18.8: Dynamic bending of a bar with two axes of symmetry

Last updated
Save as PDF

Page ID: 95339

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\dsum}{\displaystyle\sum\limits} $

$ \newcommand{\dint}{\displaystyle\int\limits} $

$ \newcommand{\dlim}{\displaystyle\lim\limits} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$\newcommand{\longvect}{\overrightarrow}$

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

18.8 Dynamic bending of a bar with two axes of symmetry

If the cross section is symmetric with respect to both the x- and y-axes through the centroid, then $x_{s c} = y_{s c} = 0$ $math-4236.png$ , I_xy = 0, $r_{x y} = 0, and s_{x y} = 0$ $math-4237.png$ . In this case of double symmetry, transverse bending is decoupled from torsion in both the inertia and stiffness terms. That is, the inertia axis and elastic axis coincide. However, the motions of the lateral displacement v(z, t) and the rotation ϕ_x(z, t) are linked because of the presence of transverse shear deformation ψ_y. The governing weak forms (18.117) and (18.119) are

$\begin{array}{l} \int_{0}^{L} [m (\ddot{v}) δ v + s_{y y} (v^{'} + ϕ_{x}) δ v^{'}] d z = \int_{0}^{L} f_{y} (z, t) δ v (z) d z + Q_{2} (t) δ v (0, t) + Q_{8} (t) δ v (L, t) . \\ \int_{0}^{L} [m (r_{x}^{2} {\ddot{ϕ}}_{x}) δ ϕ_{x} + E I_{x x} ϕ_{x}^{'} δ ϕ_{x}^{'} + s_{y y} (v^{'} + ϕ_{x}) δ ϕ_{x}] d z \\ = \int_{0}^{L} m_{x} (z, t) δ ϕ_{x} (z, t) d z + Q_{4} (t) δ ϕ_{x} (0, t) + Q_{10} (t) δ ϕ_{x} (L, t) \end{array}$ $math-4238.png$

The previous two equations are combined to the matrix form

$\begin{align} \int_{0}^{L} \{[δ ϕ_{x} δ v] [\begin{matrix} m r_{x}^{2} & 0 \\ 0 & m \end{matrix}] [\begin{matrix} {\ddot{ϕ}}_{x} \\ \ddot{v} \end{matrix}] + δ ϕ_{x}^{'} E I_{x x} ϕ_{x}^{'} + (δ v^{'} + δ ϕ_{x}) s_{y y} (v^{'} + ϕ_{x})\} d z = δ W_{n . c}, & (18.137) \end{align}$ $math-4239.png$

where the virtual work of the non-conservative forces is

$\begin{align} δ W_{n.c.} = \int_{0}^{L} [δ ϕ x δ v] [\begin{matrix} m_{z} (z, t) \\ f_{y} (z, t) \end{matrix}] d z + [δ ϕ_{x} (0, t) δ v (0, t)] [\begin{matrix} Q_{2} (t) \\ Q_{4} (t) \end{matrix}] + [\begin{matrix} δ ϕ_{x} (L, t) δ v (L, t) \end{matrix}] [\begin{matrix} Q_{8} (t) \\ Q_{10} (t) \end{matrix}] . & (18.138) \end{align}$ $math-4240.png$

18.8.1 Finite element

The development of the generalized displacements in Ω_k follows the discussion in article 17.3.1 on page 462. The lateral displacement of the kth element is denoted by v^(k)(ζ, t) and the rotation by $ϕ_{x}^{(k)} (ζ, t)$ $math-4241.png$ . Define the generalized external displacements in terms of the rotation and displacement at the nodes by

$\begin{align} v^{(k)} (- 1, t) = u_{2 k - 1} (t) ϕ_{x}^{(k)} (- 1, t) = u_{2 k} (t) v^{(k)} (1, t) = u_{2 k + 1} (t) ϕ_{x}^{(k)} (1, t) = u_{2 k + 2} (t) . & (18.139) \end{align}$ $math-4242.png$

See figure 18.16. The 7X1 displacement vector of element Ω_k is denoted by,

The global z axis is aligned with the beam of length h sub k, with the vertical y axis and v displacements positive up, normal to the beam length. Rotation phi sub x is measured positive clockwise. The left edge is the node at z sub k, with vertical displacement u sub 2 k minus 1, shear force cap Q sub 4, rotation u sub 2 k, and applied moment cap Q sub 2. Within the element, the distributed shear load is denoted as f sub y of eta sub 1 times z sub k plus eta sub 2 times z sub k plus 1, all at time t. The distributed moment is denoted as m sub x of eta sub 1 times z sub k plus eta sub 2 times z sub k plus 1, all at time t. The right edge is the node at z sub k plus 1, with vertical displacement u sub 2 k plus 1, shear cap Q sub 10, rotation u sub 2 k plus 2, and applied moment cap Q sub 8. — Fig. 18.16 Beam element Ω_k.

$\begin{align} \{u^{(k)} (t)\} = {[u_{2 k - 1} (t) u_{2 k} (t) u_{2 k + 1} (t) u_{2 k + 2} (t) u_{1}^{(k)} (t) u_{2}^{(k)} (t) u_{3}^{(k)} (t)]}^{T}, & (18.140) \end{align}$ $math-4243.png$

where $u_{1}^{(k)} (t)$ $math-4244.png$ , $u_{2}^{(k)} (t), and u_{3}^{(k)} (t)$ $math-4245.png$ are internal generalized displacement degrees of freedom. Rotation and displacement functions within the element are expressed in terms of the 2X7 shape function matrix and the 7X1 displacement vector:

$\begin{align} [\begin{matrix} ϕ_{x}^{(k)} (ζ, t) \\ v^{(k)} (ζ, t) \end{matrix}] = [N (ζ)] \{u^{(k)} (t)\}, & (18.141) \end{align}$ $math-4246.png$

where the shape function matrix is

$\begin{align} [N (ζ)] = [\begin{matrix} 0 & η_{1} (ζ) & 0 & η_{2} (ζ) & 0 & η_{3} (ζ) & 0 \\ η_{1} (η) & 0 & η_{2} (ζ) & 0 & η_{3} (ζ) & 0 & η_{4} (ζ) \end{matrix}] . & (18.142) \end{align}$ $math-4247.png$

The basis functions for the element are

$\begin{align} η_{1} (ζ) = (1 - ζ) / 2 η_{2} (ζ) = (1 + ζ) / 2 η_{3} (ζ) = (\frac{1}{2} \sqrt{\frac{3}{2}}) (ζ^{2} - 1) η_{4} (ζ) = (\frac{1}{2} \sqrt{\frac{5}{2}}) ζ (ζ^{2} - 1) . & (18.143) \end{align}$ $math-4248.png$

The virtual rotation and displacement within the element are

$\begin{align} [\begin{matrix} δ ϕ_{x} \\ δ v \end{matrix}] = [N (ζ)] \{b^{(k)}\}, & (18.144) \end{align}$ $math-4249.png$

where the 7X1 vector $\{b^{(k)}\} = {[\begin{matrix} b_{2 k - 1} & b_{2 k} & b_{2 k + 1} & b_{2 k + 2} & b_{1}^{(k)} & b_{2}^{(k)} & b_{3}^{(k)} \end{matrix}]}^{T}$ $math-4250.png$ . The virtual generalized displacements $b_{1}^{(k)}$ $math-4251.png$ , $b_{2}^{(k)}, and b_{3}^{(k)}$ $math-4252.png$ correspond to the internal degrees of freedom $u_{1}^{(k)} (t)$ $math-4253.png$ , $u_{2}^{(k)} (t), and u_{3}^{(k)} (t)$ $math-4254.png$ , respectively. The virtual generalized displacement vector {b^(k)} is independent of the physical generalized displacement vector {u^(k)}. The external virtual work of the non-conservative forces (18.138) for the elements is given by

$\begin{align} δ W_{n.c.} = \sum_{k = 1}^{M} b_{2 k - 1} Q_{2}^{(k)} + b_{2 k} Q_{4}^{(k)} + b_{2 k + 1} Q_{8}^{(k)} + b_{2 k + 2} Q_{10}^{(k)} + \sum_{k = 1}^{M} \{b^{(k)}\} \int_{- 1}^{1} {[N]}^{T} [\begin{matrix} m_{x} \\ f_{y} \end{matrix}] \frac{h_{k}}{2} d ζ . & (18.145) \end{align}$ $math-4255.png$

At the common node z_k between elements Ω_k − 1 and Ω_k there is an equilibrium relation between the externally applied force Q_2k − 1 and the externally applied moment Q_2k and the internal actions at the end of element Ω_k − 1 and the beginning of element Ω_k. Refer figure 18.17. These relations are

At a node shared by elements cap Omega sub k minus 1 and cap Omega sub k, the node experiences an equal and opposite reaction from the cap Q sub 8 and cap Q sub 10 of element k minus 1, and cap Q sub 2 and cap Q sub 4 of element k. These forces sum to oppose the applied moment Q sub 2 k and applied shear force cap Q sub 2 k minus 1 at the node. — Fig. 18.17 Free body diagram at a node between two elements

$\begin{matrix} Q_{2 k - 1} = Q_{8}^{(k - 1)} + Q_{2}^{(k)} Q_{2 k} = Q_{10}^{(k - 1)} + Q_{4}^{(k)} k = 2, 3, \dots, M - 1, and & (18.146) \\ Q_{1} = Q_{2}^{(1)} Q_{2} = Q_{4}^{(1)} Q_{2 M + 1} = Q_{8}^{(M)} Q_{2 M + 2} = Q_{10}^{(M)} . & (18.147) \end{matrix}$ $math-4256.png$

We now write the virtual work of the non-conservative generalized forces as

$\begin{align} δ W_{n.c.} = \sum_{k = 1}^{M} \{b^{(k)}\} (\{Q^{(k)}\} + \{F^{(k)}\}), & (18.148) \end{align}$ $math-4257.png$

where

$\begin{align} \{Q^{(k)}\} = & {[Q_{2 k - 1} Q_{2 k} Q_{2 k + 1} Q_{2 k + 2} 0 0 0]}^{T}, and \\ \{F^{(k)} (t)\} = \int_{- 1}^{1} {[N]}^{T} [\begin{matrix} m_{x} [η_{1} (ζ) z_{k} + η_{2} (ζ) z_{k + 1}, t] \\ f_{y} [η_{1} (ζ) z_{k} + η_{2} (ζ) z_{k + 1}, t] \end{matrix}] \frac{h_{k}}{2} d ζ . & (18.149) \end{align}$ $math-4258.png$

The partial derivatives with respect to coordinate z in eq. (18.137) are replaced by derivatives with respect to dimensionless coordinate ζ using the chain rule. That is,

$\begin{align} \frac{\partial ϕ_{x}}{\partial z} = \frac{\partial ϕ_{x}}{\partial ζ} \frac{d ζ}{d z} = \frac{2}{h_{k}} \frac{\partial ϕ_{x}}{\partial ζ} = \frac{2}{h_{k}} ϕ_{x}^{'} d z = \frac{d z}{d ζ} d ζ = \frac{h_{k}}{2} d ζ . & (18.150) \end{align}$ $math-4259.png$

Note that in the following finite element development the prime superscript denotes a derivative with respect to ζ. The derivative of the rotation for element Ω_k and the virtual rotation are

$\begin{align} ϕ_{x}^{'} = [N_{ϕ} (ζ)] \{u^{(k)}\} δ ϕ_{x}^{'} = [N_{ϕ} (ζ)] \{b^{(k)}\}, & (18.151) \end{align}$ $math-4260.png$

where the 1X7 matrix is given by

$\begin{align} [N_{ϕ} (ζ)] = [\begin{matrix} 0 & η_{1}^{'} & 0 & η_{2}^{'} & 0 & η_{3}^{'} & 0 \end{matrix}] . & (18.152) \end{align}$ $math-4261.png$

The shear strain for element Ω_k and the virtual shear strain are

$\begin{align} (\frac{2}{h_{k}} v^{'} + ϕ_{x}) = [N_{ψ} (ζ)] \{u^{(k)}\} (\frac{2}{h_{k}} δ v^{'} + δ ϕ_{x}) = [N_{ψ} (ζ)] {b}, & (18.153) \end{align}$ $math-4262.png$

where the 1X7 matrix is given by

$\begin{align} [N_{ψ} (ζ)] = [\begin{matrix} \frac{2}{h_{k}} η_{1}^{'} & η_{1} & \frac{2}{h_{k}} η_{2}^{'} & η_{2} & \frac{2}{h_{k}} η_{3}^{'} & η_{3} & \frac{2}{h_{k}} η_{4}^{'} \end{matrix}] . & (18.154) \end{align}$ $math-4263.png$

Substitute eqs. (18.151) and (18.153) into the finite element representation of eq. (18.137), and substitute eq. (18.148) for the virtual work, to get

$\begin{align} \sum_{k = 1}^{M} ({b^{(k)}}^{T} \int_{- 1}^{1} ({[N]}^{T} [\begin{matrix} m r_{x}^{2} & 0 \\ 0 & m \end{matrix}] [N] {ü^{(k)}} + \frac{2}{h_{k}} {[N_{ϕ}]}^{T} E I_{x x} \frac{2}{h_{k}} [N_{ϕ}] {u^{(k)}} \\ + \frac{2}{h_{k}} {[N_{ψ}]}^{T} s_{y y} [N_{ψ}] {u^{(k)}}) \frac{h_{k}}{2} d ζ) = \sum_{k = 1}^{M} {b^{(k)}}^{T} ({Q^{(k)}} + {F^{(k)}}) . & (18.155) \end{align}$ $math-4264.png$

We satisfy eq. (18.155) for each element in the mesh by

$\begin{align} {\{b^{(k)}\}}^{T} ([M] \{ü^{(k)}\} + [K] \{u^{(k)}\} - (\{Q^{(k)}\} + \{F^{(k)}\})) = 0 \forall (\{b^{(k)}\} \neq 0_{7 X 1}) . & (18.156) \end{align}$ $math-4265.png$

Hence, the equation of motion for element Ω_k is

$\begin{align} [M] \{ü^{(k)}\} + [K] \{ü^{(k)}\} = \{Q^{(k)}\} + \{F^{(k)}\} . & (18.157) \end{align}$ $math-4266.png$

The mass, and stiffness matrices are

$\begin{align} [M] = \int_{- 1}^{1} {[N]}^{T} [\begin{matrix} m r_{x}^{2} & 0 \\ 0 & m \end{matrix}] [N] \frac{h_{k}}{2} d ζ, and [K] = \int_{- 1}^{1} (\frac{2}{h_{k}} {[N_{ϕ}]}^{T} E I_{x x} \frac{2}{h_{k}} [N_{ϕ}] + \frac{2}{h_{k}} {[N_{ψ}]}^{T} s_{y y} [N_{ψ}]) \frac{h_{k}}{2} d ζ . & (18.158) \end{align}$ $math-4267.png$

Perform the matrix algebra in eq. (18.158) to find the 7X7 mass matrix

$\begin{align} [M] = m h_{k} [\begin{matrix} 1 / 3 & 0 & 1 / 6 & 0 & - 1 / 2 \sqrt{6} & 0 & 1 / 6 \sqrt{10} \\ 0 & r_{x}^{2} / 3 & 0 & r_{x}^{2} / 6 & 0 & - r_{x}^{2} / 2 \sqrt{\sqrt{6}} & 0 \\ 1 / 6 & 0 & 1 / 3 & 0 & - 1 / 2 \sqrt{6} & 0 & - 1 / 6 \sqrt{10} \\ 0 & r_{x}^{2} / 6 & 0 & r_{x}^{2} / 3 & 0 & - r_{x}^{2} / 2 \sqrt{6} & 0 \\ - 1 / 2 \sqrt{6} & 0 & - 1 / 2 \sqrt{6} & 0 & 1 / 5 & 0 & 0 \\ 0 & - r_{x}^{2} / (2 \sqrt{6}) & 0 & - r_{x}^{2} / 2 \sqrt{6} & 0 & r_{x}^{2} / 5 & 0 \\ 1 / (6 \sqrt{10}) & 0 & - 1 / 6 \sqrt{10} & 0 & 0 & 0 & 1 / 21 \end{matrix}] . & (18.159) \end{align}$ $math-4268.png$

Perform the matrix algebra in of eq. (18.158) to find the 7X7 stiffness matrix

$\begin{align} [K] = [\begin{matrix} s_{y y} / h_{k} & - s_{y y} / 2 & - s_{y y} / h_{k} & - s_{y y} / 2 & 0 & s_{y y} / \sqrt{6} & 0 \\ - s_{y y} / 2 & E I_{x x} / h_{k} + h_{k} s_{y y} / 3 & s_{y y} / 2 & - E I_{x x} / h_{k} + h_{k} s_{y y} / 6 & - s_{y y} / \sqrt{6} & - h_{k} s_{y y} / 2 \sqrt{6} & 0 \\ - s_{y y} / h_{k} & s_{y y} / 2 & s_{y y} / h_{k} & s_{y y} / 2 & 0 & - s_{y y} / \sqrt{6} & 0 \\ - s_{y y} / 2 & - E I_{x x} / h_{k} + h_{k} s_{y y} / 6 & s_{y y} / 2 & E I_{x x} / h_{k} + h_{k} s_{y y} / 3 & s_{y y} / \sqrt{6} & - h_{k} s_{y y} / 2 \sqrt{6} & 0 \\ 0 & - s_{y y} / \sqrt{6} & 0 & s_{y y} / \sqrt{6} & 2 s_{y y} / h_{k} & 0 & 0 \\ s_{y y} / \sqrt{6} & - h_{k} s_{y y} / 2 \sqrt{6} & - s_{y y} / \sqrt{6} & - h_{k} s_{y y} / 2 \sqrt{6} & 0 & 2 E I_{x x} / h_{k} + h_{k} s_{y y} / 5 & s_{y y} / \sqrt{15} \\ 0 & 0 & 0 & 0 & 0 & s_{y y} / \sqrt{15} & 2 s_{y y} / h_{k} \end{matrix}] . & (18.160) \end{align}$ $math-4269.png$

Search

Text Color

Text Size

Margin Size

Font Type

18.8 Dynamic bending of a bar with two axes of symmetry

18.8.1 Finite element