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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 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) \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.

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}

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}

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.

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}

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}

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 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) \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) = s_0 \dfrac{h (x-\ell)}{2 EJ} F \end{equation}

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}

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}

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 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) \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.

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}

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}

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}

Details. 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} Now you need to solve the problem

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 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) \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) = s_0 \dfrac{h (x-\ell)}{2 EJ} F \end{equation}

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}

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}