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446 lines
23 KiB
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446 lines
23 KiB
Markdown
<!-- source-page: 691 -->
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$$
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\begin{array}{l} \times \left\{ \begin{array}{c} 0 \\ 0. 8 5 8 \times 1 0 ^ {- 3} \end{array} \right\} - \frac {0 . 0 7 3}{2} \left[ \begin{array}{c c} 2 & 0 \\ 0 & 1 \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0 \end{array} \right\} \Bigg \} \\ = \frac {2}{0 . 0 7 3} \left[ \begin{array}{c c} \frac {1}{2} & 0 \\ 0 & 1 \end{array} \right] \left[ \left\{ \begin{array}{c} 0 \\ 0. 0 6 2 5 \times 1 0 ^ {- 3} \end{array} \right\} + \left\{ \begin{array}{c} 0. 0 1 6 1 \times 1 0 ^ {- 3} \\ 0. 0 4 6 6 \times 1 0 ^ {- 3} \end{array} \right\} \right] \\ \end{array}
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$$
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Simplifying, we obtain the nodal displacements at time $t = 0.50 \times 10^{-3}$ s as
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$$
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\left\{ \begin{array}{l} d _ {2 x} \\ d _ {3 x} \end{array} \right\} _ {2} = \left\{ \begin{array}{l} 0. 2 2 1 \times 1 0 ^ {- 3} \\ 2. 9 9 \times 1 0 ^ {- 3} \end{array} \right\} \text { in. } \quad \text {(at} t = 0. 5 0 \times 1 0 ^ {- 3} \text { s) } \tag {16.5.13}
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$$
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# Step 6
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Solve for the nodal accelerations $\ddot{d}_{1}$ again using Eq. (16.3.5) as
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$$
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\ddot {\underline {{d}}} _ {1} = \frac {2}{0 . 0 7 3} \left[ \begin{array}{c c} \frac {1}{2} & 0 \\ 0 & 1 \end{array} \right] \left[ \left\{ \begin{array}{c} 0 \\ 1 0 0 0 \end{array} \right\} - (3 0 \times 1 0 ^ {4}) \left[ \begin{array}{c c} 2 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{c} 0 \\ 0. 8 5 8 \times 1 0 ^ {- 3} \end{array} \right\} \right]
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$$
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Simplifying, we then obtain the nodal accelerations at time $t = 0.25 \times 10^{-3}$ s as
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$$
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\left\{ \begin{array}{l} \ddot {d} _ {2 x} \\ \ddot {d} _ {3 x} \end{array} \right\} _ {1} = \left\{ \begin{array}{l} 3 5 2 6 \\ 2 0, 3 4 5 \end{array} \right\} \text {in.} / \mathrm{s} ^ {2} \quad \text {(at} t = 0. 2 5 \times 1 0 ^ {- 3} \mathrm{s}) \tag {16.5.14}
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$$
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The reaction $R_{1}$ could be found by using the results of Eqs. (16.5.12) and (16.5.14) in Eq. (16.5.5).
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# Step 7
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Using Eq. (16.5.13) from step 5 and the boundary condition for $\underline{d}_0$ given in step 1, we obtain $\underline{d}_1$ as
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$$
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\underline {{d}} _ {1} = \frac {\left[ \left\{ \begin{array}{l} 0 . 2 2 1 \times 1 0 ^ {- 3} \\ 2 . 9 9 \times 1 0 ^ {- 3} \end{array} \right\} - \left\{ \begin{array}{l} 0 \\ 0 \end{array} \right\} \right]}{2 (0 . 2 5 \times 1 0 ^ {- 3})}
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$$
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Simplifying, we obtain
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$$
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\left\{ \begin{array}{l} \dot {d} _ {2 x} \\ \dot {d} _ {3 x} \end{array} \right\} = \left\{ \begin{array}{c} 0. 4 4 2 \\ 5. 9 8 \end{array} \right\} \text { in. / s } \quad (\text { at } t = 0. 2 5 \times 1 0 ^ {- 3} \mathrm{s})
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$$
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# Step 8
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We now use steps 5–7 repeatedly to obtain the displacement, acceleration, and velocity for all other time steps. For simplicity, we calculate the acceleration only.
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Repeating step 6 with $t = 0.50 \times 10^{-3}$ s, we obtain the nodal accelerations as
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$$
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\ddot {\underline {{d}}} _ {2} = \frac {2}{0 . 0 7 3} \left[ \begin{array}{c c} \frac {1}{2} & 0 \\ 0 & 1 \end{array} \right] \left[ \left\{ \begin{array}{c} 0 \\ 1 0 0 0 \end{array} \right\} - 3 0 \times 1 0 ^ {4} \left[ \begin{array}{c c} 2 & - 1 \\ - 1 & 1 \end{array} \right] \left\{ \begin{array}{c} 0. 2 2 1 \times 1 0 ^ {- 3} \\ 2. 9 9 \times 1 0 ^ {- 3} \end{array} \right\} \right]
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$$
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<!-- source-page: 692 -->
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On simplifying, the nodal accelerations at $t = 0 . 5 0 \times 1 0 ^ { - 3 }$ s are
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$$
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\begin{array}{l} \left\{ \begin{array}{l} \ddot {d} _ {2 x} \\ \ddot {d} _ {3 x} \end{array} \right\} _ {2} = \left\{ \begin{array}{c} 0 \\ 2 7, 4 0 0 \end{array} \right\} + \left\{ \begin{array}{c} 1 0, 5 0 0 \\ - 2 2, 8 0 0 \end{array} \right\} \\ = \left\{ \begin{array}{c} 1 0, 5 0 0 \\ 4 6 0 0 \end{array} \right\} \text { in. } / \mathrm{s} ^ {2} \quad \text {(at} t = 0. 5 \times 1 0 ^ {- 3} \mathrm{s}) \tag {16.5.15} \\ \end{array}
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$$
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# 16.6 Beam Element Mass Matrices and Natural Frequencies
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We now consider the lumped- and consistent-mass matrices appropriate for timedependent beam analysis. The development of the element equations follows the same general steps as used in Section 16.2 for the bar element.
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The beam element with the associated nodal degrees of freedom (transverse displacement and rotation) is shown in Figure 16–15.
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The basic element equations are given by the general form, Eq. (16.2.10), with the appropriate nodal force, stiffness, and mass matrices for a beam element. The stiffness matrix for the beam element is that given by Eq. (4.1.14). A lumped-mass matrix is obtained as
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$$
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[ \hat {m} ] = \frac {\rho A L}{2} \left[ \begin{array}{c c c c} \hat {d} _ {1 y} & \hat {\phi} _ {1} & \hat {d} _ {2 y} & \hat {\phi} _ {2} \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \tag {16.6.1}
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$$
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where one-half of the total beam mass has been lumped at each node, corresponding to the translational degrees of freedom. In the lumped mass approach, the inertial effect associated with possible rotational degrees of freedom has been assumed to be zero in obtaining Eq. (16.6.1), although a value may be assigned to these rotational degrees of freedom by calculating the mass moment of inertia of a fraction of the beam segment about the nodal points. For a uniform beam we could then calculate the mass moment of inertia of half of the beam segment about each end node using
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<details>
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<summary>text_image</summary>
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d̂₁ᵧ
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φ̂₁
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1 → x̂
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L
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d̂₂ᵧ
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φ̂₂
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2
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</details>
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Figure 16–15 Beam element with nodal degrees of freedom
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<!-- source-page: 693 -->
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basic dynamics as
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$$
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I = \frac {1}{3} (\rho A L / 2) (L / 2) ^ {2}
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$$
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Again, the lumped-mass matrix given by Eq. (16.6.1) is a diagonal matrix, making matrix numerical calculations easier to perform than when using the consistent-mass matrix. The consistent-mass matrix can be obtained by applying the general Eq. (16.2.19) for the beam element, where the shape functions are now given by Eqs. (4.1.7). Therefore,
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$$
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[ \hat {m} ] = \iint_ {V} \rho [ N ] ^ {T} [ N ] d V \tag {16.6.2}
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$$
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$$
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[ \hat {m} ] = \int_ {0} ^ {L} \iint_ {A} \rho \left\{ \begin{array}{l} N _ {1} \\ N _ {2} \\ N _ {3} \\ N _ {4} \end{array} \right\} \left[ \begin{array}{l l l l} N _ {1} & N _ {2} & N _ {3} & N _ {4} \end{array} \right] d A d \hat {x} \tag {16.6.3}
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$$
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with $N _ { 1 } = \frac { 1 } { L ^ { 3 } } ( 2 \hat { x } ^ { 3 } - 3 \hat { x } ^ { 2 } L + L ^ { 3 } )$
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$$
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N _ {2} = \frac {1}{L ^ {3}} \left(\hat {x} ^ {3} L - 2 \hat {x} ^ {2} L ^ {2} + \hat {x} L ^ {3}\right) \tag {16.6.4}
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$$
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$$
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N _ {3} = \frac {1}{L ^ {3}} \left(- 2 \hat {x} ^ {3} + 3 \hat {x} ^ {2} L\right)
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$$
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$$
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N _ {4} = \frac {1}{L ^ {3}} \left(\hat {x} ^ {3} L - \hat {x} ^ {2} L ^ {2}\right)
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$$
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On substituting the shape function Eqs. (16.6.4) into Eq. (16.6.3) and performing the integration, the consistent-mass matrix becomes
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$$
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[ \hat {m} ] = \frac {\rho A L}{4 2 0} \left[ \begin{array}{c c c c} 1 5 6 & 2 2 L & 5 4 & - 1 3 L \\ 2 2 L & 4 L ^ {2} & 1 3 L & - 3 L ^ {2} \\ 5 4 & 1 3 L & 1 5 6 & - 2 2 L \\ - 1 3 L & - 3 L ^ {2} & - 2 2 L & 4 L ^ {2} \end{array} \right] \tag {16.6.5}
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$$
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Having obtained the mass matrix for the beam element, we could proceed to formulate the global stiffness and mass matrices and equations of the form given by Eq. (16.2.24) to solve the problem of a beam subjected to a time-dependent load. We will not illustrate the procedure for solution here because it is tedious and similar to that used to solve the one-dimensional bar problem in Section 16.5. However, a computer program can be used for the analysis of beams and frames subjected to timedependent forces. Section 16.7 provides descriptions of plane frame and other element mass matrices, and Section 16.9 describes some computer program results for dynamics analysis of bars, beams, and frames.
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To clarify the procedure for beam analysis, we will now determine the natural frequencies of a beam.
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<!-- source-page: 694 -->
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We now consider the determination of the natural frequencies of vibration for a beam fixed at both ends as shown in Figure 16–16. The beam has mass density $\rho ,$ modulus of elasticity $E ,$ cross-sectional area A, area moment of inertia I, and length 2L. For simplicity of the longhand calculations, the beam is discretized into (a) two beam elements of length L (Figure 16–16(a)) and then (b) three beam elements of length L each (Figure 16–16(b)).
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<details>
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<summary>text_image</summary>
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1
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2
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3
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L
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L
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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1
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2
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3
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4
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L
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L
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L
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</details>
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Figure 16–16 Beam for determination of natural frequencies
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# (a) Two-Element Solution
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We can obtain the natural frequencies by using the general Eq. (16.4.7). First, we assemble the global stiffness and mass matrices (using the boundary conditions $d _ { 1 y } = 0 , \phi _ { 1 } = 0 , d _ { 3 y } = 0$ , and $\phi _ { 3 } = 0$ to reduce the matrices) as
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$$
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\underline {{K}} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c} 2 4 & 0 \\ 0 & 8 L ^ {2} \end{array} \right] \quad \underline {{M}} = \frac {\rho A L}{2} \left[ \begin{array}{c c} 2 & 0 \\ 0 & 0 \end{array} \right] \tag {16.6.6}
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$$
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where Eq. (4.1.14) has been used to obtain each element stiffness matrix and Eq. (16.6.1) has been used to calculate the lumped-mass matrix. On substituting Eqs. (16.6.6) into Eq. (16.4.7), we obtain
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$$
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\left| \frac {E I}{L ^ {3}} \left[ \begin{array}{c c} 2 4 & 0 \\ 0 & 8 L ^ {2} \end{array} \right] - \omega^ {2} \rho A L \left[ \begin{array}{c c} 1 & 0 \\ 0 & 0 \end{array} \right] \right| = 0 \tag {16.6.7}
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$$
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Dividing Eq. (16.6.7) by rAL and simplifying, we obtain
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$$
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\omega^ {2} = \frac {2 4 E I}{\rho A L ^ {4}}
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$$
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or $\omega = \frac { 4 . 9 0 } { L ^ { 2 } } \left( \frac { E I } { A \rho } \right) ^ { 1 / 2 }$ EI
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1=2 ð16:6:8Þ
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The exact solution for the first natural frequency, from simple beam theory, is given by Reference [6]. It is
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$$
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\omega = \frac {5 . 5 9}{L ^ {2}} \left(\frac {E I}{A \rho}\right) ^ {1 / 2} \tag {16.6.9}
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$$
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The large discrepancy between the exact solution and the finite element solution is assumed to be accounted for by the coarseness of the finite element model. In Example 16.6 we show for a clamped-free beam that as the number of degrees of freedom increases, convergence to the exact solution results. Furthermore, if we had used the consistent-mass matrix for the beam [Eq. (16.6.5)], the results would have been more
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<!-- source-page: 695 -->
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accurate than with the lumped-mass matrix as consistent-mass matrices yield more accurate results for flexural elements such as beams.
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(b) Three-Element Solution:
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Using Eq. (16.6.1), we calculate each element mass matrix as follows:
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$$
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\begin{array}{l} [ \hat {m} ^ {(1)} ] = \frac {\rho A L}{2} \left[ \begin{array}{c c c c} d _ {1 x} & \varphi_ {1} & d _ {2 x} & \varphi_ {2} \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \\ [ \hat {m} ^ {(2)} ] = \frac {\rho A L}{2} \left[ \begin{array}{c c c c} d _ {2 x} & \varphi_ {2} & d _ {3 x} & \varphi_ {3} \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \\ \begin{array}{c c c c} d _ {3 x} & \varphi_ {3} & d _ {4 x} & \varphi_ {4} \end{array} \\ [ \hat {m} ^ {(3)} ] = \frac {\rho A L}{2} \left[ \begin{array}{c c c c} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \tag {16.6.10} \\ \end{array}
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$$
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Knowing that $d _ { 1 y } = \varphi _ { 1 } = d _ { 4 y } = \varphi _ { 4 }$ , we obtain the global mass matrix as
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$$
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\underline {{{M}}} = \rho A L \left[ \begin{array}{c c c c} d _ {2 y} & \varphi_ {2} & d _ {3 y} & \varphi_ {3} \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \tag {16.6.11}
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$$
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Using Eq. (4.1.14), we obtain each element stiffness matrix as
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$$
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\underline {{k}} ^ {(1)} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \underline {{k}} ^ {(2)} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & 2 L ^ {\mathrm{2}} & - 6 L & 4 L ^ {2} \end{array} \right]
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$$
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$$
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\underline {{{{k}}}} ^ {(3)} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} d _ {3 y} & \varphi_ {3} & d _ {4 y} & \varphi_ {4} \\ 1 2 & 6 L & - 1 2 & 6 L \\ 6 L & 4 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 & - 6 L \\ 6 L & 2 L ^ {2} & - 6 L & 4 L ^ {2} \end{array} \right] \tag {16.6.12}
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$$
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Using Eq. (16.6.12), we asemble the global stiffness matrix as
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$$
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\underline {{{K}}} = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 1 2 - 1 2 & 6 L + 6 L & - 1 2 & 6 L \\ 6 L - 6 L & 4 L ^ {2} + 2 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 1 2 + 1 2 & - 6 L + 6 L \\ 6 L & 2 L ^ {2} & - 6 L + 6 L & 4 L ^ {2} + 4 L ^ {2} \end{array} \right] = \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 0 & 1 2 L & - 1 2 & 6 L \\ 0 & 6 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 2 4 & 0 \\ 6 L & 2 L ^ {2} & 0 & 8 L ^ {2} \end{array} \right] \tag {16.6.13}
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$$
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<!-- source-page: 696 -->
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Using the general Eq. (16.4.7), we obtain the frequency equation as
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$$
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\left| \begin{array}{c c c c} 0 & 1 2 L & - 1 2 & 6 L \\ \frac {E I}{L ^ {3}} \left[ \begin{array}{c c c c} 0 & 6 L ^ {2} & - 6 L & 2 L ^ {2} \\ - 1 2 & - 6 L & 2 4 & 0 \\ 6 L & 2 L ^ {2} & 0 & 8 L ^ {2} \end{array} \right] - \omega^ {2} \rho A L \left[ \begin{array}{c c c c} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \end{array} \right|
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$$
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$$
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= \left| \begin{array}{c c c c} \omega^ {2} \rho A L & 1 2 E I / L ^ {2} & - 1 2 E I / L ^ {3} & 6 E I / L ^ {2} \\ 0 & 6 E I / L & - 6 E I / L ^ {2} & 2 E I / L \\ - 1 2 E I / L ^ {3} & - 6 E I / L ^ {2} & 2 4 E I / L ^ {3} - \omega^ {2} \rho A L & 0 \\ 6 E I / L ^ {2} & 2 E I / L & 0 & 8 E I / L \end{array} \right| = 0 \tag {16.6.14}
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$$
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Simplifying Eq. (16.6.14), we have
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$$
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\left| \begin{array}{c c c c} - \omega^ {2} \beta & 1 2 E I / L ^ {2} & - 1 2 E I / L ^ {3} & 6 E I / L ^ {2} \\ 0 & 6 E I / L & - 6 E I / L ^ {2} & 2 E I / L \\ - 1 2 E I / L ^ {3} & - 6 E I / L ^ {2} & 2 4 E I / L ^ {3} - \omega^ {2} \beta & 0 \\ 6 E I / L ^ {2} & 2 E I / L & 0 & 8 E I / L \end{array} \right| = 0 \tag {16.6.15}
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$$
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where $\beta = \rho AL$
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Upon evaluating the four-by-four determinant in Eq. (16.6.15), we obtain
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$$
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\begin{array}{l} \frac {- 1 1 5 2 \omega^ {2} E ^ {3} I ^ {3} \beta}{L ^ {5}} + \frac {4 8 \omega^ {4} E ^ {2} I ^ {2} \beta^ {2}}{L ^ {2}} + \frac {5 7 6 E ^ {4} I ^ {4}}{L ^ {8}} - \frac {1 2 9 6 E ^ {4} I ^ {4}}{L ^ {8}} \\ + \frac {9 6 \omega^ {2} E ^ {3} I ^ {3} \beta}{L ^ {5}} - \frac {4 \omega^ {4} \beta^ {2} E ^ {2} I ^ {2}}{L ^ {2}} - \frac {6 9 1 2 E ^ {4} I ^ {4}}{L ^ {8}} = 0 \\ \end{array}
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$$
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$$
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\frac {4 4 \omega^ {4} \beta^ {2} E ^ {2} I ^ {2}}{L ^ {2}} - \frac {1 0 5 6 \omega^ {2} \beta E ^ {3} I ^ {3}}{L ^ {5}} - \frac {7 6 3 2 E ^ {4} I ^ {4}}{L ^ {8}} = 0 \tag {16.6.16}
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$$
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$$
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1 1 \omega^ {4} \beta^ {2} - \frac {2 6 4 \omega^ {2} \beta E I}{L ^ {3}} - \frac {1 9 0 8 E ^ {2} I ^ {2}}{L ^ {6}} = 0
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$$
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Dividing Eq. (16.6.16) by $\frac{4E^2I^2}{L^2}$ , we obtain two roots for $\omega_1^2\beta$ as
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$$
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\omega_ {1} ^ {2} \beta = \frac {- 5 . 8 1 7 2 5 4 E I}{L ^ {3}} \quad \omega_ {1} ^ {2} \beta = \frac {2 9 . 8 1 7 2 5 4 E I}{L ^ {3}} \tag {16.6.17}
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$$
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Ignoring the negative root as it is not physically possible and solving explicitly for $\omega_{1}$ , we have
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or
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$$
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\omega_ {1} ^ {2} = \frac {2 9 . 8 1 7 2 5 4 E I}{\beta L ^ {3}}
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$$
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$$
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\omega_ {1} = \sqrt {\frac {2 9 . 8 1 7 2 5 4 E I}{\beta L ^ {3}}} = \frac {5 . 4 6}{L ^ {2}} \sqrt {\frac {E I}{A \rho}} \tag {16.6.18}
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$$
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<!-- source-page: 697 -->
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In summary, comparing Eqs (16.6.8) and (16.6.18) with the exact solution, Eq. (16.6.9), for the first natural frequency, we have
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Two Beam Elements: $\omega = { \frac { 4 . 9 0 } { L ^ { 2 } } } { \sqrt { \frac { E I } { A \rho } } }$
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Three Beam Elements: $\omega = { \frac { 5 . 4 6 } { L ^ { 2 } } } { \sqrt { \frac { E I } { A \rho } } }$ ð16:6:19Þ
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Exact solution: $\omega = \frac { 5 . 5 9 } { L ^ { 2 } } \left( \frac { E I } { A \rho } \right) ^ { 1 / 2 }$
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We can observe that with just three elements the accuracy has significantly increased
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# Example 16.6
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Determine the first natural frequency of vibration of the cantilever beam shown in Figure 16–17 with the following data:
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<details>
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<summary>text_image</summary>
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15 in.
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30 in.
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</details>
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Figure 16–17 Fixed-free beam (two-element model, lumped-mass matrix)
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Length of the beam: $L = 3 0 { \mathrm { i n } } .$
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Modulus of elasticity: $E = 3 \times 1 0 ^ { 7 }$ psi
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Moment of inertia: $I = 0 . 0 8 3 3 \mathrm { i n } ^ { 4 }$
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Cross-sectional area: $A = 1 \mathrm { i n } ^ { 2 }$
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Mass density: $\rho = 0 . 0 0 0 7 3 ~ \mathrm { { l b } } \mathrm { { - s } } ^ { 2 } / \mathrm { { i n } } ^ { 4 }$
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Poisson’s ratio: n ¼ 0:3
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The finite element longhand solution result for the first natural frequency is obtained similarly to that of Example 16.5 as
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$$
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\omega = \frac {3 . 1 4 8}{L ^ {2}} \left(\frac {E I}{A \rho}\right) ^ {1 / 2}
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$$
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The exact solution according to beam theory [1] is
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$$
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\omega = \frac {3 . 5 1 6}{L ^ {2}} \left(\frac {E I}{\rho A}\right) ^ {1 / 2}
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$$
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<!-- source-page: 698 -->
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Figure 16–18 First, second, and third mode shapes of flexural vibration for a cantilever beam
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According to vibration theory for a clamped-free beam [1], we relate the second and third natural frequencies to the first natural frequency by
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$$
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\frac {\omega_ {2}}{\omega_ {1}} = 6. 2 6 6 9 \quad \frac {\omega_ {3}}{\omega_ {1}} = 1 7. 5 4 7 5
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$$
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Figure 16–18 shows the first, second, and third mode shapes corresponding to the first three natural frequencies for the cantilever beam of Example 16.6 as obtained from a computer program. Note that each mode shape has one fewer node where a node is a
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Table 16–3 Finite element computer solution compared to exact solution for Example 16.6
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<table><tr><td></td><td> $\omega_1$ (rad/s)</td><td> $\omega_2$ (rad/s)</td></tr><tr><td>Exact solution from beam theory</td><td>228</td><td>1434</td></tr><tr><td>Finite element solution</td><td></td><td></td></tr><tr><td>Using 2 elements</td><td>205</td><td>1286</td></tr><tr><td>Using 6 elements</td><td>226</td><td>1372</td></tr><tr><td>Using 10 elements</td><td>227.5</td><td>1410</td></tr><tr><td>Using 30 elements</td><td>228.5</td><td>1430</td></tr><tr><td>Using 60 elements</td><td>228.5</td><td>1432</td></tr></table>
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<!-- source-page: 699 -->
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point of zero displacement. That is, the first mode has all the elements of the beam of the same sign [Figure 16–18(a)], the second mode has one sign change and at some point along the beam the displacement is zero [Figure 16–18(b)], and the third mode has two sign changes and at two points along the beam the displacement is zero [Figure 16–18(c)].
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Table 16–3 shows the computer solution compared with the exact solution.
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|
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# 16.7 Truss, Plane Frame, Plane Stress=Strain, Axisymmetric, and Solid Element Mass Matrices
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The dynamic analysis of the truss and that of the plane frame are performed by extending the concepts presented in Sections 16.2 and 16.6 to the truss and plane frame, as has previously been done for the static analysis of trusses and frames.
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# Truss Element
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|
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The truss analysis requires the same transformation of the mass matrix from local to global coordinates as in Eq. (3.4.22) for the stiffness matrix; that is, the global mass matrix for a truss element is given by
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|
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$$
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\underline {{m}} = \underline {{T}} ^ {T} \underline {{\hat {m}}} \underline {{T}} \tag {16.7.1}
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$$
|
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|
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We are now dealing with motion in two or three dimensions. Therefore, we must reformulate a bar element mass matrix with both axial and transverse inertial properties because mass is included in both the global x and y directions in plane truss analysis (Figure 16–19). Considering two-dimensional motion, we express both local axial displacement u^ and transverse displacement v^ for the element in terms of the local axial and transverse nodal displacements as
|
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|
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$$
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\left\{ \begin{array}{l} \hat {u} \\ \hat {v} \end{array} \right\} = \frac {1}{L} \left[ \begin{array}{c c c c} L - \hat {x} & 0 & \hat {x} & 0 \\ 0 & L - \hat {x} & 0 & \hat {x} \end{array} \right] \left\{ \begin{array}{l} \hat {d} _ {1 x} \\ \hat {d} _ {1 y} \\ \hat {d} _ {2 x} \\ \hat {d} _ {2 y} \end{array} \right\} \tag {16.7.2}
|
||
$$
|
||
|
||
In general, $\underline { { \hat { \psi } } } = \underline { { N } } \underline { { \hat { d } } } ;$ therefore, the shape function matrix from Eq. (16.7.2) is
|
||
|
||
$$
|
||
[ N ] = \frac {1}{L} \left[ \begin{array}{c c c c} L - \hat {x} & 0 & \hat {x} & 0 \\ 0 & L - \hat {x} & 0 & \hat {x} \end{array} \right] \tag {16.7.3}
|
||
$$
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
d₂y
|
||
d̂₂x
|
||
2
|
||
d₂x
|
||
L
|
||
d₁y
|
||
d₁x
|
||
1
|
||
x
|
||
d̂₁x
|
||
</details>
|
||
|
||
Figure 16–19 Truss element arbitrarily oriented in x-y plane showing nodal degrees of freedom
|
||
|
||
<!-- source-page: 700 -->
|
||
|
||
We can then substitute Eq. (16.7.3) into the general expression given by Eq. (16.2.19) to evaluate the local truss element consistent-mass matrix as
|
||
|
||
$$
|
||
[ \hat {m} ] = \frac {\rho A L}{6} \left[ \begin{array}{c c c c} 2 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 \end{array} \right] \tag {16.7.4}
|
||
$$
|
||
|
||
The truss element lumped-mass matrix for two-dimensional motion is obtained by simply lumping mass at each node and remembering that mass is the same in both the x^ and y^ directions. The local truss element lumped-mass matrix is then
|
||
|
||
$$
|
||
[ \hat {m} ] = \frac {\rho A L}{2} \left[ \begin{array}{c c c c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \tag {16.7.5}
|
||
$$
|
||
|
||
# Plane Frame Element
|
||
|
||
The plane frame analysis requires first expanding and then combining the bar and beam mass matrices to obtain the local mass matrix. Because we recall there are six total degrees of freedom associated with a plane frame element (Figure 16–20), the bar and beam mass matrices are expanded to order $6 \times 6$ and superimposed. On combining the local axes consistent-mass matrices for the bar and beam from Eqs. (16.2.23) and (16.6.5), we obtain
|
||
|
||
$$
|
||
\underline {{\hat {m}}} = \rho A L \left[ \begin{array}{c c c c c c} 2 / 6 & 0 & 0 & 1 / 6 & 0 & 0 \\ & 1 5 6 / 4 2 0 & 2 2 L / 4 2 0 & 0 & 5 4 / 4 2 0 & - 1 3 L / 4 2 0 \\ & & 4 L ^ {2} / 4 2 0 & 0 & 1 3 L / 4 2 0 & - 3 L ^ {2} / 4 2 0 \\ & & & 2 / 6 & 0 & 0 \\ & & & & 1 5 6 / 4 2 0 & - 2 2 L / 4 2 0 \\ \text { Symmetry } & & & & & 4 L ^ {2} / 4 2 0 \end{array} \right] \tag {16.7.6}
|
||
$$
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y
|
||
\hat{d}_{1y}
|
||
\hat{d}_{1x}
|
||
\hat{d}_{2y}
|
||
\hat{d}_{2x}
|
||
\hat{d}_{1x}
|
||
\hat{d}_{2y}
|
||
\hat{d}_{1y}
|
||
\hat{d}_{1x}
|
||
\hat{d}_{2y}
|
||
\hat{d}_{2x}
|
||
\hat{x}
|
||
\hat{x}
|
||
\hat{φ}_{1}
|
||
\hat{φ}_{2}
|
||
</details>
|
||
|
||
Figure 16–20 Frame element arbitrarily oriented in local coordinate system showing nodal degrees of freedom
|