Past exam of the physics B course of the University of Cambridge / 2017 / Paper 1 / Section C / C13 / Solution

Constitutive law and Poisson ratio. Use positive stress for tension. In an isotropic linearly elastic medium, Hooke's law givesThe analogous equations follow by cyclic permutation. Under uniaxial tension in direction 1, $e_1={\tau }_1/E$ and $e_2=e_3=-\sigma {\tau }_1/E$. Thus Poisson's ratio $\sigma =-e_2/e_1$ measures transverse contraction relative to longitudinal extension when the transverse stresses are zero. It is dimensionless; it is not the ratio of transverse stress to longitudinal stress.

Radial and hoop strains. Let a material point originally at radius $r$ move to $r+R(r)$, with small axisymmetric displacement. Two neighbouring radii separated by $dr$ move apart by $dr+dR$, so the radial fractional extension is $dR/dr$. A circle's circumference changes from $2\pi r$ to $2\pi (r+R)$, so its fractional extension is $R/r$. ConsequentlyThe absence of longitudinal stress is a plane stress condition, not a claim of zero longitudinal strain.

Equilibrium of an annular wedge. Consider a slice of unit axial length, radial thickness $dr$ and small angular width $d\varphi$. The difference of the outward and inward radial face forces is $(r{\tau }_r{)}^{\prime }drd\varphi$ to leading order. The two tangential face tensions have inward radial resultant ${\tau }_{\varphi }drd\varphi$: each contributes ${\tau }_{\varphi }dr\sin (d\varphi /2)$. With no body force, their balance gives the radial equilibrium of a thick elastic cylinder,This hoop contribution is why the radial stress need not be spatially constant in a thick wall.

Displacement equation. With ${\tau }_z=0$, the two Hooke's law equations readInvert them:Substitute in ${\tau }_{\varphi }=(r{\tau }_r{)}^{\prime }$. After cancelling the common factor, the right-hand side is $r{R}^{\prime \prime }+(1+\sigma )R^{\prime }$, soThe cancellation explains why this displacement equation does not depend on the value of Poisson's ratio.

Power laws and wall pressure. Try $R\propto r^n$. The indicial equation is $n(n-1)+n-1=n^2-1=0$, soThe resulting Lamé stress distribution in a pressurized cylindrical wall can be writtenThe pressure boundary tractions are ${\tau }_r(a)=-p_1$ and ${\tau }_r(b)=-p_0$. Solving them yieldsHence the positive compressive radial pressure in the wall isIt equals $p_1$ at $a$ and $p_0$ at $b$. This directional radial pressure is not an isotropic fluid pressure: radial and tangential stresses differ. The associated displacement is $R=[(1-\sigma )Pr+(1+\sigma )Q/r]/E$. Also $e_z=-\sigma ({\tau }_r+{\tau }_{\varphi })/E=-2\sigma P/E$, generally nonzero even though ${\tau }_z=0$.

Maximum absolute tangential stress. For $p_1>p_0$, $Q>0$ and ${\tau }_{\varphi }^{\prime }=-2Q/r^3<0$. Therefore the maximum of its absolute value must be at one of the two surfaces. Keeping the specified external pressure gives the fully general answerFor the usual high internal pressure with negligible external pressure, the largest magnitude is the tensile stress on the inner wall:For finite $p_0$, the inner-wall expression is $[p_1(a^2+b^2)-2p_0{b}^2]/(b^2-a^2)$, but its absolute value is not always the larger endpoint value. For example, nearly equal internal and external pressures can leave the outer surface more compressive. With $p_1=p_0=p$, both stresses correctly reduce to $-p$ everywhere. These results use a homogeneous linear elastic wall without axial loading; a closed-ended pipe would require a different longitudinal-stress condition.