$*1\mathrm{.}17$ Consider $u_r$ to be the displacement component in the radial direction and $u_{}$ to be the

displacement component in the transverse $()$ direction. Show that:

$lo{n}_{}=\frac{1}{r}\frac{del{u}_{}}{\text{d}\text{e}\text{l}}+\frac{u_r}{r}$

As for ${}_r,$ observe that the shaded element $($Fig$.1\mathrm{.}33)$ is in the undeformed geometry

while the unshaded element is in the deformed geometry. What is the change in right

angles at corner $A$ in terms of angles $,,$ and $?$ Show that:

${}_r=\frac{del{u}_{}}{\text{d}\text{e}\text{l}\text{r}}-\frac{u_{}}{r}+\frac{1}{r}\frac{del{u}_r}{\text{d}\text{e}\text{l}}$

What is $lo{n}_{rr}?$