Elasticity/Sample final 1

Sample Final Exam Problem 1
A spherical shell with an outer radius of 5 cm has a spherical inclusion of radius 4 cm embedded in it. The shell is made of AISI-4340 steel which has a Young's modulus of 200 GPa, a Poisson's ratio of 0.3 and a tensile yield stress of 860 MPa. The inclusion is made of a rubbery material with a Young's modulus of 1 GPa, a Poisson's ratio of 0.49 and a tensile yield stress of 10 MPa. Due to a reaction in the rubbery material, the radius of the inclusion increases by 0.4 mm. Assume linear elastic behavior.
 * (a) What is the change in the outer radius of the shell caused by this expansion of the inclusion?
 * (b) Will the shell yield in shear due to this expansion ?

Solution
Calculate bulk modulus ($$K$$) and shear modulus ($$\mu$$) of the shell.
 * $$\begin{align}

K & = \frac{E}{3(1-2\nu)} = \frac{200}{3(1-2\times 0.3)} = 167~\text{GPa} \\ \mu & = \frac{E}{2(1+\nu)} = \frac{200}{2(1+0.3)} = 77~\text{GPa} \end{align}$$ Assume that there is perfect bonding at the interface. The increase in the radius of the inclusion causes an internal pressure in the shell.

The radial displacement of the inside of the shell is $$ka$$ where

k = \frac{0.4 \text{(mm)}}{40 \text{(mm)}} = 0.01 $$ From the equation for the radial displacement in a hollow spherical shell, we have (assuming that the internal pressure is $$p_i$$ and the external pressure is zero),

u_r = \frac{r}{2\mu+3\lambda}\left(\frac{a^3p_i}{b^3-a^3}\right) + \frac{a^3b^3}{4\mu r^2}\left(\frac{p_i}{b^3-a^3}\right) = \frac{r}{3K}\left(\frac{a^3p_i}{b^3-a^3}\right) + \frac{a^3b^3}{4\mu r^2}\left(\frac{p_i}{b^3-a^3}\right) $$ At $$r = a$$,

\left.u_r\right|_{r=a} = ka = \left[\frac{a^4}{3K}+\frac{ab^3}{4\mu}\right] \left(\frac{p_i}{b^3-a^3}\right) \Rightarrow \left[\frac{a^3}{3K}+\frac{b^3}{4\mu}\right] \left(\frac{p_i}{b^3-a^3}\right) = k $$ Therefore,

p_i = (12 K\mu k)\left(\frac{b^3-a^3}{4\mu a^3 + 3K b^3}\right) = (4\mu k)\left(        \cfrac{\cfrac{b^3}{a^3}-1}{\cfrac{b^3}{a^3}+\cfrac{4\mu}{3K}}\right) $$ Plugging in values (with $$b/a = 1.25$$)

p_i = (4)(77)(0.01)\left(        \cfrac{1.25^3-1}{1.25^3+\cfrac{(4)(77)}{(3)(167)}}         \right) = (3.08)\left( \cfrac{1.953-1}{1.953+0.615} \right) = (3.08)(0.371) = 1.14~\text{GPa} = 1140~\text{MPa} $$ Therefore, the displacement of the shell at $$r = b$$ is

\left.u_r\right|_{r=b} = \frac{b}{3K}\left(\frac{a^3p_i}{b^3-a^3}\right) + \frac{a^3b^3}{4\mu b^2}\left(\frac{p_i}{b^3-a^3}\right) = \left[\frac{1}{3K}+\frac{1}{4\mu}\right] \left(\cfrac{p_i b}{\cfrac{b^3}{a^3}-1}\right) $$ Plugging in values,

\left.u_r\right|_{r=b} = \left[\frac{1}{(3)(167)}+\frac{1}{(4)(77)}\right] \left(\cfrac{(1.14)(5)}{(1.25)^3-1}\right) = [0.002 + 0.0032] \left(\cfrac{5.7}{0.953}\right) = 0.031~\text{cm} = 0.31~\text{mm} $$

(a) The change in the outer radius of the shell is 0.31 mm.

The non-zero stresses in the shell are given by

\sigma_{rr} = \left[\frac{p_i}{(b/a)^3 - 1}\right] - \frac{b^3}{r^3} \left[\frac{p_i}{(b/a)^3 - 1}\right] = \left(1- \frac{b^3}{r^3}\right)\left[\frac{p_i}{(b/a)^3 - 1}\right] $$ and

\sigma_{\theta\theta} = \sigma_{\phi\phi} = \left[\frac{p_i}{(b/a)^3 - 1}\right] + \frac{b^3}{2 r^3} \left[\frac{p_i}{(b/a)^3 - 1}\right] = \left(1+ \frac{b^3}{2r^3}\right)\left[\frac{p_i}{(b/a)^3 - 1}\right] $$ Therefore, the maximum shear stress is given by

\tau_{\text{max}} = \frac{1}{2}\left|\sigma_{rr}-\sigma_{\theta\theta}\right| = \frac{1}{2} \left(\frac{3b^3}{2r^3}\right)\left[\frac{p_i}{(b/a)^3 - 1}\right] $$ The magnitude of the maximum shear stress is largest at $$r = a$$. Hence,

\left.\tau_{\text{max}}\right|_{r=a} = \left(\frac{3}{4}\right)\left[\frac{p_i}{1-(a/b)^3}\right] = \left(\frac{3}{4}\right)\left[\frac{1.14}{1-(0.8)^3}\right] = 1.75~\text{GPa} = 1750~\text{MPa} $$ The yield stress in shear is $$\tau_{Y} = \sigma_{Y}/2 = 430$$ MPa which is much lower than the maximum shear stress.

(b) The shell will yield in shear due to the given expansion.