Why shouldn't we use Bernoulli beam theory if deformations and rotations are large?

Question

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?

Answer 1

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

Answer 2

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:

$$\kappa(x)=|f''(x)|$$

for large displacements the above simplification cannot be assumed.

Answer 3

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.

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