Derive the beam element of the previous example starting from the standard quadratic Timoshenko beam element in Section 2.7.
Question : Derive the beam element of the previous example starting fro...
The curvature and transverse shear strain fields for the 3-noded quadratic Timoshenko beam element of Figure 2.7 are
\kappa =\frac{\mathrm{d} \theta }{\mathrm{d} x}=\frac{\theta _{3}-\theta _{1}}{l^{(e)}}+\frac{2}{l^{(e)}}\left ( \theta _{1}-2\theta _{2}+\theta _{3} \right )\xi
\gamma _{xz}=\frac{\partial w}{\partial x}-\theta =\frac{1}{l^{(e)}}\left ( w_{3}-w_{1}-l^{(e)}\theta _{2} \right )+\frac{1}{l^{(e)}}\left ( 2w_{1}-4w_{2}+2w_{3}+\frac{l^{(e)}}{2}\theta _{1}-\frac{l^{(e)}}{2}\theta _{3} \right )\xi +\left ( \theta _{2}-\frac{\theta _{1}}{2}-\frac{\theta _{3}}{2} \right )\xi ^{2}=A+B\xi +C\xi ^{2}
Note that the bending strain field coincides with that obtained in the previous example.
The condition \gamma _{xz}= constant requires B = 0 and C = 0. From the later we deduce \theta _{2}=\frac{\theta _{1}+\theta _{3}}{2} . Substituting this value into the interpolation for the bending strain gives \frac{\mathrm{d} \theta }{\mathrm{d} x}=\frac{\theta _{3}-\theta _{1}}{l^{(e)}} , and therefore the required linear bending distribution can not be obtained.
The alternative is to impose \gamma _{xz}= A+C\xi ^{2}. The condition that must be
satisfied now is
B=0\Rightarrow w_{2}=\frac{w_{1}+w_{3}}{2}+\frac{l^{(e)}}{8}\left ( \theta _{1}-\theta _{3} \right )
The deflection field is
w=\frac{1}{2}\left ( 1-\xi \right )w_{1}+\frac{1}{2}\left ( 1+\xi \right )w_{3}+\frac{l^{(e)}}{8}\left ( \theta _{1}-\theta _{3} \right )
=\begin{bmatrix}\frac{1}{2}\left ( 1-\xi \right ) , \frac{l^{(e)}}{8},\frac{1}{2}\left ( 1+\xi \right ), -\frac{l^{(e)}}{8}, 0\end{bmatrix}a^{(e)}
with a^{(e)}=\begin{bmatrix}w_{1},\theta _{1},w_{3},\theta _{3},\theta _{2}\end{bmatrix}^{T} , and the shear strain distribution is
\gamma _{xz}=\frac{w_{3}-w_{1}}{l^{(e)}}-\left (\frac{\theta _{1}+\theta _{3}}{2} \right )\xi ^{2}-\left ( 1-\xi ^{2} \right )\theta _{2}
It is interesting that at \xi = \pm \frac{1}{\sqrt{3}} the transverse shear strain is
\left ( \gamma _{xz} \right )_{\xi } =\pm \frac{1}{\sqrt{3}}=\frac{w_{3}-w_{1}}{l^{(e)}}-\frac{\theta _{1}+4\theta _{2}+\theta _{3}}{6}
which coincides with the constant transverse shear strain obtained in the previous example. Therefore, an “effective” constant transverse shear field can be obtained by using a two-point Gauss quadrature for the shear stiffness terms. This coincidence is a consequence of the properties of the Gauss points; i.e. the constant transverse shear distribution is the least square approximation of the quadratic field chosen here and both fields take the same value at the two Gauss points (Section 6.7 of [On4]).
