chapter 9 - design for strength and endurance | carl zorowski · design for strength and endurance...

Design for Strength and Endurance – Chapter 9

_____________________________________________________________________________________Fluctuating Load Analysis - - C.F. Zorowski 2002 189

Chapter 9 Fluctuating Load Analysis

Screen Titles

Fluctuating Stresses Generic Stress-Time Behavior Stress-Time Relations Modified Goodman Diagram Mean/Fluctuating Stress Diagram Soderberg Failure Theory Goodman Failure Theory Gerber Failure Theory Sample Problem – 1 Problem – 1 Solution Torsional Fatigue Combined Loading Modes Sample Problem – 2 Problem – 2 Solution Cumulative Fatigue Damage Palmgren-Minor Theory Sample Problem – 3 Problem – 3 Solution Mason’s Modification Review Exercise Off Line Exercise



1. Title page Chapter 9 continues the study of the fatigue analysis and behavior of mechanical parts. Attention is now directed to elements subjected to general fluctuating loads as contrasted to complete reversed stress states treated in Chapter 8. The presentation begins with defining generic parameters to represent fluctuating loads. General fatigue failure under fluctuating normal stress loading is then discussed and three specific failure theories are introduced together with fatigue under fluctuating torsional stresses. A method of fatigue analysis for combined stress states is also covered. The chapter concludes with two methods of determining the effects of cumulative fatigue damage due to multiple loads applied for different durations of cyclic application. Several exercise problems and extended sample problems are included to demonstrate the application of the subject content.

2. Page Index Listed on this page are all the individual pages in Chapter 9 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 9.



3. Fluctuating Stresses Chapter 8 considered the fatigue behavior of mechanical elements subjected to complete reversed stress states in which the stress level varied from some specific negative value to the same positive value and then repeated this behavior continuously. This chapter will deal with repeated stress states that may vary from zero to some specific value and then back to zero continuously or even more generally from some minimum value to a different numeric positive value and then back to the minimum value repeatedly. In this instance the variation will possess some mean value of stress that will lie between the minimum and maximum value of the applied stress state.

4. Generic Stress Time Behavior Depicted in the figure is an idealized general fluctuating stress versus time behavior curve. Although this is represented as a smooth sinusoidal function the parameters that will be used to generalize its description will be the same irrespective of the specific shape of the curve provided it is periodic in behavior. In other words the shape could be saw tooth or a square wave, it would still be characterized for fatigue analysis in the same way. The four parameters used to define the curves characteristics for the fatigue discussions and analyses to follow are its maximum value, sigma max, its minimum value, sigma min, its mean value, sigma m and its alternating value about the mean, sigma a.



5. Stress –Time Relations Of primary interest in the developments to follow are the mean stress and the alternating stress components. The mean stress is of course just the average of the two extreme values or one half the sum of the maximum and minimum values. The alternating stress, which is sometimes referred to as the stress amplitude, is simply one half the difference between the maximum and minimum values. Of occasional value is the stress ratio, R, defined as the minimum stress divided by the maximum stress and the amplitude ratio, A, which is the alternating component divided by the mean component.



6. Modified Goodman Diagram Fatigue failure behavior from experiments conducted under different combinations of fluctuating normal stress loading states are conveniently presented and described on a “Modified Goodman Diagram”. In this representation the mean stress is plotted on the horizontal axis and all other stresses are plotted vertically with tension to the right and up. A 45-degree line from the origin in the first quadrant represents the mean stress of an applied stress state. When the mean stress is zero fatigue failure is represented by plus and minus the endurance stress on the vertical axis. This is indicated by the red dot at the origin and the green dot on the vertical axis. A green dot on the negative vertical axis is not shown due to space limitations. Now consider the case where the mean stress plus the stress amplitude are equal to the yield stress of the material. This is illustrated by the second set of red and green dots. Assuming that yielding is considered as undesirable, that is equivalent to “failure”, then lines drawn from the second set of green dots to the positive and negative endurance limits on the vertical axes corresponds to a proposed criteria for defining fatigue failure for a generalized fluctuating stress loading. The failure boundary is completed by extending the line from the upper green dot on the maximum stress line to the value of the yield stress on the mean stress line and from that point to the lower green dot on the minimum stress line. Note that when the mean stress is equal to the yield stress no alternating stress is necessary to produce yielding. However, as the mean stress decreases the corresponding value of alternating stress required to define fatigue

failure increases as indicated by the third set of red and green dots. Thus the area enclosed by the red boundary in the first quadrant proposes states of stress for which fatigue failure will not occur. Experimental results have verified that this is a good approximation for defining fatigue behavior under a general fluctuating state of stress.



7. Mean/Fluctuating Stress Diagram A second way of depicting fatigue failure under the action of a fluctuating stress state is obtained by plotting the mean stress horizontally and the alternating stress vertically. The yield stress of the material is first plotted on both the vertical axis and the positive and negative axes. Then diagonal lines are used to join these three points as illustrated on the graphic. Next the endurance limit is plotted on the vertical axis together with the ultimate tensile strength on the horizontal axis. In the first quadrant a line is drawn from the endurance limit to the tensile strength. Any combination of alternating stress and mean stress that falls on the lower of the fatigue line or yield line, indicated by the heavy red boundary, defines a condition of failure either by fatigue or yielding. It has been determined experimentally that compressive mean stresses have virtually no effect on reducing the fatigue strength of a mechanical element. Hence in the second quadrant a horizontal line is extended from the endurance limit until it intersects with the yield line. Any combination of alternating stress and compressive mean stress that falls on the lower of this fatigue line or the yield line again indicated by the heavy red boundary defines a condition of failure either by fatigue or yielding. Since this is just another way of presenting the proposed behavior depicted by the modified Goodman diagram experimental test results indicates that it is a reasonable approximation of fatigue behavior under application of a general fluctuating stress state.

8. Exercise Problem 1 Application of the behavior depicted on the previous page to a design analysis process requires a mathematical model. This exercise provides an introduction to the development of this type of model. Following the exercise the presentation will introduce and discuss three theories of fatigue failure based on this type of development. After completing the exercise click on the solution button to check your result.

(Solution on Page 215)



9. Soderberg Failure Theory Three distinct proposed failure theories for fatigue design application will now be presented and discussed. The first is the Soderberg Theory. Using the graphic on the previous page this theory proposes that designs for fluctuating normal stress states should be based on a limiting condition defined by a straight line drawn from the endurance limit on the vertical axis to the yield point on the horizontal axis in the first quadrant. This is defined analytically by the equation that the ratio of the alternating stress, sigma a, to the endurance limit, sigma e, plus the ratio of the mean stress, sigma m, to the yield stress, sigma y, is equal to 1. A factor of safety, n, can be introduced into this equation by dividing the right side of the equation by n. This can be seen to be a fairly conservative design approach.

10. Goodman Failure Theory The second proposed failure theory for fatigue design application under a general fluctuating normal stress loading is the Goodman Theory. It proposes a failure line that extends from the endurance limit on the vertical axis to the tensile strength on the horizontal axis. In effect it discounts yielding as a failure condition and is less conservative than the Soderberg theory particularly for mean stress values in excess of the yield strength. Analytically it is defined by the equation that the ratio of the alternating stress, sigma a, to the endurance limit, sigma e, plus the ratio of the mean stress, sigma m, to the tensile stress, sigma u, is equal to one. Again a factor of safety, n, can be introduced by dividing the right side of the equation by n.



11. Gerber Failure Theory.) The Gerber Failure Theory differs from the Soderberg and the Goodman theories in that it represents the failure line as a quadratic curve that passes through the endurance limit and the tensile stress. Of the three theories it is the least conservative and is considered by many to be the more accurate of the true behavior and impact of fluctuating loads on fatigue strength. Analytically it is represented by the equation the ratio of the alternate stress, sigma a, to the endurance limit, sigma e, plus the square of the ratio of the mean stress, sigma m, to the tensile strength, sigma u, is equal to one. To introduce a factor of safety into this express n is added to the numerator of the two stress ratios on the left side of the equation. Of the three theories presented preference for design applications and analysis is generally accorded to the Goodman theory. As such it will be used throughout the remainder of this chapter.

12. Exercise Problem - 2 In this exercise you are asked to determine numerically the magnitude of the difference between the Goodman and Gerber theories of failure over the range of sigma m over sigma u from zero to one. Since both theories pass through the endurance limit and the tensile strength there of course is no difference between them at these two points. When you have completed your analysis click on the solution button to check your answer.

(Solution on Pages 215 and 216)



13. Sample Problem 1 A sample problem will now be solved to illustrate a typical application of the Goodman theory of failure analysis. In this instance a flat cantilevered spring is subjected to a fluctuating end load. The dimensions of the element are obtained by clicking on the hot word in red at the beginning of the problem statement. The tensile strength of the spring material is specified as well as a stress concentration at the end where it is clamped. With these specifications and the load variation it is desired to determine the factor of safety for three specific conditions. The first requires that the mean stress be held constant, the second requires that the alternating stress be held constant and the third specifies that the amplitude ratio of the alternating to the mean stress be held constant.

(Linked Figure on Page 219)

14. Prob. – 1 Solution – Page 1 Since only the tensile strength of the material is given the first task is to obtain an estimate for the endurance limit of the steel. A value for the uncorrected or test specimen endurance limit is obtained by multiplying the tensile strength, sigma u, by 0.504 to give a value of 42.8 kpsi. Next the correction factors need to be determined to reduce the uncorrected endurance limit to what it will be for the mechanical element. Using the appropriate constants for a ground surface finish the surface factor, ka, is calculated to be 0.918.



15. Prob. – 1 Solution – Page 2 The size factor is determined by first substituting the cross section dimensions of the spring into the expression for the effective diameter. This gives a value of 0.1429 inches. This effective diameter is then substituted into the size factor equation for shafts less than 2 inches in diameter. The result of the calculation is a value for kb of 1.08. Since size factors are normally considered to be less than one kb will be taken as one to be conservative. With the spring subjected to a bending load the load factor, kc , will also be one as will the temperature factor, kd, since the spring will operate at room temperature. Finally the factor ke due to the stress concentration at the clamp, where the stress loading will be the greatest is one over 1.2 giving a value of 0.835.

16. Prob. – 1 Solution – Page 3 The part endurance limit is now obtained by multiplying sigma e by the product of all the correction factors. This results in a value of 32.73 kpsi for the corrected endurance limit sigma e prime. Next the loads and stresses need to be calculated. The mean load of the variation from 6 to 14 lbs is just 10 lbs while the alternating load component is 4 lbs. The bending stresses due to these loads will be a maximum at the clamp and will be given by the expression Mc over I. I is determined from the cross sectional dimensions and the expression bh cubed over twelve as 1.02 times ten to the minus fifth power. Substituting the loads, beam length, I and c as 1 /32 inch into the stress formula gives a mean stress of 30.7 kpsi and an alternating stress of 12.3 kpsi.



17. Prob. – 1 Solution – Page 4 Now consider the construction of a Goodman theory diagram with the numerical values calculated so far. Sigma e prime equal to 32.73 kpsi is plotted on the vertical axis while the tensile strength sigma u equal to 85 kpsi is plotted on the horizontal axis. A straight line is then drawn between these two points depicting the Goodman theory of failure line. Next the coordinates sigma m of 30.7 kpsi and sigma a of 12.3 kpsi define the point marked by the light gray dot. The solution to the first part of the problem is obtained by holding sigma m constant and allowing sigma a to increase to the value of sigma a super a at the red dot and then dividing sigma a super a by sigma a to obtain a factor of safety. A similar procedure is used to solve the second part of the problem with the exception that sigma a is held constant and sigma m is permitted to increase to sigma m super b at the green dot. For the third part the ratio of sigma a to sigma m is held constant and both values are permitted to increase along the diagonal to the blue dot at point c.

18. Prob. – 1 Solution – Page 5 The ratios of the sides of similar triangles will now be used to determine the unknown mean and alternating stress values for failure in the three scenarios described on the previous page. For part a the two triangles to be used are shown on the Goodman diagram by clicking on the graphic button. An equation to solve for sigma a super a is obtained by equating the vertical to the base on the green triangle to the vertical to the base on the red triangle. Now close the popup window. By substituting the appropriate numbers for all the known quantities sigma a super a can be determined to be 20.91 kpsi. Dividing this value by sigma a from the original loading gives a factor of safety of 1.70. Another way of interpreting this result is that the alternating value of the load can be increased by 70 % while keeping the mean load constant before failure due to fatigue will take place. For part b click on the graphic button to see that the ration of the vertical side of the green triangle to its base is similar to the vertical side of the red triangle to its base. This results in an equation in which the only unknown is sigma m super b. Now close the pop up window. Substituting the known parameters into the equation of side ratios permits sigma m super b to be determined as 53.06 kpsi. Dividing this value by sigma m equal to 30.7 gives a factor of safety of 1.73 if sigma a is held constant.



19. Prob. –1 Solution – Page 6 For part b the figure shows that the ratio of the vertical side of the green triangle to its base is similar to the vertical side of the red triangle to its base. This results in an equation in which the only unknown is sigma m super b. Substituting the known parameters into the equation of side ratios permits sigma m super b to be determined as 53.06 kpsi. Dividing this value by sigma m equal to 30.7 gives a factor of safety of 1.73 if sigma a is held constant.

20. Prob. –1 Solution – Page 7 In part c the equation involving the unknown stress parameters that will bring on fatigue failure is obtained using the similar triangles in the figure obtained by again clicking on the graphic button. Once more the equation becomes the ratio of the vertical side of the green triangle to its base equated to the ratio of the vertical side to the base of the red triangle. Now go back to the presentation by closing the pop up window. It is observed that the equation involves two unknowns, sigma a super c and sigma m super c. However, the ratio of sigma a to sigma m is to be held constant in this part of the solution so sigma a super c can be replaced by 0.401 times sigma m super c. Solving for sigma m super c gives a value of 41.64 kpsi. The factor of safety is again determined by dividing sigma m super c by sigma m to give value of 1.36. Why is this factor of safety significantly less than that calculated for parts a and b?



21.Exercise -3 This exercise will give you practice with solving a problem similar to the sample just concluded. Note that the applied tensile load consists of two components, one that is constant and a second component that is fluctuating. Also note that the ratio of the endurance stress to the tensile strength is not 0.504. When you obtain your result click on the solution button to check your answer before going on to the rest of the presentation.

(Linked Sketch on Page 219) (Solution on Pages 216, 217 and 218)

22. Fluctuating Torsional Fatigue All the subject material covered to this point in Chapter 9 has dealt with fatigue due to a fluctuating normal stress loading condition. The question is now raised as to how to treat this type of problem when the applied stress is due to a torsional loading. A Goodman diagram is again employed with the endurance limit replaced by a corrected torsional endurance limit and the tensile strength is replaced by an ultimate shear strength. The process of analysis remains the same after the alternating and mean shear stress components are determined from the torsional loading. In determining the corrected shear endurance limit from the endurance limit the load factor kc is always taken to be 0.577 for torsion. The ultimate strength in shear is estimated as 0.67 times the tensile strength after the work of Joerres at Associated Spring.



23. Combined Loading Mode The next problem of interest is how to perform a fatigue analysis when the loading system results in both normal and shear fluctuating stress in a combined mode of application. On this and the following page are the series of steps generally accepted as appropriate to analyze this type of problem. The first step is to determine the mean and alternating stress components for all applied loads. The second step is to apply stress concentration factors to the alternating components of all resultant stresses. The third step is to multiply any alternating axial stress components by 1.083. With all stress components known, modified and combined appropriately use Mohr’s circle to determine the principal mean stresses from the mean stress component system and the principal alternating stresses from the alternating stress component system.

24. Combined Loading Mode (continued) With the mean and alternating principal stresses determined effective values of mean stress and alternating stress are calculated using the equation in step 5. This effective stress is called the von Misses stress and is a consequence of a distortion energy consideration. Step 6 specifies that the material properties to be used are the endurance limit in bending and the tensile strength. However, the endurance limit should not be corrected for stress concentration effects, as these were included in step 2. Finally apply the Goodman fatigue analysis using the effective alternating stress, sigma a super e, and the effective mean stress, sigma m super e.



25. Sample Problem - 2 A sample problem involving combined loading will now be solved using the process outlined on the last two pages. A rotating shaft transmits a constant torque that generates a shear stress of 8 kpsi. The shaft is also subjected to an axial load that produces a tension of 10 kpsi. In addition the shaft carries a bending load that results in a maximum alternating bending stress of +/- 23 kpsi. Determine the factor of safety for this shaft it its material has a tensile strength of 75 kpsi and its corrected endurance limit is 37 kpsi.

26. Prob. 2 Solution Page 1 All corrections for axial load and stress concentrations are neglected in this problem for simplicity. Begin by considering the stressed element on the upper left of the page. It is subjected to a fluctuating bending stress and constant axial stress as well as a constant shear. The alternating normal stress component is one half the difference between the maximum and minimum normal stresses combing the components from bending and axial tension. This gives a value for sigma a equal to 23 kpsi. The mean normal stress is the average of the maximum and minimum normal stress which gives a value of 10 kpsi. These results could have been seen by inspection since the axial tension is constant and the bending contribution is completely reversed. The mean shear stress, tau m is 8 kpsi since there is no alternating component which in turn means that tau a is zero. The principal mean and alternating components are now calculated by substituting the mean and alternating stress components into the equations that define principal stress. This gives values of 14.43 kpsi for sigma one super m and –4.43 for sigma 2 super m.



27. Prob. 2 Solution Page 2 The principal alternating stresses are simple to calculate since there is no alternating shear stress component. Hence sigma one super a is just 23 kpsi and sigma two super a is zero. Next the effective mean and alternating von Mises stresses are calculated using the formula that comes from distortion energy considerations. Substituting the values of sigma one super m and sigma two super m into the equation gives a value of 17.7 kpsi for the effective mean stress, sigma e super m.

28. Prob. 2 Solution Page 3 The effective alternating stresses are again easy to calculate since there is no alternating shear stress component. Consequently sigma e super a is just 23 kpsi. The Goodman diagram with all the appropriate stress parameters is depicted on the lower portion of the page. Again the ratios of the sides of similar triangles are used to establish an equation relating sigma e super m prime to sigma e super a prime that designates a failure point on the Goodman diagram assuming the ratio of sigma e super a to sigma e super m remains constant.



29. Prob. 2 Solution Page 4 The final step in the solution is to solve the equation for the failure condition from the previous slide for one of the unknown primed effective stresses. This is accomplished by first recognizing that holding constant the ratio of the applied fluctuating effective stress to the applied effective mean stress permits sigma e super a prime to be replaced by 1.35 times sigma e super m prime. The resulting equation is then solved for sigma e super m prime to give a value of 20.1 kpsi. The factor of safety is then given by the ratio of sigma e super m prime divided by sigma e super m. The final factor of safety is 1.17. Although this sample was somewhat simplified to shorten the required calculations the fatigue analysis process used to include the effect of a combined stress state has been demonstrated.

30. Cumulative Fatigue Damage If a reversed stress in excess of the endurance limit is applied for a finite number of cycles less than the fatigue life associated with that stress level the material will undergo fatigue damage. That is, the endurance limit for the material will be lowered as a consequence of the damage so that further application of another level of reversed stress will reduce its fatigue life from what it could be if applied initially. A method of analytically treating this problem called the Palmgren-Minor Summation Theory will now be discussed. Basically this theory is expressed analytically by the equation that the sum of the ratios of the number of cycles of applied load to the original fatigue life at that stress level is equal to a constant. That is n1 over cap N1 plus n2 over cap N2 plus additional like terms is equal to a constant C. Experimentally the value of this constant appears to vary from 0.7 to 2.2. For simplicity and since it will in general be conservative the constant is taken to be unity for fatigue design analysis purposes. The theory is then expressed as the sum of the ratios is simply equal to one.



31. Palmgren-Minor Theory Application The application of the Palmgren-Minor theory can be demonstrated graphically on the S-N diagram shown on this page. Both of the axes are log scales so that the fatigue life versus cycles of reversed load application can be represented as a straight line over the range of 103 to 106 cycles starting at 0.9 of the tensile strength, sigma u, and decreasing to the endurance limit of the original material, sigma e super 0, before it becomes a horizontal line. Now consider the application of a reversed stress of magnitude sigma 1 applied for n1 cycles as indicated. At this stress level the normal fatigue life would be cap N 1cycles as designated by the point sigma f super 0. The difference between cap N 1and n 1 is plotted as a point at the sigma 1 stress level as shown by the point sigma f super 1. To apply the Palmgren-Minor theory a straight line parallel to the original fatigue life line is drawn through the (cap N 1 –n 1) point to 106 cycles before it becomes horizontal. This is indicated by the dotted red curve. It is observed that the intersection of this line with a vertical from the 106 cycle point on the horizontal line is at a stress level less than the original endurance limit, sigma e super 0. The reduced endurance limit due to the fatigue damage of the material

sustained from the application of sigma one for n 1 cycles is designated sigma e super 1. Although it won’t be shown here the dotted red line can be used to derive the generic Palmgren-Minor summation equation equal to 1. Note that n 2 on the graph which is the reduced life at the original endurance limit due to the damage can be solved for directly with the equation that n 2 is equal to the 1 minus the ratio of n 1 to cap N 1 with the entire quantity multiplied by cap N 2. This will all be demonstrated in a numerical answer that follows immediately.



32. Sample Problem 3 To demonstrate the numerical application of the Palmgren-Minor theory consider the following example problem. A part with a tensile strength of 90 kpsi and an endurance limit of 40 kpsi is subjected to a reversed normal stress of 65 kpsi for 3000 cycles. For these conditions find: a. the remaining life of the part if the stress is maintained at 65 kpsi., b. the remaining life if the stress is reduced to 40 kpsi from 65 kpsi. and c. the endurance limit after being subjected to 65 kpsi for 3000 cycles.

33. Prob. – 3 Solution – page 1 To determine the remaining fatigue life at 65 kpsi it is first necessary to find the total fatigue life at this stress level. Again the ratio of the sides of similar triangles on the figure on the previous page will be used for this purpose. Click on the graphic button to pop up a figure illustrating these similar triangles. The first equation on this page in which cap N 1 is the only unknown is developed using the green and red triangles depicted. Keep in mind that the equation is written in terms of the logs of the parameters since the axes on the figure are to a log scale both horizontally and vertically. Substituting the parameters values into the equation and taking logs as indicated leads to a numerical value for the log of cap N 1. Taking the anti log of this number gives the fatigue life at 65 kpsi as 8729 cycles. Hence the remaining life after 3000 cycles at 65 kpsi is 5729 cycles. This is the solution to part a of the problem.

(Graphic on Page 220)



34. Prob. – 3 Solution – page 2 For part b the Palmgren-Minor summation equation can be used to solve for n 2 the number of cycles the damaged material will under go after the stress level is reduced to 40 kpsi which was the endurance limit of the undamaged material. This is simply a matter of substituting the appropriate values into the equation for n 2 given by one minus the ratio of n 1 to cap N1 with the entire quantity multiplied by cap N 2 which would be 106 cycles at the original endurance limit. Carrying out this calculation indicates that the fatigue life is reduced to a finite value of 656 thousand cycles from an initial fatigue life of 1 million cycles or more.

35. Prob. – 3 Solution – page 3 To determine the reduced endurance limit of the damaged material as requested in part c again the ratio of the sides of similar triangles will be used to obtain an equation in which sigma e super 1 is the only unknown. Click on the graphic button to uncover a pop up window that shows the triangles used to obtain this equation. Again note that both the vertical and horizontals axes are prescribed in log scale divisions. The green triangle is used for the left side of the equation and the red triangle for the right side. Once again the known parameters are inserted in the equation log are taken as required and the log of sigma e super 1 is determined to be 4.582. Taking the anti log gives a value of 38.2 kpsi for the endurance limit of the damaged material. Note that his represents a 4.5 %decrease from the original value of 40 kpsi.




36. Mason’s Modification. It should be recognized that the Palmgren-Minor theory in proposing a parallel line to the original fatigue life line to represent the reduced fatigue life of the damaged material introduces an error at 103 cycles since it also reduces the value of the ultimate strength used to establish the starting point of the fatigue life line. Mason has proposed a modification to correct this error. In this modification the reduced fatigue life line of the damaged material still passes through the point sigma f super 1 at cap N 1minus n 1 cycles as already calculated but the line starts at 0.9 sigma u as on the original fatigue life line. This is illustrated by the graphic on this page. The consequence of this is that the finite life n 2 at the original endurance limit stress will be shorter than that predicted by Palmgren-Minor and the endurance limit of the damaged material, sigma e super 1, will be smaller than that calculated on the previous slide.

37. Prob. – 3 Solution – page 4 The reduction in n 2 and sigma e super 1 will now be recalculated incorporating the Mason modification. An equation for n 2m is again developed using the ratio of the sides of two similar triangles. Click on the graphic button to see these triangles on the Mason modified diagram. The green triangle is used for the left side of the equation and the red triangle for the right side. The only unknown in this equation is n 2m the finite life at the original endurance limit. Substituting the known parameters and carrying out the indicated numerical manipulations gives a result for the log of n 2m equal to 5.42. This converts to a finite life of 263 thousand cycles or a reduction of almost 60 % from that predicted by the Palmgren-Minor theory. This large reduction is a consequence of the effect of the log scale.

(Graphic on 1)



38. Prob. – 3 Solution – page 5 To determine the reduced endurance limit as affected by the Mason modification the ratio of the sides of similar triangles are again used from the damaged S-N diagram. Click on the graphic button to bring up a diagram showing these triangles. The green triangle is used for the left side of the equation and the red triangle is used for the right side. The only unknown in this equation is the reduced endurance limit, sigma e super 1. Substituting the appropriate parameter values into the equation and solving for the log of sigma e super 1 gives a value of 4.53. Hence the value of the reduced endurance limit is 33,880 psi. This represents an 11 percent reduction in the value as predicted by the Palmgren-Minor method. The use of the Mason modification is considered to be a more accurate method of analyzing cumulative damage and is generally recommended and used in fatigue design considerations where appropriate.


39. Review Exercise In this review exercise the items in the list on the left are to be matched with the 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.



40. Off Line Exercise A vertical power screw used to raise and lower a load is supported by a single bearing as shown in the linked drawing. Below the bearing the shaft has a reduced diameter with a fillet radius at the transition. The applied axial load varies between 12,000 and 8,000 lbs in phase with a driving torque that fluctuates between 1000 in lbs and –500 in lbs. The material has a tensile strength of 150 kpsi. At the fillet the stress concentration for the axial load is 1.23 and for the torsional load is 1.35. Estimate the factor of safety for this design. When you have finished with this problem statement click on the exit button or main menu button to leave the chapter.

(Linked Drawing on Page 220) (Solution in Appendix)


_____________________________________________________________________________________Fluctuating Load Analysis - 213 - C.F. Zorowski 2002

Chapter 9 Fluctuating Load Analysis

Problem Solutions

Screen Titles

Problem 1 Solution Problem 2 Solution Problem 2 Solution (cont.) Problem 3 Solution Problem 3 Solution (cont.) Problem 3 Solution (cont.) Problem 3 Solution (cont.)



1. Problem 1 - Solution The equation of a straight line can be expressed as x over a constant "a" plus y over a constant "b" is equal to one. In this case the variable x is sigma m and y is sigma a. To determine the constants "a" and "b" use the two conditions that x or sigma m is equal to sigma u, the tensile strength, when y or sigma a is zero and y or sigma a is equal to sigma e, the endurance limit, when x or sigma m is zero. This gives "a" equal to sigma u and "b" equal to sigma e. The final equation becomes the ratio of sigma m to sigma u plus the ratio of sigma a to sigma e equal to one.

2. Problem 2 - Solution Begin with the Goodman equation for failure and the Gerber equation including a factor of safety. Since n as a function of the ratio of sigma m, the mean stress, to sigma u, the tensile strength, is desired it is necessary to eliminate sigma a over sigma e between the two equations. This is done by solving the Goodman equation for sigma a over sigma e and substituting into the Gerber equation which contains the factor of safety.



3. Problem 2- Solution (cont.) The substitution discussed on the previous page results in a quadratic equation in n with coefficients that are dependent on the ratio of sigma m to sigma u. The table gives values of n for values of sigma m over sigma u from zero to one at increments of 0.2. It is observed that using the Gerber equation provides at most a factor of safety of 20 % over the failure prediction by the Goodman diagram.

4. Problem 3 – Solution At the transition point the maximum axial force is 1300 lbs while the minimum axial force is -300 lbs due to the two axial components. Thus the mean value of fluctuating force is the average of these two values or 800 lbs. and the alternating component is 500 lbs. The area of the cross section acted on by the forces is calculated using the larger diameter of 3/16 inch. This gives an area of .026 square inches. Then the mean and alternating normal stress are calculated to be 30,769 psi and 19,230 psi respectively.



5. Problem 3 – Solution (cont.) The correction factors for the part endurance limit are now determined. With a ground finish ka is calculated to be .902. The size factor Kb for a 3/16-diameter rod is given by 1.05 and is taken to be 1 to be conservative. The load factor kc for an axial load is 0.923 since sigma u is less than 220 kpsi. The temperature factor kd is unity and the ke factor due to the stress concentration factor of 1.15 is .87. Multiplying these factors together times the uncorrected endurance limit gives a part endurance limit of 47.1 kpsi.

6. Problem 3 – Solution (cont.) Shown on this page is the Goodman diagram for this problem with the relevant parameters already calculated. To determine the largest possible alternating component of stress, sigma a prime, with sigma m held constant an equation can be developed from the geometry of the diagram making use of the vertical and horizontal sides of similar triangles. Thus the left side of the equation involving sigma a prime is obtained from the small green triangle. The left side of the equation comes from the large red triangle. This equation can then be solved for sigma a prime since all the other parameters are known.



7. Problem 3 – Solution (cont.) By substituting the known values from the diagram into the equation for sigma a prime gives a result for this unknown quantity of 33.33 kpsi. This value of sigma a prime, which is the maximum value the alternating stress can achieve, is 1.73 times the actual value of the alternating stress. Then F a prime becomes 865 lbs. resulting in an F max of 1665 lbs and an F min of – 65 lbs. Finally, with F1 held constant at 500 lbs. F 2 prime can vary from +1165 lbs. to – 565 lbs. before failure will occur.



Figure for Screen 13





Graphic for Screen 33

chapter 9 - design for strength and endurance | carl zorowski · design for strength and endurance...

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