Figure 6–22a shows a rotating shaft simply supported in ball bearings at A and D and loaded by a nonrotating force F of 6.8 kN. Using ASTM “minimum” strengths, estimate the life of the part.

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Accepted Answer

From Fig. 6–22b we learn that failure will probably occur at B rather than at C or at the point of maximum moment. Point B has a smaller cross section, a higher bending moment, and a higher stress-concentration factor than C, and the location of maximum moment has a larger size and no stress-concentration factor.

We shall solve the problem by first estimating the strength at point B, since the strength will be different elsewhere, and comparing this strength with the stress at the same point.

From Table A–20 we find [latex]S_{u t}=690 MPa \text { and } S_{y}=580 MPa[/latex]. The endurance limit [latex]S_{e}^{\prime}[/latex] is estimated as

[latex]S_{e}^{\prime}=0.5(690)=345 MPa[/latex]

From Eq. (6–19) and Table 6–2,

[latex]k_{a}=a S_{u t}^{b}[/latex] (6–19)

Table 6–2 Parameters for Marin Surface Modification Factor, Eq. (6–19)
Surface Finish	Factor a		Exponent b
Surface Finish	[latex]S_{u t r} \text { kpsi }[/latex]	[latex]S_{\text {utr }} MPa[/latex]	Exponent b
Ground	1.34	1.58	−0.085
Machined or cold-drawn	2.7	4.51	−0.265
Hot-rolled	14.4	57.7	−0.718
As-forged	39.9	272	−0.995

From Eq. (6–20),

[latex]k_{b}=\left\{\begin{array}{ll}(d / 0.3)^{-0.107}=0.879 d^{-0.107} & 0.11 \leq d \leq 2 \text { in } \0.91 d^{-0.157} & 2

[latex]k_{b}=(32 / 7.62)^{-0.107}=0.858[/latex]

[latex]\text { Since } k_{c}=k_{d}=k_{e}=k_{f}=1[/latex]

[latex]S_{e}=0.798(0.858) 345=236 MPa[/latex]

To find the geometric stress-concentration factor [latex]K_{t}[/latex] we enter Fig. A–15–9 with D/d = 38/32 = 1.1875 and r/d = 3/32 = 0.093 75 and read [latex]K_{t} \doteq 1.65[/latex]. Substituting [latex] S_{u t}=690 / 6.89=100[/latex] kpsi into Eq. (6–35a) yields [latex]\sqrt{a}=0.0622 \sqrt{i n}=0.313 \sqrt{ mm }[/latex]. Substituting this into Eq. (6–33) gives

[latex]\text { Bending or axial: } \quad \sqrt{a}=0.246-3.08\left(10^{-3}\right) S_{u t}+1.51\left(10^{-5}\right) S_{u t}^{2}-2.67\left(10^{-8}\right) S_{u t}^{3}[/latex] (6–35a)

[latex]K_{f}=1+\frac{K_{t}-1}{1+\sqrt{a / r}}[/latex] (6–33)

[latex]K_{f}=1+\frac{K_{t}-1}{1+\sqrt{a / r}}=1+\frac{1.65-1}{1+0.313 / \sqrt{3}}=1.55[/latex]

The next step is to estimate the bending stress at point B. The bending moment

[latex]M_{B}=R_{1} x=\frac{225 F}{550} 250=\frac{225(6.8)}{550} 250=695.5 N \cdot m[/latex]

Just to the left of B the section modulus is [latex]I / c=\pi d^{3} / 32=\pi 32^{3} / 32=3.217\left(10^{3}\right) mm ^{3}[/latex]. The reversing bending stress is, assuming infinite life,

[latex]\sigma_{ rev }=K_{f} \frac{M_{B}}{I / c}=1.55 \frac{695.5}{3.217}(10)^{-6}=335.1\left(10^{6}\right) Pa =335.1 MPa[/latex]

This stress is greater than [latex]S_{e}[/latex] and less than [latex] S_{y}[/latex]. This means we have both finite life and no yielding on the first cycle.

For finite life, we will need to use Eq. (6–16). The ultimate strength, [latex]S_{u t}=690[/latex] MPa = 100 kpsi. From Fig. 6–18, f = 0.844. From Eq. (6–14)

[latex]N=\left(\frac{\sigma_{ rev }}{a}\right)^{1 / b}[/latex] (6–16)

[latex]a=\frac{\left(f S_{u t}\right)^{2}}{S_{e}}[/latex] (6–14)

[latex]a=\frac{\left(f S_{u t}\right)^{2}}{S_{e}}=\frac{[0.844(690)]^{2}}{236}=1437 MPa[/latex]

and from Eq. (6–15)

[latex]b=-\frac{1}{3} \log \left(\frac{f S_{u t}}{S_{e}}\right)[/latex] (6–15)

[latex]b=-\frac{1}{3} \log \left(\frac{f S_{u t}}{S_{e}}\right)=-\frac{1}{3} \log \left[\frac{0.844(690)}{236}\right]=-0.1308[/latex]

From Eq. (6–16),

[latex]N=\left(\frac{\sigma_{ rev }}{a}\right)^{1 / b}=\left(\frac{335.1}{1437}\right)^{-1 / 0.1308}=68\left(10^{3}\right) cycles[/latex]

Question 6.9: Figure 6–22a shows a rotating shaft simply supported in ball...

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