Question 16.S-P.3: Using the aluminum alloy 2014-T6, determine the smallest dia...
Using the aluminum alloy 2014-T6, determine the smallest diameter rod which can be used to support the centric load P = 60 kN if (a) L = 750 mm, (b) L = 300 mm.
For the cross section of a solid circular rod, we have
I = \frac{π}{4} c^{4} A = πc^{2} r = \sqrt{\frac{I}{A} }= \sqrt{\frac{πc^{4}/4}{πc^{2}}} = \frac{c}{2}
a. Length of 750 mm. Since the diameter of the rod is not known, a value of Lyr must be assumed; we assume that L/r > 55 and use Eq. (16.30). For the centric load P, we have σ = P/A and write
L/r ≥ 55: σ_{all} = \frac{54,000 ksi}{(L/r)^{2}} =\frac{372 × 10^{3} MPa}{(L/r)^{2}} (16.30)
\frac{P}{A} = σ_{all} =\frac{372 × 10^{3} MPa}{(L/r)^{2}}
\frac{60 × 10^{3} N}{πc^{2}} = \frac{372 × 10^{9} Pa}{\left( \frac{0.750 m}{c/2} \right)^{2}}
c^{4} = 115.5 × 10^{-9} m^{4} c = 18.44 mm
For c = 18.44 mm, the slenderness ratio is
\frac{L}{r} =\frac{L}{c/2} = \frac{750 mm}{(18.44 mm)/2} = 81.3 > 55
Our assumption is correct, and for L = 750 mm, the required diameter is
d = 2c = 2(18.44 mm) d = 36.9 mm
b. Length of 300 mm. We again assume that L/r > 55. Using Eq. (16.30), and following the procedure used in part a, we find that c = 11.66 mm and L/r = 51.5. Since L/r is less than 55, our assumption is wrong; we now assume that L/r < 55 and use Eq. (16.29^{′}) for the design of this rod.
L/r < 55: σ_{all} = [212 – 1.585(L/r) ] MPa (16.29^{′})
\frac{P}{A} = σ_{all} = \left[ 212 – 1.585 \left(\frac{L}{r}\right) \right] MPa
\frac{60 × 10^{3} N}{πc^{2}} = \left[ 212 – 1.585 \left( \frac{ 0.3 m}{c/2} \right) \right] 10^{6} Pa
c = 12.00 mm
For c = 12.00 mm, the slenderness ratio is
\frac{L}{r} = \frac{L}{c/2} = \frac{300 mm}{(12.00 mm)/2}= 50
Our second assumption that L/r < 55 is correct. For L = 300 mm, the required diameter is
d = 2c = 2(12.00 mm) d = 24.0 mm
