A right solid circular cone standing on its base on a horizontal surface is of height H and base radius R. The cone is made of a material with specific weight w and elastic modulus E. The vertical deflection at the mid-height of the cone due to self-weight is given by

The correct option is wH²/8E.

Analysis of the Cone Diagram:
The provided diagram shows a right circular solid cone standing on its base on a horizontal surface:
1. The apex of the cone is labeled as point A, the midpoint is labeled as point B (at height H/2 from the base), and the base center is labeled as point C.
2. The total height of the cone is H and the base radius is R. Thus, the radius of the cross-section at the midpoint B is R/2.
3. An element of thickness dx is considered at a depth of x from the apex A. The cross-sectional area of this element is denoted as a.

Step-by-step Derivation:

1. Cross-sectional Area and Volume:
At any distance x from the apex A, the radius of the cross-section is:

$r_x=\frac{Rx}{H}$

The area a of this cross-section is:

$a=\pi r_x^2=\pi {\left(\frac{Rx}{H}\right)}^2$

2. Self-Weight of the Shaded Region:
The axial force acting on the element of thickness dx at a distance x from the apex is due to the weight of the portion of the cone above it (shaded region in the diagram). The volume of this upper cone of height x and base area a is:

$V=\frac{1}{3}ax$

Given the specific weight of the material is w, the downward force (axial compressive force) is:

$P\left(x\right)=-w\left(\frac{1}{3}ax\right)=-\frac{wax}{3}$

3. Element Deflection:
The vertical deformation of the element of thickness dx under this compressive load is:

$d\delta =\frac{P\left(x\right)dx}{aE}=\frac{-\frac{wax}{3}dx}{aE}=-\frac{wx}{3E}dx$

4. Mid-height Deflection Determination:
To determine the total vertical deflection at the mid-height point B (where x = H/2) relative to the fixed base C (where x = H), we integrate the deformations from x = H/2 to x = H:

${\delta }_B={\int }_{H/2}^H-\frac{wx}{3E}dx$

${\delta }_B=-\frac{w}{3E}{\left[\frac{x^2}{2}\right]}_{H/2}^H$

${\delta }_B=-\frac{w}{3E}\left[\frac{H^2}{2}-\frac{H^2}{8}\right]$

${\delta }_B=-\frac{w}{3E}\left[\frac{3H^2}{8}\right]$

${\delta }_B=-\frac{w{H}^2}{8E}$

The negative sign indicates the direction of deflection is downward. Thus, the magnitude of the vertical deflection at the mid-height of the cone is:

${\delta }_B=\frac{w{H}^2}{8E}$