Solving Eq. (16.8.29) for $t _ { 2 }$ and $t _ { 3 } ,$ , we obtain $$ t _ {2} = 2 9. 7 3 2 ^ {\circ} \mathrm{C} \quad t _ {3} = 2 1. 7 5 2 ^ {\circ} \mathrm{C} $$ The results through a time of 3 s are tabulated in Table 16–4 and plotted in Figure 16–24. # 16.9 Computer Program Example Solutions for Structural Dynamics In this section, we report some results of structural dynamics from a computer program. We report the results of the natural frequencies of a fixed-fixed beam using the plane stress element in Algor [15] and compare how many elements of this type are necessary to obtain correct results. We also report the results of three structural dynamics problems, a bar, a beam, and a frame subjected to time-dependent loadings. Finally, we show two additional models, one of a time-dependent three-dimensional gantry crane made of beam elements and subjected to an impact loading, and the other of a cab frame that travels along the underside of a crane beam. Figure 16–25 shows a fixed-fixed steel beam used for natural frequency determination using plane stress elements. Table 16–5 shows the results of the first five natural frequencies using 100 elements and then using 1000 elements. Comparisons to the analytical solutions from beam theory are shown. We observe that it takes a large number of plane stress elements to accurately predict the natural frequencies whereas it 

text_image

Cross section 100 in. 1 in. 1 in.

Figure 16–25 Fixed-fixed beam for natural frequency determination modeled using plane stress element Table 16–5 Results for first five frequencies using 100 and 1000 elements and exact solution

ω (rad/s)	Analytical	100 Elements	1000 Elements
1	130.8	130.7	130.6
2	360.8	359.8	359.7
3	707.3	704.7	704.1
4	1169.2	1163.3	1161.6
5	1746.6	1734.5	1731.0


text_image

1 2 3 100 in 100 in F(t) F(t), lb 1000 0.001 0.002 t, s

Figure 16–26 Bar subjected to forcing function shown only took a few beam elements to accurately predict natural frequencies (see Example 16.6 and Table 16–3). Figure 16–26 shows a steel bar subjected to a time-dependent forcing function. Using two elements in the model, the nodal displacements at nodes 2 and 3 are presented in Table 16–6. A time step of integration of 0.00025 s was used. This time step is based on that recommended by Eq. (16.5.1) and determined in Example 16.4, as the bar has the same properties as that of Example 16.4. Figure 16–27 shows a fixed-fixed beam subjected to a forcing function. Here $E = 6 . 5 8 \times 1 0 ^ { 6 }$ psi, $I = 1 0 0 \mathrm { i n } . ^ { 4 }$ , mass density of ${ \dot { 0 } } . 1 1 \mathrm { b } . \mathrm { s } ^ { 2 } / \mathrm { i n } . ^ { 4 }$ and a time step of integration of 0.01 s were used for the beam. The natural frequencies and displacement-time history for nodes 2 and 3 are shown in Table 16–7. Table 16–7 lists the first six natural frequencies for the fixed-fixed beam and the vertical displacement versus time for nodes 2 and 3 of the beam. The natural frequencies 1, 2, 3, and 6 are flexural modes, while mode 5 is an axial mode. These modes are seen by looking at the modes from a frequency analysis. The maximum displacement under the load (at node 3) compares with the solution in Reference [14]. This maximum displacement is at node 3 at a time of 0.08 s with a value of 1.207 in. The minimum displacement at node 3 is 0.2028 in. at time 0.16 s. The static deflection for the beam with a concentrated load at mid-span is 0.633 in. as obtained from the classical solution of $y = P L ^ { 3 } / 1 9 2 E I$ . The time-dependent response oscillates about the static deflection. A time step of 0.01 was used in the fixed-fixed beam as it meets the recommended time step as suggested in Section 16.3. That is, $\Delta t < T _ { n } / 1 0$ to $T _ { n } / 2 0$ is recommended to provide accurate results for Wilson’s direct integration scheme as used in the Algor program. From the frequency analysis (see the output in Table 16–7), the circular frequency $\omega _ { 6 } = 1 9 7 . 5 2$ or the natural frequency is $f _ { 6 } = \omega _ { 6 } / ( 2 \pi ) = 3 1 . 4 4$ cycles/s or Hertz (Hz). Now we use $\Delta t = T _ { n } / 2 0 = 1 / ( 2 0 f _ { 6 } ) = 1 / [ 2 0 ( 3 1 . 4 3 ) ] = 0 . 0 1 5 \mathrm { ~ s ~ }$ . Therefore, $\Delta t = 0 . 0 1$ s is acceptable. Using a time step greater than $T _ { n } / 1 0$ may result in loss of accuracy as some of the higher mode response contributions to the solution may be missed. Often times a cut-off period or frequency is used to decide what largest natural frequency to use in the analysis. In many applications only a few lower modes contribute significantly to the response. The higher modes are then not necessary. The highest frequency used in the analysis is called the cut-off frequency. For machinery parts, the cut-off frequency is often taken as high as 250 Hz. In the fixed-fixed beam, we have Table 16–6 Displacement time history, nodes 2 and 3 of Figure 16–26

	*NODE NUMBER* - (COMPONENT NUMBER)
TIME	2-(2)	3-(2)
.00025	4.410E-06	6.156E-05
.00050	4.600E-05	4.668E-04
.00075	2.147E-04	1.425E-03
.00100	6.507E-04	2.967E-03
.00125	1.481E-03	4.873E-03
.00150	2.699E-03	6.439E-03
.00175	4.061E-03	7.143E-03
.00200	5.109E-03	6.860E-03
.00225	5.349E-03	5.793E-03
.00250	4.501E-03	4.385E-03
.00275	2.670E-03	2.862E-03
.00300	3.265E-04	1.141E-03
.00325	-1.907E-03	-9.441E-04
.00350	-3.538E-03	-3.354E-03
.00375	-4.376E-03	-5.694E-03
.00400	-4.530E-03	-7.319E-03
.00425	-4.232E-03	-7.646E-03
.00450	-3.645E-03	-6.463E-03
.00475	-2.772E-03	-4.057E-03
.00500	-1.514E-03	-1.083E-03
.00525	1.599E-04	1.740E-03
.00550	2.082E-03	3.921E-03
.00575	3.867E-03	5.313E-03
.00600	5.055E-03	6.021E-03
.00625	5.312E-03	6.185E-03
.00650	4.583E-03	5.814E-03
.00675	3.106E-03	4.776E-03
.00700	1.282E-03	2.947E-03
.00725	-5.031E-04	4.073E-04
.00750	-2.015E-03	-2.460E-03
.00775	-3.183E-03	-5.051E-03
.00800	-4.013E-03	-6.763E-03
.00825	-4.477E-03	-7.233E-03
.00850	-4.466E-03	-6.464E-03
.00875	-3.838E-03	-4.770E-03
.00900	-2.542E-03	-2.594E-03
.00925	-7.098E-04	-3.179E-04
MAXIMUM ABSOLUTE VALUES
MAXIMUM	5.349E-03	7.646E-03
TIME	2.250E-03	4.250E-03

selected a cut-off frequency of f6 ¼ 31:44 Hz in determining the time step of integration. This frequency is the highest flexural mode frequency computed for the fourelement beam model. 

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| time (s) | F(t) | | -------- | -------- | | 0 | 10,000 lb | | 0.1 | 10,000 lb | | 0.2 | 0 |

Figure 16–27 Fixed-fixed beam subjected to forcing function Table 16–7 Natural frequencies and displacement time history (nodes 2 and 3, Figure 16–27)

Frequencies = mode number	6 circular frequency (rad/sec)
1	4.52276232074113D+01
2	4.52276232074113D+01
3	1.20159893475319D+02
4	1.20159893475319D+02
5	1.24168832797688D+02
6	1.97518763916263D+02
Y-DISPLACEMENT
*NODE NUMBER* - (COMPONENT NUMBER)
TIME	2-(2)	3-(2)
.01000	1.791E-02	4.050E-03
.02000	1.203E-01	3.458E-02
.03000	2.987E-01	1.197E-01
.04000	5.201E-01	2.542E-01
.05000	7.624E-01	3.978E-01
.06000	9.907E-01	5.143E-01
.07000	1.152E+00	5.916E-01
.08000	1.207E+00	6.246E-01
.09000	1.150E+00	6.024E-01
.10000	1.003E+00	5.217E-01
.11000	7.873E-01	3.989E-01
.12000	5.270E-01	2.601E-01
.13000	2.601E-01	1.241E-01
.14000	3.174E-02	4.247E-03
.15000	-1.267E-01	-8.361E-02
.16000	-2.028E-01	-1.244E-01
.17000	-1.962E-01	-1.153E-01
MAXIMUM ABSOLUTE VALUES
MAXIMUM	1.207E+00	6.246E-01
TIME	8.000E-02	8.000E-02

Damping will not be considered in any examples. However, Algor allows you to consider damping using Rayleigh damping in the direct integration method. For Rayleigh damping, the damping matrix is $$ [ C ] = \alpha [ M ] + \beta [ K ] \tag {16.9.1} $$ where the constants a and b are calculated from the system equations $$ \alpha + \beta \omega_ {i} ^ {2} = 2 \omega_ {i} \zeta_ {i} \tag {16.9.2} $$ Table 16–8 Forces and moments versus time for elements 1 and 2 of Figure 16–27

1**** BEAM ELEMENT FORCES AND MOMENTS
ELEMENT NO.	CASE (MODE)	AXIAL FORCE R1	SHEAR FORCE R2	SHEAR FORCE R3	TORSION MOMENT M1	BENDING MOMENT M2	BENDING MOMENT M3
1	1	0.000E+00	1.685E+02	0.000E+00	0.000E+00	0.000E+00	6.764E+02
		0.000E+00	-1.685E+02	0.000E+00	0.000E+00	0.000E+00	7.748E+03
1	2	0.000E+00	6.662E+02	0.000E+00	0.000E+00	0.000E+00	-7.100E+03
		0.000E+00	-6.662E+02	0.000E+00	0.000E+00	0.000E+00	4.041E+04
1	3	0.000E+00	-4.880E+02	0.000E+00	0.000E+00	0.000E+00	-7.116E+04
		0.000E+00	4.880E+02	0.000E+00	0.000E+00	0.000E+00	4.676E+04
1	4	0.000E+00	-3.738E+03	0.000E+00	0.000E+00	0.000E+00	-1.961E+05
		0.000E+00	3.738E+03	0.000E+00	0.000E+00	0.000E+00	9.226E+03
1	5	0.000E+00	-7.069E+03	0.000E+00	0.000E+00	0.000E+00	-3.272E+05
		0.000E+00	7.069E+03	0.000E+00	0.000E+00	0.000E+00	-2.624E+04
1	6	0.000E+00	-9.022E+03	0.000E+00	0.000E+00	0.000E+00	-4.211E+05
		0.000E+00	9.022E+03	0.000E+00	0.000E+00	0.000E+00	-2.998E+04
1	7	0.000E+00	-1.008E+04	0.000E+00	0.000E+00	0.000E+00	-4.794E+05
		0.000E+00	1.008E+04	0.000E+00	0.000E+00	0.000E+00	-2.448E+04
1	8	0.000E+00	-1.086E+04	0.000E+00	0.000E+00	0.000E+00	-5.098E+05
		0.000E+00	1.086E+04	0.000E+00	0.000E+00	0.000E+00	-3.335E+04
2	1	0.000E+00	-4.514E+02	0.000E+00	0.000E+00	0.000E+00	-7.748E+03
		0.000E+00	4.514E+02	0.000E+00	0.000E+00	0.000E+00	-1.482E+04
2	2	0.000E+00	-2.566E+03	0.000E+00	0.000E+00	0.000E+00	-4.041E+04
		0.000E+00	2.566E+03	0.000E+00	0.000E+00	0.000E+00	-8.791E+04
2	3	0.000E+00	-4.229E+03	0.000E+00	0.000E+00	0.000E+00	-4.676E+04
		0.000E+00	4.229E+03	0.000E+00	0.000E+00	0.000E+00	-1.647E+05
2	4	0.000E+00	-4.476E+03	0.000E+00	0.000E+00	0.000E+00	-9.226E+03
		0.000E+00	4.476E+03	0.000E+00	0.000E+00	0.000E+00	-2.146E+05
2	5	0.000E+00	-4.970E+03	0.000E+00	0.000E+00	0.000E+00	2.624E+04
		0.000E+00	4.970E+03	0.000E+00	0.000E+00	0.000E+00	-2.747E+05
2	6	0.000E+00	-6.623E+03	0.000E+00	0.000E+00	0.000E+00	2.998E+04
		0.000E+00	6.623E+03	0.000E+00	0.000E+00	0.000E+00	-3.611E+05
2	7	0.000E+00	-8.118E+03	0.000E+00	0.000E+00	0.000E+00	2.448E+04
		0.000E+00	8.118E+03	0.000E+00	0.000E+00	0.000E+00	-4.304E+05
2	8	0.000E+00	-8.196E+03	0.000E+00	0.000E+00	0.000E+00	3.335E+04
		0.000E+00	8.196E+03	0.000E+00	0.000E+00	0.000E+00	-4.431E+05

where $\omega _ { i }$ are circular natural frequencies obtained through modal analysis, and $\zeta _ { i }$ are damping ratios specified by the analyst. For instance, assume we assign damping ratios $\zeta _ { 1 }$ and $\zeta _ { 2 } .$ , from the above Eq. (16.9.2), we can show that a and $\beta$ are $$ \alpha = \frac {2 \omega_ {1} \omega_ {2}}{\omega_ {2} ^ {2} - \omega_ {1} ^ {2}} (\omega_ {2} \zeta_ {1} - \omega_ {1} \zeta_ {2}) \quad \beta = \frac {2}{\omega_ {2} ^ {2} - \omega_ {1} ^ {2}} (\omega_ {2} \zeta_ {2} - \omega_ {1} \zeta_ {1}) \tag {16.9.3} $$ For $\beta = 0 , [ C ] = \alpha [ M ]$ and the higher modes are only slightly damped, while for $\alpha = 0 , [ C ] = \beta [ K ]$ and higher modes are heavily damped. To obtain a and $\beta ,$ we then necessarily run the modal analysis program first to obtain the frequencies. For instance, in the fixed-fixed beam, the first two different frequencies are $\omega _ { 1 } = 4 5 . 2 3$ rad/s and $\omega _ { 3 } = 1 2 0 . 1 6$ rad/s $( \omega _ { 2 }$ is the same as o3, so use o3). Now assume light damping 

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L 3L 3L y 1 z x L L 3L L 2L 2F(t) 2 ① ② ③ ④ ⑤ ⑥ ⑦ 6 F(t) (a)


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| t | F(t) | | --------- | ---- | | 0 | 0 | | t₁ = 17.5 ms | F₁ | | t₂ = 35 ms | 0 |

Figure 16–28 (a) Six-member plane frame; (b) dynamic load $\left( \zeta \leq 0 . 0 5 \right)$ . Therefore, let $\zeta _ { 1 } = \zeta _ { 2 } = 0 . 0 5$ : Using these o’s and $\zeta ^ { \gamma } \mathbf { s } ,$ , in Eqs. (16.9.3), we obtain $\alpha = 3 . 2 8 6$ and $\beta = 0 . 0 0 0 6 0 5$ : These values could be used for a and b if you want to include 5% damping $( \zeta = 0 . 0 5 )$ : Table 16–8 lists the element forces and moments for elements 1 and 2 up to time 0.08 s. This time corresponds to when the maximum displacement occurs and is also when the maximum moments occur. The largest element 1 bending moment is $\mathbf { M } 3 = - 5 0 9 , 8 0 0$ lb in. at the wall (node 1) at a time of 0.08 s (see the column ‘‘Case (Mode),’’ number 8). The largest element 2 bending moment is $\mathbf { M } 3 = - 4 4 3 , 1 0 0$ lb in. Figure 16–28(a) shows a plane frame consisting of six rigidly connected prismatic members with dynamic forces $F ( t )$ and $2 { \cal F } ( t )$ applied in the x direction at joints 6 and 4, respectively. The time variation of $F ( t )$ is shown in Figure 1 6–28(b). The results are for steel with cross-sectional area of $3 0 ~ \mathrm { i n } ^ { 2 } .$ , moment of inertia of 1000 in4, L ¼ 50 in., and $F _ { 1 } = 1 0 , 0 0 0$ lb. Figure 16–29 shows the displaced frame for the worst stress at time of 0.035 s. The largest x displacement of node 6 for the time of 0.035 s is 0.1551 inch. This value compares closely with the solution in Reference [16]. Finally, Figures 16–30(a) and 16–31(a) show models of a gantry crane and a cab frame subjected to dynamic loading functions (Figures 16–30(b) and (16–31(b)). For details of these design solutions consult [17–18]. 

line

| Beam-Truss | Value | | ---------- | ------ | | 8127.9 | 8127.9 | | 5751.8 | 5751.8 | | 3375.8 | 3375.8 | | 999.75 | 999.75 | | -1376.3 | -1376.3 | | -3752.3 | -3752.3 | | -6128.4 | -6128.4 | | -8504.4 | -8504.4 |

Figure 16–29 Displaced frame with worst stress at time 0.035 s 

other

| Point | Value | |-------|-------| | 1 | 2 | | 2 | 3 | | 3 | 4 | | 4 | 5 | | 5 | 6 | | 6 | 7 | | 7 | 8 | | 8 | 9 | | 9 | 10 | | 10 | 11 | | 11 | 12 | | 12 | 13 | | 13 | 14 | | 14 | 15 | | 15 | 16 | | 16 | 17 | | 17 | 18 | | 18 | 19 | | 19 | 20 | | 20 | 21 | | 21 | 22 | | 22 | 23 | | 23 | 24 | | 24 | 25 | | 25 | 26 | | 26 | 27 | | 27 | 28 | | 28 | 29 | | 29 | 30 | | 30 | 31 | | 31 | 32 | | 32 | 33 | | 33 | 34 | | 34 | 35 | | 35 | 36 | | 36 | 37 | | 37 | 38 | | 38 | 39 | | 39 | 40 | | 40 | 41 | | 41 | 42 | | 42 | 43 | | 43 | 44 | | 44 | 45 | | 45 | 46 | | 46 | 47 | | 47 | 48 | | 48 | 49 | | 49 | 50 | | 50 | 51 | | 51 | 52 | | 52 | 53 | | 53 | 54 | | 54 | 55 | | 55 | 56 | | 56 | 57 | | 57 | 58 | | 58 | 59 | | 59 | 60 | | 60 | 61 | | 61 | 62 | | 62 | 63 | | 63 | 64 | | 64 | 65 | | 65 | 66 | | 66 | 67 | | 67 | 68 | | 68 | 69 | | 69 | 70 | | 70 | 71 | | 71 | 72 | | 72 | 73 | | 73 | 74 |

(a) 

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| Time, t (sec) | F(t), lb | | ------------- | -------- | | 0.0 | 0 | | 0.1 | 5400 | | 0.4 | 5400 | | 0.6 | 0 |

Figure 16–30 (a) Gantry crane model composed of 73 beam elements and (b) the time-dependent trapezoidal loading function applied to the top edge of the crane [17] 

geo

| Node | Element | Small Number - Node | Big Number - Element | |---|---|---|---| | 1 | 1 | 1 | 1 | | 2 | 1 | 1 | 1 | | 3 | 1 | 1 | 1 | | 4 | 1 | 1 | 1 | | 5 | 1 | 1 | 1 | | 6 | 1 | 1 | 1 | | 7 | 1 | 1 | 1 | | 8 | 1 | 1 | 1 | | 9 | 1 | 1 | 1 | | 10 | 1 | 1 | 1 | | 11 | 1 | 1 | 1 | | 12 | 1 | 1 | 1 | | 13 | 1 | 1 | 1 | | 14 | 1 | 1 | 1 | | 15 | 1 | 1 | 1 | | 16 | 1 | 1 | 1 | | 17 | 1 | 1 | 1 | | 18 | 1 | 1 | 1 | | 19 | 1 | 1 | 1 | | 20 | 1 | 1 | 1 | | 21 | 1 | 1 | 1 | | 22 | 1 | 1 | 1 | | 23 | 1 | 1 | 1 | | 24 | 1 | 1 | 1 |


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| Time, s | F(t), lb | | ------- | -------- | | 0.1 | 0 | | 0.2 | 4054 | | 0.3 | 4054 | | 0.4 | 0 |

Figure 16–31 (a) Finite element model of a cab with 8 plate elements (upper right triangular elements) and 15 beam elements and (b) the time-dependent trapezoidal loading applied to node 10 [18] # References [1] Thompson, W. T., and Dahleh, M. D., Theory of Vibrations with Applications, 5th ed., Prentice-Hall, Englewood Cliffs, NJ, 1998. [2] Archer, J. S., ‘‘Consistent Matrix Formulations for Structural Analysis Using Finite Element Techniques,’’ Journal of the American Institute of Aeronautics and Astronautics, Vol. 3, No. 10, pp. 1910–1918, 1965. [3] James, M. L., Smith, G. M., and Wolford, J. C., Applied Numerical Methods for Digital Computation, 3rd ed., Harper & Row, New York, 1985. [4] Biggs, J. M., Introduction to Structural Dynamics, McGraw-Hill, New York, 1964. [5] Newmark, N. M., ‘‘A Method of Computation for Structural Dynamics,’’ Journal of the Engineering Mechanics Division, American Society of Civil Engineers, Vol. 85, No. EM3, pp. 67–94, 1959. [6] Clark, S. K., Dynamics of Continuous Elements, Prentice-Hall, Englewood Cliffs, NJ, 1972. [7] Bathe, K. J., Finite Element Procedures in Engineering Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1982. [8] Bathe, K. J., and Wilson, E. L., Numerical Methods in Finite Element Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1976. [9] Fujii, H., ‘‘Finite Element Schemes: Stability and Convergence,’’ Advances in Computational Methods in Structural Mechanics and Design, J. T. Oden, R. W. Clough, and Y. Yamamoto, Eds., University of Alabama Press, Tuscaloosa, AL, pp. 201–218, 1972. [10] Krieg, R. D., and Key, S. W., ‘‘Transient Shell Response by Numerical Time Integration,’’ International Journal of Numerical Methods in Engineering, Vol. 17, pp. 273–286, 1973. [11] Belytschko, T., ‘‘Transient Analysis,’’ Structural Mechanics Computer Programs, Surveys, Assessments, and Availability, W. Pilkey, K. Saczalski, and H. Schaeffer, Eds., University of Virginia Press, Charlottesville, VA, pp. 255–276, 1974. [12] Hughes, T. J. R., ‘‘Unconditionally Stable Algorithms for Nonlinear Heat Conduction,’’ Computational Methods in Applied Mechanical Engineering, Vol. 10, No. 2, pp. 135–139, 1977. [13] Hilber, H. M., Hughes, T. J. R., and Taylor, R. L., ‘‘Improved Numerical Dissipation for Time Integration Algorithms in Structural Dynamics,’’ Earthquake Engineering in Structural Dynamics, Vol. 5, No. 3, pp. 283–292, 1977. [14] Paz, M., Structural Dynamics Theory and Computation, 3rd ed., Van Nostrand Reinhold, New York, 1991. [15] Linear Stress and Dynamics Reference Division, Docutech On-line Documentation, Algor, Inc., Pittsburgh, PA, 1999. [16] Weaver, W., Jr., and Johnston, P. R., Structural Dynamics by Finite Elements, Prentice-Hall, Englewood Cliffs, NJ, 1987. [17] Salemganesan, Hari, Finite Element Analysis of a Gantry Crane, M. S. Thesis, Rose-Hulman Institute of Technology, Terre Haute, IN, September 1992. [18] Leong Cheow Fook, The Dynamic Analysis of a Cab Using The Finite Element Method, M. S. Thesis, Rose-Hulman Institute of Technology, Terre Haute, IN, January 1988. # Problems 16.1 Determine the consistent-mass matrix for the one-dimensional bar discretized into two elements as shown in Figure P16–1. Let the bar have modulus of elasticity $E ,$ mass density $\rho ,$ and cross-sectional area A.