Estimate the Marin loading factor kcfor a 1–in-diameter bar that is used as follows.(a) In bending. It is made of steel with Sut= 100LN(1,0.035) kpsi, and the designer intends to use the correlation S?e= ?0.30¯Sutto predict S?e.(b) In bending, but endurance testing gave S?e= 55LN(1,0.081) kpsi.

Question

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

Accepted Answer

(a) Since the bar is in bending,

[latex]k_c=\left(1,0\right) [/latex]

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

[latex]k_c=\left(1,0\right) [/latex]

(c) From Eq. (6–73),

[latex]\left(k_c\right) _{axial} = 1.23\bar{S} ^{-0.0778}_{ut} LN\left(1,0.125\right) [/latex]

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

[latex]\bar{k} _c=1.23\left(86.2\right) ^{-0.0778}\left(1\right) =0.870[/latex]

[latex]\hat{\sigma }_{kc} =Ck_c=0.125\left(0.870\right) =0.109[/latex]

(d) From Table 6–15,

Table 6–15 Heywood’s Parameter √a and coefficients ofvariation [latex]C_{Kf}[/latex] for steels
	[latex]\sqrt{a} \left(\sqrt{in} \right) [/latex]	[latex]\sqrt{a} \left(\sqrt{mm} \right) [/latex]
Notch Type	[latex]S_{ut} \ in \ kpsi[/latex]	[latex]S_{ut} \ in MPa[/latex]	Coefficient of Variation CKf
Transverse hole	[latex]{5}/{S_{ut}}[/latex]	[latex]{174}/{S_{ut}}[/latex]	0.10
Shoulder	[latex]{4}/{S_{ut}}[/latex]	[latex]{139}/{S_{ut}}[/latex]	0.11
Groove	[latex]{3}/{S_{ut}}[/latex]	[latex]{104}/{S_{ut}}[/latex]	0.15

[latex]\bar{k}_c=0.90 \ , \ \hat{\sigma }_{kc} =0.07 \ , \ and \ C_{kc}=\frac{0.07}{0.90} =0.08[/latex]

Table 6–15 Heywood’s Parameter √a and coefficients ofvariation $C_{Kf}$ for steels
	$\sqrt{a} \left(\sqrt{in} \right)$	$\sqrt{a} \left(\sqrt{mm} \right)$
Notch Type	$S_{ut} \ in \ kpsi$	$S_{ut} \ in MPa$	Coefficient of Variation CKf
Transverse hole	${5}/{S_{ut}}$	${174}/{S_{ut}}$	0.10
Shoulder	${4}/{S_{ut}}$	${139}/{S_{ut}}$	0.11
Groove	${3}/{S_{ut}}$	${104}/{S_{ut}}$	0.15

Question 2.6.17: Estimate the Marin loading factor kcfor a 1–in-diameter bar ...

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