Consider a unit cube; vertex $O$ is at the origin of the $xyz-$coordinates, the opposite vertex $A$ is at $,$

$y=1,z=1,$ and the point $H$ is halfway between at the center of the cube $(x=0\mathrm{.}5,y=0\mathrm{.}5,z=0\mathrm{.}5).$ The

cube then experiences the following displacement:

$u=0\mathrm{.}2+0\mathrm{.}1x-0\mathrm{.}3y+0\mathrm{.}4xz$

$v=0\mathrm{.}3-0\mathrm{.}4x+0\mathrm{.}2y^2-0\mathrm{.}1xz$

$w=-0\mathrm{.}1+0\mathrm{.}3x+0\mathrm{.}1z+0\mathrm{.}3xy$

Compute both the:

xyz$-$coordinates of $O^{\text{'}},A^{\text{'}},$ and $H^{\text{'}},$ the new locations of $O,A,$ and $H,$ respectively, after the displacement

is applied

the three normal strains and three shear strains, leaving your answers in terms of $x,y,$ and $z$