A shaft made of hot-rolled and normalized SAE 1045 steel is loaded in bending and has a diameter change, as in Fig. A.12(b) of Appendix A. The stress concentration factor for the fillet radius is kt = 3.00, and the member is repeatedly subjected to the history of net section nominal stress shown in

Question

A shaft made of hot-rolled and normalized SAE 1045 steel is loaded in bending and has a diameter change, as in Fig. A.12(b) of Appendix A. The stress concentration factor for the fillet radius is [latex]k_{t} = 3.00[/latex], and the member is repeatedly subjected to the history of net section nominal stress shown in Fig. E14.4(a).[latex]^{1}[/latex] How many times can this loading history be applied before fatigue cracking is expecte

[latex]^{1}[/latex] Note: The peak nominal stress in this history is close to the yield strength of the material. However, with reference to Fig. A.15(c), gross yielding will not occur, as [latex]M_{o}/M_{i} = 1.7[/latex] indicates that the fully plastic bending moment [latex]M_{o}[/latex] is 70% above the moment [latex]M_{i}[/latex] that corresponds to [latex]S = σ_{o}[/latex].

Accepted Answer

The constants for this material’s cyclic stress–strain and strain–life curves, Eqs. 14.1 and 14.4,

[latex]\varepsilon_a=\frac{\sigma_a}{E}+\left(\frac{\sigma_a}{H^{\prime}} ight)^{1 / n^{\prime}}[/latex] (14.1)

[latex]\varepsilon_a=\frac{\sigma_f^{\prime}}{E}\left(2 N_f ight)^b+\varepsilon_f^{\prime}\left(2 N_f ight)^c[/latex] (14.4)

are listed in Table 14.1. We will employ [latex]k_{t}[/latex] with Neuber’s rule and the cyclic stress–strain curve to estimate the stress–strain response for the given load history and, on this basis, make the life estimate.

To begin, the first peak and valley in the history are moved to the end, and the second peak is repeated at the end, so that the highest absolute value of S occurs first and last. This is shown on the left in Fig. E14.4(b), as is rainflow cycle counting of the history according to the procedure of Section 9.9.2. There are 50 cycles B-C, one cycle E-F, and the major cycle A-D. Since these cycles correspond to closed stress-strain hysteresis loops, the stress–strain response is as shown on the right in Fig. E14.4(b). The stress and strain values for each peak and valley are calculated in a manner similar to Ex. 13.5, with the details given in Table E14.4(a).

For peak A, the value of [latex]S_{A}[/latex] is substituted into Eq. 13.61,

[latex]S=\frac{1}{k_t} \sqrt{\sigma^2+\sigma E\left(\frac{\sigma}{H} ight)^{1 / n}}, \quad k_t S=\sqrt{\sigma^2+\sigma E\left(\frac{\sigma}{H} ight)^{1 / n}}[/latex] (13.61)

and the equation is solved iteratively for stress [latex]σ_{A}[/latex]. Then [latex]σ_{A}[/latex] is substituted into the cyclic stress–strain curve, Eq. 14.1, to calculate [latex]ε_{A}[/latex]. The specific forms used are

[latex]S_A=\psi \frac{1}{k_t} \sqrt{\sigma_A^2+\psi \sigma_A E\left(\frac{\psi \sigma_A}{H^{\prime}} ight)^{1 / n^{\prime}}}, \quad \varepsilon_A=\frac{\sigma_A}{E}+\psi\left(\frac{\psi \sigma_A}{H^{\prime}} ight)^{1 / n^{\prime}}[/latex]

where ψ allows either tensile or compressive initial loading to be handled. Note that the cyclic stress–strain curve is assumed to give the initial monotonic stress–strain path, ε = f (σ ).

Next, stress and strain ranges [latex]Δσ_{XY} ext{ and } Δε_{XY}[/latex] are calculated from nominal stress ranges [latex]ΔS_{XY}[/latex] corresponding to smooth hysteresis loop curves that follow Δε/2 = f (Δσ/2). In particular, this is done for the closed stress–strain hysteresis loops, B-C, E-F, and A-D, and also for ranges A-B and D-E that locate starting points for closed loops. Equations 13.61 and 14.1 are again employed, now for half-ranges as follows:

[latex]\Delta S_{X Y}=\left|S_Y-S_X ight|, \quad \frac{\Delta S_{X Y}}{2}=\frac{1}{k_t} \sqrt{\left(\frac{\Delta \sigma_{X Y}}{2} ight)^2+\frac{\Delta \sigma_{X Y} E}{2}\left(\frac{\Delta \sigma_{X Y}}{2 H^{\prime}} ight)^{1 / n^{\prime}}}[/latex]

[latex]\frac{\Delta \varepsilon_{X Y}}{2}=\frac{\Delta \sigma_{X Y}}{2 E}+\left(\frac{\Delta \sigma_{X Y}}{2 H^{\prime}} ight)^{1 / n^{\prime}}[/latex]

The value of the range [latex]ΔS_{XY}[/latex] is first calculated. Then [latex]ΔS_{XY} /2[/latex] is substituted into the second equation, allowing [latex]Δσ_{XY}/2[/latex] to be obtained from an iterative calculation, after which the result is substituted into the third equation to give [latex]Δε_{XY} /2[/latex]. In Table E14.4(a), the resulting full ranges [latex]ΔS_{XY} , \Delta σ_{XY} , ext{ and } \Delta ε_{XY}[/latex] are tabulated.

The stress–strain response calculations are completed by starting from the initial point A and computing the stress and strain at each subsequent point by adding or subtracting the appropriate ranges:

[latex]\sigma_Y=\sigma_X+\psi \Delta \sigma_{X Y}, \quad \varepsilon_Y=\varepsilon_X+\psi \Delta \varepsilon_{X Y}[/latex]

Here, ψ causes addition or subtraction, depending on whether S is increasing or decreasing. In these calculations, the memory effect is observed. For example, we locate point D relative to its origin point A by subtracting ranges A-D, with cycles B -C not affecting the calculation.

As stress and strain values are now available for each peak and valley in the load history, a life calculation based on the Palmgren–Miner rule can proceed in a straightforward manner. If the Morrow equation is used for the mean stress effect, values are needed for the strain amplitude and mean stress, [latex]ε_{a} ext{ and } σ_{m}[/latex], the quantity N*, and finally the life [latex]N_{f}[/latex] . This is done for each cycle, B-C, E-F, and A-D. Thus, Eqs. 14.20

[latex]\varepsilon_a=\frac{\sigma_f^{\prime}}{E}\left(2 N^* ight)^b+\varepsilon_f^{\prime}\left(2 N^* ight)^c[/latex] (14.20)

and 14.23

[latex]N_f=N_{m i}^*\left(1-\frac{\sigma_m}{\sigma_f^{\prime}} ight)^{-1 / b}[/latex] (14.23)

are needed and are applied as follows, with [latex]N_{m i}^*=N^*:[/latex]

[latex]\varepsilon_a=\frac{\Delta \varepsilon}{2}, \quad \sigma_m=\frac{\sigma_{\max }+\sigma_{\min }}{2}[/latex]

[latex]\varepsilon_a=\frac{\sigma_f^{\prime}}{E}\left(2 N^* ight)^b+\varepsilon_f^{\prime}\left(2 N^* ight)^c, \quad N_f=N_{m i}^*\left(1-\frac{\sigma_m}{\sigma_f^{\prime}} ight)^{-1 / b}[/latex]

Table E14.4(b) gives the results of these calculations. Each [latex]ε_{a}[/latex] is obtained from the corresponding Δε in Table E14.4(a), and each [latex]σ_{m}[/latex] from the two appropriate stress values in the same table. For example, for cycle B-C, the value of [latex]ε_{a}[/latex] is half of the Δε on the third line of Table E14.4(a)—that is, the line for point C with origin B. For calculating [latex]σ_{m}[/latex] for cycle B-C, our [latex]σ_{max}[/latex] is the σ value from the third (point C) line of Table E14.4(a), and [latex]σ_{min}[/latex] is the σ value from the second (point B) line. Note that an iterative calculation is needed to calculate N* from [latex]ε_{a}[/latex], and then [latex]N_{f}[/latex] as affected by [latex]σ_{m}[/latex] follows.

In Table E14.4(a), the [latex]N_{j}[/latex] and [latex]N_{f j}[/latex] values are employed to calculate cycle ratios [latex]N_{j} /N_{f j}[/latex] , and the sum of these is computed. Finally, the estimated number of repetitions to failure is obtained by substituting this sum into the Palmgren–Miner rule in the form of Eq. 9.34,

[latex]B_f\left[\sum \frac{N_j}{N_{f j}} ight]_{ ext {one rep. }}=1[/latex] (9.34)

with the result being

[latex]B_f=1 /\left[\sum \frac{N_j}{N_{f j}} ight]_{ ext {one rep. }}=1 / 1.362 imes 10^{-3}=734 ext { repetitions }[/latex]

Another option is to use the SWT equation. In this case, the [latex]ε_{a} ext{ and } σ_{max}[/latex] values for each cycle give the product [latex]σ_{max}ε_{a}[/latex], which is then substituted into Eq. 14.30

[latex]\sigma_{\max } \varepsilon_a=\frac{\left(\sigma_f^{\prime} ight)^2}{E}\left(2 N_f ight)^{2 b}+\sigma_f^{\prime} \varepsilon_f^{\prime}\left(2 N_f ight)^{b+c}[/latex] (14.30)

to obtain the [latex]N_{f}[/latex] value. Details are given in Table E14.4(c).

Question 14.4: A shaft made of hot-rolled and normalized SAE 1045 steel is ...

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