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I want to work out how much I can expand a piece of rubber tubing before it ruptures. The wall is 0.75mm and the ID 5mm. Hence I am working out the stress/strain from thick-walled cylinder theory using these pages.

In Lame's equations (derived on those pages), are the radiuses in the formula the original radiuses before expansion, or those resulting from expansion? Does the same go for the radius under investigation?

many thanks for your help

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  • $\begingroup$ Why class rubber that is about to rupture as thick-walled ? $\endgroup$ – Solar Mike Jul 3 '18 at 17:39
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    $\begingroup$ The pages you link to use small strain theory. Your question amounts to "what happens if you use small strain theory for large strains" - and the answer is "if you want the right answers, you don't do that". $\endgroup$ – alephzero Jul 3 '18 at 19:05
  • $\begingroup$ I think you link is broken. I have attempted to fix the link. If not please fix the link. $\endgroup$ – 706Astor Jul 3 '18 at 19:29
  • $\begingroup$ @SolarMike Any wall thickness of rubber tube will fail under enough pressure. It doesn't have to stretch until it becomes "thin" before it fails. The common formulas for the stress distribution in a thin walled tube are just simplified approximations to the general thick walled case, and using them as a failure criterion assumes some stress components are negligible. $\endgroup$ – alephzero Jul 3 '18 at 22:42
  • $\begingroup$ @alephzero any tube will fail under enough pressure... but I was after how the OP was justifying the choice ... $\endgroup$ – Solar Mike Jul 3 '18 at 22:45
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A good place to start is Ogden's book Nonlinear elastic deformations, section 5.2.3: "Extension and inflation of a circular cylindrical tube".

For a material with a strain energy function: $$ W = F(I_1) - A I_2 + B I_3 $$ where $I_1$, $I_2$, $I_3$ are the three principal invariants of the stress tensor, Ogden shows that for the situation where a stretch $\lambda_3$ is applied along the cylindrical axis and there are no internal or external pressures, the final radius ($r$) of the cylinder is given by $$ r(R) = \tfrac{1}{2} (\alpha - \lambda_3) R + \frac{\beta}{R} $$ where $R$ is the initial location of a point in the radial direction and $\beta = 0$. The term $\alpha$ is obtained by solving the equation $$ F'(\alpha) - A\alpha + \tfrac{1}{2}(A + B\lambda_3)(\alpha - \lambda_3) = 0 $$ The tube fails when $r(R_o) - r(R_i)$ reaches a critical value ($R_i, R_o$ are the initial inner and outer radii of the cylinder).

You can follow the approach used by Ogden to solve the problem for other strain energy functions, boundary conditions, and constraints such as incompressibility.

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  • $\begingroup$ +1 for excellent answer addressing the non-small deformation complication of rubber tubing. $\endgroup$ – Mark Jul 4 '18 at 5:33
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According to J. E. Gordon's book "Structures or why things don't fall down" he explains The concept of Work of Fracture. During his exposition he explains that cylindrical shapes have a specific relationship of tensile strength to the thickness of the cylinder weigh is what a tube is. The general rule is that the tensile strength of any material is determined by the surface area perpindicular to the plane in which the strain exists multiplied by the length of the cylinder. On page 77 he states rubber has about 300% working strain. On p122 a diagram displays an image of circumferential stress in a cylinder with the formula of $s_2=\frac{rp}t$ where s=stress, r=radius, p=pressure and t=cylinder thickness.

In almost all cases where internal pressure exceeds the strain capacity of the cylinder, a longitudinal rupture occurs. Since your question asks about rubber specifically there is another issue in that the cylindrical thickness decreases with pressure increases proportionally to the radial expansion. In simple terms, the rubber is thinnest where the diameter is greatest but the strain energy is identical throughout the tube because the internal pressure is constant.

In your example, the thickness of the tube is 0.75mm with an internal diameter of 5mm. Since we know rubber had approximately 300% working strain, the pressure required to break it must therefore be $$ s=\frac {rp}t$$ $$p=\frac {st}{r}$$ $$p=\frac {3 \cdot 0.75}{5}$$ $$p=0.45$$

So pressure must exceed $0.45 units of pressure per {mm^2}$ to rupture the tube by which the length of the unexpanded section to be expanded must be multiplied.

book cover

The book is a great read and humorous in many places. As well as being an invaluable source of education, it is a recommended course book by the university of Wolverhampton for civil engineering honours students.

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  • $\begingroup$ Thanks for this - however this is I believe thin-walled cylinder theory, which I think is judged unapplicable when the wall thickness is 10% oif the radius of the cylinder, as you then have different strains acting on the inside and outside surfaces of the wall according to the radius under investigation. $\endgroup$ – Oliver Walters Jul 4 '18 at 10:34
  • $\begingroup$ See this [wikipedia link] (en.wikipedia.org/wiki/Cylinder_stress) "Thin-walled assumption: For the thin-walled assumption to be valid the vessel must have a wall thickness of no more than about one-tenth (often cited as Diameter / t > 20) of its radius.[2] " $\endgroup$ – Oliver Walters Jul 4 '18 at 10:35
  • $\begingroup$ I'm also unsure how to use it, when I know only the expansion of the tubing, but not the pressure? $\endgroup$ – Oliver Walters Jul 4 '18 at 10:37
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    $\begingroup$ Since we know that the maximum pressure is going to be o.45 lb.mm^2, and you have the maximum distension measurement... It is possible therefore to calculate the pressure by volume. If rubber expands up to 800%, you need only measure the account of cylindrical expansion as a proportion of theoretical maximum expansion. If the tube has a diameter of 10mm and it has expanded to 8 cm, it is safe to assume that any further expansion would result in catastrophic failure? $\endgroup$ – Rhodie Jul 11 '18 at 1:46

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