PLATES is GIVEN by: \(u = \frac{1}{{2\mu }}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)(dy - {y^2})\)Now\(\frac{{du}}{{dy}} = \frac{1}{{2\mu }}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left( {d - 2y} \right)\)Shear stress distribution, τ is given by\(\tau = \frac{{\mu \;du}}{{dy}}\)\(\tau = \frac{1}{2}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left( {d - 2y} \right)\)From above, following can be concluded:1. Shear Stress distribution, ‘τ’ is linear.2. At y = d/2 i.e. mid point; τ = 0 i.e. At center shear stress = 0.3. At y = 0 i.e. at boundary; \(\tau = {\tau _{MAX}} = \frac{1}{2}\left( {\frac{{ - \partial P}}{{\partial x}}} \right)\left( d \right)\) i.e. shear stress is maximum at boundary.∴ Shear stress is maximum at the plate boundaries and zero at plane \(\frac{d}{2}\) away from each plate