Derive a 2-noded beam element by imposing the condition of zero transverse shear strain at the two Gauss points \xi =\pm \frac{1}{\sqrt{3}} in the quadratic Timoshenko beam element. Verify that the stiffness matrix of the new element coincides with that of the 2-noded Euler-Bernoulli beam element.
Question : Derive a 2-noded beam element by imposing the condition of z...
The transverse shear strain distribution for the quadratic Timoshenko beam element can be seen in Example 2.4 and in Eq.(2.62). The condition of zero transverse shear strain at the two Gauss points \xi =\pm \frac{1}{\sqrt{3}} is written as
\left (\gamma _{xz} \right )_{\xi }=\frac{1}{\sqrt{3}}=0 and \left (\gamma _{xz} \right )_{\xi }=-\frac{1}{\sqrt{3}}=0
Eq.(2.62):B_{s}=\begin{bmatrix}\frac{dN_{1}}{d\xi }\frac{d\xi}{dx}, -N_{1} , \frac{dN_{2}}{d\xi }\frac{d\xi}{dx} , -N_{2} ,\frac{dN_{3}}{d\xi }\frac{d\xi}{dx}, – N_{3}\end{bmatrix}= \frac{2}{l^{(e)}}\begin{bmatrix}\xi -\frac{1}{2},-\frac{l^{(e)}}{4}\left ( \xi ^{2}-\xi \right ),-2\xi , -\frac{l^{(e)}}{2}\left ( 1-\xi ^{2} \right ), \xi +\frac{1}{2},\frac{l^{(e)}}{4}\left ( \xi ^{2} +\xi \right )\end{bmatrix}
Substituting the expression for \gamma _{xz} from Example 2.3, the following two equations are obtained
-\frac{\left ( 1+2a \right )}{l^{(e)}}w_{1}+\frac{4a}{l^{(e)}}w_{2}+\frac{\left ( 1-2a \right )}{l^{(e)}}w_{3}-\frac{a\left ( a+1 \right )}{2}\theta _{1}--\left ( 1-a^{2} \right )\theta _{2}-\frac{a\left ( a-1 \right )}{2}\theta _{3}=0
-\frac{\left ( 1-2a \right )}{l^{(e)}}w_{1}-\frac{4a}{l^{(e)}}+w_{2}+\frac{\left ( 1+2a \right )}{l^{(e)}}w_{3}+\frac{a\left ( 1-a \right )}{l^{(e)}}\theta _{1}-
-\left ( 1-a^{2} \right )\theta _{2}-\frac{a\left ( a+1 \right )}{2}\theta _{1}-\frac{a\left ( a+1 \right )}{2}\theta_{3}=0
with a=\frac{1}{\sqrt{3}}.Eliminating w_{2} and \theta _{2} from the above equations gives
\theta _{2}=-\frac{1}{4}\left ( \theta _{1}+\theta _{3} \right )+\frac{3}{2l^{(e)}}\left ( w_{3}-w_{1} \right )
w_{2}=\frac{1}{2}\left ( w_{1}+w_{3} \right )+\frac{l^{(e)}}{8}\left ( \theta _{1}-\theta _{3} \right )
Substituting these values into the original displacement field gives
\theta =\frac{1}{4}\left ( 3\xi ^{2}-2\xi -1 \right )\theta _{1}+\frac{1}{4}\left ( 3\xi ^{2}+2\xi -1 \right )\theta _{3}--\frac{3}{2l^{(e)}}\left ( 1-\xi ^{2} \right )w_{1}+\frac{3}{2l^{(e)}}\left ( 1-\xi ^{2} \right )w_{3}
w=\frac{1}{2}\left ( 1-\xi \right )w_{1}+\frac{1}{2}\left ( 1+\xi \right )w_{3}+\frac{l^{(e)}}{8}\left ( 1-\xi ^{2} \right )\theta _{3}
The bending strain and the transverse shear strain fields are obtained from the new displacement field as
\kappa =\frac{\mathrm{d} \theta }{\mathrm{d} x}=\frac{6\xi }{l^{(e)^{2}}}w_{1}+\frac{\left ( 3\xi -1 \right )}{l^{(e)}}\theta _{1}-\frac{6\xi }{\left (l^{(e)} \right )^{2}}w_{3}+\frac{\left ( 3\xi +1 \right )}{l^{(e)}}\theta _{3}
\gamma _{xz}=\frac{\partial w}{\partial x}-\theta =\frac{\left ( 1-3\xi ^{2} \right )}{2l^{(e)}}w_{1}+\frac{\left ( 3\xi ^{2}-1 \right )}{4}\theta _{1}+\frac{\left ( 3\xi ^{2}-1 \right )}{2l^{(e)}}w_{3}+\frac{\left ( 3\xi ^{2}-1 \right )}{4}\theta _{3}
The bending strain field is identical to the curvature field for the 2-noded Euler-Bernoulli beam element (Eq.(1.16a)). It can also be verified that \gamma _{xz}=0 at \xi =\pm \frac{1}{\sqrt{3}}. The resultant generalized strain matrix is
B=\begin{Bmatrix}B_{b}\\ B_{s}\end{Bmatrix} = \begin{bmatrix}\frac{6\xi }{\left (l^{(e)} \right )^{2}} \frac{\left ( 3\xi -1 \right )}{l^{(e)}}-\frac{6\xi }{\left (l^{(e)} \right )^{2}} \frac{\left ( 3\xi +1 \right )}{l^{(e)}}\\ \frac{\left ( 1-3\xi ^{2} \right )}{2l^{(e)}} \frac{\left ( 3\xi ^{2}-1 \right )}{4} \frac{\left ( 3\xi ^{2}-1 \right )}{2l^{(e)}} \frac{\left ( 3\xi ^{2}-1 \right )}{4}\end{bmatrix}
(Eq.(1.16a)): \kappa = \frac{\mathrm{d^{2}}w }{\mathrm{d} x^{2}}=\frac{4}{\left (l^{(e)} \right )^{2}}\left ( \frac{d^{2}N_{1}}{d\xi ^{2}}w_{1}+\frac{l^{(e)}}{2}\frac{d^{2}\overline{N_{1}}}{d\xi ^{2}}\left ( \frac{\mathrm{d} w}{\mathrm{d} x} \right )_{1}+ \frac{d^{2}N_{2}}{d\xi ^{2}}w_{2}+\frac{l^{(e)}}{2}\frac{d^{2}\overline{N_{2}}}{d\xi ^{2}}\left ( \frac{\mathrm{d} w}{\mathrm{d} x} \right )_{2}\right ) =\begin{bmatrix}\frac{6\xi }{\left ( l^{(e)} \right )^{2}},\frac{\left ( -1+3\xi \right )}{l^{(e)}},\frac{-6\xi }{\left ( l^{(e)} \right)^{2}} , \frac{\left (1+3\xi \right )}{l^{(e)}}\end{bmatrix}\begin{Bmatrix}w_{1}\\ \left ( \frac{\mathrm{d} w}{\mathrm{d} x}\right )_{1}\\ w_{2}\\ \left ( \frac{\mathrm{d} w}{\mathrm{d} x}\right )_{2}\end{Bmatrix}=B_{b}a^{(e)}
and the element stiffness matrix is
K^{(e)}=\int_{-1}^{+1}\begin{bmatrix}B_{b}^{T}\left ( \hat{D_{b}} \right )B_{b}+ B_{s}^{T}\left ( \hat{D_{s}} \right )B_{s}\end{bmatrix} \frac{l^{(e)}}{2}d\xi =K_{b}^{(e)}+K_{s}^{(e)}
The expression for K_{b}^{(e)} can be integrated exactly using a two-point Gauss quadrature as
K_{b}^{(e)}=\left ( \frac{\hat{D}_{b}}{l^{3}} \right )^{(e)} \begin{bmatrix}12 &6l &-12 &6l \\ & 4l^{2} &-6l &2l^{2}\\ & &12 &-6l \\ Symm. & & & 4l^{2}\end{bmatrix}
which coincides with the stiffness matrix for the 2-noded Euler-Bernoulli beam element (Eq.(1.20)).
The exact integration of K_{s}^{(e)} requires a three-point Gauss quadrature (Appendix C). For the two-point reduced quadrature, B_{s} , and therefore K_{s}^{(e)} , vanish and the stiffness matrix is the same as for the 2-noded Euler-Bernoulli beam element.
This coincidence is logical as the quadratic rotation field obtained yields a linear bending field, as for the 2-noded Euler-Bernoulli element. Also \gamma _{xz}=0 at the two Gauss points that integrate the bending stiffness exactly. Therefore, the shear stiffness contribution is zero and the stiffness matrix coincides with that for the 2-noded Euler-Bernoulli beam element.
(Eq.(1.20)):k^{(e)}=\left ( \frac{EI_{y}}{l^{3}} \right )^{(e)}\begin{bmatrix}12 &6l^{(e)} & -12 &6l^{(e)} \\ \ddots & 4\left ( l^{(e)} \right )^{2} &-6l^{(e)} &2\left ( l^{(e)} \right )^{2} \\ & \ddots & 12 &-6l^{(e)} \\ & & \ddots & 4\left ( l^{(e)} \right )^{2}\end{bmatrix}