For a cantilever beam loaded in bending, prove that β = 2/3 for the S^β/ρ guidelines in Fig. 2–19.

Question

For a cantilever beam loaded in bending, prove that [latex]\beta=2 / 3[/latex] for the [latex]S^{\beta} / \rho[/latex] guidelines in Fig. 2-19.

Accepted Answer

For strength,

[latex]\sigma=F l / Z=S[/latex] (1)

where [latex]F l[/latex] is the bending moment and [latex]Z[/latex] is the section modulus [see Eq. (3-26b), p. 90[latex]][/latex]. The section modulus is strictly a function of the dimensions of the cross section and has the units in [latex]^{3}[/latex] (ips) or [latex]\mathrm{m}^{3}[/latex] (SI). Thus, for a given cross section, [latex]Z=C(A)^{3 / 2}[/latex], where [latex]C[/latex] is a number. For example, for a circular cross section, [latex]C=(4 \sqrt{\pi})^{-1}[/latex]. Then, for strength, Eq. (1) is

[latex]\frac{F l}{C A^{3 / 2}}=S \quad \Rightarrow \quad A=\left(\frac{F l}{C S}\right)^{2 / 3}[/latex] (2)

For mass,

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

So, [latex]f_{3}(M)=\rho / S^{2 / 3}[/latex], and maximize [latex]S^{2 / 3} / \rho[/latex]. Thus, [latex]\beta=2 / 3[/latex]. .

Eq. (3-26b)

[latex]\sigma_{\max }=\frac{M}{Z}[/latex]

Question 2.33: For a cantilever beam loaded in bending, prove that β = 2/3 ...

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