How Much Does a Marble Column Compress Under Load?

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SUMMARY

The discussion focuses on calculating the change in length (ΔL) of a marble column under a specified load. Using the formula ΔL = (FL) / (AY), where F is the force, L is the height, A is the cross-sectional area, and Y is Young's modulus, the correct calculation yields ΔL = 9.44 x 10^-4 meters. The initial confusion arose from the incorrect conversion of the cross-sectional area from 45 cm² to 0.0045 m². The compressive strength of 1.8 x 10^8 Pa was mentioned as a potential factor for assessing the column's structural integrity.

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  • Understanding of Young's modulus and its application in material science.
  • Knowledge of basic mechanics, specifically force and area calculations.
  • Familiarity with unit conversions, particularly between cm² and m².
  • Basic principles of structural engineering and material strength.
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  • Study the implications of compressive strength in structural materials.
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Homework Statement



A marble column with cross sectional area of 45cm^2 supports a load of 8.5 x 10^4 Newtons.
The marble has a Young's modulus of 5.5 x 10^10 N/m^2 and a compressive strength of 1.8 x 10^8 Pa. If the column is 2.75m high, what is the change in length of the column?


Homework Equations


Y(ΔL/L) = F/A

Where Y is young modulus.

The Attempt at a Solution



Y(ΔL/L) = F/A
Solve the equation for ΔL and you get.
ΔL = (FL)/ (AY)
So plugging in the values you get.
ΔL = ((8.5 x 10^4 N)(2.75m)) / ((.45m^2)(5.5 x 10^10 N/m^2))
So ΔL = 9.44 x 10^-6
Can someone double check me? I don't understand why my instructor gave me this number
1.8 x 10^8 Pa.
Please confirm or deny my answer. Thanks very much.
 
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45cm^2 is not .45m^2.
Can someone double check me? I don't understand why my instructor gave me this number
1.8 x 10^8 Pa.
Maybe to check if the column breaks down.
 
OK. So with that correction

ΔL = ((8.5 x 10^4 N)(2.75m)) / ((0.0045 m^2)(5.5 x 10^10 N/m^2))
So ΔL = 9.44 x 10^-4

Is this correct now? Thanks for the help.
 

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