Bodies in Equilibrium, elasticity and fracture

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SUMMARY

The pressure required to compress a steel block by 0.14 percent is calculated using the bulk modulus of steel, which is approximately 160 GPa. The formula applied is dP = G(dV/V), resulting in a pressure change (dP) of 224 MPa. This pressure is significant as it approaches the yield stress of low carbon steels, which is around 220 MPa, beyond which permanent deformation occurs. The elastic behavior of steel allows it to return to its original volume once the pressure is removed, except when yield stress is exceeded.

PREREQUISITES
  • Understanding of bulk modulus and its application in material science
  • Familiarity with stress-strain relationships in materials
  • Knowledge of SI units, particularly pressure measurements in Pascals
  • Basic principles of elasticity and plastic deformation in metals
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  • Research the properties and applications of low carbon steel in engineering
  • Explore the concept of yield stress and its implications in material selection
  • Learn about the relationship between pressure and volume in elastic materials
  • Investigate the effects of different types of steel on compressive strength and deformation
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pupatel
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can anyone help me with this please...? :eek:

How much pressure is needed to compress the volume of a steel block by 0.14 percent?
 
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I'm not sure what units you are used to but I'll use SI here :

The bulk modulus of steel is about
[tex]G=160 * 10^9 N/m^2 = dP/(dV/V)[/tex]
[tex]=>dP = G(dV/V)[/tex]
[tex]=160*10^9*0.0014[/tex]
[tex]=224 * 10^6 N/m^2[/tex]
[tex]= 224 MPa[/tex]

Of course, this change in volume exists only so long as the pressure is applied, and the steel springs back once the pressure is removed. However, in a very small number of low carbon steels, at about 220 MPa, you reach the yield stress and the steel will begin to plastically deform. Once this starts, the volume change becomes permanent.
 
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