In this entry:
- Definition of stress
- Formula for normal stress from tension/compression
- Tensile stiffness
- Sample solution of stretching/squeezing task
Axial tension and compression
Before we discuss tension and compression, let's define what tension is.
| Stress - a quantity defined as the force per unit area. The SI unit of stress is [Pa]. Imperial unit is [psi]. |
| SI units |
![1[Pa]=1[\frac {N}{mm^2}].](https://s0.wp.com/latex.php?latex=1%5BPa%5D%3D1%5B%5Cfrac+%7BN%7D%7Bmm%5E2%7D%5D+&bg=ffffff&fg=000&s=2&c=20201002) Most often you will encounter a unit: ![1[MPa]=10^6 [Pa].](https://s0.wp.com/latex.php?latex=1%5BMPa%5D%3D10%5E6+%5BPa%5D+&bg=ffffff&fg=000&s=2&c=20201002) Practical conversion rate ![1[{kN}{cm^2}]=10[MPa].](https://s0.wp.com/latex.php?latex=1%5B%5Cfrac+%7BkN%7D%7Bcm%5E2%7D%5D%3D10%5BMPa%5D++&bg=ffffff&fg=000&s=2&c=20201002) |
| Imperial units |
![1[psi]=1[\frac {lbf}{inch^2}].](https://s0.wp.com/latex.php?latex=1%5Bpsi%5D%3D1%5B%5Cfrac+%7Blbf%7D%7Binch%5E2%7D%5D+&bg=ffffff&fg=000&s=2&c=20201002) Most often you will encounter a unit: ![1[ksi]=10^3 [psi].](https://s0.wp.com/latex.php?latex=1%5Bksi%5D%3D10%5E3+%5Bpsi%5D+&bg=ffffff&fg=000&s=2&c=20201002) |
Tension and compression is the simplest state of stress we will encounter in strength of materials. In textbooks, this is generally the section where everything starts.
Well, just stretching or squeezing it state of stress induced in a simple prismatic bar ( that is, in a bar with a constant cross-section) by such a load that causes only normal stress. If the stresses are positive then we have stretching on the other hand, if they are negative then compression.
Formula for normal stress from tension/compression
Below you will find the formula for normal stress:
 where: N-normal force A-section area |
Tensile stiffness
We define tensile stiffness as the product of Young's modulus "E" and the area of the cross-section being stretched "A". That is, this stiffness depends on two factors:
- on what our stretched component is made of. Different materials have different longitudinal moduli of elasticity
- From the cross-sectional dimensions, the larger the cross-sectional area , the greater the stiffness
Formula for elongation/shortening from tension/compression
Below you will find the formula for the elongation or shortening of an element. Elongation if we are dealing with tension and shortening if we are talking about compression.
 where: N-normal force L-initial length E- Young's module A-section area |
The greater the force and the initial length of the element, the greater the change in length will be. The unit of elongation is [m]. The higher the tensile stiffness, the lower the elongation will be.
Sample solution of stretching/squeezing task
Below you will find a diagram of a bar loaded with two normal forces. Let's solve this task together. This is a prismatic rod with a cross section o:
- A=20 [mm^2].
- length L=8 [m]
- Young's modulus E=200,000 [MPa].
| All examples used in this post were created in the Normal Forces Calculator. You are invited to try it out. In this application you will determine the normal forces, normal stress, and elongation or shortening of the bar. |
We will first determine the "R" reaction in the restraint.
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 step, we will calculate the axial forces, normal stress and elongation in each compartment. We determine the compartments where the normal force changes. In our example, we have two compartments.
Compartment one
For each compartment, we will determine the normal force "N", writing the equilibrium equation of axial forces. In the first section of the bar, this force is equal to the restraint 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 length of the first fragment. Here we will make use of elongation formula. The element will be extended by 0.05 [mm].
Compartment two
In the next section of the bar, the normal force is equal to the sum of R + F1 = 100 [N].
The normal stress is 5 [MPa]. And as in the first compartment we are dealing with tension so the stress is positive.
As for the change in length, we have an elongation of 0.10 [mm].
Finally, let's still determine the total elongation as the sum of the elongation of the individual fragments.
In the last step, we will present the determined values of normal forces, normal stresses and the change in length on graphs.
At this point we will finish solving our example. More examples more complex in the next post.