船舶靜力學(xué)復(fù)習(xí)資料(一)

第一章某海洋客貨輪排水體積V=9750m3,長寬比L/B=8,寬度吃水比B/d=2.63,船型系數(shù)Cm=0?9,Cp=0.66,Cvp=0?78，試求：(1)船長L；(2)船寬B;(3)吃水d；(4)水線面系數(shù)Cw；(5)方形系數(shù)Cb；(6)水線面面積Aw。I解：Cp=V/CmBdL=V/CmB(B/2.63)(8B)=2.63V/8CmB3???B=((9750*2.63)/(8*0.66*0.9))13=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=0.762Aw=0.762*140.32*175.54=1875.44m2第二章1.1-JPJ°,E11-JPJ°,E1o806040200某船水線長L=100m在正浮狀態(tài)時，各站號的橫剖面面積如下表所列：站號012345678910面積013.330.444.453.857.354.344.730.113.50(1)以適當(dāng)比例畫出該船的橫剖面面積曲線;"2)用梯形法和辛氏第一法按表格計算排水體積V,浮心縱向坐標(biāo)XB；(3)求縱向棱形系數(shù)Cp。梯形法：V=L/10*(Sy-e)=100/10*(0+13.3+30.4+44.4+53.8+57.3+54.3+44.7+30.1+13.5+0)=3418m3Myoz=L/10*(xy-￡)=100/10*(-5*0-4*13.3-3*30.4-2*44.4-1*53.8+0*57.3+1*54.3+2*44.7+3*30.1+4*13.5+5*0)*10=100m4Xb=MyozA^=100/3418=0.029m

y)=100/30*(0+4*13.3+2*30.4+4*44.4+2*53.8+4*57.3+2*54.3+4*44.7+2*30.1+4*13.5+0)=3433m3=100/30*(-5*0-4*4*13.3-2*3*30.4-4*2*44.4-2*1*53.8+4*0*5731+2*1*543+4*2*44.7+2*3*30.1+4*4*13.5m4Xb=Myoz/V=160/3433=0?046mCp=V/AmL=3433/5731/100=0.5992?|某內(nèi)河船吃水d=2.4m,方形系數(shù)Cb=|654,水線面系數(shù)Cw=0.785，卸下貨物p=80%排水量，求船舶的平均吃水(設(shè)在吃水變化范圍內(nèi)船體為直舷)。解：p=0.08^=0.08?CbLBd6d=p/3CwLB=0.08?CbLBd/?CwLB=0.08dCb/Cw=0.08*2.4*0.654/0.785=0.16md1=d-6d=2.4—0.16=2.24m某船L=60m,其水線以下橫剖面為等邊三角形，從尾向首其半寬為0.3,1.6,43,5.0,4.6,3.3m,試求水線面漂心xF,排水體積▽,浮心Xb,zb，方形系數(shù)Cb。用梯形法解：§L二60/5二12my0.3001.6004.3005.0004.6003.30019.100X-30.000-18.000-6.0006.00018.00030.000yx-9.000-28.800-25.80030.00082.80099.000148.200A0.1564.42931.98843.25036-6071&840135.269z0.1730.9252.4862.8902.6591.908Az0.0274.09679.507125.00097.33635.937341.903Ax-4.671-79.718-191.926259.500658.922565.1911207.298Aw=2*12*(X-e)=2*12*(19.1-(0.3+3.3)^^<^m2xF=12*(424.8-(18+0)/2)/207.6=24.035a逹竺竺V=12*(135.269-(0.156+18.84)/2)=810.678m3Zb,=12*(341.903-(0.027+35.937)/2)/810.678=4.795mxb=12*(1207.298-(-4.671+565.191)/2)/810.678=13.722m第三章1.某巡洋艦排水量厶=102001,船長L=200m，當(dāng)尾傾為1.3m時，水線面面積縱向慣性矩IL=420X104m4，重心的縱向坐標(biāo)xG=-4.23m，xB=-4?25m，水的重度3=1.025t/m3。求縱穩(wěn)性高。解：Ttg。=(xb-xg)/(zg-zb)=t/L?：(zg—zb)=L*(xb—xg)/t=200*(—4.25+4?23)/(T?3)

=3.078mBML=IL/(△/3)=420*10*1.025/10200=422.059m???GML=BML—(zg—zb)=422?059—3.078=419m某內(nèi)河船排水量4=8201,It=2380m4,GM=1?7m,求重心在浮心上的高度。解：BM=It/▽=2380/820=2.902mGM=zb+BM-zg???zg-zb=BM-GM=2?902一1.7=1.202m=72001，■水d=6m，水線面面積Aw=1320m2,移動載已知某方形河船船長L=100m，寬B=12m，吃水d=6m,重心垂向坐標(biāo)zg=3.6m,船中縱剖面兩側(cè)各有一淡水艙，其長l=10m，寬b=6m,深=72001，■水d=6m，水線面面積Aw=1320m2,移動載2.某船正浮時浮心垂向坐標(biāo)zB=2?9m,重心垂向坐標(biāo)zg=1?5m,橫傾角?=40o時的浮心橫向、垂向坐標(biāo)分別為yB40=1.75m和zB40=3?加，求此時的靜穩(wěn)性臂140。解：1=YbCos40°+[(Zb-Zbo)-(Zg-Zbo)]Sin40°=YbCos40°+(Zb-Zg)Sin40°=1?75Cos40°+(3?2—4?5)Sin40°=0.505某船在橫傾巾=4亍時動穩(wěn)性臂Ld=0.7mt這時浮心橫向坐標(biāo)yb=2.55mJ求船在該橫傾角時的復(fù)原力臂，船正浮時重心在浮心上a=2.1nic解：橫傾前浮心到重心的距離忍=2Jm,橫傾45o時浮心到重心的距離增加了Ld=0.7m,所以：BZ^2.8m,Ls=(yb-BZcos45)/cos45=yb/cos45-BZ第五章2-55/cos45-2.8=0.806m渥釋艙壁甲板：用于核算不沉性的連續(xù)水密甲板。限界線：艙壁甲板邊線下76mm處的界線。滲透率：艙室進水后浸水體積與艙室容積之比的百分?jǐn)?shù)?？山L度：使艙室浸水后平衡水線恰好與限界線相切的艙室長度。分艙因數(shù)：按抗沉性規(guī)范要求的小于1的分艙系數(shù)。許用艙長：分艙因數(shù)與可浸長度的積。1?某內(nèi)河船的數(shù)據(jù)為：4=8001,d=2?2m,TPC=5?7t/cm,GM=1?75m,求該船破損后的浮態(tài)和穩(wěn)性高?已知進水艙的進水體積v=90m2,x=xF,y=2?4m,z=1?2m,自由液面積為1*b=12?5*5m.第二類艙：(增加重量法)p=wv=1.0*90=90t;p+A=90+800=890tix=lb3/12=12.6*53/12=131.25m4iy=bl3/12=5*12.63/12=833.49m46d=p/Tpc=90/5.7=15.8cm=0.158mGiMi=GM+p/(p+A)(d+'d/2-GM-Z)=1.75+90/890(2.2+0.158/2-1.75-1.2)=1.682m

GiMli=AGMl/(p+A)=800*GMl/890=0.899GMlm堰①=py/(p+A)G1M1=90*2.4/(890*1.682)=0.144(8.2°)tg"=p(x-xf)/(p+A)GiMli=90*(^xf)/(890*GiMli)=0(x=xf)d*=d+§d=2.2+0.158=2.358m2.已知某船的數(shù)據(jù)為：L=95m,B=12.4m,dF=5?8m,dA=6?3m,Cb=0?7,Cw=0.78,xF=1?4m,GM=0.42m,GML=125m。因船體損傷，雙層底艙淹水，該艙的體積v=60m3,形心坐標(biāo)x=20m,y=2.7m,z=0.4m。求該船損傷后的橫傾角和首尾吃水。解：第一類艙：d=0.5(dF+dA)=0.5*(6.3+5.8)=6.05mA=qLBd=1.025*0.7*95*12.4*6.05=5113.6tAw=CwLB=0?78*95*12?4=918?8m2P=1?025*60=61.5t;p+A=61.5+5113.6=5175?1t5d=p/oAw=61.5/(1.025*918.8)=0.065mG1M1=GM+p/(p+A)(d+5d/2-GM-Z)=0.42+61.5/5175.1(6.05+0.065/2-0.4-0.42)=0.482mG1ML1=AGML/(p+A)=5113.6*125/5175.1=123.5mtg?=py/(p+A)G1M1=61.5*2.7/(5175?1*0.482)=0.067(3?8°)tg0=p(x-xf)/(p+A)G1ML1=61?5*(20—1?4)/(5175.1*123.5)=0.0018(0?1°)解：初始狀況：△=3LBd=lX100X12X6=7200tzb=d/2=6/2=3mI BM=(LB*I*3/12)/LBd=B2/12d=122/12*6=2mI GM=zb+BM-zg=34-2-3.6=1.4mI單個水艙：ix=lb3/12=10*63/12=180m4*6■內(nèi)河船做傾斜試驗時，■水量?荷p=50t,移動距離l=9?25m,擺錘長入=3.96m,擺動距離k=0?214m。試驗后還須加裝850t的燃油，燃油重心z=5?18m,31=0?86t/m3，自由液面ix=490m2。求最后的橫穩(wěn)性高G1M1。解：tg?-1=pl/^GM=50*9.25/(7200GM)=k/入=0.214/3.96???GM二0.214*7200/(3.96*50*9.25)=0.841m加裝燃油p1=8501,6d=p1/3Aw=850/1*1320=0.644mG1M1=0.841+850(6+0.644/2-5.18-0.841)/(7200+850)-0.86*490/(7200+850)=0.820m若船靠岸時有80名乘客集中一舷，已知乘客移動的距離l=4m，每乘客重60kg,船橫傾1度力矩