Md-05 Repeated Loading

7/31/2019 Md-05 Repeated Loading

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330:148 (g)Machine Design

Nageswara Rao Posinasetti

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5. Repeated LoadingObjectives Identify the various kinds of loading encountered on a

part and learn to combine them as appropriate.

Determine the level of stress in a material at which ahypothesized defect would not propagate.

Recognize what types of factors affect thisendurance limit for the material.

Factor in the effect of shapes and discontinuities asthey affect the stress concentration factors.

Gain experience using fatigue equations whendesigning parts subject to repeated loads.

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Dynamic Loads

Forces that vary frequently in magnitudeand direction are called dynamic loads.

Varying load: Magnitude changes but not the direction

Reversing load: Both magnitude and direction

Shock load: Impact

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Dynamic Strength

Fatigue failure in a material is when itfails even when the stress may not gobeyond the proportional limit.

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Dynamic Strength

Courtesy Esposito - Machine Design,

1991, Delmar PublishersAugust 15,

2007

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Dynamic Strength


1991, Delmar Publishers


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Fatigue failureTensile stress

Compressive stress

Time, t0

T

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Fatigue failure

After some number of stress reversal cycles acrack is initiated.

This crack propagates towards the centre.

Many more cracks are formed around theperiphery.

Ultimately the shaft breaks when the stress inthe remaining solid area exceeds the ultimatestrength, as shown in Fig. 5.4.

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Fatigue failure

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Fatigue failure

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Endurance Limit and EnduranceStrength

It is the strength of a material to resistthe propagation of cracks under stress

reversals. Endurance Limit (Sn): Is the stress value

below which an infinite number of cycleswill not cause failure

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EnduranceLimit



Sn = 0.5 Su


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Modifying factors for endurancelimits

The actual value of endurance limit tobe applied needs to be modifieddepending upon the application

Type of stress

Ctype = 0.8 axial

Ctype = 0.8 shear or torsion

Ctype = 1.0 bending

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Modifying factors for endurancelimits

Size Csize = 1.00 for < inch

Csize = 0.85 for < to 2 inches

Csize = 0.75 for < 2 inches and over

Surface

Rough surfaces have stress raisers

Sn = Ctype Csize Csurface Sn

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Surface Effect

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Surface Effect

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Variation of Stresses

Mean stress, Sm Alternating stress, Sa

2

SSS minmax

m

+=

2

SSS minmaxa

=

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Theories of failure

Soderberg equationn

a

SS

N1 +=

y

m

SS

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Example Problem 5-1Theories of Failure: Soderberg Equation

For a smooth rotating shaft with no

sharp corners or change in shape,determine the required diameter underthe loading condition shown.

Ignore any torque.

The surface of the shaft is highlypolished.

The shaft is made from annealed AISI4140 steel.

Use a safety factor of 2.

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Example Problem 5-1Theories of Failure: Soderberg Equation (contd.)

For the first trial, assume the resultant diameter will be between and 2 inches:

Z

lbin

Z

MS

Z

lbin

Z

MS

lbinFL

M

lbininlb

M

FLM

==

==

==

==

=

5000

5000

50004

50004

40500

4

minmin

maxmax

min

max

max

Su = 95 ksi Sy = 60 ksi(Appendix 4)

First calculate Smax and Smin:

(Appendix 2)

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Example Problem 5-1Theories of Failure: SoderbergEquation (contd.)

Find Smean and Salternating:

Z

lbinZ

lbin

Z

lbin

S

SSS

Z

inlb

Z

lbin

S

SSS

alt

alt

mean

mean

=

=

=

=

=

+=

5000

2

50005000

2

02

50005000

2

minmax

minmax

(5-2)

(5-3)

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Find the endurance limit modifying factors:

Sn= Csize Csurface Ctype Sn

As surface is polished: Csurface = 1

As Sn is unknown, use .5 Su:

Use Csize = . 85 assuming .5 < D < 2

Bending Ctype = 1

Sn = .85 (1) (1) (.5) 95 ksi

Sn = 40.375 ksi

(5-1)

(Figure 5-7b)

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1

N=

SmSy

+SaSn

1

2=

5000 in-lb

Z

40,375 lb/in2

Z = .247 in

Z =D32

D3

=32 Z

D =3 32 (.247 in3)

D = 1.36 in

Use D = 1-3/8 or 1.375 in


(5-5)

Appendix 3

Substituting into the

Soderberg equation:


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Example Problem 5-2Theories of Failure: SoderbergEquation

A 12-inch-long round cantilever beam is loaded repeatedly but no load reversal occurs.

If the bar is made from annealed AISI 302 stainless steel and the surface is polished fora safety factor of 1.6, find the required diameter using the Soderberg method.

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Su = 90 ksi Sy = 37 ksi Sn = 34 ksi

0

000,12

0

000,12

121000

min

maxmax

min

max

max

max

=

==

=

=

=

=

S

Z

lbin

Z

MS

M

lbinM

inlbM

FLM

(Appendix 8)

(Appendix 2) Find Smax and Smin:

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(5-2)

Smean =Smax + Smin

2

Smean =

12,000 in-lb

Z+ 0

2

Smean =6,000 in-lb

Z

(5-3)

Salt =Smax Smin

2

Salt =

12,000 in-lb

Z 0

2

Salt =6,000 in-lb

Z


Find Smean and Salternating:

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Csize = .85 (assume < D < 2 inches)

Ctype = 1 (bending)

Csurface = 1 (polished surface)

Sn = Csize Csurface Ctype Sn'

Sn = .85 (1) (1) 34 ksi

Sn = 28.9 ksi


Find the endurance limit modifying factors:

(5-1)

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(5-4)

1

N=

SmSy

+SaSn

1

1.6=

6,000 in-lb

Z

37,000 lb/in2=

6,000 in-lb

Z

28,900 lb/in2

Z = .592 in3

(Appendix 3)

Z =D

3

32

D3 =32 Z

D =

3 32 (.592 in3)

D = 1.82 inches


Substituting into the Soderbergequation:

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Stress Concentration:Discontinuities in part contour

Shafts with shoulders to accommodatethe seating of bearings

Key ways in shafts that use key tosecure pulleys, cams and gears

Threads on one end of a bolt and a

head on the other end

Fillet at the base of gear teeth


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Stress Concentration

Physical discontinuity is called a stressraiser or a region of stressconcentration.

Stress concentration factor, K =avg_calc

max_actual

S

S

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Stress Concentration

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Stressconcentration

HB

avgA-A

FS =

KSS avgmax =

K = Stressconcentration factor



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Stress concentration



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Sample problem 2.1: Fig. 2.18 shows a rectangular barcontaining a hole and undergoing a tensile force of10,000 pounds. In this case t = 2 in., d = 0.5 in., and w= 1.5 in.a) Find the tensile stress at no hole condition;b) Find the average stress at the hole section assuming

no stress concentration; andc) Find the maximum stress at the hole section taking

into consideration the effect of the stress raiser.

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Stress concentration problem (Fig. 2.18)

2

0.5 1.5




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Solution

a) Stress with no hole = 10000/ (2 1.5)

= 333 psi

b)

= 5000 psi

c) From Fig 2.17, K = 2.3

= 11,500 psi

HB

avgA-A

FS =

0.52-1.52

10000Savg

=

KSS avgmax =

2.35000Smax =

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Reducing stress concentrations

Make all transitions as gradual aspossible. (Fig. 2.20)

Drill holes on both sides of key way (Fig.2.19)

Thread root diameter equals theadjacent shank diameter. (Fig. 2.21)

Add shoulders for press fittedcomponents. (Fig. 2.22)

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Stress concentration Fig. 2.19)

Courtesy Esposito - Machine Design,1991, Delmar Publishers

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Stress concentration Fig. 2.20)


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Stress concentration (Fig. 2.21)


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Stress concentration (Fig. 2.22)




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Allowable stress &Factor of Safety

terialductile_ma

ngthyield_stre

FSS

terialbrittle_ma

trengthultimate_s

FS

SSallowable =

Sallowable =

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Uncertainity

Variability of material composition andproperties

Effect of processing on properties

Effect of nearby assemblies such asweldments

Effect of thermo-mechanical treatment

on properties

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Uncertainity

Intensity and distribution of loading

Validity of mathematical models used

Intensity of stress concentrations

Influence of time on strength andgeometry

Effect of corrosion

Effect of wear

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Unknowns Exact type and magnitude of all loads

Material property variations

Precise stress concentration effects

Extremes of environmental conditions,such as heat and moisture

Approximate stress analysis formulae

Residual stresses produced duringmanufacturing

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Factor of safety

Determining a realistic factor of safety isvery difficult.

Higher factor increases cost and lowerfactor is likely to cause a prematurefailure.

Hence

FS = a b c d

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Factor a

Considers the variability of appliedloads.

a = 1 for loads that are constant inmagnitude and directions

a = 2 for complete load reversals in order totake fatigue into account


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Factor b

Considers the abruptness with whichloads are applied.

b = 1 for gradually applied loads

b = 2 for suddenly applied loads

b = 3 or more (depending on the severity) forimpact loads

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Factor c

Considers the consequences of failurerelated to human safety, cost and so on.

c normally varies between 1.2 and 2depending upon the seriousness of thefailure

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Factor d

Differentiates FS ductile material frombrittle material.

d =strengthYield

strengthUltimate

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