Question 16.3: Using the aluminum alloy 2014-T6 for the circular rod shown,...

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 \quad \quad A=\pi c^2 \quad \quad \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. (16.32). For the centric load P, σ = P/A and write

\frac{P}{A}=\sigma_{\text{all}}=\frac{356 \times 10^3 \text{ MPa}}{(L/r)^2} \\ \frac{60 \times 10^3 \text{ N}}{\pi c^2}=\frac{356 \times 10^9 \text{ Pa}}{\left(\frac{0.750 \text{ m}}{c/2} \right)^2 } \\ c^4=120.7 \times 10^{-9}\text{ m}^4 \quad \quad \quad c=18.64 \text{ mm}

For c = 18.64 mm, the slenderness ratio is

\frac{L}{r} =\frac{L}{c/2}=\frac{750 \text{ mm}}{(18.64 \text{ mm})/2}=80.5 \gt 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. (16.32) and following the procedure used in part a, c = 11.79 mm and L/r = 50.9.

L/r\ge 52.7: \ \ \ \ \ \ \ \ \sigma_{\text{all}}=\frac{51,400\text{ ksi}}{(L/r)^2} \ \ \ \ \ \ \ \ \sigma_{\text{all}}=\frac{356 \times 10^3\text{ MPa}}{(L/r)^2}                    (16.32a,b)

Since L/r is less than 52.7, this assumption is wrong. Now assume that L/r < 52.7 and use Eq. (16.31b) 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] \text{ MPa} \\ \frac{60 \times 10^3 \text{ N}}{\pi c^2}=\left[273.6-3.205\left(\frac{0.3m}{c/2}\right)+0.00836\left(\frac{0.3m}{c/2}\right)^2 \right] 10^6 \text{ Pa}

c = 11.95 mm

For c = 11.95 mm, the slenderness ratio is

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

The second assumption that L/r < 52.7 is correct. For L = 300 mm, the required diameter is

d = 2c = 2(11.95 mm)                           d = 23.9 mm