Timoshenko Beams and Frames

Abstract

This chapter introduces first the theory to derive the elemental stiffness matrix of Timoshenko beam elements for an arbitrary number of nodes and assumptions for the displacement and rotation fields. Then, the principal finite element equation of such beam elements and their arrangements as plane frame structures are briefly covered. Furthermore, a combination of Timoshenko beam and rod element is introduced as a generalized beam and frame element. Comments on the computer implementation of the corresponding Maxima modules are provided. The chapter includes detailed Maxima examples which allow an easy transfer to other problems.

5.1 Introductory Remarks

In the case of the Timoshenko beam, also called the shear-flexible or thick beam, the shear deformation is considered in addition to the bending deformation. As a consequence, Bernoulli’s hypothesis is partly no longer fulfilled for the Timoshenko beam: plane cross sections remain plane after the deformation. However, a cross section which stood at right angles to the beam axis before the deformation is not at right angles to the beam axis after the deformation. Furthermore, the real parabolic distribution of the shear stress over the cross section is simplified to an equivalent constant shear stress. As a consequence, a so-called shear correction factor $(k_\text {s})$ is introduced [4].

5.2 Calculation of the Elemental Stiffness Matrix

5.2.1 Elements with Two Nodes

The simplest Timoshenko element is based on two nodes where the displacement field and the rotation field are interpolated based on linear functions. This gives for the nodal approaches of the displacement $(u_{z}^\text {e}(x))$ and rotation $(\phi _{y}^\text {e}(x))$ field, for example in the x-z plane, the following representations

$$\begin{aligned} u_{z}^\text {e}(x) = \begin{bmatrix} N_{1u}&0&N_{2u}&0 \end{bmatrix} \begin{bmatrix} u_{1z}\\ \phi _{1y}\\ u_{2z}\\ \phi _{2y} \end{bmatrix} =\varvec{N}_{u}^\text {T}\varvec{u}_\text {p}\,, \end{aligned}$$

(5.1)

$$\begin{aligned} \phi _{y}^\text {e}(x) = \begin{bmatrix} 0&N_{1\phi }&0&N_{2\phi } \end{bmatrix} \begin{bmatrix} u_{1z}\\ \phi _{1y}\\ u_{2z}\\ \phi _{2y} \end{bmatrix} =\varvec{N}_{\phi }^\text {T}\varvec{u}_\text {p}\,, \end{aligned}$$

(5.2)

where the interpolation functions are shown in Fig. 5.1.

Fig. 5.1

Linear interpolation functions $N_1=N_{1u}=N_{1\phi }$ and $N_2=N_{2u}=N_{2\phi }$ for the Timoshenko element: a physical coordinate (x); b natural coordinate $(\xi )$

Based on this assumption of two linear fields, the general expression for the elemental stiffness matrix $\varvec{K}^\text {e}$ of a two-noded linear Timoshenko beam element is obtained as [19, 22]

$$\begin{aligned} \varvec{K}^\text {e}_\text {}&= \varvec{K}^\text {e}_\text {b} +\varvec{K}^\text {e}_\text {s}\end{aligned}$$

(5.3)

$$\begin{aligned}&=\int \limits _0^L \begin{bmatrix} 0&0&0&0 \\ 0&EI_y\tfrac{\mathrm {d}N_{1\phi }}{\mathrm {d}x}\tfrac{\mathrm {d}N_{1\phi }}{\mathrm {d}x}&0&EI_y\tfrac{\mathrm {d}N_{1\phi }}{\mathrm {d}x}\tfrac{\mathrm {d}N_{2\phi }}{\mathrm {d}x}\\ 0&0&0&0 \\ 0&EI_y\tfrac{\mathrm {d}N_{2\phi }}{\mathrm {d}x}\tfrac{\mathrm {d}N_{1\phi }}{\mathrm {d}x}&0&EI_y \tfrac{\mathrm {d}N_{2\phi }}{\mathrm {d}x}\tfrac{\mathrm {d}N_{2\phi }}{\mathrm {d}x}\\ \end{bmatrix} \mathrm {d}x \nonumber \\&+\int \limits _0^L \begin{bmatrix} k_\text {s}GA\tfrac{\mathrm {d}N_{1u}}{\mathrm {d}x}\tfrac{\mathrm {d}N_{1u}}{\mathrm {d}x}&k_\text {s}GA \tfrac{\mathrm {d}N_{1u}}{\mathrm {d}x}N_{1\phi }&k_\text {s}GA\tfrac{\mathrm {d}N_{1u}}{\mathrm {d}x}\tfrac{\mathrm {d}N_{2u}}{\mathrm {d}x}&k_\text {s}GA\tfrac{\mathrm {d}N_{1u}}{\mathrm {d}x}N_{2\phi }\\ k_\text {s}GAN_{1\phi }\tfrac{\mathrm {d}N_{1u}}{\mathrm {d}x}&k_\text {s}GAN_{1\phi }N_{1\phi }&k_\text {s}GAN_{1\phi }\tfrac{\mathrm {d}N_{2u}}{\mathrm {d}x}&k_\text {s}GAN_{1\phi }N_{2\phi } \\ k_\text {s}GA\tfrac{\mathrm {d}N_{2u}}{\mathrm {d}x}\tfrac{\mathrm {d}N_{1u}}{\mathrm {d}x}&k_\text {s}GA\tfrac{\mathrm {d}N_{2u}}{\mathrm {d}x}N_{1\phi }&k_\text {s}GA\tfrac{\mathrm {d}N_{2u}}{\mathrm {d}x}\tfrac{\mathrm {d}N_{2u}}{\mathrm {d}x}&k_\text {s}GA\tfrac{\mathrm {d}N_{2u}}{\mathrm {d}x}N_{2\phi }\\ k_\text {s}GAN_{2\phi }\tfrac{\mathrm {d}N_{1u}}{\mathrm {d}x}&k_\text {s}GAN_{2\phi }N_{1\phi }&k_\text {s}GAN_{2\phi }\tfrac{\mathrm {d}N_{2u}}{\mathrm {d}x}&k_\text {s}GAN_{2\phi }N_{2\phi } \\ \end{bmatrix} \!\mathrm {d}x\,. \end{aligned}$$

(5.4)

The two expressions for the bending $(\varvec{K}^\text {e}_\text {b})$ and shear $(\varvec{K}^\text {e}_\text {s})$ parts of the element stiffness matrix can be superposed for the principal finite element equation of the Timoshenko beam on the element level, see Eq. (5.5). In case that the bending stiffness $EI_y$ and the shear stiffness $k_\text {s}GA$ are constant, the simplified representation given in Eq. (5.6) is obtained.

5.2.2 Higher-Order Elements

This subsection is adapted from the derivations presented in [19] and develops first a general approach for a Timoshenko element with an arbitrary number of n nodes. In generalization of Eqs. (5.1) and (5.2), the following representation results for the nodal approaches:

$$\begin{aligned} u_{z}^\text {e}(x)&=\sum _{i\,=\, 1}^n N_{iu}(x)u_{iz}\,,\end{aligned}$$

(5.7)

$$\begin{aligned} \phi _{y}^\text {e}(x)&=\sum _{i\,=\, 1}^n N_{i\phi }(x)\phi _{iy}\,, \end{aligned}$$

(5.8)

or alternatively in matrix notation as

$$\begin{aligned} u_{z}^\text {e}(x) = \begin{bmatrix} N_{1u}&\ldots&N_{nu}&0&\ldots&0 \end{bmatrix} \begin{bmatrix} u_{1z}\\ \vdots \\ u_{nz}\\ \phi _{1y}\\ \vdots \\ \phi _{ny} \end{bmatrix} =\varvec{N}_{u}^\text {T}\varvec{u}_\text {p}\,, \end{aligned}$$

(5.9)

$$\begin{aligned} \phi _{y}^\text {e}(x) = \begin{bmatrix} 0&\ldots&0&N_{1\phi }&\ldots&N_{n\phi } \end{bmatrix} \begin{bmatrix} u_{1z}\\ \vdots \\ u_{nz}\\ \phi _{1y}\\ \vdots \\ \phi _{ny} \end{bmatrix} =\varvec{N}_{\phi }^\text {T}\varvec{u}_\text {p}\,, \end{aligned}$$

(5.10)

where first all the displacements and only after all the rotations are collected.

With this generalized approach, the deflection and the rotation can be evaluated at n nodes. For the interpolation functions $N_i$ usually Lagrange polynomials¹ are used, which in general are calculated as follows:

$$\begin{aligned} \nonumber N_i=\,&\prod _{j\,=\, 1\wedge j\,\ne \, i}^{n}\frac{x_j-x}{x_j-x_i}\\ =\,&\frac{(x_1-x)(x_2-x)\cdots [x_i-x]\cdots (x_n-x)}{(x_1-x_i)(x_2-x_i)\cdots [x_i-x_i]\cdots (x_n-x_i)}\,, \end{aligned}$$

(5.11)

whereupon the expressions in the square brackets for the ith interpolation function remains unconsidered. The abscissa values $x_1,\ldots , x_n$ represent the x-coordinates of the n nodes.

For the derivation of the general stiffness matrix, the weighted residual method, for example, can be considered and the bending stiffness matrix results

(5.12)

and a corresponding procedure yields the shear stiffness matrix

(5.13)

These two stiffness matrices can be be superposed additively at this point and the following general structure for the total stiffness matrix is obtained:

$$\begin{aligned} \varvec{K}^\text {e} = \begin{bmatrix} \varvec{K}^{11}&\varvec{K}^{12}\\ \varvec{K}^{21}&\varvec{K}^{22} \end{bmatrix}\,, \end{aligned}$$

(5.14)

with

$$\begin{aligned} \varvec{K}_{kl}^{11}&=\int \limits _{0}^Lk_\text {s}AG \frac{\mathrm {d}N_{ku}}{\mathrm {d}x}\frac{\mathrm {d}N_{lu}}{\mathrm {d}x}\,\mathrm {d}x\,,\end{aligned}$$

(5.15)

$$\begin{aligned} \varvec{K}_{kl}^{12}&=\int \limits _{0}^Lk_\text {s}AG \frac{\mathrm {d}N_{ku}}{\mathrm {d}x}N_{l\phi }\mathrm {d}x\,,\end{aligned}$$

(5.16)

$$\begin{aligned} \varvec{K}_{kl}^{21}&=\varvec{k}_{kl}^{12,\text {T}}=\int \limits _{0}^Lk_\text {s}AG N_{k\phi }\frac{\mathrm {d}N_{lu}}{\mathrm {d}x}\,\mathrm {d}x\,,\end{aligned}$$

(5.17)

$$\begin{aligned} \varvec{K}_{kl}^{22}&=\int \limits _{0}^L\left( k_\text {s}AG N_{k\phi }N_{l\phi } +EI_y\frac{\mathrm {d}N_{k\phi }}{\mathrm {d}x}\frac{\mathrm {d}N_{l\phi }}{\mathrm {d}x} \right) \mathrm {d}x\,. \end{aligned}$$

(5.18)

Let us now simplify this approach for the case of two nodes, i.e. $n=2$. The nodal approaches given in Eqs. (5.9)–(5.10) simplify to:

$$\begin{aligned} u_{z}^\text {e}(x) = \begin{bmatrix} N_{1u}&N_{2u}&0&0 \end{bmatrix} \begin{bmatrix} u_{1z}\\ u_{2z}\\ \phi _{1y}\\ \phi _{2y} \end{bmatrix} =\varvec{N}_{u}^\text {T}\varvec{u}_\text {p}\,, \end{aligned}$$

(5.19)

and for the rotation:

$$\begin{aligned} \phi _{y}^\text {e}(x) = \begin{bmatrix} 0&0&N_{1\phi }&N_{2\phi } \end{bmatrix} \begin{bmatrix} u_{1z}\\ u_{2z}\\ \phi _{1y}\\ \phi _{2y} \end{bmatrix} =\varvec{N}_{\phi }^\text {T}\varvec{u}_\text {p}\,. \end{aligned}$$

(5.20)

In the next step, we simplify the general expressions for the stiffness matrices given in Eqs. (5.12)–(5.13):

(5.21)

and a corresponding procedure yields the simplified shear stiffness matrix:

(5.22)

These two stiffness matrices can be be superposed additively and the following total stiffness matrix is obtained:

(5.23)

If we arrange the column matrix in a more common way, i.e. collecting the unknown by nodes, we obtain the following representation (as in Eq. (5.5)):

It is also possible to further generalize the approach represented in Eqs. (5.7)–(5.8). If we evaluate the displacement field at m nodes but the rotation field at a different number of n nodes, we obtain the following representation:

$$\begin{aligned} u_{z}^\text {e}(x)&=\sum _{i\,=\, 1}^m N_{iu}(x)u_{iz}\,,\end{aligned}$$

(5.25)

$$\begin{aligned} \phi _{y}^\text {e}(x)&=\sum _{i\,=\, 1}^n N_{i\phi }(x)\phi _{iy}\,, \end{aligned}$$

(5.26)

or alternatively in matrix notation as

$$\begin{aligned} u_{z}^\text {e}(x) = \begin{bmatrix} N_{1u}&\ldots&N_{mu}&0&\ldots&0 \end{bmatrix} \begin{bmatrix} u_{1z}\\ \vdots \\ u_{mz}\\ \phi _{1y}\\ \vdots \\ \phi _{ny} \end{bmatrix} =\varvec{N}_{u}^\text {T}\varvec{u}_\text {p}\,, \end{aligned}$$

(5.27)

$$\begin{aligned} \phi _{y}^\text {e}(x) = \begin{bmatrix} 0&\ldots&0&N_{1\phi }&\ldots&N_{n\phi } \end{bmatrix} \begin{bmatrix} u_{1z}\\ \vdots \\ u_{mz}\\ \phi _{1y}\\ \vdots \\ \phi _{ny} \end{bmatrix} =\varvec{N}_{\phi }^\text {T}\varvec{u}_\text {p}\,. \end{aligned}$$

(5.28)

The bending and shear part of the elemental stiffness matrices are obtained in generalization of Eqs. (5.12)–(5.13) as:

(5.29)

and for the shear stiffness matrix

(5.30)

These two stiffness matrices can be superposed additively at this point to obtain the total stiffness matrix.

5.3 Maxima Modules

To automatize the solution procedure explained in the previous section, several Maxima modules were written. The following sections explain a few elements. The entire source code is provided in Chap. 6.

Fig. 5.2

Flowchart for the analytical integration of the elemental stiffness matrix ${\varvec{K}}_{XZ}^\text {e}$ of a Timoshenko beam element in the X-Z plane

The module K_el_Timoshenko_beam_xz_AnalyticalInt_mnNodes(ncoor, m, n) allows the analytical integration (see Fig. 5.2) whereas the module K_el_Timos henko_beam_xz_NumericalInt_mnNodes(ncoor,lint,m, n) is provided to execute numerical integration (see Fig. 5.3).

Fig. 5.3

Flowchart for the numerical integration of the elemental stiffness matrix ${\varvec{K}}_{XZ}^\text {e}$ of a Timoshenko beam element in the X-Z plane

5.4 Examples Based on Maxima

5.1

Example: Timoshenko beam element with quadratic interpolation for the displacement field and linear interpolation for the rotation field

Derive the elemental stiffness matrix for a Timoshenko beam element in the x-z plane with quadratic interpolation for the displacement field $(u_z(x))$ and linear interpolation for the rotation field $(\phi _{y}(x))$, see Fig. 5.4.

Fig. 5.4

Timoshenko beam element with quadratic interpolation functions for the deflection and linear interpolation functions for the rotation: a deformation parameters; b load parameters

Fig. 5.5

Timoshenko beam element with quadratic interpolation functions for the deflection and the rotation: a deformation parameters; b load parameters

Derive the solution first based on the analytical integration and then choose a sufficient number of integration points for the numerical integration of the bending as well as the shear part of the stiffness matrix.

Solution 5.1

The following listing 5.1 shows the entire wxMaxima code for the derivation of elemental stiffness matrix based on analytical integration. It should be noted here that the column matrix of nodal unknowns has in this case the following form:

$$\begin{aligned} \varvec{u}_\text {p} = \begin{bmatrix} u_{1z}&u_{2z}&u_{3z}&\phi _{1y}&\phi _{2y} \end{bmatrix}\,. \end{aligned}$$

(5.31)

Module 5.1

Timoshenko beam element with quadratic interpolation for the displacement field and linear interpolation for the rotation field based on analytical integration

The following listing 5.2 shows the entire wxMaxima code for the derivation of elemental stiffness matrix based on numerical integration.

Module 5.2

Timoshenko beam element with quadratic interpolation for the displacement field and linear interpolation for the rotation field based on numerical integration

5.2

Example: Timoshenko beam element with quadratic interpolation for the displacement field and the rotation field

Derive the elemental stiffness matrix for a Timoshenko beam element in the x-z plane with quadratic interpolation for the displacement field $(u_z(x))$ and quadratic interpolation for the rotation field $(\phi _{y}(x))$, see Fig. 5.5

Solution 5.2

The following listing 5.3 shows the entire wxMaxima code for the derivation of elemental stiffness matrix based on analytical integration. It should be noted here that the column matrix of nodal unknowns has in this case the following form:

$$\begin{aligned} \varvec{u}_\text {p} = \begin{bmatrix} u_{1z}&u_{2z}&u_{3z}&\phi _{1y}&\phi _{2y}&\phi _{3y} \end{bmatrix}\,. \end{aligned}$$

(5.32)

Module 5.3

Timoshenko beam element with quadratic interpolation for the displacement and the rotation field based on analytical integration

The following listing 5.4 shows the entire wxMaxima code for the derivation of elemental stiffness matrix based on numerical integration.

Module 5.4

Timoshenko beam element with quadratic interpolation for the displacement and the rotation field based on numerical integration

5.5 Principal Finite Element Equation

5.5.1 Timoshenko Beam Elements

Bending is again considered in two perpendicular planes. We use in the following the x-y and x-z planes under the assumption that the x-axis is aligned to the longitudinal axis of the beam. Then, one can state the principal finite element equation for a single linear two-noded beam element of length L with constant bending (EI) and shear stiffness $(k_\text {s}AG)$ (see [13, 15] for details of the derivation) in the x-y plane as

$$\begin{aligned} \frac{k_\text {s}AG}{4L} \begin{bmatrix} 4&2L&-4&2L \\ 2L&\tfrac{4}{3}L^2+\alpha&-2L&\tfrac{4}{6}L^2-\alpha \\ -4&-2L&4&-2L \\ 2L&\tfrac{4}{6}L^2-\alpha&-2L&\tfrac{4}{3}L^2+\alpha \\ \end{bmatrix} \begin{bmatrix} u_{1y} \\ \phi _{1z} \\ u_{2y} \\ \phi _{2z} \\ \end{bmatrix} = \begin{bmatrix} F_{1y} \\ M_{1z} \\ F_{2y} \\ M_{2z} \\ \end{bmatrix} + \int \limits _0^L q_y(x) \begin{bmatrix} N_{1u} \\ 0 \\ N_{2u} \\ 0 \\ \end{bmatrix}\mathrm {d}x \,, \end{aligned}$$

(5.33)

where the abbreviation $\alpha =\tfrac{4EI_z}{k_\text {s}AG}$ is used. The corresponding equation for the x-z plane reads as

$$\begin{aligned} \frac{k_\text {s}AG}{4L} \begin{bmatrix} 4&-2L&-4&-2L \\ -2L&\tfrac{4}{3}L^2+\alpha&2L&\tfrac{4}{6}L^2-\alpha \\ -4&2L&4&2L \\ -2L&\tfrac{4}{6}L^2-\alpha&2L&\tfrac{4}{3}L^2+\alpha \\ \end{bmatrix} \begin{bmatrix} u_{1z} \\ \phi _{1y} \\ u_{2z} \\ \phi _{2y} \\ \end{bmatrix} = \begin{bmatrix} F_{1z} \\ M_{1y} \\ F_{2z} \\ M_{2y} \\ \end{bmatrix} + \int \limits _0^L q_z(x) \begin{bmatrix} N_{1u} \\ 0 \\ N_{2u} \\ 0 \\ \end{bmatrix}\mathrm {d}x\,, \end{aligned}$$

(5.34)

where the abbreviation $\alpha =\tfrac{4EI_y}{k_\text {s}AG}$ is used. Both equations can be stated in abbreviated form, i.e in matrix notation, as

$$\begin{aligned} {\varvec{K}}^\text {e} {\varvec{u}}^\text {e}_\text {p}= {\varvec{f}}^\text {e}\,, \end{aligned}$$

(5.35)

where ${\varvec{K}}^\text {e}$ is the elemental stiffness matrix, ${\varvec{u}}^\text {e}_\text {p}$ is the column matrix of nodal unknowns, and ${\varvec{f}}^\text {e}$ is the column matrix of nodal loads. The interpolation functions $N_{1u}(x)=1-\tfrac{x}{L}$ and $N_{2u}(x)=\tfrac{x}{L}$ in Eqs. (5.33) and (5.34) are used to calculate the equivalent nodal loads for a given distributed load q(x). It should be noted here that also other types of interpolation functions can be found in literature, see [15].

Once the nodal displacements and rotations are known, e.g. based on ${\varvec{u}}^\text {e}_\text {p}=({\varvec{K}}^\text {e})^{-1} {\varvec{f}}^\text {e}$, further quantities can be calculated based on this result (the so-called post-processing, see Tables 5.1 and 5.2). Looking at Tables 5.1 and 5.2, it can be concluded that this type of Timoshenko beam element has a linear displacement and linear rotation distribution between the nodes as wells as a linear shear strain distribution within an element. In regard to the internal reactions, the following can be stated: the bending moment distribution is linear while the shear force distribution is constant.

Table 5.1 Displacement, rotation, curvature, shear strain, shear force and bending moment distribution for a linear Timoshenko beam element given as a function of the nodal values in Cartesian coordinates (bending occurs in the x-y plane)

Table 5.2 Displacement, rotation, curvature, shear strain, shear force and bending moment distribution for a linear Timoshenko beam element given as a function of the nodal values in Cartesian coordinates (bending occurs in the x-z plane)

5.5.2 Rotation of Beam Elements

The modeling of more realistic structures requires the arrangement of single beam elements in the three-dimensional space as a connecting mesh. For educational simplicity, we restrict ourselves to plane, i.e. two-dimensional, arrangements of beam structures, see Fig. 4.7 with the corresponding figure for an Euler–Bernoulli beam. Following the same approach as in Sect. 4.5.2, the transversal displacements at each node can be transformed into components parallel to the global axes (X-Y or X-Z plane). Application of the corresponding transformation matrix to the elemental principal finite element Eq. (5.33) gives the following expression of the elemental principal finite element equation of a Timoshenko beam element which is rotated by the rotation angle $\alpha$ in the X-Y plane:

(5.36)

or for the X-Z plane:

(5.37)

It should be noted here that the sine (‘$\text {s}( \alpha )$’) and cosine (‘$\text {c} (\alpha )$’) values of the rotation angle $\alpha$ as well as the element length L can be calculated through the global node coordinates via Eqs. (4.77)–(4.78) and (4.80)–(4.81).

5.6 Generalized Beam Elements

To further increase the modeling ability of one-dimensional elements, it is possible to combine the rod (see Sect. 3.5) and Timoshenko beam element (see Sect. 5.5.1) as a so-called generalized beam element which can now deform along and perpendicular to its longitudinal axis. The independent superposition of both stiffness matrices gives the following formulation for the elemental stiffness matrix in the x-y plane

$$\begin{aligned} \begin{bmatrix} \tfrac{EA}{L}&0&0&-\tfrac{EA}{L}&0&0\\ 0&\tfrac{k_\text {s}AG}{L}&\tfrac{k_\text {s}AG}{2}&0&-\tfrac{k_\text {s}AG}{L}&\tfrac{k_\text {s}AG}{2} \\ 0&\tfrac{k_\text {s}AG}{2}&\tfrac{k_\text {s}AGL}{3}+\tfrac{EI_z}{L}&0&-\tfrac{k_\text {s}AG}{2}&\tfrac{k_\text {s}AGL}{6}-\tfrac{EI_z}{L} \\ -\tfrac{EA}{L}&0&0&\tfrac{EA}{L}&0&0\\ 0&-\tfrac{k_\text {s}AG}{L}&-\tfrac{k_\text {s}AG}{2}&0&\tfrac{k_\text {s}AG}{L}&-\tfrac{k_\text {s}AG}{2} \\ 0&\tfrac{k_\text {s}AG}{2}&\tfrac{k_\text {s}AGL}{6}-\tfrac{EI_z}{L}&0&-\tfrac{k_\text {s}AG}{2}&\tfrac{k_\text {s}AGL}{3}+\tfrac{EI_z}{L} \\ \end{bmatrix} \begin{bmatrix} u_{1x} \\ u_{1y} \\ \phi _{1z} \\ u_{3x} \\ u_{2y} \\ \phi _{2z} \\ \end{bmatrix} = \begin{bmatrix} F_{1x} \\ F_{1y} \\ M_{1z} \\ F_{2x} \\ F_{2y} \\ M_{2z} \\ \end{bmatrix}\,, \end{aligned}$$

(5.38)

or in the x-z plane accordingly:

$$\begin{aligned} \begin{bmatrix} \tfrac{EA}{L}&0&0&-\tfrac{EA}{L}&0&0\\ 0&\tfrac{k_\text {s}AG}{L}&-\tfrac{k_\text {s}AG}{2}&0&-\tfrac{k_\text {s}AG}{L}&-\tfrac{k_\text {s}AG}{2} \\ 0&-\tfrac{k_\text {s}AG}{2}&\tfrac{k_\text {s}AGL}{3}+\tfrac{EI_y}{L}&0&\tfrac{k_\text {s}AG}{2}&\tfrac{k_\text {s}AGL}{6}-\tfrac{EI_y}{L} \\ -\tfrac{EA}{L}&0&0&\tfrac{EA}{L}&0&0\\ 0&-\tfrac{k_\text {s}AG}{L}&\tfrac{k_\text {s}AG}{2}&0&\tfrac{k_\text {s}AG}{L}&\tfrac{k_\text {s}AG}{2} \\ 0&-\tfrac{k_\text {s}AG}{2}&\tfrac{k_\text {s}AGL}{6}-\tfrac{EI_y}{L}&0&\tfrac{k_\text {s}AG}{2}&\tfrac{k_\text {s}AGL}{3}+\tfrac{EI_y}{L} \\ \end{bmatrix} \begin{bmatrix} u_{1x} \\ u_{1z} \\ \phi _{1y} \\ u_{2x} \\ u_{2z} \\ \phi _{2y} \\ \end{bmatrix} = \begin{bmatrix} F_{1y} \\ F_{1z} \\ M_{1y} \\ F_{2x} \\ F_{2z} \\ M_{2y} \\ \end{bmatrix}\,, \end{aligned}$$

(5.39)

The fact that the rod and beam deformation modes are decoupled is represented by the set of zero-entries in Eqs. (5.38)–(5.39).

As in Sect. 4.6, we are going to rotate this generalized beam element in a single plane as a two-dimensional frame structure. The corresponding expressions of the elemental principal finite element equations of a generalized beam element which is either rotated in the X-Y or the X-Z plane is given in Eqs. (5.40) and (5.41).

5.7 Maxima Modules

The solution steps for Timoshenko beam and frame structures are basically the same as outlined for the Euler–Bernoulli beam in Sect. 4.7. The major difference is that the user must define now in addition the shear stiffness $(k_\text {s}AG)$. The entire source code is again provided in Chap. 6.

5.8 Examples Based on Maxima

5.3

Example: Cantilever beam based on one element

Use a single Timoshenko frame element to model the cantilever beam as shown in Fig. 5.6. The member has the properties $A, I, k_\text {s}, E, G$, and the length is equal to L.

Fig. 5.6

Modeling of a cantilever beam with end loads: a single moment and b single force load case

Determine for both load cases

• the deformations of the free node,

• the reactions at the support, and

• the elemental normal and shear forces as well as the bending moments.

Solution 5.3

The solution starts with the free-body diagram of the problem, including a global coordinate system, subdivision of the geometry into the finite elements, indication of the node and element numbers, and local coordinate systems, see Fig. 5.7.

Fig. 5.7

Modeling of a cantilever beam with end load: free-body diagram with node and element numbering: a single moment and b single force load case

The next step is to generate the required nodal arrays (see Table 5.3) and elemental definitions (see Table 5.4) to facilitate the model definition in Maxima.

Table 5.3 Different nodal arrays for the problem shown in Fig. 5.7

Table 5.4 Element definitions for the problem shown in Fig. 5.7

The following listings 5.5–5.6 in the gray boxes show the entire wxMaxima code, including the obtained results.

(a) Single moment load case, see listing 5.5

(b) Single force load case, see listing 5.6.

Module 5.5

Cantilever beam based on one element: single moment load case

Module 5.6

Cantilever beam based on one element: single force load case

5.4

Example: Tensile rod based on one element

Use a single Timoshenko frame element to model the simple tensile rod as shown in Fig. 5.8. The member has a uniform cross-sectional area $A_\text {}$, Young’s modulus E, and the length is equal to L.

Fig. 5.8

Modeling of a simple tensile rod

Fig. 5.9

Modeling of a simple tensile rod: free-body diagram with node and element numbering

Table 5.5 Different nodal arrays for the problem shown in Fig. 5.9

Table 5.6 Element definitions for the problem shown in Fig. 5.9

Determine

• the displacement of the free node,

• the reaction forces at the support, and

• the stress, and normal force in the element.

Solution 5.4

The solution starts with the free-body diagram of the problem, indication of the node and element numbers, and local coordinate systems, see Fig. 5.9.

The next step is to generate the required nodal arrays (see Table 5.5) and elemental definitions (see Table 5.6) to facilitate the model definition in Maxima.

The following listing 5.7 shows the entire wxMaxima code, including the obtained results.

Module 5.7

Tensile rod mode of Timoshenko frame element based on one element

5.5

Example: Timoshenko frame element supported by a rod

Given is a vertical Timoshenko frame element as shown in Fig. 5.10 where a horizontal rod supports the member at the upper end. Both members have the length L whereas the frame has the properties $A, I, k_\text {s}, E, G$ and the rod is characterized by E, A. The structure is loaded at the upper right corner by a prescribed horizontal displacement $u_0$ and vertical force $F_0$.

Consider two finite elements and determine

• the deformations of the nodes,

• the reactions at the supports and nodes where displacements are prescribed, and

• the elemental normal and shear force as well as the bending moment.

Fig. 5.10

Timoshenko frame element supported by a rod: overview

Fig. 5.11

Timoshenko frame element supported by a rod: free-body diagram with node and element numbering

Table 5.7 Different nodal arrays for the problem shown in Fig. 5.11

Table 5.8 Element definitions for the problem shown in Fig. 5.11

Module 5.8

Timoshenko frame element supported by a rod

Fig. 5.12

Generalized cantilever Timoshenko beam with two types of distributed loads: overview

Fig. 5.13

Generalized cantilever Timoshenko beam with two types of distributed loads: free-body diagram with node and element numbering

Solution 5.5

The solution starts with the free-body diagram of the problem, including a global coordinate system, subdivision of the geometry into the finite elements, indication of the node and element numbers, and local coordinate systems, see Fig. 5.11.

The next step is to generate the required nodal arrays (see Table 5.7) and elemental definitions (see Table 5.8) to facilitate the model definition in Maxima.

The following listing 5.8 shows the entire wxMaxima code, including the obtained results.

5.6

Generalized cantilever Timoshenko beam with two types of distributed loads

The generalized Timoshenko beam shown in Fig. 5.12 is loaded by a constant vertical distributed load $q_0$ in the range $0 \le X \le L$ and a constant horizontal load $p_0$ in the range $L \le X \le 2L$. The material constants (E, G) and the geometrical properties $(A,I, k_\text {s})$ are constant and the total length of the beam is equal to 2L. Model the member with two generalized Timoshenko beam finite elements of length L to determine:

• the unknowns at the nodes,

• the reactions at the supports, and

• the internal reactions (normal force, shear force and bending moment) in each element.

Solution 5.6

The solution starts with the free-body diagram of the problem, including a global coordinate system, subdivision of the geometry into the finite elements, indication of the node and element numbers, and local coordinate systems, see Fig. 5.13.

The next step is to generate the required nodal arrays (see Table 5.9) and elemental definitions (see Table 5.10) to facilitate the model definition in Maxima.

The following listing 5.9 shows the entire wxMaxima code, including the obtained results.

Table 5.9 Different nodal arrays for the problem shown in Fig. 5.13

Table 5.10 Element definitions for the problem shown in Fig. 5.13

Fig. 5.14

Plane frame Timoshenko structure composed of generalized beam elements: overview

Fig. 5.15

Plane frame Timoshenko structure composed of generalized beam elements: free-body diagram with node and element numbering

Module 5.9

Generalized cantilever Timoshenko beam with two types of distributed loads

Module 5.10

Plane frame Timoshenko structure composed of generalized beam elements

Table 5.11 Different nodal arrays for the problem shown in Fig. 5.15

Table 5.12 Element definitions for the problem shown in Fig. 5.15

5.7

Plane frame Timoshenko structure composed of generalized beam elements

The plane frame Timoshenko structure shown in Fig. 5.14 is composed of generalized beams which are arranged in a T-shaped formation. The structure is loaded by a single force $F_0$ in the middle of the structure. The material constants (E, G) and the geometrical properties $(A,I, k_\text {s})$ are constant and the horizontal length of the beam is equal to L while the vertical dimension is equal to $\tfrac{L}{2}$. Model the structure with three generalized Timoshenko beam finite elements of length $\tfrac{L}{2}$ to determine:

• the unknowns at the nodes,

• the reactions at the supports, and

• the internal reactions (normal force, shear force and bending moment) in each element.

Solution 5.7

The solution starts with the free-body diagram of the problem, including a global coordinate system, subdivision of the geometry into the finite elements, indication of the node and element numbers, and local coordinate systems, see Fig. 5.15.

The next step is to generate the required nodal arrays (see Table 5.11) and elemental definitions (see Table 5.12) to facilitate the model definition in Maxima.

The following listing 5.10 shows the entire wxMaxima code, including the obtained results.