Dimensional Formula for Stress
The dimensional formula of Stress is represented as:
[M1 L-1 T-2]
Where,
M denotes the mass
L denotes the length
T denotes the time
What is Stress?
Stress is the external restoring force that acts on per unit area and it is denoted by σ. It is denoted as N/m². It is used to find stress which is applied on any given body when force and area on which it is exerted is given in a problem.
Therefore, \[σ = \frac{F}{A}\]
σ is the amount of stress on the object
F is the force that is acting on the object.
A is the cross-sectional area
or it can also be said that stress is the force applied on an object making it completely to deform. In Physics terminology, we found how the derivation of the stress formula is done. It is known that larger objects can withstand a large number of forces. By using stress as an alternative to force we are able to use the same yield stress for the same material, it doesn’t matter how large the object is.
Also, stress and strain are directly related with each other and if one increases, the other automatically increases. And, the more is the stress of the object, the more deformation it experiences.
Derivation of Dimension of Stress
Stress = Force × Area⁻¹ . . .. . (i)
The dimensional formula of area = [M0 L² T0] . . .. (ii)
Since we know,
Force = M × a = MM × [M0 L¹ T⁻²]
Therefore, the dimensional formula of force = M¹L¹T⁻²M¹L¹T⁻² . . . . (iii)
On substituting equation (ii) and (iii) in equation (i) we get,
Stress = Force × AreaArea⁻¹
Or, Stress = M¹L¹T⁻²M¹L¹T⁻² × [M0 L ² T0]⁻¹ = M¹L⁻¹T⁻²M¹L⁻¹T⁻²
Therefore, dimensional formula of stress is represented as M¹L⁻¹T⁻²M¹L⁻¹T⁻².
Example of a Solved Problem
Question: If an elastic spring is given a force of 1000 N over an area of 0.2 m2. Calculate the amount of stress?
Ans: As we know from the above problem,
F (Force) = 1000 N,
A (Area) = 0.2 m ²
σ = \[\frac{F}{A}\]
= \[\frac{1000N}{0.2}\]
σ = 5000N/m2
Types of Stress
1. Normal Stress
When an object is loaded by an axial force, it is called normal stress. Normal stress is represented when axial force is divided by the cross-sectional area. It will occur when an object is placed in compression.
2. Longitudinal Stress
When the length of the body changes its length by the applied normal stress, it is called longitudinal stress.
Longitudinal stress is represented by the division of the deforming force by the area of cross section.
3. Bulk Stress or Volume Stress
It is the stress in which the body volume changes due to the stress. Normal stress on an object changes its length or volume and the tangential stress leads to change in the shape of the body which is called volume stress.
Shearing Stress is the force applied tangentially over the plane’s surface area. When the forces being applied to the surface is parallel to it and the stress acting on the surface plots a tangent, it is known as shearing stress.
4. Tensile Stress
It is the force per unit area and the stress when applied and increases the length of the body because of the force, it is called tensile stress. It is observed when a rod is stretched under motion’s third law. A common example of tensile stress is rubber and stretching is its quantity associated with it.
5. Compression Stress
It occurs when we apply a tangential force on the body and the shape plus volume of the object change. Compression stress results in a decrease of the length of the object. It is opposite to the tensile stress.
Importance of Stress
Stress analysis is an important part of applied Physics and it helps in classification of the internal distribution of internal forces in solid objects.
An important part of engineering, stress helps in studying and designing of structures such as structural frames, dams, runners and mechanical parts. It is also important in other regimentation, for instance, in Geology, studying theory like plate tectonics, volcanism and avalanches, and in Biology, it is important to understand the anatomy of living beings.
Stress is the force we apply to an object in order for it to completely deform. Furthermore, we are aware of human stress, but stress in physics is a little more difficult to comprehend. Furthermore, in this topic, we will talk about stress, the stress formula, its derivation, and a solved example.
What Happens when Objects React?
To make an origami crane, fold a piece of paper. Also, even after you relief the paper, the origami retains its new shape after completion. When you stretch a rubber band, however, it will snap back when you let go. Furthermore, some objects retain some of their original shapes when stretched.
Furthermore, some things change shape when forces are applied to them, but not all items change in the same way, even when we apply the same amount of force.
Stress
There are a few concepts that must be addressed before discussing stress. Furthermore, stress is defined as the amount of force (strength or energy) we apply to an object divided by the cross-section area on which the force acts.
Furthermore, larger objects can withstand greater forces. Furthermore, by using stress instead of force, we can use the same yield stress for the same material regardless of how large the object is.
Most importantly, stress and strain are inextricably linked, and as one rises, the other follows suit. Furthermore, the more stress an object is subjected to, the more it deforms until it fails.
Furthermore, all objects undergo elastic deformation at first, but when the stress on the object exceeds a certain threshold, it undergoes plastic deformation, and this is when the switch occurs and the object has reached its yield stress.
Furthermore, the stress and strain bond exists in all materials, even if the size of each portion varies. Furthermore, the elastic deformity is linear. Furthermore, the slope line is affected by the materials used to construct the object. Furthermore, because plastic deformation is not linear, it is more difficult to model.
Formula for Stress
The stress formula is the force divided by the cross-section area.
Force/Area = Stress
\[σ = \frac{F} {A}\]
The Stress Formula's Derivation
The amount of stress on the object is denoted by =.σ
F = denotes the force acting on the object.
A denotes the cross-sectional area.
Example of a Stress Formula that has been solved
To help you understand the formula better, let's look at some examples.
Exemplification 1
What is the stress of an object with an acting force of 50 Newtons (N) and a cross-section area of 5 mm2?
Solution:
Let's jot down what's in the question.
F (force) = 50 N (Newtons)
A = 5 mm2 cross-sectional area
Now, enter the values from the question into the formula.
\[σ = \frac{F }{A}\]
\[σ= \frac{50N}{5\times10^{-6}}\]
\[σ = 10×10^6\] Nm2
As a result, the object's stress is \[10×10^6\] Nm2.
Example No. 2
A force of 1000 N was applied to an elastic spring over an area of 0.2 m2. So, how do you calculate the stress on the elastic spring?
Solution:
First, write down everything we've learned from the formula.
Force (F) = 1000 N
A is equal to 0.2 m2.
Now, enter the values into the formula.
\[σ = \frac{F}{A}\]
\[σ = \frac{1000N}{0.2}\]
σ = 5000 Nm2
As a result, the elastic spring is stressed to the tune of 5000Nm2.
Young’s Modulus of Elasticity
Young's modulus is a numerical constant named after the 18th-century English physician and physicist Thomas Young that describes the elastic properties of a solid undergoing tension or compression in only one direction, such as a metal rod that returns to its original length after being stretched or compressed lengthwise. Young's modulus is a measure of a material's ability to withstand changes in length when subjected to lengthwise tension or compression. Young's modulus, also known as the modulus of elasticity, is equal to the longitudinal stress divided by the strain. In the case of a metal bar under tension, stress and strain can be described as follows.
When a metal bar with cross-sectional area A is pulled at both ends by a force F, the bar stretches from its original length L0 to a new length Ln. (At the same time, the cross-section decreases.) The stress is defined as the quotient of tensile force divided by cross-sectional area, or F/A. The strain or relative deformation is defined as the difference in length, Ln L0, divided by the original length, or (Ln L0)/L0. (Strength has no dimensions.) As a result, Young's modulus can be expressed mathematically as,
Stress/strain = (FL0)/A(Ln L0) = Young's modulus
This is a variant of Hooke's law of elasticity. Young's modulus is measured in pounds per square inch (psi) in the English system and newtons per square metre (N/m2) in the metric system. Aluminium has Young's modulus of about 1.0 107 psi or 7.0 1010 N/m2. Steel has a value that is approximately three times greater, which means that it takes three times as much force to stretch a steel bar as a similarly shaped aluminium bar.
Young's modulus is only useful when the stress is proportional to the strain and the material returns to its original dimensions when the external force is removed. As stresses increase, the material may flow, deform permanently, or finally break.
When a tensioned metal bar is elongated, its width is slightly reduced. This lateral shrinkage results in a transverse strain equal to the width change divided by the original width. Poisson's ratio is the ratio of transverse strain to longitudinal strain. The average Poisson's ratio for steel is 0.28, and 0.33 for aluminium alloys. Materials with Poisson's ratios less than 0.50 expand under longitudinal tension and contract under longitudinal compression.
