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2D Elastic Shear-Beam Element Formulations

3.3 Hierarchical Form of the Five DOFs Element

The preceding section dealt with the conventional form of the five DOFs shear-beam element. The key step in going to the hierarchical form of this element is to express the internal DOF at the midpoint, now denoted ˜𝜃𝑦3, as the deviationin bending rotation from the value interpolated from the external DOFs at the endpoints. The external DOFs themselves stay the same. Thus, the hierarchical finite element configuration becomes as shown in Fig. 3.7.

𝑥 𝑧

𝐿𝑜/2 𝐿𝑜/2

𝑣𝑧1 𝜃𝑦1

𝑣𝑧2

𝜃𝑦2 𝜃˜𝑦3

1 3 2

Figure 3.7: Hierarchical Version of the Five DOFs Shear-Beam Element The reasons for converting to the hierarchical form are twofold: In large displace-ment analysis internal rotational DOFs should be ‘small’ quantities, which is better justified by having them on hierarchical form. Secondly; it is now possible to isolate the effect of shear deformation to the internal DOF only so that setting ˜𝜃𝑦3 = 0 as an internal constraint, will lead to recovery of the ordinary Bernoulli-Euler beam el-ement. Thus, the effect of shear on a particular problem can be evaluated by making comparative runs without and with this internal constraint.

The derivation of the hierarchical form may now start by expressing the interpo-lation of 𝑤𝑜 and 𝛾𝑥𝑧 in terms of the two different sets of DOFs

𝑤𝑜 = 𝑁𝑤𝑜 𝑣 = 𝑁˜𝑤𝑜 𝑣˜ (3.58)

𝛾𝑥𝑧 = 𝑁𝛾𝑥𝑧𝑣 = 𝑁˜𝛾𝑥𝑧𝑣˜ (3.59) where the conventional DOFs and shape functions are ordered as in Eqs.(3.12-3.14), and the corresponding hierarchical siblings are

˜ 𝑣𝑇 =

[︂

𝑣𝑧1 𝜃𝑦1 𝑣𝑧2 𝜃𝑦2 𝜃˜𝑦3

]︂

(3.60) 𝑁˜𝑤𝑜 =

[︂

𝑁˜𝑤(𝑣𝑜1) 𝑁˜𝑤(𝜃𝑜1) 𝑁˜𝑤(𝑣𝑜2) 𝑁˜𝑤(𝜃𝑜2) 𝑁˜𝑤(𝜃𝑜3)

]︂

(3.61) 𝑁˜𝛾𝑥𝑧 =

[︂

𝑁˜𝛾(𝑣1)

𝑥𝑧

𝑁˜𝛾(𝜃1)

𝑥𝑧

𝑁˜𝛾(𝑣2)

𝑥𝑧

𝑁˜𝛾(𝜃2)

𝑥𝑧

𝑁˜𝛾(𝜃3)

𝑥𝑧

]︂

(3.62)

Furthermore, the following transformation can be established that relates the con-ventional DOFs to the hierarchical

𝑣 = 𝑇˜𝑣˜ (3.63)

with the transformation matrix given by

𝑇˜ =

Here the unknown entries 𝑡51 through 𝑡54 will be determined from the succeeding shear strain constraints. Insertion of Eq.(3.63) into Eq.(3.59) gives the following relationship between the two sets of shear strain shape functions

𝑁˜𝛾𝑥𝑧 = 𝑁𝛾𝑥𝑧 𝑇˜ (3.65)

where the entries of 𝑁𝛾𝑥𝑧 are given by Eqs.(3.20-3.24). By requiring zero shear strain contribution from each of the four external DOFs of the hierarchical element, the foregoing relationship yields

𝑁˜𝛾(𝑣𝑥𝑧1) = 𝑁𝛾(𝑣𝑥𝑧1) + 𝑡51𝑁𝛾(𝜃𝑥𝑧3) = 0 → 𝑡51 = 3

and with the fifth equation yielding the only nonzero shear strain shape function of the hierarchical element

𝑁˜𝛾(𝜃𝑥𝑧3) = 𝑁𝛾(𝜃𝑥𝑧3) = 2

3 (3.70)

Then the transformation matrix is in final form

𝑇˜ =

Now the shape functions of the hierarchical element pertaining to transverse displace-ment can be found. By inserting Eq.(3.63) into Eq.(3.58), the relationship between the two sets of displacement shape functions becomes

𝑁˜𝑤𝑜 = 𝑁𝑤𝑜 𝑇˜ (3.72)

where the entries of 𝑁𝑤𝑜 are given by Eqs.(3.15-3.19). Carrying out the matrix product yields

𝑁˜𝑤(𝑣𝑜1) = 𝑁𝑤(𝑣𝑜1) + 3

2𝐿𝑜 𝑁𝑤(𝜃𝑜3) = 1

4(2 + 𝜉) (1𝜉)2 (3.73) 𝑁˜𝑤(𝜃𝑜1) = 𝑁𝑤(𝜃𝑜1) − 1

4𝑁𝑤(𝜃𝑜3) = −𝐿𝑜

8 (1 + 𝜉) (1𝜉)2 (3.74) 𝑁˜𝑤(𝑣𝑜2) = 𝑁𝑤(𝑣𝑜2) − 3

2𝐿𝑜𝑁𝑤(𝜃𝑜3) = 1

4(2 − 𝜉) (1 + 𝜉)2 (3.75) 𝑁˜𝑤(𝜃𝑜2) = 𝑁𝑤(𝜃𝑜2) − 1

4𝑁𝑤(𝜃𝑜3) = 𝐿𝑜

8 (1 − 𝜉) (1 + 𝜉)2 (3.76) 𝑁˜𝑤(𝜃𝑜3) = 𝑁𝑤(𝜃𝑜3) = −𝐿𝑜

6 𝜉(1 − 𝜉) (1 + 𝜉) (3.77) As indeed was the expected result; the displacement shape functions pertaining to the four external DOFs of the hierarchical element have now become theHermitiancubic shape functions of the ordinary Bernoulli-Euler beam element. The shape function pertaining to the hierarchical internal DOF stays the same as for the conventional shear-beam element. Fig. 3.8 shows a summary of the hierarchical shape functions.

The remaining steps to form the material stiffness matrix 𝑘˜𝑚 and the consistent node-force vector 𝑝˜ of the hierarchical element may now follow two optional lines:

Since the final expressions for the conventional element already exist, the hierarchical counterparts may be found from the transformations

𝑘˜𝑚 = 𝑇˜𝑇𝑘𝑚𝑇˜ (3.78)

˜

𝑝 = 𝑇˜𝑇𝑝 (3.79)

The alternative way is to form˜𝑘𝑚and𝑝˜directly based on the hierarchical shape func-tions by proceeding as outlined in the previous section for the conventional element.

The latter will be the main option here in order to be consistent with the presenta-tion given in Chapter 4when going to the nonlinear 3D formulation (where only the hierarchical version will be treated). Thus, the strain-displacement relationship now becomes

𝜖𝜖𝜖

𝜖 = 𝐵˜ 𝑣˜ (3.80)

where𝜖𝜖𝜖𝜖 is from Eq.(3.27) and with the new strain-displacement matrix given by 𝐵˜ =

−𝑧𝑑2𝑁˜𝑤𝑜

𝑑𝑥2

𝑁˜𝛾𝑥𝑧

=

𝐿62

𝑜𝑧𝜉 𝐿1

𝑜𝑧(3𝜉−1) 𝐿62

𝑜𝑧𝜉 𝐿1

𝑜𝑧(3𝜉+ 1) −𝐿4

𝑜𝑧𝜉

0 0 0 0 23

(3.81)

𝑁˜𝑤(𝑣𝑜1):

𝑁˜𝑤(𝜃𝑜1) :

𝑁˜𝑤(𝑣𝑜2):

𝑁˜𝑤(𝜃𝑜2) :

𝑁˜𝑤(𝜃𝑜3) :

𝑁˜𝛾(𝑣𝑥𝑧1) = 0

𝑁˜𝛾(𝜃𝑥𝑧1) = 0

𝑁˜𝛾(𝑣𝑥𝑧2) = 0

𝑁˜𝛾(𝜃𝑥𝑧2) = 0

𝑁˜𝛾(𝜃𝑥𝑧3) = 23 1

1

1

1

1

Figure 3.8: Shape Functions of the Hierarchical 2D Shear-Beam Element

Application of the principle of virtual displacements as stated in Eq.(3.34), insert-ing Eqs.(3.31,3.58(last part),3.80), yields the stiffness equations of the hierarchical element

Because of the favorable hierarchical form, the shear energy will contribute to one en-try of the material stiffness matrix only. Hence, the final expressions for𝑘˜𝑚 may now conveniently be given as a single matrix (rather than making a split as in Eq.(3.39) for the conventional element). By introducing Eqs.(3.25,3.33,3.81), Eq.(3.83) converts to

˜𝑘𝑚 = 𝐸𝐼𝑦𝐿𝑜 Then by carrying out the matrix products and integrating the first part over the length of the beam, the material stiffness matrix of the hierarchical element finally becomes

As also previously indicated; the submatrix of ˜𝑘𝑚 pertaining to the four external DOFs (i.e. first four rows and columns) can be recognized as the stiffness matrix of the ordinary Bernoulli-Euler beam element.

The consistent node-force vector of the hierarchical element will then be developed for the same loading conditions as the conventional element, i.e. for the uniformly and triangularly distributed transverse loads given by Eqs.(3.45,3.46), respectively. By applying Eqs.(3.25,3.73-3.77), Eq.(3.84) takes the form for the two loading conditions

˜

Similarly to the stiffness matrix; here the subvectors pertaining to the four external DOFs are the same as the consistent node-force vectors of the Bernoulli-Euler element.

Furthermore,𝑝˜0 and 𝑝0 become identical.

Finally, the expressions for the internal bending moment 𝑀𝑦 and shear force 𝐹𝑧 will be given. Correspondingly to Eqs.(3.49,3.50), these now become

𝑀𝑦 = −𝐸 𝐼𝑦 𝑑2𝑤𝑜

𝑑𝑥2 = −𝐸 𝐼𝑦 𝑑2𝑁˜𝑤𝑜

𝑑𝑥2 𝑣˜ (3.90)

𝐹𝑧 = 𝐺 𝐴𝑠𝛾𝑥𝑧 = 𝐺 𝐴𝑠𝑁˜𝛾𝑥𝑧𝑣˜ = 2

3𝐺 𝐴𝑠𝜃˜𝑦3 (3.91) where the shape function derivatives may be taken from Eq.(3.81).

Examples:

The cantilever beam examples from the preceding section will now be revisited in order to demonstrate that the hierarchical and conventional versions of the five DOFs shear-beam element indeed yield identical results. Enforcing the boundary conditions 𝑣𝑧1 = 0 and 𝜃𝑦1 = 0, the constrained stiffness equations of the hierarchical element become

where𝛼 is from Eq.(3.87), and the load vector on the right hand side in turn will be substituted by

for tip load, uniform load and triangular load, respectively. Inversion of Eq.(3.92) yields

with the determinant of the constrained stiffness matrix given by

|˜𝑘| = 16

and the adjoints of the entries of 𝑘˜

Compared with the corresponding expressions in Eqs.(3.54,3.55) for the conventional element, only terms pertaining to the internal DOF have changed. Furthermore, the internal bending moment can be recovered from

𝑀𝑦 = 𝐸𝐼𝑦 while the shear force expression is given by the last part of Eq.(3.91). Finally, for the sake of making a complete comparison with the conventional element solutions, the total bending rotation at the midpoint of the hierarchical element can be recovered from the last transformation in Eq.(3.63). Thus

𝜃𝑦3 = − 3

2𝐿𝑜 𝑣𝑧2 − 1

4𝜃𝑦2 + ˜𝜃𝑦3 (3.98)

Tip Load Uniform Load Triangular Load 𝑣𝑧2 13𝑃𝐸𝐼𝑧𝐿3𝑜

Table 3.3: Results with One Hierarchical Element for Cantilevered Beam The results with one hierarchical element for the cantilevered beam are sum-marized in Tab. 3.3. A comparison with the one conventional element solutions as presented in Tab. 3.1, confirms that the results are identical in every respect.

Chapter 4