Young's modulus in microscopic terms

In summary, the Young's modulus can be expressed as Y = - (1/x_0) * (df/dx)|x=x_0, where x_0 is the equilibrium separation of atoms in a solid. This is derived from the force-separation curve, f(x), and the equation f(x) = - (A/x)^7 + (B/x)^13. To find x_0, f(x) is set equal to 0 and solved for x. Additionally, the Young's modulus can also be calculated as the gradient of the stress/strain graph, or E = F L_0/A_0 \Delta L. A reasonable estimate for A_0 on the atomic scale is approximately 1 ang
  • #1
Mathsboi
3
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
:(
 
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  • #2
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.
 
  • #3
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?
 

FAQ: Young's modulus in microscopic terms

1. What is Young's modulus in microscopic terms?

Young's modulus, also known as the modulus of elasticity, is a measure of a material's stiffness or resistance to deformation when subjected to an applied force. In microscopic terms, it is a measure of the strength of the bonds between atoms in a material.

2. How is Young's modulus calculated at the microscopic level?

Young's modulus is calculated by dividing the stress (force per unit area) by the strain (change in length per unit length) of a material under an applied force. At the microscopic level, it is calculated by measuring the atomic forces and displacements within a material.

3. What factors influence Young's modulus at the microscopic level?

The primary factors that influence Young's modulus at the microscopic level include the type of atoms present, their arrangement in the material, and the strength of their bonds. Other factors such as temperature and impurities can also affect the modulus of a material.

4. How does Young's modulus relate to a material's mechanical properties?

Young's modulus is a fundamental mechanical property of a material and is closely related to its other mechanical properties, such as tensile strength, compressive strength, and shear modulus. It provides insight into how a material will respond to external forces and deformations.

5. Why is Young's modulus important in materials science?

Young's modulus is important in materials science because it allows us to compare and predict the behavior of different materials under stress. It is also a crucial parameter in the design and engineering of structures and products, as it determines the maximum amount of stress a material can withstand before it deforms or breaks.

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