Question 2.6.17: Estimate the Marin loading factor kcfor 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 Sut=100LN(1,0.035)S{ut}= 100LN \left(1,0.035\right) kpsi, and the designer intends to use the correlation Sˊe=Φ0.30Sˉut\acute{S} _e=\Phi _{0.30}\bar{S} _{ut} to predict Sˊe\acute{S} _e .
(b) In bending, but endurance testing gave Sˊe=55LN(1,0.081)\acute{S} _e=55LN \left(1,0.081 \right) kpsi.
(c) In push-pull (axial) fatigue, Sut=LN(86.2,3.92)kpsiS_{ut}= LN\left(86.2,3.92\right) kpsi , and the designer intended to use the correlation Sˊe=Φ0.30Sˉut\acute{S} _e=\Phi _{0.30}\bar{S} _{ut}.
(d) In torsional fatigue. The material is cast iron, and  Sˊe\acute{S} _e is known by test.

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(a) Since the bar is in bending,

kc=(1,0)k_c=\left(1,0\right)

 

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

 

kc=(1,0)k_c=\left(1,0\right)

 

(c) From Eq. (6–73),

(kc)axial=1.23Sˉut0.0778LN(1,0.125)\left(k_c\right) _{axial} = 1.23\bar{S} ^{-0.0778}_{ut} LN\left(1,0.125\right)

 

(kc)axial=1.23(86.2)0.0778LN(1,0.125)\left(k_c\right) _{axial}=1.23\left(86.2^{}\right) ^{-0.0778}LN\left(1,0.125\right)

 

kˉc=1.23(86.2)0.0778(1)=0.870\bar{k} _c=1.23\left(86.2\right) ^{-0.0778}\left(1\right) =0.870

 

σ^kc=Ckc=0.125(0.870)=0.109\hat{\sigma }_{kc} =Ck_c=0.125\left(0.870\right) =0.109

 

(d) From Table 6–15,

Table 6–15 Heywood’s Parameter √a and coefficients ofvariation CKfC_{Kf} for steels
a(in)\sqrt{a} \left(\sqrt{in} \right) a(mm)\sqrt{a} \left(\sqrt{mm} \right)
Notch Type Sut in kpsiS_{ut} \ in \ kpsi Sut in  MPaS_{ut} \ in   MPa Coefficient of
Variation CKf
Transverse hole 5/Sut{5}/{S_{ut}} 174/Sut{174}/{S_{ut}} 0.10
Shoulder 4/Sut{4}/{S_{ut}} 139/Sut{139}/{S_{ut}} 0.11
Groove 3/Sut{3}/{S_{ut}} 104/Sut{104}/{S_{ut}} 0.15
kˉc=0.90 , σ^kc=0.07 , and Ckc=0.070.90=0.08\bar{k}_c=0.90 \ , \ \hat{\sigma }_{kc} =0.07 \ , \ and \ C_{kc}=\frac{0.07}{0.90} =0.08

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