Calculating Young's Modulus with Force and Extension Measurements

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SUMMARY

This discussion focuses on calculating Young's Modulus using force and extension measurements. The relevant formulas include stress (σ = F/A) and strain (ε = e/l), leading to Young's Modulus (E = σ/ε). Participants clarify the importance of using specific data points from the provided table and suggest plotting a graph for better visualization. The area is specified as 50mm by 5mm, which is crucial for accurate calculations.

PREREQUISITES
  • Understanding of stress and strain concepts
  • Familiarity with Young's Modulus calculations
  • Basic knowledge of graph plotting techniques
  • Ability to interpret force and extension data
NEXT STEPS
  • Learn how to plot stress-strain graphs for material properties
  • Study the derivation and application of Young's Modulus
  • Explore the significance of area in stress calculations
  • Investigate the relationship between force, extension, and material elasticity
USEFUL FOR

Students in physics or engineering, materials scientists, and anyone involved in mechanical testing or material property analysis will benefit from this discussion.

ACCC12
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hi

i have been given these results in order to work out the young's modulus can any 1 tell me how i go about doing this

Force (kN) Extension (mm)
0.5 0.003
1 0.011
1.5 0.019
2 0.021
2.5 0.024
3 0.032
3.5 0.040
4 0.043
4.5 0.050
5 0.060
5.5 0.064
6 0.070
6.5 0.076
7 0.081

Area 50mm by 5

thanks for all your help
 
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E=\frac{stress}{strain}


Write the formula for stress and the formula fopr strain and then simply the formula for Young's Modulus,E.
 
i know the equation but i didnt know which part of the data to use, do i just take any part of the data from the table or do i have to plot a graph! thanks for all you help!
 
ACCC12 said:
i know the equation but i didnt know which part of the data to use, do i just take any part of the data from the table or do i have to plot a graph! thanks for all you help!

what is meant by "area 50mm by 5 " in your data ? please explain.
 
E=\frac{\frac{F}{A}}{\frac{e}{l}}


E=\frac{Fl}{Ae}

Right?


From your graph, what does the fraction F/e represent?
 
50mm length, 5mm diameter
 
got the answer to be 5.3 is that correct!

equation used

2.5 x 50
----------
981.75 x 0.024
 
any takers! thanks
 

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