Question 10.SP.4: Using the aluminum alloy 2014-T6, determine the smallest dia...

For the cross section of a solid circular rod, we have

I=\frac{\pi}{4} c^{4} \quad A=\pi c^{2} \quad r=\sqrt{\frac{I}{A}}=\sqrt{\frac{\pi c^{4} / 4}{\pi c^{2}}}=\frac{c}{2}

a. Length of 750 \mathrm{~mm}. Since the diameter of the rod is not known, a value of L / r must be assumed; we assume that L / r>55 and use Eq. (10.46).

L / r \geq 55: \quad \sigma_{\text {all }}=\frac{55,400  \mathrm{ksi}}{(L / r)^2}=\frac{382 \times 10^3  \mathrm{MPa}}{(L / r)^2} \qquad (10.46)

For the centric load \mathbf{P}, we have \sigma=P / A and write

\begin{gathered} \frac{P}{A}=\sigma_{\text {all }}=\frac{382 \times 10^{3}  \mathrm{MPa}}{(L / r)^{2}} \\ \frac{60 \times 10^{3} \mathrm{~N}}{\pi c^{2}}=\frac{382 \times 10^{9} \mathrm{~Pa}}{\left(\frac{0.750 \mathrm{~m}}{c / 2}\right)^{2}} \\ c^{4}=112.5 \times 10^{-9} \mathrm{~m}^{4} \quad c=18.31 \mathrm{~mm} \end{gathered}

For c=18.44 \mathrm{~mm}, the slenderness ratio is

\frac{L}{r}=\frac{L}{c / 2}=\frac{750 \mathrm{~mm}}{(18.31 \mathrm{~mm}) / 2}=81.9>55

Our assumption is correct, and for L=750 \mathrm{~mm}, the required diameter is

d=2 c=2(18.31 \mathrm{~mm}) \qquad d=36.6 \mathrm{~mm}

b. Length of 300 \mathrm{~mm}. We again assume that L / r>55. Using Eq. (10.46), and following the procedure used in part a, we find that c= 11.58 \mathrm{~mm} and L / r=51.8. Since L / r is less than 55 , our assumption is wrong; we now assume that L / r<55 and use Eq. \left(10.45^{\prime}\right)

L / r \lt 55: \quad \sigma_{\text {all }}= [213 – 1.577(L/r)] MPa \qquad (10.45′)

for the design of this rod.

\begin{aligned} \frac{P}{A}=\sigma_{\text {all }} & =\left[213-1.577\left(\frac{L}{r}\right)\right]  \mathrm{MPa} \\ \frac{60 \times 10^{3} \mathrm{~N}}{\pi c^{2}} & =\left[213-1.577\left(\frac{0.3 \mathrm{~m}}{c / 2}\right)\right] 10^{6} \mathrm{~Pa} \\ c & =11.95 \mathrm{~mm} \end{aligned}

For c=11.95 \mathrm{~mm}, the slenderness ratio is

\frac{L}{r}=\frac{L}{c / 2}=\frac{300 \mathrm{~mm}}{(11.95 \mathrm{~mm}) / 2}=50.2

Our second assumption that L / r<55 is correct. For L=300 \mathrm{~mm}, the required diameter is

d=2 c=2(11.95 \mathrm{~mm}) \quad d=23.9 \mathrm{~mm}