叙述船舶倾斜试验的目的和基本原理以及试验方法、步骤和注意事项。⏺Exercise 3-1某内河船排水量△=820t,It=2380m4,GM=1.7m,求重心在浮心上的高度。解:BM=It/▽=2380/820= 2.902 m∵ GM=zb+BM-zg∴ zg-zb=BM-GM=2.902-1.7=1.202 mExercise 3-2*某长方形起重船的主要尺寸:船长L=15m,B=9m,型深 D=2m,起重船主体重p1=56t,其重心高度KG1=0.85 m,船的上层建筑重p2=78t,重心高度KG2=7.5m,水的重量密度为w=1.025t/m3,试计算:(1)横稳性高GM,(2)纵稳性高GML。Exercise 3-2*Δ=P1+P2=56+78=134 t d=Δ/w/(LB)=134/1.025/15/9=0.968 m KG=(56*0.85+78*7.5)/134=4.721 mIt=1/12LB3=15*93/12=911.25 m4IL=1/12BL3=9*153/12=2531.25 m4V= Δ/1.025=134/1.025=130.732 m3BM=It/V=911.25/130.732=6.970 mBML=IL/V=2531.25/130.732=19.362 mGM=Zb+BM-KG=0.484+6.970-4.721=2.733 mGML=Zb+BML-KG=0.484+19.362-4.721=15.125 mExercise 3-2*有同学将主船体与上层建筑分开算,作增加载荷处理,不妥。这时,P主=56t;P上=78t。假定均处于正浮状态。 d=56/1.025/15/9=0.405 m,δd=78/1.025/15/9=0.564 GML=Zb+BML-Zg=0.405/2+2531.25 /56/1.025-0.85=43.451G1ML1=Δ/ (Δ+p)*GML=56*43.451/(56+78)=18.159 mG1ML1=GML+Δ/(Δ+p)*(d+δd/2- GML-Z)=43.451+56/(56+78)*(0.405+0.564/2-43.451-7.5)=22.445 m(原=15.125 m)Exercise 3-3某巡洋舰排水量△=10200t,船长L=200m,当尾倾为1.3m 时,水线面面积的纵向惯性矩IL=420×104m4,重心的纵向坐标xG=-4.23m,xB=-4.25m,水的重度ω=1.025t/m3。求纵稳性高。解:∵tgθ=(xb-xg)/(zg-zb)=t/L∴(zg-zb)=L*(xb-xg) / t =200*(-4.25+4.23)/(-1.3)=3.078 m BML=IL/(△/ω)=420*10*1.025/10200=422.059 m ∴GML=BML-(zg-zb)=422.059-3.078=419 mExercise 3-4二 二-|||-e 长4m的矩形木材,边长为0.2m,重量密度ω1=0.5 t/m3, 浮于淡水上,求其稳心距水线的高度。解:其重心在水线面上, 稳心距水线高=GM二 二-|||-e ▽=l a2/2; Ix=l (√2a)3/12∴BM=Ix/▽=2*2√2 a / 12=0.094 mGM=BM - (zg-zb)=0.094 - 1/3*(√2a/2)=0.047 mExercise 3-5二 二-|||-e求圆锥体的BM值,其要素如图。解:排水体积▽=πr2d/3 圆:It=3πr4/4BM=It/▽=(3πr4/4)/(πr2d/3)=3r2/4dExercise 3-6 *二 二-|||-e某一方形剖面的匀质物体正浮于淡水中,水的重量密度为 1.0t/m3,试问该物体的重量密度w1为多少时才能保持其稳定漂浮状态?设:物重度为ω,长为L,宽及型深均为B。∵ ωLBB=LBd ∴d=ωB二 二-|||-e Zb=0.5ωB; KG=0.5B; BM=1/12*LB3/LBd=B/(12ω) GM=0.5ωB+ B/(12ω)-0.5B≥0 解方程 ω2-ω+1/6≥二 二-|||-e∴ ω< 0.212, ω>0.785Exercise 3-7某箱形双体船横剖面如图所示,其重心在基线以上 3.875m处,吃水d=2.0m,如果要求初稳性高GM≮2m,试问两单体中心线a的最小值为多少? 解:▽=2 Lbd=12L m3Zb=d/2=2/2=1 mIx=2(Lb3/12+Lb(a/2)2)=3L(3+a2)/2GM=zb+Ix/▽-zg= 1+3L(3+a2)/(2*12L)=(3+a2)/8 - 2.876 ≥2 a2≥(2+2.875)*8-3=36∴ a≥6 mExercise 3-8 *巳知某内河船的数据为船长L=48m,船宽B=8.2m,吃水 d=1.2m。方形系数Cb=0.68,横稳性高GM=1.8m,纵稳性高GML=92m,试求:(1) 横倾1o力矩;(2) 纵倾1厘米力矩;(3) 如果把船上10t重物横向移动2m,纵向移动5m(往船尾方向移动),求重物移动后的横倾角、纵倾角及首尾吃水。假定水线面漂心xF的位置在船中央。Exercise 3-8 *Δ=ωCbLBd=1*0.68*48*8.2*1.2=321.178 tMo= ΔGM/57.3=321.178*1.8/57.3=10.089 t.mMTC= ΔGML/100L=321.178*92/(100*48)=6.156 t.m Φ=pδy/Mo=10*2/10.089=1.98° t=pδx/MTC=-10*5/6.156=-8.112 cm θ=t/L=-0.081/48= -0.00172= -0.09° d’F=dF+(L/2-xf).t/L=1.2-0.018/2=1.159 m d’A=dA-(L/2+xf).t/L=1.2+0.018/2=1.241 mExercise 3-9某船正浮时初稳性高GM=0.6 m, 排水量△=10000t,把船内p=100t的货物向上移动3m,向右移动10m,求货物移动后船的横倾角。解:G1M=GM-p(z2-z1)/△=0.6-100*3/10000=0.57 m φ=tg-1(p(y2-y1)/△G1M)= tg-1(100*10/(10000*0.57))=9.95oExercise 3-10某船有初始横倾角φ=2o36’,现将重量为3%排水量的货物横向移动,使船回到正浮。已知船的初稳性高GM=1.3m,求货物移动的距离。解:p=0.03△△GMsin(2.6)=pl∴l= △GMsin(2.6)/p= △*1.3*sin(2.6)/(0.03△)=1.966 mExercise 3-11某巡洋舰的首吃水dF=5.65m, 尾吃水dA=5.97m,每厘米纵倾力矩MTC=272t.m,试问从后舱要抽出多少吨油到前舱方可使船正浮。两舱间距l=156m。解:pl=(dA-dF).MTCp=100*(5.97-5.65)*272/156 = 55.8 tExercise 3-12L=135m,B=14.2m,dF=5.2m,dA=4.8m, Δ=5200t, GM=0.95m, GML=150m,Tpc=13.8t/cm,xf=-3.5m,在x=-35m,y=1m,z=9m处装货p=200t,求浮态? δd=200/13.8=14.5 cm=0.145 m G1M1=GM+p/(Δ+p)(d+δd/2-GM-Z)=0.95+200/(5200+200)(5.0+0.078-0.95-9)=0.770 mG1ML1=pGML/(Δ+p)=(150*5200)/(5200+200)=144.44 mExercise 3-12tgΦ=py/(Δ+p)G1M1 =(200*1)/(5400*0.770)=0.048 (=2.7°) tgθ=p(x-xf)/(Δ+p)G1ML1 =200*(-35+3.5)/(5400*144.44)=-0.008 (=0.5°尾倾) d’F=dF+δd+(L/2-xf)tgθ=5.2+0.145+(135/2+3.5)*(-0.008)= 4.777 m, d’A= =dA+δd-(L/2+xf)tgθ=4.8+0.145-(135/2-3.5)*(-0.008)= 5.457 m,Exercise 3-13 *某船长L=100m,首吃水dF=4.2 m,尾吃水dA=4.8m,每厘米吃水吨数TPC=80t/cm,每厘米纵倾力矩MTC=75tm,漂心纵向坐标xF=4.0 m。今在船上装120t货物。问货物装在何处才能使船的首吃水和尾吃水相等。原纵倾力矩:ML=100*(dF-dA) MTC =-60*75=4500 tm(x-xf)p=4500 tm∴x=4500/120+4=41.5 mExercise 3-14已知某方形河船船长L=100m,宽B=12m,吃水d=6m,重心垂向坐标zg=3.6m,船中纵剖面两侧各有一淡水舱,其长l=10m,宽b=6m,深h=4m。初始状态两舱都装满淡水。试求一舱内淡水消耗一半时船的横倾角。解:初始状况: △=ωLBd=1×100×12×6=7200 t zb=d/2=6/2=3 mBM=(LB3/12) /LBd=B2/12d=122/12*6=2 m GM=zb+BM-zg =3+2-3.6 =1.4 m 单个水舱:ix=lb3/12=10*63/12=180 m4Exercise 3-14卸载: p=-lbh/2=-10*6*4/2=-120 t, y=b/2=6/2=3 m, z=3 m δd=p/ωLB=-120/(1*100*12)=-0.1 m G1M1=GM+p/(△+p).(d+δd/2-GM-z) -2*ix/▽=1.4-120*(6-0.1/2-1.4-3)/(7200-120) -2*180/7080=1.323 mφ=57.3*py/(△+p)G1M1=57.3*120*3/(7200-120)/1.323=2.2 oExercise 3-15已知某内河船L=58m,B=9.6m,dF=1m,dA=1.3m, Cb=0.72m,GML=65m,为了通过浅水航道,必须移动多重的货物,才能使船处于正浮。(货物移动距离l=28m)纵倾角θ=(dF-dA)/L=(1-1.3)/58=-0.0052 rad Δ=ωCbLB(dF+dA)/2=1*0.72*58*9.6*1.15=461.03 t p= ΔGMLθ/ l =461.03*65*0.0052/28=5.57 tExercise 3-16 *某内河客船的一舷受到风的作用,受风面积Af=410m2,受风面积的中心在基线以上的高度zf=4.7m,风压p=490Pa,已知船的要素为:船长L=75m,船宽B=8.1m,吃水 d=2.2m,方形系数Cb=0.645,初稳性高GM=1.4m,假定水阻力中心在其水线处,求该船受风力作用时的横倾角。Δ=ωCbLBd=1.0*0.645*75*8.1*2.2=862.04 tMf=pAf(Zf-d)=490*410*(4.7-2.2)=502250 NmΦ=acsin(Mf/ΔGM)=acsin(502250/(9800*862.04*1.4))=2.4°(0.0419 rad)Exercise 3-17若船靠岸时有80名乘客集中一舷,已知乘客移动的距离l=4m,每乘客重60kg,船横倾1度力矩Mo=8.2t.m,求船的横倾角。M客=80*0.06*4=19.2 tmΦ= M客/Mo=19.2/8.2=2.34°Exercise 3-18某内河驳船△=1100t,平均吃水d=2m,每厘米吃水吨数TPC=6.5t/cm,在6个同样的矩形舱内装有重度ω1= 0.93t/m3的石油,每舱都有自由液面,油舱l=15m, b=6m, 这时船的初稳性高GM=1.86m,若把右舷中间的一个舱中重量 p=120t的油完全抽出,其重心垂向Zg=0.8m,求船的横倾角。解:单个油舱ix=lb3/12=15*63/12=270 m4 , y=3 m δd=p/TPC=-120/6.5= -18.46 cm= -0.185 mG1M1=GM+p(d+δd /2-GM-z)-5ω1ix/▽ 二 二-|||-e=1.86-120(2-0.185/2-1.86-0.8)/(1100-120)-5*0.93*270/980=0.671 m φ=tg-1(py/(△-p)G1M1)=tg-1(120*3/(980*0.671))=28.7 o⏺Exercise 3-19 *某长方形船在港内进行倾斜试验,其主尺度和主要数据为:船长L=32m,船宽B=9.15m,首吃水dF=1.83 m,尾吃水dA=3.66m,移动重量p=3t,横移距离l=4.6m,摆锤长 λ=4.6m,摆动距离k=0.1m,试验后尚需从船上x= -8.2m,z=2.4m处卸去50t重量。求该重量卸去后的重心高度和首、尾吃水。计算浮心(精确解): M = LBdF二 二-|||-e(dAdF 二 二-|||-e∴Zb = M = AdF + (dA−3dF )2 = 3.66*1.83+ 1.833 2 = 1.42m dLB dA + dF 3.66+1.83二 二-|||-ed=(dF+dA)/2=(1.83+3.66)/2=2.745 m Zb=0.5d=0.5*2.745=1.373 m (精确解1.42 m)Δ=ωLBd=1.0*32*9.15*2.745=803.74 tΦ=actg(k/λ)= actg(0.1/4.6)=1.24° (=0.0216 rad)GM=pl /ΔsinΦ=(3*4.6)/(803.74*sin1.24)=0.793 mBM=1/12*LB3/LBd=B2/12d=9.152/(12*2.745)=2.542 m∴Zg=Zb+BM-GM=1.373+2.542-0.793=3.122 m=1.42+2.542-0.793=3.169 m BML=1/12*BL3/LBd=L2/12d=322/(12*2.745)=31.087 mGML=KB+BM-KG=1.373+31.087-3.122=29.338 mGM’L=Δ/(Δ+p)*GML=(803.74*29.338)/(803.74-50)=31.284 m δd =p/ωAw=-50/(1.0*32*9.15)=-0.171 m tgθ=px/(Δ’GM’L)=-50*(-8.2)/(753.74*31.284)=0.0174θ=1od’F=dF+δd+(L/2-xf)tgθ=1.83-0.171+(32/2)*(0.0174)= 1.937 m d’A= =dA+δd-(L/2+xf)tgθ =3.66-0.171-0.278= 3.211 mZg’=(ΔZg-pz)/(Δ-p)=(3.122*803.74-50*2.4)/753.74=3.170 m=(3.169*803.74-50*2.4)/753.74=3.220 m⏺V=1/3(πr2)*2r=2/3 πr3∴Cm=Am/(2r*r)= π/4=0.785Cp=V/(1/2πr2*4r)= 1/3=0.333Cwp=Aw /(4r*2r)= 1/2=0.500Cb=V/(4r*2r*r)=π/12=0.261 Cvp=V/(4r2*r)=π/6=0.5224/4 二 二-|||-eExercise 1-3某海洋客货轮排水体积V=9750m3,长宽比L/B=8,宽度吃水比B/d=2.63,船型系数Cm=0.9, Cp=0.66, Cvp=0.78,试求:(1)船长L;(2)船宽B;(3)吃水d;(4)水线面系数Cw;(5)方形系数Cb;(6)水线面面积Aw。解:Cp=V/CmBdL=V/CmB(B/2.63)(8B)=2.63V/8CmB3∴B=((9750*2.63)/(8*0.66*0.9))1/3=17.54mL=8*17.54=140.32m d=17.54/2.63=6.67mCb=Cm.Cp=0.9*0.66=0.594Cw=Cb/Cvp=0.594/0.78=0.762A. w=0.762*140.32*175.54=1875.44m2 B. 1-4 C. L/B=6.7,B/d=2.46,Cb=0.53。求排水体积。 D. =B/d*d=2.46*2.05=5.043 E. L=L/B*B=6.7*5.043=33.788 F. V=CbLBd=0.53*33.788*5.043*2.05=185.13 t G. 1-5V=25m3,L/B=5,B/d=2.7,Cb=0.52。求该艇的主尺度。b=V/LBd=V/(5B*B*B/2.7)3-20w=1320m2,移动载荷p=50t,移动距离 l=9.25m,摆锤长λ=3.96m,摆动距离k=0.214m。试验后还须加装850t的燃油,燃油重心z=5.18m,ω1=0.86t/m3,自由液面ix=490m2。求最后的横稳性高G1M1。φ-1=pl/△GM=50*9.25/(7200GM)=k/λ=0.214/3.96 ∴GM=0.214*7200/(3.96*50*9.25)=0.841 m加装燃油p1=850t,w=850/1*1320=0.644 m1M1=0.841+850(6+0.644/2-5.18-0.841)/(7200+850) 0.86*490/(7200+850)= 0.820 m