Repeat Prob. 2–28, except that the design situation is failure by excessive deflection, and it is
desired to minimize the weight.
Question 2.29: Repeat Prob. 2–28, except that the design situation is fail...
Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape it can be shown that I=C A^{2}, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h=c b, where c is a
constant. The moment of inertia is I=b h^{3} / 12, and the area is A=b h. Then I=h\left(b h^{2}\right) / 12 =c b\left(b h^{2}\right) / 12=(c / 12)(b h)^{2}=C A^{2}, where C=c / 12 (a constant)
Thus, Eq. (2-27) becomes
A=\left(\frac{k l^{3}}{3 C E}\right)^{1 / 2}and Eq. (2-29) becomes
m=A l \rho=\left(\frac{k}{3 C}\right)^{1 / 2} l^{5 / 2}\left(\frac{\rho}{E^{1 / 2}}\right)Thus, minimize f_{3}(M)=\frac{\rho}{E^{1 / 2}}, or maximize M=\frac{E^{1 / 2}}{\rho}. From Fig. (2-16)
From the list of materials given, aluminum alloys are clearly the best followed by steels and tungsten carbide. Polycarbonate polymer is not a good choice compared to the other candidate materials. .
Eq. (2-26)
I=\frac{\pi D^{4}}{64}=\frac{A^{2}}{4 \pi}Eq. (2-27)
A=\left(\frac{4 \pi k l^{3}}{3 E}\right)^{1 / 2}Eq. (2-29)
m=2 \sqrt{\frac{\pi}{3}}\left(k^{1 / 2}\right)\left(l^{5 / 2}\right)\left(\frac{\rho}{E^{1 / 2}}\right)
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