# Bulk Modulus of Elasticity Definition Formula

## Introduction

Bulk modulus is one of the most important properties in all of materials science. It’s defined as the ratio of stress to strain within a material and can be used to calculate things like how much stress it takes to fracture the material or how much strain it will undergo when you apply a given amount of force to it. If you’re working with any type of elastic object in your day-to-day life, then you’ll definitely want to brush up on this important material property.

## Bulk Modulus of Elasticity

Before I go into detail on what bulk modulus of elasticity is, it's important that we first go over some basic terminology. The elastic modulus refers to how much a material can stretch without breaking or deforming. A material with a higher elastic modulus will tend to be harder, more rigid, and less flexible than one with a lower one.

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Generally speaking, materials with high values of elastic moduli will experience large amounts of stress in response to small deformations while those with low values will see small deformations when stressed by large amounts of force.

With that said, we can now take a look at bulk modulus of elasticity. The value of the bulk modulus of elasticity is denoted by the K symbol and the unit per square inch (psi) or newtons per square meter (N/m2).

For example, gold has a bulkmodu[s]s that measures approximately 71 gigapascals (GPas), while PVC has less than 1 GPas. Depending on application needs, you may need something more flexible or something more rigid...like solid gold!

StatesSolidsBulk modulus ($10^9 Nm^{-2}$ or GPa)
Solid

Aluminum

Brass

Copper

Glass

Iron

Nickel

72

61

140

37

100

260

Liquid

Water

Ethanol

Carbon-disulfide

Glycerine

Mercury

2.2

0.9

1.56

4.76

25

GasesAir10*10-4

## Use Bulk Modulus of Elasticity in Daily Life

• When dealing with substances that are in a solid-state, such as ice, you can use a bulk modulus of elasticity. This gives a good approximation of whether or not you will be able to damage or crack the ice. If it has a high bulk modulus of elasticity, then it is unlikely that you will break it.

• If you are designing a shoe or handbag, it is important that you know how easily it will stretch. This is something that designers regularly use to make use of the bulk modulus of elasticity.

In order for bulk moduli of elasticity to be used effectively in your life, though, there are several things that you need to know about it first. First and foremost is how simple it really is.

Secondly, though, you must know what exactly bulk moduli are in order for them to work properly. If they know at an early stage that there are going to be issues with elasticity, they can change their design and avoid wasted time, energy, and resources.

## Bulk Modulus of Elasticity Formula

The bulk modulus of elasticity is shown by the ratio of pressure applied to the corresponding relative reduction in the volume of the material.

The bulk modulus of elasticity in mathematical format

$\mathrm{B \:= \:\Delta P /(\Delta V/V)}$

Where:

• B represents the bulk modulus

• $\mathrm{\Delta P}$ represents the change in pressure that is appliance on the material

• $\mathrm{\Delta V}$ represents the change in volume due to applied pressure

• V represents the initial volume of material

The bulk modulus of elasticity is a measure of stiffness. For rubber, it represents how hard you have to push on an object in order for it to flex.

## Examples of Bulk Modulus of Elasticity

Example 1. Calculate the bulk modulus of the elasticity of the medium that changes from 1.01 x 105 Pa to 1.165 * 105 Pa under applied pressure and at the constant temperature change in volume is keeping 10%.

Solution:

As per bulk modulus,

$\mathrm{B \:= \:\Delta P /(\Delta V/V)}$

Where,

• B = Bulk Modulus of elasticity =?

• $\mathrm{\Delta P}$ = Change in pressure

• $\mathrm{\Delta V/V}$ = Change in Volume

Given, $\mathrm{\Delta P}$ = p2 - p1

$\mathrm{\Delta P}$ = (1.165 * 105 - 1.01 * 105) Pa = 0.155 * 105Pa

and $\mathrm{(\Delta V/V)}$ = 10/100 = 0.1

So,

B or K = 0.155 * 105/0.1

= 1.55 * 105

= 1.6 * 105.Pa

Example 2: Find the change in the pressure on a litre of water in which volume is changed by 0.10%? The bulk modulus of water = 2.2 * 109Pa

Solution:

As we know

Bulk modulus $\mathrm{B \:= -\:\Delta P /(\Delta V/V)}$

Given, $\mathrm{(\Delta V/V)}$ = 0.10% (compression) = -0.10/100 =-1.0 * 10-3

$\therefore$ change in pressure, $\mathrm{\Delta p = B(-\Delta V/V)}$

= 2.2 x 109(1.0 x 10-3)

= 2.2* 106pa.

## FAQs

Q1. What is the effect of temperature on bulk modulus?

Ans: In general, bulk modulus increases with increasing temperature. At low temperatures, water and other liquids are less compressible than they are at higher temperatures because of intramolecular forces such as hydrogen bonding between neighboring molecules. But above a certain temperature range called thermal depolymerization, or melting point, bulk modulus starts decreasing again (because of intermolecular interactions) as it approaches another liquid-gas phase transition line at about 32 degrees Celsius for water.

Q2. Why is bulk modulus negative?

Ans: Simply put, the bulk modulus is a measure of how much something expands under pressure. In other words, it’s a measure of elasticity and deformation under pressure. The higher its bulk modulus, the stiffer an object will be in response to applied stress. That stiffness causes it to resist deformation under load.

Q3. Which is more elastic water or air?

Ans: The quick answer is that water is more elastic than air. But let’s go into a little more detail about why it’s true and why it matters. In any material, elasticity refers to how much its shape or size changes when a force is applied. Air, for example, is less elastic than water: put your hand in a glass of water (or stand outside in a rainstorm) and you’ll feel how much your hand stretches when you move it around.

Q4. Does bulk modulus increase with pressure?

Ans: Generally speaking, yes, bulk modulus does increase with pressure. A good rule of thumb is that for every 2.5 times greater pressure you apply, the bulk modulus will rise by a factor of 10 times.