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.
Let us take each a priori decision and note the consequence:
| A Priori Decision | Consequence |
| Use 1018 CD steel Function: | S_{ut} = 87.6 kpsi, C_{Sut} = 0.0655 |
| Carry axial load | C_F= 0.12 \ , \ C_{kc}= 0.125 |
| R ≥ 0.999 95 | z = -3.891 |
| Machined surfaces | C_{ka}= 0.058 |
| Hole critical | C_{Kf}= 0.10 \ , C_{\acute{\sigma_a } }=\left(0.10^2+0.12^2\right) ^{{1}/{2}}=0.156 |
| Ambient temperature | C_{kd}= 0 |
| Correlation method | C_{\acute{S} _{e}}=0.138 |
| Hole drilled | C_{Se}= \left(0.0582+ 0.1252+ 0.1382\right)^{{1}/{2}}= 0.195 |
| 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 | |
| \begin{aligned}\bar{n} &=\exp \left[-(-3.891) \sqrt{\ln \left(1+0.2467^{2}\right)}+\ln \sqrt{1+0.2467^{2}}\right] \\<br /> &=2.65\end{aligned} |
These eight a priori decisions have quantified the mean design factor as\bar{n} = 2.65.
Proceeding deterministically hereafter we write
from which
t= \frac{\bar{K_f}\bar{n}\bar{F} }{\left(w-d\right)\bar{S_e} } (1)
To evaluate the preceding equation we need \bar{S_e} \ and \ \bar{K_f} .
The Marin factors are
k_a=2.67\bar{S}^{-0.265}_{ut} LN\left(1,0.058\right) =2.67\left(87.6\right) ^{-0.265}LN\left(1,0.058\right)\\\bar{k} _a=0.816\\k_b=1\\k_c=1.23\bar{S}^{-0.078}_{ut}LN\left(1,0.125\right) =0.868LN\left(1,0.125\right) \\\bar{k} _c=0.868\\\bar{k} _d=\bar{k}_f =1
and the endurance strength is
\bar{S}_e=0.816\left(1\right) \left(0.868\right) \left(1\right)0.506 \left(87.6\right) =31.4
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. \sigma_0= {F}/{A} \ , \ where \ A = \left(w − d\right) \ t and t is the thickness. |
 |
we find {d}/{w}=0.50 \ , \ therefore \ K_t=2.15.
From Table 6–15
| 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 C_{Kf} |
| 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 |
\sqrt{a} ={5}/{\bar{S}_{ut} }={5}/{87.6}=0.0571 \ , \ r=0.1875 in.
From Eq. (6–78)
\bar{K}_f=\frac{K_t}{1+\frac{2\left(K_t-1\right) }{K_t}\frac{\sqrt{a} }{\sqrt{r} } }the fatigue stress-concentration factor is
\bar{K}_f=\frac{2.18}{1+\frac{2\left(2.18-1\right) }{2.18}\frac{{0.0571} }{\sqrt{0.1875} } } =1.91The thickness t can now be determined from Eq. (1)
t \ge\frac{\bar{K}_f\bar{n}\bar{F}}{\left(w-d\right)S_e } =\frac{1.91\left(2.65\right)1000 }{0.75-0.375}31400 =0.430 in
Use \frac{1}{2} in-thick strap for the workpiece.
The \frac{1}{2} in thickness attains and, in the rounding
to available nominal size, exceeds the reliability goal.
