A strap to be made from a cold-drawn steel strip workpiece is to carry a fully reversed axial load F = LN(1000, 120) lbf as shown in Fig. 6–39. Consideration of adjacent parts established the geometry as shown in the figure, except for the thickness t.

Question

A strap to be made from a cold-drawn steel strip workpiece is to carry a fully reversed axial load F = LN(1000, 120) lbf as shown in Fig. 6–39. Consideration of adjacent parts established the geometry as shown in the figure, except for the thickness t. Make a decision as to the magnitude of the design factor if the reliability goal is to be 0.999 95,
then make a decision as to the workpiece thickness t.

Accepted Answer

Let us take each a priori decision and note the consequence:

Consequence	A Priori Decision
[latex]\overline{S}_{ut} = 87.6 kpsi, C_{Sut} = 0.0655[/latex]	Use 1018 CD steel
Function:
[latex]C_{F}= 0.12, C_{kc}= 0.125[/latex]	Carry axial load
z =-3.891	R ≥ 0.999 95
[latex]C_{ka}= 0.058[/latex]	Machined surfaces
[latex]C_{Kf}= 0.10, C_{σ′_{a}}= (0.10^{2} 0.12^{2})^{1/2} = 0.156[/latex]	Hole critical
[latex]C_{kd}= 0[/latex]	Ambient temperature
[latex]C_{S′_{e}}=0.138[/latex]	Correlation method
[latex]C_{Se}= (0.058^{2} + 0.125^{2} + 0.138^{2})^{1/2} = 0.195[/latex]	Hole drilled
[latex]C_{n}=\sqrt{ \frac {C^{2}_{Se} +C^{2}_{σ′a}}{1 + C^{2}_{σ′a}} }=\sqrt {\frac {0.195^{2} + 0.156^{2}}{1 +0.156^{2}}}= 0.2467[/latex] [latex]\overline{n}= exp[ − (−3.891) \sqrt {ln(1 + 0.2467^{2})} +ln \sqrt {1 + 0.2467^{2}}] [/latex] =2.65

These eight a priori decisions have quantified the mean design factor as [latex]\overline{n}[/latex] = 2.65. Proceeding deterministically hereafter we write

[latex]σ′_{a} =\frac {\overline{S}_{e}}{\overline{n}} = \overline{K}_{f} \frac {\overline{F}}{(w − d)t}[/latex]

from which

[latex]t =\frac {\overline{K}_{f} \overline{n} \overline{F}}{(w − d) \overline{S}_{e}} [/latex] (1)

To evaluate the preceding equation we need [latex]\overline{S}_{e}[/latex] and [latex]\overline{K}_{f}[/latex] . The Marin factors are

[latex]k_{a} = 2.67 \overline{S}^{−0.265}_{ut} LN(1, 0.058) = 2.67(87.6)^{−0.265} LN(1, 0.058)[/latex]
[latex]\overline{K}_{a}= 0.816[/latex]
[latex]k_{b} = 1[/latex]
[latex]k_{c} = 1.23 \overline{S}^{−0.078}_{ut} LN(1, 0.125) = 0.868LN(1, 0.125)[/latex]
[latex]\overline {k}_{c} = 0.868[/latex]
[latex]\overline {k}_{d}= \overline {k}_{f} = 1[/latex]

and the endurance strength is

[latex]\overline{S}_{e}= 0.816(1)(0.868)(1)(1)0.506(87.6) = 31.4 kpsi[/latex]

The hole governs. From Table A–15–1 we find d/w = 0.50, therefore [latex]K_{t}[/latex] = 2.18. From Table 6–15 [latex]\sqrt {a} = 5/ \overline{S}_{ut}[/latex] = 5/87.6 = 0.0571, r = 0.1875 in. From Eq. (6–78) the fatigue stress concentration factor is

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

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

[latex]\overline{K}_{f} =\frac {K_{t}}{1 + \frac {2(K_{t} − 1)}{K_{t}} \frac {\sqrt{a}}{\sqrt{r}}}[/latex] (6–78)

[latex]\overline{K}_{f} =\frac {2.18}{1 + \frac {2(2.18 − 1)}{2.18} \frac {0.0571}{\sqrt{0.1875}}}[/latex]

The thickness t can now be determined from Eq. (1)

[latex]t ≥\frac {\overline{K}_{f} \overline{n} \overline{F}}{(w − d)S_{e}}=\frac {1.91(2.65)1000}{(0.75 − 0.375)31 400} =0.430 in[/latex]

Use [latex]\frac {1}{2}[/latex] -in-thick strap for the workpiece. The 1[latex]\frac {1}{2}[/latex] -in thickness attains and, in the rounding to available nominal size, exceeds the reliability goal.

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

Question 6.21: A strap to be made from a cold-drawn steel strip workpiece i...

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