Question 1.5: A solid conical rod of diameter D at base is suspended from ...
A solid conical rod of diameter D at base is suspended from top as shown in Figure 1.20. Find out the deflection of its free end due to the self-weight of the cone. Consider specific gravity of the cone material as γ and its modulus of elasticity as E.
Let us take a thin section of thickness dx at a distance x from the free end as shown in Figure 1.20.
Diameter of this thin element may be considered to be uniform, ‘d ’. Thus, area of the element is (π /4)d². Now, self-weight acting on this element is
\frac{\pi d^2}{4} \times \frac{x}{3} \times \gamma
where \left\lgroup \frac{\pi}{4} d^2 \times \frac{x}{3} \right\rgroup denotes volume of the lower part.
Now, elongation of the thin element is PL/AE, where P is load, A is the cross-sectional area, L is the length of the element and E is the modulus of elasticity. Substituting the expressions, we get
\begin{aligned} \text { Elongation of the thin element } &=\frac{\left\lgroup\frac{\pi d^2}{4} \times \frac{x}{3} \times \gamma \right\rgroup \times d x}{\frac{\pi d^2}{4} \times E} \\ &=\frac{1}{3} \times \frac{\gamma x}{E} d x \end{aligned}
Therefore, \text { Total elongation }=\int_0^L \frac{1}{3} \times \frac{\gamma x}{E} d x=\frac{\gamma L^2}{6 E}
Therefore, deflection of the free end is γL²/6E .
Note: Observe that the deflection becomes independent of cone diameter, D.