Question 6.17: Estimate the Marin loading factor kc for a 1–in-diameter bar...

Estimate the Marin loading factor k_{c} for a 1–in-diameter bar that is used as follows.
(a) In bending. It is made of steel with S_{ut} = 100LN(1, 0.035) kpsi, and the designer intends to use the correlation S′_{e} =\Phi_{0.30} \overline{S}_{ut} to predict S′_{e} .
(b) In bending, but endurance testing gave S′_{e} = 55LN(1, 0.081) kpsi.
(c) In push-pull (axial) fatigue, S_{ut} = LN(86.2, 3.92) kpsi, and the designer intended to use the correlation S′_{e} =\Phi_{0.30} \overline{S}_{ut} .
(d) In torsional fatigue. The material is cast iron, and S′_{e} is known by test.

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(a) Since the bar is in bending,

k_{c} = (1, 0)

(b) Since the test is in bending and use is in bending,

k_{c} = (1, 0)

(c) From Eq. (6–73),

(k_{c})_{axial} = 1.23 \overline{S}^{−0.0778}_{ut} LN(1, 0.125)                       (6–73)

(k_{c})_{ax} = 1.23(86.2)^{−0.0778}LN(1, 0.125)

\overline {k}_{c} = 1.23(86.2)^{−0.0778}(1) = 0.870

\hat {σ}_{kc} = C \overline {k}_{c} = 0.125(0.870) = 0.109

(d) From Table 6–15, \overline {k}_{c}= 0.90, \hat {σ}_{kc} = 0.07,and

C_{kc} =\frac{0.07}{0.90} = 0.08

Table 6–15  Heywood’s Parameter \sqrt {a} and coefficients of variation C_{Kf} for steels

Coefficient of Variation C_{Kf} \sqrt{a}(\sqrt {mm}) ,S_{ut} in MPa \sqrt{a}(\sqrt {in}) ,S_{ut} in kpsi Notch Type
0.10 174/S_{ut} 5/S_{ut} Transverse hole
0.11 139/S_{ut} 4/S_{ut} Shoulder
0.15 104/S_{ut} 3/S_{ut} Groove

 

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