# Compressive Stress and it’s overview with important facts

## Compressive stress definition:

The tensile and compressive property of the material represent the axial loads along the orthogonal axes. Loads that are stretched at the system boundaries are described as tensile loads, while those compressed at the system boundaries described as compressive loads.

The externally applied force on the body deforms the body in such a way that the body decreases in volume, and length is called compressive stress.

It is the restored strain of the body to deform when applied to external compressive load. An increase in Compressive stress to slender, long cylinders tend to undergo structural failure due to buckling of columns. When the material fails to withstand the compression, stress buckling occurs. compressive stress

## Compressive stress formula:

The normal force is applied to the unit area. Where,

Compressive force (F): compression force is the load required to compress the material to put the material together.

## Compressive stress unit:

The SI unit of it is same as unit of the force to that of the area.

So, it is represented as N/m2 or Pa.

## Dimension of Compressive stress:

Compressive stress dimension is [ML-1T-2].

## Is compressive stress positive or negative?

Answer: compressive stress is negative as it is compressed since change in dimension (dL) has the opposite direction.

## Are yield strength and compressive strength the same ?

Answer: No, yielding in tension and compression is not the same. Value will change as per applicability.

## Compressive strength:

This is the capacity of the material to withstand the compression occur due to compressive stress. There are some materials that can withstand the only tension, some materials can withstand the only compression, and there are some materials that can withstand both tension and compression. The ultimate compressive strength is the value obtained when the material goes through its complete failure. The compression test is done the same as the tensile test. Only difference is the load used is compressive load.

Compressive strength is higher in rock and concrete. Image credit : Ironpole at en.wikipedia, Engineering stress strain, CC BY-SA 3.0

## Compressive stress of mild steel | low carbon steel:

Material that undergoes large strains before failure is ductile materials such as mild steel, aluminum and its alloys. Brittle materials, when undergoes compressive stress, the occurrence of rupture due to the sudden release of the stored energy. Whereas when the ductile material undergoes compressive stress, the material will compress, and deformation takes place without any failure.

## Stress-strain diagram: Compression stress Image credit: Wei SUN et al

The stress-strain diagram for compression is different from tension.

Under compression test, the stress-strain curve is a straight line till an elastic limit. Beyond that point, a distinct bend in the curve representing the onset of plasticity; the point shows the composite compressive yield stress, which is directly related to residual stress. The increase in residual stress increases compressive stress.

In the compression test, the linear region is an elastic region following Hooke’s law. Hence the region can be represented as,

E= Young’s modulus

In this region, the material behaves elastically and returns to its original position by the removal of stress.

## Yield point:

This is the point where elasticity terminates, and plasticity region initiate. So, after yield point, material will not able to return in its actual shaped after the removal of stress.

It is found if crystalline material goes through compression, the stress-strain curve is opposite to tension applications in the elastic region. The tension and compression curves vary at larger deformations (strains) as there is compression at the compressed material, and at the tension, the material undergoes plastic deformation.

## Stress-strain in tension | tensile test: Image credit: Nicoguaro, Stress strain ductile, CC BY 4.0

## Line OA: Proportional limit

Line OA represents a proportional limit. The proportional limit is the limit till when the stress is proportionate to strain following the Hooks Law. As stress increases, the deformation of the material increases.

### Point A: Elastic limit:

In this point maximum stress within a solid material has been applied. This point is called elastic limit. The material within elastic limit, will undergo deformation, and after stress removal, material will back to its actual position.

## What is Elasto-plastic region?

### Elasto-plastic region:

It is the region between yield point and elastic point.

### Point B: Upper yield point

Plastic deformation initiates with dis-location from its crystalline structure. This displacement becomes higher after upper yield point, and it limits the movement of it,  this characteristics known known as strain hardening.

### Point C: Lower yield point

This is the point after which the characteristics like strain hardening initiates. And it is observed that beyond elastic limit, the property like plastic deformation happens.

## Permanent deformation:

### Upper yield point:

A point at which maximum load or stress is applied to initiate plastic deformation.

The upper yield point is unstable due to crystalline dislocations movement.

### Lower yield point:

The limit of min load or stress essential to preserve plastic behavior.

The lower yield point is stable as there is no movement of crystalline.

Stress is the resistance offered by the material when applied to an external load, and strain hardening is an increase in resistance slowly due to an increase of dislocations in the material.

### Point D: ultimate stress point

It represents the ultimate stress point. The maximum stress can withstand the ultimate stress. After the increase of load, failure occurs.

### Point E: Rupture point

It represents the breaking or rupture point. When the material undergoes rapid deformation after the ultimate stress point, it leads to failure of the material. It the maximum deformation occurred in the material.

## Compressive stress example problems| Applications

• Aerospace and Automotive Industry: Actuation tests and spring tests
• Construction Industry: The construction industry directly depends on the compressive strength of the materials. The pillar, the roofing is built by using compressive stress.
• Concrete pillar: In a concrete pillar, the material is squeezed together by compressive stress.
• The material is compressed bonded, such as to avoid failure of the building. It has a sustainable amount of strained stored energy.
• Cosmetic Industry: compaction of compact powder, eyeliners, lip balms, lipsticks, eye shadows is made by applying the compressive stress.
• Packaging Industry: Cardboard packaging, compressed bottles, PET bottles.
• Pharmaceutical Industry: In the pharmaceutical Industry, compressive stress is mostly used.
• The breaking, compacting, crumbling is done in the making of tablets. The hardness and compression strength is a major part of the pharmaceutical Industry.
• Sports industry: cricket ball, tennis ball, basketball ball are compressed to make it tougher.

## Compression test:

The compression test is determination of the behavior of a material under compressive load.

The Compression test is usually used for rock and concrete. Compression test gives the stress and deformation of the material. The experimental result has to validate of the theoretical findings.

### Types of compression testing:

• Flexure test
• Spring test
• Crushing test

Compression test is to determine the integrity and safety parameter of the material by enduring compressive stress. It also provides the safety of finished products, components, manufactured tools. It determines whether the material is fit for the purpose and manufactured accordingly.

The compression tests provide data for the following purposes:

• To measure the batch quality
• To understand the consistency in manufacture
• To assist in the design procedure
• To decrease material price
• To guarantee international standards quality etc.

## The compressive strength testing machine:

Compression testing machines comprises the measurements of material properties as Young’s modulus, ultimate compression strength, yield strength, etc., hence overall static compressive strength characteristics of materials.

The compression apparatus is configured for multiple applications. Due to machine design, it can perform tensile, cyclic, shear, flexure tests.

The compression test is operated the same as tensile testing. Only the load variation occurs in both the testing. Tensile test machines use tensile loads, whereas compression test machines use compressive loads.

## What is allowable compressive stress for steel?

Answer:  The allowable stresses are commonly measured by structure codes of that metal such as steel, and aluminum. It is represented by the fraction of its yields stress (strength)

## What is compressive strength of concrete at various ages?

It is the minimum compressive strength were material in standard test of 28-day-old concrete cylinder.

The concrete compressive strength measurements necessitate around 28 to 35MPa at 28 days.

## Solution:

δ= δ=δ cast iron=δ steel=0.7mm

δ cast iron = = 0.7

P cast iron = 50306.66 πN

δ steel= = 0.7

P steel=57166.66πN

ΣFV=0

P= P cast iron +P steel

P=50306.66π+57166.66π

P=107473.32πN

P=337.63kN

## Solution :

The volume of the tower segment with height

H=3.0m and cross-sectional area A=0.2m2 is

V= A*H= 0.3*0.2=0.6m^3

Density ρ=2.7×10^3 kg/m3, (graphite)

Mass of tower segment

m= ρV =(2.7×10^3 *0.60m3)=1.60×10^3 kg.

The weight of the tower segment is

Wp = mg=(1.60×103*9.8)=15.68KN.

The weight of the sculpture is

Ws =10KN,

normal force 3m below the sculpture,

F⊥= wp  + ws  =(1.568+1.0)×104N=25.68KN.

Therefore, the stress is calculated by F/A

=2.568×104*0.20

=1.284×10^5Pa=128.4 kPa.

Y=4.5×10^10Pa = 4.5×10^7kPa.

So, the compressive strain calculated at that position is

Y=128.4/4.5×107

=2.85×10−6.

## E= 2.1*10^5MPa. L1=1000mm, L2=1500mm, L3=800mm.A1=500mm2,A2=1000mm2,A3=700mm2.

From equilibrium: = 0

+8000-10000+P-5000=0

P=7000N

For more articles click here

About Sulochana Dorve I am Sulochana. I am a Mechanical Design Engineer—M.tech in design Engineering, B.tech in Mechanical Engineering. I have worked as an intern at Hindustan Aeronautics limited in the design of the armament department. I have experience working in R&D and design. I am skilled in CAD/CAM/CAE: CATIA | CREO | ANSYS Apdl | ANSYS Workbench | HYPER MESH | Nastran Patran as well as in Programming languages Python, MATLAB and SQL.
I have expertise on Finite Element Analysis, Design for Manufacturing and Assembly(DFMEA), Optimization, Advanced Vibrations, Mechanics of Composite Materials, Computer-Aided Design.
I am passionate about work and a keen learner. My purpose in life is to get a life of purpose, and I believe in hard work. I am here to excel in the field of Engineering by working in a challenging, enjoyable & professionally bright environment where I can fully utilize my technical and logical skills, constantly upgrade myself & benchmark against the best.
Looking forward to connect you through LinkedIn -
https://www.linkedin.com/in/sulochana-dorve-a80a0bab/