For a cantilever beam loaded in bending, prove that β= 1/2 for the E^β/ρ guidelines in Fig. 2–16.

Question

[latex] ext { For a cantilever beam loaded in bending, prove that } \beta=1 / 2 ext { for the } E^{\beta} / ho ext { guidelines in Fig. 2-16. }[/latex]

Accepted Answer

Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape it can be shown that [latex]I=C A^{2}[/latex], where [latex]C[/latex] is a constant. For example, consider a rectangular section of height [latex]h[/latex] and width [latex]b[/latex], where for a given scaled shape, [latex]h=c b[/latex], where [latex]c[/latex] is a constant. The moment of inertia is [latex]I=b h^{3} / 12[/latex], and the area is [latex]A=b h[/latex]. Then [latex]I=h\left(b h^{2}\right) / 12[/latex] [latex]=c b\left(b h^{2}\right) / 12=(c / 12)(b h)^{2}=C A^{2}[/latex], where [latex]C=c / 12[/latex]

Thus, Eq. [latex](2-27)[/latex] becomes

[latex]A=\left(\frac{k l^{3}}{3 C E}\right)^{1 / 2}[/latex]

and Eq. (2-29) becomes

[latex]m=A l \rho=\left(\frac{k}{3 C}\right)^{1 / 2} l^{5 / 2}\left(\frac{\rho}{E^{1 / 2}}\right)[/latex]

So, minimize [latex]f_{3}(M)=\frac{\rho}{E^{1 / 2}}[/latex], or maximize [latex]M=\frac{E^{1 / 2}}{\rho} .[/latex] Thus, [latex]\beta=1 / 2 .[/latex] .

2-26

[latex]I=\frac{\pi D^{4}}{64}=\frac{A^{2}}{4 \pi}[/latex]

2-27

[latex]A=\left(\frac{4 \pi k l^{3}}{3 E}\right)^{1 / 2}[/latex]

2-29

[latex]m=2 \sqrt{\frac{\pi}{3}}\left(k^{1 / 2}\right)\left(l^{5 / 2}\right)\left(\frac{\rho}{E^{1 / 2}}\right)[/latex]

Question 2.32: For a cantilever beam loaded in bending, prove that β= 1/2 f...

Related Answered Questions