In this entry:

### Axial tension and compression

Before we discuss tension and compression, let's define what stress is.

Stress – a quantity defined as force per unit area. The SI unit of stress is [Pa]. In Imperial units stress is measured in pound-force per square inch, which is often shortened to "psi". |

Tension and compression are the simplest states of stress that we encounter in the strength of materials. In textbooks, this is usually the section where everything starts.

Tension or compression is **a state of stress** caused in a simple prismatic rod (i.e. a rod with a constant cross-section) by a load that causes only normal stress. If the stresses are positive, we have **tension** , if they are negative, we have **compression.**

### Formula for normal stresses due to tension/compression

Below you will find the formula for normal stress:

### Tension stiffness

Tension stiffness is defined as the product of Young's modulus "E" and the tension cross-sectional area "A". This means that stiffness depends on two factors:

on what our stretched element is made of. Different materials have different moduli of longitudinal elasticity

on the cross-sectional dimensions, the larger the cross-sectional area, the greater the stiffness

### Formula for elongation/shortening due to tension/compression

Below you will find a formula for extending or shortening an element. Elongation if we are talking about tension and shortening if we are talking about compression.

The greater the force and the initial length of the element, the greater the change in length will be. The SI unit of elongation is [m] in Imperial units it is [in]. The greater the tension stiffness, the smaller the elongation.

### An example solution to a tension/compression problem

Below you will find a diagram of a rod loaded with two normal forces. We will solve this task together. It is a prismatic rod with a cross-section of:

A=20 [mm^2]

length L=8 [m]

Young's modulus E=200,000 [MPa]

All examples used in this post were created in Tension/Compresion Calculator. I invite you to try it out. In this application, you will determine normal forces, normal stress and extension or shortening of the bar. |

First, we will determine the reaction "R" in the support.

To determine the reaction, we only need one equilibrium equation. The sum of the forces in the horizontal direction must be zero.

In the next stage, we will calculate the axial forces, normal stress and elongation in individual compartments. We mark the intervals where the normal force changes. In our example we have two compartments.

#### First part

For each interval we will determine the normal force "N" by writing the axial force balance equation. In the first fragment of the bar, this force is equal to the force reaction R = 50 [N].

Normal stress 2.5 [MPa]. We are dealing with tension, so the stress is positive.

In the last step, we will calculate the change in the length of the first fragment. We will use the elongation formula here. The element will be extended by 0.05 [mm].

#### Second part

In the next fragment of the rod, the normal force is equal to the sum of R + F1 = 100 [N].

The normal stress is 5 [MPa]. And just like in the first part, we are dealing with stretching, so the stress is positive.

As for the change in length, we have an extension of 0.10 [mm].

Finally, let us determine the total elongation as the sum of the elongations of individual fragments.

In the last step, the determined values of normal forces, normal stresses and length changes will be presented on diagr.

This concludes our example. More more complex examples in the next entry.

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