Computing the strain in a cantilever beam under a known deflection

Question

I intend to use a cantilever beam to calibrate a strain gauge, in a setup similar to the picture.

My idea is to introduce a small known vertical displacement $\delta$ on one side of the cantilever through a micrometric screw and to compute the strain in the upper surface of the cantilever to compare it with the strain gauge's resistance variation.

I have found that the deflection at any point $x$ along the span of an end-loaded cantilever beam can be calculated by:

$$y\left(x\right)=\frac{F}{6EI}x^2\left(x-3L\right) \tag{1}$$

with $F$ a force acting on the tipo of the beam, $L$ the span of the beam, $E$ the modulus of elasticity, and $I$ the area moment of inertia of the beam's cross-section. So, to find the variation in length, I initially thought naively about finding the arc length integrating the profile $y(x)$ of the beam:

$$s = \int _0 ^{\tilde{L}} \sqrt{1+(y'(x))^2}dx \tag{2}$$

However, I am unsure about this approach, since, as I understand it, equation (1) would give the deflection of the neutral axis of the beam, and it wouldn't account for the strains that are present in the upper and lower surfaces due to stresses present in those surfaces. Moreover, the upper limit of the integral, $\tilde{L}$, is unknown.

Therefore, what could be an approach to compute the strain in the external surfaces of the beam from the displacement $\delta$ on the free side?

Answer 1

Short answer. For a elastic beam described by Euler-Bernoulli model, the internal bending moment can be related to the second derivative of the transverse displacement of the beam, through the elastic constant and the inertia of the section \begin{equation} M(x) = EJ w''(x) \ . \end{equation} In your problem you can easily evaluate the internal bending moment with the equilibrium of the section of the truss including the free end, \begin{equation} M(x) = - F (x - \ell) \ . \end{equation} For beams with symmetric sections, it's possible to relate axial strain $\varepsilon(x,y)$ and the maximum displacement $\varepsilon_{max}$ (the one you measure with the strain gauge) with $w''(x)$ and the thickness $h$ of the section \begin{equation} \varepsilon_{max}(x) = \dfrac{h}{2} w''(x) \qquad , \qquad \varepsilon(x,y) = \frac{2 y }{h} \varepsilon_{max}(x) \end{equation} Keep using this first term theory, the elongation of the strain gauge then reads \begin{equation} \Delta s = s - s_0 = s_0 \, \varepsilon_{max}(x) = s_0 \dfrac{h}{2} w''(x) = s_0 \dfrac{h}{2 EJ} M(x) \ , \end{equation} so that the link between the elongation and the external force eventually reads \begin{equation} \Delta s = s_0 \dfrac{h (\ell-x)}{2 EJ} F \ , \end{equation} being $s_0$ the initial length of the strain gauge, $h$ the section thickness, $EJ$ the bending stiffness and $F$ the external force. The factor $\ell-x$ is the arm lever of the force, and tells you that the best point to put the strain gauge for this static measure (the point providing the best sensitivity) is close to the root, where $\ell - x \sim \ell$. Usually it's better to leave some space from the clamped end, to avoid local stress due to the constraint that are hard to predict and rapidly fade as you move away from these points along the beam.

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Some details. If the Euler-Bernoulli beam model is good enough for your problem, here it is an answer.

A truss subject to bending, with the constraints and forcing of your problem is governed by the problem \begin{equation} \begin{aligned} & EJ w'''' = 0 \qquad \qquad x \in [0,\ell] \\ & w(0) = 0 \\ & w'(0) = 0 \\ & EJw''(\ell) = 0 \qquad \text{(no external moment at the free end)} \\ & EJw'''(\ell) = F \qquad \text{(external force $F$ at the end)} \end{aligned} \end{equation} while the displacement and stress distribution is provided by a linear distribution (for a symmetric section, with thickness $h$), \begin{equation} \varepsilon_z = \varepsilon_{max} \dfrac{2 y}{h} \qquad \sigma_z = E \varepsilon_z = E\varepsilon_{max} \dfrac{2 y}{h} \ . \end{equation} It's possible to compute the maximum stress and thus the maximum displacement from the equivalence of the internal moment and the resultant of the stress distribution, \begin{equation} M = \int_A y \sigma_z dA = \int_A 2 \dfrac{y^2}{h} \sigma_{max} dA = 2 \dfrac{J}{h} \sigma_{max} = 2 \dfrac{EJ}{h} \varepsilon_{max} \end{equation} Using the constitutive equation of a linear elastic medium, the internal bending moment reads $M = EJ w''$, and thus you can find the relationship between the maximum strain $\varepsilon_{max}(z)$ (what you measure with the strain gauge on the surface) and the displacement of the meanline $w(z)$ of the beam, \begin{equation} \varepsilon_{max}(z) = \dfrac{h}{2} w''(z) \end{equation} Once you have the solution in terms of the displacement \begin{equation} w(x) = \dfrac{F}{6EJ} x^2(x-3\ell) \ , \end{equation} you could find \begin{equation}\begin{aligned} w'(x) & = \dfrac{F}{6EJ} \left(3 x^2 - 6 x \ell \right) \\ w''(x) & = \dfrac{F}{EJ} \left( x - \ell \right) \end{aligned}\end{equation}

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Some extra details about Euler-Bernoulli beam. In static conditions,

• the indefinite equilibrium for bending for a beam with distributed moment load $m(x)$ reads $M' = m$, being $M$ the internal bending moment, $M = EJ \theta' = EJ w''$,
• shear equilibrium reads $T' = f$, with distributed force $f(x)$
• relation between bending moment and shear force reads $M' = T$ so that you can put everything together to get the equation \begin{equation} f = T' = M'' = (EJ w'')'' = EJ w'''' \ . \end{equation}