김경종
22 KiB
10.7.2 Spherical shell example
The shell, ^{(8)} shown in Fig. 10.3, is subjected to a distributed step pressure of 600 lb/in ^{2} . The material is assumed to obey the Von Mises yield condition with linear isotropic hardening. The dimensions and properties of the shell are given as follows:
| Internal radius | $R = 22.27 \text{ in}$ |
| Thickness of shell | $t = 0.41 \text{ in}$ |
| Semi angle | $\alpha = 26.67^{\circ}$ |
| Elastic modulus | $E = 10.5 \times 10^{6} \text{ lb/in}^{2}$ |
| Poisson's ratio | $\nu = 0.3$ |
| Yield stress | $\sigma_{Y} = 0.024 \times 10^{6} \text{ lb/in}^{2}$ |
| Tangent hardening modulus | $E_{T} = 0.21 \times 10^{6} \text{ lb/}^{2}$ |
| Mass density | $\rho = 2.45 \times 10^{-4} \text{ lb-sec}^{2}/\text{in}^{4}$ |
| Step distributed pressure | $p = 600 \text{ lb/in}^{2}$ |
text_image
P R α
Fig. 10.3 Spherical shell and finite element mesh.
The shell is divided into ten, 8-noded, axisymmetric, isoparametric elements. The fundamental period of the shell is T_{f} = 0.55 \times 10^{-3} sec, (Reference 8). For explicit central difference analysis, the time step is taken as 0.4 \times 10^{-6} sec.
In order to illustrate the versatility of program DYNPAK we consider the following three cases:
(i) Small elastic displacements
(ii) Large elastic displacements
(iii) Small elasto-viscoplastic displacements (with a fluidity parameter value of \gamma = 100.0 ).
line
| t × 10⁻³ secs | Small elastic displacement | Large elastic displacement |
|---|---|---|
| 0 | 0.00 | 0.00 |
| 2 | 0.03 | 0.04 |
| 4 | 0.07 | 0.08 |
| 6 | -0.04 | -0.03 |
| 8 | 0.08 | 0.07 |
| 10 | 0.02 | 0.05 |
Fig. 10.4(a) Results of the transient dynamic analysis of a spherical shell cap. Cases (i) and (ii).
Figure 10.4(a) shows the vertical displacement of the crown lower point for the analyses based on both small and large elastic displacement assumptions. The results show that the inclusion of geometrically nonlinear effects in the analysis elongates the period. Figure 10.4(b) shows the small displacement, elasto-viscoplastic response (Case (iii)) of the spherical shell cap in which the value of the fluidity parameter is taken as \gamma = 100.0 . It should be noted that permanent viscoplastic deflections occur thus providing a completely different response to either of the elastic responses shown in Fig. 10.4(a).
In Chapter 11 this problem is repeated using an elasto-plastic material model. It should be noted that in order to simulate elasto-plastic behaviour with DYNPAK a high value of the fluidity parameter (say \gamma = 10000.0 )
line
| Time (secs) | Vertical displacement |
|---|---|
| 0 | 0.00 |
| 2 | 0.06 |
| 4 | 0.08 |
| 6 | 0.03 |
| 8 | 0.07 |
| 10 | 0.06 |
Fig. 10.4(b) Results of the transient dynamic analysis of a spherical shell cap. Case (iii).
should be adopted. Interested readers may like to compare DYNPAK and MIXDYN for elasto-plastic behaviour using a high fluidity parameter. However, care should be taken since the use of high fluidity parameter values requires the use of a smaller time step when an Euler scheme is used to evaluate the viscoplastic strains (see Section 8.3). Typical input data for Case (ii) are given in Appendix IV.
At this stage it is probably worth mentioning the important problem of combining material and geometric nonlinearities. Among the several papers on this topic in the existing literature we suggest that the interested reader could profitably refer to the following as a starting point for further study:
McMEEKING, R. M. and RICE, J. R., Finite element formulations for problems of large elastic-plastic deformation, Int. J. Solids Structures, 11, 601–616 (1975).
HIBBITT, H. D., MARCAL, P. V. and RICE, J. R., A finite element formulation for problems of large strain and large displacement, Int. J. Solids Structures, 6, 1069–1086 (1970).
BATHE, K. J., RAMM, E. and WILSON, E. L., Finite element formulations for large deformation analysis, Int. J. Num. Meth. Engng., 9, 353–386 (1975).
10.7.3 Gravity dam example
The geometry of the dam, the seismic acceleration history, the water level and material properties for both dam and foundation are arbitrary.
other
| Dimension | Value |
|---|---|
| Height | 81.45 |
| Width | 107.00 |
| E (tsec²) | 3164000 t/m² |
| ν (sec²) | 0.20 |
| ρ (sec²) | 0.269 tsec²/m⁴ |
| Width | 15.24 m |
| E (tsec²) | 1800000 t/m² |
| ν (sec²) | 0.20 |
| ρ (sec²) | 0.183 tsec²/m⁴ |
| Width | -50.00 |
Fig. 10.5(a) Concrete gravity dam.
area
| Added masses | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
|---|---|---|---|---|---|---|---|
| Row 1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Row 2 | 18 | 19 | 20 | 23 | 26 | 25 | 28 |
| Row 3 | 15 | 16 | 17 | 22 | 23 | 25 | 26 |
| Row 4 | 30 | 31 | 32 | 27 | 28 | 29 | 24 |
| The chart displays a stacked area chart with each row representing a mass value. The total added mass is labeled on the left side. |
Fig. 10.5(b) Finite element mesh for concrete gravity dam.
Both the gravity dam and the foundation shown in Fig. 10.5(a) are idealized with two-dimensional, plane-strain, 8-noded isoparametric elements as shown in Fig. 10.5(b), using a 2 \times 2 Gauss integration rule for the stiffness evaluation, and using a special mass lumping scheme with a 3 \times 3 Gauss integration rule. The adopted 2 \times 2 Gauss integration rule for the stiffness terms ensures that no locking behaviour will occur in the mesh, whereas the 3 \times 3 Gauss integration rule for the lumped mass matrix terms renders better mass representation. The model base is assumed to be fixed, i.e. u = v = 0, and side boundaries are represented by horizontal rollers, i.e. v = 0.
A short duration analytic earthquake (sinesweep) ^{(9)} with a maximum acceleration level 0.33 g (developed as an equivalent to the El Centro NS accelerogram) will be used as a prescribed horizontal acceleration history at the model base level. It is assumed that this signal is the result of the deconvolution process of a prescribed signal at the foundation level. The displacements obtained in the solution process are relative to the model base.
Both the concrete and rock are assumed to behave as elasto-viscoplastic materials with no hardening. The Mohr–Coulomb yield surface is adopted, and the parameters c and \phi are obtained from the uniaxial properties f_{cu} and f_{t} as indicated in Table 10.3.
f_{t}, f_{cu} = \text{tensile, compressive strengths of concrete,}
\alpha = \frac {f _ {t}}{f _ {c u}} = \frac {1 - \sin \phi}{1 + \sin \phi},
\phi = \operatorname{arc} \sin \left(\frac {1 - a}{1 + a}\right),
c = \frac {(a) ^ {- 1 / 2}}{2} f _ {c u},
F _ {0} (\text { Mohr - Coulomb }) = c \cos \phi .
$$<table><tr><td></td><td> $f_{cu}$ $(t/m^{2})$ </td><td> $f_{t}$ $(t/m^{2})$ </td><td> $\alpha$ </td><td> $c$ $(t/m^{2})$ </td><td> $\phi$ </td><td> $F_{0}=c\cos\phi$ $(t/m^{2})$ </td></tr><tr><td>concrete</td><td>4000</td><td>500</td><td>0.125</td><td>707.11</td><td>62.73</td><td>323.94</td></tr><tr><td>rock</td><td>3600</td><td>400</td><td>0.133</td><td>547.72</td><td>61.93</td><td>257.75</td></tr></table>
Table 10.3 Mohr–Coulomb yield surface parameters for concrete dam example.
The values of the fluidity parameters $\gamma$ are considered to be the same for both the concrete and rock materials. Values of $\gamma = 0.00001$ and $\gamma = 0.001$ have been used for the two analyses presented. The stress level in the structure prior to the seismic excitation is assumed to be due to the self-weight and hydrostatic pressure of the water only.
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The influence of the reservoir water on the dynamic behaviour of the dam is considered by taking into account the mass of water attached to the upstream face of the dam. The simple representation of 'added mass' with concentrated masses is used. The adopted model could be improved significantly with transmitting boundaries, better 'added mass' representation, a more realistic signal and a finer mesh.
The choice of the time step length depends on two criteria. For the explicit central difference integration scheme of the dynamic equilibrium equations, the highest mesh frequency defines the critical time step length
\Delta t _ {C D} = \frac {2}{\omega_ {\max}} \simeq \mu L \left(\frac {\rho (1 + \nu) (1 - 2 \nu)}{E (1 - \nu)}\right) ^ {1 / 2}. \tag {10.51}
For the integration of the equations, which govern viscoplastic straining using the Euler method, the critical time step for the Mohr-Coulomb viscoplastic material is defined as
\Delta t _ {M C} = \frac {4 (1 + \nu) (1 - 2 \nu) c \cos \phi}{\gamma (1 - 2 \nu + \sin^ {2} \phi)}. \tag {10.52}
For the mathematical model under consideration, $(L = 2.4665 \, \text{m})$ , the choice of the time step is governed by the $\Delta t_{CD}$ criterion for both analyses. Note that since
\Delta t _ {C D} = 0. 0 0 0 4 7 8 \mathrm{sec} \tag {10.53}
the adopted time step length is $\Delta t = 0.0004$ sec.
On the basis of the adopted mathematical model, (Fig. 10.5), input data can be prepared following the user notes, given in the Appendix III.
<details>
<summary>line</summary>
| Time (s) | Acceleration Level 0.5 M/SEC2 |
| -------- | ----------------------------- |
| 0 | 0 |
| 1 | 0.5 |
| 2 | 1.5 |
| 3 | 3.0 |
| 4 | 2.0 |
| 5 | -5.0 |
| 6 | 6.0 |
| 7 | -6.0 |
| 8 | 5.0 |
| 9 | -4.0 |
| 10 | 3.0 |
| 11 | -3.0 |
| 12 | 2.0 |
| 13 | -2.0 |
| 14 | 1.0 |
| 15 | -1.0 |
</details>
JOHNSON/EPSTEIN SINESWEEP EARTHQUAKE 0.20 SEC
Fig. 10.6(a) Johnson/Epstein sinesweep earthquake.
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<table><tr><td colspan="7">SINESWEEP DT 0.01 SEC 300 ENTRIES</td></tr><tr><td>0.0034</td><td>0.0069</td><td>0.0104</td><td>0.0140</td><td>0.0177</td><td>0.0215</td><td>0.0255</td></tr><tr><td>0.0296</td><td>0.0339</td><td>0.0385</td><td>0.0433</td><td>0.0484</td><td>0.0539</td><td>0.0597</td></tr><tr><td>0.0659</td><td>0.0725</td><td>0.0795</td><td>0.0871</td><td>0.0951</td><td>0.1038</td><td>0.1130</td></tr><tr><td>0.1229</td><td>0.1335</td><td>0.1449</td><td>0.1570</td><td>0.1700</td><td>0.1838</td><td>0.1986</td></tr><tr><td>0.2144</td><td>0.2312</td><td>0.2491</td><td>0.2681</td><td>0.2884</td><td>0.3098</td><td>0.3326</td></tr><tr><td>0.3567</td><td>0.3823</td><td>0.4092</td><td>0.4377</td><td>0.4677</td><td>0.4992</td><td>0.5324</td></tr><tr><td>0.5672</td><td>0.6036</td><td>0.6417</td><td>0.6815</td><td>0.7229</td><td>0.7660</td><td>0.8106</td></tr><tr><td>0.8568</td><td>0.9046</td><td>0.9537</td><td>1.0042</td><td>1.0558</td><td>1.1086</td><td>1.1622</td></tr><tr><td>1.2165</td><td>1.2713</td><td>1.3263</td><td>1.3812</td><td>1.4357</td><td>1.4894</td><td>1.5420</td></tr><tr><td>1.5930</td><td>1.6419</td><td>1.6881</td><td>1.7312</td><td>1.7705</td><td>1.8054</td><td>1.8351</td></tr><tr><td>1.8589</td><td>1.8761</td><td>1.8859</td><td>1.8874</td><td>1.8797</td><td>1.8621</td><td>1.8337</td></tr><tr><td>1.7935</td><td>1.7408</td><td>1.6747</td><td>1.5945</td><td>1.4993</td><td>1.3887</td><td>1.2621</td></tr><tr><td>1.1191</td><td>0.9594</td><td>0.7829</td><td>0.5899</td><td>0.3805</td><td>0.1554</td><td>-0.0845</td></tr><tr><td>-0.3381</td><td>-0.6038</td><td>-0.8798</td><td>-1.1638</td><td>-1.4533</td><td>-1.7372</td><td>-1.9899</td></tr><tr><td>-2.2286</td><td>-2.4500</td><td>-2.6507</td><td>-2.8273</td><td>-2.9764</td><td>-3.0948</td><td>-3.1793</td></tr><tr><td>-3.2271</td><td>-3.2356</td><td>-3.2025</td><td>-3.1262</td><td>-3.0056</td><td>-2.8402</td><td>-2.6303</td></tr><tr><td>-2.3768</td><td>-2.0819</td><td>-1.7485</td><td>-1.3804</td><td>-0.9825</td><td>-0.5607</td><td>-0.1220</td></tr><tr><td>0.3258</td><td>0.7742</td><td>1.2139</td><td>1.6349</td><td>2.0272</td><td>2.3806</td><td>2.6849</td></tr><tr><td>2.9306</td><td>3.1090</td><td>3.2125</td><td>3.2351</td><td>3.1726</td><td>3.0230</td><td>2.7867</td></tr><tr><td>2.4670</td><td>2.0698</td><td>1.6041</td><td>1.0818</td><td>0.5176</td><td>-0.0715</td><td>-0.6660</td></tr><tr><td>-1.2454</td><td>-1.7879</td><td>-2.2718</td><td>-2.6762</td><td>-2.9821</td><td>-3.1733</td><td>-3.2373</td></tr><tr><td>-3.1663</td><td>-2.9582</td><td>-2.6169</td><td>-2.1529</td><td>-1.5831</td><td>-0.9256</td><td>-0.2197</td></tr><tr><td>0.4796</td><td>1.1375</td><td>1.7207</td><td>2.1988</td><td>2.5461</td><td>2.7439</td><td>2.7810</td></tr><tr><td>2.6553</td><td>2.3743</td><td>1.9551</td><td>1.4235</td><td>0.8133</td><td>0.1640</td><td>-0.4813</td></tr><tr><td>-1.0789</td><td>-1.5873</td><td>-1.9703</td><td>-2.2002</td><td>-2.2599</td><td>-2.1450</td><td>-1.8645</td></tr><tr><td>-1.4408</td><td>-0.9084</td><td>-0.3116</td><td>0.2988</td><td>0.8699</td><td>1.3510</td><td>1.6985</td></tr><tr><td>1.8803</td><td>1.8793</td><td>1.6956</td><td>1.3476</td><td>0.8703</td><td>0.3129</td><td>-0.2659</td></tr><tr><td>-0.8041</td><td>-1.2427</td><td>-1.5330</td><td>-1.6417</td><td>-1.5565</td><td>-1.2875</td><td>-0.8674</td></tr><tr><td>-0.3482</td><td>0.2047</td><td>0.7201</td><td>1.1307</td><td>1.3817</td><td>1.4390</td><td>1.2948</td></tr><tr><td>0.9696</td><td>0.5104</td><td>-0.0151</td><td>-0.5283</td><td>-0.9507</td><td>-1.2170</td><td>-1.2848</td></tr><tr><td>-1.1436</td><td>-0.8166</td><td>-0.3588</td><td>0.1518</td><td>0.6265</td><td>0.9814</td><td>1.1528</td></tr><tr><td>1.1094</td><td>0.8596</td><td>0.4508</td><td>-0.0377</td><td>-0.5100</td><td>-0.8715</td><td>-1.0488</td></tr><tr><td>-1.0054</td><td>-0.7506</td><td>-0.3392</td><td>0.1389</td><td>0.5778</td><td>0.8784</td><td>0.9720</td></tr><tr><td>0.8371</td><td>0.5057</td><td>0.0580</td><td>-0.3962</td><td>-0.7437</td><td>-0.8966</td><td>-0.8159</td></tr><tr><td>-0.5228</td><td>-0.0956</td><td>0.3502</td><td>0.6922</td><td>0.8352</td><td>0.7389</td><td>0.4312</td></tr><tr><td>0.0025</td><td>-0.4198</td><td>-0.7084</td><td>-0.7749</td><td>-0.5989</td><td>-0.2364</td><td>0.1960</td></tr><tr><td>0.5575</td><td>0.7285</td><td>0.6519</td><td>0.3541</td><td>-0.0615</td><td>-0.4488</td><td>-0.6697</td></tr><tr><td>-0.6446</td><td>-0.3831</td><td>0.0171</td><td>0.4045</td><td>0.6304</td><td>0.6073</td><td>0.3446</td></tr><tr><td>-0.0521</td><td>-0.4214</td><td>-0.6111</td><td>-0.5422</td><td>-0.2444</td><td>0.1539</td><td>0.4789</td></tr><tr><td>0.5870</td><td>0.4302</td><td>0.0801</td><td>-0.3015</td><td>-0.5364</td><td>-0.5135</td><td>-0.2443</td></tr><tr><td>0.1398</td><td>0.4490</td><td>0.5290</td><td>0.3396</td><td>-0.0213</td><td>-0.3657</td><td>-0.5121</td></tr><tr><td>-0.3826</td><td>-0.0479</td><td>0.3081</td><td>0.4876</td><td>0.3899</td><td>0.0713</td><td>-0.2837</td></tr><tr><td>-0.4675</td><td>-0.3716</td><td>-0.0541</td><td>0.2916</td><td>0.4528</td><td>0.3295</td><td></td></tr></table>
Fig. 10.6(b) Digital form of Johnson/Epstein sinesweep earthquake.
Prior to the dynamic analysis, the initial stresses $\sigma_{0}$ must be evaluated using some static finite element program. Nodal loads and the stress state for every Gauss integration point are recorded, and added to the input data for the dynamic analysis. The sinesweep accelerogram and 300 readings for $\Delta t = 0.01$ sec are given in Fig. 10.6. The accelerogram information is read in from a separate input unit (here tape 7, the assumed seismic excitation in the horizontal direction).
The displacement histories for selected nodal points and stress histories for selected Gauss integration points are written on separate output units (tape 10, tape 11) and may be used later for plotting the results. The displacement histories for nodal points 51 (structure base level) and 127 (dam crest) are given in Fig. 10.7.
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<details>
<summary>line</summary>
| time (sec) | γ = 0.00001 | γ = 0.01 |
| ---------- | ----------- | -------- |
| 0 | 0 | 0 |
| 1 | ~0.5 | ~0.3 |
| 2 | ~1.0 | ~0.8 |
| 3 | ~0.5 | ~0.3 |
| 4 | ~0.8 | ~0.6 |
</details>
Fig. 10.7 Results of transient dynamic analysis of a concerte gravity dam.
# 10.8 Problems
10.1 A simply supported beam is subjected to a step uniformly distributed load. The dimensions and material properties of the beam are shown in Fig. 10.8(a). Only one quarter of the beam needs to be analysed as shown in Fig. 10.8(b). Use DYNPAK to find the midspan lateral deflection when the step lateral load is 0.75 $p_{0}$ where $p_{0}$ is the static collapse load. Note that this problem has been solved by Liu and Lin $^{(10)}$ , Bathe et al. $^{(11)}$ and Nagarajan and Popov. $^{(12)}$ Use the Von Mises yield criterion, a high value of the fluidity parameter $\gamma$ and 8-node elements.
10.2 Repeat Problem 10.1 using the Tresca yield criterion.
10.3 Repeat Problem 10.1 using loads of intensity 0.625 $p_{0}$ and 0.50 $p_{0}$ . Compare your results with those of Liu and Lin. $^{(10)}$
10.4 For a step lateral load of 0.625 $p_{0}$ , repeat Problem 10.1 for various degrees of hardening. Compare your results with those of Liu and Lin. $^{(10)}$
10.5 Solve the problem given in Chapters 7 and 8 using dynamic relaxation. $^{(13,14)}$
10.6 Implement an explicit elasto-plastic, transient dynamic, Mindlin plate program based on DYNPAK. Typical examples are given elsewhere. $^{(15,16)}$
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<details>
<summary>text_image</summary>
0.75 p₀ lb/in²
2 in
30 in
1 in
15 in
</details>
Fig. 10.8 Simply supported beam example (a) Geometry and loading, (b) Finite element idealisation.
# 10.9 References
1. NICKELL, R. E., Nonlinear dynamics by mode superposition, Comp. Meth. Appl. Mech. Engng. 7, 107–129 (1976).
2. BATHE, K. J. and WILSON, E. L., Numerical Methods in Finite Element Analysis, Prentice-Hall, Englewood Cliffs, New Jersey, 308-344 (1976).
3. NAYAK, G. C., Plasticity and large deformation by the finite element method, Ph.D. Thesis, University College of Swansea (1971).
4. IRONS, B. M. and AHMAD, S., Finite Element Techniques, Ellis Horwood, Chichester (1980).
5. BICANIĆ, N., Nonlinear finite element transient response of concrete structures, Ph.D. Thesis, University College of Swansea (1978).
6. HINTON, E., ROCK, T. A. and ZIENKIEWICZ, O. C., A note on mass lumping and related processes in the finite element method, Int. J. Earthquake Engng. Struct. Dynamics, 4, 246–249 (1976).
7. HINTON, E. and OWEN, D. R. J., Finite Element Programming, Academic Press, London (1977).
8. BATHE, K. J. and OZDEMIR, H., Elasto-plastic large deformation static and dynamic analysis, Computers and Structures, 6, 81-92 (1976).
9. JOHNSON, G. R. and EPSTEIN, H. I., Short duration analytic earthquake, J. ASCE, Struct. Div., 102, No. ST5, 993-1000 (1976).
10. Liu, S. C. and Lin, T. H., Elastic-plastic dynamic analysis of structures using known elastic solutions, Int. J. Earthquake Engng. Struct. Dynamics, 7, 147-159 (1979).
11. BATHE, K. J., OZDEMIR, H. and WILSON, E. L., Static and geometric and material nonlinear analysis, Structures and Materials Research Report No. UC SESM 74-4, University of California, Berkeley (1974).
12. NAGARAJAN, S. and POPOV, E. P., Elastic-plastic dynamic analysis of axisymmetric solid, Computers and Structures, 4, 1117-1134 (1974).
13. PICA, A. and HINTON, E., Transient and pseudo-transient analysis of Mindlin plates, Int. J. Num. Meth. Engng. 15, 189–208 (1980).
14. BREW, J. S. and BROTTON, D. M., Nonlinear structural analysis by dynamic relaxation, Int. J. Num. Meth. Engng. 3, 463–483 (1971).
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15. HINTON, E., OWEN, D. R. J. and SHANTARAM, D., Dynamic transient nonlinear behaviour of thick and thin plates, In: The Mathematics of Finite Elements and Applications, MAFELAP II, 1975, Ed. J. R. Whiteman, Academic Press, London, 423–438 (1977).
16. RAO, S. S. and RAGHAVAN, K. S., Dynamic response of inelastic thick plates, AIAA J., 17, 85–90 (1979).