Derive the 2-noded Euler-Bernoulli beam element by imposing the condition of zero transverse shear strain over the

Question

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.

Accepted Answer

The transverse shear strain is obtained as

[latex]\gamma _{xz}=\frac{\mathrm{d} w}{\mathrm{d} x}- heta =A+B\xi +C\xi ^{2}[/latex]

The condition [latex]\gamma _{xz} = 0[/latex] 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 [latex]w_{2} , w_{3}[/latex] and [latex] heta _{5}[/latex] as

[latex]A=0 \Rightarrow heta _{5}=\frac{1}{8l^{(e)}}\left ( w_{1}-27w_{2}+27w_{3}-w_{4} ight )[/latex]

[latex]B=0 \Rightarrow w_{2}=w_{1}-w_{3}+w_{4}+\frac{2l^{(e)}}{9}\left ( heta _{1}- heta _{4} ight )[/latex]

Substituting these values into the equation for C = 0 gives

[latex]w_{3}=\frac{1}{27}\begin{bmatrix}7w_{1}+20w_{4}+2l^{(e)}\left ( heta _{1}-2 heta _{4} ight )\end{bmatrix}[/latex]

and consequently

[latex]w_{2}=\frac{1}{27}\begin{bmatrix}20w_{1}-7w_{4}+2l^{(e)}\left ( heta _{1}-2 heta _{4} ight )\end{bmatrix}[/latex]

Substituting [latex]w_{2}[/latex] and [latex]w_{3}[/latex] into the original cubic deflection field gives

[latex]w=\frac{1}{4}\left ( \xi ^{3}-3\xi +2 ight )w_{1}+\frac{1}{4}\left ( 2+3\xi -\xi ^{3} ight )w_{4}+\frac{l^{(e)}}{8}\left ( \xi ^{2}-1 ight )\left ( \xi -1 ight ) heta _{1}+\frac{l^{(e)}}{8}\left ( \xi ^{2}-1 ight )\left ( \xi +1 ight ) heta _{4}=N_{1}w_{1}+\bar{N}_{1} heta _{1}+N_{4}w_{4}+\bar{N}_{4} heta _{4}[/latex]

where[latex] N_{1}[/latex] , [latex]\overline{N}_{1} , N_{4} , \overline{N}_{4}[/latex] coincide with the Hermite shape functions for the 2-
noded Euler-Bernoulli beam element (Eqs.(1.11a)). On the other hand, the condition[latex] \gamma _{xz}=0[/latex] implies [latex] heta =\frac{\mathrm{d} w}{\mathrm{d} x}[/latex] over the element and the deflection field can be written as

[latex]w=N_{1}w_{1}+\overline{N}_{1}\left ( \frac{\mathrm{d} w}{\mathrm{d} x} ight )_{1}+N_{4}w_{4}+\overline{N}_{4}\left ( \frac{\mathrm{d} w}{\mathrm{d} x} ight )_{4}[/latex]

It can clearly be seen that the element has[latex] C^{1}[/latex] continuity and it coincides with the 2-noded Euler-Bernoulli beam element of Section 1.3.

Question : Derive the 2-noded Euler-Bernoulli beam element by imposing ...

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