Compression concept and formulas, calculation, examples, exercises

The compression or compressive stress is the force per unit area whose result is to push, squeeze or compress an object, tending to shorten it. Mathematically it is:

E = F / A

Here AND denotes effort, F the magnitude of the force and TO the area on which it acts, the unit in the SI International System being the newton / m^two or pascal (Pa). Compressive stress is a normal effort, because the force that produces it is perpendicular to the area on which it is exerted.

Such effort can compress the object or, on the contrary, tension and stretch it, as applied. In the case of compressive stress, the forces are applied in the opposite direction to exert the effect of squeezing and shortening the object.

Once the forces cease, many materials return to their original dimensions. This property is known by the name of elasticity. But while that happens, the elastic unit deformation suffered by a material subjected to a stress is:

Strain = (Final Size - Initial Size) / Initial Size

Strain can be linear, surface, or volumetric, although strain is unitless. However, the information it provides is very important, since it is not the same to deform a 10 m long bar by 1 cm, to deform another 1 m long bar by 1 cm..

In an elastic material, deformation and stress are proportional, fulfilling Hooke's law:

Effort ∝ Unit deformation

Article index

• 1 How to calculate compression?
  • 1.1 Modulus of elasticity of different materials
• 2 Examples
  • 2.1 Columns and pillars
  • 2.2 Chairs and benches
• 3 Exercises
  • 3.1 - Exercise 1
  • 3.2 - Exercise 2
• 4 References

¿How to calculate compression?

The compressive stress causes the particles of the material to get closer and closer, reducing their size. Depending on the direction in which the effort is applied, there will be a shortening or reduction in some of its dimensions.

Let's start by assuming a thin bar of original length L,  to which normal stress of magnitude AND. If the stress is compressive, the bar experiences a reduction in its length, denoted by δ. If it is tension, the bar will lengthen.

Naturally, the material from which the element is made is decisive in its ability to withstand stress..

These elastic characteristics of the material are included in the aforementioned constant of proportionality. Is named modulus of elasticity or Young's modulus and is denoted as Y. Each material has a modulus of elasticity, which is determined experimentally by laboratory tests.

With this in mind, the effort AND is expressed in mathematical form like this:

Effort ∝ Unit deformation

Finally, to establish this condition as an equation, a constant of proportionality is required to substitute the symbol of proportionality ∝ and substitute it for equality, like this:

Stress = Constant of proportionality x Unit strain        

E = Y. (δ / L)

The quotient (δ / L) is the strain, denoted as ε and with δ = Final length - Initial length. In this way, the effort AND looks like:

E = Y. ε

Since strain is dimensionless, the units of Y are the same as those of AND: N / m^two o Pa in the SI system, pounds / in^two or psi in the British system, as well as other combinations of force and area, such as kg / cm^two.

Modulus of elasticity of different materials

Y values ​​are determined experimentally in the laboratory, under controlled conditions. Next, the modulus of elasticity for materials widely used in construction and also that of bones:

Table 1

Material	Modulus of elasticity Y (Pa) x 109
Steel	200
Iron	100
Brass	100
Bronze	90
Aluminum	70
Marble	fifty
Granite	Four. Five
Concrete	twenty
Bone	fifteen
Pinewood	10

Examples

Compressive forces act on various structures; They are subject to the action of forces such as the weight of each of the elements that compose them, as well as forces from external agents: wind, snow, other structures and more..

It is usual for most structures to be designed to withstand stress of all kinds without deforming. Therefore, the compression stress must be taken into account to prevent the part or object from losing its shape..

Also the bones of the skeleton are structures subjected to various stresses. Although the bones are resistant to them, when the elastic limit is exceeded by accident, fissures and fractures are originated.

Columns and pillars

Columns and piers in buildings must be made to resist compression, otherwise they tend to bow. This is known as lateral bending or buckling.

The columns (see figure 1) are elements whose length is considerably greater compared to their cross-sectional area..

A cylindrical element is a column when its length is equal to or greater than ten times the diameter of the cross section. But if the cross section is not constant, its smaller diameter will be taken in order to classify the element as a column.

Chairs and benches

When people sit on furniture such as chairs and benches, or add objects on top, then the legs are subjected to compressive stresses that tend to decrease their height.

Furniture is usually made to withstand weight quite well and returns to its natural state once it is removed. But if heavy weight is placed on fragile chairs or benches, the legs give way to compression and break..

Training

- Exercise 1

There is a rod that originally measures 12 m in length, to which it is subjected to a compressive stress such that its unit deformation is -0.0004. What is the new length of the rod?

Solution

Starting from the equation given above:

ε = (δ / L) = - 0.0004

Yes L_F is the final length and L_or the initial length, since δ = L_F - L_or  you have:

(L_F - L_or) / L_or = -0.0004

Therefore: L_F - L_or = -0.0004 x 12 m = -0.0048 m. And finally:

L_F  = (12 - 0.0048) m = 11.9952 m.

- Exercise 2

A solid steel bar, cylindrical in shape, is 6 m long and 8 cm in diameter. If the bar is compressed by a load of 90,000 kg, find:

a) The magnitude of the compressive stress in megapascals (MPa)

b) By how much did the length of the bar decrease?

Solution to

First we find the area A of the cross section of the bar, which depends on its diameter D, resulting in:

A = π. D^two / 4 = π. (0.08 m)^two / 4 = 5.03 x 10^-3 m^two

The force is immediately found, through F = m.g = 90,000 kg x 9.8 m / s^two= 882,000 N.

Finally the average effort is calculated like this:

E = F / A = 882,000 N / 5.03 x 10^-3 m^two = 1.75 x 10⁸ Pa = 175 MPa

Solution b

Now the equation for stress is used, knowing that the material has an elastic response:

E = Y. (δ / L)

The Young's modulus of steel is found in Table 1:

δ = E.L / Y = 6 m x 1.75 x 10⁸ Pa / 200 x 10 ⁹ Pa = 5.25 x 10 ^-3 m = 5.25 mm.

References

1. Beer, F. 2010. Mechanics of materials. 5th. Edition. Mcgraw hill.
2. Giancoli, D. 2006. Physics: Principles with Applications. 6^tth  Ed. Prentice Hall.
3. Hibbeler, R.C. 2006. Mechanics of materials. 6th. Edition. Pearson Education.
4. Tippens, P. 2011. Physics: Concepts and Applications. 7th Edition. Mcgraw hill
5. Wikipedia. Stress (Mechanics). Recovered from: wikipedia.org.