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Design for Strength and Endurance – Chapter 7

_____________________________________________________________________________________Factors of Safety - - C.F. Zorowski 2002 145

Chapter 7 Factors of Safety

Screen Titles

Purpose and Definition Items Effecting Factor of Safety Variation Scenario – 1 Variation Scenario – 2 Modified Theories of Failure Generic Factors of Safety Actual Load Distribution Load Capability Distribution Difference of Distribution Gaussian (Normal) Distribution Transformation of Variables Transformation of Normal Distribution Application to Lc-L Distribution Determination of tf

Sample Problem Problem Solution Alternate form for n No Failure Scenario

Review Exercise Off Line Exercises Off Line Exercises



1. Title page Chapter 7 deals with the subject of the factor of safety in design from both a classical and statistical point of view. The presentation begins with a consideration of the purpose of introducing this concept and its definition in a classical sense. This is followed by looking at factors that influence its impact on a design and how it is included in the established theories of static failure. Some recommendations for factors of safety based on generically defined design scenarios are presented and discussed. The chapter content then takes up how a statistical treatment can be used in very large part populations to predict required factors of safety in terms of the properties of the sample’s statistical representation. A number of exercise problems are included to demonstrate the application of the concepts and principles discussed.

2. Page Index Listed on this page are all the individual pages in Chapter 7 with the exception of the exercise problems. Each page title is hyperlinked to its specific page and can be accessed by clicking on the title. It is suggested that the reader first proceed through all pages sequentially. Clicking on the text button at the bottom of the page provides a pop up window with the text for that page. The text page is closed by clicking on the x in the top right corner of the frame. Clicking on the index button returns the presentation to the page index of chapter 7.



3. Purpose and Definition The term “factor of safety” is ubiquitous in the practice of design. Everyone seems to know and appreciate what it means generically but similarly find it difficult to specify what it should be in a given situation. It has been called the “factor of ignorance” or the “factor of uncertainty” by many. The second designation is probably the more accurate. Another sense of its importance is summarized in the couplet “If in doubt make it stout”. One of the better ways to begin its study is to look at what might be considered its principal purpose in design. This can be simply stated as “the concept of factor of safety is introduced into design for the purpose of minimizing the risk of potential part failure”. This risk of potential failure is a direct consequence of the approximate analytical techniques used to determine load and strength levels and the uncertainty and variability in the values used in the numerical calculations involved. Classically this is incorporated in a simple analytical definition that states that the factor of safety is equal to the material strength of a part divided by the working stress that the part is designed to carry. Thus, if the factor of safety is two it simply means that it is anticipated that the part can be subjected to effectively twice the load it was designed for before the design criteria of either yielding or fracture, whichever has been chosen, will be exceeded in a static application.

4. Items Effecting Factor of Safety The natural variations and the level of confidence in understanding and knowing the extent of these variations in four specific areas give rise to the need and utility of a factor of safety to meet the purpose cited on the previous page. The first is the mathematical model with its included assumptions used to calculate the working stress in the part. For example, is simple bending theory based on elastic behavior appropriate to determine the bending stress in the part under consideration? The second area is the geometry of the part. This is characterized by the variability in the dimensions of the part as dictated by how it was physically processed into its final shape. In other words how close are the final dimensions to those used in calculating the working stress? The third area is the degree of certainty of the magnitude of the applied loads. Is there some question about the level of overloading the part may be subjected to or the direction in which the load may be applied? Finally, there is the area of the strength properties of the material from which the part will be made. How accurately are the material properties known and what is the degree of their variability?



5. Variation Scenario - 1 To determine the effect of variations of parameters associated with the areas discussed on the previous page consider the following numerical example. A bending moment of 10,000 in. lbs. plus or minus 500 in. lbs. acts on a rectangular cross section beam with nominal dimensions of 2 in. by 1 in. The width is only known to within plus or minus 1/16 inch and the height may vary by as much as plus or minus 1/8 inch. Using the nominal load and dimensions the nominal bending stress at the top of the section is calculated to be 14,992 psi. However, using the maximum possible bending moment together with the minimum possible dimensions results in a comparative maximum working stress in bending of 19,114 psi at the top of the section. This is 1.27 times the nominal working stress due to the uncertainty of the loading and the geometry.

6. Variation Scenario - 2 Now assume that the yield stress for the beam material is 15,000 psi plus or minus 10% or 1500 psi as established from a series of tensile tests. Therefore the minimum yield stress will be 13,500 psi or 9/10ths of sigma y nominal. Applying the criteria of the maximum normal stress theory can be stated as the maximum working stress must be less than or equal to the minimum yield stress. Substituting 1.27 times the nominal working stress for the maximum working stress and .9 nominal yield stress for the minimum yield stress gives the inequality that 1.27 sigma working nominal must be equal to or less than .9 sigma yield nominal. The concept of the factor of safety is now introduced by the equation that the nominal working stress should be equal to the nominal yield stress divided by the factor of safety. Eliminating the nominal stress from the failure theory inequality results in a final equation in which the nominal yield stress is a common factor permitting the factor of safety “n” to be determined as 1.4. This is the value of “n” that should be applied in the nominal design calculations to account for the uncertainty of the parameter values of load, geometry and material property to insure that yielding will not take place.



7. Modified Theories of Failure - 1 Incorporating the concept of a factor of safety into the equations for the static theories of failure for both ductile and brittle materials is a simple modification of the expressions developed in Chapter 6. It is only necessary to divide the strength property representation of the material designated by either the yield or ultimate stress that normally appears on the right side of the various ductile theory equations by the factor of safety “n”. In the equations shown here the material strength parameter in the first three theory relations has been placed in the denominator of the left side so that the right side simply becomes one over the factor of safety “n”. It should be noted that in the modified Coulomb Mohr theory where the material properties in tension and compression already appear on the left side of the equation that the right side is again simply one over the factor of safety “n”. This implies that the same factor of safety is being applied to both the tensile and compressive properties.

8. Exercise Problem In this first interactive exercise problem you are to determine the factor of safety for a shaft in pure torsion applying the maximum shear stress theory where the twisting torque may vary by plus 30%, the diameter is known to within 2% and the yield stress in tension is assumed to be within plus or minus 7% of the nominal value listed in a table of properties. When you have completed your calculation click on the solution button to check your answer. Then continue on by use of the return button.

(Solution on Page 165)



9. Generic Factors of Safety Unfortunately, variations in the parameters representing the loading, geometry and material properties are seldom known precisely at the beginning of the design process. This makes it virtually impossible to establish apriori what the specific magnitude of the factor of safety should be. To circumvent this difficulty accumulated design experience and judgment over many years has lead to suggested recommendations for possible factors of safety to be used based on generic classification of such general characteristics as material properties, loading and application environment. One such table of recommendations is shown on this page taken from the 20th edition of Machinery’s Handbook. Note how broad and generic the classifications are. Their precise interpretation is of course left up to the judgment of the designer. It is to be observed that in the two numerical examples already covered both would seem to fit into the category of the materials properties being know to a level of high reliability and the load and environment as being not severe, whatever that means, and that the weight of the part has significant importance. You can be sure that these kinds of recommendations are very strongly conservative irrespective of how they are interpreted.

10. Exercise Problem –2 In this exercise the parameters of the specified part design are expressed in nominal terms for the loading, geometry and the yield property of the material. Your task is to determine whether the part will yield where the stresses are maximum and if yielding does not take place what minimum factor of safety is provided by the design. Assume that the part is made of a ductile material and that the load is constant. After you have completed your analysis click on the solution button to check your conclusions. When you are satisfied with the solution click on the return button to continue to the next page of the chapter.

(Solution on Pages 166 and 167)



11. Actual Load Distribution When a part is designed and fabricated in very large numbers for the same application it is of value to apply statistical considerations to the concept of a factor of safety. The remainder of this chapter will be devoted to this subject. Begin by considering that the loads to be carried by the total population of parts will have some probability distribution around a mean value of the load designated by L bar. Any specific load L in this distribution will fall between the limits of L bar plus or minus delta L representing the maximum variation of the load from the mean value. In general the natural probability distribution of L will look something like the bell shaped curve in the figure. That is, it is expected that the probability of L will be highest around the mean value and will tail off in both the positive and negative directions from L bar approaching zero at L bar plus or minus delta L.

12. Load capability In a similar fashion the load capability of the part, that is, the ability of the part to carry a load L will also possesses a probability distribution due to variances in its geometric and material property values. Again the distribution is defined in terms of a mean load capability Lc bar and variance of plus or minus delta Lc as depicted in the figure. The height and breadth of the distribution may be different from that of the actual load distribution but its appearance will again may be expected to be bell shaped for a very large number of parts.



13. Comparison of Distributions Now consider the interpretation of these two distributions plotted beside each other on the same axis system. It is expected that the mean value of the load capability Lc bar will be greater than the actual mean load L bar to prevent part failure. However, if any portion of the two distributions overlap as shown in the figure it means that there are parts that will be subjected to loads that are greater than the available load capability of some parts. Hence, the region of overlap represents potential failure. Whether such a region exists in a specific instance is a function of the mean values of the actual load and the load capability and the variances in both these parameters. If the mean actual load and load capability are interpreted as nominal load and load capability then it is appropriate to define a factor of safety as the ratio of Lc bar to L bar as was done earlier in the chapter.

14. Difference of Distributions It is now convenient to consider the interpretation of the difference of these two distributions. This difference will also be a bell shaped probability distribution as depicted in the figure. The distance from the origin to the peak value of the distribution will just be equal to the difference in the mean values of the separate distributions. If a portion of the distribution extends over into the negative side off the origin the area under that portion of the curve represents the portion of parts subject to failure from the total population, which is represented by the total area under the total combined distribution curve.



15. Gaussian (Normal) Distribution) The assumption is now made that the bell shaped probability distribution of the difference between the load capability and the actual load curve can be represented as a Gaussian or normal distribution. This permits some of the mathematics of statistics to be introduced into this analysis to create an analytic model for predicting numerical factors of safety where large populations are involved. The normal probability density distribution y of a variable x is defined by the exponential function adjacent to the curve in the figure. In this exponential function x bar is the arithmatic mean of the variable x as N approaches infinity as indicated by the accompanying equation. The symbol Dx is called the standard deviation and is defined as the square root of one over N minus 1 times the sum over N of the quantity the difference of the variable x and its mean value squared. The most important property of the standard deviation in this development is that the area under the probability density curve within three standard deviations on either side of the mean value of x represents 99.73 percent of the total population under the entire probability curve from minus to plus infinity.

16. Transformation of Variables A transformation of variables is now introduced to put this statistical model into a form in which the results of integrating the normal distribution over some portion of its variable can be obtained from available standard tables of numerical results. The importance of this will be clarified shortly. The transformation introduced is a new variable t that is equal to x minus x bar divided by the standard deviation Dx. It is demonstrated on this page that the mean value t bar of the variable t is zero and that its standard deviation Dt is simply 1. Hence, the transformed probability distribution of t, which is also normal, is uniformly distributed around the origin of the axis system. The transformed normal distribution of the variable t is depicted in the figure. It is seen as centered on the origin since its mean value is zero. The area under the portion of the total distribution represented by the horizontal coordinates plus and minus 3 constitutes 99.73 percent of the total population. In this instance the total area under the curve is unity. The shaded area that corresponds to the integral from minus infinite to minus tf represents the decimal percentage of all points in the population with values less than minus tf. The importance of this shaded area as it relates to factor of safety will be made clear on the next page.



17. Transformed Normal Distribution The transformed probability distribution is now expressed more simply as y equal to one over the square root of two Pi multiplied by e raised to the power of minus t squared over two. The curve is now observed centered with respect to the origin with its mean value at t equal to zero. With a standard deviation of one the area under the portion of the curve between t equal to –3 to t equal to +3 represents 99.73 % of the total area under the curve from minus infinity to plus infinity. The total area under this transformed distribution is simply unity. The shaded area under the portion of the curve from minus infinity to –tf represents the decimal % of all points with values less than –tf. The parameter Tf plays a very important role in the statistical definition of factor of safety as will be demonstrated on the next two slides.

18. Application to Lc-L distribution To relate these properties of the unit normal distribution to factor of safety the variable t is now written in terms of the parameters of the difference of the load capability and actual load probability distribution assuming this distribution is also normal. That is, t is now defined as the quantity Lc minus L minus the quantity Lc bar minus L bar all divided by the standard deviation of the Lc minus L distribution. This standard deviation can itself be written as the square root of the sum of the squares of the standard deviations of the individual Lc and L distributions. For proof of this refer to any standard text on statistics. Now set t equal to -tf for which Lc minus L is equal to zero. Click on the recall button to pop the previous graph of Lc minus L to see that this is true. Having satisfied yourself that this is the case substitute these conditions into the general expression for tf, solve the result for Lc bar and divide both sides of the equation by L bar. Thus the left side of the resulting equation is simply the factor of safety “n” defined as the mean load capability divided by the mean actual load and the right side of the equation contains terms which define the characteristics of the Lc and L distributions multiplied by tf. Appropriate application of this equation permits the determination of a factor of safety for a large population of parts whose load characteristics approximate normal probability distributions.



19. Determination of tf To apply the equation from the previous page to calculate a factor of safety it is first necessary to determine the appropriate value of tf. Recall that tf is the negative upper limit of the integral of the area from minus infinity of the unit normal distribution that now represents the decimal percentage of failure due to Lc being less than L. Unfortunately this integration cannot be carried out in closed form. Hence, it is necessary to refer to numerical tables that list the value of the integral for specific values of tf. A portion of such a table is presented on this page in which values of the integral are given for values of tf from 2.00 to 3.08 in intervals of .02. As an example if a specific design instance will permit a failure potential of 2% or two parts per hundred the value to be used for the integral of the area representing this failure level is 0.02. The closest value of tf from the table corresponding to this integral value is approximately equal to 2.04. This is the value of tf that would be used in the equation for determining the associated factor of safety depending of course on the values of other distribution parameters in the equation. by the direction of epsilon 2 and an axis perpendicular to it. Note that the construction results in angles of two theta of 90 degrees between the diameter for axes 13 and the direction of epsilon three, which agrees with the fact that on the rosette axes 1,2 and 3 are separated by 45 degrees.

20. Approximate values of tf For low values of acceptable failure rates a convenient approximation for tf can be employed. First define Fd as the percent decimal failure. That is, Fd is equal to the integral of the unity distribution function from minus infinity to minus tf. Then for Fd between .001 and .015 that represents failure rates from .1% to 1.5% tf can be approximated by the equation tf equal to 1.29 divided by Fd raised to the 0.128 power. This equation for the limits previously specified is accurate to within 2% error. This approximation is then substituted into the factor of safety equation resulting in a relationship that includes the decimal percent failure rate directly along with the other parameters describing the actual load and load capability distributions.



21. Sample Problem This statistical approach for determining a factor of safety will now be applied to a specific problem. Assume that a bar in tension is to carry a load of 1000 lbs. with a possible variation of +/- 25 %. Thus the maximum load capability must be at least 1250 lbs. However, the load capability can only be known to within +/- 10 % variation. Assume that a sufficient number of parts are to be produced such that these variables can be considered normally distributed. Calculate the necessary factor of safety to be applied to the mean values of load and load capability if the failure rate is not to exceed 2 %.

22. Problem Solution From the conditions of the problem L bar is1000 lbs. and delta L is 250 lbs. The value of Lc bar should be at least 1250 lbs. To be a little conservative it will be assumed to be 1300 lbs. Its variance will then be 130 lbs. The standard deviation for the actual load will be delta L over three or 83.3 lbs. while the standard deviation for the load capacity will be delta Lc over three or 43.3 lbs. A value of tf is now needed for a failure rate of 2%. Click on the % failure button to pop up the numerical chart from which it is seen that tf is just about 2.04. Substituting all these values into the equation for the factor of safety and carrying out the indicated mathematical manipulations gives a final answer of about 1.2. This value would be applied to design calculations based on the nominal values of actual load and load capability.



23. Alternate form for n An alternate formulation for n in terms of just the variations in actual load and load capability can be developed by manipulating the form of the equation used in the previous problem. Starting with the equation on the first line on this page the standard deviations are first replaced with delta Lc over 3 and delta L over 3. The L bar from the denominator of the second term is taken inside the square root term. It is then recognized the term delta Lc over L bar can be rewritten as delta Lc over Lc bar times the factor of safety n. This permits the starting equation for n to be written in the form given at the bottom of the page.

24. Alternate form for n (continued) The last equation from the previous page is rearranged so that only the square root term remains on the right. This expression is now squared and the common terms of n and its powers are combined. This results in the quadratic equation for n at the bottom of the page in which the coefficients are only functions of the variations of the actual load and the load capability along with a measure of the failure rate in the parameter tf. One further modification that could be made to this equation would be to replace tf by it equivalent representation in terms of the decimal percentage failure Fd provided Fd is limited to a failure rate between .1 to 1.5 %. This is left for the reader to do if desired.



25. Exercise Problem - 3 This exercise deals with determining the factor of safety “n” for the sample problem just solved using the alternate form for “n” as a quadratic equation. Recall that the mean applied load was 1000 lbs. with a variation of 25% and the mean load capability was taken to be 1300 lbs. with a variation of +/- 10 %. The acceptable failure rate was 2% for which tf was 2.04. Compare the result with the solution for “n” obtained in the sample problem. When you have calculated your answer click on the solution button to check your result. When finished with the solution click on the return button to continue on to the next page.

(Solution on Page 167)

26. “No Failure” Scenario A relationship will now be developed for “n” for the case in which the failure rate is to be zero. This condition is defined by the inequality that L bar plus delta L must be less than or equal to Lc bar minus delta Lc. In other words the maximum actual load must be less than the minimum load capability. This inequality is first solved for Lc bar. Then both sides of the equation are divided by L bar. The last term of delta Lc over L bar is rewritten as delta Lc over Lc bar times n the factor of safety. The resulting expression is then solved for n giving the final relation that “n” is greater than or equal to the ratio of one plus delta L over L bar to one minus delta Lc over Lc bar. Note that this equation only involves the percentage variation of the actual load and the percentage variation of the load capability.



27. Example Problem –4 In this final chapter 7 exercise determine the factor of safety for the previous problem assuming the “no failure scenario”. Recall again that the mean applied load is 1000 lbs. with a variation of +/- 25% and the mean load capability was taken to be 1300 lbs. with a variation of +/- 10 %. Compare this “no failure” factor of safety with the value calculated for a 2 % failure rate. When you have calculated your answer click on the solution button to check your result. When finished with the solution click on the return button to continue on to the next page.

28. Review Exercise In this exercise the items in the list on the left are to be matched with the symbols and mathematical relationships on the right. Place the cursor over an item on the left and hold down the left button. A pencil will appear that can be dragged to one of the green dots on the right. If the right choice is made the arrow will remain. If the selection is incorrect the arrow will disappear. After the exercise is completed proceed to the next page.



29. Off Line Exercise An established design procedure calls for the use of a factor of safety of 2 for a class of specific non-moving parts used in an automotive suspension system. If the material properties variation is expected to be 10% determine the allowable projected load variation for zero failure potential based on the “no failure” scenario. How does this load variation change if 1 part per 100 or 1 part per 1000 is the prescribed failure rate? When you have finished with this statement click on the exit or main menu button to leave the chapter.

(Solution in Appendix)



Chapter 7 Factors of Safety

Screen Titles

Problem 1 - Solution Problem 1 – Solution (cont.) Problem 2 - Solution Problem 2 – Solution (cont.) Problem 2 – Solution (cont.) Problem 2 – Solution (cont.) Problem 3 - Solution Problem 3 – Solution (cont.) Problem 4 - Solution


_____________________________________________________________________________________Factors of Safety - 164 - C.F. Zorowski 2002



1. Problem 1 - Solution For a shaft in pure torsion the nominal principal stress sigma one is equal to the maximum shear stress defined by TR over J with J equal to Pi R fourth over two. This give sigma one nominal equal to two T over Pi times R cubed. It then follows that sigma one max is equal to two T max divided by Pi times R min cubed. Substituting T max and R min into this expression gives sigma one max equal to 1.34 times sigma one nominal. The problem also states that sigma yield min is equal to .9 sigma yield nominal.

2. Problem 1 Solution (cont.) From the maximum shear stress theory of failure sigma one max must be less than or equal to sigma yield min over two. The previous expressions for sigma one max and sigma yield min are now substituted into this inequality. The factor of safety is introduced with the equation that sigma one nominal is given by sigma yield nominal divided by 2n to satisfy the maximum shear stress theory of failure. Using this equation to eliminate sigma one nominal from the previous inequality allows n to be determined as the ratio of 1.34 to .9 for a final result of 1.49. When finished with this solution click on the return button to go to the next page in Chapter 7.



3. Problem 2- Solution The location of the maximum bending stress in the bar is where the bar and rod are welded together. This stress is equal to Mc over I where I is the moment of inertia of the rectangular cross section given by b times h cubed over 12. The magnitude of the moment M is the force, 30 lbs. times the length of the bar, two inches, giving 60 in. lbs. With the given dimensions of the cross section the I is calculated to be 1.1 times ten to the minus three inches to the fourth power. Taking c to be half the height of the section the maximum bending stress becomes 10,230 psi. Since this is the only normal stress in the bar the applicable theory of failure is the normal stress theory. With the yield stress in tension for the material in the bar being 50,000 psi the factor of safety for this portion of the part is determined to be 4.9

4. Problem 2 – Solution (cont.) Now consider the stresses at the fixed end of the rod. These will consist of a shear stress due to the twisting torque created by the force F and a bending stress resulting from the bending moment due to the force F. Both the twisting torque T and the bending moment M have the same numerical value of 60 in. lbs. Since the rod is a quarter inch in diameter its polar moment of inertia J is calculated to be .35 time ten to the minus three inches to the fourth. The moment of inertia needed for the bending stress calculation is just half the value of J since the cross section is a solid circle. Substituting these values into the Mc over I and Tr over J equations for the stress gives a shear stress of 21,420 psi and a bending stress of 42,840 psi.



5. Problem 2 – Solution (cont.) Next the principal stresses must be calculated before applying an appropriate theory of failure to determine the factor of safety for this portion of the part. Since there is no normal stress perpendicular to the bending stress the equation for the maximum principal stress is given by the equation at the top of the page. Substituting the appropriate values for the normal stress and the shear stress into this equation gives sigma one equal to 51,700 psi. In a similar fashion the minimum principal stress is calculated to be –8,900 psi. With the two principal stresses being opposite in sign the appropriate theory of failure to apply is the maximum shear stress theory. Substituting the numerical values of sigma 1, sigma 2 and the yield stress of 75,000 psi into the equation for the maximum shear stress with the factor of safety included results in a numerical value for the factor of safety of 1.24. This is less than the value calculated for the bar section and thus is the minimum factor of safety for the part. When you have finished with this solution click on the return button to go to the next page in Chapter 7.

6. Problem 3 – Solution The solution is begun by recalling the equation for n in quadratic form. The fractional variations for the load capability of .10 corresponding to 10% and for the actual load of .25 representing 25 % are substituted into this equation along with the value of tf of 2.04 from the sample problem. This permits the coefficients of the n terms in the quadratic equation to be numerically determined. The resulting equation is then divided through by the coefficient of the n squared term and the quadratic formula is applied to determine a final value for n of 1.19 or approximately 1.2. It is observed that this is the same result as determined in the sample problem without an assumption being required as to the magnitude of the mean value of the load capability. When you have finished with this page click on the return button to go to the next page in Chapter 7.



7. Problem 4 – Solution To calculate the factor of safety for the sample problem for the condition of a zero failure rate requires the application of the equation for the “No failure “ scenario. The only parameters needed to carry out this application numerically are the fractional variations of the actual load and the load capability. Respectively these values are .25 and .10. Substituting these values into the “No failure” scenario relationship gives a numerical value for the factor of safety of 1.39. Compared with the value of 1.2 calculated for n with a 2 % acceptable failure rate this represents a 16 % increase in the factor of safety. When you have finished with this page click on the return button to go to the next page in Chapter 7.

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