I was going through the basic assumptions made by Bernoulli beam theory. I realized that one of the most important assumption in this theory is that the deformations and rotations of the beams are supposed to be small, not large. I couldn't understand that how is this important and if it could invalidate the Bernoulli beam theory if they are not small? In some places, its written that it is to make the calculations easier and reach to a more understandable form, while some sites write that it is because if the deformations/rotations are large then significant shear deformations develop on the cross section. I don't know what is the actual reason behind it. I mean even if the shear deformations develop, then isn't it possible to super impose the deformations from Bernoulli beam equation onto the ones coming from shear? Or does it completely invalidates the Bernoulli beam theory results if deformations are large?

Plus, how would we decide that this deformation is small and this deformation is large?

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    $\begingroup$ Most beams are made out of strong stuff that we don't want to bend. $\endgroup$
    – Tiger Guy
    Commented Oct 26, 2021 at 1:41

3 Answers 3


Apart from the other answers another reason why the Bernoulli-Euler beam theory falls apart at large deformation is due to the approximation about the radius of curvature.

The Bernoulli-Euler beam theory is usually encountered in the following equation form:

$$EI w''(x) = M(x)\qquad\text{ or } \qquad w''(x) = \frac{M(x)}{EI }$$ where:

  • E: elastic modulus
  • I: second moment of area of the crosssection
  • w(x): the function of deflection of the beam along its length ($w''(x)$ is its second derivative wrt x)
  • M(x) : the function of the bending moment along the length of the beam x

However, $w''(x)$ is an approximation of the curvature of the beam $\kappa$ (and it's inverse, the radius of curvature $\rho$). i.e. the equation is normally:

$$ \kappa(x) = \frac{1}{\rho}= \frac{M(x)}{EI}$$

However, the curvature of a function f(x) is given by:

$$\kappa(x) = \frac{|f''(x)|}{(1+f'(x)^2)^{\frac{3}{2}}}$$

This is where the assumption of is important. For small deformations the $f'(x)$ is small, so $f'(x)^2)$ is even smaller, therefore it can be neglected. So:

$$\kappa(x) = \frac{|f''(x)|}{(1+f'(x)^2)^{\frac{3}{2}}}\rightarrow \kappa(x)=\frac{|f''(x)|}{(1+0)^{\frac{3}{2}}}\rightarrow \kappa(x)=\frac{|f''(x)|}{1}$$

and therefore:


for large displacements the above simplification cannot be assumed.

  • $\begingroup$ so what you mean is that we use small displacement assumption because the large deformations still has no relation to the development of shear deformations on the plane, but is just for the simplicity of mathematics behind it, right? $\endgroup$ Commented Oct 26, 2021 at 12:53
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    $\begingroup$ The shear deformations is another issue. however as you said in another answer you can superimpose the shear deformation. Regarding the simplicity that's one way to look at it, however I think it's more correct to say that if the assumption do not hold then the theory is invalidated. $\endgroup$
    – NMech
    Commented Oct 26, 2021 at 12:58
  • $\begingroup$ I will try to tell you my actual concern here. So, you explained very well why the actual Bernoulli equation cannot be used if deformations are large. Now, what I am confused about is if the deformations are large, then can we just use the complete equation of the curvature and try to equate it to the M(x)/EI and then solve for the displacements over a length of a beam, or we cannot use this approach at all and we then have to go for other methods, like Timoshenko beam theory (where it takes into concern the shear deformations as well). I hope you understood my query. $\endgroup$ Commented Oct 26, 2021 at 13:15
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    $\begingroup$ To be perfectly honest regarding the timoshenko beam theory I am not adequately familiar to comment. Regarding the complete curvature, I can comment that you can use --at least-- numerical methods to solve it iteratively (the problem is that you have $w'(x)$ and $w''(x)$ and you need to make an approximation). In some cases there might be some analytical solutions also. $\endgroup$
    – NMech
    Commented Oct 26, 2021 at 13:45

Shear deformation causes the cross-section to twist, thus the plane no longer remains plane, which is one of the cornerstones of the Bernoulli Beam Theory.

Also, the large deflection invalidates the solution of differential equations of deflection for a beam with small angles of rotation, which was solved with some simplifications. When the slopes and deflections become large, these simplifications are not valid, and the exact solution needs to be found. (See Text by Timoshenko on "Shear deformation; and large Deflection")

Deep Beam Deformation and Stresses Diagrams

enter image description here

  • $\begingroup$ So, finding the exact solution without using the simplifications is possible or not? If it is, then why do we need to include these simplifications in the first place when we can solve for it without incorporating these simplifications. $\endgroup$ Commented Oct 26, 2021 at 8:47
  • $\begingroup$ So your answered appeared as if the emergence of shear deformations (and hence the shear stresses) on the cross sections won't be a problem for large deflections, but its just that the simplified mathematics won't be valid for it. Is it correct? Or large deformations inherently brings in significant shear deformations, which needs to be taken into account now? $\endgroup$ Commented Oct 26, 2021 at 8:49
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    $\begingroup$ You should realize that large deflection and shear deformation are not directly related. Shear deformation is a concern of deep beams, as its name suggests, it does add to deflection, however, the focus of this phenomenon is more on the increase in shear and non-linear stress distribution across the cross-section rather than deflection. Note, the shear deformation is ignored in the non-deep beams as its influence is immaterial. Beam size is not a concern of large deflection problems, as it is usually related to the properties of the materia and loadings. $\endgroup$
    – r13
    Commented Oct 26, 2021 at 15:24
  • $\begingroup$ Except for the academic, "exact" solution is usually unnecessary for most of the practical concerns (especially prior to the computer age). Even today, it is prone to make mistakes by using advanced theories/solutions (especially computer programs) without a solid understanding of the underlay fundamentals. $\endgroup$
    – r13
    Commented Oct 26, 2021 at 15:37

Consider a point along the length of the beam at coordinate $(u_0, z)$ let's call the x of the point U and the Y of the neutral axis w.

After deformation:

$$U_{x,y}= (U_{0(x)} - z*sin(x)) $$, and

$$W_{x y}=W_{(x)}+z*cos\theta-\theta$$

refer to the diagram. $\theta$ is the rotation of the neutral axis.

Under the assumption of small deformation we have,

$Sin(\theta)\approx\theta \quad and\ cos(\theta)\approx1$

So the displacement simplifies to:

$U(x y)\approx U_{0(x)}-z\theta, \ and\quad W(x y)\approx W_{0(x)}$

And $\theta\approx \frac{dW(x)}{dx}$

That is the basis of our integration. If the angle is not small our assumptions are wrong.


Euler- Bernoulli beam

  • $\begingroup$ so what should we do if the deformations are not small? Use a different method? $\endgroup$ Commented Oct 26, 2021 at 13:17
  • $\begingroup$ @RameezUlHaq large deformations are solved by the Timoshenko method. $\endgroup$
    – kamran
    Commented Oct 26, 2021 at 14:23
  • $\begingroup$ but Timoshenko method involves shear deformation and hence shear stresses on the cross section. But I don't want that. I just want the large deformation effect to be taken into consideration. $\endgroup$ Commented Oct 26, 2021 at 14:49
  • $\begingroup$ this MIT course is useful, but it is for moderately large deflections. ocw.mit.edu/courses/mechanical-engineering/… $\endgroup$
    – kamran
    Commented Oct 26, 2021 at 14:59

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