Derive the linked interpolation of Eq.(2.68) for the 2-noded Timoshenko beam element.
Eq.(2.68) : w=\frac{1}{2}\left ( 1-\xi \right )w_{1}+\frac{1}{2}\left ( 1+\xi \right )w_{2}+\left ( 1-\xi ^{2} \right )\frac{l^{(e)}}{8}\left ( \theta _{1}-\theta _{2} \right )
The starting point is the 3-noded Timoshenko beam element with node numbers 1,3,2 where node number 3 corresponds to the mid-node. The original quadratic interpolation for the deflection and the rotation is
w=\frac{1}{2}\left ( \xi -1 \right )\xi w_{1}+\left ( 1-\xi ^{2} \right )w_{3}+\frac{1}{2}\left ( \xi +1 \right )\xi w_{2}
\theta =\frac{1}{2}\left ( \xi -1 \right )\xi \theta_{1}+\left ( 1-\xi ^{2} \right )\theta_{3}+\frac{1}{2}\left ( \xi +1 \right )\xi \theta_{2}
The transverse shear strain is obtained by
\gamma _{xz}=\frac{\mathrm{d} w}{\mathrm{d} x}-\theta =\frac{2}{l^{(e)}}\left ( \xi -\frac{1}{2} \right )w_{1}-\frac{\varepsilon }{2}w_{3}+\frac{2}{l^{(e)}}\left ( \xi +\frac{1}{2} \right )w_{2}-\frac{1}{2}\left ( \xi -1 \right )\xi \theta _{1}-\left ( 1-\xi ^{2} \right )\theta _{3}-\frac{1}{2}\left ( \xi +1 \right )\xi \theta _{2}=\frac{w_{2}-w_{1}}{l^{(e)}}+\frac{2}{l^{(e)}}\xi \left ( w_{1}-2w_{3}+w_{2} \right )-\theta _{3}-\frac{\xi }{2}\left ( \theta _{1} -\theta _{2}\right )-\frac{\xi ^{2}}{2}\left ( \theta _{1}-2\theta _{3} +\theta _{2}\right )=0
Clearly, \gamma _{xz} should vanish for slender (Euler-Bernoulli) beams. This is achievable for the above interpolation if the linear and quadratic terms in \xi are zero and \gamma _{xz} is simply expressed as
\gamma _{xz}=\frac{w_{2}-w_{1}}{l^{(e)}}-\theta _{3}
The condition \gamma _{xz}=0 implies \frac{w_{2}-w_{1}}{l^{(e)}}=\theta _{3} , i.e. the average slope equals the rotation at the mid-node, which is a physical condition for slender beams.
The vanishing of the linear and quadratic terms in the original quadratic expression for \gamma _{xz} yields the following two conditions
\frac{2}{l^{(e)}}\left ( w_{1}-2w_{3}+w_{2} \right )-\frac{\theta _{1}-\theta _{2}}{2}=0
\theta _{1}-2\theta _{3}+\theta _{2}=0
From the above we obtain
w_{3}=\frac{w_{1}+w_{3}}{2}-\frac{\theta _{1}-\theta _{2}}{8}l^{(e)}
\theta _{3}=\frac{\theta _{1}+\theta _{2}}{2}
Substituting these values into the original quadratic interpolation gives, after some algebra
w=\frac{1}{2}\left ( 1-\xi \right )w_{1}+\frac{1}{2}\left ( 1+\xi \right )w_{2}+\left ( 1-\xi ^{2} \right )\frac{l^{(e)}}{8}\left ( \theta _{1}-\theta _{2} \right )
\theta =\frac{1}{2}\left ( 1-\xi \right )\theta _{1}+\frac{1}{2}\left ( 1+\xi \right )\theta _{2}
which is the linked interpolation (2.68) we are looking for. It can be verified that the above interpolation yields the constant transverse shear strain field of Eq.(2.68).