Hooke's law formulas

${}_x=\frac{E}{(1+v)(1-2v)}[(1-v)lo{n}_x+$

$u(lo{n}_y+lo{n}_z]]$

$lo{n}_x=\frac{1}{E}[{}_x-$

$u({}_y+{}_z)]$

${}_{xy}=G_{{}_{yy}}=\frac{E}{2(1+v{)}^2}{}_{yy}$

${}_y=\frac{E}{(1+)}$

$u(1-2v)[(1-$

$u)lo{n}_y+$

$u(lo{n}_z+lo{n}_x)]$

$lo{n}_y=\frac{1}{E}[{}_y-$

$u({}_x+{}_z)]$

${}_r=G_{y_z}=\frac{E}{2(1+v{)}^2}{}_{r_z}$

${}_z=\frac{E}{(1+v)(1-2v)}[(1-v)lo{n}_z+$

$u(lo{n}_z+lon)]$

$lo{n}_z=\frac{1}{E}[{}_z-$

$u({}_x+{}_y)]$

${}_{}=G_{{}_x}=\frac{E}{2(1+v{)}^2}{}_{}$

Assume a polymeric material has a Young's modulus of

$3$ GPa and Poisson's ratio of $0\mathrm{.}41.$ The material is in a

biaxial state of stress in the $x-y$ plane with normal

strains measured as $lo{n}_x=-156\frac{m}{m}$ and $lo{n}_y=323$

$\frac{m}{m}.$ Calculate the strain in the z$-$direction, $lo{n}_z.$ Answer

in units of $\frac{m}{m}$ to one decimal place.

HINT: Think about what equations are available, what

stresses are present and what stress is zero. Don't make

this harder than it really is$.$