A steel bar is 150 mm square and has a hot-rolled finish. It will be used in a fully reversed bending application. Sut for the steel used in the bar is 600 MPa. (a) Using a 99.9% reliability factor, determine the endurance limit of the bar material. (b) How many cycles can it survive if alternating stress is 100 MPa

See explanation

Explanation:

GivenThe bar is square and has a hot-rolled finish. The loading is fully reversed bending.

Tensile Strength

Sut:600MPa

Maximum temperature

Tmax:500°C

Bar side dimension

b:150mm

Alternating stress

σa:100MPa

Reliability

R:0.999 Note 1.

AssumptionsInfinite life is required and is obtainable since this ductile steel will have an endurance limit. A reliability factor of 99.9% will be used.

SolutionSee Excel file Ex06-01.xls.

1Since 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:300MPa = 0.5 * Sut

2The loading is bending so the load factor from equation 6.7a is

Cload:1

3The 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

A95:1125mm2 = 0.05 * b * b Note 2.

dequiv:121.2mm = 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

4The 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

5The temperature factor is found from equation 6.7f :

Ctemp:0.710 = 1 - 0.0058 * (Tmax - 450)

6The reliability factor is taken from Table 6-4 for R = 0.999 and is

Creliab:0.753

7The corrected endurance limit Se can now be calculated from equation 6.6:

Se:69.94MPa = Cload * Csize * Csurf * Ctemp *

Creliab * Sprme

Let

Se:70MPa

8To 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:540MPa = 0.9 * Sut

9The 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

NSn (MPa)

1000540 =aa*B73^bb

2000440

4000358

8000292

16000238

32000194

64000158

128000129

256000105

51200085

100000070

FIGURE 6-34. S-N Diagram and Alternating Stress Line Showing Failure Point

10The 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,

Sn3:540MPa = aa * 1000^bb

At N = 106 cycles,

Sn6:70MPa = 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.