# Determine the bulk modulus of the material

1. Jan 7, 2017

### electr

1. The problem statement, all variables and given/known data

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

3. 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?

2. Jan 7, 2017

### Staff: Mentor

If the linear strain is 0.001, what is the volumetric strain?

3. Jan 7, 2017

### electr

i dont 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?

4. Jan 7, 2017

### Staff: Mentor

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?

5. Jan 8, 2017

### electr

Sorry still dont 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

6. Jan 8, 2017

### Staff: Mentor

$$\Delta V=\frac{4}{3}\pi (r-\Delta r)^3-\frac{4}{3}\pi r^3\tag{1}$$
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.

7. Jan 8, 2017

### electr

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

8. Jan 8, 2017

### Staff: Mentor

Yes, but don't forget the minus sign in the definition of K.
Only for small strains.
For large strains, $$K=-\frac{\Delta \ln V}{\Delta P}$$ and $$\Delta \ln V=3\Delta \ln r$$

9. Jan 8, 2017

### electr

No i wont 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

10. Jan 8, 2017

### Staff: Mentor

The answer is 1.333 GPa or 1333 MPa

11. Jan 8, 2017

### electr

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