 Solution

Write $W_i=\partial W/\partial {\alpha }_i$, evaluated at $({\alpha }^{-1/2},{\alpha }^{-1/2},\alpha )$. Applying the chain rule to the reduced strain energy density givesMultiply by $\alpha$ and use ${\alpha }_1={\alpha }_2={\alpha }^{-1/2}$:The transverse symmetry specified in the hint has $W_1=W_2$. ThusThe lateral faces have zero traction, so their diagonal Cauchy stress tensor components satisfy $t_1=t_2=0$. The hydrostatic pressure enforcing elastic incompressibility is therefore $p={\alpha }_1{W}_1={\alpha }_2{W}_2$, and the axial component simplifies toThis elimination of the Lagrange multiplier $p$ is what makes the reduced strain energy density useful for the spring balance.