Lecture $20$ Matrix methods for Uniform Combined Axial & Shear Stress & Strain, True Strain, and PDEs for Static Equilibrium

Problem $20\mathrm{.}1$

Suppose a large rectangular prism is made of an elastic material with $E=80$MPa $($note: MPa not GPa $)$ and $v=0\mathrm{.}4$ and has dimensions $Lx=1m,Ly=1m,$ and $Lz=2m,$ extending from one corner at the origin to an opposite corner at $(Lx,Ly,Lz).$ The prism is put into a uniform shear stress & strain condition by applying appropriate normal and shear forces on its external faces so that the internal axial stress vector is $=(0,0,40$kPa$)$ and the internal shear stress vector is $=($:${}_{yz},{}_{xz},{}_{xy}$:$)=($:$0,40$kPa,$0$:$).$

If the corner at the origin is fixed in place and the z $-$displacement of the entire $z=0$ surface is $0($i$.$e$.,$ the surface can't rotate in any direction or translate in z $),$ find an expression for the $x-,y-,$ and z $-$displacement of the prism induced by this loading at any point in the prism.

Hint: with axial and shear strain happening at once, displacement in the $x-$direction due to strain is $($from a total derivative, with small strains$)du=\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{x}}dx+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{y}}dy+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{z}}dz$

If the partial derivatives are constant over the whole object $($as they would be in a uniform strain situation$),$ then integrating gives: $u=\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{x}}x+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{y}}y+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{z}}z,$ i$.$e$.,$

$u(x,y,z)-u(x_0,y_0,z_0)=\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{x}}(x-x_0)+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{y}}(y-y_0)+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{z}}(z-z_0)$

and similarly for $v$ and $w.$

Hint $2$: In your answer, explain why this set of equations

$\{[du=\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{x}}dx+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{y}}dy+\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{z}}dz],[dv=\frac{\text{d}\text{e}\text{l}\text{v}}{\text{d}\text{e}\text{l}\text{x}}dx+\frac{\text{d}\text{e}\text{l}\text{v}}{\text{d}\text{e}\text{l}\text{y}}dy+\frac{\text{d}\text{e}\text{l}\text{v}}{\text{d}\text{e}\text{l}\text{z}}dz],[dw=\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{x}}dx+\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{y}}dy+\frac{\text{d}\text{e}\text{l}\text{w}}{\text{d}\text{e}\text{l}\text{z}}dz]\}$

simplifies to $\{[du={}_{x}dx+{}_{zx}dz],[dv={}_{y}dy],[dw={}_{z}dz]\}$; in particular, explain why there's no ${}_{xz}dx$ contribution to $dw$