Determine the bulk modulus of the material

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Discussion Overview

The discussion revolves around determining the bulk modulus of a material formed into a solid sphere, based on changes in pressure and diameter. Participants explore the relationship between linear strain and volumetric strain, as well as the application of relevant equations in the context of the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant presents a calculation for the bulk modulus using the formula K=− Δp/ΔV/V0, providing specific values for pressure change and volume change.
  • Another participant questions the relationship between linear strain and volumetric strain, seeking clarification on their connection.
  • A participant introduces the concept of using the binomial expansion to derive the relationship between changes in radius and volume, suggesting that for small strains, volumetric strain is three times the linear strain.
  • Further discussion includes the application of the derived relationship for small strains and inquiries about its validity for larger strains.
  • Participants express uncertainty about the calculations and the implications of the derived equations, particularly regarding units and signs in the context of the bulk modulus.
  • Some participants confirm calculations and express appreciation for the assistance received in understanding the problem and related concepts.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical relationships discussed, but there remains uncertainty regarding the application of these relationships for larger strains and the correctness of initial calculations. No consensus is reached on the final value of the bulk modulus, as different calculations yield varying results.

Contextual Notes

Participants note limitations in their understanding of the relationships between linear and volumetric strains, as well as the applicability of certain equations under different strain conditions. There is also mention of potential confusion arising from the use of a Socratic method in the discussion.

electr
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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?
thank you in advance
 
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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?
thank you in advance
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
 
  • #10
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
 
  • #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
 

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