Solid Mechanics: Calculating Stress for Aluminium & Steel

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SUMMARY

The discussion focuses on calculating stress (\sigma) for aluminum and steel using the formula \sigma=F/A while considering material properties such as elasticity. The participants reference Hooke's Law and the relationship between stress and strain, specifically \sigma=\epsilon*E, where E represents the modulus of elasticity. They establish that the strains for both materials are the same and derive equations based on the forces acting on each material, concluding that stress is independent of the height of the element.

PREREQUISITES
  • Understanding of Hooke's Law and its application in solid mechanics
  • Knowledge of stress and strain relationships in materials
  • Familiarity with the modulus of elasticity for aluminum (70 GPa) and steel (200 GPa)
  • Basic principles of force distribution in structural components
NEXT STEPS
  • Study the application of Hooke's Law in different materials
  • Learn about the calculation of stress in composite materials
  • Research the effects of cross-sectional area on stress distribution
  • Explore advanced topics in solid mechanics, such as finite element analysis (FEA)
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Mechanical engineers, materials scientists, and students studying solid mechanics who need to understand stress calculations for different materials under load.

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now i am looking for the stress \sigma for the aluminium and for the steel
i would want to say simply \sigma=F/A but i need to somehow take into account the types of materials and their elasticity,
i think that hookes law could help me here but i don't know the strains
also how do i take the volume into account? the only equation i have for volume is \Delta=\epsilonx + \epsilony + \epsilonz, but i doubt that's going to help
 
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You know a little about the strains, since the rigid top plate supplies a constraint on the components' deformation. What can you say about the strain of the steel and aluminum?

Each component will have its own stress, cross-sectional area, and force. Can you relate the forces to the applied load?
 
i would think that the aluminium and steel have the same strains.
i can find the ratio of forces on each material as their area of the rigid plate? is this correct?

F(Al) = 4P/7
F(Steel)= 3P/7

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can i say that since i only have forces on the y axis

\sigma=\epsilon*E

now since the strains are the same

\sigmaAl=\epsilon*70*109
\sigmaSteel=\epsilon*200*109

now i have t equations and 3 unknowns, but i know

\sigmaAl*2*10-3 + \sigmaSteel*1.5*10-3=385*103



thanks for the help
 
does that mean that the stress has nothing to do with the height of the element
 
That's what it looks like.
 

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