# Solucionario faires

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14-Jul-2015Category

## Engineering

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SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 1 of 131

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

.

SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 2 of 131

inbh16155.1 ==

(b) Based on yield strength

N = factor of safety = 3 for repeated but not reversed load (Table 1.1)

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=

SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 3 of 131

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

6000,55

b=

inb 763.0= say in87

.

inbh16515.1 ==

(b) Based on yield strength

N = factor of safety = 3 for repeated but not reversed load (Table 1.1)

AF

Ns

s ud ==

25.18000

3500,36

b=

inb 622.0= say in1611

.

inbh32115.1 ==

(c) Elongation = AEFL

= where,

in005.0= lbF 8000=

psiE 61025= inL 15=

25.1 bA = then,

SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 4 of 131

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

bh 5.1= therefore 25.1 bA =

(a) Based on ultimate strength

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 ==

(c) Elongation = AEFL

= where,

in005.0= lbF 8000=

psiE 6105.14 =

SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 5 of 131

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

(a) Based on ultimate strength

u

u

s

FNA =

( )( )5.15227.398

42

=dpi

ind 62.1= say in851

(b) Based on yield strength

y

y

s

FNA =

( )( )5.13227.394

42

=dpi

SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 6 of 131

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

SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 7 of 131

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 61025=

( )( )( )( )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

SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 8 of 131

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.

SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 9 of 131

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

===

SECTION 1 DESIGN FOR SIMPLE STRESSES

Page 10 of 131

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