron, dead load

318

hM

Ns

s u ==

( )3

65.2118620

h=

inh 89.4=

inhb 63.13

==

say inh415= , inb

431=

(b) For malleable cast iron, ASTM A47-32 grade 32510 ksisu 52= , ksisy 34= 4~3=N , say 4 for ductile, dead load

318

hM

Ns

s u ==

( )3

65.21184

52h

=

inh 11.3=

inhb 04.13

==

say inh 3= , inb 1= (c) For AISI C1035, as rolled

ksisu 85= , ksisy 55= 4=N , based on ultimate strength


27

318

hM

Ns

s u ==

( )3

65.21184

85h

=

inh 64.2=

inhb 88.03

==

say inh852= , inb

87

=

(d) For AISI C1035, as rolled ksissu 64=

4=N , lbRBV 9382= lbFFR HBH 125030sin2500sin ====

( ) ( )22222 12509382 +=+= BHBVB RRR lbRB 9465=

AR

Ns

s Bsus ==

22

242 DDA pipi =

=

2

2

465.94

64

Dss pi

==

inD 614.0=

say inD85

=

(e) Limit Design

4

2bhsM y=

For AISI C1035 steel, ksisy 55=

3hb =

( )4

35565.21

2hh

M

==

inh 68.1=

inhb 56.03

==

say ininh871875.1 == , inb

85

=


28

22. A cast-iron beam, ASTM 50, as shown, is 30 in. long and supports two gradually applied, repeated loads (in phase), one of 2000 lb. at ine 10= from the free end, and one of 1000 lb at the free end. (a) Determine the dimensions of the cross section if acb 3= . (b) The same as (a) except that the top of the tee is below.

Problem 22. Solution:

For cast iron, ASTM 50 ksisu 50= , ksisuc 164=

For gradually applied, repeated load 8~7=N , say 8

( )edFdFM ++= 21 where:

lbF 20001 = lbF 10002 =

ind 201030 == ined 30=+

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

IMc

s =

Solving for I , moment of inertia

( )( ) ( )( ) ( )( ) ( )( )[ ]yaaaaaaaaaa 332

532

3 +=

+

23ay =


29

( )( ) ( )( )( ) ( )( ) ( )( )( )2

1731233

123 42

32

3a

aaaaa

aaaaaI =+++=

(a)

23a

ct =

25a

cc =

Based on tension

IMc

Ns

s tut ==

( )

=

217

2370

850

4a

a

ina 255.1= Based on compression

IMc

Ns

s cucc ==

( )

=

217

2570

8164

4a

a

ina 001.1= Therefore ina 255.1=

Or say ina411=

And ( ) inacb 75.325.133 ====


30

Or incb433==

(b) If the top of the tee is below

25a

ct =

23a

cc =

217 4aI =

inkipsM = 70

Based on tension

IMc

Ns

s tut ==

( )

=

217

2570

850

4a

a

ina 488.1= Based on compression

IMc

Ns

s cucc ==

( )

=

217

2370

8164

4a

a

ina 845.0= Therefore ina 488.1=

Or say ina211=

And inacb2143 ===

CHECK PROBLEMS


31

23. An I-beam is made of structural steel, AISI C1020, as rolled. It has a depth of 3 in. and is subjected to two loads; 1F and 12 2FF = ; 1F is 5 in. from one end and

2F is 5 in. from the other ends. The beam is 25 in. long; flange width is inb 509.2= ; 49.2 inI x = . Determine (a) the approximate values of the load to

cause elastic failure, (b) the safe loads for a factor of safety of 3 based on the yield strength, (c) the safe load allowing for flange buckling (i1.24), (f) the maximum deflection caused by the safe loads.

Problems 23 25. Solution:

[ ] = 0AM ( ) BRFF 252205 11 =+ 18.1 FRB = [ ] = 0VF BA RRFF +=+ 11 2 111 2.18.13 FFFRA == Shear Diagram

Moment Diagram


32

19FM = = maximum moment For AISI C1020, as rolled

ksisy 48=

(a) I

Mcsy =

where indc 5.123

2===

( )( )9.2

5.1948 1Fsy ==

kipsF 31.101 = kipsFF 62.202 12 ==

(b) I

McNs

sy

==

( )( )9.2

5.19348 1Fs ==

kipsF 44.31 = kipsFF 88.62 12 ==

(c) 1596.9509.225

or

( ) 23max 33

+=

aLbEILFay , ab >


33

maxy caused by 1F

( ) 231111max 331

+=

aLbEIL

aFy , 11 ab >

where ksiE 000,30= ina 51 = inb 201 = inL 25=

49.2 inI =

( )( )( )( )

( )1

23

1max 0022.03

52520259.2000,303

51

FFy =

+=

maxy caused by 2F

( ) 232222max 332

+=

bLaEIL

bFy , 22 ba >

where inb 52 = ina 202 =

( )( )( )( )

( )1

23

1max 0043.03

52520259.2000,303

522

FFy =

+=

Total deflection = 111maxmax 0065.00043.0022.021 FFFyy =+=+=

Deflection caused by the safe loads in (a) ( ) ina 067.031.100065.0 ==

Deflection caused by the safe loads in (b) ( ) inb 022.044.30065.0 ==

Deflection caused by the safe loads in (c) ( ) inc 028.030.40065.0 ==

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

Mcsy =


34

( )( )9.2

5.1947 1Fsy ==

kipsF 10.101 = kipsFF 20.202 12 ==

(b) I

McNs

sy

==

( )( )9.2

5.193

47 1Fs ==

kipsF 36.31 = kipsFF 72.62 12 ==

(c) 1596.9509.225


35

Solution:

(a) For C1020, as rolled, ksisu 65= Consider flange buckling

3066.2

80==

bL

since 4015


36

addl ( ) inkipsinlbwLM ==

== 513.0513

880

127.7

8

22

513.020 += FM = total moment

IMc

ss 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 -in. pulley? The blade thickness is 0.0265 in. (Additional stresses arise from the initial tension and forces of sawing.)

Solution:

intc 01325.00265.02

===

inr 76325.1301325.075.13 =+= Using Eq. (1.4) page 11 (Text)

r

Ecs =

where psiE 61030= ( )( ) psis 881,28

76325.1301325.01030 6

=

=

27. A cantilever beam of rectangular cross section is tapered so that the depth varies uniformly from 4 in. at the fixed end to 1 in. at the free end. The width is 2 in. and the length 30 in. What safe load, acting repeated with minor shock, may be applied to the free end? The material is AISI C1020, as rolled.

Solution: For AISI C1020, as rolled

ksisu 65= (Table AT 7) Designing based on ultimate strength,

6=N , for repeated, minor shock load


37

ksiNs

s u 8.10665

===

Loading Diagram

x

h 130

14 =

110.0 += xh

12

3whI =

2h

c =

FxM =

( )( )2223 110.0

3326

12

2+

===

==

x

FxhFx

hFx

wh

hFx

IMc

s

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

( ) ( ) ( )( )( )( ) 0110.0

10.0110.021110.03 42

=

+

++=

x

xxxFdxds

( ) 010.02110.0 =+ xx inx 10=

( ) inh 211010.0 =+= 2

3hFx

Ns

s u ==

( )( )22

1038.10 F=

kipsF 44.1=

TORSIONAL STRESSES

DESIGN PROBLEMS


38

28. A centrifugal pump is to be driven by a 15-hp electric motor at 1750 rpm. What should be the diameter of the pump shaft if it is made of AISI C1045 as rolled? Consider the load as gradually repeated.

Solution:

For C1045 as rolled, ksisy 59= ksisus 72=

Designing based on ultimate strength

Ns

s us= , 6=N (Table 1.1)

ksis 12672

==

Torque, ( )( ) kipsinlbinlbftnhpT ===== 540.054045

1750215000,33

2000,33

pipi

For diameter,

316

dT

spi

=

( )3

540.01612dpi

=

ind 612.0=

say ind85

=

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: ( )

( ) kipsinlbftnhpT ==== 276036,23

57022500000,33

2000,33

pipi

For AISI 1137, annealed ksisy 50= (Table AT 8)

ksiss yys 306.0 ==

Designing based on yield strength 3=N for medium shock, one direction


39

Design stress

ksiNs

sys 10

330

===

(a) Let D = shaft diameter

JTc

s =

32

4DJ pi=

2D

c =

316

DT

spi

=

( )3

2761610Dpi

=

inD 20.5=

say inD415=

(b) ( ) ( )[ ]32

1532

232

44444iiiio DDDDDJ pipipi ===

iio DDDc ===

22

2

34 1532

3215 ii

i

DT

DTD

spipi

=

=

( )315

2763210iDpi

=

inDi 66.2= inDD io 32.52 ==

say

inDi 852=

inDo 415=

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


40

For solid shaft =w weight per foot of length

( )( ) ftlbDDw 8.7325.5284.0334

12 222 ===

= pipipi

For hollow shaft

( ) ( ) ( ) ( ) ( )[ ] ftlbDDDDw ioio 3.55625.225.5284.033412 222222 ==== pipipi

Therefore hollow shaft is lighter

Percentage lightness = ( ) %5.33%1003.55

3.558.73=

(d) Torsional Deflection

JGTL

=

where inftL 12010 ==

ksiG 3105.11 =

For solid shaft, 32

4DJ pi=

( )( )( ) ( ) ( )

o2.2180039.0039.0105.1125.5

32

12027634

=

==

=

pipi rad

For hollow shaft, ( )32

44io DDJ = pi

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

o4.2180041.0041.0105.11625.225.5

32

120276344

=

==

=

pipi 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: ( )

( ) kipsinlbftnhpT ==== 276036,23

57022500000,33

2000,33

pipi

For AISI 4340, OQT 1200 F ksisy 130=

( ) ksiss yys 781306.06.0 ===

Designing based on yield strength


41

3=N for mild shock

Design stress

ksiNs

sys 26

378

===

(a) Let D = shaft diameter

JTc

s =

32

4DJ pi=

2D

c =

316

DT

spi

=

( )3

2761626Dpi

=

inD 78.3=

say inD433=

(b) ( ) ( )[ ]32

1532

232

44444iiiio DDDDDJ pipipi ===

iio DDDc ===

22

2

34 1532

3215 ii

i

DT

DTD

spipi

=

=

( )315

2763226iDpi

=

inDi 93.1= inDD io 86.32 ==

say inDi 2= inDo 4=

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


42

For solid shaft =w weight per foot of length

( )( ) ftlbDDw 6.3775.3284.0334

12 222 ===

= pipipi

For hollow shaft

( ) ( ) ( ) ( ) ( )[ ] ftlbDDDDw ioio 1.3224284.033412 222222 ==== pipipi

Therefore hollow shaft is lighter

Percentage lightness = ( ) %1.17%1001.32

1.326.37=

(d) Torsional Deflection

JGTL

=

where inftL 12010 ==

ksiG 3105.11 =

For solid shaft, 32

4DJ pi=

( )( )( ) ( ) ( )

o48.8180148.0148.0105.1175.3

32

12027634

=

==

=

pipi rad

For hollow shaft, ( )32

44io DDJ = pi

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

o99.6180122.0122.0105.1124

32

120276344

=

==

=

pipi 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) ( )( ) kipsinlbftnhpT ==== 04.5420

500240000,33

2000,33

pipi

ksiG 3105.11 = DL 20=


43

rad180

1 pi == o

JGTL

=

( )( )( )34 105.11

32

2004.5180

=

DD

pi

pi

inD 72.1=

say inD431=

(b) ( )( ) ksiDT

s 8.475.104.51616

33 === pipi

Based on yield strength 3=N

( )( ) ksiNssys 4.148.43 === ksi

ss

ysy 246.0

4.146.0

===

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

ZT

Ns

sys

==

where, bh =

92

92 32 bhbZ == (Table AT 1)

kipsinlbinT == 2.11200 ( )

3292.1

345

bs ==

inhb 71.0==

say inhb43

==


44

CHECK PROBLEMS

33. A punch press is designed to exert a force sufficient to shear a 15/16-in. hole in a -in. steel plate, AISI C1020, as rolled. This force is exerted on the shaft at a radius of -in. (a) Compute the torsional stress in the 3.5-in. shaft (bending neglected). (b) What will be the corresponding design factor if the shaft is made of cold-rolled AISI 1035 steel (Table AT 10)? Considering the shock loading that is characteristics of this machine, do you thick the design is safe enough?

Solution: For AISI C1020, as rolled

ksisus 49=

( )DtsF us pi= where inD

1615

=

int21

=

( ) kipsF 2.7221

161549 =

= pi

FrT =

where inr43

=

( ) kipsinT =

= 2.54

432.72

(a) 316

dT

spi

=

where ind 5.3= ( )( ) ksis 44.65.3

2.54163 ==

pi

(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 ( 10N )

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

Solution:


45

ind 75.2=

(a) 316

dT

spi

=

( )( ) ksis 3.1375.2

2.54163 ==

pi

(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


46

STRESS ANALYSIS

DESIGN PROBLEMS

36. A hook is attached to a plate as shown and supports a static load of 12,000 lb. The material is to be AISI C1020, as rolled. (a) Set up strength equations for dimensions d , D , h , and t . Assume that the bending in the plate is negligible. (b) Determine the minimum permissible value of these dimensions. In estimating the strength of the nut, let dD 2.11 = . (c) Choose standard fractional dimensions which you think would be satisfactory.

Problems 36 38.

Solution: s = axial stress

ss = shear stress

(a)

22

4

41 d

F

d

Fs

pipi==

Equation (1) s

Fdpi

4=


47

( ) ( ) ( )[ ] ( )2222212212 44.14

2.144

41 dD

FdD

FDD

F

DD

Fs

=

=

=

=

pipipipi

Equation (2) 244.14 ds

FD +=pi

dhF

hDF

sspipi 2.11

==

Equation (3) sds

Fhpi2.1

=

DtF

sspi

=

Equation (4) sDs

Ft

pi=

(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

ksiNs

s u 16465

===

ksiNs

s uss 12449

===

kipslbF 12000,12 ==

From Equation (1) ( )( ) ins

Fd 98.0161244

===

pipi

From Equation (2) ( )( ) ( ) inds

FD 53.198.044.11612444.14 22 =+=+=

pipi

From Equation (3)

( )( ) indsFh

s

27.01298.02.1

122.1

===

pipi

From Equation (4)

( )( ) inDsF

ts

21.01253.1

12===

pipi


48

(c) Standard fractional dimensions

ind 1=

inD211=

inh41

=

int41

=

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.

ksiNs

s u 41565

===

ksiNs

s uss 31549

===

kipslbF 44000 ==

From Equation (1) ( )( ) ins

Fd 13.14444

===

pipi

From Equation (2) ( )( ) ( ) inds

FD 76.113.144.144444.14 22 =+=+=

pipi

From Equation (3)

( )( ) indsFh

s

31.0313.12.1

42.1

===

pipi

From Equation (4)

( )( ) inDsF

ts

24.0376.1

4===

pipi


49

(c) Standard fractional dimensions

ind811=

inD431=

inh83

=

int41

=

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

41 d

Fs

pi=

sdF 241

pi=

(1)

=

NsdF u2

41

pi


50

( ) ( )22212 44.141

41 dD

F

DD

Fs

=

=

pipi

( )sdDF 22 44.141

= pi

(2) ( )

=

NsdDF u22 44.1

41

pi

dhF

hDF

sspipi 2.11

==

sdhsF pi2.1=

=

=

Nsdh

NsdhF uus 75.02.12.1 pipi

(3)

=

NsdhF u59.0 pi

DtF

sspi

=

sDtsF pi=

=

=

NsDt

NsDtF uus 75.0pipi

(4)

=

NsDtF upi75.0

Equate (2) and (1) ( )

=

=

Nsd

NsdDF uu 222

4144.1

41

pipi

22 44.2 dD = dD 562.1=

Equate (3) and (1)

=

=

Nsd

NsdhF uu 2

419.0 pipi

( ) ddh 278.0

9.04==

Equate (4) and (1)

=

=

Nsd

NsDtF uu 2

4175.0 pipi

( )( )

=

=

Nsd

Ns

tdF uu 241562.175.0 pipi

( )( ) dd

t 214.0562.175.04

==


51

39. (a) For the connection shown, set up strength equations representing the various methods by which it might fail. Neglect bending effects. (b) Design this connection for a load of 2500 lb. Both plates and rivets are of AISI C1020, as rolled. The load is repeated and reversed with mild shock. Make the connection equally strong on the basis of yield strengths in tension, shear, and compression.

Problems 39, 40 Solution:

(a)

=

2

415 D

Fss

pi

Equation (1) ss

FDpi54

=

( )DbtF

s2

=

Equation (2) Dts

Fb 2+=

DtF

s5

=

Equation (3) DsF

t5

=

(b) For AISI C1020, as rolled ksisy 48= (Table AT 7)

ksiss yys 286.0 ==

4=N for repeated and reversed load (mild shock) based on yield strength ksis 12

448

==

ksiss 7428

==

From Equation (1)


52

ss

FDpi54

=

where kipslbF 5.22500 ==

( )( ) ins

FDs

30.0755.24

54

===

pipi say in

165

From Equation (3)

( )in

DsF

t 13.012

1655

5.25

=

== say in

325

From Equation (2)

( )inD

ts

Fb 96.11652

12325

5.22 =

+

=+= say in2

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

Solution: (a) Same as 39.

(b) ) For 2024-T4, aluminum alloy ksisy 47= (Table AT 3)

ksiss yys 2555.0 ==

4=N for repeated and reversed load (mild shock) based on yield strength ksis 12

447

==

ksiss 6425

==

From Equation (1)

ss

FDpi54

=

where kipslbF 5.22500 ==

( )( ) ins

FDs

33.0655.24

54

===

pipi say in

83

From Equation (3)


53

( )in

DsF

t 11.012

835

5.25

=

== say in

81

From Equation (2)

( )inD

ts

Fb 42.2832

1281

5.22 =

+

=+= say in

212

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.


(a)

( )DbtF

sP

= or ( )Dbt

FsP 2

43

= Equation (1)

( )2414 2

=

D

FssR

pi

Equation (2)


54

DtF

sR 4= Equation (3)

(b) For AISI C1015, as rolled

ksisuR 61= , ksiss uRusR 4575.0 == For AISI C1020, as rolled

ksisuP 65=

6=N , based on ultimate strength

ksiNs

s uPP 8.10665

===

ksiNs

s uRR 1.10661

===

ksiN

ss usRsR 5.76

45===

kipslbF 88000 == Solving for D

22 DF

ssRpi

=

( ) insFD

sR

412.05.72

82

===

pipi say in

167

Solving for t

DtF

sR 4=

( )in

DsF

tR

453.01.10

1674

84

=

== say in

21

Solving for b

Using ( )DbtF

sP

=

( )inD

ts

FbP

92.1167

8.1021

8=+

=+= say in2

Using ( )DbtF

sP 243

=


55

( )( )

inDts

FbP

99.11672

8.10214

83243

=

+

=+= say in2

Therefore inb 2=

inD167

=

int21

=

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

212

1

4

41 D

F

D

Fs

pipi== Equation (1)

( )eDLF

s2

= Equation (2)


56

eDF

s2

= Equation (3)

( ) ( )2222224

41 Da

F

Da

Fs

=

=

pipi Equation (4)

eDDF

eDD

Fs

222

222

44

41

=

=

pipi Equation (5)

Shear Stresses

ebF

ss 2= Equation (6)

( )teDLF

ss+

=

22 Equation (7)


57

at

Fss

pi= Equation (8)

mDF

ss1pi

= Equation (9)

hDF

ss22

= Equation (10)

43. A steel rod, as-rolled AISI C1035, is fastened to a 7/8-in., as-rolled C1020 plate by means of a cotter that is made of as-rolled C1020, in the manner shown. (a) Determine all dimensions of this joint if it is to withstand a reversed shock load

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

Solution: (See figure of Prob. 42) inint 875.0

87

== , 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)

21

41DF

Ns

sy

pi==

( )21

104755

Dpi=

inD 27.11 =

say inD4111 =


58

From Equation (9)

mDF

Ns

ssy

s

1

1

pi==

m

=

411

10733

pi

inm 54.0=

say inm169

=

From Equation (3)

eDF

Ns

sy

2

1==

eDs

2

10755

==

273.12 =eD From Equation (5)

eDDF

Ns

sy

222 441

==

pi

( )( )273.14

104755

22

=

Dpi

inD 80.12 =

say inD4312 =

and 273.12 =eD

273.1431 =

e

ine 73.0=

say ine43

=

By further adjustment Say inD 22 = , ine 8

5=

From Equation (8)

at

FN

ss

sys

pi==

2

( )875.010

728

api=

ina 91.0= say ina 1=


59

From Equation (4)

( )22242

DaF

Ns

sy

==

pi

( )( )22 2

104748

=

api

ina 42.2=

say ina212=

use ina212=

From Equation (7)

( )teDLF

Ns

ssy

s+

==

222

( )875.08522

10728

+

=

L

inL 80.2= say inL 3= From Equation (6)

ebF

Ns

ssy

s 22

==

b

=

852

10728

inb 2= From Equation (10)

hDF

Ns

ssy

s

222

==

( )h2210

728

=

ininh85625.0 ==

Summary of Dimensions inL 3=

inh85

=

inb 2=

int87

=


60

inm169

=

ina212=

inD4111 =

inD 22 =

ine85

=

(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

inb0000.00030.0

0.2

+= inb

0058.00040.0

0.2

=

allowance = 0.0040 in

inD0000.00030.0

0.22

+= inD

0058.00040.0

0.22

=


ine0000.00016.0

625.0

+= ine

0030.00020.0

625.0

=


44. A 1-in. ( 1D ) steel rod (as-rolled AISI C1035) is to be anchored to a 1-in. steel plate (as-rolled C1020) by means of a cotter (as rolled C1035) as shown. (a) Determine all the dimensions for this connection so that all parts have the same ultimate strength as the rod. The load F reverses direction. (b) Decide upon tolerances and allowances for loose-running fits.

Solution: (Refer to Prob. 42)

(a) For AISI C1035, as rolled ksisu 851 = ksisus 641 =

For AISI C1020, as rolled


61

ksisu 652 = ksisus 482 =

Ultimate strength Use Equation (1)

( ) ( ) kipsDsF uu 8.6614185

41 22

11 =

=

= pipi

Equation (9) mDsF usu 11pi=

( )( )( )m1648.66 pi= inm 33.0=

say inm83

=

From Equation (3) eDsF uu 21=

( ) eD2858.66 = 7859.02 =eD

From Equation (5)

= eDDsF uu 2

224

11

pi

( )

= 7859.0

41858.66 22Dpi

inD 42.12 =

say inD8312 =

7859.08312 =

= eeD

ine 57.0=

say ine169

=

From Equation (4) ( )

=

22

2

41

2DasF uu pi

( )

=

22

831

41658.66 api

ina 79.1=

say ina431=

From Equation (8)


62

atsF usu pi2= ( )( )( )( )1488.66 api=

ina 44.0=

say ina21

=

use ina431=

From Equation (2) ( )eDLsF uu 22 =

( )

=

169

831658.66 L

inL 20.3=

say inL413=

From Equation (7) ( )teDLsF usu = 222

( ) ( )1169

8314828.66

= L

inL 51.1=

say inL211=

use inL413=

From Equation (6) ebsF usu 12=

( ) b

=

1696428.66

inb 93.0= say inb 1= From Equation (10)

hDsF usu 212=

( ) h

=

8316428.66

inh 38.0=

say inh83

=

Dimensions

inL413=


63

inh83

=

inb 1= int 1=

inm83

=

ina431=

inD 11 =

inD8312 =

ine169

=

(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

inb0000.00035.0

0.1

+= inb

0065.00045.0

0.1

=


inD0000.00040.0

375.12

+= inD

0075.00050.0

375.12

=


ine0000.00028.0

5625.0

+= ine

0051.00035.0

5625.0

=


45. Give all the simple strength equations for the connection shown. (b) Determine the ratio of the dimensions a , b , c , d , m , and n to the dimension D so that the connection will be equally strong in tension, shear, and compression. Base the calculations on ultimate strengths and assume uus ss 75.0= .


64

Problems 45 47. Solution: (a) Neglecting bending

Equation (1):

=

2

41 DsF pi

Equation (2):

=

2

412 csF s pi

Equation (3): ( )bcsF 2= Equation (4): ( )acsF = Equation (5): ( )[ ]bcdsF = 2 Equation (6): ( )mbsF s 4= Equation (7): ( )nbsF s 2= Equation (8): ( )acdsF =

(b) Ns

s u= and Ns

s uss =

Therefore sss 75.0=

Equate (2) and (1)

=

=

22

41

412 DscsF s pipi

=

2241

2175.0 Dscs

Dc 8165.0= Equate (3) and (1)

( )

==

2

412 DsbcsF pi

( ) 2418165.02 DDb pi=

Db 4810.0=


65

Equate (4) and (1)

==

2

41 DssacF pi

( ) 2418165.0 DDa pi=

Da 9619.0= Equate (5) and (1)

( )[ ]

==

2

412 DsbcdsF pi

( )( ) 2414810.08165.02 DDd pi=

Dd 6329.1= Equate (6) and (1)

( )

==

2

414 DsmbsF s pi

( )( ) 2414810.0475.0 DDm pi=

Dm 5443.0= Equate (7) and (1)

( )

==

2

412 DsnbsF s pi

( )( ) 2414810.0275.0 DDn pi=

Dn 0886.1= Equate (8) and (1)

( )

==

2

41 DsacdsF pi

( ) 2418165.06329.1 DaDD pi=

Da 9620.0=

Summary Da 9620.0= Db 4810.0= Dc 8165.0= Dd 6329.1= Dm 5443.0=

Dn 0886.1=


66

46. The same as 45, except that the calculations are to be based on yield strengths. Let ysy ss 6.0= .

Solution: (Refer to Prob. 45) (a) Neglecting bending

Equation (1):

=

2

41 DsF pi

Equation (2):

=

2

412 csF s pi

Equation (3): ( )bcsF 2= Equation (4): ( )acsF = Equation (5): ( )[ ]bcdsF = 2 Equation (6): ( )mbsF s 4= Equation (7): ( )nbsF s 2= Equation (8): ( )acdsF =

(b) Ns

sy

= and Ns

ssy

s =

Therefore sss 6.0=

Equate (2) and (1)

=

=

22

41

412 DscsF s pipi

=

2241

216.0 Dscs

Dc 9129.0= Equate (3) and (1)

( )

==

2

412 DsbcsF pi

( ) 2419129.02 DDb pi=

Db 4302.0=

Equate (4) and (1)

==

2

41 DssacF pi

( ) 2419129.0 DDa pi=

Da 8603.0=


67

Equate (5) and (1) ( )[ ]

==

2

412 DsbcdsF pi

( )( ) 2414302.09129.02 DDd pi=

Dd 8257.1= Equate (6) and (1)

( )

==

2

414 DsmbsF s pi

( )( ) 2414302.046.0 DDm pi=

Dm 7607.0= Equate (7) and (1)

( )

==

2

412 DsnbsF s pi

( )( ) 2414302.026.0 DDn pi=

Dn 5214.1= Equate (8) and (1)

( )

==

2

41 DsacdsF pi

( ) 2419129.08257.1 DaDD pi=

Da 8604.0=

Summary Da 8604.0= Db 4302.0= Dc 9129.0= Dd 8257.1= Dm 7607.0=

Dn 5214.1=

47. Design a connection similar to the one shown for a gradually applied and reversed load of 12 kips. Base design stresses on yield strengths and let the material be AISI C1040 steel, annealed. Examine the computed dimensions for proportion, making changes that you deem advisable.

Solution: (See figure in Prob. 45 and refer to Prob. 46)


68

4=N based on yield strength for gradually applied and reversed load. For AISI C1040, annealed

ksisy 47= (Fig. AF 7) ksiss ysy 286.0 ==

ksiNs

sy 75.11

447

===

=

2

41 DsF pi

=

2

4175.1112 Dpi

inD 14.1=

say inD811=

inDa 97.08118604.08604.0 =

==

but Da >

say ina411=

inb 48.08114302.0 =

=

say inb21

=

inc 030.18119129.0 =

=

say inc 1=

ind 05.28118257.1 =

=

say ind 2=

inm 86.08117607.0 =

=

say inm87

=

inn 71.18115214.1 =

=

say inn431=

Dimension:

ina411=


69

inb21

=

inc 1= ind 2=

inm87

=

inn431=

inD811=

48. Give all the strength equations for the union of rods shown.

Problems 48 68. Solution:

=

2

41 dsF pi Equation (1)

( )adsF s pi= Equation (2)

( )tcsF s 2= Equation (3)


70

( )[ ]beDsF s = 2 Equation (4)

setF = Equation (5)

( )teDsF = Equation (6)

( )

=

22

41

eksF pi Equation (7)

( ) ( )

= tememsF 22

41

pi Equation (8)

( )efsF s 2= Equation (9)


71

= etesF 2

41

pi Equation (10)

49-68. Design a union-of-rods joint similar to that shown for a reversing load and material given in the accompanying table. The taper of cotter is to be in. in 12 in. (see 172). (a) Using design stresses based on yield strengths determine all dimensions to satisfy the necessary strength equations. (b) Modify dimensions as necessary for good proportions, being careful not to weaken the joint. (c) Decide upon tolerances and allowances for loose fits. (d) Sketch to scale each part of the joint showing all dimensions needed for manufacture, with tolerances and allowances.

Prob. No. Load, lb. AISI No., As Rolled

49 3000 1020 50 3500 1030 51 4000 1117 52 4500 1020

52 5000 1015 54 5500 1035 55 6000 1040 56 6500 1020

57 7000 1015 58 7500 1118 59 8000 1022 60 8500 1035

61 9000 1040 62 9500 1117 63 10,000 1035 64 10,500 1022

65 11,000 1137 66 11,500 1035 67 12,000 1045 68 12,500 1030


72

Solution: (For Prob. 49 only)

(a) For AISI 1020, as rolled

ksisy 48= ( ) ksiss yys 8.28486.06.0 ===

For reversing load, 4=N based on yield strength

ksiNs

sy 12

448

===

ksiNs

sys

s 2.748.28

===

kipslbF 33000 ==

Equation (1)

=

2

41 dsF pi

=

2

41123 dpi

ind 5642.0=

say ind169

=

Equation (2) ( )adsF s pi= ( )

=

1692.73 api

ina 236.0=

say ina41

=

Equation (5) setF =

et123 = 25.0=et

Equation (10)

= etesF 2

41

pi

= 25.0

41123 2epi

ine 798.0=

say ine1613

=

25.0=et


73

25.01613

=

t

int 308..0=

say int165

=

Equation (6) ( )teDsF =

=

165

1613123 D

inD 6125.1=

say inD851=

Equation (4) ( )[ ]beDsF s = 2

= b

1613

85122.73

inb 256.0=

say inb41

=

Equation (7) ( )

=

22

41

eksF pi

=

22

1613

41123 kpi

ink 989.0= say ink 1= Equation (9)

( )efsF s 2= ( ) f

=

161322.73

inf 256.0= say inf

41

=

Equation (8) ( ) ( )

= tememsF 22

41

pi

=

165

1613

1613

41123

22 mmpi


74

2539.03125.05185.07854.025.0 2 += mm 05146.03125.07854.0 2 = mm

06552.03979.02 = mm inm 032.1=

say inm 1= Equation (3)

( )tcsF s 2= ( ) c

=

16522.73

inc 667.0=

say inc1611

=

DIMENSIONS:

ind169

=

ina41

=

inb41

=

inc1611

=

inf41

=

ine1613

=

int165

=

ink 1=

inD851=

inm 1=

(b) Modified dimensions ind

169

=

ina41

=

inb43

=

inc1611

=


75

inf21

=

ine1613

=

int165

=

ink 1=

inD851=

inm411=

(c) Tolerances and allowances No fit, in010.0

ind 010.05625.0 = ina 010.0250.0 = inf 010.0500.0 = inD 010.0625.1 = ink 010.0000.1 = inm 010.0250.1 =

Fits, Table 3.1, loose-running fits, say RC 8

Female Male

inb0000.00035.0

750.0

+= inb

0065.00045.0

750.0

=


inc0000.00028.0

6875.0

+= inc

0051.00035.0

6875.0

=


ine0000.00035.0

8125.0

+= ine

0065.00045.0

8125.0

=

allowance = 0.0045

int0000.00022.0

3125.0

+= int

0040.00030.0

3125.0

=



76

(d)

ROD

COTTER


77

SOCKET

CHECK PROBLEMS

69. The connection shown has the following dimensions: ind411= , inD

212= ,

inD2111 = , inh 8

5= , int

21

= ; it supports a load of 15 kips. Compute the tensile,

compressive, and shear stresses induced in the connection. What is the corresponding design factor based on the yield strength if the rod and nut are made of AISI C1045, as rolled, and the plate is structural steel (1020)?


78

Problem 69. Solution: Tensile Stresses

(1) ksid

Fs 22.12

411

41

15

41 221

=

==

pipi

(2) ksiD

Fs 4.8

211

41

15

41 22

1

2 =

==

pipi

Compressive Stress

(3) ( ) ksiDDF

s 78.4

211

212

41

15

41 222

12

3 =

=

=

pipi

Shear Stresses

(4) ksiDtF

ss 82.3

21

212

154

=

==

pipi

(5) ksihD

Fss 09.5

85

211

151

5=

==

pipi

For AISI C1045, as rolled (rod and nut) ksisy 591 =

( ) ksiss yys 4.35596.06.01 === For structural steel plate (1020)

ksisy 482 = ( ) ksiss yys 8.28486.06.01 ===

Solving for design factor


79

(1) 83.422.12

591

11

===

s

sN y

(2) 95.649.8

592

21

===

s

sN y

(3) 04.1078.4

483

32

===

s

sN y

(4) 54.782.3

8.28

4

24 ===

s

ys

s

sN

(5) 96.609.5

4.35

5

15 ===

s

ys

s

sN

The corresponding design factor is 83.4=N

70. In the figure, let inD43

= , int167

= , inb433= , and let the load, which is applied

centrally so that it tends to pull the plates apart, be 15 kips. (a) Compute the stresses in the various parts of the connection. (b) If the material is AISI C1020, as rolled, what is the design factor of the connection based on yield strengths?


(a) Tensile stresses

( ) ksiDbtF

s 43.11

43

433

167

151 =

=

=

( )( )

ksiDbt

Fs 43.11

432

433

167

1543

243

2 =

=

=

Compressive bearing stress

ksiDtF

s 43.11

167

434

1543

=

==


80

Shearing stress

( ) ( )ksi

D

Fss 24.4

24315

2414

22

4 =

=

=

pipi

(b) For AISI C1020, as rolled ksisy 48=

ksiss yys 8.286.0 ==

s

sN y= or

s

ys

s

sN =

Using s

sN y=

2.443.11

48===

s

sN y

Using s

ys

s

sN =

8.624.4

8.28===

s

ys

s

sN

Therefore the design factor is 2.4=N

71. For the connection shown, let ina1615

= , inb169

= , inc43

= , ind211= ,

inD43

= , innm1615

== . The material is AISI C1040, annealed (see Fig. AF 1). (a) For a load of 7500 lb., compute the various tensile, compressive, and shear stresses. Determine the factor of safety based on (b) ultimate strength, (c) yield strengths.

Problem 71.

Solution:

(a) Tensile stresses


81

ksiD

Fs 98.16

43

41

5.7

41 221

=

==

pipi

( ) ksicdbF

s 89.8

43

211

1692

5.722

=

=

=

( ) ksicdaF

s 67.10

43

211

1615

5.73 =

=

=

Compressive Stresses (Bearing) ksi

bcF

s 89.8

43

1692

5.724

=

==

ksiac

Fs 67.10

43

1615

5.75 =

==

Shearing Stresses

ksimbF

ss 56.3

169

16154

5.746

=

==

ksinbF

ss 11.7

169

16152

5.727

=

==

For AISI C1040, annealed,Fig. AF 1 ksisy 47= ksisu 79=

ksiss yys 286.0 == ksiss uus 44760 .. ==

(b) Based on ultimate strength 65.4

98.1679

1

===

s

sN u

(c) Based on yield strength 77.2

98.1647

1

===

s

sN y

72. The upper head of a 60,000-lb. tensile-testing machine is supported by two steel rods, one of which A is shown. These rods A are attached to the head B by split rings C. The test specimen is attached to the upper head B so that the tensile force


82

in the specimen pulls down on the head and exerts a compressive force on the rods A. When the machine is exerting the full load, compute (a) the compressive stress in the rods, (b) the bearing stress between the rods and the rings, (c) the shearing stress in the rings,


lbsF 000,60=

(a) ( )( )

ksipsisc 75.11753,113

213

41

2000,602

2==

=

pi

(b) ( )( )

ksipsisb 19.10186,10

2134

41

2000,602

2

==

=

pi

(c) ( )( )( ) ksipsisc 18.3183,3132000,60

===

pi

DEFORMATIONS

73. A load of 22,000 lb. is gradually applied to a 2-in. round rod, 10 ft. long. The total elongation is observed to be 0.03 in. If the stretching is entirely elastic, (a) what is the modulus of elasticity, and (b) what material would you judge it to be, wrought iron or stainless steel (from information available in the tables)? (c) How much energy is absorbed by the rod? (d) Suppose that the material is aluminum alloy 3003-H14; compute its elongation for the same load. Is this within elastic action?

Solution: lbsF 000,22=

inD 2= inftL 12010 ==

in03.0=


83

(a) EAFL

= ( )( )( )( )( ) psiD

FLA

FLE 622 1028203.0120000,2244

====pipi

(b) Use both stainless steel, Table AT 4, psiE 61028= and wrought iron , Table AT 7, psiE 61028= .

(c) Energy absorbed = ( )( ) inlbF == 33003.0000,2221

21

(d) For Aluminum alloy, 3003-H14

psiE 61010= ksisy 21=

( )( )( )( )( ) inDE

FLEAFL 084.0

21010120000,2244

262 ====

pipi

( )( )( ) ysksipsiD

Fs


84

( )( )( ) ysksipsiD

Fs >==== 0.28011,28

2000,8844

22 pipi, not within the elastic limit, therefore

not valid.

75. (a) A square bar of SAE 1020, as rolled, is to carry a tensile load of 40 kips. The bar is to be 4 ft. long. A design factor of 5 based on the ultimate stress is desired. Moreover, the total deformation should not exceed 0.024 in. What should be the dimensions of the section? (b) Using SAE 1045, as rolled, but with the other data the same, find the dimensions. (c) Using SAE 4640, OQT 1000 F, but with other data the same as in (a), find the dimensions. Is there a change in dimensions as compared with part (b)? Explain the difference or the lack of difference in the answers.

Solution: inftL 484 ==

(a) For SAE 1020, as rolled ksisu 65= , ksiE 000,30=

AF

Ns

s u ==

240

565

x=

inx 754.1=

EAFL

= ( )( )

( ) 2000,304840024.0

x=

inx 633.1=

Therefore say inx431=

(b) For SAE 1045, as rolled ksisu 96= , ksiE 000,30=

AF

Ns

s u ==

240

596

x=

inx 443.1=

EAFL

= ( )( )

( ) 2000,304840024.0

x=

inx 633.1=


85


(c) For SAE 4640, as rolled ksisu 152= , ksiE 000,30=

AF

Ns

s u ==

240

5152

x=

inx 15.1=

EAFL

= ( )( )

( ) 2000,304840024.0

x=

inx 633.1=


There is lack of difference in the answers due to same dimensions required to satisfy the required elongation.

76. The steel rails on a railroad track are laid when the temperature is 40 F. The rails are welded together and held in place by the ties so that no expansion is possible due to temperature changes. What will be the stress in the rails when heated by

the sun to 120 F (i1.29)? Solution:

LtL

LEs

==

For steel Finin = 000007.0 psiE 61030=

( )( )( )6103040120000007.0 == tEs psis 800,16=

77. Two steel rivets are inserted in a riveted connection. One rivet connects plates that have a total thickness of 2 in., while the other connects plates with a total thickness of 3 in. If it is assumed that, after heading, the rivets cool from 600 F and that the coefficient of expansion as given in the Text applies, compute the stresses in each rivet after it has cooled to a temperature of 70 F, (no external load). See i1.29. Also assume that the plates are not deformed under load. Is such a stress likely? Why is the actual stress smaller?


86

Solution: tEs =

For steel Finin = 000007.0 psiE 61030=

( )( )( ) ksis 30.111000,3070600000007.0 ==

The stress is unlikely because it is near the ultimate strength of steel. Actual stress must be smaller to allow for safety.

78. Three flat plates are assembled as shown; the center one B of chromium steel, AISI 5140 OQT 1000 F, and the outer two A and C of aluminum alloy 3003-H14, are fastened together so that they will stretch equal amounts. The steel plate is 2 x in., the aluminum plates are each 2 x 1/8 in., inL 30= ., and the load is 24,000 lb. Determine (a) the stress in each plate, (b) the total elongation, (c) the energy absorbed by the steel plate if the load is gradually applied, (d) the energy absorbed by the aluminum plate. (e) What will be the stress in each plate if in addition to the load of 24,000 lb. the temperature of the assembly is increased by 100 F?

Problem 78, 79. Solution: For chromium steel, AISI 5140 OQT 1000F (Table AT 7)

Finin = 000007.01 ksipsiE 000,301030 61 ==

For aluminum alloy, 3003-H14 (Table AT 3) Finin = 0000129.02

ksipsiE 000,101010 62 ==

(a) CA PP = FPPP CBA =++

(1) kipsFPP BA 242 ==+ ( ) 22 25.08

12 inA =

=

( ) 21 1212 inA =

=


87

BA =

1122 EALP

EALP BA

=

( )( ) ( )( )000,301000,1025.0BA PP

=

(2) AB PP 12=

(1) kipsPP AA 24122 =+ kipsPA 714.1=

( ) kipsPB 568.20714.112 ==

Stresses: Aluminum plate

ksiAP

ss ACA 856.625.0714.1

2

====

Chromium steel plate

ksiAP

s BB 568.201568.20

1

===

(b) ( )( )( )( ) inEALPA 021.0

000,1025.030714.1

22

===

(c) Energy absorbed by steel plate ( )( ) inkipsPB === 216.0021.0568.2021

21

(d) Energy absorbed by aluminum plate ( )( ) inkipsPA === 018.0021.0714.121

21

(e) kipsFPP BA 242 ==+ BTAT BA

+=+ tL

AT= 2

tLBT

= 1 ( )( )( ) in

AT0387.0100300000129.0 ==

( )( )( ) inBT

021.010030000007.0 == ( )

( )( ) AAA

A PP

EALP 012.0

000,1025.030

22

===

( )( )( ) B

BBB P

PEALP 001.0

000,30130

11

===

Then BA PP 001.0021.0012.00387.0 +=+

( )AA PP 224001.0012.00177.0 =+ AA PP 002.0024.0012.00177.0 =+


88

kipsPA 45.0= ( ) kipsPB 1.2345.0224 ==


ksiAP

s AA 8.125.045.0

2

===


ksiAP

s BB 1.2311.23

1

===

79. The same as 78, except that the outer plates are aluminum bronze, B150-1, annealed.

Solution:

For aluminum bronze, B150-1, annealed (Table AT 3) ksiE 000,152 =

Finin = 0000092.02

(a)

(1) kipsFPP BA 242 ==+ BA =

1122 EALP

EALP BA

=

( )( ) ( )( )000,301000,1525.0BA PP

=

(2) AB PP 8= kipsPP AA 2482 =+

kipsPA 4.2= ( ) kipsPB 2.194.28 ==


ksiAP

ss ACA 6.925.04.2

2

====


ksiAP

s BB 2.1912.19

1

===


89

(b) ( )( )( )( ) inEALPA 019.0

000,1525.0304.2

22

===

(c) Energy absorbed by steel plate ( )( ) inkipsPB === 182.0019.02.1921

21

(d) Energy absorbed by aluminum plate ( )( ) inkipsPA === 023.0019.04.221

21

(e) kipsFPP BA 242 ==+ BTAT BA

+=+ tL

AT= 2

tLBT

= 1 ( )( )( ) in

AT0276.0100300000092.0 ==

( )( )( ) inBT

021.010030000007.0 == ( )

( )( ) AAA

A PP

EALP 008.0

000,1525.030

22

===

( )( )( ) B

BBB P

PEALP 001.0

000,30130

11

===

Then BA PP 001.0021.0008.00276.0 +=+

( )AA PP 224001.0008.00066.0 =+ AA PP 002.0024.0008.00066.0 =+

kipsPA 74.1= ( ) kipsPB 52.2074.1224 ==


ksiAP

s AA 96.625.074.1

2

===


ksiAP

s BB 52.20152.20

1

===

80. A machine part shown is made of AISI C1040, annealed steel; inL 151 = .,

inL 62 = ., inD 43

1 = ., and inD 21

2 = . Determine (a) the elongation due to a force lbF 6000= ., (b) the energy absorbed by each section of the part if the load is

gradually applied.


90

Problems 80, 81 Solution: For AISI C1040, annealed steel

psiE 61030=

(a) 21 += ( )( )( )

inEA

FL 0068.01030

43

4

1560006

21

11 =

==

pi

( )( )( )

inEA

FL 0061.01030

21

4

660006

22

22 =

==

pi

in0129.00061.00068.021 =+=+=

(b) Energy absorbed ( )( ) inlbFU ==== 4.200068.06000

21

21

11

( )( ) inlbFU ==== 3.180061.0600021

21

22

81. A rod as shown is made of AISI 2340 steel, OQT 1000 F, and has the following dimensions: inL 201 = ., inL 122 = ., inD 8

71 = ., and inD 4

32 = . The unit strain at

point A is measured with a strain gage and found to be 0.0025 in./in. Determine (a) the total elongation, and (b) the force on the rod.

Solution:

For steel ksiE 30000=

(a) EA

FL 22

2==


91

EAF 2=

1

2

1

21

1

2

1

121 LD

DLAA

EAELA

=

==

( ) inLLDD

T 067.0122087430025.0

2

21

2

1

221 =

+

=

+

=+=

(b) ( ) kipsEAF 13.33000,3043

40025.0

2

2 =

==

pi

82. A rigid bar H is supported as shown in a horizontal position by the two rods (aluminum 2024 T4, and steel AISI 1045, as rolled), whose ends were both in contact with H before loading was applied. The ground and block B are also to be considered rigid. What must be the cross-sectional area of the steel rod if, for the assembly, 2=N based on the yield strengths?


For aluminum 2024-T4 (Table AT 3) ksisy 471 = , ksiE 600101 ,=

For steel AISI 1045, as rolled (Table AT 7) ksisy 592 = , ksiE 000,302 =

[ ]0= GM ( ) ( ) ( )20241224 21 =+ RR 402 21 =+ RR Equation (1)


92

122421

=

21 2 =

11

111 AE

LR=

22

222 AE

LR=

inftL 9681 == inftL 144122 ==

21 5.0 inA =

21 2 =

22

22

11

11 2AELR

AELR

=

( )( )( )

( )( ) 2

21

000301442

506001096

A

RR

,.,=

2

21

530A

RR

.=

But Ns

AR

sy2

2

22 ==

5.292

592

2==

AR

( ) kipsR 64155295301 ... ==

Ns

AR

sy1

1

11 ==

247

5.01

=R

kipsR 75.111 = use kipsR 75.111 =

( ) kipsR 5.1675.112402 == 22

2 56.05.295.16

5.29inRA ===


93

83. The bar shown supports a static load kipsF 5.2= with 0= ; ind 3= .,

inL 10= ., inh432= . inb 1= . It is made of AISI 1035, as rolled. (a) How far

does point C move upon gradual application of the load if the movement of A and B is negligible? (b) How much energy is absorbed?


[ ] = 0AM ( )FLddRB += ( )( )5.21033 +=BR kipsRB 83.10= [ ] = 0BM LFdRA =

( )5.2103 =AR kipsRA 33.8=


94

3= xRxRM BA

383.1033.822

== xxdy

ydEIM

122 3415.5165.4 Cxx

dydyEI +=

2133 3805.1388.1 CxCxxEIy ++=

When 0=x , 0=y ( ) ( ) ( ) 2133 00805.10388.10 CCEI ++=

02 =C When 3=x , 0=y

( ) ( ) ( ) 030805.13388.10 133 ++= CEI 492.121 =C

xxxEIy 492.123805.1388.1 33 = When inLdx 13=+=

( ) ( ) 108213492.1210805.113388.1 33 ==EIy For AISI 1035, as rolled , ksiE 000,30=

( )( ) 433 7331.112

75.2112

inbhI ===

1082=EIy ( )( ) 10827331.1000,30 =y

iny 021.0= , upward.


95

PRESSURE VESSELS

84. A storage tank for air, 36 in. in diameter, is to withstand an internal pressure of 200 psi with a design factor of 4 based on us . The steel has the strength equivalent of C1020 annealed and the welded joints should have a relative strength (efficiency) of 90 %. Determine a suitable plate thickness. Compute the stress on a diametral section and compare it with the longitudinal stress.

Solution: For C1020 annealed

ksisu 57=

ksiNs

s u 25.144

57===

Solving for plate thickness

tpD

s2

=

ksipsip 2.0200 == inD 36=

( )( )( )9.02

362.525.14t

s ==

int 281.0=

say int165

=

Stress on diametral section ( )( )

( )ksi

t

pDs 40.6

9.01654

362.04

=

==

Stress on longitudinal section ( )( )

( )ksi

t

pDs 80.12

9.01652

362.02

=

==

Stress on diametral section < stress on longitudinal section

85. A spherical air tank stores air at 3000 psig. The tank is to have an inside diameter of 7 in. (a) What should be the wall thickness and weight of the tank if it is made of 301, -hard, stainless steel, with a design factor of 1.5 based on the yield strength and a joint efficiency of 90 %. (b) Compute the wall thickness and weight if annealed titanium (B265, gr. 5) is used? (c) What is the additional saving in weight if the titanium is hardened? Can you think of circumstances for which the higher cost of titanium would be justified?


96

Solution:

(a) For 301, hard, stainless steel ksisy 75= (Table AT 4)

ksiNs

sy 50

5.175

===

ksipsip 33000 ==

tpD

s4

=

( )( )( )90.04

7350t

=

int 117.0=

3286.0 inlb= ( ) ( )( ) lbtDtrW 2.5286.0117.074 222 ==== pipipi

(b) For annealed titanium B265, gr. 5 ksisy 130= (Table AT 3)

ksiNs

sy 67.86

5.1130

===

ksipsip 33000 ==

tpD

s4

=

( )( )( )90.04

7367.86t

=

int 061.0=


(c) For hardened titanium ksisy 158= (Table AT 3)

ksiNs

sy 105

5.1158

===

ksipsip 33000 ==

tpD

s4

=

( )( )( )90.04

73105t

=


97

int 056.0=


Savings in weight = ( ) %8%10050.1

38.150.1=

Circumstances: less in weight and small thickness.

86. Decide upon a material and estimate a safe wall thickness of a cylindrical vessel to contain helium at 300 F and 2750 psi. The welded joint should have a relative strength 87 %, and the initial computations are to be for a 12-in.-diameter, 30-ft.-long tank. (Note: Mechanical properties of metals at this low temperature are not available in the Text. Refer to INCO Nickel Topics, vol. 16, no. 7, 1963, or elsewhere.)

Solution: From Kents Handbook, Table 8 Material Hot Rolled Nickel At 300 F, ksisu 100= , 4=N (Table 1.1)

ksiNs

s u 254

100===

tpD

s2

=

ksipsip 75.22750 == inD 12=

%87= ( )( )

( )87.021275.225

ts ==

int 759.0=

say int43

=

CONTACT STRESSES

87. (a) A 0.75-in. diameter roller is in contact with a plate-cam surface whose width is 0.5-in. The maximum load is 2.5 kips where the radius of curvature of the cam surface is 3.333 in. Compute the Hertz compressive stress. (b) The same as (a) except that the follower has a plane flat face. (c) The same as (a) except that the roller runs in a grooved face and contacts the concave surface. (d) What is the maximum shear stress for part (a) and how far below the surface does it exist?

Solution:


98

(a) inr 75.02 1 = , inr 375.01 = inr 333.32 =

kipsF 5.2= inb 5.0=

21

21

21max 11

1135.0

+

+

=

EEb

rrF

sc

ksiE 000,30=

( )ksisc 279

000,3025.0

333.31

375.015.235.0

21

max =

+

=

(b) ( )

ksisc 126

000,3025.0

333.31

333.315.235.0

21

max =

+

=

(c) ( )

ksisc 249

000,3025.0

333.31

375.015.235.0

21

max =

=

(d) Maximum shear stress ( ) ksiss cs 842793.03.0 maxmax ===

Location:

( ) ( )( )in

rr

EEs

w

c

023.0

333.31

375.01

000,3023.012794

11

1114 2

21

21

2max

=

+

=

+

+

=

88. Two 20o involute teeth are in contact along a line where the radii of curvature of the profiles are respectively 1.03 and 3.42 in. The face width of the gears is 3 in. If the maximum permissible contact stress for carburized teeth is 200 ksi, what normal load may these teeth support?


99

Solution: inr 03.11 = inr 42.32 =

inb 3= ksisc 200max =

21

21

21max 11

1135.0

+

+

=

EEb

rrF

sc

ksiE 000,30=

21

max

000,3023

42.31

03.1135.0

200

+

==

Fsc

kipsF 18=

TOLERANCES AND ALLOWANCES

89. The pin for a yoke connection has a diameter of D of in., a total length of 2 in., with a head that is 1 in. in diameter and 3/8 in. thick. The tolerance on D (both pin and hole) is 0.003 in., with an allowance of 0.001 in., basic-hole system. Sketch the pin showing all dimensions with appropriate tolerances.

Solution: inD 75.0=

For pin

inD003.0000.0

749.0

+=

For hole

inD000.0003.0

750.0

+=

Sketch


100

90. A shaft with a nominal diameter of 8 in. is to fit in a hole. Specify the allowance, tolerances, and the limit diameters of the shaft and hole on a sketch for: (a) a close sliding fit, (b) a precision-running fit, (c) medium-running fit, (d) a loose-running fit.

Solution: inD 8=

(a) For close-sliding fit, RC 1

Hole, in Shaft, in +0.0008 - 0.0006 -0.0000 -0.0012

Allowance = 0.0006 in With tolerances,

Hole inD0000.00008.0

0000.8

+=

Shaft inD0006.00000.0

9994.7

+=

Limit dimension, Hole intoD 0008.80000.8= Shaft intoD 9988.79994.7= Sketch


101

(b) For a precision-running fit, RC 3 Hole, in Shaft, in +0.0012 -0.0020 -0.0000 -0.0032 Allowance = 0.0020 in With tolerances,

Hole inD0000.00012.0

0000.8

+=

Shaft inD0012.00000.0

9980.7

+=


(c) For medium-running fit, RC5, RC 6. Say RC 5

Hole, in Shaft, in +0.0018 -0.0040 -0.0000 -0.0058


Hole inD0000.00018.0

0000.8

+=

Shaft inD0018.00000.0

9960.7

+=



102

(d) For loose-running fit, RC 8, RC 9. Say RC 8

Hole, in Shaft, in +0.0070 -0.0100 -0.0000 -0.0145


Hole inD0000.00070.0

0000.8

+=

Shaft inD0045.00000.0

9900.7

+=


91. The same as 90, except that the nominal diameter is 4 in.

Solution: inD 4=

(a) For close-sliding fit, RC 1


103

Hole, in Shaft, in +0.0006 -0.0005 -0.0000 -0.0009

Documents

Design for Simple Stresses