Young's Modulus: Difference between revisions
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==The Main Idea== | ==The Main Idea== |
Revision as of 08:37, 2 April 2017
This page discusses Young's Modulus and examples of how it is used.
In 2017: Claimed by Ali Azadi (aazadi3)
The Main Idea
Young's Modulus is a macroscopic property of a material that measures how stretchy a solid material is. It is independent of size or weight, and it will change depending on the material. The uses of Young's modulus extend to two main sets of relationships, macroscopic springs and microscopic springs. In the case of macroscopic springs, Young's modulus is a measure of the stretchiness of a solid material outside of considerations of size and shape. In microscopic strings, when a solid object is modeled as a system of balls (atoms) connected by springs (an image of which can be seen as the cover image of the physics textbook Matter and Interactions I 3rd Edition), the Young's modulus constant of the material can be used to determine the 'interatomic spring stiffness' constant Ksi of a material in order to determine the stiffness and stretchiness of interatomic bonds.
A Mathematical Model
The definition of Young's Modulus can be expressed as: [math]\displaystyle{ {Y = \frac{stress}{strain}} = \frac{\frac{{F}_{T}}{A}}{\frac{{ΔL}}{L_o}} }[/math] where [math]\displaystyle{ {F}_{T} }[/math] is equal to the tension force, [math]\displaystyle{ A }[/math] is equal to the cross sectional area, [math]\displaystyle{ ΔL }[/math] is equal to the change in length due to the tension force, and [math]\displaystyle{ L_o }[/math] is equal to the initial length of the material.
Young's Modulus can also be expressed as: [math]\displaystyle{ {Y = \frac{Ksi}{d}} }[/math] where Y is the Young's modulus of the material being examined, Ksi is the value that represents the interatomic bond stiffness (how much a bond between two atoms will stretch), and d is the distance from the center of one atom to the center of another atom.
A Computational Model
It's hard to demonstrate Young's Modulus through programming, but this photo does a great job of demonstrating the concept. A force causes a solid material to stretch by a constant certain amount. This relationship is named Young's Modulus and is independent of mass of the object, and varies based on material.
Examples
Be sure to show all steps in your solution and include diagrams whenever possible
Simple
Example 1. A cylinder of wood has a stress of 800 and a strain of [math]\displaystyle{ 8*10^-7 }[/math]. What is Young's modulus for wood?
First we lay out the equation for the problem:
[math]\displaystyle{ Y = \frac{stress}{strain} }[/math]
We then plug in using the numbers given to us.
[math]\displaystyle{ Y = \frac{800}{8*10^-7} }[/math]
and so Y = 10^9 N/m^2
Example 2.
The Young's Modulus of Collagen in Bone is about 6 GPa [1]. Given this, determine the force applied to a cylindrical segment of collagen of radius 1 cm and length 0.75 m to cause it to deform (stretch) 1 mm.
First, consider the equation of Young's modulus most useful for this problem:
[math]\displaystyle{ {Y = \frac{stress}{strain}} = \frac{\frac{{F}_{T}}{A}}{\frac{{ΔL}}{L_o}} }[/math]
Substitute the appropriate values:
[math]\displaystyle{ 6000000000 = \frac{\frac{{F}_{T}}{3.14*0.01^2}}{\frac{{0.001}}{0.75}} }[/math]
Solve for F:
[math]\displaystyle{ {F = 2.55*10^{10} Newtons} }[/math]
Does this seem reasonable? Yes! because collagen that makes up bone is a very hard material, and does not stretch easily. Making bone stretch 1 mm longitudinally (using the bone as a vertical spring system) will take a lot of force.
Middling
Example 1.
A flan created by Dr. Schatz has a strawberry placed on it, stretching the flan from a length of 0.15 m to 0.2 m. The flan has a cross sectional area of .01. With the knowledge that flan has a Young’s modulus of ~ 1.6e4 in tension, what force was used to stretch the flan?
First we would write out the equation for Youngs modulus: [math]\displaystyle{ {Y = \frac{stress}{strain}} = \frac{\frac{{F}_{T}}{A}}{\frac{{ΔL}}{L_o}} }[/math]
Next we would fill in the equation for Young's Modulus: [math]\displaystyle{ {1.6e4 = \frac{\frac{{F}_{T}}{.01}}{\frac{{.05}}{.15}}} }[/math]
We can then fill in every thing that we know: [math]\displaystyle{ {16000 = 300*F_T} }[/math]
And so,[math]\displaystyle{ {F_T} }[/math] is equal to 53.3!
Example 2.
A man of weight 100 kg gets onto a bungee jump ride at a carnival. He is suspended in air by one rubber band (Young's Modulus of 0.01 GPa [2] of diameter 5 cm and original length of 10 m. Calculate the stretch of the band when the man reaches the bottom of the ride.
First, use the appropriate Young's Modulus equation:
[math]\displaystyle{ Y = \frac{\frac{{F}_{T}}{A}}{\frac{{ΔL}}{L_o}} }[/math]
Then, substitute the appropriate values:
[math]\displaystyle{ {1e7 = \frac{{\frac{100*9.8}{3.14*0.05^2}}}{{\frac{ΔL}{10}}}} }[/math]
Solve:
[math]\displaystyle{ {ΔL=.125 m} }[/math]
Difficult
Example 1.
A new cylandrical flan is created by Dr. Schatz which has an orange placed on it which has a mass of .15 kg, compressing the flan a certain amount. Knowing that the initial length of the flan was .15 m and the flan has a diameter of .2 meters. With the knowledge that flan has a Young’s modulus of ~ 2e5 in tension, what length is the flan now?
First we need to write down the equation:
First we would write out the equation for Youngs modulus: [math]\displaystyle{ {Y = \frac{stress}{strain}} = \frac{\frac{{F}_{T}}{A}}{\frac{{ΔL}}{L_o}} }[/math]
Next, let's solve for the unknowns: Cross sectional area:
[math]\displaystyle{ {A} = pi*r^2 }[/math] [math]\displaystyle{ {A} = pi*.1^2 }[/math] [math]\displaystyle{ {A} = .0314 meters^2 }[/math]
Force of the orange:
[math]\displaystyle{ {F}_{T} = m*g }[/math] [math]\displaystyle{ {F}_{T} = .15*-9.8 }[/math] [math]\displaystyle{ {F}_{T} = -1.47 Newtons }[/math]
Next we would write out the equation for Youngs modulus: [math]\displaystyle{ {2e5 = \frac{\frac{-1.47}{.0314}}{\frac{{ΔL}}{.15}}} }[/math]
We can solve the equation for : [math]\displaystyle{ {-7/ΔL = 200000} }[/math]
And so,[math]\displaystyle{ {ΔL } }[/math] is equal to -3.5e-5! It was compressed 3.5e-5 meters!
Example 2.
Using the Young's modulus of Tungsten ([math]\displaystyle{ 4*10^11 }[/math][3]) determine the interatomic "spring" stiffness of Tungsten.
First, find the center-to-center distance between 2 tungsten atoms.
[math]\displaystyle{ \frac{\frac{184 grams/mole}{6.022*10^23 atoms/mol}}{d^3}=19.25 g/cm^3 }[/math] where d is the interatomic distance between Tungsten atoms. (19.25 g/cm^3 is the density of Tungsten[4]).
Solving, we get:
[math]\displaystyle{ d = 2.51e-8 }[/math]
Using the appropriate formula:
[math]\displaystyle{ {Y = \frac{Ksi}{d}} }[/math]
[math]\displaystyle{ {4*10^11 = \frac{Ksi}{2.51e-8}} }[/math]
[math]\displaystyle{ {Ksi = 10040 N/m} }[/math]
Connectedness
- Young's modulus is connected to all solid material, and it highlights the slight impression given to everything showing that things do push down ever so slightly on stuff that seems stationary.
- My major is CS, so not a ton in physics directly applies to it, but Young's modulus is guaranteed to be in every physics simulation built which attempts to simulate the building integrity and usage of materials.
- Young's modulus is used all the time in civil engineering and it is often used to help determine structural integrity of certain materials when deciding on a building.
- Young's modulus has several biomedical applications in prosthetics and in human disease. It is used to determine the structural characteristics of prosthetic material used for implants. Young's modulus of tissues changes with aging and is being studied as a factor to evaluate mortality related to vascular stiffness from aging.
History
Young's Modulus was first developed in 1727 by the famous Leonhard Euler in Switzerland, but it was further expanded upon by Italian scientist Giordano Riccati in 1782. Finally, it was given a name by the British Scientist Thomas Young who finished work on it in the 1800s. It is used in order to discover the elasticity of solid materials and shows the stress per strain of a solid material.
See also
Further reading
The VERY in-depth wiki page which goes for beyond applications in physics 1.[5]
External links
HyperPhysics[6] which is a great tool for just about any entry level physics.
References
Image: [7]
History: [8]