0
$\begingroup$

The general derivation of the Flexural formula uses the pure bending case, where the distance from a curved section to the neutral axis is assumed to be constant even after bending, denoted by y. The Flexural formula, as we all know, is used to determine the bending stresses at a distance y from the neutral axis.

Now my question is that can we use this general Flexural formula to find the bending stresses at a distance y from the neutral axis, when a shear/transverse load is applied to it? Because I was thinking that the distance y from the neutral axis of that section (whose bending stresses I am trying to compute) will not remain the same after bending due to this transverse loading. So why do we still use it? [Because literally the derivation of the Flexural formula assumes that the distance from the neutral section to any of the section at distance y remains constant after bending in pure bending case].

$\endgroup$
1
  • 1
    $\begingroup$ Not looking down on you, but my sincere advice to you is: if you are a student, get a copy of "Mechanics of Materials" by Gere & Timoshenko (the best) from the school library; if you are a practitioner, buy a used copy that is available at Amazon.com, and review the relevant topics, as the derivation and transformation are quite lengthy, thus prone to make mistakes at the limit space thus as this forum. $\endgroup$
    – r13
    Nov 14, 2021 at 22:59

3 Answers 3

0
$\begingroup$

External shear stress parallel to neutral axis of a beam is known as eccentric column loading and can be handled several ways.

Some cases are delt with as P delta bending stress, some use secant formula.

But internal shear stress parralel to the beam axis is countered by shear flow and Mohr circle stresses.

$\endgroup$
0
$\begingroup$

If you meant the generalized flexural formula:

  • $\sigma_x = \dfrac{(M_y*I_z + M_y*I_{yz})z - (M_z*I_y + M_y*I_{yz})y}{I_y*I_z - I{yz}^2}$

then, "YES", the formula is applicable to all shapes of beams that follow Hook's Law, with the shape that is doubly symmetrical about its geometric axes (the neutral axes are coincident with the geometric axes) as a special case, for which the product of the moment of inertia $I_{yz} = 0$, and the formula is reduced to:

  • $\sigma_x = \dfrac{M_yz}{I_y} - \dfrac{M_zy}{I_z}$
$\endgroup$
3
  • $\begingroup$ I was not talking about the use of formula for various shapes. I was asking that can this formula be used in the case of transverse loading? Because this formula is literally derived from assumption of pure bending case for a beam. I was reading some documents where it was said that this formula can also be applicable to the case of transverse loading if stresses due to bending needs to be found. However, I don't know if bending due to pure bending and bending due to transverse loading could be assumed to be 'kinda' similar or not. $\endgroup$ Nov 15, 2021 at 8:33
  • $\begingroup$ Because according to my understanding, the distance y between the neutral section and a section located up or below this neutral axis, should not remain the same before and after the deformation due to bending arising from transverse loading. However, this distance will be the same before and after the deformation due to pure bending. And the flexural formula is literally derived from assuming this distance y remains constant before and after the deformation. So should this be applicable for the bending due to transverse loading case as well? I hope you understood my query. $\endgroup$ Nov 15, 2021 at 8:36
  • $\begingroup$ I understand your question - "How a formula derived from pure bending be used for other loading cases?". Well, again refer to the textbook, "Mechanics of Material",, see the chapter on "Shear Center", in which the effect on bending moment due to transverse loadings is explained. $\endgroup$
    – r13
    Nov 15, 2021 at 14:05
0
$\begingroup$

Even derivation in pure bending assumes small deformations. If you do not use this assumption, the transverse distances from neutral axis change also in pure bending. So if you take into account transverse shear assuming small deformations, the distances are assumed to be constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.