Review Problem $6($Stress Concentration Factor$)$ Solution

Objective Compute the maximum tensile stress.

Given F $=12500lb$;$D=1\mathrm{.}50in$;$d=0\mathrm{.}75in$;$r=0\mathrm{.}060in$

Analysis Because of the change in diameter, use Equation $(3-10).$

Use the chart in Appendix A$-18-2$ to find the value of Kt using r$/$d and D$/$d as parameters.

Results

${}_{\text{m}\text{a}\text{x}}=K_t{}_{\text{n}\text{o}\text{m}}$

${}_{\text{n}\text{o}\text{m}}={}_2=\frac{F}{A_2}=\frac{F}{\frac{d^2}{4}}=\frac{12500lb}{\frac{(0\mathrm{.}75in{)}^2}{4}}$

${}_{\text{n}\text{o}\text{m}}=28,294l\frac{b}{i}{n}^2$

$\frac{r}{d}=\frac{0\mathrm{.}06}{0\mathrm{.}75}=0\mathrm{.}080$ and $\frac{D}{d}=\frac{1\mathrm{.}50}{0\mathrm{.}75}=2\mathrm{.}00$

Read $K_t=2\mathrm{.}12$ from Appendix A$-18-2.$

Then ${}_{\text{m}\text{a}\text{x}}=K_t{}_{\text{n}\text{o}\text{m}}=2\mathrm{.}12(28294)=59,983.$ What is the Maximum Tensile Stress in $($psi$)$ due to these new parameters?

Note:

Give your answer rounded to $3$ significant figures i$.$e$.43452\mathrm{.}23=43500$ or $43442\mathrm{.}23=43400$

Do NOT include any units or other text. The size of your numeric value should be given in pounds per square inch $($psi$).$ i$.$e$.$ if your answer is $43,400($psi$)$ then enter

this as $"43400"($Do NOT enter $"43400$ psi"$)$

The stepped bar shown in the above figure is subjected to an axial tensile force of $12500$ lb$.$ Compute the maximum tensile stress in the bar for the following dimensions:

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Solution

Objective Compute the maximum tensile stress.

Given F $=12500$ lb; D $=1\mathrm{.}50$ in; d $=0\mathrm{.}75$ in; r $=0\mathrm{.}060$ in

Analysis Because of the change in diameter, use Equation $(310).$

Use the chart in Appendix A$182$ to find the value of Kt using r$/$d and D$/$d as parameters.

Results

$\backslash$sigma max $=$ Kt $\backslash$sigma nom

$\backslash$sigma nom $=\backslash$sigma $2=$ F$/$A$2=$ F$/(\backslash$pi d$2/4)=(12500$ lb$)/[\backslash$pi $(0\mathrm{.}75$ in$)2/4]$

$\backslash$sigma nom $=28,294$ lb$/$in$2$

r$/$d $=0\mathrm{.}06/0\mathrm{.}75=0\mathrm{.}080$ and D$/$d $=1\mathrm{.}50/0\mathrm{.}75=2\mathrm{.}00$

Read Kt $=2\mathrm{.}12$ from Appendix A$182.$

Then $\backslash$sigma max $=$ Kt$\backslash$sigma nom $=2\mathrm{.}12(28294$psi$)=59,983$ psi.