Derive the generalized constitutive matrix Dˆ‘ for an arbitrary position of the beam axis.

Question

Derive the generalized constitutive matrix [latex]\hat{D}{}' [/latex]for an arbitrary position of the beam axis.

Accepted Answer

Let us assume that the [latex] {x}'[/latex] axis does not coincide with the neutral axis and [latex]{y}'[/latex] , [latex]{z}'[/latex] are not principal inertia axes. Substituting the expression for the axial stress [latex]\varepsilon _{{x}'}[/latex] (Eq.(4.26)) into the integral for the resultant stress vector (Eq.(4.32)gives

[latex]\hat{\sigma {}'}=\int \int_{A}^{}\begin{Bmatrix}E\left ( \frac{\partial u_{0{}'}}{\partial {x}'} +z{}'\frac{\partial\theta _{x{}'}}{\partial {x}'}-{y}'\frac{\partial \theta _{x{}'}}{\partial {x}'}\right )\ G_{{y}'}\left ( \frac{\partial {v}'_{c}}{\partial x}-\theta _{{x}'} \right )\ G_{{z}'}\left ( \frac{\partial {w}'_{c}}{\partial x}-\theta _{{y}'} \right )\ {z}'E\left ( \frac{\partial u_{0{}'}}{\partial {x}'} +z{}'\frac{\partial \theta _{x{}'}}{\partial {x}'}-{y}'\frac{\partial \theta _{x{}'}}{\partial {x}'}\right )\ \end{Bmatrix}dA[/latex]

where [latex]D_{t}[/latex] is defined in Eq.(4.33). Simple algebra gives

[latex]\hat{{\sigma}' }=\hat{{D}'}\hat{{\varepsilon }'}[/latex] with [latex]\hat{{D}'}=\int \int_{A}^{}\begin{bmatrix}E &0 &0 &{z}'E &{-y}'E &0 \ 0 &G_{{y}'} &0 &0 &0 &0 \ 0 &0 &G_{{z}'} &0 &0 &0 \ {z}'E &0 &0 &{z}'^{2}E &-{y}'{z}'E &0 \ -{y}'E &0 &0 &-{y}'{z}'E &{y}'^{2}E &0 \ 0 &0 &0 &0 &0 &D_{t} \end{bmatrix}[/latex]

with [latex]{\hat{\varepsilon} }'[/latex] given by Eq.(4.27b).

We see that [latex]\hat{{D}'}[/latex] is now a full (still symmetric) matrix. This implies that axial and flexural effects are coupled, i.e. an axial force induces flexural (bending and shear) effects and viceversa. The off-diagonal terms in [latex]\hat{{D}'}[/latex] vanish if [latex]{x}'[/latex] is the neutral axis and [latex]{y}'[/latex] ,[latex] {z}'[/latex] are the principal inertial axes .
In practice, either the full form of [latex]\hat{{D}'}[/latex] given above or the diagonal form of Eq.(4.35a) can be used and both yield identical results. The simpler diagonal form requires the “a priori” computation of the position of the neutral axis and the principal inertia axes.

Above considerations do not affect the value of the torsional stiffness [latex]\hat{D_{t}}[/latex].

\(Eq.(4.26)): [latex]{\varepsilon }'=\begin{Bmatrix}\varepsilon _{{x}'}\ \gamma _{{x}'{y}'}\ \gamma _{{x}'{z}'}\end{Bmatrix}=[/latex]

[latex]\overset{\begin{Bmatrix}\frac{\partial {u}'_{0}}{\partial x}\ 0\ 0\end{Bmatrix}+\begin{Bmatrix}{z}'\frac{\partial \theta _{{y}'}}{\partial {x}'}\ 0\ \frac{\partial {w}'_{c}}{\partial {x}'+\theta _{{y}'}}\end{Bmatrix}+\begin{Bmatrix}-{y}'\frac{\partial \theta _{{z}'}}{\partial {x}'}\ \frac{\partial {v}'_{c}}{\partial {x}'}-\theta _{{z}'}\ 0\end{Bmatrix}+\begin{Bmatrix}0\ \left ( \frac{\partial w}{\partial {y}'}-\left ( {z}'-{z}'_{c} \right ) \right )\frac{\partial \theta _{\hat{x}}{}'}{\partial x}\ \left ( \frac{\partial w}{\partial {z}'}-\left ( {y}'-{y}'_{c} \right ) \right )\frac{\partial \theta _{\hat{x}}{}'}{\partial x}\end{Bmatrix}}{\begin{matrix}axial & bending in &bending in && & & &free torsion \ & plane{x}'{z}' &plane{x}'{y}' & & & & & \end{matrix}}[/latex]

(Eq.(4.32)): [latex]\hat{\sigma}{}'=\iint_{A}^{}\begin{Bmatrix}E\varepsilon _{{x}'}\ G_{{y}'}\left ( \frac{\partial {v}'_{c}}{\partial {x}'}-\theta _{{x}'} \right )\ G_{{z}'}\left ( \frac{\partial {w}'_{c}}{\partial {z}'}-\theta _{{y}'} \right )\ {z}'E\varepsilon _{{x}'}\ -{y}'E\varepsilon _{{x}'}\ D_{t}\end{Bmatrix}dA[/latex]

(Eq.(4.33)): [latex]D_{t}=\begin{bmatrix}G_{{z}'}\left ( \frac{\partial w}{\partial {z}'}+{y}'-{y}'_{c} \right )\left ( {y}'-{y}'_{c}\right )-G_{{y}'}\left ( \frac{\partial w}{\partial {y}'}+{z}'-{z}'_{c} \right ) \left ( {z}'-{z}'_{c} \right )\end{bmatrix}\frac{\partial \theta _{\hat{x}{}'}}{\partial {x}'}[/latex]

(Eq.(4.27b)): [latex]\hat{\varepsilon} {}'=\begin{bmatrix}\frac{\partial {u}'_{c}}{\partial {x}'},\left ( \frac{\partial {v}'_{c}}{\partial {x}'}-\theta _{{z}'} \right ),\left ( \frac{\partial {w}'_{c}}{\partial {x}'}-\theta _{{y}'} \right),\frac{\partial \theta _{{y}'}}{\partial {x}'}, \frac{\partial \theta _{{z}'}}{\partial {x}'},\frac{\partial\theta _{{\hat{x}}'}}{\partial {x}'}\end{bmatrix}^{T}[/latex]

(Eq.(4.35a)): [latex]\hat{D}{}'=\begin{bmatrix}\hat{D}_{a} &\vdots & 0 &0 &0 &0 &\vdots &0 \ \cdots &\cdots &\cdots &\cdots &\cdots &\cdots &\cdots &\cdots \ 0& \vdots &\hat{D}_{s_{{y}'}} &0 &0 &0 &\vdots &0 \ 0 &\vdots &0 &\hat{D}_{s_{{z}'}} &0 &0 &\vdots &0 \ 0 &\vdots &0 & 0 &\hat{D}_{b_{{y}'}} & 0 &\vdots &0 \ 0 &\vdots &0 &0 &0 &\hat{D}_{b_{{z}'}} &\vdots &0 \ \cdots &\cdots &\cdots &\cdots &\cdots &\cdots &\cdots &\cdots \ 0 &\vdots &0 &0 &0 &0 &\vdots &\hat{D_{t}} \end{bmatrix}=\begin{bmatrix}\hat{D}_{a} &0 &0 \ 0 &\hat{D{}'}_{f} & 0\ 0&0 &\hat{D}_{t} \end{bmatrix}[/latex]

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