Question 10.SP.4: Using the aluminum alloy 2014-T6 for the circular rod shown,......

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.

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