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Solid mechanics

  • Thread starter Dell
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  • #1
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now i am looking for the stress [tex]\sigma[/tex] for the aluminium and for the steel
i would want to say simply [tex]\sigma[/tex]=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 dont know the strains
also how do i take the volume into account? the only equation i have for volume is [tex]\Delta[/tex]=[tex]\epsilon[/tex]x + [tex]\epsilon[/tex]y + [tex]\epsilon[/tex]z, but i doubt thats going to help
 

Answers and Replies

  • #2
Mapes
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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?
 
  • #3
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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

[tex]\sigma[/tex]=[tex]\epsilon[/tex]*E

now since the strains are the same

[tex]\sigma[/tex]Al=[tex]\epsilon[/tex]*70*109
[tex]\sigma[/tex]Steel=[tex]\epsilon[/tex]*200*109

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

[tex]\sigma[/tex]Al*2*10-3 + [tex]\sigma[/tex]Steel*1.5*10-3=385*103



thanks for the help
 
  • #4
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does that mean that the stress has nothing to do with the height of the element
 
  • #5
Mapes
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That's what it looks like.
 

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