The uniform bar of length L, mass per unit length m, and cross sectional area A shown in Figure E19.1 is clamped at its left hand end and free at the right hand end. It is allowed to vibrate in the axial direction. The bar is divided into two components and both components are further subdivided

Question

The uniform bar of length [latex]\mathrm{L}[/latex], mass per unit length [latex]\mathrm{m}[/latex], and cross sectional area A shown in Figure E19.1 is clamped at its left hand end and free at the right hand end. It is allowed to vibrate in the axial direction. The bar is divided into two components and both components are further subdivided into 4 finite elements as shown. For each component, obtain reduced stiffness and mass matrices using two fixed interface normal modes and appropriate number of constraint modes. Couple the characteristic matrices by enforcing compatibility of the displacement at the interface and obtain the frequencies and mode shapes of the coupled system. The modulus of elasticity is [latex]\mathrm{E}[/latex].

Accepted Answer

For component 1 the stiffness matrix and consistent mass matrix are given by

[latex]\mathbf{K}^{(1)}=\frac{E A}{l}\left[\begin{array}{rrrr}2 & -1 & 0 & 0 \-1 & 2 & -1 & 0 \0 & -1 & 2 & -1 \0 & 0 & -1 & 1\end{array} ight] \quad \mathbf{M}^{(1)}=m l\left[\begin{array}{cccc}\frac{2}{3} & \frac{1}{6} & 0 & 0 \\frac{1}{6} & \frac{2}{3} & \frac{1}{6} & 0 \0 & \frac{1}{6} & \frac{2}{3} & \frac{1}{6} \0 & 0 & \frac{1}{6} & \frac{1}{3}\end{array} ight] \qquad ext { (a) }[/latex]

where [latex]l=L / 8[/latex].

The partitions of the two matrices along the free d.o.f. are given by

[latex]\mathbf{K}_{f f}^{(1)}=\frac{E A}{l}\left[\begin{array}{rrr}2 & -1 & 0 \-1 & 2 & -1 \0 & -1 & 2\end{array} ight] \quad \quad \mathbf{M}_{f f}^{(1)}=m l\left[\begin{array}{ccc}\frac{2}{3} & \frac{1}{6} & 0 \\frac{1}{6} & \frac{2}{3} & \frac{1}{6} \0 & \frac{1}{6} & \frac{2}{3}\end{array} ight]\qquad ext { (b) }[/latex]

An eigenvalue solution using the matrices in Equation (b) gives the following two lower order mode shapes

[latex]\boldsymbol{\Phi}_{k}^{(1)}=\left[\begin{array}{cc}0.5264 & -0.8666 \0.7444 & 0 \0.5264 & 0.8666 \0 & 0\end{array} ight] \qquad ext { (c) }[/latex]

The single constrained mode obtained from Equation [latex]19.6[/latex]

[latex]\boldsymbol{\Psi}_{f c} =-\mathbf{K}_{f f}^{-1} \mathbf{K}_{f c}[/latex] (19.6)

with [latex]\mathbf{K}_{f_{c}}^{T}=\left[\begin{array}{lll}0 & 0 & -1\end{array} ight][/latex] is

[latex]\boldsymbol{\Psi}_{c}^{T}=\left[\begin{array}{llll}0.25 & 0.50 & 0.75 & 1.0\end{array} ight] \qquad ext { (d) }[/latex]

Using the transformation matrix [latex]\mathbf{T}[/latex] given in Equation [latex]19.8[/latex]

[latex]\begin{array}{c} \left\{\begin{array}{l} \mathbf{u}_f \ \mathbf{u}_c \end{array} ight\} \end{array} {\left[\begin{array}{cc} \boldsymbol{\Phi}_{f k} & \boldsymbol{\Psi}_{f c} \ 0 & \mathbf{I}\end{array} ight]\left\{\begin{array}{l} \mathbf{p}_k \ \mathbf{p}_c \end{array} ight\} } [/latex]

(19.8)

the transformed mass and stiffness matrices work out to

[latex] ilde{\mathbf{M}}^{(1)}=m l\left[\begin{array}{ccc}1.0000 & 0 & 0.8985 \0 & 1.0000 & 0.4330 \0.8985 & 0.4330 & 1.3333\end{array} ight] \quad ilde{\mathbf{K}}^{(1)}=\frac{E A}{l}\left[\begin{array}{ccc}0.6492 & 0 & 0 \0 & 3.00 & 0 \0 & 0 & 0.25\end{array} ight] \qquad ext { (e) }[/latex]

For component 2 the stiffness matrix and consistent mass matrix are given by

[latex]\mathbf{K}^{(2)}=\frac{E A}{l}\left[\begin{array}{rrrrr}2 & -1 & 0 & -1 & 0 \-1 & 2 & -1 & 0 & 0 \0 & -1 & 2 & 0 & -1 \-1 & 0 & 0 & 1 & 0 \0 & 0 & -1 & 0 & 1\end{array} ight] \quad \mathbf{M}^{(2)}=m l\left[\begin{array}{ccccc}\frac{2}{3} & \frac{1}{6} & 0 & \frac{1}{6} & 0 \\frac{1}{6} & \frac{2}{3} & \frac{1}{6} & 0 & 0 \0 & \frac{1}{6} & \frac{2}{3} & 0 & \frac{1}{6} \\frac{1}{6} & 0 & 0 & \frac{1}{3} & 0 \0 & 0 & \frac{1}{6} & 0 & \frac{1}{3}\end{array} ight] \qquad ext { (f) }[/latex]

The mode shapes and frequencies obtained from the 3 by 3 inner partitions of the stiffness and mass matrices along the free d.o.f. of the component are identical to those for component 1. The two constrained mode, the first related to the interface d.o.f. 4 and the other related to the boundary d.o.f. 5, are obtained from

[latex]\begin{aligned}&\boldsymbol{\Psi}_{f c}^{(2)}=-\left[\begin{array}{rrr}2 & -1 & 0 \-1 & 2 & -1 \0 & -1 & 2\end{array} ight]^{-1}\left[\begin{array}{rr}-1 & 0 \0 & 0 \0 & -1\end{array} ight]=\left[\begin{array}{ll}0.75 & 0.25 \0.50 & 0.50 \0.25 & 0.75\end{array} ight]\qquad (\mathrm{g}) \&\boldsymbol{\Psi}_{c}^{(2)}=\left[\begin{array}{c}\boldsymbol{\Psi}_{f c} \\mathbf{I}\end{array} ight]^{(2)}=\left[\begin{array}{cc}0.75 & 0.25 \0.50 & 0.50 \0.25 & 0.75 \1.00 & 0.00 \0.00 & 1.00\end{array} ight]\qquad (\mathrm{h})\end{aligned}[/latex]

The transformed mass and stiffness matrices as obtained from Equation [latex]19.15[/latex]

[latex]\begin{aligned} ilde{\mathbf{M}}^{(2)} &=\left[\begin{array}{ccc}\mathbf{I}_{k k} & ilde{\mathbf{M}}_{k a} & ilde{\mathbf{M}}_{k r} \ ilde{\mathbf{M}}_{a k} & ilde{\mathbf{M}}_{a a} & ilde{\mathbf{M}}_{a r}^{(2)} \ ilde{\mathbf{M}}_{r k} & ilde{\mathbf{M}}_{r a} & ilde{\mathbf{M}}_{r r}\end{array} ight], \quad ilde{\mathbf{K}}^{(2)}=\left[\begin{array}{ccc}\Lambda_{k k} & 0 & 0 \0 & ilde{\mathbf{K}}_{a a} & ilde{\mathbf{K}}_{a r} \0 & ilde{\mathbf{K}}_{r a} & ilde{\mathbf{K}}_{r r}\end{array} ight]^{(2)}\end{aligned}[/latex] (19.15)

are

[latex]\begin{aligned}& ilde{\mathbf{M}}^{(2)}=m l\left[\begin{array}{cccc}1.0000 & 0 & 0.8985 & 0.8985 \0 & 1.0000 & -0.4330 & 0.4330 \0.8985 & -0.4330 & 1.3333 & 0.6667 \0.8985 & 0.4330 & 0.6667 & 1.3333\end{array} ight] \& ilde{\mathbf{K}}^{(2)}=\frac{E A}{l}\left[\begin{array}{cccc}0.6492 & 0 & 0 & 0 \0 & 3.0000 & 0 & 0 \0 & 0 & 0.2500 & -0.2500 \0 & 0 & -0.2500 & 0.2500\end{array} ight]\end{aligned} \qquad ext{(i)}[/latex]

The component matrices can be coupled by noting the correspondence between the local generalized coordinates and global generalized coordinates shown in Table E19.1.

Table E19.1 Correspondence between the local and global generalized coordinates.

[latex]\begin{array}{lllll}\hline ext{Local coordinate} & 1 & 2 & 3 & 4 \\hline & ext{Global}& ext{coordinate }\\hline ext{Component }1 & 1 & 2 & 3 & \ ext{Component }2 & 4 & 5 & 3 & 6 \\hline\end{array}[/latex]

The assembled stiffness and mass matrices are given by

[latex]\begin{aligned} ilde{\mathbf{K}} =\frac{E A}{l}\left[\begin{array}{cccccc}0.6492 & 0 & 0 & 0 & 0 & 0 \0 & 3.00 & 0 & 0 & 0 & 0 \0 & 0 & 0.50 & 0 & 0 & -0.25 \0 & 0 & 0 & 0.6492 & 0 & 0 \0 & 0 & 0 & 0 & 3.00 & 0 \0 & 0 & -0.25 & 0 & 0 & 0.25\end{array} ight] \quad(\mathrm{j}) \ ilde{\mathbf{M}}=m l {\left[\begin{array}{cccccc}1.0000 & 0 & 0.8985 & 0 & 0 & 0 \0 & 1.0000 & 0.4330 & 0 & 0 & 0 \0.8965 & 0.4330 & 2.6667 & 0.8985 & -0.4330 & 0.6667 \0 & 0 & 0.8985 & 1.0000 & 0 & 0.8985 \0 & 0 & -0.4330 & 0 & 1.0000 & 0.4330 \0 & 0 & 0.6667 & 0.8985 & 0.4330 & 1.3333\end{array} ight](\mathrm{k}) }\end{aligned}[/latex]

An eigenvalue solution using the coupled mass and stiffness matrices gives the following frequencies

[latex]\omega^{T}=\left[\begin{array}{llllll}1.5735 & 4.7817 & 8.1772 & 12.001 & 16.798 & 21.603\end{array} ight] \sqrt{\frac{E A}{m L^{2}}}\qquad (\mathrm{l})[/latex]

These may be compared with the following exact values for the axial vibrations of a uniform clamped free bar obtained from Equation [latex]15.113[/latex].

[latex]\omega^{T}=\left[\begin{array}{llllll}1.5708 & 4.7124 & 7.854 & 10.996 & 14.137 & 17.279\end{array} ight] \sqrt{\frac{E A}{m L^{2}}} \qquad (\mathrm{m})[/latex]

Question 19.1: The uniform bar of length L, mass per unit length m, and cro...

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