Determine the cumulative damage experienced by a ground circular rod, 38 mm diameter, subjected to the combination of the cycles of loading and varying levels of reversed, repeated bending stress shown in Table 5–5.

Question

The bar is made from SAE 4340, HB275 alloy steel with an ultimate strength of 1048 MPa and an endurance limit of 430 MPa. Use Figure 5–10 for the [latex]S–N[/latex] curve.

TABLE 5–5 Loading Pattern for Example Problem 5–5
Stress Level		Number of Cycles
MPa	ksi	[latex]n_i[/latex]
650	94.3	2000
600	87	2000
500	72.5	10000
350	50.8	25000
300	43.5	15000

Accepted Answer

Given: SAE 4340, HB 275 alloy steel rod. [latex]s_u[/latex] = 1048 MPa [latex]D[/latex] = 38 mm. Ground surface.
Endurance strength data ([latex]S-N[/latex]) shown in Figure 5–10 [latex]s_n[/latex] = 430 MPa.
Loading is reversed, repeated bending. Load history shown in Table 5–5.

Analysis: First adjust the [latex]S-N[/latex] data for actual conditions using methods of Section 5–6. Use Miner’s rule to estimate the portion of life used by the loading pattern.

Results:

For SAE 4340, HB 275, [latex]s_{u}=1048 \mathrm{MPa}[/latex]
From Figure 5-11, basic [latex]s_{n}=475 \mathrm{MPa}[/latex] for ground surface
Material factor, [latex]C_{m}=1.00[/latex] for wrought steel
Type-of-stress factor, [latex]C_{s t}=1.0[/latex] for reversed, rotating bending stress
Reliability factor, [latex]C_{R}=0.81[/latex] (Table 5-3) for [latex]R=0.99[/latex] (Design decision)
Size factor, [latex]C_{s}=0.84[/latex] (Figure 5-12 and Table 5-4 for [latex]D=38 \mathrm{~mm}[/latex] )
Estimated actual endurance limit, [latex]s_{n}^{\prime}-[/latex] Computed:
[latex]s_{n}^{\prime}=s_{n} C_{m} C_{s t} C_{R} C_{s}=(475 \mathrm{MPa})(1.0)(1.0)(0.81)(0.84)=323 \mathrm{MPa}[/latex]
This is the estimate for the actual endurance limit of the steel. In Figure 5-10, the endurance limit for the standard specimen is [latex]430 \mathrm{MPa}[/latex]. The ratio of the actual to the standard data is [latex]323 / 430=0.75[/latex]. We can now adjust the entire [latex]S-N[/latex] curve by this factor. The result is shown in Figure 5-23.

Now we can read the number of cycles of life, [latex]N_{i}[/latex], corresponding to each of the given loading levels from Table 5-5. The combined data for the number of applied load cycles, [latex]n_{j}[/latex], and the life cycles, [latex]N_{j}[/latex], are now used in Miner's rule, Equation 5-35, to determine the cumulative damage, [latex]D_{c}[/latex]. Results are shown in Table 5-6.

Comment: We can conclude from this number that approximately 57% of the life of the component has been accumulated by the given loading. For these data, the greatest damage occurs from the 650 MPa loading for 2000 cycles. An almost equal amount of damage is caused by the 500 MPa loading for 10 000 cycles. Note that the cycles of loading at 300 MPa contributed nothing to the damage because they are below the endurance limit of the steel.

Table 5–3 Approximate Reliability Factors, [latex]C_R[/latex]
Desired reliability	[latex]C_R[/latex]
0.5	1
0.9	0.91
0.99	0.81
0.999	0.75

TABLE 5–4 Size Factors
U.S. customary units
Size rangeFor D in inches
[latex]D \leq 0.30[/latex]	[latex]C_{s}=1.0[/latex]
[latex]0.30	[latex]C_{s}=(D / 0.3)^{-0.11}[/latex]
[latex]2.0	[latex]C_{s}=0.859-0.02125 D[/latex]
SI units
Size range	For D in mm
[latex]D \leq 7.62[/latex]	[latex]C_{s}=1.0[/latex]
[latex]7.62	[latex]C_{s}=(D / 7.62)^{-0.11}[/latex]
[latex]50	[latex]C_{s}=0.859-0.000837 D[/latex]

TABLE 5–6 Cumulative Life Calculations for Example Problem 5–5
Stress Level		Number of Cycles	Life Cycles
MPa	ksi	[latex]n_i[/latex]	[latex]N_i[/latex]	[latex]n_i/N_i[/latex]
650	94.3	2000	[latex]1.1 imes 10^{4}[/latex]	0.182
600	87	3000	[latex]1.8 imes 10^{4}[/latex]	0.167
500	72.5	10000	[latex]5.8 imes 10^{4}[/latex]	0.172
350	50.8	25000	[latex]5.6 imes 10^{5}[/latex]	0.045
300	43.5	15000	[latex]\infty[/latex]	0
				Total 0.566

Question 5.5: Determine the cumulative damage experienced by a ground circ...

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