# Plastic Properties – Bending

The previous articles have shown how to calculate the second moment of area, I, and the elastic modulus, Z, of different shapes and have shown that this can be used to calculate the elastic stress on a section.

Other articles on this website also discuss the difference between elasticity and plasticity and have shown how elastoplastic materials behave differently above and below a materials yield stress.

In this article we will look at what happens if you load an elastoplastic beam beyond its elastic limit.

#### Plastic Stress Blocks

If we take a beam made out of an elastoplastic material such as steel and bend it, stresses will occur in the top and bottom of a section. We can keep bending the section until one of these outer fibres reaches the yield stress, but then what happens?

When the stress in one of the outer fibres of the beam reaches the material’s yield stress the material in that location begins to deform plastically. If strain hardening is ignored then the stress in the yielded location remains at the yield stress and cannot develop a higher stress, but it can deform.

If bending continues to increase then sections closer to the elastic neutral axis also being to reach the materials yield stress and also deform plastically. Bending can increase until all points on the beam’s cross section have the yield stress in either tension or compression.

Once all sections of the beam have yielded then it is not possible to increase the bending moment on the element, maintaining the load will simply cause plastic strain to occur and the beam to deflect, this is sometimes known as a “plastic hinge”.

The diagrams above show that as parts of the beam begin to yield the neutral axis (the point of zero stress) of the beam begins to move, this is to ensure that the total compressive force from the area undergoing compressive stress and the total tensile force from the areas undergoing tensile stress are equal, such that force equilibrium is satisfied in the cross section.

The position of the neutral axis when the section has fully yielded is called the “plastic neutral axis”. It is worth noting that the elastic neutral axis and the plastic neutral axis are often at different positions within the beam, as a beam is loaded past its elastic moment of resistance the neutral axis begins to shift towards the plastic neutral axis.

#### Finding the Plastic Neutral Axis

The plastic neutral axis divides the section into two, ensuring that the tensile force from the half in tension is equal to the compressive force from the half in compression. This ensures force equilibrium at the cross section.

As force is equal to a stress times an area and at the plastic limit the full cross section is at the materials yield stress, then the only way to ensure force equilibrium is have equal areas above and below the plastic neutral axis.

Therefore, the plastic neutral axis is defined at the axis that divides the cross section into two equal areas.

#### Plastic Neutral Axis Example

In the article on area and centroid, the elastic neutral axis, which lies at the centroid of a section, was found:

The elastic neutral axis was found as:

Element Id | Breadth | Depth | Area | y | Ay |
---|---|---|---|---|---|

1 | 600 | 60 | 36,000 | 30 | 1,080,000 |

2 | 20 | 800 | 16,000 | 460 | 7,360,000 |

3 | 320 | 40 | 12,800 | 880 | 11,264,000 |

Sum | 64,800 | 19,704,000 |

$$\text{Using the values of } \sum A_{i}\overline{y,i} \text{ and } \sum A_{i} \text{ from the above table, we can find:}$$ $$\overline{y}_{elastic} = \frac{\sum A_{i}\overline{y,i}}{\sum A_{i}}= \frac{19,704,000}{64,800} = 304mm$$

The plastic neutral axis can be found as follows:

$$\text{The plastic neutral axis divides the section into two equal areas}$$ $$\text{Total area of section, } A = 64,800mm^{2}$$ $$\text{Half of area, } = \frac{A}{2} = 32,400mm^{2}$$ $$\text{By inspection the plastic neutral axis lies in Element 1, the bottom flange}$$ $$\text{Expressions for the area above and below the plastic neutral axis can be found} $$ $$\text{The plastic neutral axis height will be denoted } \overline{y}_{plastic}$$ $$\text{Area below } = A_{below} = 32,400mm^{2} = 600 \overline{y}_{plastic}$$ $$\text{Area above } = A_{above} = 32,400mm^{2} = 12,800mm^{2} + 16,000mm^{2} + 600(60- \overline{y}_{plastic})$$ $$\text{Therefore}$$ $$\overline{y}_{plastic} = 54mm$$ $$\text{This compares with}$$ $$\overline{y}_{elastic} = 304mm$$

It’s worth noting that for the above asymmetrical section the neutral axis and plastic axis are at two different locations. However for sections that are symmetrical about the axis of bending the elastic neutral axis and plastic neutral axis will be at the same point.

#### Finding the Plastic Modulus

In the article on section moduli, moment and stress, the concept of the elastic section moduli was discussed. The elastic section moduli are typically calculated at the top and bottom of a section, using the second moment of area of a section, I, and the distance from the elastic neutral axis to the top or bottom of section, y, as appropriate.

$$\qquad Z_{top}= \frac{I}{y_{top}}$$ $$\qquad Z_{bottom}= \frac{I}{y_{bottom}}, $$ $$and$$ $$\qquad \sigma= \frac{My}{I} \rightarrow M = \frac{\sigma I}{y} $$ $$\text{elastic moment resistance can be calculated}$$ $$\qquad M_{elastic,resistance}= min(Z_{top}, Z_{bottom})\times\sigma_{y}$$

A similar variable can be derived for plastic sections, this is called the plastic section modulus. The plastic section modulus can be used to calculate a moment resistance in the same way as the elastic section moduli:

$$\text{Plastic moment resistance can be calculated}$$ $$\qquad M_{plastic,resistance}= Z_{plastic}\times\sigma_{y}$$

The plastic section modulus is calculated by adding the area in compression multiplied by the lever arm to the centroid of the compression area to the area in tension multiplied by the lever at to the centroid of the tensile area:

$$Z_{plastic} = A_{C} y_{C} + A_{T} y_{T}$$

Where \( Z_{plastic}\) is the plastic section modulus (click for unit info) , \( A_{C}\) is the area of the section in compression, \( y_{C} \) is the lever arm from the plastic neutral axis to the centroid of the area in compression, \( A_{T}\) is the area of the section in tension and \( y_{T} \) is the lever arm from the plastic neutral axis to the centroid of the area in tension.

For sections which comprise of complex shapes it is often possible to break a section down into smaller simple shapes and add up the individual contribution to the plastic modulus from each shape. This is shown in the example below.

#### Plastic Modulus Example

The plastic modulus of the section used in the previous example will be calculated.

$$\text{The plastic neutral axis height was calculated previously:}$$ $$\overline{y}_{plastic} = 54mm$$

A table will be used to calculate the contributions from the sections above and below the plastic neutral axis:

Elements above the plastic neutral axis:

Element Id | Breadth | Depth | Area | y to plastic neutral axis | Ay |
---|---|---|---|---|---|

1 | 600 | 6 | 3600 | 3 | 10,800 |

2 | 20 | 800 | 16,000 | 406 | 6,496,000 |

3 | 320 | 40 | 12,800 | 826 | 10,572,800 |

Sum | 17,079,600 |

Elements below the plastic neutral axis:

Element Id | Breadth | Depth | Area | y to plastic neutral axis | Ay |
---|---|---|---|---|---|

1 | 600 | 54 | 32,400 | 27 | 874,800 |

Sum | 874,800 |

The plastic section modulus is now found by adding the contributions from above and below the neutral axis:

$$Z_{plastic} = 17,079,600+874,800$$ $$Z_{plastic} = 17,954,400 = 17.95\times10^{6}mm^3$$