Derive a beam element with linear bending and constant transverse shear fields starting from the cubic/quadratic Timoshenko beam element of Figure 2.10.
Question : Derive a beam element with linear bending and constant trans...
Figure 2.10 shows the original element and the cubic N_{i}^{4} and quadratic N_{i}^{3} shape functions for the deflection and the rotation, respectively. The quadratic rotation field automatically guarantees a linear bending field. The constant transverse shear field is obtained as follows. From the original displacement approximation the transverse shear strain is found as
\gamma _{xz}=\frac{\partial w}{\partial x}-\theta =\sum_{i=1}^{4}\frac{\partial N_{i}^{4}}{\partial x}w_{i}-N_{1}^{3}\theta _{1}-N_{4}^{3}\theta _{4}-N_{5}^{3}\theta _{5}=A+B\xi +c\xi ^{2}
with
A=\frac{1}{8l^{(e)}}\left ( w_{1}-27w_{2}+27w_{3}-w_{4}-8l^{(e)}\theta _{5} \right )
B=\frac{9}{4l^{(e)}}\begin{bmatrix}w_{1}-w_{2}-w_{3}+w_{4}+\frac{2l^{(e)}}{9}\left ( \theta _{1}-\theta _{4} \right )\end{bmatrix}
C=\frac{27}{8l^{(e)}}\begin{bmatrix}-w_{1}+3w_{2}-3w_{3}+w_{4}-\frac{4l^{(e)}}{27}\left ( \theta _{1}+\theta _{4} \right )+\frac{8l^{(e)}}{27}\theta _{5}\end{bmatrix}
For \gamma _{xz} be constant it is required that
B = C = 0
These conditions lead to two equations from which two nodal DOFs can be eliminated. Selecting the intermediate deflections w_{2} and w_{3} gives
w_{2}=\frac{2w_{1}+w_{4}}{3}+\frac{l^{(e)}}{81}\left ( 11\theta _{1}-7\theta _{4}-4\theta _{5} \right )
w_{3}=\frac{w_{1}+2w_{4}}{3}+\frac{l^{(e)}}{81}\left ( 7\theta _{1}-1\theta _{4}-4\theta _{5} \right )
Substituting w_{2} and w_{3} into the original cubic field for w yields
w=\frac{1}{2}\left ( 1-\xi \right )w_{1}+\frac{1}{2}\left ( 1+\xi \right )w_{4}+\frac{l^{(e)}}{24}\left ( 2\xi ^{3}-2\xi -3\xi ^{2}+3 \right )\theta _{1}+\frac{l^{(e)}}{24}\left ( 2\xi ^{3}-2\xi +3\xi ^{2}-3 \right )\theta _{4}+\frac{l^{(e)}}{6}\xi \left ( 1-\xi ^{2} \right )\theta _{5}=\sum_{i=1}^{5}N_{i}a_{i}=N^{(e)}a^{(e)}with
a^{(e)}=\begin{bmatrix}w_{1}, &\theta _{1}, &w_{4}, &\theta _{4}, & \theta _{5}\end{bmatrix}^{(e)}
whereas the original quadratic interpolation is kept for the rotation. Note that the new interpolation for the deflection involves also the nodal rotations. This is another example of “linked” interpolation similar to those described in Section 2.8.3. It can be verified that the rigid body condition \left ( \sum_{i=1}^{5}N_{i}=1\right ) still holds in this case.
We can easily check that the required conditions are satisfied, i.e.
\kappa =\frac{\mathrm{d} \theta }{\mathrm{d} x}=\frac{2}{l^{(e)}}\left ( 2\xi -1 \right )\theta _{1}-\frac{4\xi }{l^{(e)}}\theta _{4}+\frac{1}{l^{(e)}}\left ( 2\xi +1 \right )\theta _{5}=\frac{1}{l^{(e)}}\left ( \theta _{5}-\theta _{1} \right )+\frac{2}{l^{(e)}}\left ( \theta _{1}-2\theta _{4}+\theta _{5} \right )\xi =\frac{1}{l^{(e)}}\begin{bmatrix}0,\left ( 2\xi -1 \right ),0,-4,\left ( 2\xi +1 \right )\end{bmatrix}a^{(e)}=B_{b}a^{(e)}
\gamma _{xz}=\frac{\mathrm{d} w}{\mathrm{d} x}-\theta =\frac{w_{4}-w_{1}}{l^{e}}-\frac{\theta _{1}+4\theta _{5}+\theta _{4}}{6}=\begin{bmatrix}-\frac{1}{l^{(e)}}, \frac{1}{6},\frac{1}{l^{(e)}}, -\frac{1}{6},\frac{2}{3}\end{bmatrix}a^{(e)}=B_{s}a^{(e)}