What is the Plastic Neutral Axis?

If you enjoyed this post or know someone who would, please share! (It really helps me grow)

Most engineering students can readily find the centroid of a shape, where the elastic neutral axes cross. This concept is thoroughly covered in any introductory Mechanics of Materials or Deformable Body Mechanics class.

What’s not typically covered well is plastic behavior in bending, and in particular, locating the plastic neutral axis. The Plastic Neutral Axis (PNA) is the dividing line between the tension and compression zones of a shape that has developed full plasticity. The sum of the yield strength times areas above the PNA is equal to that below it.

What trips up many engineering students (including me) is the quick and crass introduction to plastic behavior in their steel design course later. In a 2016 study of students at Western Kentucky University, only 27% of students could correctly explain what the plastic neutral axis was, compared to 73% correctly explaining the elastic neutral axis.

For shapes that are both symmetric and composed of a single material, the Elastic Neutral Axis (ENA) and PNA are the same, as in Example 1 of the video below. When the shape is not symmetric about the x-axis, as in Example 2, the ENA and PNA differ. The same is true for shapes that are geometrically symmetric, but whose material makeup is asymmetric, as in Example 3.

Worked Examples Scans

Click here to download the PDF scan of my full worked examples from the video.

Video Option: Three Worked Examples

Background Mechanics

The PNA defines the line between tension and compression regions of a fully-developed plastic hinge section subjected to pure bending. Because there is no resultant axial force, the compressive and tensile forces must be equal.

Full development of a plastic hinge occurs when every portion of the material hits yield stress. This can only occur in very ductile materials with a large yield plateau on their stress-strain curves, like mild steel.

Traditional elastic design relies on keeping the extreme fiber stresses below the yield strength of the material. For a brittle material, like concrete, yield is basically non-existent: once the extreme fibers give out, the section cracks.

But for a ductile material, like A36 Steel, the increased strain does not produce additional stress until strain-hardening kicks in. The plateau of Zone II below shows this plastic deformation region. Once the extreme fibers hit Point C, they can be squished or stretched without producing additional reaction load.

By Toiyabe (talk) (Uploads) – Own work, CC BY-SA 3.0, https://en.wikipedia.org/w/index.php?curid=3351645

Meanwhile, the next fibers in have hit yield, followed by the ones below them, all the way to the plastic neutral axis. This state can be described by the general equation:

Under pure bending, the resultant forces above and below the PNA must be equal, and those forces are equal to the integral (or summation if you’re dealing with convenient shapes) of the areas times their yield stresses.

When solving for the location of the ENA of geometric shapes, a single formula can be used, with no assumptions, to calculate the weighted average of the centroids of each component shape.

This isn’t the case with the PNA, which will instead often require an assumption be made, and various a set of assumptions and various cases be tested. Whenever you make an assumption, always remember to go back and check that it was correct. I drop myself a large checkbox as a reminder to confirm my assumptions at the end.

Symmetric Shapes of One Material

In doubly-symmetric shapes made of one material, the ENA and PNA will both occur in the same location, where the axes of symmetry cross.

So long as only one material is used, the yield stress will be the same for the entire shape. That allows division of both sides of the general equation (1) by the constant yield strength, reducing the equation to:

From equation (2), the location of the PNA must fall at the mid-height of any symmetric section made up of one material.

Asymmetric Shapes of One Material

In asymmetric sections of one material, equation (2) from above holds, but the ENA and PNA are no longer coincident.

The Elastic Neutral Axis is based on a weighted average of the centroids of the component areas, whereas the Plastic Neutral Axis of a mono-material shape is based simply on the line which halves the area.

Picture an I-beam with extra flanges coming off the web, one flange just above the lower flange, and one just above the mid-height of the section. So long as these asymmetrically-located flanges have equal area above and below the mid-height of the section, the PNA will still be at the mid-height. But the ENA will be pulled down towards the lower half by the concentration of area lower in the section.

Similarly, in a built-up T-section made of a pair of plates of equal dimension, the PNA will occur at the junction of flange and stem, where half the area is above and half is below. The ENA will instead be located lower down the stem, drawn by the spread of the area vertically along the stem.

Shapes of Asymmetric Materials

When multiple materials are used, whether different metals or different alloys of the same metal, the simpler Equation (2) is no longer valid, and Equation (1) must be used.

In finding centroids of composite shapes, a modular ratio is used to transform dissimilar materials into an equivalent section, based on the comparative Young’s Moduli of the materials. In the case of the PNA, the relevant material property is the yield strength, and a yield ratio could be constructed, though it’s not really necessary.

For example, picture an I-section constructed from three plates of identical dimensions, using A36 steel for the top flange and web, and a Grade 50 steel for the bottom flange. A36 and Grade 50 have identical Young’s Moduli, so the ENA is unaffected by the use of the stronger plate, and remains at the mid-height of the section.

The PNA, however, would be drawn downward towards the stronger material. A36 has a yield strength of 36 ksi, while the Grade 50 has a yield strength of 50 ksi. The higher yield strength of the Grade 50 effectively serves as a multiplier on the area of that plate for the plastic section.

Finding the Plastic Moment

The plastic moment can be found by multiplying the yield strength of the material by the plastic section modulus, equivalent to the elastic section modulus, S, frequently used in elastic design.

The plastic section modulus, Z, is one half the area of the total shape times the distance from the centroid of the upper half of the area to the centroid of the lower half of the area. There is only one plastic section modulus about a given axis, unlike the two elastic section moduli for the same.

Elastic section moduli are calculated by dividing the moment of inertia, I, by the distance from the centroid of the shape to the extreme compressive or tensile fibers.

Don’t forget to swing by PPI2Pass for all your FE, PE, and SE prep needs!

Engineer Eric

Eric is a licensed Professional Engineer working as a structural engineer for an architectural facade manufacturer, which straddles the line between structural and mechanical engineering. He holds an MS in Structural Engineering from the University of Minnesota.

Recent Posts