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 within the base. The elastic torque caused because of this bending about the central axis of the tree is given by \frac{\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.

Answers (1)

Let us consider triangle ABC which is a right-angled triangle. We use Pythagoras theorem here.

R^{2}=(R-d)^{2}+\left (\frac{h}{2} \right )^{2}

2Rd=\frac{h^{2}}{4}

d=\frac{h^{2}}{8R} ------------(1)

we assume W to be the weight of trunk per unit volume.

Weight of trunk = total vol. x W = \pi r^{2}h\times W

Torque exerted by bending the trunk, T = \pi r^{2}h\times W \times d

Given value of torque = \frac{\pi r^{4}Y}{4R}

Equating both values we get,

\pi r^{2}h \times W \times \frac{h^{2}}{8R}=\frac{\pi r^{4}Y}{4R}

h^{3} = \pi r^{4}Y \times \frac{8R}{4R\pi r^{2}W}

h^{3} = [\frac{2Y}{W}]^{\frac{1}{3}}r^{2/3}

 

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