Solution to Problem 219 Axial Deformation

For the differential strip shown:
δ = dδ
P = weight of segment y carried by the strip
L = dy
A = area of the strip
 

For weight of segment y (Frustum of a cone):
$P = wV_y$
 

From section along the axis:
$\dfrac{x}{y} = \dfrac{D - d}{L}$

$x = \dfrac{D - d}{L} y$
 

Volume for frustum of cone
$V = \frac{1}{3} \pi h \, (R^2 + r^2 + Rr)$

$V_y = \frac{1}{3} \pi h \, [ \, \frac{1}{4} (x + d)^2 + \frac{1}{4} d^2 + \frac{1}{2} (x + d) (\frac{1}{2} d) \, ]$

$V_y = \frac{1}{12} \pi y \, [ \, (x + d)^2 + d^2 + (x + d)d \, ]$
 

$P = \frac{1}{12} \pi w \, [ \, (x + d)^2 + d^2 + (x + d)d \, ] \, y$

$P = \frac{1}{12} \pi w \, [ \, x^2 + 2xd + d^2 + d^2 + xd + d^2 \, ] \, y$

$P = \frac{1}{12} \pi w \, [ \, x^2 + 3xd + 3d^2 \, ] \, y$

$P = \dfrac{\pi w}{12} \left[ \dfrac{(D - d)^2}{L^2} y^2 + \dfrac{3d (D - d)}{L} y + 3d^3 \right] \, y$
 

Area of the strip:
$A = \frac{1}{4} \pi (x + d)^2 = \dfrac{\pi}{4} \left( \dfrac{D - d}{L} y + d \right)^2$
 

Thus,
$\delta = \dfrac{PL}{AE}$

$d\delta = \dfrac{\dfrac{\pi w}{12} \left[ \dfrac{(D - d)^2}{L^2} y^2 + \dfrac{3d (D - d)}{L} y + 3d^3 \right] \, y \, \, dy}{\dfrac{\pi}{4} \left( \dfrac{D - d}{L} y + d \right)^2 \, E}$

$d\delta = \dfrac{4w}{12E} \left[ \dfrac{\dfrac{(D - d)^2}{L^2}y^2 + \dfrac{3d(D - d)}{L}y + 3d^2}{\dfrac{(D - d)^2}{L^2}y^2 + \dfrac{2d(D - d)}{L}y + d^2} \right] \, y \,\, dy$

$d\delta = \dfrac{w}{3E} \left[ \dfrac{\dfrac{(D - d)^2 y^2 + 3Ld (D - d)y + 3L^2 d^2}{L^2}}{\dfrac{(D - d)^2 y^2 + 2Ld (D - d)y + L^2 d^2}{L^2}} \right] \, y \,\, dy$

$d\delta = \dfrac{w}{3E} \left[ \dfrac{(D - d)^2 y^2 + 3Ld (D - d)y + 3L^2 d^2}{(D - d)^2 y^2 + 2Ld (D - d)y + L^2 d^2} \right] \, y \,\, dy$
 

Let:
$a = D - d$ and
$b = Ld$
 

$d\delta = \dfrac{w}{3E} \left[ \dfrac{a^2 y^2 + 3ab\,y + 3b^2}{a^2 y^2 + 2ab\,y + b^2} \right] \, y \,\, dy$

$d\delta = \dfrac{w}{3E} \left[ \dfrac{a^2 y^2 + 3ab\,y + 3b^2}{(ay)^2 + 2(ay)b + b^2} \times \dfrac{a}{a} \right] \, y \,\, dy$

$d\delta = \dfrac{w}{3aE} \left[ \dfrac{a^3 y^3 + 3(a^2 y^2) b + 3(ay) b^2}{(ay + b)^2} \right] \, dy$

$d\delta = \dfrac{w}{3aE} \left\{ \dfrac{ [ \, (ay)^3 + 3(ay)^2 b + 3(ay) b^2 + b^3 \, ] - b^3}{(ay + b)^2} \right\} \, dy$
 

The quantity
$(ay)^3 + 3(ay)^2 b + 3(ay)b^2 + b^3 = (ay + b)^3$
 

$d\delta = \dfrac{w}{3aE} \left[ \dfrac{(ay + b)^3 - b^3}{(ay + b)^2} \right] \, dy$

$d\delta = \dfrac{w}{3aE} \left[ \dfrac{(ay + b)^3}{(ay + b)^2} - \dfrac{b^3}{(ay + b)^2} \right] \, dy$

$d\delta = \dfrac{w}{3aE} \, [ \, (ay + b) - b^3 (ay + b)^{-2} \, ] \, dy$
 

$\delta = \dfrac{w}{3aE} \displaystyle \int_0^L [ \, (ay + b) - b^3 (ay + b)^{-2} \, ] \, dy$

$\delta = \dfrac{w}{3aE} \left[ \dfrac{(ay + b)^2}{2a} - \dfrac{b^3 (ay + b)^{-1}}{-a} \right]_0^L$

$\delta = \dfrac{w}{3a^2 E} \left[ \dfrac{(ay + b)^2}{2} + \dfrac{b^3}{ay + b} \right]_0^L$

$\delta = \dfrac{w}{3a^2 E} \left\{ \, \left[ \frac{1}{2} (aL + b)^2 + \dfrac{b^3}{aL + b} \right] - \left[ \frac{1}{2}b^2 + \dfrac{b^3}{b} \right] \, \right\}$

$\delta = \dfrac{w}{3a^2 E} \left\{ \, \frac{1}{2} (aL + b)^2 + \dfrac{b^3}{aL + b} - \frac{3}{2} b^2 \, \right\}$

$\delta = \dfrac{w}{3a^2 E} \left[ \dfrac{(aL + b)^3 + 2b^3 - 3b^2 (aL + b)}{2(aL + b)} \right]$

$\delta = \dfrac{w}{6a^2 E} \left[ \dfrac{(aL)^3 + 3(aL)^2 b + 3(aL)b^2 + b^3 + 2b^3 - 3ab^2 L - 3b^3}{aL + b} \right]$

$\delta = \dfrac{w}{6a^2 E} \left[ \dfrac{a^3 L^3 + 3a^2 b L^2}{aL + b} \right]$
 

Note that we let
$a = D - d$ and
$b = Ld$
 

$\delta = \dfrac{w}{6(D - d)^2 E} \left[ \dfrac{(D - d)^3 L^3 + 3(D - d)^2 (Ld) L^2}{(D - d)L + Ld} \right]$

$\delta = \dfrac{w}{6(D - d)^2 E} \left\{ \, \dfrac{(D - d)L^3 \, [ \, (D - d)^2 + 3d(D - d) \, ]}{LD - Ld + Ld} \, \right\}$

$\delta = \dfrac{wL^3}{6(D - d) E} \left[ \dfrac{(D - d)^2 + 3d(D - d)}{LD} \right]$

$\delta = \dfrac{wL^3}{6(D - d) E} \left[ \dfrac{D^2 - 2Dd + d^2 + 3Dd - 3d^2}{LD} \right]$

$\delta = \dfrac{wL^3}{6(D - d) E} \left[ \dfrac{D^2 + Dd - 2d^2}{LD} \right]$

$\delta = \dfrac{wL^3}{6(D - d) E} \left[ \dfrac{D(D + d) - 2d^2}{LD} \right]$

$\delta = \dfrac{wL^3}{6(D - d) E} \left[ \dfrac{D(D + d)}{LD} \right] - \dfrac{wL^3}{6(D - d) E} \left[ \dfrac{2d^2}{LD} \right]$

$\delta = \dfrac{wL^2 (D + d)}{6E (D - d)} - \dfrac{wL^2 d^2}{3ED(D - d)}$           answer
 

For a cone:
$D = D$ and
$d = 0$
 

$\delta = \dfrac{wL^2 (D + 0)}{6E (D - 0)} - \dfrac{wL^2 (0^2)}{3ED(D - 0)}$

$\delta = \dfrac{wL^2}{6E}$           answer