# Determine the bulk modulus of the material

• electr
In summary, the conversation discusses how to determine the bulk modulus of a material based on changes in pressure and volume. The formula K=-Δp/ΔV/V0 is used, where Δp is the change in pressure, ΔV is the volume change due to pressure, and V0 is the original volume. The conversation also touches on the relationship between linear and volumetric strain and how to use it to calculate the volumetric strain for small strains. For large strains, a different formula is used: K=-ΔlnV/ΔP. The final answer is determined to be 1.333 GPa or 1333 MPa.
electr

## Homework Statement

A material is formed into a solid sphere and has a diameter of 100mm when at a
pressure of 2MPa. If the diameter of the sphere reduces by 0.1mm when the
pressure is increased to 6MPa, determine the bulk modulus of the material

## Homework Equations

K=− Δp/ΔV/V0 (Δp is pressure change,ΔV is volume change due to pressure,V0 is original volume)

## The Attempt at a Solution

volume of sphere 4/3*pi*r^3
so original volume or V0=4/3*pi.(50x10^-3)^3=5.23 m^3
and the volume with the new diameter and pressure (100mm-0.1mm) is 5.22 m^3
and change in pressure or Δp is 6MPa - 2MPa = 4MPa
so K=- 4X10^6/5.22-5.23/5.23= 2.092 GPa

is this correct or not even close?

electr said:

## Homework Statement

A material is formed into a solid sphere and has a diameter of 100mm when at a
pressure of 2MPa. If the diameter of the sphere reduces by 0.1mm when the
pressure is increased to 6MPa, determine the bulk modulus of the material

## Homework Equations

K=− Δp/ΔV/V0 (Δp is pressure change,ΔV is volume change due to pressure,V0 is original volume)

## The Attempt at a Solution

volume of sphere 4/3*pi*r^3
so original volume or V0=4/3*pi.(50x10^-3)^3=5.23 m^3
and the volume with the new diameter and pressure (100mm-0.1mm) is 5.22 m^3
and change in pressure or Δp is 6MPa - 2MPa = 4MPa
so K=- 4X10^6/5.22-5.23/5.23= 2.092 GPa

is this correct or not even close?
If the linear strain is 0.001, what is the volumetric strain?

i don't understand the question,linear is the ratio of the change in the length of a body to its initial length and volumetric for volume,is there any relationship between them?coudnt find something in my notes ,i would appreciate if u could explain me ,my attempt is completelly wrong and i need something else?another equation?

Algebraically, what is ##\frac{d(r^3)}{r^3}##? If ##\frac{dr}{r}=0.001##, what is ##\frac{d(r^3)}{r^3}## equal to? What is ##(1.001)^3-1## equal to? Do you see how all this relates to your problem?

Sorry still don't understand, ,the 0.001 is a random example or it has to do with the problem and if so how you get it?the volume for sphere is 4/3 *pi*r^3,how the dr/r is going to help?what do i need to do?or if you can send me a link to read so i can understand what you are trying to say, and thank you for your time

$$\Delta V=\frac{4}{3}\pi (r-\Delta r)^3-\frac{4}{3}\pi r^3\tag{1}$$
Factoring: $$\Delta V=\frac{4}{3}\pi [(r-\Delta r)^3-r^3]\tag{2}$$
But, using the binomial expansion: $$(r-\Delta r)^3=r^3-3r^2\Delta r+3r(\Delta r)^2-(\Delta r)^3\tag{3}$$
Substituting Eqn. 3 into Eqn. 2 yields: $$\Delta V=-\frac{4}{3}\pi [3r^2\Delta r-3r(\Delta r)^2+(\Delta r)^3]\tag{4}$$
Dividing by V yields:$$\frac{\Delta V}{V}=-\left[3\frac{\Delta r}{r}-3\left(\frac{\Delta r}{r}\right)^2+\left(\frac{\Delta r}{r}\right)^3\right]\tag{5}$$
In the limit of small ##\frac{\Delta r}{r}## (as in your problem), this becomes:$$\frac{\Delta V}{V}=-3\frac{\Delta r}{r}\tag{6}$$
You can see from this that, for small strains, the volumetric strain is equal to 3 times the linear strain. In your problem, the linear strain is -0.001, so the volumetric strain is equal to -0.003.

Wow nothing in my notes...so i divide the 4MPa with -0.003?the units are Gpa? the ΔV/V = - 3 Δr/r is general and i should use it or only for the small strains?for bigger ones what? And again thank you

electr said:
Wow nothing in my notes...so i divide the 4MPa with -0.003?the units are Gpa?
Yes, but don't forget the minus sign in the definition of K.
the ΔV/V = - 3 Δr/r is general and i should use it or only for the small strains?
Only for small strains.
for bigger ones what?
For large strains, $$K=-\frac{\Delta \ln V}{\Delta P}$$ and $$\Delta \ln V=3\Delta \ln r$$

No i won't forget the minus in the calculation,the units are Gpa?Really appreciate your help not only you helped me sovle the problem but learned something usefull

electr said:
No i won't forget the minus in the calculation,the units are Gpa?Really appreciate your help not only you helped me sovle the problem but learned something usefull
The answer is 1.333 GPa or 1333 MPa

Yes i got the same ,your Socratic method with the questions so i can get the right answers comfused me at first because i didnt know about these equations,good to know now,thank you again for your time,this is a very helpfull forum

## 1. What is the bulk modulus of a material?

The bulk modulus of a material is a measure of its resistance to changes in volume under pressure. It is the ratio of the change in pressure to the change in volume.

## 2. How is the bulk modulus determined?

The bulk modulus is determined through experiments where pressure is applied to a material and the corresponding change in volume is measured. This data is then used to calculate the bulk modulus using the formula: K = -V(dP/dV), where K is the bulk modulus, V is the initial volume, and dP/dV is the change in pressure over the change in volume.

## 3. What factors affect the bulk modulus of a material?

The bulk modulus of a material is affected by its chemical composition, density, and temperature. Generally, materials with stronger intermolecular forces and higher density have higher bulk moduli. Temperature can also affect the bulk modulus, with most materials becoming less stiff at higher temperatures.

## 4. How is the bulk modulus related to other material properties?

The bulk modulus is closely related to other material properties such as stiffness, compressibility, and elasticity. It is also related to the Young's modulus, which measures a material's resistance to changes in length under tension.

## 5. Why is the bulk modulus important?

The bulk modulus is an important material property for engineers and scientists, as it helps to determine a material's ability to withstand changes in pressure and volume. It is commonly used in the design of structures and materials for applications such as shock absorption, pressure vessels, and hydraulic systems.

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