Question 16.SP.3: Using the aluminum alloy 2014-T6 for the circular rod shown,...
Using the aluminum alloy 2014-T6 for the circular rod shown, determine the smallest diameter that can be used to support the centric load P = 60 kN if (a) L = 750 mm, (b) L = 300 mm.
STRATEGY: Use the aluminum allowable stress equations to design the column, i.e., to determine the smallest diameter that can be used. Because there are two design equations based on L/r, it is first necessary to assume which governs. Then check the assumption.
MODELING: For the cross section of the solid circular rod shown in Fig. 1,
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}ANALYSIS:
a. Length of 750 mm. Because the diameter of the rod is not known, L/r must be assumed. Assume that L/r > 52.7 and use Eq. (16.32). For the centric load P, σ = P /A and write
L / r \geq 52.7: \quad \sigma_{\text {all }}=\frac{51,400 \text{ ksi}}{(L / r)^2} \quad \sigma_{\text {all }}=\frac{356 \times 10^3 MPa}{(L / r)^2} (16.32a,b)
\begin{aligned}&\frac{P}{A}=\sigma_{\text {all }}=\frac{356 \times 10^3 MPa}{(L / r)^2} \\&\frac{60 \times 10^3 N}{\pi c^2}=\frac{356 \times 10^9 Pa}{\left(\frac{0.750 m}{c / 2}\right)^2} \\&c^4=120.7 \times 10^{-9} m^4 \quad c=18.64 mm\end{aligned}For c = 18.64 mm, the slenderness ratio is
\frac{L}{r}=\frac{L}{c / 2}=\frac{750 mm}{(18.64 mm) / 2}=80.5 > 52.7The assumption that L/r is greater than 52.7 is correct. For L = 750 mm, the required diameter is
d = 2c = 2(18.64 mm)
d = 37.3 mm
b. Length of 300 mm. Assume that L/r > 52.7. Using Eq. (16.32b) and following the procedure used in part a, c = 11.79 mm and L/r = 50.9. Because L/r is less than 52.7, this assumption is wrong. Now assume that L/r is between 17.0 and 52.7 and use Eq. (16.31b) for the design of this rod.
\sigma_{\text {all }}=\left[273.6-3.205(L / r)+0.00836(L / r)^2\right] MPa (16.31b)
\begin{aligned}\frac{P}{A}=\sigma_{\text {all }} &=\left[273.6-30.205\left(\frac{L}{r}\right)+0.00836\left(\frac{L}{r}\right)^2\right] MPa \\\frac{60 \times 10^3 N}{\pi c^2} &=\left[237.6-3.205\left(\frac{0.3 m}{c / 2}\right)+0.00836\left(\frac{0.3 m}{c / 2}\right)^2\right] 10^6 Pa \\c &=11.95 mm\end{aligned}For c = 11.95 mm, the slenderness ratio is
\frac{L}{r}=\frac{L}{c / 2}=\frac{300 mm}{(11.95 mm) / 2}=50.2The second assumption that L/r is between 17.0 and 52.7 is correct. For L = 300 mm, the required diameter is
d = 2c = 2(11.95 mm)
d = 23 mm
