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TENSION / COMPRESSION

TENSION / COMPRESSION. y. x. N= P. z. M x =0, M y =0, M z =0. Formal definition: the case when set of internal forces reduces to the sum which is tangent to the bar axis. N ≠ 0 , Q y =0 , Q z =0. Example: a straight bar loaded by concentrated forces at its ends (truss strut or tie). N.

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TENSION / COMPRESSION

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  1. TENSION / COMPRESSION

  2. y x N= P z Mx=0, My=0, Mz=0 Formal definition: the case when set of internal forces reduces to the sum which is tangent to the bar axis N ≠ 0, Qy=0, Qz=0 Example: a straight bar loaded by concentrated forces at its ends (truss strut or tie) N P P  P P M=P· N(x)=P

  3. Early experiments R.Hooke (1635-1703),„De Potentia Restitutiva” ,1678 E.Mariotte (1620-1684)

  4. Modern testing machine

  5. K’ B’ C’ D’ E’ Experimental approach K B C D E P x For constant P Axial displacement is linear function of x! Normal strain is constant along xaxis!

  6. K’ B’ C’ D’ E’ Experimental approach Bernoulli hypothesis K B C D E P Axial displacement does not depend on y i z variables Normal strain does not depend on y i z variables x For constant P Normal stress is constant over the whole cross section Axial displacement is linear function of x! Normal strain is constant function of x!

  7. y z Distribution of Distribution of Experimental approach x N The condition of equivalence of internal cross-sectional forces with its sum yields:

  8. y Experimental approach For This is NOT a full solution z w v

  9. BVP approach Solution validity range De Saint-Venant hypothesis: As stress tensor is given in principal axes with x axis which coincides with the cross-section centre of gravity thus axes y and z – are arbitrary orthogonal axes with origin at the centre of gravity Z Poisson coefficient

  10. BVP approach BVP approachmakes it easy to notice that at any other than principal axes the angular stresses arise . The maximum value of these appears – as we know from 3D strain analysis – on planes inclined by 45o to principal axes. These stresses are equal to half of the difference between two consecutive principal normal stresses, whereas the normal stress is equal to the half of sum of normal stresses there.

  11. 2 · · 3 1 3 2 P P 1

  12. BVP approach BVP solution demonstrates also that for the case of non-prismatic bar the stress tensor as found for prismatic bar does not satisfy Static Boundary Conditions if the side surface of the bar is free of traction. Therefore, the normal stresses in x,y,z, axes are not the principal stresses.

  13. z y x BVP approach ? P P SBC ?

  14. stop

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