Pipe strength and the mysterious "Y"

Posted in real world problems on Tuesday, May 20 2014 Updated on Tue 07 June 2016

Recently, at work, there was some work where someone needed to break out ASME B31.3 and figure out the allowable pressure for a given length of pipe -- back calculating from what was there to what was "OK".

This is neither here nor there, but since I was around I read through what they were doing. I have only vague memories of the intro mechanical engineering courses required for chemical engineers so I was interested in seeing what that entailed and where those equations came from.

In B31.3 is the basic minimum thickness formula: $$ t = \frac{ p D}{2 \left( SEW + pY \right)} $$

Or in terms of stress $$ SEW = \sigma = p \left( \frac{D}{2t} - Y \right) $$


  1. t is the thickness of the pipe,
  2. p is pressure,
  3. D is the outside diameter of the pipe,
  4. S is the allowable stress, with E and W efficiency factors for the joint and weld, I am just going to lump it all together into $\sigma$,
  5. Y is the "Y factor"... okay...

I asked around at work but none of the people knew what the physical significance of the Y factor was, the best answer I got was that it was a fitting parameter to make the equation line up with empirical data.

Well, let's rewind and do some math. Suppose we have a pipe with internal pressure p at inner radius ri and zero pressure at the outer radius ro. First consider a small element of a cylindrical shell within this pipe. If we do a force balance: pipe wall stress

We end up with: $$ \sigma_c - \sigma_r = r \frac{d \sigma_r}{dr} $$

Where σc is the stress in the circumferential direction, i.e. the hoop stress, and σr is the stress in the radial direction. This is the setup for the Lamé equations.

If we assume that the stress in the longitudinal direction is zero, we have a system of plane stress, and the stress invariant is: $$ \sigma_r + \sigma_c = \mbox{a constant} = C_1 $$

Which gives us: $$ C_1 - 2 \sigma_r = r \frac{d \sigma_r}{dr} $$

Separating and integrating: $$ \ln r = -\frac{1}{2} \ln \left( C_1 - 2 \sigma_r \right) + C_2 $$

Where C2 is a constant of integration, if we suppose $ C_2 = -\frac{1}{2} \ln C_3$ we can clean things up quite a bit:

$$ -2 \ln r = \ln \left( C_3 \left( C_1 - 2 \sigma_r \right) \right) $$

$$ \sigma_r = \frac{1}{2} \left( \frac{1}{C_3 r^2} - C_1 \right) $$

At the outer boundary the radial stress is zero (from the boundary conditions) $$ \sigma_r = 0 = \frac{1}{2 C_3 r_o^2} - \frac{C_1}{2} $$ $$ C_1 = \frac{1}{C_3 r_o^2} $$

At the inner boundary the radial stress equals -p (from the boundary conditions) $$ \sigma_r = -p = \frac{1}{2 C_3} \left( \frac{1}{r_i^2} - \frac{1}{r_o^2} \right) $$ $$ C_3 = - \frac{1}{2p} \left( \frac{r_o^2 - r_i^2}{r_o^2 r_i^2} \right) $$

Finally, after some algebra: $$ \sigma_r = \frac{p r_i^2}{r_o^2 - r_i^2} \left(1 - \frac{r_o^2}{r^2} \right) $$ and $$ \sigma_c = \frac{p r_i^2}{r_o^2 - r_i^2} \left(1 + \frac{r_o^2}{r^2} \right) $$

Now the maximum shear stress at any given point in the pipe wall is (if you don't trust me look at Mohr's circle for this situation): $$ \tau_{max} = \frac{\sigma_c - \sigma_r}{2} $$

Assuming the pipe is a ductile material, Tresca's failure criterion is appropriate, which is: $$ \tau_{max} \le \frac{\sigma_{failure}}{2} $$ $$ \sigma = \sigma_c - \sigma_r \le \sigma_{failure} $$

Note that the shear stress is at a maximum at the inside wall of the pipe. Consider at $r_i$:

$$ \sigma = p \frac{ r_o^2 + r_i^2 }{r_o^2 - r_i^2} + p $$ $$ \sigma = p \frac{ 2 r_o^2 }{r_o^2 - r_i^2} $$

Letting $r_o = \frac{D}{2}$ and $r_i = \frac{D}{2} - t$:

$$ \sigma = p \frac{ D^2 }{2t \left( D - t \right)} $$

We can further simplify to: $$ \sigma = p \left( \frac{ D }{2t} - \frac{D}{2 \left( D - t \right)} \right) $$

If we assume $ t \ll D$ then $\frac{D}{2 \left( D-t \right)} \approx \frac{D}{2D} = \frac{1}{2}$ $$ \sigma = p \left( \frac{ D }{2t} - \frac{1}{2} \right) = p \left( \frac{ D }{2t} - Y \right) $$

Where Y = 0.5.

Apparently, according to the internets, the ASME task group for this decided to use Y = 0.4 as a safety factor in 1943. Later editions of B31.3 and B31.1 replaced this with a generic fudge factor, Y, which is based on burst test results. The Y factor is now a function of temperature as the previous version of the equation lead to excessively thick pipes in high temperature steam service, with all the associated difficulties in that.