Young's modulus in microscopic terms

Click For Summary
SUMMARY

The discussion focuses on deriving Young's modulus (Y) at the microscopic level using the force-separation curve for adjacent atoms in a solid. The formula established is Y = - (1/x_0) (df/dx) | (x=x_0), where x_0 represents the equilibrium separation of atoms. The force function f(x) is defined as f(x) = - (A/x)^7 + (B/x)^13, with A and B set to 1 angstrom (1E-10 m). The relationship between tensile stress and strain is expressed as E = F L_0 / (A_0 ΔL), indicating that Y is the gradient of the stress-strain graph.

PREREQUISITES
  • Understanding of Young's modulus and its significance in material science
  • Familiarity with force-separation curves in atomic physics
  • Knowledge of tensile stress and strain calculations
  • Basic grasp of differential calculus for interpreting derivatives
NEXT STEPS
  • Explore the derivation of Young's modulus from atomic interactions in solids
  • Study the implications of the Lennard-Jones potential in force-separation relationships
  • Investigate the relationship between atomic bond lengths and Young's modulus in different materials
  • Learn about the application of differential calculus in material deformation analysis
USEFUL FOR

Students and researchers in materials science, physicists studying atomic interactions, and engineers focusing on material properties and stress analysis.

Mathsboi
Messages
2
Reaction score
0

Homework Statement


By considering the force-separation curve for two adjacent atoms in a solid, f(x), show that the Young’s modulus can be expressed on the microscopic scale as:
[tex]Y = - \frac{1}{x_0} \frac{df}{dx}\right| |_{x=x_0}[/tex]
(the | is meant to go allt he way form the top to bottom of df/dx)
where [tex]x_0[/tex] is the equilibrium separation of the atoms

Homework Equations



[tex]f(x) = - \left(\frac{A}{x}\right)^7 + \left(\frac{B}{x}\right)^{13}[/tex]
(I'm assuming A and B should be 1 angstrom so 1E-10m)

The Attempt at a Solution


[tex]x_0[/tex] is found by doing f(x).dx = 0 to find where f(x) crosses the x axis.
[tex]E \equiv \frac{\mbox {tensile stress}}{\mbox {tensile strain}} = \frac{\sigma}{\varepsilon}= \frac{F/A_0}{\Delta L/L_0} = \frac{F L_0} {A_0 \Delta L}[/tex]
Y is the gradient of stress/straing graph

hmmmm
:(
 
Physics news on Phys.org
I can't exactly read your equations but a long time ago I remember doing an analysis based on a typical diamond lattice structure whereby you could show that Posson's ration came out correctly ( approximatley 1/3) if you assumed that the bond lengths stayed the same and only the bond angles were deformed.
 
Mathsboi said:

Homework Statement


By considering the force-separation curve for two adjacent atoms in a solid, f(x), show that the Young’s modulus can be expressed on the microscopic scale as:
[tex]Y = - \frac{1}{x_0} \frac{df}{dx}\right| |_{x=x_0}[/tex]
(the | is meant to go allt he way form the top to bottom of df/dx)
where [tex]x_0[/tex] is the equilibrium separation of the atoms

Homework Equations



[tex]f(x) = - \left(\frac{A}{x}\right)^7 + \left(\frac{B}{x}\right)^{13}[/tex]
(I'm assuming A and B should be 1 angstrom so 1E-10m)

The Attempt at a Solution


[tex]x_0[/tex] is found by doing f(x).dx = 0 to find where f(x) crosses the x axis.
[tex]E \equiv \frac{\mbox {tensile stress}}{\mbox {tensile strain}} = \frac{\sigma}{\varepsilon}= \frac{F/A_0}{\Delta L/L_0} = \frac{F L_0} {A_0 \Delta L}[/tex]
Y is the gradient of stress/straing graph

hmmmm
:(

What's [itex]E =F L_0/A_0 \Delta L[/itex] in differential form? What's a reasonable estimate for [itex]A_0[/itex] on the atomic scale?
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
41
Views
15K
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
29K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
14K
Replies
6
Views
2K