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B E N D I N G

B E N D I N G. y. x. z. M x =0, M y ≠ 0 , M z =0. Formal definition: the case when set of internal forces reduces solely to the moment vector which is perpendicular to the bar axis. N=0, Q y =0 , Q z =0. N=0, Q y =0 , Q z =0. M y.

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B E N D I N G

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  1. B E N D I N G

  2. y x z Mx=0, My ≠ 0, Mz=0 Formal definition: the case when set of internal forces reduces solely to the moment vector which is perpendicular to the bar axis N=0, Qy=0, Qz=0 N=0, Qy=0, Qz=0 My Example: a straight bar loaded by concentrated moments applied at its ends. M My(x)=M=const M Qz(x)=0 „Pure” bending My= M or Mx=0, My =0, Mz ≠ 0 N=0, Qy=0, Qz=0

  3. Remarks on terminology

  4. y y x x z z NORMAL (proste) INCLINED (ukośne) 90o <90o 90o <90o Mx=0, My ≠ 0, Mz=0 Mx=0, My ≠ 0, Mz ≠0

  5. y y x x z z PURE (czyste) IMPURE („nie-czyste”) NON-UNIFORM (poprzeczne) Mx=0, My ≠ 0, Mz=0 Mx=0, My ≠ 0, Mz=0 N=0, Qy=0, Qz=0 N=0, Qy=0, Qz ≠0

  6. End of remarks

  7. EXPERIMENTAL approach E.Mariotte (1620-1684) Galileo (1564-1642)

  8. P uD wD x z EXPERIMENTAL approach Jacob Bernoulli (1654-1705) Galileo (1564-1642) D h D’ l u is linear function of z ! M=M(x) Mx=0, My ≠ 0, Mz=0 is linear function of z and does not depend on x if M=const|x Q=Q(x) N=0, Qy=0, Qz ≠0 For h<<l shear forces can be neglected N=0, Qy=0, Qz =0

  9. y x z Continuum Mechanics application tension compression Hooke law:

  10. z y My Continuum Mechanics application z x ?

  11. z y My Equilibrium conditions z x ? y-axis is the central inertia axis of cross-section area y-z axes are central principal inertia axesof cross-section area

  12. Continuum Mechanics application Axes x – which coincides with bar axis y,z – which are central principal inertia axes of the bar cross-section area are principal axes of strain and stress matrices

  13. z y My Pure plane bending z Neutral axis x Wy is called section modulus where For z=0 (i.e. along y-axis) there is and section of y-axis within bar cross-section is called neutral axis (for normal stress and strain) Neutral axis coincides with only non-zero bending moment component My

  14. Important remarks 1. All above formulas are valid only for principal central axes of cross-section inertia 2. If moment vector coincides with any of two principal axes we have to deal with plane bending. If this is not the case – we have to deal with inclined bending and derived formulas cannot be used. 3. Bar axis (x-axis) is one of the principal axis of strain and stress matrices. As two remaining principal stress and strains are equal therefore any two perpendicular axes lying in the plane of bar cross-section are also principal axes. 4.The neutral axis for normal stress and strain coincides with bending moment vector.

  15. stop

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