1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Determine the bulk modulus of the material

  1. Jan 7, 2017 #1
    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?
    thank you in advance
  2. jcsd
  3. Jan 7, 2017 #2
    If the linear strain is 0.001, what is the volumetric strain?
  4. Jan 7, 2017 #3
    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?
  5. Jan 7, 2017 #4
    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?
  6. Jan 8, 2017 #5
    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
  7. Jan 8, 2017 #6
    $$\Delta V=\frac{4}{3}\pi (r-\Delta r)^3-\frac{4}{3}\pi r^3\tag{1}$$
    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.
  8. Jan 8, 2017 #7
    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
  9. Jan 8, 2017 #8
    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 $$\Delta \ln V=3\Delta \ln r$$
  10. Jan 8, 2017 #9
    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
  11. Jan 8, 2017 #10
    The answer is 1.333 GPa or 1333 MPa
  12. Jan 8, 2017 #11
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted