mechanical design an integrated approach Part 6 ppsx

Use the maximum shear stress and the energy of distortion failure criteria.. Determine the vessel wall thickness on the basis of the maximum shear stress and the energy of distortion fai

EXAMPLE 7.5 Finding the Safety Factor in the Yielding of Material under

Combined Static Loading

"Phe stresses shown in Figure 7.7 occur at a point ina ductile structural member of yield strength Sy

Determine the factor of safety n with respect to inelastic deformation

Design Decisions: The member is made of steel of 5, = 36 ksi Use the maximum shear stress

and the energy of distortion failure criteria

Solution: Following the usual sign convention, we write the stress components o, = ~10 ksi,

oy = 2ksi, and try = 8 ksi Substituting into Eq: (3.33),

es i( +82 = —4 + lŨ

oy = ~ 1A ksi

or

ơi = 6 ksi, Maximum shear stress theory Through the use of Eq (7.7),

36 6+ l4= —

A thin-walled conical vessel, or tank, is supported on its

edge and filled with a liquid, as depicted in Figure 7.8

Expressions for stress in this shell are developed in Case Study 16-2 Determine the vessel wall thickness on

the basis of the maximum shear stress and the energy of distortion failure theories

2 The factor of safety against yielding is n

The vessel is taken to be simply supported from the top

Solution: The tangential stress og = o; and meridional stress os = 02 in the tank are expressed by Eqs (16.76a) and (16.77a) as follows

ft = vessel wall thickness

œ = half angle at the apex of cone

y = specific weight of liquid

FAILURE ANALYSIS OF A CONICAL TANK

The largest magnitudes of these principal stresses are given by

equation in the form

2

ra0astt n tana

(7.19a)

Sy cose

Maximum energy of distortion criterion Inasmuch

as the maximum magnitudes of o; and a2 occur at differ- ent locations, we must first determine the section at which combined stresses are at a critical value For this purpose,

we substitute Eq (a) into Eq (7.14):

San hy) tan œ " h 2 tanœ TỶ

TH ⁄ 37 P dt cosa

¢ ) tan œ A 2 tan œ

m cosa T33)” 27cosa

(e)

Differentiation of Eq (c) with respect to the variable y and

equating the result to 0 gives [20],

y = 0.52h

Introducing this value of y back into Eq (c), the thickness

of the vessel is found to bé

yhˆn tana

= 0,2

Comment: The thickness according to the maximum

shear stress criterion is therefore 10% larger than that based on the maximum distortion energy criterion

7.9 COMPARISON OF THE YIELDING THEORIES

Two approaches may be used to compare the theories of yielding heretofore discussed The first comparison equates, for each theory, the critical values corresponding to uniaxial loading and torsion Referring to Table 7.3,

Maximum shearing stress theory: Sys = 0.505,

equivalent, the octahedral shear stress theory: Sys = 0.5776,

We observe that the difference in strength predicted by these criteria is not substantial A second comparison may be made by means of superposition of Figures 7.4 and 7.6 This is left as an exercise for the reader

Experiment shows that, for ductile materials, the yield stress obtained from a torsion test is 0.5 to 0.6 times that determined from simple tension test We conclude, therefore, that the energy of distortion criterion or octahedral shearing stress criterion is most suitable

for ductile materials However, the shear stress theory, which results in Sy, = 0.50S,, is

simple to apply and offers a conservative result in design

As a third comparison, consider a solid shaft of diameter D and tensile yield strength

Sy subjected to combined loading consisting of tension P and torque T The yield criteria based on the maximum shear stress and energy of distortion theories, for n = 1, are given

Figure 7.9 Yield curves for torsion-tension shaft The points shown in this

CHAPTER 7 ° FAIHLURE CRITERIA AND RELIABILITY

Adimensionless plot of Eqs (a) and some experimental results are shown in Figure 7.9 We note again particularly good agreement between the maximum energy of distortion crite- rion and experimental data for ductile materials The difference in results is not very great, however, and both theories are widely used in design of members

283

7.10 MAXIMUM PRINCIPAL STRESS THEORY

In accordance with the maximum principal stress theory, credited to W J M Rankine

(1820-1872), a material fails by fracturing when the maximum principal stress reaches the

ultimate strength S, in a simple tension test Thus, at the beginning of the fracture,

For the case of plane stress (03 = 0), Eq (7.21), the fracture condition, is given by

leu ` or Tới L& (7.22)

principal stress

_ EXAMPLE 7.6 The Failure of a Pipe of Brittle Material under Static Torsion Loading

‘A cast pipe'of outer diaméter D and inner diameter d is made of an aluminum alloy having ultimate

strengths in tension and compression S, and S,., respectively Determine the maximum torque that

can be applied without causing rupture

Given: D = 100 mm, d= 60 mm, Sy = 200 MPa, Sue = 600 MPa Design Decision: Use the maximum principal stress theory and a safety factor of n = 2

Solution © The torque and the maximum shear stress are related by the torsion formula:

Jo TK 50.054 0.03") ¢ =1709x1 -6

+ : 3005 170.9 x 107°r {a}

The state ‘of stréss is described by

OL = —=Øy =f, 03 = 0 From Eqs (7.22) and the preceding, we have t = S,./n Then, Eq, (a) results in

that material This criterion uses Moht’s circles of stress, Using the extreme values of prin-

cipal stress enables one to apply the Mohr approach to either two- or three-dimensional

cases

Experiments are performed on a given material to determine the states of stress that

result in failure Each such state of stress defines a Mohr circle, When the data describing

states limiting stress are derived from only simple tension, compression, and torsion tests,

CHAPTER7 © FAILURE CRITERIA AND RELIABILITY

the three resulting circles are sufficient to construct the envelope, labeled by lines AB and

A’B' in Figure 7.11

Note that the Mohr envelope represents the locus of all possible failure states Many solids, particularly those that are brittle, show greater resistance to compression than to tension As a result, higher limiting shear stresses, for these materials, are found to the left

of the origin as depicted in the figure

285

7.12 THE COULOMB-MOHR THEORY

The Coulomb-Mohr theory, like the Mohr criterion, may be employed to predict the effect

of a given state of stress on a brittle material having different properties in tension and in

compression The Mohr’s circles for the uniaxial tension and compression tests are used to

predict failure by Coulomb-Mohr theory as shown in Figure 7.12a The points of contact of the straight-line envelopes (AB and A’B’) with the stress circles define the state of stress at

a fracture For example, if such points are C and C’, the stresses and the planes on which

they act can be obtained using the established procedure for Mohr’s circle of stress

In the case of plane stress, we have o3 = 0 When o; and oy have opposite signs (that is, one is tensile and the other is compressive), it can be verified that [22] the onset of fracture is expressed by

ơi — o2/Su/Sucl

Relationships for the case where the principal stresses have the same sign may be deduced from Figure 7.12a In the case of biaxial tension, the corresponding circle is rep- resented by diameter OE Hence, fracture occurs if either of the two tensile stresses achieves the value S,,; that is,

The foregoing expressions are depicted in Figure 7.12b, for the case in which n = 1

Lines ab and af represent Eq (7.26), and lines de and de, Eq (7.27) The boundary be is obtained by applying Eq (7.24) Line ef completes the hexagon in a way similar to Fig- ure 7.4 Points within the shaded area represent states of nonfailure according to the Coulomb-Moht theory The boundary of the figure depicts the onset of failure due to fracture

In the case of pure shear, the corresponding limiting point g represents the ultimate shear strength S,5 At point g, 0) = Sys and o2 = —o, = —S,; Substituting for oj, 02, and a=: | into Eq (7.24), we have

_ EXAMPLE 7.7 m Rework Example 7.6, Employing the Coulomb-Mohr Theory

Solution.” We have the following results, from Example 7.6:

Te 1109 X 10 $y and 0) = —03\= t So, applying Eq: (7.24) with n = 2,

+ —£ 1

200 6108 = 600 x 108 2

;Sölving,r = 75 MPa The first equation then gives 7 = 12.82 kN - m

Comments: On the basis of the maximum principal stress theory, the torque that can be applied to

“the pipe 17.09 KN - m obtained in Example 7.6 is thus 25% larger than on the basis of the Coulomb-

intension and compression S, and S,,, respectively Calculate the allowable load P

sign of cast iron machine elements Hence, the stress distribution across the section is taken to be linear

The distances from the neutral axis to the extreme fibers are c, = 110 mm and cg = 190 mm

The greatest tensile and compressive stresses occur at points A and B, respectively:

0<KR<tl

For instance, reliability of R = 0.98 means that there is 98% chance, under certain operat-

ing conditions, that the part will perform its proper function without failure; that is, if

100 parts are put into service and an average of 2 parts fail, then the parts proved to be 98% reliable

Recall from Section 1:6 that, in the conventional design of members, the possibility

of failure is reduced to acceptable levels by factor of safety based on judgment derived from past performances In contrast, in the reliability method, the variability of material properties, fabrication-size tolerances, as well as uncertainties in loading and even design approximations can be appraised on a statistical basis As far as possible, the proposed cri- teria are calibrated against well-established cases The reliability method has the advantage

of consistency in the safety factor, not only for individual members but also for complex structures and machines Important risk analyses of complete engineering systems are based

on the same premises Clearly, the usefulness of the reliability approach depends on ade- quate information on the statistical distribution of loading applied to parts in service, from which can be calculated the significant stress and significant strength of production runs of manufactured parts

Note that the reliability method to design is relatively new and analyses leading to an assessment of reliability address uncertainties Of course, this approach is more expensive and time consuming than the factor of safety method of design, because a larger quantity

of data must be obtained by testing However, in certain industries, designing to a desig- nated reliability is necessary We use the factor of safety for most of the problems in this text For an interactive statistics program from the engineering software database, see the

To obtain quantitative estimates of the percentages of anticipated failures from a study, we

must look into the nature of the distribution curves for significant quantities involved We

consider only the case involving the normal or Gauss distribution, credited to K FE Gauss (1777-1855) This is the most widely used model for approximating the dispersion of the observed data in applied probability [23]

Several other distributions might prove useful in situations where random variables have only positive values or asymmetrical distributions A formula, introduced by

W Weibull, is often used in mechanical design This formula does not arise from classical statistics and is flexible to apply The Weibull distribution is used in work dealing with experimental data, particularly reliability The distribution of bearing failures at a con- stant load can be best approximated by the Weibull distribution (see Section 10.14)

In analytical form, the Gaussian, or normal, distribution is given as follows:

The standard deviation is widely used and regarded as the usual index of dispersion or

scatter of the particular quantity The mean value and standard deviations are defined by

where n is the total number of elements, called the population

; A plot of Eq (7.29), the standard normal distribution [24], is shown in Figure 7.14

This bell-shaped curve is symmetrical about the mean value jz Since the probability that any value of x will fall between plus and minus infinity is 1, the area under the curve in the

figure is unity Note that about 68% of the population represented fall within the band

+ lo, 95% fall within the band 2 + 20, and so on

*The symbol o used ti lati i

the Mã lở Ha standard deviation here should not be confused with the symbol of stress, although often

289

Number of standard deviations z

Figure 7.15 Reliability chart: Generalized normal distribution curve plotted on special probability paper

The reliability, or rate of survival, R is a function of the number of standard deviations z, also referred to as the safety index:

{7.31}

The straight line of Figure 7.15, drawn on probability-chart paper, plots percentages of re- liability R as an increasing function of number of standard deviations z A larger z results in fewer failures; hence, a more conservative design We note that the percentage of the pop- ulation corresponding to any portion of the standard normal distribution (Figure 7.14) can

be read from the reliability chart

CHAPTER? 8 — FAfLURE CRITERIA AND RELIABILITY 291

7.15 THE RELIABILITY METHOD AND

THE MARGIN OF SAFETY

In the reliability method of design, for a given member, the distribution of loads and

strengths are determined, then these two are related to achieve an acceptable success rate

The designer’s task is to make a judicious selection of materials, processes, and geometry (size) to achieve a reliability goal The approach of reliability finds considerably more

application with members subjected to wear and fatigue loading We here introduce it in

the simpler context of static loading Section 8.6 discusses the reliability factor for fatigue endurance strength of materials

Consider the distribution curves for the two main random variables, load L and strength 5S, which is also called capacity or resistance (Figure 7.16) For a given member, the frequency functions p(/) and p(s) define the behavior of critical parameters load and strength The mean value of strength is denoted by „ and the mean value of the load by

tị So, based on mean values, there would be a margin of safety

However, the “interference” or shaded area of overlap in the figure indicates some possibility

of a weak part in which failure could occur, The preceding margin of safety must not be con- fused with that used in aerospace industry (see Section 1.8)

Figure 7.17 shows a corresponding plot of the distribution of margin of safety In this diagram, the probability of failure is given by the (shaded) area under the tail of the curve

to the left of the origin The member would survive in all instances to the right of the origin By statistical theory, the difference between two variables with normal distribu- tions has itself a normal distribution Therefore, if the strength Š and the load L are nor- mally distributed, then the margin of safety m also has a normal distribution, as shown in Figures 7.16 and 7.17

The margin of safety has a mean value f,, and standard deviation om expressed as follows (25, 26]:

henge a Les Se KH (7.33a)

(7.3349)

Figure 7.16 Normal distribution curves of Figure 7.17 Normal distribution curve of

Flere o, and ơ are, respecfively, the standard deviations for the strength S and the load L

When S > L, mis positive The designer is interested in the probability that m > 0; that is, the area to the right of 0 in Figure 7.17 At x = m = 0, Eq (7.31) becomes z = /ơ The number of standard deviations, on introducing Eq (7.32) may now be written in the following form:

(7.34

For the prescribed mean and deviation values of the strength and load, Eq (7.34) is solved to yield the number of standard deviations z Then, the probability that a margin of safety exists may be read as the reliability R from the chart of Figure 7.15 Equation (7.34)

is therefore called the coupling equation, because it relates the reliability, through z, to the statistical parameters of the normally distributed strength and load For example, when the mean values of S and L are equal (that is, us = 4) it follows that z = 0 and the reliability

of a part is 50% Reliability of an assembly or system of parts may be found from their individual reliability values [27]

Application of the reliability theory is illustrated in the solution of the following

numerical problems

EXAMPLE 7:9 A Shipment of Control Rods

Ina shipment of 600: control rods, the mean tensile strength is found to be 35 ksi and the standard deviation 5 ksi: How many rods can be expécted to have

(ay A’stréngth of less than 29.5 ksi

(by A’ strength of between 29.5 and 48.5 ksi, Given? © jz; = 29.5 and 48.5 ksi, jy = 35 ksi, Om = 5 ksi

Assumption Both loading and strength have normal distributions

The corresponding reliability, obtained”: from: Figure: 7.15 is 86.5% Note that

= 0.865 = 0.135 represents the proportion of the total rods having a strength less than

= 29.5 ksi Hence, the number of: rods: with: a” strength “less than 29.5 ksi is

in service? : ae

Given: uu, =S50ksi, wy = 30k, a, S4ksi, 0 SS ksi

Assumption: Both loading and stréngth have ‘normal distribution

Solution: Through the use of Eqs (7.33), we have

‘The Iwisting-off Strength of Bolts EXAMPLE 7.11

Bolts, each of which fas a:mean: twisting-off strength of 25 Nm: with a’ standard deviation of 1.5N m, are fishfened with automatic wrenches’ on a production line (see Section 15.7) If the auto-

matic wrenches havea standard’ deviation of: 2 N’:'m, calculate’ mean value of wrench torque setting

that results in an estimated bolt in'400 twisting off during assembly

Solution: Substitution of oy = 1.5 Nim and oy = 2.N:m in Eq, (7.33) gives on, = 2.5N-m

Figure 7.15 shows that a reliability of 399/400 = 0.9975, or 99.75%, corresponds to 2.8 standard de-

“Aiation: The: mean: value is then [ig = 20m = 2.8(2.5) = 7N-m Since us = 25 N-m, we have, from Bq (7.33a), uy = 18 Nm This i¢ the réquired value of wrench setting

REFERENCES

1 Nadai, A Theory of Flew and Fracture of Solids New York: McGraw-Hill, 1950

2 Marin, J Mechanical Behavior of Materials Upper Saddle River, NJ: Prentice Hall, 1962

3 Van Vlack, L H Elements of Material Science and Engineering, 6th ed Reading, MA: Addison-

Wesley, 1989

4 Dowling, N E Mechanical Behavior of Materials Upper Saddle River, NJ: Prentice

Hall, 1993

5, American Society of Metals Metals Handbook Metals Park, OH: ASM, 1961

6 Griffith, A A “The Phenomena on Rupture and Flow of Solids.” Philosophical Transactions of the Royal Society (London) A221 (1921), pp 163-98

7 Irwin, G R “Practure Mechanics.” Proceedings, First Symposium on Naval Structural

Mechanics New York: Pergamon, 1958, p 557

8 Cook, R D., et al Concepts and Applications of Finite Element Analysis, 4th ed New York:

Wiley, 2001

9, Boresi, A P., and R C Schmidt Advanced Mechanics of Materials, 6th ed New York:

Wiley, 2003

10 Knott, J R Fundamentals of Fracture Mechanics New York: Wiley, 1973

- 11, Brock, D The Practical Use of Fracture Mechanics Dordrecht, the Netherlands: Kiuwer, 1988

12, Gdoutos, E E Fracture Mechanics Criteria and Applications, Dordrecht, the Netherlands:

Kluwer, 1990

13 Meguid, S A Engineering Fracture Mechanics London: Elsevier, 1989

14 Brown, W F, ed Review of Developments in Plane-Strain Fracture Toughness Testing,

Tech Pub 463 Philadelphia: ATM, 1970

15 Irwin, G R “Mechanical Testing.” Metals Handbook, 9th ed., vol 8 Metals Park, OF: ASM,

1989, pp 437-493

16 Dieter, G E Mechanical Metallurgy, 3rd ed New York: McGraw-Hill, 1986

17 Norton, R L Machine Design: An Integrated Approach, 2nd ed Upper Saddle River, NI:

Prentice Hali, 2000

18 Damage Tolerant Design Handbook, MCIC-HB-Ol Kettering, OH: Air Force Materials

Laboratory, Wright-Patterson Air Force Base, December 1972

19 ASM International Guide in Selecting Engineering Materials Metals Park, OH: American Soci-

ety of Metals, 1989

CHAPTER 7 8 FAILURE CRITERIA AND RELIABILITY

20 Fauppel, J H., and FB Fisher Engineering Design, 2nd ed New York: Wiley, 1981

21 Taylor, G 1, and H Quinney The Plastic Distortion of Metals, Philosophical Transactions of

the Royal Society, Section A, No 230, 1932, p 323 :

22 Ugural, A.C., and S K Fenster Advanced Strength and Applied Elasticity, 4th ed Upper Saddie River, NJ: Prentice Hall, 2003

23 Shigley, J E., and C.R Mischke Mechanical Engineering Design, 6th ed New York: McGraw-

Hill; 2001

24 Juvinall, R C., and K M Marshek Fundamentals of Machine Component Design, 3rd ed

New York: Wiley, 2000

25 Lipson, C., and J Sheth Statistical Design and Analysis of Engineering Experiments

New York: McGraw-Hill, 1973,

26 Kennedy, J B., and A M Neville Basie Statistical Methods for Engineers and Scientists,

3rd ed New York: Harper and Row, 1986

27 Burr, A H., and J B Cheatham Mechanical Analysis and Design, 2nd ed Upper Saddle River, NJ: Prentice Hall, 1995

Given: f = 25 mm, w se 250 mm, a = 25 mm 7,2 Estimate the maximum load P that the plate shown in Case B of Table 7.1 can carry What is the mode of failure?

Given: S, = 650MPa, K = 100Pa/m, w=350mm, =25mm, 2= l5 mm,

factor of safety # = 1.2 73° A2024-T851 aluminum-alloy plate of width w and thickness 1 is subjected to a tensile loading

It contains a transverse crack of length a on one edge (Figure 7.2) There is concern that the plate will undergo sudden fracture Calculate the maximum allowable axial load P What is the nominal stress at fracture?

Given: w = 125mm, t=: 25 mm, a= 20mm 7.4 An AISH-4340 steel pipe of diameter d and wail thickness ¢ contains a crack of length 2a

Estimate the pressure p that will cause fracture when

(a) The crack is longitudinal as in Figure P7.4

(b) The crack is circumferential

Given: d = 50 mm, +=4mm, a=5mm Assumption: A factor of safety n = 1.5 and geometry factor A = 1.01 are used (Table 7.1)

75 A 7075-17351 aluminum alloy beam containing an edge crack of length a is in pure bending

as shown in Case D of Table 7.1 Determine the maximum moment M that can be applied without causing sudden fracture

Given: a = 40 mm, w = 100 mm, ? = 25 mm 7.6 An AIS]-4340 steel plate of width w = 5 in and thickness ? = Ì in ts under uniaxial tension

A crack of length a is present on the edge of the plate, as shown in Figure 7.2 Determine (a) The axial load possible Py, for the case in which a = ‡ in,

(b) The critical crack length a, if the plate is made of Ti-6A1-6V titanium alloy and subjected

to the Py, calculated in part a

7.7 Rework Example 7.3, for the case in which the bracket is made of AISI-403 stainless steel and a=0.6in., đ = 6.25 in., w = L5 Ìn., = } in, anda = 2

7.8 An AISI-4340 sieel ship deck-panel of width w and thickness ? is under tension Calculate the maximum load P that can be applied without causing fracture when double-edge cracks grow

to a length of a (Case C of Table 7.1)

Given: f = Lin., wz=21n., a=0.221n., n L4

Sections 7.5 through 7.12

7.9 A solid steel bar having yield strength S, and diameter Ð carries end loads P, M, and T Calculate the factor of safety n, assuming that failure occurs according to the following criteria:

(a) Maximum energy of distortion

(b) Maximum shear stress,

7.10 A steel bar AB of diameter D and yield strength S, supports an axial load P and vertical load F acting at the end of the arm BC (Figure P7.10) Determine the largest value of F according to the maximum energy of distortion theory of failure

Given: D = 40 mm, Sy == 250 MPa, P=20F Assumptions: The effect of the direct shear is negligible and the factor of safety n = 1.4

7.41 Acantilever WF aluminum alloy beam of yield strength S, is loaded as shown in Figure P7.11

Using a factor of safety of n, determine whether failure occurs according to the maximum shear stress criterion

Resolve Problem 7.11 applying the maximum energy of distortion theory

A thin-walled cylindrical pressure vessel of diameter d and constructed of structural steel with

yield strength S, must withstand an internal pressure p Calculate the wall thickness t required

Given: $, = 36 ksi, d= 20in., p = 500 psi, n= 1S Design Decision: Use the following criteria:

(a) Maximum shear stress

(b) Maximum energy of distortion

Redo Problem 7.13, if the vessel is made of a material having S, = 50 ksi and 5„¿ = 90 ksi

Design Decision: Apply the following theories:

(a) Maximum principal stress

(b) Coulomb-Mohr

A cantilever WF cast iron beam of ultimate tensile strength S, and ultimate compression strength S,,, is subjected to a concentrated load at its free end (Figure P7.11) What is the fac- tor of safety 2?

Given: S, = 280 MPa, Sye = 620 MPa Assumption: Failure occurs in accordance with the following theories:

(a) Maximura principal stress

(6) Coulomb-Mohr

7.16 and 7.17 The state of stress shown (Figures P7.16 and P7.17) occurs at a critical point in an ASTM A-48 gray cast iron (Table B.1) component of a lawn mower Calculate the factor of safety n with respect to fracture

7.23

® FUNDAMENTALS Design Decision: Apply the following criteria:

(a) Maximum principal stress

(6) Coulomb-Mohr

A thin-walled steel spherical storage tank is filled with liquid of density y and supported on a

ring, as shown in Figure P7.18 According to the maximum shear stress and maximum energy

of distortion criteria, determine the factor of safety 1, in terms of thickness ¢, mean radius a, yield strength Šy, and y, as required

Figure P7.18 Given: Tangential stress og = 0, and meridian stress og = 0 ato = 150°, are calculated from Eqs (16.27) as follows:

o = 3.5672, = ~1.70 2

where the negative sign indicates compression

Resolve Problem 7.18 for the condition that the tank is made of cast iron having ultimate com- pression strength S,, five times ultimate tensile strength S,,

Design Decision: Maximum principal stress and Coulomb-Mohr criteria are used

A cylindrical rod of diameter D is made of ASTM-A36 steel (Table B.1) Use the maximum

shear stress criterion to determine the maximum end torque T that can be applied to the rod

simultaneously with an axial load of P = 100 kips (Figure 7.5)

Given: D = 2 in

Assumption: 2 = 1.5

Redo Problem 7,20, applying the maximum energy of distortion criterion

An ASTM-20 gray cast iron rod (Table B.2) is under pure torsion Determine, with a factor of safety n = 1.4, the maximum shear stress t that may be expected at impending rupture using

(a) The Coulomb-Mohr criterion

(b) The principal stress criterion

The state of stress shown in Figure P7.23 occurs at a critical point in a machine component

made of ASTM-A47 malleable cast iron (Table B.1) Apply the Coulomb-Mohr theory to

calculate the maximum value of the shear stress t for a safety factor of n = 2

CHAPTER7 ® FAILURE CRITERIA AND RELIABILITY

| 10 MPa Figure P7.23

7.24 Resolve Problem 7.23 for the condition that the machine component is made of an ASTM- A242 high-strength steel (Table B.1) Use

(a) The maximum energy of distortion criterion

(b) The maximum shear stress criterion

7.25 An ASTM-A36 steel shaft of length L carries a torque T and its own weight per unit length w

(see Table B.1), as depicted in Figure P7.25, Determine the required shaft diameter D, using the maximum energy of distortion criterion with a safety factor of = 2.1

Given: L = 6m, T = 400N-m Assumption: The bearings at the ends act as simple supports

Figure P7.25

Sections 7.13 through 7.15

7.26 Ata critical location in a component in tension, the load induced stresses are jz; = 250 MPa

and a =: 35 MPa What is the reliability R against yielding?

Given: The material yield strengths are jz, = 400 MPa and o, = 30 MPa

7.27 Calculate the diameter d of a bar subjected to an axial tensile load P for a desired reliability of

y 782 821 603 915 846 707 684 908 750 925 618 80.1

i Notes: n = number of students, x = final examination grade, y = course grade.

Search and download the statistics shareware program on the website at www.mecheng.asme

org/database/STAT/MASTER.HTML for computing the mean and standard deviations for a normal distribution Resolve Problem 7.28 using this program

A total of 68 cold-drawn steel bars have been tested to obtain the 0.2% offset yield strength S,

in ksi The results are as follows:

Sy 74 66 62 78 81 82 85 86 89 94

" 7 2 5 5 10 18 8 3 6 4

Based on normal distribution, determine

(a) The mean yz and standard deviation o of the population

(b) The reliability for a yield strength of S, = 75 ksi

A bar under a maximum load of 5 kips was designed to carry a load of 6 kips The maximum load is applied with standard deviation of 600 Ib and shaft strength standard deviation of

400 Ib, both are normally distributed Calculate the expected reliability

A structural member is subjected to a maximum load of 20 kN Assume that the load and

strength have normal distributions with standard variations of 3 and 2.5 KN, respectively If the member is designed to withstand a load of 25 KN, determine failure percentage that would be expected

85 8.6 8.7 8.8 8.9 8.10 8.14 8.12 8.13 8.14 8.15

Modified Endurance Litnit Endurance Limit Reduction Factors Fluctiating Stresses

Theories of Fatigue Failure

Comparison of the Fatigue Criteria Design for Simple Fluctuating Loads

Design for Combined Fluctuating Loads

Prediction of Cumulative Fatigue Damage Fracture Mechanics Approach to Fatigue Surface Fatigue Failure: Wear

302 PARTE ® FUNDAMENTALS CHAPTER 8 © FATIGUE 303

A member may fail at stress levels substantially below the yield strength of the material if

it is subjected to time-varying loads rather than static loading The phenomenon of pro- gressive fracture due to repeated loading is called fatigue Its occurrence is a function of the magnitude of stress and a number of repetitions, so it is called fatigue failure We observe

throughout this chapter that the fatigue strength of a component is significantly affected by

a variety of factors A fatigue crack most often is initiated at a point of high stress concen- tration, such as at the edge of a notch, or by minute flows in the material Fatigue failure is

of a brittle nature even for materials that normally behave in a ductile manner The usual fracture occurs under tensile stress and with no warning For combined fluctuating loading conditions, it is common practice to modify the static failure theories and material strength for the purposes of design

The fatigue failure phenomenon was first recognized in the 1800s when railroad axles fractured after only a limited time in service Until about the middle of the 19th century, re- peated and static loadings were treated alike, with the exception of the use of safety factors

Poncelet’s book in 1839 used the term fatigue owing to the fluctuating stress At the present time, the development of modern high-speed transportation and machinery has increased the importance of the fatigue properties of materials In spite of periodic inspection of parts for cracks and other flaws, numerous major railroad and aircraft accidents have been caused by fatigue failures The basic mechanism associated with fatigue failure is now rea- sonably well understood, although research continues on its many details [1-31] The com-

Sudden fracture zone

(b) reversed bending [1]

High nominal stress Low nominal stress

Smooth | _ Notched Smooth [Notched

High nominal stress Low nominal stress

Smooth Notched Smooth Notched

| 8.2 THE NATURE OF FATIGUE FAILURES

The type of fracture produced in ductile metals subjected to fatigue loading differs greatly from that of fracture under static loading, considered in Section 2.3 In fatigue fractures, two regions of failure can be detected: the beachmarks (so termed because they resemble

’ ripples left on sand by retracting waves) zone produced by the gradual development of the crack and the sudden fracture zone As the name suggests, the latter region is the portion that fails suddenly when a crack reaches its size limit Figure 8.{ depicts fatigue fracture surfaces of two common cross sections under high nominal stress conditions [1] Note that the curvature of the beachmarks serves to indicate where the failure originates The beach- marked area, also referred to as the fatigue zone, has a smooth, velvety texture This con- trasts with the sudden-fracture region, which is relatively dull and rough, looking like a

Microscopic examinations of ductile metal specimen subjected to fatigue stressing re- veal that little if any distortion occurs, whereas failure due to static overload causes exces~

sive distortion The appearance of the surfaces of fracture greatly aids in identifying the

cause of crack initiation to be corrected in redesign For this purpose, numerous pho- tographs and schematic representations of failed surfaces have been published in technical gin, appearance, and location of fatigue fracture for variously loaded sections, For all axial

and bending stress conditions, as well as the high-torsion smooth stress condition, the crack growth in the beachmarks region is indicated by curved vectors, starting from the point of crack initiation,

Note that the sudden fracture region can be a small portion of the original cross section, particularly under bending and torsion fatigue stressing Unless interrupted by notches, the

304 PARTI) @ ~ FUNDAMENTALS CHAPTERS ® EATIGUE 305

fatigue crack under bending is normal to the tensile stresses; that is perpendicular to the axis

of a shaft In torsional fatigue failures, the crack is at a 45° angle to the axis of a notch-free shaft (or spring wire) under high nominal stress conditions (Figure 8.2) Finally, note that, if cracks initiate at several circumferential points, the sudden fracture zone is more centered,

at speeds in ranges of 500 10,000 rpm The device can apply a moment up to 200 Ib-in

to the specimen

REVERSED BENDING TEST

In the rotating-beam test, the machine applies a pure bending moment to the highly pol- ished, so-called mirror finish, specimen of circular cross section (Figure 8.3b) As the

| 83 FATIGUETESTS ị specimen rotates at a point on its outer surface, the bending stress varies continuously from

maximum tension to maximum compression This fully or completely reversed bending stress can be represented on the stress S-cycles N axes by the curves of Figure 8.3c It is obvious that the highest level of stress is at the center, where the smallest diameter is about

0.3 in The large radius of curvature avoids stress concentration Various standard types of

fatigue specimens used, including those for axial, torsion, and bending stresses described

in the ASTM manual on fatigue testing

In some fatigue testing machines, constant-speed (usually 1750 rpm) motors are used, which give the sinusoidal type or fully reversed cyclic stress variation shown in the figure

It takes about one-half of a day to reach 10° cycles and about 40 days to reach 10° cycles

on one specimen A series of tests performed with various weights and using multiple spec-

imens, carefully made to be nearly the same as possible, gives results or the fatigue data

To determine the strength of materials under the action of fatigue loads, four types of tests are performed: tension, torsion, bending, and combinations of these In each test, speci- mens are subjected to repeated forces at specified magnitudes while the cycles or stress reversals to rupture are counted A widely used fatigue testing device is the R R Moore high-speed rotating-beam machine (Figure 8.3a) To perform a test, the specimen is loaded with a selected weight W Note that turning on the motor rotates the specimen, however, not the weight There are various other types of fatigue testing machines [3] A typical rotating-beam fatigue testing machine has an adjustable-speed spindle, operating

ị Fen, | @ logarithmic scale; that is, S against log N Sometimes data are represented by plotting S ver-

sus N or log S versus log N Inasmuch as fatigue failures originate at local points of relative

Polis) olished surface (0.300 in) sutFa 7.62 mm End : weakness, usually the data contain a Jarge amount of scatter In any case, an average curve, } Ƒ

tending to conform to certain generalized pattern, is drawn to represent the test results

Figure 8.4 shows two typical S-N diagrams corresponding to rotating beam tests on a series of identical round steel and aluminum specimens subjected to reversed flexural loads

THe ENDURANCE LIMIT AND FATIGUE STRENGTH

The endurance limit and fatigue strength are two important cyclic properties of the materi- als The fatigue strength (S,,), sometimes also termed endurance strength, is the completely reversed stress under which a material fails after a specified number of cycles Therefore, when a value for the fatigue strength of a material is stated, it must be accompanied by the number of stress cycles The endurance limit (S’) or fatigue limit is usually defined as the maximum completely reversed stress a material can withstand “indefinitely” without frac- ture The endurance limit is therefore stated with no associated number of cycles to failure

Bending Fatigue Strength

For ferrous materials, such as steels, the stress where the curve levels off is the endurance limit S, (Figure 8.4) Note that the curve for steel displays a decided break or “knee” oc- curring before or near | x 10° cycles This value is often used as the basis of the endurance limit for steel Beyond the point(S,, N.), failure does not occur, even for an infinitely large number of loading cycles At N = Ny cycles, rupture occurs at approximately static frac- ture stress Sy * 0.95, where S,, is the ultimate strength in tension

On the other hand, for nonferrous metals, notably aluminum alloys, the typical S-N curve indicates that the stress at failure continue to decrease as the number of cycles increases (Figure 8.4) That is, nonferrous materials do not show a break in their S-N curves, and as a result, a distinct endurance limit cannot truly be specified For such materials, the stress corresponding to some arbitrary number of 5(10°) cycles is commonly

assigned as the endurance strength, S,

It is necessary to make the assumption that most ferrous materials must not be stressed above the endurance limit S) if about 10° or more cycles to failure is required This is illustrated in Figure 8.5, presenting test results for wrought steels having ultimate strength

S, < 200 ksi Note the large scatter in fatigue life N corresponding to a given stress level and the small scatter in fatigue stress corresponding to a prescribed life The preceding is

Cycles to failure, V (log)

Figure 8.5 Fully reversed rotating beam S-N curve for wrought steels of S, < 200 ksi with superimposed data points (20]

typical of fatigue strength tests The figure also depicts that samples run at higher reversed stress levels break after fewer cycles, some (labeled in the dotted circle) do not fail at all

prior to their tests being stopped (here at 10” cycles) The data are bracketed by solid lines

{nterestingly, at the lower bound of the scatter band, the endurance limit can be conserva- tively estimated as 0.55, for design purposes We mention that, for most wrought steels, the endurance limit varies between 0.45 and 0.60 of the ultimate strength

Axial Fatigue Strength Various types of fatigue servohydraulic testing machines have been developed for applying fluctuating axial compression A specimen similar to that used in static tensile tests (see Section 2.3) is used The most common types apply an axial reversed sinusoidal stress as shown in Figure 8.3c A comparison of the strengths obtained for uniaxial fatigue stresses and bending fatigue stresses indicate that, in some cases, the former strengths are about 10

to 30% lower than the latter strengths for the same material [10-14] Data for completely reversed axial loading test on AISI steel (S,, = 125 ksi) are shown in Figure 8.6 Observe the slope at around 10° cycles and the change to basically no slope at about 10° cycles corresponding to the endurance limit S)

Torsional Fatigue Strength

A limited number of investigations have been made to determine the torsional fatigue strengths of materials using circular or cylindrical specimens subjected to complete stress reversal For ductile metals and alloys, it was found that the torsional fatigue strength

CHAPTER 8 ° FAtIGUE 307

FaTicue REGIMES

The stress-life (S-N) diagrams indicate different types of behavior as the cycles to failure in- crease, Two essential regimes are the low-cycle fatigue (1 < N < 103) and the high-cycle fatigue (10° < N) Note that there is no sharp dividing line between the two regions In this text, we assume high-cycle fatigue starting at around N = 10° cycles The infinite life be- gins at about 10° cycles, where failures may occur with only negligibly small plastic strains

The finite life portion of the curve is below about 107 cycles The boundary between the in- finite life and finite life lies somewhere between 10° and 10’ cycles for steels, as shown in Figure 8.6 For low-cycle fatigue, the stresses are high enough to cause local yielding

A fracture mechanics approach is applied to finite life problems in Section 8.14 We use the stress-life data in treating the high-cycle fatigue of components under any type of

loading For further details, see [5~7], which also provide discussion of the strain-life ap-

proach to fatigue analysis

8.5 ESTIMATING THE ENDURANCE LIMIT

AND FATIGUE STRENGTH

Many criteria have been suggested for interpreting fatigue data No correlation exists be- tween the endurance limit and such mechanical properties as yield strength, ductility, and so

on However, experiments show that the endurance limit, endurance strength, endurance limit in shear, and ultimate strength in shear can be related to the ultimate strength in tension

Experimental values of ultimate strengths in tension S, and shear should be used if they are available (see Appendix B) Recall from Section 2.10 that S,, can be estimated from a nondestructive hardness test For reference purposes, Table 8,1 presents the rela- tionships among the preceding quantities for a number of commonly encountered loadings and materials {12, 14] Steel product manufacturers customarily present stress data of this kind in terms of the ultimate tensile strength, because S,, is the easiest to obtain and most reliable experimental measure of part strength

If necessary, the ultimate strength of a steel can be estimated by Eq (2.22) as

S, = 3500Hg in MPa or S, == 500Hg in ksi Here Hg denotes the Brineli hardness num- ber (Bhn) It should be noted, however, that Eq (8.1) can be relied on only up to Bhn val- ues of about 400 Test data show that the endurance limit S’ may or may not continue to increase for greater hardness, contingent on the composition of the steel [14]

In the absence of test data, the values given in the table can be used for preliminary

design calculations The relations are based on testing a polished laboratory specimen of

a fixed size and geometric shape and on a 50% survival rate Therefore, these data must

Sy = 700 MPa (200 ksi) [S, = 1400 MPa (200 ksj)]

` Š = 0.46 si

Si = 160 MPa (24 ksi) (S, = 400 MPa (60 ksi)]

Aluminums , i Ộ [Sy < 330 MPa (48 kg] (8.2b)

Si, = 130 MPa (19 ksi) [Su 2 330 MPa (48 ksi)}

Si, = 0.46 si

Copper alloys " f [Sy < 280 MPa (40 ksi) (@.20

Si = 100 MPa (14 ksi} [Su = 280 MPa (40 ksi)]

Axial Loading

Steels Si, = 0.458, (8.3) Torsional Loading

8.6 MODIFIED ENDURANCE LIMIT

The specimen used in the laboratory to determine the endurance limit is prepared very carefully and tested under closely controlled conditions However, it is unrealistic to expect the endurance limit of a machine or structural member to match the values obtained in the laboratory, Material, manufacturing, environmental, and design conditions influence fatigue

Typical effects include the size, shape, and composition of the material; heat treatment;

mechanical treatment; stress concentration; residual stresses; corrosion; temperature; speed;

type of stress; and fe of the member [3]

To account for the most important of these effects, various endurance limit modifying factors are used These empirical factors, when applied to steel parts, lead to results of good accuracy, because most of the data on which they are based are obtained from testing steel