Answer:
See explanation
Explanation:
 
Given	The bar is square and has a hot-rolled finish. The loading is fully reversed bending. 
 
 Tensile Strength 
 Sut:	600	MPa 
 Maximum temperature 
 Tmax:	500	°C 
 Bar side dimension 
 b:	150	mm 
 Alternating stress 
 σa:	100	MPa 
 Reliability 
 R:	0.999 Note 1.	
 
Assumptions	Infinite life is required and is obtainable since this ductile steel will have an endurance limit. A reliability factor of 99.9% will be used. 
 
Solution	See Excel file Ex06-01.xls. 
 
1	Since no endurance-limit or fatigue strength information is given, we will estimate S'e based on the ultimate tensile strength using equation 6.5a. 
 S'e:	300	MPa = 0.5 * Sut 
 
2	The loading is bending so the load factor from equation 6.7a is 
 
 Cload:	1 
 
3	The part size is greater than the test specimen and the part is not round, so an equivalent diameter based on its 95% stressed area must be determined and used to find the size factor. For a rectangular section in nonrotating bending, the A95 area is defined in Figure 6-25c and the equivalent diameter is found from equation 6.7d 
 
 A95:	1125	mm2 = 0.05 * b * b Note 2.	
 dequiv:	121.2	mm = SQRT(A95val / 0.0766) 
 
 and the size factor is found for this equivalent diameter from equation 6.7b, to be 
 
 Csize:	0.747 = 1.189 * dequiv^-0.097 
 
4	The surface factor is found from equation 6.7e and the data in Table 6-3 for the specified hot-rolled finish. 
 Table 6-3 constants 
 A:	57.7 
 b:	-0.718 Note 3.	
 
 Csurf:	0.584 = Acoeff * Sut^bCoeff 
 
5	The temperature factor is found from equation 6.7f : 
 
 Ctemp:	0.710 = 1 - 0.0058 * (Tmax - 450) 
 
6	The reliability factor is taken from Table 6-4 for R = 0.999 and is 
 
 Creliab:	0.753 
 
7	The corrected endurance limit Se can now be calculated from equation 6.6: 
 
 Se:	69.94	MPa = Cload * Csize * Csurf * Ctemp * 
 Creliab * Sprme 
 Let 
 Se:	70	MPa 
 
8	To create the S-N diagram, we also need a value for the estimated strength Sm at 103 cycles based on equation 6.9 for bending loading. 
 
 Sm:	540	MPa = 0.9 * Sut 
 
9	The estimated S-N diagram is shown in Figure 6-34 with the above values of Sm and Se. The expressions of the two lines are found from equations 6.10a through 6.10c assuming that Se begins at 106 cycles. 
 
 b:	-0.2958 Note 4.	
 a:	4165.7 
 
 Plotting Sn as a function of N from equation 6.10a 
 
 N	Sn (MPa) 
 1000	540 =aa*B73^bb 
 2000	440 
 4000	358 
 8000	292 
 16000	238 
 32000	194 
 64000	158 
 128000	129 
 256000	105 
 512000	85 
 1000000	70 
 
 
 
 
 
 
 FIGURE 6-34. S-N Diagram and Alternating Stress Line Showing Failure Point 
 
10	The number of cycles of life for any alternating stress level can now be found from equation 6.10a by replacing σa for Sn. 
 At N = 103 cycles, 
 Sn3:	540	MPa = aa * 1000^bb 
 
 At N = 106 cycles, 
 Sn6:	70	MPa = aa * 1000000^bb 
 
 The figure above shows the intersection of the alternating stress line (σa = 100 MPa) with the failure line at N = 3.0 x 105 cycles.