2-lectures lec 38

8/12/2019 2-Lectures LEC 38

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CH-8 LEC 38 Slide 1Dr. A. Aziz Bazoune Chapter 8: Screws, Fasteners and the Design of Nonpermanent Joints


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8-1 Thread Standards and Definitions

8-2 The Mechanics of Power Screws8-3 Strength Constraints8-4 Joints-Fasteners Stiffness8-5 Joints-Member Stiffness8-6 Bolt Strength

8-7 Tension Joints-The External Load8-8 Relating Bolt Torque to Bolt Tension8-9 Statically Loaded Tension Joint with Preload8-10 Gasketed Joints8-11 Fatigue Loading of Tension Joints

8-12 Shear Joints8-13 Setscrews8-14 Keys and Pins8-15 Stochastic Considerations



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8-11 Fatigue Loading of Tension Joints8-12 Shear Joints8-13 Setscrews8-14 Keys and Pins



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Design I Example 5 (Example 8-4 Textbook)

(a) The grip is l = 1.50 in. From Table A-

31, the nut thickness is 35/64 in.

Adding two threads beyond the nut

of 2/11ingives a bolt length of

35 21.50 2.229 in

64 11

L



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(a) From Table A-17, the next fraction size bolt is 2 1/4 in.



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2 0.25 in, L 6inT

L d



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Fatigue Analysisa) External Load

8-7 Tension Joints-The External Load



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Design I8-11 Fatigue Loading of Tension Joints


Fatigue Analysis

In general, bolted joints are subject to 0-Pmax, e.g pressure vessels, flanges, pipes,

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Table 8-16 lists average fatigue stress-concentration factors for the fillet under

the bolt head and also at the beginning of the threads on the bolt shank.

These are already corrected for notch sensitivity and for surface finish.

Useof rolled threads is the predominant method of thread-forming in screwfasteners, where Table 8-16 applies.

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In thread-rolling the amount of cold-work and strain strengthening is unknown to

the designer; therefore, fully corrected (including Kf) axial endurance strength isreported in Table 8-17.

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Fatigue-loaded bolted joints subjected toFatigue action can be analyzed directly by

the methods of Chapter 7.

Fatigue loading is the one in which the

externally applied load fluctuates between

zeroand some maximum force P.

Fmax= Fb and Fmin= Fi Fa= (Fmax Fmin)/2 = (FbFi)/2

a= Fa/At


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2

2

b iba

b ibm

F FF

F FF

( )

2 2 2

( )

2 2 2

b i i iba

t t t

b i i i i

bmt t t t

bm ba i

F F CP F F CP

A A A

F F CP F F F CP

A A A A

Fmax= Fb andFmin= Fi Fa= (Fmax Fmin)/2 = (FbFi)/2

Fm= (Fmax + Fmin)/2 = (Fb+ Fi)/2

a= Fa/At

(8-34)

(8-35)

(8-36)

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On the designers fatigue


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On the designer s fatigue

diagram, shown in Figure 8-20,

the load line is .

High Preload is especially

important in fatigue. iis aconstant the load line at Fi/Athas

a unit slope, r = 1.0

m a i

Figure 8-20Designers fatigue diagram showing a Goodman

failure locus and how a load line is used to definefailure and safety in preloaded bolted joints infatigue.

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8 i di f i i


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Next, find the strength components Saand Smof

the fatigue failure locus. These depend on thefailure Criteria.

Goodman1a m

e ut

S S

S S

Gerber 2

1a m

e ut

S S

S S

ASME-Elliptic 22

1a me p

S S

S S

(8-39)

(8-38)

(8-37)

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For simultaneous solution between Eq. (8-36), as Sa= Sm+ i and each of Eqs. (8-37) to (8-39)


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Goodman

e ut ia

ut e

m a i

S SS

S SS S

Gerber

2 21

4 2

2a ut ut e e i ut i e

e

m a i

S S S S S S S

SS S

ASME-Elliptic

2 2

2 2e

a p p e i i e

p e

m a i

SS S S S S

S SS S

(8-43)

(8-42)

(8-40)

(8-41)

q ( ), a m i q ( ) ( )

gives

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When using relations of this section, be sure to use K

ffor both

a

and m

.

Otherwise, the slope of the load line will not remain1 to 1.

The factor of safety guarding against fatigue is given by

Applying this to the Goodman criterion, for example, with Eqs. (8-34) and

(8-40) and gives

af

a

Sn

i i tF A

2e ut t i

f

ut e

S S A F n

CP S S

(8-44)

(8-45)

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With no preload C=1 F =0 and Eq (8-45) becomes


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With no preload, C=1,Fi=0, and Eq. (8-45) becomes

Preload is beneficial for resisting fatigue when is greater than unity. For

Goodman, Eqs.(8-45) and (8-46) with puts an upper bound on the

preloadFiof

If this cannot be achieved, and nfis unsatisfactory, use the Gerber or the ASME-elliptic criterion to obtain a less conservative assessment.

After solving Eq. (8-44), you should also check the possibility of yielding, using

the proof strength

0f f

n n

(8-46)

(8-47)

02

e ut t

f

ut e

S S An

P S S

0 1

f fn n

1i ut tF C S A

pp

m a

Sn

(8-48)

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Example 6 (Example 8-5 Textbook)

Figure 8-21Pressure-cone frustum member model for a

cap screw. For this model the significant sizesare

Where l = effective grip. The solutions are for=30oand dw=1.5d.

2 2

2

1

2

2

2

tan 1.5 0.577

1.5

w

w

h t t d l

h D t d

D d l d l

D d d

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Figure 8-22

Designers fatigue diagram forpreloaded bolts, dr showing themodified Goodman locus, theGerber locus and the Langer proofstrength locus, with an explodedview of the area of interest, The

strengths used areSp= 85 kpsi, Se= 18.6 kpsi, andSut=120 kpsi. The coordinates are

A: i=63.72 kpsi

B: a=3.1 kpsi, m=66.82 kpsiC: Sa=7.55 kpsi, Sm=71.29 kpsi

D: Sa=10.64 kpsi, Sm=74.35 kpsiE: Sa=11.32 kpsi, Sm=75.04 kpsi

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