Question 10.SP.4: 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. Since 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. Since the diameter of the rod is not known, L/r must be assumed. Assume that L/r > 52.7 and use Eq. (10.48b). For the centric load P, σ = P/A and write
\frac{P}{A}=\sigma_{\text {all }}=\frac{356 \times 10^3 MPa }{(L / r)^2} (10.48b)
\begin{gathered} \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{gathered}
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>52.7
The 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. (10.48b) and following the procedure used in part a, c = 11.79 mm and L/r = 50.9.
Since 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. (10.47b) for the design of this rod.
\frac{P}{A}=\sigma_{\text {all }}=\left[273.6-3.205\left(\frac{L}{r}\right)+0.00836\left(\frac{L}{r}\right)^2\right] MPa (10.47b)
\begin{aligned} \frac{60 \times 10^3 N }{\pi c^2} & =\left[273.6-3.205\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}=50.2
The 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 = 23.9 mm
