Question 5.1.3: A plug of length L and radius a, made from elastic material ......

The problem becomes non-trivial because of the non-zero Poisson’s ratio of the plug material. Indeed, in the case of σ = 0 the longitudinal stresses caused by the pushing force have no effect on the stresses in the plane transverse to the z-axis. Thus, the pressure acting on the lateral surface of the plug would remain equal to p, making the required pushing force simply equal to F_{min} = 2πaLκp. However, for σ > 0 the pushing force results in an increased lateral pressure, which makes F_{min} much bigger for a long plug with L ≫ a. By using the cylindrical coordinates (r, θ, z), consider now the r and z components of the equilibium equation (5.9):

\frac{∂σ_{ik}}{∂x_{k}} + ρg_{i} = 0,        (5.9)

\frac{∂(rσ_{rr})}{r∂r} − \frac{σ_{θθ}}{r} + \frac{∂σ_{zr}}{∂z} = 0        (5.13)

\frac{∂(rσ_{zr})}{r∂r} + \frac{∂σ_{zz}}{∂z} = 0        (5.14)

(in this case \vec{g} = 0, and the azimuthal equilibrium is satisfied identically due to the azimuthal symmetry). In the absence of the pushing force there are only two non-zero components of the stress tensor:

σ_{θθ} = σ_{rr} = −p,

where the first equality follows from the radial equilibrium condition (5.13), and the second one comes from the boundary condition at the plug’s lateral surface. Thus, the plug is uniformly deformed and, according to the Hooke’s law (5.8),

u_{ik} = \frac{1}{E} [(1 + σ)σ_{ik} − σσ_{ll}δ_{ik}]        (5.8)

u_{rr} = u_{θθ} = −p(1 − σ)/E, u_{zz} = 2σp/E        (5.15)

Under the action of the pushing force the stress tensor components σ_{zz} and σ_{zr} also arise. However, if κ ≪ 1, one may expect (and it is confirmed by what follows) that the component σ_{zr} remains small. Therefore, the respective term in equation (5.13) can be neglected and, hence, as before, σ_{rr} ≈ σ_{θθ}, but now they are z-dependent. Since the channel is absolutely rigid, the radial deformation of the plug remains unchanged and equal to the one given in (5.15): u_{rr} = −p(1 − σ)/E. On the other hand, from Hooke’s law (5.8),

u_{rr} = \frac{1}{E} [(1 + σ)σ_{rr} − σ(σ_{rr} + σ_{θθ} + σ_{zz}] =

\frac{1}{E} [(1 − σ)σ_{rr} − σσ_{zz} ],

which yields the following relation between σ_{rr} and σ_{zz}:

σ_{rr} = −p + \frac{σ}{(1 − σ)} σ_{zz}        (5.16)

Then, integration of equation (5.14) over r with the requirement that σ_{zr} is finite at r = 0 gives

σ_{zr} = − \frac{1}{2} r\frac{∂σ_{zz}}{∂z},

and hence, by using the boundary condition σ_{zr}(a) = κσ_{rr} and expression (5.16), one gets the following equation for σ_{zz}(z):

− \frac{a}{2} \frac{∂σ_{zz}}{∂z} = κ \left(−p + \frac{σ}{(1 − σ)} σ_{zz} \right)

The solution with the required boundary condition σ_{zz}(0) = −F/πa^{2} reads:

σ_{zz}(z) = \frac{(1 − σ)}{σ} p \left[\exp  \left(\frac{2κσz}{a(1 − σ)} \right) − 1 \right] − \frac{F}{πa^{2}}        (5.17)

Consider now the variation of σ_{zz} along the plug under different magnitudes of the pushing force F. If the latter is not too big, the absolute value of σ_{zz} decreases monotonically from the initial one at z = 0 (which is equal to F/πa^{2}), down to zero at some point z = z_{0} < L. In this case Equation (5.17) becomes not applicable at z_{0} < z < L, where one can put σ_{zz} = 0. In physical terms it means that the friction force exerted on the 0 ≤ z ≤ z_{0} portion of the plug is sufficient to withstand the pushing force; hence, the rest of the plug plays no role and can be cut off and removed without making any difference.

When the force F is increasing, so is the value of z_{0}, and the critical point is reached when z_{0} = L. This, according to expression (5.17), occurs when

F = F_{0} = πa^{2}p \frac{(1 − σ)}{σ} \left[\exp \left(\frac{2κσL}{a(1 − σ)} \right) − 1\right]      (5.18)

Clearly, there is no equilibrium if the pushing force exceeds F_{0}, which, therefore, is the sought after minimum force required to move the plug. As seen from (5.18), this force becomes exponentially large when the plug’s aspect ratio L/a ≫ 1. The reason lies in the non-zero Poisson’s ratio. Indeed, in the limit σ → 0 expression (5.18) reduces to the expected one: F_{min}(σ → 0) = 2πLaκp.