In nature the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall withinthe base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{{Y\pi {r^4}}}{{4R}}$ $Y$ is the Young’s modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
In nature the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall withinthe base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{{Y\pi {r^4}}}{{4R}}$ $Y$ is the Young’s modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
Let us see the diagram of the given situation according to the problem the bending torque on the trunk of radius $r$ of tree $=\frac{\mathrm{Y} \pi r^{4}}{4 \mathrm{R}}$ where $\mathrm{R}$ is the radius of curvature of the bent surface,
when trunk is bend then, $\mathrm{Wd}=\frac{\mathrm{Y} \pi r^{4}}{4 \mathrm{R}}$
If $\mathrm{R}>>h$, then the centre of gravity of tree is at a height $l=\frac{1}{2} h$ from the ground.
From $\Delta \mathrm{ABCR}^{2}=(\mathrm{R}-d)^{2}+\left(\frac{1}{2} h\right)^{2}$
If $d<\mathrm{R}, \mathrm{R}^{2}=\mathrm{R}^{2}-2 \mathrm{R} d+\frac{1}{4} h^{2}$
$\therefore d=\frac{h^{2}}{8 \mathrm{R}}$
If $\omega_{0} \frac{\text { weight }}{\text { volume }}$, then
$\frac{\mathrm{Y} \pi r^{4}}{4 \mathrm{R}}=\omega_{0}\left(\pi r^{2} h\right) \frac{h^{2}}{8 \mathrm{R}}$
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