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Euler-Bernoulli beam equation is given by

$$ EI \frac{\mathrm d^2 u}{\mathrm d x^2} = M'(x) \\ EI \frac{\mathrm d u}{\mathrm d x} = xM'(x) + C_1 $$ Where, $E$ is modulus, $I$ is second moment of area, $u$ is displacement as function of $x$, $M'(x)$ is (moment) loading.

From classical beam bending theory, $$ \sigma = \frac{My}{I} \\ \epsilon = \frac{\sigma}{E}=\frac{My}{EI} $$ $\sigma$ is normal stress, $M$ is bending moment, $y$ is distance of fiber from neutral axis, $I$ is second moment of area, $\epsilon$ is normal strain, $E$ is elastic modulus.

I am confused here. With infinitesimal strain theory (http://en.wikipedia.org/wiki/Infinitesimal_strain_theory)

$$ \epsilon = \frac{du}{dx} \implies \epsilon = \frac{xM'(x) + C_1}{EI} $$ (from second equation)

Which strain relation is correct? I am messing up somewhere, but couldn't find.

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  • $\begingroup$ I want to know whether both fourth and fifth equations are correct or only one of them is correct. And importantly, why. $\endgroup$ – gyeox29ns Nov 5 '14 at 2:56
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The "vertical" (lateral) co-ordinate has two different meanings. In the static Euler Bernoulli equations (your first set), $u$ is the lateral displacement of the centre of the beam from its zero strain position, i.e. its position when there are no stresses or forces on the beam. Since the beam cross section is assumed not to deform, $u$ is the lateral displacement of any point in the cross section from its no-strain position.

The second set analyses the stress in a given cross section as a function of position in the cross section. Here $y$ is not the lateral displacement but is instead the orthogonal distance of the point in question from the neutral axis for the cross section.

EB beam theory models the beam as a bundle of fibres with no friction and only normal forces between them. If there is a constant moment on the beam, the fibres bend into circular arcs, and the shape of any fibre is indeed an arc of a circle. On the "inside" edge of the beam, i.e. that nearest to the axis made up of all the centres of curvature of the fibres, the fibres are in compression; on the other edge they are in tension. There is an axis through the cross section along which all of the fibres are in a zero strain state. This is the neutral axis.

One thing you seem to be having trouble grasping is that there is no one strain for each cross section. The strain varies over the cross section, as my derivation below should show. There is zero strain on the neutral axis of each cross section. So the fifth equation would be wrong even if the second were right.

Now for your equations:

The third and fourth are correct.

The first is incorrect. It should be $M(x) =-E\,I\,\frac{{\rm d}^2\, u(x)}{{\rm d}\,x^2}$, where $u(x)$ is the lateral displacement of any point in a cross section from its zero strain position. I shouldn't worry about the difference in sign too much as long as you are consistent (although the negative sign shows the wonted convention), but the second derivative of $u$ is the moment, not its first derivative. You can derive this equation as follows: for small gradients, $\frac{{\rm d}^2\, u(x)}{{\rm d}\,x^2}\approx \kappa(x)=\frac{1}{R(x)}$ is approximately the curvature $\kappa(x)$ of the beam as measured at the neutral axis. One radian of the circular beam arc at this point therefore has a length $R(x)$. This is the length of the beam as measured at the neutral axis, and therefore this represents an unstrained length. The length of one radian of the fibre which an orthogonal distance $y$ from the neutral axis is $(R + y)$. Therefore the strain, the fractional change relative to the unstrained length, is $((R + y)/R) - 1=y\,/\,R$. The stress normal to the cross section at this point is therefore $E\,y/R$. We then integrate the moment of this stress across the whole beam's cross section to get:

$$M(x) = -\frac{E}{R(x)}\,\int_\mathscr{S}\,y^2\,{\rm d} A = -\frac{E\,I}{R(x)}\tag{1}$$

As we have seen, $1/R(x)\approx \frac{{\rm d}^2\, u(x)}{{\rm d}\,x^2}$, whence the correct equation:

$$M(x) = -E\,I\,\frac{{\rm d}^2\, u(x)}{{\rm d}\,x^2}\tag{2}$$

Since your first equation is wrong, so is the second.

Now, the fourth equation is correct as follows. As we have seen, the strain $\epsilon(x,\,y)$ an orthogonal distance $y$ from the neutral axis is $\frac{y}{R(x)}$, where $R(x)$ is the radius of curvature of the beam (at the neutral axis). We have just seen (eq (1)) that $\frac{1}{R(x)} = -\frac{M}{E\,I}$ therefore $\epsilon(x,\,y)=\frac{y}{R(x)} =-\frac{M(x)\,y}{E\,I}$, whence your fourth equation follows. The third simply follows by multiplying the strain $\epsilon(x,\,y)$ by the Young's modulus to get the stress.

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  • $\begingroup$ What about the last equation? Is it correct? $\endgroup$ – gyeox29ns Nov 5 '14 at 1:16
  • $\begingroup$ @gyeox29ns Please see my updated answer, beginning at "One thing you seem to be having trouble grasping ..." $\endgroup$ – WetSavannaAnimal Nov 5 '14 at 6:56

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