Is there a formula or method to calculate the pressure of failure of a cylindrical container of known material, thickness, and size?


In this answer, I aim to extend the discussion to thicker walled cylinders too. If you want to skip the maths, the pressures are summarised at the bottom.


$\sigma_{rr}$ - radial component of stress

$\sigma_{\theta\theta}$ - circumferential component of stress

$\sigma_{zz}$ - axial component of stress

In this answer, I will assume that failure occurs in the curved walls of the cylinder. Different failure modes, could be considered, but that is beyond the scope of my answer. It is in my intuition that the curved walls are the weak points of failure, but I will leave the discussion of that to more qualified answers.


When it comes to determining the failure of thick walled cylinders (defining failure to occur the instance yield occurs anywhere in the cylinder), Lamé's Equations are useful. These equations give the stresses that lie in the plane of the circular cross section of the cylinder:

enter image description here

Note the rotational symmetry axis is moving out of the page, normal to the cross section.

Lamé's Equations are derived from the governing equilibrium equation for an axisymmetric solid:

$$\frac{\partial}{\partial r} \left( r \sigma_{rr}\right) = \sigma_{\theta\theta}$$

Where $r$ is the distance of the point where stresses are being considered from the axis of rotational symmetry.

Lamé's equations themselves are given as:

$$\sigma_{rr} = A - \frac{B}{r^2}$$ $$\sigma_{\theta\theta} = A + \frac{B}{r^2}$$

Where $A$ and $B$ are arbitrary constants.

For a cylinder with internal pressure $p$, no external pressure, an internal radius of $a$ and thickness $t$, these equations become:

$$\sigma_{rr} = \frac{p}{m^2-1}\left(1-\frac{m^2a^2}{r^2}\right)$$ $$\sigma_{\theta\theta} = \frac{p}{m^2-1}\left(1+\frac{m^2a^2}{r^2}\right)$$

Where $m=1 + \frac{t}{a}$ (i.e. ratio of outer to inner radii). Note that as $t$ gets smaller, $m$ approaches 1 and the stresses approach that

To determine the final stress component, it is reasonable to assume that, away from the edges of the cylinder, the axial stress is constant in the walls. This stress will oppose the pressure acting on the cylinder ends:

$$\sigma_{zz} = \frac{p}{m^2-1}$$

Here is a rough plot of the three stresses as function of $r$:

enter image description here

With the three components of stress known, a suitable yield criterion should be used. Usually Tresca or von Mises are suitable for metals like steel. Here, I will assume Tresca, which states that yield first occurs at a point on the cylinder whenever the following is satisfied:

$$\max\left(|\sigma_{rr}-\sigma_{\theta\theta}|,|\sigma_{rr}-\sigma_{zz}|, |\sigma_{\theta\theta}-\sigma_{zz}|\right)=\sigma_y$$

Where $\sigma_y$ is the yield stress of the material.

By looking at the plot, it can be seen that yield will occur on the inside of the cylinder, so that at yield:

$$|\sigma_{rr} - \sigma_{\theta\theta}| = \frac{2m^2}{m^2-1} p=\sigma_y$$

Therefore, failure by first yield occurs when:


The maximum pressure before first yield increases with thickness, up to a maximum of $\frac{\sigma_y}{2}$ at infinite thickness.


Even when yield first occurs, the walls of the cylinder won't burst open, i.e. collapse has yet to occur. The point at which collapse occurs should occur after first yield: would you ever expect an infinitely thick cylinder to burst? The collapse pressure is useful to determine since it is at this point that the cylinder fails to retain its pressurised contents.

When first yield occurs, there is still regions in the cylinder that have yet to yield, and so the internals are retained. As you increase the pressure beyond that of first yield, you get a plastic region of yielded material that forms at the inner face of the cylinder and creeps along the cylinder thickness towards the outer face:

enter image description here

Collapse occurs as soon as the plastic zone reaches the outer face of the cylinder. The issue in trying to calculate this load is that elasticity no longer applies: Lamé's equations, which are only valid for linear elastic cylinders, will not be valid. However, the governing equilibrium equation still applies: equilibrium must be upheld regardless whether the material law is linear or not. Since the plastic zone has extended through the entire cross section, less us assume that the stresses abide by Tresca's criterion throughout the cylinder (this assumption is valid according to the Lower Bound Theorem, which returns an estimate of the collapse pressure which must be less than the actual collapse pressure, giving a conservative estimate):

$$\sigma_{\theta\theta} - \sigma_{rr} = \sigma_y$$

We can now substitute this into the equilibrium equation to give:

$$r \frac{\partial \sigma_{rr}}{\partial r} = \sigma_y$$

Solving this and applying boundary conditions then gives a collapse pressure of:

$$p = \ln{(m)} \sigma_y$$

Interestingly, for all positive $m$, $\ln (m)$ is greater than or equal to $\frac{m^2-1}{2m^2}$, so first yield must always occur before collapse. Only when $m$ approaches 1 (thin-wall) do the pressures for first yield and collapse approach being equal.


For failure by first yield, the pressure must be:


For failure by collapse, the pressure must be:

$$p=\ln{(m)} \sigma_y$$

Where $m$ is the ratio of external to internal radii.

Here is a rough plot for both:

enter image description here

As $m$ increases, the first-yield limit increases to a maximum of $\frac{\sigma_y}{2}$, while the collapse limit increases to infinity. For the collapse limit to be double that of the first-yield limit, $m$ needs to be around 2.2.


There is a simple formula for calculating the stress in thin-walled cylindrical pressure vessels. In the circumferential direction, the formula is:

$$ \sigma_c=\frac{pr}{t} $$

and in the longitudinal direction (along the length of the cylinder):

$$ \sigma_x=\frac{pr}{2t} $$


$p =$ the pressure inside the cylinder,

$r =$ the inner radius of the cylinder, and

$t =$ the thickness of the wall $(r/t \geq 10)$.

With this formula, it is a simple matter to calculate at what pressure the vessel will fail, based on its material properties and whatever factor of safety you like.

Note that $\sigma_c$ and $\sigma_x$ are assumed to be constant throughout the wall of the cylinder, and that the walls of the cylinder are subjected to tension.

Source: Mechanics of Materials Ninth Edition by R.C. Hibbeler.

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    $\begingroup$ The $(r/t \geq 10)$ comment implies this, but it's worth explicitly stating that this only works for thin-walled vessels. If that ratio is not satisfied, then the math gets more complicated. $\endgroup$ – Wasabi Aug 6 '16 at 5:14
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    $\begingroup$ Note also that this is strictly for the cylindrical section - it does not apply to the ends. $\endgroup$ – Mark Aug 6 '16 at 12:47
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    $\begingroup$ This only applies where the internal pressure is higher than the external pressure, and the vessel is in tension. If the external pressure is higher, a thin-walled structure will usually fail by buckling, which is more complicated to predict accurately and may occur at a much smaller pressure difference. $\endgroup$ – alephzero Aug 6 '16 at 18:54
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    $\begingroup$ @kruschk How can I factor the material's physical properties into this formula? This formula appears to be material independant (unless I've misinterpreted the purpose of the $\sigma$ variable) $\endgroup$ – Hypnos Stratagem Aug 7 '16 at 1:26
  • $\begingroup$ Are you familiar with stress-strain curves? Engineers typically design components to the yield stress of the material, which we would represent as $\sigma_c$ or $\sigma_x$. The yield stress is a property of the material. $\endgroup$ – kruschk Aug 7 '16 at 2:49

If you draw the free body diagram of half of cylinder like an arch and draw pressure force vectors, it has pressure force radiating around the entire surface out with equal magnitude but fanning out 180 degrees. The upward projection of these vectors will sum up to: $p\cdot 2r = \text{lateral force}$.

This force has to be carried by 2x thickness of the wall x unit length. Hence we get $$pr = \sigma t$$ But for the ends we have $$\frac{p\pi r^2}{2rt}= \frac{p\pi r}{2}$$

However we have to calculate for buckling due to higher exterior pressure as well.

A very rough estimate for buckling is to assume the entire external pressure is supported at the edges of the container acting in fixed end moment joints as the abutments of an arch if the container is long ($l>4r$).

If there are attachments or stiffeners buckling has to be modeled considering the total geometry. We have seen one to many twisted frozen can of coke left in freezer.


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