11th Class Physics Mechanical Properties of Solids / ठोसों के यांत्रिक गुण

  • question_answer 49)
                      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 \[\frac{Y\pi {{r}^{4}}}{4R}\,Y\] 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 natural surface). Estimate the critical height of a tree for a given radius of the trunk.                

    Answer:

                      When tree bends, torque acting on the tree due to its weight \[W=Wd\]             \[\therefore \]\[Wd=\frac{Y\pi {{r}^{4}}}{4R}\]                 From figure, \[{{R}^{2}}={{\left( \frac{h}{2} \right)}^{2}}+{{(R-d)}^{2}}\]                 \[=\frac{{{h}^{4}}}{4}+{{R}^{2}}+{{d}^{2}}-2Rd\]                 Since \[d<<R,\] so \[{{d}^{2}}\] can be neglected                 \[\therefore \]\[2Rd=\frac{{{h}^{2}}}{4}\]or \[d=\frac{{{h}^{2}}}{8R}\]                 Put this value in eqn. (i), we get                 \[\frac{W{{h}^{2}}}{8R}=\frac{Y\pi {{r}^{4}}}{4R}\] or \[h=\sqrt{\frac{2Y\pi {{r}^{3}}}{W}}\]                


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