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

Estimate the Marin loading factor kck_{c} for a 1–in-diameter bar that is used as follows.
(a) In bending. It is made of steel with SutS_{ut} = 100LN(1, 0.035) kpsi, and the designer intends to use the correlation Se=Φ0.30SutS′_{e} =\Phi_{0.30} \overline{S}_{ut} to predict SeS′_{e} .
(b) In bending, but endurance testing gave SeS′_{e} = 55LN(1, 0.081) kpsi.
(c) In push-pull (axial) fatigue, SutS_{ut} = LN(86.2, 3.92) kpsi, and the designer intended to use the correlation Se=Φ0.30SutS′_{e} =\Phi_{0.30} \overline{S}_{ut} .
(d) In torsional fatigue. The material is cast iron, and SeS′_{e} is known by test.

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(a) Since the bar is in bending,

kc=(1,0)k_{c} = (1, 0)

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

kc=(1,0)k_{c} = (1, 0)

(c) From Eq. (6–73),

(kc)axial=1.23Sut0.0778LN(1,0.125) (k_{c})_{axial} = 1.23 \overline{S}^{−0.0778}_{ut} LN(1, 0.125)                       (6–73)

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

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

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

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

Ckc=0.070.90=0.08C_{kc} =\frac{0.07}{0.90} = 0.08

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

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

 

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