Derive the 2-noded Euler-Bernoulli beam element by imposing the condition of zero transverse shear strain over the cubic/quadratic Timoshenko beam element of Figure 2.10.
Question : Derive the 2-noded Euler-Bernoulli beam element by imposing ...
The transverse shear strain is obtained as
\gamma _{xz}=\frac{\mathrm{d} w}{\mathrm{d} x}-\theta =A+B\xi +C\xi ^{2}The condition \gamma _{xz} = 0 over the element is satisfied if
A = B = C = 0
This leads to a system of three equations which allows us to eliminate the internal nodal DOFs w_{2} , w_{3} and \theta _{5} as
A=0 \Rightarrow \theta _{5}=\frac{1}{8l^{(e)}}\left ( w_{1}-27w_{2}+27w_{3}-w_{4} \right ) B=0 \Rightarrow w_{2}=w_{1}-w_{3}+w_{4}+\frac{2l^{(e)}}{9}\left ( \theta _{1}-\theta _{4} \right )Substituting these values into the equation for C = 0 gives
w_{3}=\frac{1}{27}\begin{bmatrix}7w_{1}+20w_{4}+2l^{(e)}\left ( \theta _{1}-2\theta _{4} \right )\end{bmatrix}and consequently
w_{2}=\frac{1}{27}\begin{bmatrix}20w_{1}-7w_{4}+2l^{(e)}\left ( \theta _{1}-2\theta _{4} \right )\end{bmatrix}Substituting w_{2} and w_{3} into the original cubic deflection field gives
w=\frac{1}{4}\left ( \xi ^{3}-3\xi +2 \right )w_{1}+\frac{1}{4}\left ( 2+3\xi -\xi ^{3} \right )w_{4}+\frac{l^{(e)}}{8}\left ( \xi ^{2}-1 \right )\left ( \xi -1 \right )\theta _{1}+\frac{l^{(e)}}{8}\left ( \xi ^{2}-1 \right )\left ( \xi +1 \right )\theta _{4}=N_{1}w_{1}+\bar{N}_{1}\theta _{1}+N_{4}w_{4}+\bar{N}_{4}\theta _{4}where N_{1} , \overline{N}_{1} , N_{4} , \overline{N}_{4} coincide with the Hermite shape functions for the 2-
noded Euler-Bernoulli beam element (Eqs.(1.11a)). On the other hand, the condition \gamma _{xz}=0 implies \theta =\frac{\mathrm{d} w}{\mathrm{d} x} over the element and the deflection field can be written as
It can clearly be seen that the element has C^{1} continuity and it coincides with the 2-noded Euler-Bernoulli beam element of Section 1.3.