- How is it calculated?
- Hooke's law and normal stress
- Importance of stress in the strength of materials and geology
- Examples
- -Exercise 1
- Solution
- -Exercise 2
- Solution
- References
The normal stress applied to a certain material, also called uniaxial stress, is the relationship that exists between the force applied perpendicularly on a certain surface and the cross-sectional area on which it acts, or the load per unit area. Mathematically, if P is the magnitude of the force and A is the area where it is applied, the stress σ is the quotient: σ = P / A.
The units of normal stress in the International System are newton / meter 2, known as Pascals and abbreviated Pa. These are the same units of pressure. Other units that appear frequently in the literature are pounds / inch 2 or psi.
Figure 1. Rocks are constantly stressed due to tectonic activity, causing deformations in the earth's crust. Source: Pixabay.
In Figure 2 two forces of equal magnitude are applied perpendicular to the cross-sectional area, exerting a very light traction on the bar that tends to elongate it.
These forces produce a normal stress that is also called centered axial load, because its line of action coincides with the axial axis, on which the centroid is located.
Figure 2. The bar shown is subjected to tensile forces. Source: self made.
Efforts, whether normal or otherwise, continually appear in nature. In the lithosphere, rocks are subjected to gravity and tectonic activity, undergoing deformations.
In this way, structures such as folds and faults originate, the study of which is important in the exploitation of minerals and in civil engineering, for the construction of buildings and roads, to name a few examples.
How is it calculated?
The equation given at the beginning σ = P / A allows to calculate the average normal stress over the area in question. The value of P is the magnitude of the resultant force on the area applied to the centroid and is sufficient for many simple situations.
In this case, the distribution of forces is uniform, especially at points far from where the bar is subject to tension or compression. But if you need to calculate the stress at a specific point or the forces are not uniformly distributed, you should use the following definition:
So in general, the value of the stress at a particular point can be different from the average value. In fact the effort may vary depending on the section to be considered.
This is illustrated in the following figure, in which the tensile forces F try to separate the equilibrium bar in sections mm and nn.
Figure 3. Distribution of normal forces in different sections of a bar. Source:
As section nn is very close to where the downward force F is applied, the distribution of forces on the surface is not completely homogeneous, the lower the force the further away from that point is. The distribution is a little more homogeneous in the mm section.
In any case, normal effort always tends to stretch or compress the two parts of the body that are on both sides of the plane on which they act. On the other hand, other different forces, such as that of shear, tend to displace and separate these parts.
Hooke's law and normal stress
Hooke's law states that within elastic limits, the normal stress is directly proportional to the deformation experienced by the bar or object. In that case:
The constant of proportionality being Young's modulus (Y):
σ = Y. ε
With ε = ΔL / L, where ΔL is the difference between the final and initial length, which is L.
Young's modulus or modulus of elasticity is a characteristic of the material, whose dimensions are the same as those of stress, since the unit strain is dimensionless.
Importance of stress in the strength of materials and geology
Determining how resistant materials are to stress is very important. For the structures used in the construction of buildings, as well as in the design of parts for different devices, it must be ensured that the chosen materials adequately fulfill their function.
For this reason, materials are exhaustively analyzed in laboratories through tests aimed at knowing how much force they can withstand before deforming and breaking, thus losing their functions. Based on this, the decision is made as to whether or not they are suitable to manufacture a certain part or form part of a device.
The first scientist to systematically study the strength of materials is believed to have been Leonardo Da Vinci. He left evidence of tests in which he determined the resistance of wires by hanging stones of different weights on them.
In the efforts, both the magnitude of the force as well as the dimensions of the structure and in what way it is applied is important, in order to establish the limits within which the material has an elastic behavior; that is, it returns to its original form when the effort ceases.
With the results of these tests, stress-strain curves are made for different types of materials, such as steel, concrete, aluminum and many more.
Examples
In the following examples it is assumed that the forces are uniformly distributed, and that the material is homogeneous and isotropic. This means that their properties are the same in either direction. Therefore it is valid to apply the equation σ = P / A to find the forces.
-Exercise 1
In figure 3, it is known that the average normal stress acting on section AB has magnitude 48 kPa. Find: a) The magnitude of the force F acting on CB, b) The effort on the section BC.
Figure 4. Normal stresses on the structure of example 1..
Solution
Since the structure is in static equilibrium, according to Newton's second law:
PF = 0
The normal stress on section AB has magnitude:
σ AB = P / A AB
From where P = σ AB. A AB = 48000 Pa. (40 x 10 -2 m) 2 = 7680 N
Therefore F = 7680 N
The normal stress on section BC is the quotient between the magnitude of F and the cross-sectional area of that side:
σ BC = F / A BC = 7680 N / (30 x 10 -2 m) 2 = 85.3 kPa.
-Exercise 2
A wire 150 m long and 2.5 mm in diameter is stretched by a force of 500 N. Find:
a) The longitudinal stress σ.
b) The unit deformation, knowing that the final length is 150.125 m.
c) The modulus of elasticity Y of this wire.
Solution
a) σ = F / A = F / π.r 2
The radius of the wire is half the diameter:
r = 1.25 mm = 1.25 x 10 -3 m.
The cross-sectional area is π.r 2, so the stress is:
σ = F / π.r 2 = 500 / (π. (1.25 x 10 -3) 2 Pa = 101859.2 Pa
b) ε = Δ L / L = (Final length - Initial length) / Initial length
Thus:
ε = (150.125 - 150) / 150 = 0.125 / 150 = 0.000833
c) The Young's modulus of the wire is solved knowing the values of ε and σ previously calculated:
Y = σ / ε = 101859.2 Pa / 0.000833 = 1.22 x 10 8 Pa = 122 MPa.
References
- Beer, F. 2010. Mechanics of materials. 5th. Edition. McGraw Hill. 7 - 9.
- Giancoli, D. 2006. Physics: Principles with Applications. 6 t th Ed. Prentice Hall. 238-242.
- Hibbeler, RC 2006. Mechanics of materials. 6th. Edition. Pearson Education. 22 -25
- Valera Negrete, J. 2005. Notes on General Physics. UNAM. 87-98.
- Wikipedia. Stress (Mechanics). Recovered from: wikipedia.org.