Question 6.11: A circular log of timber has diameter d. Find the ratio of b...

Refer to Figure 6.26 wherein a rectangular beam section of breadth b and depth h has been cut from a circular log of diameter d. Now, the strongest section of beam against bending implies maximum value of the section modulus of the rectangular section:

Z=\frac{1}{6} b h^3                 (1)

where      d² = b² + h²            (2)

Therefore, from Eqs. (1) and (2), we can write

Z=\frac{1}{6} h^3 \sqrt{d^2-h^2}=\frac{1}{6} \sqrt{d^2 h^6-h^8}

Here, maximum z implies maximum value of d^2 h^6-h^8 .So,

\begin{aligned} & \frac{ d }{ d h}\left(d^2 h^6-h^8\right)=0 \\ & \Rightarrow 6 d^2 h^5-8 h^7=0 \\ & \Rightarrow h^2=\frac{6}{8} d^2=\frac{3}{4} d^2 \end{aligned}

Thus, h=\sqrt{3} d / 2 . From Eq. (2), we obtain by putting the value of h:

b^2=d^2-h^2=d^2-\frac{3}{4} d^2=\frac{d^2}{4}

which gives b = d/2. Therefore, for the strongest rectangular section cut out from the log, we get

b: h=1: \sqrt{3}