A strap to be made from a cold-drawn steel strip workpiece is to carry a fully reversed axial load $F = LN \left(1000,120\right)$ 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,athen make a decision as to the workpiece thickness t.

Question

A strap to be made from a cold-drawn steel strip workpiece is to carry a fully reversed axial load [latex]F = LN \left(1000,120\right)[/latex] 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,athen make a decision as to the workpiece thickness t.

Accepted Answer

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

A Priori Decision	Consequence
Use 1018 CD steel Function:	[latex]S_{ut} = 87.6 kpsi, C_{Sut} = 0.0655[/latex]
Carry axial load	[latex]C_F= 0.12 \ , \ C_{kc}= 0.125[/latex]
R ≥ 0.999 95	z = -3.891
Machined surfaces	[latex]C_{ka}= 0.058[/latex]
Hole critical	[latex]C_{Kf}= 0.10 \ , C_{\acute{\sigma_a } }=\left(0.10^2+0.12^2 ight) ^{{1}/{2}}=0.156[/latex]
Ambient temperature	[latex]C_{kd}= 0[/latex]
Correlation method	[latex]C_{\acute{S} _{e}}=0.138[/latex]
Hole drilled	[latex]C_{Se}= \left(0.0582+ 0.1252+ 0.1382 ight)^{{1}/{2}}= 0.195[/latex]
	[latex]C_n=\sqrt{\frac{C^2_{Se}+C^2_{\acute{\sigma}_a}}{1+C^2_{\acute{\sigma}_a}} } =\sqrt{\frac{0.195^2+0.156^2}{1+0.156^2} } =0.2467[/latex]
	[latex]\begin{aligned}\bar{n} &=\exp \left[-(-3.891) \sqrt{\ln \left(1+0.2467^{2} ight)}+\ln \sqrt{1+0.2467^{2}} ight] \ &=2.65\end{aligned}[/latex]

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

[latex]\acute{\sigma } _a=\frac{\bar{S_e} }{\bar{n} } =\bar{K_f} \frac{\bar{F} }{\left(w-d ight)t } [/latex]

from which

[latex]t= \frac{\bar{K_f}\bar{n}\bar{F} }{\left(w-d ight)\bar{S_e} } [/latex] (1)

To evaluate the preceding equation we need [latex]\bar{S_e} \ and \ \bar{K_f} [/latex].

The Marin factors are

[latex]k_a=2.67\bar{S}^{-0.265}_{ut} LN\left(1,0.058 ight) =2.67\left(87.6 ight) ^{-0.265}LN\left(1,0.058 ight)\\bar{k} _a=0.816\k_b=1\k_c=1.23\bar{S}^{-0.078}_{ut}LN\left(1,0.125 ight) =0.868LN\left(1,0.125 ight) \\bar{k} _c=0.868\\bar{k} _d=\bar{k}_f =1[/latex]

and the endurance strength is

[latex]\bar{S}_e=0.816\left(1 ight) \left(0.868 ight) \left(1 ight)0.506 \left(87.6 ight) =31.4[/latex]

The hole governs. From Table A–15–1

Table A–15 Charts of Theoretical Stress-Concentration Factors K*

Figure A–15–1 Bar in tension or simple compression with a transverse hole.
[latex]\sigma_0= {F}/{A} \ , \ where \ A = \left(w − d ight) \ t [/latex] and t is the thickness.

we find [latex]{d}/{w}=0.50 \ , \ therefore \ K_t=2.15[/latex].

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} ight) [/latex]	[latex]\sqrt{a} \left(\sqrt{mm} ight) [/latex]
Notch Type	[latex]S_{ut} \ in \ kpsi[/latex]	[latex]S_{ut} in MPa[/latex]	Coefficient of Variation [latex]C_{Kf}[/latex]
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]\sqrt{a} ={5}/{\bar{S}_{ut} }={5}/{87.6}=0.0571 \ , \ r=0.1875[/latex] in.

From Eq. (6–78)

[latex]\bar{K}_f=\frac{K_t}{1+\frac{2\left(K_t-1 ight) }{K_t}\frac{\sqrt{a} }{\sqrt{r} } } [/latex]

the fatigue stress-concentration factor is

[latex]\bar{K}_f=\frac{2.18}{1+\frac{2\left(2.18-1 ight) }{2.18}\frac{{0.0571} }{\sqrt{0.1875} } } =1.91[/latex]

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

[latex]t \ge\frac{\bar{K}_f\bar{n}\bar{F}}{\left(w-d ight)S_e } =\frac{1.91\left(2.65 ight)1000 }{0.75-0.375}31400 =0.430[/latex] in

Use [latex]\frac{1}{2} [/latex] in-thick strap for the workpiece.

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

Question 2.6.21: A strap to be made from a cold-drawn steel strip workpiece i...

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