The modulus of elasticity measures a material's resistance to elastic deformation under external forces. Understanding this property is important for designing structures with materials like metals, concrete, and polymers to ensure they can withstand stresses without permanent deformation.
The modulus of elasticity is categorised into three types: Bulk Modulus, Young's Modulus, and Shear Modulus, each describing material behaviour under different stresses. This article focuses on Bulk Modulus, which measures a material's response to uniform compression, defined as the ratio of compressive stress to volumetric strain. Here, we will discuss the concept of Bulk modulus in detail, mathematical formulation, factors, and practical applications of Bulk Modulus in various fields.
What is Bulk Modulus?
Bulk Modulus, also known as the modulus of compressibility, is a constant that describes a material's ability to resist changes in its volume when subjected to uniform pressure. It is the ratio of compressive stress to the resulting volumetric strain. Bulk Modulus measures how much a material's volume decreases under a compressive force and how well it returns to its original volume once the force is removed.
When external pressure is applied, the material undergoes a change in volume, but once the pressure is released, it gradually recovers its initial volume. The higher the Bulk Modulus, the less compressible the material is, meaning it resists volume changes more effectively. Bulk Modulus is important for understanding how solids and liquids behave under pressure and plays a key role in material design and analysis across various industries.
- The S.I unit of Bulk Modulus is Pascal ( Pa) or N/m2
- The Dimensional Formula of Bulk Modulus is [ M1 L-1T-2]
The formula of Bulk modulus is,
K = -ΔP (/ΔV/V)
Where,
- K = Bulk modulus (Pascal)
- V = Actual volume of object (m3)
- ΔP = Change in pressure (Pascal)
- ΔV = Change in volume (m3)
⁛The negative sign in the expression indicates that the volume of the system decreases as pressure is applied.
Below is the derivation of the Bulk Modulus, which quantifies a material’s resistance to uniform compression.
By Hooke's Law,
Stress is directly proportional to Strain
Hydraulic stress α Volume strain
Hydraulic stress = B × Volume strain
B is the bulk modulus of elasticity, and the proportionality constant is B.
p = B × ΔV / V
∴ B = ΔP × V / ΔV
Factors Affecting Bulk Modulus
The Bulk Modulus of a material is influenced by several factors, including its inherent properties and the conditions under which it is subjected to pressure. These factors determine how resistant a material is to compression.
- Material Composition: Different materials have varying atomic structures, affecting their resistance to compression.
- Temperature: Higher temperatures generally reduce a material's Bulk Modulus, making it more compressible.
- Phase of the Material: Solids, liquids, and gases all have different Bulk Moduli due to their distinct molecular arrangements.
- Pressure: At very high pressures, the Bulk Modulus may change, as materials can undergo phase transitions or structural changes.
- Density: Materials with higher density typically exhibit higher Bulk Moduli, as their particles are more tightly packed.
Applications of Bulk Modulus
Bulk Modulus has various practical applications, especially in fields that involve the behavior of materials under pressure. Some of the key applications include:
- Engineering and Structural Design: Understanding Bulk Modulus helps engineers design materials and structures that can withstand high-pressure environments, such as dams, submarines, and aircraft.
- Geophysics: Bulk Modulus is used to study the Earth's interior, as it helps in understanding how seismic waves propagate through different layers of the Earth.
- Hydraulics and Fluid Mechanics: In fluid systems, Bulk Modulus is crucial for analyzing the compressibility of liquids and gases, ensuring the proper functioning of hydraulic systems.
- Material Science: It aids in the development of new materials by predicting how they will behave under different pressure conditions, crucial for manufacturing materials used in extreme environments.
- Oceanography: The Bulk Modulus of seawater is important for studying underwater acoustics and sonar technologies, as sound travels differently in water depending on its compressibility.
- Aerospace: In spacecraft and high-altitude aviation, Bulk Modulus helps to understand how materials behave under varying atmospheric pressures.
Sample Problems
Problem 1: What is the bulk modulus of elasticity of a liquid compressed in a cylinder from 0.0125 m3 volume at 80 N/cm2 pressure to 0.0124 m3 volume at 150 N/cm2 pressure?
Solution:
Given: ΔP = 150 - 80 = 70 N/cm2, ΔV = 0.0124 - 0.0125 = -0.0001 m3, V = 0.0125 m3
Since,
K = ΔP × V / ΔV
∴ K = 70 × 0.0125 / 0.0001
∴ K = 8750 N/cm2
Problem 2: If the pressure of the liquid is increased from 70 N/cm2 to 130 N/cm2, calculate the bulk modulus of elasticity. The amount of liquid in the container shrinks by 0.15 %.
Solution:
Given: ΔV = V × 0.15 /100 = 15V × 10-4 m3,
Volume of liquid = V m3, ΔP = 130 - 70 = 60 N/cm2
Volumetric Strain = 15 × 10-4
K = ΔP × V / ΔV
∴ K = 60 / 15 × 10-4
∴ K = 4 × 104 N/cm2
Problem 3: Assume that the change in pressure is 80 N/Cm2, that the object's actual volume is 0.128 m3, and that the bulk modulus is 7390 N/cm2. Calculate the volume change.
Solution:
Given: ΔP = 80 N/cm2,
K = 7390 N/cm2,
V = 0.128 m3
Since,
K = ΔP × V / ΔV
∴ ΔV = ΔP × V / K
∴ ΔV = 80 × 0.128 / 7390
∴ ΔV = 10.24 / 7390
∴ ΔV = 0.001384 m3
Problem 4: A rubber ball's volume falls by 0.1 percent when transported to a depth of 200 meters in a pool. The volume elasticity in N/m2 will be if the density of water is 1 × 103 kg/m3 and g = 10 m/s2.
Solution:
Given: ΔP = ρgh = 103 × 10 × 200 = 20000000
ΔV = 0.1, V = 100
Since,
K = ΔP × V / ΔV
∴ K = 20000000 × 100 / 0.1
∴ K = 2 × 109 N/m2
Problem 5: If the pressure change is 78 N/Cm2, the volume change is 0.128 m3, and the bulk modulus is 7390 N/cm2. Calculate the object's actual volume?
Solution:
Given: ΔP = 78 N/cm2, K = 6390 N/cm2
ΔV = 0.1 m3
Since,
K = ΔP × V / ΔV
∴ V = K × ΔV / ΔP
∴ V = 6390 × 0.1 / 78
∴ V = 8.192 m3
Problem 6: When a liquid is compressed in a cylinder from 0.0225 m3 volume at 50 N/cm2 pressure to 0.0124 m3 volume at 110 N/cm2 pressure, what is the bulk modulus of elasticity?
Solution:
Given: ΔP = 110 - 50 = 60 N/cm2
ΔV = 0.0124 - 0.0225 = -0.0101 m3
V = 0.0225 m3
Since,
K = ΔP × V / ΔV
∴ K = 60 × 0.0225 / 0.0101
∴ K = 133.66 N/cm2
Conclusion
The bulk modulus measures a substance's resistance to uniform compression, indicating how much pressure is needed to produce a change in volume. It is defined as the ratio of the change in pressure to the relative change in volume. A higher bulk modulus means the material is more resistant to compression and less prone to volume changes.
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