Element with Five DOFs

Consider a 2D beam lying in the (𝑥, 𝑧)-plane of a right handed Cartesian coordi-nate system with the 𝑥-axis coinciding the longitudinal centroidal axis of the beam.

To account for transverse shear deformation in addition to bending, plane sections that initially were normal to the longitudinal axis, are assumed to remain plane during deformation but not necessarily normal anymore. This assumption complies with the Timoshenko beam theory and deviates from the classical bending theory of Bernoulli-Euler by leaving out the normality constraint for the sections. The assumed deformational behavior is depicted in Fig. 3.1 where 𝜃_𝑦 is the rotation of the cross section (bending rotation) and subscript ‘𝑜’ on displacements refers to values at the centroid. Thus, the longitudinal displacement can be expressed

𝑢(𝑥, 𝑧) ≈ 𝑢_𝑜(𝑥) + 𝑧 𝜃_𝑦(𝑥) (3.1) While the transverse displacement is taken to be same over the depth of the beam, i.e.

𝑤(𝑥, 𝑧) ≈ 𝑤𝑜(𝑥) (3.2)

The derivations in the following will consider linear effects only. By neglecting the quadratic terms of the Green strain components in Eq.(2.6), the infinitesimal normal strain 𝜖𝑥 becomes

𝜖_𝑥 = 𝜕𝑢

𝜕𝑥 = 𝑑𝑢_𝑜

𝑑𝑥 + 𝑧𝑑𝜃_𝑦

𝑑𝑥 (3.3)

When also applying Eq.(2.10), the corresponding expression for the infinitesimal shear strain 𝛾_𝑥𝑧 takes the form

𝛾_𝑥𝑧 = 𝜕𝑢

𝜕𝑧 + 𝜕𝑤

𝜕𝑥 = 𝜃_𝑦 + 𝑑𝑤_𝑜

𝑑𝑥 (3.4)

Thus, the bending rotation is connected to the infinitesimal shear strain and the slope of the transverse displacement through

𝜃𝑦 = 𝛾𝑥𝑧 − 𝑑𝑤_𝑜

𝑑𝑥 (3.5)

𝑥, 𝑢 𝑧, 𝑤

𝑑𝑥

𝑢𝑜

𝑤_𝑜

𝑑𝑤𝑜/𝑑𝑥 𝜃_𝑦

𝛾_𝑥𝑧

Figure 3.1: Deformations of a Beam ‘Slice’

As already stated in the preceding section, the attempt now is to develop a lower order shear-beam element that can model linear variation of bending moment and constant transverse shear for the linear elastic case. Consequently, the natural choice should then be to subject the transverse displacement and the shear strain to trial expansions, respectively a cubic polynomial and a constant. Furthermore, the longi-tudinal centroidal displacement can be left out in this pure bending-shear problem.

Thus

𝑢𝑜 ≈ 0 (3.6)

𝑤_𝑜 ≈ 𝑎₀ + 𝑎₁𝑥 + 𝑎₂𝑥² + 𝑎₃𝑥³ (3.7)

𝛾_𝑥𝑧 ≈ 𝑏₀ (3.8)

Then the expression for the normal strain in Eq.(3.3) simplifies to 𝜖_𝑥 = −𝑧 𝑑²𝑤_𝑜

𝑑𝑥² (3.9)

The problem of solving the displacement and strain fields of the beam element has now five unknowns, i.e.𝑎₀through𝑎₃ and𝑏₀. To seek a configuration of corresponding five physical DOFs, transverse displacement and bending rotation will be selected at locations that can give rise to states of constant shear and linear bending inside the element. The displacement and rotation at each endpoint of the element are obvious candidates. As opposed to the displacement at the midpoint, the rotation there is

𝑥 𝑧

𝐿𝑜/2 𝐿𝑜/2

𝑣𝑧1

𝜃_𝑦1

𝑣𝑧2

𝜃_𝑦2 𝜃_𝑦3

1 3 2

Figure 3.2: 2D Shear-Beam Element with Five DOFs

also creating constant shear, and thus the latter will be the selected fifth DOF. The finite element configuration then becomes as shown in Fig. 3.2.

The shape functions pertaining to each DOF can be found by imposing one DOF at a time equal to unity and the others equal to zero, and then solve for the five unknowns (𝑎₀−𝑎₃, 𝑏₀) in turn. Note that rotations are bending rotations as given by Eq.(3.5). The results can be written on the form

𝑤𝑜 = 𝑁𝑤𝑜 𝑣 (3.10)

𝛾_𝑥𝑧 = 𝑁_𝛾𝑥𝑧 𝑣 (3.11)

where

𝑣^𝑇 =

[︂

𝑣_𝑧1 𝜃_𝑦1 𝑣_𝑧2 𝜃_𝑦2 𝜃_𝑦3

]︂

(3.12) 𝑁_𝑤𝑜 =

[︂

𝑁_𝑤^(𝑣_𝑜¹⁾ 𝑁_𝑤^(𝜃_𝑜¹⁾ 𝑁_𝑤^(𝑣_𝑜²⁾ 𝑁_𝑤^(𝜃_𝑜²⁾ 𝑁_𝑤^(𝜃_𝑜³⁾

]︂

(3.13) 𝑁_𝛾𝑥𝑧 =

[︂

𝑁_𝛾^(𝑣_𝑥𝑧¹⁾ 𝑁_𝛾^(𝜃_𝑥𝑧¹⁾ 𝑁_𝛾^(𝑣_𝑥𝑧²⁾ 𝑁_𝛾^(𝜃_𝑥𝑧²⁾ 𝑁_𝛾^(𝜃_𝑥𝑧³⁾

]︂

(3.14) with shape functions pertaining to transverse displacement given by

𝑁_𝑤^(𝑣_𝑜¹⁾ = 1

2(1 − 𝜉) (3.15)

𝑁_𝑤^(𝜃_𝑜¹⁾ = −𝐿_𝑜

24(3 − 2𝜉) (1 − 𝜉²) (3.16)

𝑁_𝑤^(𝑣_𝑜²⁾ = 1

2(1 + 𝜉) (3.17)

𝑁_𝑤^(𝜃_𝑜²⁾ = 𝐿_𝑜

24(3 + 2𝜉) (1 − 𝜉²) (3.18)

𝑁_𝑤^(𝜃_𝑜³⁾ = −𝐿_𝑜

6 𝜉(1 − 𝜉²) (3.19)

and shape functions pertaining to shear strain 𝑁_𝛾^(𝑣_𝑥𝑧¹⁾ = − 1

𝐿_𝑜 (3.20)

𝑁_𝛾^(𝜃_𝑥𝑧¹⁾ = 1

6 (3.21)

𝑁_𝛾^(𝑣_𝑥𝑧²⁾ = 1

𝐿_𝑜 (3.22)

𝑁_𝛾^(𝜃2)

𝑥𝑧 = 1

6 (3.23)

𝑁_𝛾^(𝜃_𝑥𝑧³⁾ = 2

3 (3.24)

Furthermore, 𝜉 is the natural beam coordinate that takes the values -1, 1 and 0 at node 1, 2 and 3, respectively. Thus

𝜉 = 2𝑥

𝐿_𝑜 − 1 (3.25)

The shape functions are depicted in Fig. 3.3.

Having established the shape functions for the element, the next is to relate the strain field to the nodal DOFs through the strain-displacement matrix 𝐵. For shear strain the relation needed is given by Eq.(3.11) directly, while for normal strain Eq.(3.10) has to be differentiated twice before inserted into Eq.(3.9). Thus

𝜖𝜖𝜖

𝜖 = 𝐵 𝑣 (3.26)

where

𝜖𝜖𝜖 𝜖 =

⎧

⎪⎨

⎪⎩

𝜖𝑥

𝛾_𝑥𝑧

⎫

⎪⎬

⎪⎭

(3.27)

𝐵 =

⎡

⎢

⎣

𝑏_𝑏 𝑏_𝑠

⎤

⎥

⎦ (3.28)

with the subvectors pertaining to bending (i.e. normal) strain 𝑏_𝑏 and shear strain 𝑏_𝑠 given by

𝑏𝑏 = −𝑧𝑑²𝑁𝑤𝑜

𝑑𝑥² = 𝑧 𝐿_𝑜

[︂

0 −(1 − 2𝜉) 0 (1 + 2𝜉) −4𝜉

]︂

(3.29) 𝑏_𝑠 = 𝑁_𝛾𝑥𝑧 =

[︂

−_𝐿¹

𝑜

1 6

1 𝐿𝑜

1 6

2 3

]︂

(3.30) Since linear effects are the only considered here, the material law can now suffi-ciently be formulated on total form (as opposed to the incremental form in Eq.(2.32)).

Thus

𝜎𝜎𝜎

𝜎 = 𝐶𝜖𝜖𝜖𝜖 (3.31)

𝑁_𝑤^(𝑣_𝑜¹⁾:

𝑁_𝑤^(𝜃_𝑜¹⁾ :

𝑁_𝑤^(𝑣_𝑜²⁾:

𝑁_𝑤^(𝜃_𝑜²⁾ :

𝑁_𝑤^(𝜃_𝑜³⁾ :

𝑁_𝛾^(𝑣_𝑥𝑧¹⁾ = −_𝐿¹

𝑜

𝑁_𝛾^(𝜃_𝑥𝑧¹⁾ = ¹₆

𝑁_𝛾^(𝑣_𝑥𝑧²⁾ = _𝐿¹

𝑜

𝑁_𝛾^(𝜃_𝑥𝑧²⁾ = ¹₆

𝑁_𝛾^(𝜃_𝑥𝑧³⁾ = ²₃ 1

1

1

1

1

Figure 3.3: Shape Functions of the 2D Shear-Beam Element

where𝜎𝜎𝜎𝜎 is the Cauchy stress vector corresponding to the infinitesimal strains, i.e. and 𝐶 is the constitutive matrix, which for the 2D linear elastic material is given by

𝐶 =

Here𝐸 is the elastic modulus and 𝐺 is the shear modulus.

Finally, the applied load on the element will be specified in terms of transverse load per unit axial length of the beam𝑞_𝑧.

Now everything needed is defined for applying the principle of virtual displace-ments to carry out the expressions for the final element quantities. Again, since linear effects are the only considered here, the principle on total form becomes sufficient.

Thus, from Eq.(2.44)

Insertion of Eqs.(3.10,3.26,3.31) yields the discretized form 𝛿𝑣^𝑇 which must be valid for an arbitrary 𝛿𝑣. Thus

𝑘_𝑚𝑣 = 𝑝 (3.36)

are the material stiffness matrix and the consistent node-force vector of the element, respectively.

The integral of 𝑘_𝑚 can most conveniently be solved by splitting the expression into two parts; one that arises from the bending energy 𝑘_𝑏, and one from the shear energy 𝑘_𝑠

𝑘_𝑚 = 𝑘_𝑏 + 𝑘_𝑠 (3.39)

By applying Eqs.(3.25,3.28-3.30,3.33) these two contributions become 𝑘_𝑏 = 𝐸

where 𝐼𝑦 is the moment of inertia of the cross section with respect to the 𝑦-axis, i.e. the effective shear area 𝐴_𝑠 is used in Eq.(3.41) instead of 𝐴_𝑜 to compensate for the approximation lying in Eq.(3.1) that leads to a uniform shear strain (stress) distribution over the cross section rather than the correct parabolic form for the linear elastic case. By carrying out the matrix product in Eq.(3.40) and integrating over the length of the beam, the final expression for the contribution from bending energy to the material stiffness matrix takes the form

𝑘_𝑏 = 𝐸𝐼𝑦

While the final form of the contribution from shear energy becomes by carrying out the matrix product of Eq.(3.41)

𝑘_𝑠 = 𝐺𝐴_𝑠

The integral expression for the consistent node-force vector will be solved for two loading conditions; a uniformly distributed and a triangularly distributed transverse load as depicted in Fig. 3.4, i.e.

𝑞_𝑧 = 𝑞_𝑧0 (3.45)

𝑞_𝑧 = 1

2(1 + 𝜉)𝑞_𝑧1 (3.46)

By applying Eqs.(3.15-3.19,3.25), 𝑝 takes the form for the two loading conditions 𝑝₀ = 𝑞_𝑧0 𝐿_𝑜

𝑃𝑧 𝑞_𝑧0 𝑞𝑧1

𝐿_𝑜 𝐿_𝑜 𝐿_𝑜

Figure 3.4: Loading Conditions of Cantilevered Beam

Note that for uniformly distributed transverse load, the consistent node-forces coa-lesce with those of the ordinary Bernoulli-Euler beam element.

When the stiffness equations are solved for the unknown DOFs, the internal bend-ing moment𝑀_𝑦 and shear force𝐹_𝑧 can be recovered at the element level. Application of the strain expressions Eqs.(3.9-3.11) and the material law Eq.(3.31), and then in-tegration of stresses over the cross section to force resultants, yield

𝑀_𝑦 = −𝐸 𝐼_𝑦 𝑑²𝑤_𝑜

𝑑𝑥² = −𝐸 𝐼_𝑦 𝑑²𝑁_𝑤𝑜

𝑑𝑥² 𝑣 (3.49)

𝐹_𝑧 = 𝐺 𝐴_𝑠𝛾_𝑥𝑧 = 𝐺 𝐴_𝑠𝑁_𝛾𝑥𝑧𝑣 (3.50) where shape function expressions can be found from Eqs.(3.29,3.30).

Examples:

The performance of the element will now be investigated by considering a cantilevered beam, modeled with one single element, under the three different loading conditions in Fig.3.4; concentratedtip load,uniform loadandtriangular load. Since the element handles only linear variation of moment and constant shear internally, it can possibly give exact results for the tip load case only. However, it is also of importance to clarify how good approximations can be obtained under higher order load variations.

Enforcing the boundary conditions 𝑣_𝑧1 = 0 and 𝜃_𝑦1 = 0, the constrained stiffness equations become

where 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.(3.51)

with the determinant of the constrained stiffness matrix given by

|𝑘| = 16

and the adjoints of the entries of 𝑘 𝑎₁₁ = 16

Then the internal bending moment and shear force can be recovered from 𝑀_𝑦 = 𝐸𝐼_𝑦 The one element solutions for the three different loading conditions are sum-marized in Tab. 3.1, and the corresponding exact results according to Timoshenko beam theory are presented in Tab. 3.2. As can be seen; the nodal displacements (𝑣_𝑧2, 𝜃_𝑦2, 𝜃_𝑦3) are practically exact for all load cases. The only slight deviation (≈

0.5%) occurs at the midpoint rotation for triangular load. The internal forces (𝑀_𝑦, 𝐹𝑧) are exact as expected for the tip load case only. However, the linear approximations for bending moments are fairly close to the the exact solutions also for the two other load conditions, while the computed shear forces represent averages of the exact val-ues. These results are best illustrated in Figs. 3.5 and 3.6 that depict the internal force variations for uniform and triangular loads, respectively. Here all values are scaled with respect to the corresponding exact values at the built-in support. Note that increasing number of elements will lead to closer fit to exact solutions for internal forces.

Tip Load Uniform Load Triangular Load

Table 3.1: Results with One Element for Cantilevered Beam Tip Load Uniform Load Triangular Load 𝑣_𝑧2 ¹₃^𝑃_𝐸𝐼^{𝑧𝐿3𝑜}

Table 3.2: Exact Results for Cantilevered Beam

𝑀𝑦

Figure 3.5: Moment and Shear Force Variations for Cantilevered Beam withUniform Load

Figure 3.6: Moment and Shear Force Variations for Cantilevered Beam with Trian-gular Load

In document Nonlinear analysis of concrete structures based on a 3D shear-beam element formulation (sider 39-49)