18452195 wk2 failure theories

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Mechanical Design 337

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FAILURE THEORIES 2 - 2

308803 Mechanical Design 337 Semester 2, 2009

1.0 INTRODUCTION The following topics are discussed in this chapter:

- Definitions, Terminologies and Formulas - Types of Loading - Mechanical Failures - Static Failure Theories - Reliability

2.0 DEFINITIONS, TERMINOLOGIES AND FORMULAS

DUCTILITY The tendency of a material to deform significantly before fracturing is a measure of its ductility. The absence of significant deformation before fracture is called brittleness. Figure 2.1 shows a test specimen of a ductile material (steel) after fracture. The distortion (also known as necking-down) can clearly be seen at the break. The fracture surface appears torn and is laced with hills and valleys, also indicating a ductile failure. The ductility of a material is measured by its percent elongation to facture, or percent reduction in area at fracture. Materials with more than 5% elongation at fracture are considered ductile.

Figure 2.1: A tensile test specimen of ductile material [2]

BRITTLENESS Figure 2.2 shows a cast iron test specimen after fracture. The break shows no evidence of necking and has the fine surface contours typical of a brittle material. It is noted that the yield point is not clearly defined in stress-strain curve for brittle materials (Figure 2.3b). In such situations, yield strength Sy is defined by an offset method as shown in Figure 2.3b,



where a line is drawn at slope E. The line offset distance is usually 0.2 percent of the original gauge length.

Figure 2.2: Brittle material failure [2]

Figure 2.3: Stress-strain curve [3]

pl – Proportional limit (Until this point the material follows Hook’s law)

el – Elastic limit (The material is still under its elastic range but the stress-strain relationship is not linear in the region between point el and pl). After the point, pl, the material is in plastic range. If the specimen is loaded beyond this point, the deformation is said to be plastic and the material will take on a permanent deformation.



STRESS AND STRAIN Stress (σ ) is defined as load per unit area of the specimen.

oAP

=σ

Strain (ε ) is the change in length per unit length and is expressed as

olll 0−

=ε

It’s also known as elongation and it is a dimensionless number. However, it is customary to speak of it in terms of mm per mm or inch per inch.

DEFORMATION (δ ) The total change of length in a uniform body caused by an axial load is called the deformation. Mathematically,

olδε =

For most materials used in engineering, stress and strain are directly proportional. Under this condition, the material is said to follow Hook’s law. The linear relationship between stress and strain can be expressed as

Eεσ =

OR

Eσε =

OR

EAPl

o

o=δ

The constant E is called the Modulus of Elasticity, or Young’s Modulus, for the material. E is a measure of the stiffness of the material in its elastic range and has dimension of stress. It can also be visualised as the tensile stress that would cause a body to double in length, ε =1, provided the material would remain elastic with such excessive loading. Table 2.1 shows values of E for common materials used in engineering.



Table 2.1: Average values for mechanical properties of engineering materials [3]

MATERIAL STIFFNESS

Stiffness ( k ) is the resistance of an elastic body to deflection or deformation by an applied force.

δPk =

YIELD POINT At point y, as shown in Figure 2.3, the strain begins to increase very rapidly without a corresponding increase in stress. This point is called the yield point. The strength at this point is called yield strength and is denoted by Sy.

ULTIMATE STRENGTH (SU) This strength corresponds to point u in Figure 2.3 and is the maximum stress reached on the stress-strain curve.

THE COMPRESSION TEST The compression test can be conducted by the tensile test machine in reverse to apply a compressive load to a specimen, as shown in Figure 2.4. It is difficult to obtain a useful stress-strain curve from this test because ductile material will yield and increase its cross-sectional area as shown below. The ductile sample will not fracture in compression. If pressed further, it will take the shape of a pancake.



Figure 2.4: Compression test specimens [2]

Most ductile materials have compressive strengths similar to their tensile strengths, and the tensile stress-strain curve is used to represent their compressive behaviour as well. Even Material A material that has essentially equal tensile and compressive strengths is called an even material. Brittle materials will facture when compressed. Brittle materials generally have much greater strength in compression than in tension. Compressive stress-strain curves can be gathered, since the material factures rather than crushes and the cross-sectional area doesn’t change appreciably. Uneven Material A material that has different tensile and compressive strengths is called an uneven material.



3.0 LOADING & STRESSES The manner of variation of the load and the resulting variation of stress with time are the main factors to consider when specifying the type of loading to which a machine part is subjected. Stress variations can be categorised by the following four key values:

- Maximum stress, σmax - Minimum stress, σmin - Mean (average) stress, σm - Alternating stress, σa (It is also known as stress amplitude)

Stress analysis or finite element methods (FEMs) are usually employed to determine the maximum and minimum stresses. Experimental stress analysis can also be used to measure these stresses. Then the following relationships are used to compute the mean and alternating stresses:

𝜎𝜎𝑚𝑚 =𝜎𝜎𝑚𝑚𝑚𝑚𝑚𝑚 + 𝜎𝜎𝑚𝑚𝑚𝑚𝑚𝑚

2

𝜎𝜎𝑚𝑚 =

𝜎𝜎𝑚𝑚𝑚𝑚𝑚𝑚 − 𝜎𝜎𝑚𝑚𝑚𝑚𝑚𝑚2

Figure 2.5 shows the relationship between different components of the stresses.

Figure 2.5: Maximum, minimum, mean and alternating stresses [4]

Stress range (∆𝜎𝜎) is defined as the difference between the maximum and minimum stresses.

3.1 Static stress When the load applied to a part is increased gradually, without shock, and is held at a constant value (Figure 2.6), the resulting stress in the part is called static stress. Examples include column under the dead weight of materials. Figure 2.6 illustrates this.



Figure 2.6: Static stresses

3.2 Stresses in rotating machinery The following stress-time functions, as shown in Figure 2.7, are experienced by rotating machinery:

Figure 2.7: Fully reversed, repeated and fluctuating stresses [2]

FULLY REVERSED STRESSES

This type of stresses occurs when a given element of a load-carrying member is subjected to a certain level of stress followed by the same level of compressive stress. If this cycle is repeating many thousands of time, the resulting stresses are called fully revered stress. This is illustrated in Figure 2.7(a). In this case:

σmin = - σmax

σm = 0

This type of loading is also called fatigue loading. The material’s ability to withstand fatigue loads is called its endurance strength. It is referred to as the stress level that a material can



survive for a given number of cycles. In the case of infinite number of cycles, the stress level is called endurance limit.

REPEATED STRESSES:

As shown in Figure 2.7(b), the stress value ranges from zero to a maximum with a mean value equal to the alternating component.

σm = σa

FLUCTUATING STRESSES

There are applications in which a load-carrying member is subjected to an alternating stress with nonzero value of mean stress, as shown in Figure 2.7(c). This is called fluctuating stress.

3.3 Impact loading Impact loads cause materials to react in a different way to ‘normal’. Stress concentration effects are usually ignored in static load situations but must be carefully considered in both fatigue and impact loading.

Impact loads tend to change the properties of ductile materials towards those of brittle materials (think of it in terms of the material not being given the time to yield). This phenomena needs to be considered when designing components to absorb energy from impact loads.

3.4 Random loading When the applied loads are not regular in their amplitude, the loading is called random. This type of loading leads to more complex analysis. Statistical analysis is used to model the random loading for the purpose of analysis.



4.0 MECHANICAL FAILURES Why do mechanical components fail, and how do we predict when they will?

Failure of a loaded member can be regarded as any behaviour that renders it unsuitable for its intended function. Some of the most common failures are from corrosion and wear, but excluding these, components fail under load by yielding, fracture, fatigue, buckling and rapid fracture etc. These are explained below.

YIELDING If yield stress is exceeded due to bending, tension, compression, torsion or shear the material is permanently deformed – it will not return to its original length or shape if the load is removed. It usually still in one piece, but cannot normally perform its desired function, for example, a shaft is bent, a bolt or tension member is permanently stretched, a pressure vessel is bulged.

FACTURE If a material is strained beyond the point of yield, and sufficient deformation occurs that UTS is reached and passed, the components will eventually facture (i.e. it comes apart). For example, pipes burst, pressure vessels explode, bolts break. A ductile material requires considerable deformation before fracture, a brittle material almost none.

FATIGUE The above two failures are usually the result of essentially static (or slowly increasing) loads. Fatigue is caused by the application of cyclic loads in any form: tension, bending, torsion, shear etc. Multiple application of a cyclic stress can grow fatigue cracks, which start at places having some form of discontinuity. This could be caused by having a defect in the material itself, or where machine marks are left on the surface, where stresses are usually highest.

A cyclic stress at a certain level, and after a large number of cycles, will cause a crack to start and grow. There is very little deformation of the part, making the presence of a crack very difficult to detect without some form of specialised test, and this makes failure by fatigue one of the most dangerous forms.

The final failure can occur by simple overload of the remaining cross section of the part accompanied by small deformation, or in certain circumstances it can fail by the mechanism described below.



RAPID FRACTURE ‘Fracture mechanics’ is the study of how under certain conditions, normally ductile materials can fail by the rapid propagation of a crack, in a way usually associated with brittle materials.

This sort of fracture usually occurs when the length of crack reaches the ‘critical length’, and then propagates almost instantly through the remainder of the cross section. The tendency to fail in this way is governed by the ‘fracture toughness’ of the material, which is strongly temperature dependent.

BUCKLING If a component is subjected to compressive stress (even if this is caused by a bending load, i.e. in a beam or truss) it may fail by bucking well before it reaches its yield stress. A member which does not have sufficient lateral rigidity or support to prevent bucking, will fail if the load exceeds some critical value by complete and instantaneous collapse. As with fatigue or fracture, there is no large deformation prior to failure so there is no sign of imminent collapse.



5.0 PREDICTION OF FAILURE The only information available to engineers about how a material fails is usually the result of tensile test performed on one or more specimens of the material.

This test subjects the material to a ‘uni-axial’ stress (i.e. stress occurs in one direction) and the results are usually expressed as Syt, Sut and the % elongation at these values.

In many applications in engineering, components are not subjected to uniaxial stress, but have normal and shear stresses in a number of directions. If the directions of the stresses can be related to a set of orthogonal axes then they are called ‘direct’ stresses; xσ , yσ , xyτ etc., as shown in

Figure 2.8

Figure 2.8: Direct stresses

τxy and τyx indicate shear stresses. It will be positive shear stress if it tends to rotate the

stress element in clockwise direction. In Figure 2.5, τyx is positive, and τxy is negative. Their magnitudes must be equal to keep the element in equilibrium.

In many cases, these direct stresses are not the maximum stresses which exist in the component but the engineer will use their magnitudes in Mohr’s circle to determine the ‘principal stresses.’ Principal stresses may exist on the same or different planes to x. y and z planes but they have no shear stress present on that plane.

Principal stresses are denoted by 1σ , 2σ and 3σ to identify them compared to direct stresses xσ , yσand zσ .

In general, the relationship is presented on Mohr’s circle as shown in Figure 2.9.



Figure 2.9: Mohr’s Circle

Since the principal stresses represent the largest (and smallest) normal stresses on the part, it is the principal stresses which must be used to predict failure. Note also how Mohr’s circle gives the maximum and minimum value of shear stress (at places on the component 45° around from the places of maximum normal stresses.

The “theory” behind failure theories is that whatever is responsible for failure in the standard tensile test will also be responsible for failure under all other conditions of static loading.

This normally boils down to whether the material fails in a brittle way (snaps like a carrot?) or does it require deformation – i.e is it ductile – and does it fail in shear?



EXAMPLE Consider a part subjected to a combination of stresses at a critical location, which combine to give principal stresses of MPa5601 =σ , MPa2802 −=σ and 03 =σ . If the material was found to fail (i.e. yield or fracture) in a standard tensile test at a stress of 700 MPa, will the component fail in the proposed application?

SOLUTION

This is a ‘biaxial’ stress state and is quite common in engineering. Many critical locations are at the extremities of the component, i.e. on the surface, where it is not possible for a normal or shear stress to exist.

Proposed Application Tensile Test

MPa5601 =σ MPaS yt 560=

MPa2802 −=σ 03 =σ

MPa4202

280560max =

+=τ MPa

S yt 3502

7002max ===τ

Mohr’s circle representation of the given data and the maximum shear stress is shown below:

Figure 2.10: Specimen under tensile test and proposed application



The answer depends on which failure theory best fits the material of the component.

Does the material fail if the tensile stress in any direction reaches the 700 MPa it reached in the tensile test? If this were the case, it would fit the ‘maximum normal stress’ theory, and if it were applied to the example above, it would be considered ‘safe’, with a small factor of safety since the maximum tensile stress reached is only 560 MPa.

Does the material fail if the shear stress reaches the value reached in the tensile test, a value of 350 MPa? This would fit the maximum shear stress theory, and if applied in this case, failure would be predicted, since the shear stress would reach 420 MPa in service.

5.1 FAILURE PREDICTION METHODS A number of different methods of failure are available. The selection of a particular method depends on the conditions of the project. The following factors are considered when selecting a failure prediction method:

- The nature of load (Static, repeated and reverse, or fluctuating) - The type of material involved (ductile or brittle) - The amount of design effort and analysis that can be justified by the nature of the

component or product being designed

Followings are some of the widely used methods of failure prediction:

Failure Prediction Method Application

1 Maximum normal stress theory Uni-axial static stress on brittle materials

2 Maximum shear stress theory Biaxial static stress on ductile materials

3 Maximum distortion energy theory Biaxial or tri-axial stress on ductile materials

4 Modified Mohr’s circle Biaxial static stress on brittle materials

5 Goodman Fluctuating stress on ductile materials

6 Gerber Fluctuating stress on ductile materials

7 Soderberg Fluctuating stress on ductile materials

These methods are based on the relationship between the applied stress on a component and the strength of the material from which it is to be made that is most relevant to the conditions of service. The strength basis for design can be yield strength, ultimate strength, endurance strength, or some combination of these.



MAXIMUM NORMAL STRESS THEORY

This is the simplest of the failure theories to use. It states that a material will fail (fracture) when the maximum normal stress (either tension or compression) exceeds the ultimate strength of the material as obtained from a standard tensile or compressive test. Unfortunately, this theory does not correlate well with the actual failures of ductile materials (most engineering materials) although its correlation with brittle materials (under pure uni-axial static tension or compression) is fair.

According to this theory, a failure will always occur:

- Whenever the greatest tensile strength tends to exceed the uni-axial tensile strength or

- Whenever the largest compressive stress tends to exceed the uni-axial compressive strength

Figure 2.11a -Mohr’s circle – Failure is predicted for any state of stress for which the principal Mohr circle extends beyond either of the dotted vertical boundaries.

Figure 2.11b - Failure is predicted for all combinations of 1σ and 2σ falling outside the shaded area.

Figure 2.11: Maximum normal stress theory [1]



According to the maximum normal stress theory, the following equations are applied to design:

𝐾𝐾𝑡𝑡𝜎𝜎 < 𝜎𝜎𝑑𝑑 = 𝑆𝑆𝑢𝑢𝑡𝑡𝐹𝐹𝐹𝐹𝑆𝑆

(For tensile stress)

𝐾𝐾𝑡𝑡𝜎𝜎 < 𝜎𝜎𝑑𝑑 = 𝑆𝑆𝑢𝑢𝑢𝑢𝐹𝐹𝐹𝐹𝑆𝑆

(For compressive stress)

Where,

σ = Applied stress

σd = Design stress

Kt = stress concentration factor.

Sut = Material’s ultimate strength in tension

Suc = Materials ultimate strength in compression

FoS = Factor of safety

For a biaxial stress state ( 03 =σ ), the theory corresponds to the envelope shown below, which limits normal stresses at failure in the standard tensile test. Note, it reflects the possibility that for same materials (most brittle materials) Sut is usually smaller than Suc.

Figure 2.12: Maximum Normal Stress Theory



Failure is predicted if any principal stress exceeds the compression or tensile strength obtained from a standard tensile test, irrespective of the magnitude of the other principal stresses.

However, consider the Mohr’s circle representation of the stress states at A, B and C. The maximum shear stress obtained at C is two or more times the magnitude reached at A or B. Clearly, if shear stress is involved in the material failure, this failure theory will not apply, there being no clear limiting value of shear stress.

MAXIMUM SHEAR STRESS THEORY

This theory states that a material subjected to any combination of loads will fail (by yielding or fracture) if the maximum shear stress exceeds the shear strength (yield or ultimate) of the material found from a standard uni-axial tensile test. This theory correlates well with yielding of ductile materials and it is quite easy to use. It is common to assume that Syt and Sut have the same magnitude as the yield and ultimate strengths in compression for most ductile materials.

Figure 2.13 shows a graphical representation of maximum shear stress theory.

Figure 2.13: Maximum Shear Stress Theory [1]

The ‘envelope’ which describes this theory is based on combinations of principal stresses which produce the failure yield stress of the tensile test.

NOTE: According to Mohr’s circle analysis for uni-axial tension test, the maximum shear stress is one-half of the applied tensile stress. Mathematically,



𝑆𝑆𝑠𝑠𝑠𝑠 =𝑆𝑆𝑠𝑠𝑡𝑡2

Where, Ssy refers to shear yield strength.

Figure 2.14 shows graphical representation of yielding failure in a tensile test. The material fails when the maximum shear stress (τ) reaches the material’s maximum shear strength (Ssy).

Figure 2.14: Yielding failure

The following equations apply the maximum shear stress theory to design:

𝜏𝜏𝑚𝑚𝑚𝑚𝑚𝑚 < 𝜏𝜏𝑑𝑑 = 𝑆𝑆𝑠𝑠𝑠𝑠𝐹𝐹𝐹𝐹𝑆𝑆

=0.5 𝑆𝑆𝑠𝑠𝑡𝑡𝐹𝐹𝐹𝐹𝑆𝑆

This theory correlates well with yielding failure of ductile materials and its two advantages are that it is easy to use, and slightly conservative. For more precise results, the distortion-energy theory is preferred.



MAXIMUM DISTORTION ENERGY THEORY

The theory is usually known as the Van Mises distortion energy theory. This method has been shown to be the best predictor of failure for ductile materials under static loads or completely reversed normal, shear, or combined stresses. It predicts failure due to the energy absorbed by a body in distorting under the action of the applied load. It postulates that the material has a certain capacity to absorb energy of distortion. The corollary is that if there is no distortion, there is no failure – consider the case of a component immersed in a fluid and subjected to enormous hydrostatic forces. Since 1σ = 2σ = 3σ all in compression, no distortion occurs and so the part does not fail.

The theory is shown graphically by plotting the envelope of an equivalent stress eσ (Van Mises Equivalent) which if exceeded will cause failure.

Figure 2.15: Maximum distortion energy theory

In general,

[ ] 212

232

132

12 )()()(22 σσσσσσσ −+−+−=e

Note that the signs of stresses must be strictly adhered.

For a biaxial stress state (above diagram) where 03 =σ

[ ] 21

212

22

1 )()( σσσσσ −+=e



The direct stresses may also be used without first finding principal stresses

[ ] 21222 )(3)()( xyyxyxe τσσσσσ +−+=

If only xσ and xyτ are present, the equation reduces to

[ ] 2122 )(3)( xyxe τσσ +=

According to distortion-energy theory, the following equations are applied to design:

𝜎𝜎𝑒𝑒 < 𝜎𝜎𝑑𝑑 = 𝑆𝑆𝑠𝑠𝑡𝑡𝐹𝐹𝐹𝐹𝑆𝑆



COMPARING THE FAILURE THEORIES The three theories are plotted below (Figure 2.16). It can be seen that the maximum-shear stress theory is generally conservative in all the four quadrants (It is clearer in Figure 2.15). It coincides with the distortion energy ellipse at six points.

Figure 2.16 : Static Failure Theories [1]

Just like the maximum shear stress theory, the maximum normal stress theory is also conservative in the first and third quadrants. But it is important to note that the maximum normal stress theory is dangerously non-conservative in the second and fourth quadrants.



MODIFIED MOHR THEORY There is another theory, the modified Mohr theory, which is the recommended theory for use with brittle materials. This is shown in Figure 2.17.

Figure 2.17: Modified Mohr theory for biaxial stresses )(σ 03 = [1]

The theory is similar to the maximum normal stress theory, but is modified where there is a mixture of compressive and tensile principal stresses. It correlates better with test data on brittle materials than either the Mohr’s circle or the maximum normal stress theory.



6.0 RELIABILITY A factor of safety > 1 does not guarantee that components don’t fail. This is because of the fact that no physical parameter exists without uncertainty. There is always some uncertainty in a material strength – no two test results will be identical if a number of tests are performed, a spread of results will be found. The same applies to the loads that components are subjected to in service.

The reliability method of design is one in which the distribution of stresses and the distribution of strengths are obtained and then related in order to achieve an acceptable success rate.

The statistical measure of probability that a mechanical component will not fail in use is called the reliability of that element. The reliability is denoted by R and can be expressed by a number having the range (0 ≤ R ≥ 1). A reliability of R=0.85 means that there is 85 precent chance that the component will perform its proper function without failure.

The failure of 12 parts out of every 1000 parts might be considered an acceptable rate for a certain class of components. In this case, the reliability of that component will be

988.01000

121 =−=R (or 98.8%)

The usefulness of the reliability approach depends on having adequate information on the statistical distribution of:

I. loading applied to parts in service, from which can be calculated the significant stress

II. the significant strength of production runs of manufactured parts

Figure 2.19 shows two hypothetical distribution curves for significant stress and for corresponding significant strength. Consider a component subjected at some critical point to a ‘significant stress’ y, and which has a ‘significant strength’ x. If the mean value of strength μx is 40 MPa, and mean value of strength μy is 70 MPa, it can be concluded that if an “average” part from the product run were put into under “average” condition, there would be margin of safety (z) of 30. This does not mean the component can never fail – it simply has a low probability of failure. This probability is highlighted by the unshaded area indicating failures (Figure 2.20).



Figure 2.19: Distribution curves for significant strength x and significant stress y [1]

The smaller the ‘overlap’ of the region on the centre where x < y, the smaller the probability, and the frequency of failures. The more reliable the accuracy of the loading, and the material strength, the smaller this overlap, since these have the effect of changing the standard deviation and the shape of the distribution curves. Figure 2.20 shows a corresponding plot of the distribution of margin of safety (z). In most instances, interest would be focused on the size of the unshaded area at the left, indicating failures.



Figure 2.20: Distribution curve for margin of safety z[1]

NORMAL DISTRIBUTION A statistical quantity is described by its mean, standard deviation, and distribution type. The most common distribution of the parameters is the normal or Gaussian distribution and is the only distribution considered here.

THE STANDARD DEVIATION It defined mathematically as:

∑=

−−

=n

iix

n 1

2)(1

1 µσ

Where µ is the mean value and is given as

∑=

=n

iix

n 1

1µ

Varying the value of the standard deviation (σ ) changes the shape of the curve (Figure 2.21). Standard deviation can be thought as the standard index of dispersion or scatter of the particular quantity. The more dispersed the data, the higher the standard deviation, as shown in Figure 2.21 below.



Figure 2.21: Normal distribution curves having a common μ mean and various [1]

The probability density is defined as the number of occurrences divided by the total sample number. Mathematically,

−−= 2

2

2)(exp

21)(

σµ

σπxxp

Where p(x) is the probability density function.

INTERPRETATION OF DISTRIBUTION CURVE As shown in Figure 2.22 and 2.23, the probability distribution can be used to find out how many components will survive (or fail). This information is also available in tabular form. Both forms, Graph and Table, show the same data, however, the Table is more accurate. How much of the population represented fall within certain bands, described by how far they are (in terms of standard deviation) from the mean. For example, σµ 1± - 68 percent fall within this range

σµ 2± - 95 percent and so on. All normal distributions have the same properties. The distribution shows the percentage of the population within one σ bands of the normal distribution. By summing these percentages to the left or right gives cumulative



probabilities of survival or failure. The graph, as shown in Figure 2.23, indicates theses cumulative percentages on special probability coordinates. For the case being considered, the number of survivors and failures are simply obtained from the chart or tables, once it has been determined how many standard deviations is z=0 below the mean, μz. For the mean

yxz µµµ −=

22

yxz σσσ +=

By definition the probability of failure is p(z<0). Furthermore, if the strength x and stress y are normally distributed, then it can be shown that the margin of safety z also has a normal distribution.

Figure 2.22: Properties of all normal distribution curves [1]



Figure 2.23: Generalised normal distribution curve plotted on special probability

coordinates [1]



EXAMPLE [1] Bolts installed on a production line are tightened with automatic wrenches. They are to be tightened sufficiently to yield the full cross section in order to produce the highest possible initial tension. The limiting condition is twisting off the bolts during assembly. The bolts have a mean twisting-off strength of 20 N.m with a standard deviation of 1 N.m. The automatic wrenches have a standard deviation of 1.5 N.m. What mean value of wrench torque setting would result in an estimated 1 bolt in 500 twisting off during assembly? ANALYSIS AND SOLUTION For bolts: Mean Twist-off strength xµ = 20 N.m Standard Deviation xσ = 1 N.m For wrench: Standard deviation yσ = 1.5 N.m

Mean Torque setting yµ =? Probability of failure = Twisting-off 1 bolt in 500 = 0.002 = 0.2% The standard deviation for the wrench torque setting can be calculated as

8.15.11 2222 =+=+= yxz σσσ N.m

The mean value of z, zµ is not known. It can be determined by finding the number of standard deviations for a failure rate of 0.2%. As shown in Figure 2.23, a failure percentage of 0.2% corresponds to 2.9 standard deviations below the mean. mNk zz .22.58.19.20 =×==− σµ Since yxz µµµ −=

mNzxy .78.1422.520 =−=−= µµµ This is the maximum setting of the torque wrench to give 1 in 500 or less bolt failures. Look at how far below the bolt twist-off strength this is!



Safety factor = FS = 20/14.78 = 1.35 (but still 1 in 500 fails) Figure 2.24 shows the graphical representation of above results.

Figure 2.24: a) Distribution curves for x, y and z [1]



7.0 REFERENCES 1. Juvinall, R.C. and K.M. Marshek, Fundamentals of Machine Component Design. Fourth

ed. 2006: John Wiley & Sons, Inc. 2. Norton, R.L., Machine Design: An Integrated Approach. Third ed. 2006: Pearson Prentice

Hall. 3. Spotts, M.F., T.E. Shoup, and L.E. Hornberger, Design of Machine Elements. Eighth ed.

2004: Pearson Prentice Hall. 4. Mott, R.L., Machine Elements in Mechanical Design. 4th ed. 2006, Singapore: Prentice

Hall (Pearson Education South Asia Pte Ltd).

PRACTICE QUESTIONS Q1: Define even and uneven materials.

Q2: Differentiate between static and fatigue loading.

Q3: Draw sketches to illustrate stresses in rotating machine elements.

Q4: Draw a sketch to compare the failure theories.

Q5: What are the major types of mechanical failures? Explain yielding failure.

Q6: In some situations, the maximum-normal stress theory is dangerously non-conservative. Draw a sketch to highlight this fact.

Q7: Differentiate between factor of safety and margin of safety.

Q8: What method of failure prediction would be more suitable for ductile materials?

18452195 wk2 failure theories

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