nastran强迫运动

1、nastran强迫运动1概述(1) 用于分析具有地基输入加速度、速度和位移的约束结构(2) 常见例子是地震(瞬态分析)、正弦扫描器测试仿真(频响分析)(3) Nastran中将作用在未约束结构上的力转化为等效约束结构的强迫运动(4) 使用的方法有：大质量法、大刚度法、Lagrange乘子法2瞬态分析中的强迫运动(1) 如果卡片TLOADi中的第5字段选择，给定的加速度、速度、位移不会执行(2) Nastran中仅能作用力P(t)在结构上，如果选择强迫运动，即假设用户在大质量上施加了一个运动，因此作用在大质量上的力是与加速度成正比的3瞬态分析中的大质量法Structure ofInterestR

2、igid Body(1)对强迫加速度ubub P (Ml = large mass)(2) 对作用在大质量上的力 P可以得到期望的Ub,质量ML在卡片CMASSi或CONM2中输入；比例因子 ML (对力)在DAREA或DLOAD中输入(3) 对瞬态响应中期望 Ub,Ub，用卡片TLOAD1或 TLOADi第5字段进行区分(4) 大质量应该为结构质量的 103108倍，106倍效果最好(5) 根据卡片TLOADi输入的强迫加速度、速度、位移得到加速度历程a) 定义Fn = TLOADi at time N(Atn)F(sj = computed force at time N(Atn)b) 强

3、迫加速度c）强迫速度FN-FN-1)/(At)if tN = 0d）强迫位移nFM-F1J FN-1FN-2rAt1 .if tN = 0e）使用大质量法，Ub同样要乘比例因子F = MaMass Scaling for I l Calculatjon ofApplied ForceComputed fromTLOADi Input4瞬态分析中的大刚度法S円叫刀莎Structure ofInterest(1 )对强迫位移ub ,1Ub =(Kl = lrge spring)(2) 作用力P于弹簧或结构表面得到期望的 ub,弹簧Kl在卡片CELAS1或CELAS2中输入；比例因子 Kl在DARE

4、A或DLOAD中输入(3) 大质量应该为结构质量的 102104倍，其中为激励的截断频率5瞬态分析中的LAGRANGE子法(1) 通过DMIG或模拟约束的瞬态函数增加动态矩阵的行和列(2) 每行表示如下形式的约束Uj = Yj (Yj = enforced displacement)(3) 每列表示称为Lagrange乘子的附加自由度 j，其值为第j个约束自由度的约束 力(4) 在直接瞬态分析中，EPOINTs, SPOINTs或GRID用于描述Lagrange乘子，模态响应分析中仅用 EPOINTs(5) Nastran输入中需要设置卡片 SDCOMP = 326强迫位移例子Constrai

5、nedStructureUnconstrained Structurefor MSC/NASTRAN EnforcedMotion SimulationEnforced Displacement RampTimeFor this example m = 1.0, k = (2tt)2To get k, use rod with A = 39.4784, E = 1.0, L = 1.0Solution*:u(t) = dt-1/a)sin cotu(t) = dco sin(otwhere w = 6.28318d = 2.5(1)大质量法1在TLOAD的第5字段中，指定ub为强迫运动(2)大

6、刚度法m = 1.0m在TLOAD仲指定Pb为力，乘以强迫运动值 kB,在DAREA中输入Large Mass MethodSOL 109TIME 10CEND TITLE=SDOF, GROUND MOTION SUBTI= LARGE MASS LABEL = X VIA XSET 1=1 DLOAD=105 TSTEP=6 DISPL=ALL ACCEL=ALL OLOAD=1 BEGIN BULK DAREA,105,1,1,1.0E8 CONM2,3,2,1.0 CONM2,999,1,1.0E8 CROD,1,1,1,2 GRID,1,0.,0.,0.,23456 GRID,2,1

7、.,0.,0.,23456 MAT1,1,1.,0.PARAM,USETPRT,1 PROD,1,1,34784 TABLED1,8,+TAB1 +TAB1,0.,0.,.025,0.,.05,0.,.075,0.,+TAB2 +TAB2,5.075,12.5,10.075,25.0,ENDT TLOAD1,105,105,1,8 TSTEP,6,199,.025,1ENDDATALarge Stiffness MethodSOL 109 TIME 10 CEND TITLE=SDOF, GROUND MOTION SUBTI= LARGE STIFFNESS LABEL = X VIA FO

8、RCING SET 1=1 SPC = 100DLOAD=105 TSTEP=6 DISPL=ALL ACCEL=ALL OLOAD=1 BEGIN BULK DAREA,105,1,1,4.0E8CONM2,3,2,1.0CR0D,1,1,1,2GRID,1,0.,0.,0,23456GRID,2,1.,0.,0,23456GRID,99,-1.,0.,0.SPC,100,99,123456CELAS2,5,4.0E8,99,1,1,1MAT1,1,1,0.PARAM,USETPRT,1PROD,1,1,34784TABLED1,8,+TAB1+TAB1,0.,0.,.025,0.,.05,

9、0.,.075,0.,+TAB2+TAB2,5.075,12.5,10.075,25.0,ENDTTLOAD1,105,105,0,8TSTEP,6,199,.025,1ENDDATA(3) Lagrange 乘子法m = 11.0k = 39.4784f PB = P1OO1运动方程（静力部分）为394784-39.47841.0U1 IP1丨0 r-39.478439.47840 “口2卜*P2卜= 01.00X1001Y1001Y100139.4784uj 39.47842 + 対= 0-39.4784U! + 39.4784u2= 0ui+ xiooi= yiooi其中，x为一个小整数

10、，用 K2PP输入Lagrange乘子Lagra nge Multiplier MethodSOL 109TIME 10CENDTITLE=SDOF, GROUND MOTIONSUBTI= LAGRANGE MULT - DISPLK2PP=STIFSET 1=1DLOAD=105TSTEP=6DISPL=ALLACCEL=ALLOLOAD=ALLBEGIN BULKDAREA,105,1001,0,1.0CONM2,3,2,1.0CROD,1,1,1,2DMIG,STIF,0,6,1,0DMIG,STIF,1001,101,-1.0E-10DMIG,STIF,1001,1,1, .1EPO

11、INT,1001GRID,1,0.,0.,0.,23456GRID,2,1.,0.,0.,23456MAT1,1,1.,0.PARAM,USETPRT,1PROD,1,1,34784TABLED1,8,+TAB1+TAB1,0.,0.,.025,0.,.05,0.,.075,0.,+TAB2+TAB2,5.075,12.5,10.075,25.0,ENDTTLOAD1,105,105,1,8TSTEP,6,199,.025,1ENDDATA7频率响应分析中的大质量法在频率响应中，假设输入和响应都为简谐函数u = u（）eitot动力学方程为-o2M + icoB + KHu（w） = Pg施加

12、加速度如瞬态分析中的大质量法一样，使大质量移动的力和加速度为P = maiojt因此,P(w) =对施加位移u = u(W)eiwt可以得到加速度为a何=-cn2uo)eiwt因此作用力为2P(cd)= ma(o)= - m(D u(d)其中，或 2由卡片TABLED1或TABLED4俞入8频率响应分析中的大刚度法与大质量方法一样，期望的力为P = Ku对施加位移，力为P(ct) = Ku(o)对施加加速度，力为co) = Ka co)_3用 TABLED1 输入!/ 或 H -29例子：具有强迫加速度的直接瞬态响应zu(1) 问题：一 250Hz的单位加速度正弦脉冲作用在地基的z方向，100

13、01b的大质量施加在基础上，结构阻尼系数为0.06,在250Hz处将结构阻尼转化为等效粘性阻尼(2) 输入文件ID SEMINAR, PROB7SOL 109TIME 30CENDTITLE = TRANSIENT RESPONSE WITH BASE EXCITATIONSUBTITLE = USING DIRECT TRANSIENT METHOD, NO REDUCTIONECHO = UNSORTEDSPC = 200SET 111 = 23, 33DISPLACEMENT (SORT2) = 111VELOCITY (SORT2) = 111ACCELERATION (SORT2)

14、= 111SUBCASE 1DLOAD = 500TSTEP = 100$OUTPUT (XYPLOT)XGRID=YESYGRID=YESXTITLE= TIME (SEC)YTITLE= BASE ACCELERATIONXYPLOT ACCELERATION RESPONSE / 23 (T3)YTITLE= BASE DISPLACEMENTXYPLOT DISP RESPONSE / 23 (T3)YTITLE= TIP CENTER DISPLACEMENT RESPONSEXYPLOT DISP RESPONSE / 33 (T3)$BEGIN BULK$ PLATE MODEL

15、 DESCRIBED IN NORMAL MODES EXAMPLE $INCLUDE plate.bdf PARAM, COUPMASS, 1PARAM, WTMASS, 0.00259$ SPECIFY STRUCTURAL DAMPING$PARAM, G, 0.06PARAM, W3, 1571.$ APPLY EDGE CONSTRAINTS$SPC1, 200, 12456, 1, 12, 23, 34, 45$ PLACE BIG FOUNDATION MASS (BFM) AT BASE$CMASS2, 100, 1000., 23, 3$ RBE MASS TO REMAIN

16、ING BASE POINTS$RBE2, 101, 23, 3, 1, 12, 34, 45$ APPLY LOADING TO FOUNDATION MASS$TLOAD2, 500, 600, , 0, 0.0, 0.004, 250., -90.$DAREA, 600, 23, 3, 2.588$ SPECIFY INTEGRATION TIME STEPS$TSTEP, 100, 200, 2.0E-4, 1 $ENDDATA3)结果3X5込XX 19例子：具有强迫位移的直接频率响应Large Mass(100,000 lbs)(1) 问题：图示平板在角点受 0.1的强迫位移，在频率

17、范围20,1000内使用步长20Hz, 结构阻尼系数0.06(2) 输入文件ID SEMINAR, PROB8SOL 108TIME 30CENDTITLE= FREQUENCY RESPONSE DUE TO .1 DISPLACEMENT AT TIPSUBTITLE= DIRECT METHODECHO= UNSORTEDSPC= 1SET 111= 11,33, 55DISPLACEMENT(PHASE, SORT2)= 111$SDISP(PHASE, SORT2)= ALLset 222 = 11OLOAD= 222SUBCASE 1DLOAD= 500FREQUENCY=100$

18、OUTPUT (XYPLOT)$XTGRID= YESYTGRID= YESXBGRID= YESYBGRID= YESYTLOG= YESYBLOG= NOXTITLE= FREQUENCY (HZ)YTTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, MAGNITUDEYBTITLE= DISPLACEMENT RESPONSE AT LOADED CORNER, PHASE XYPLOT DISP RESPONSE / 11 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, MA

19、GNITUDEYBTITLE= DISPLACEMENT RESPONSE AT TIP CENTER, PHASEXYPLOT DISP RESPONSE / 33 (T3RM, T3IP)YTTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, MAGNITUDE YBTITLE= DISPLACEMENT RESPONSE AT OPPOSITE CORNER, PHASE XYPLOT DISP RESPONSE / 55 (T3RM, T3IP)$BEGIN BULK$ PLATE MODEL DESCRIBED IN NORMAL MOD