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船舶原理习题集(解)1船舶原理习题集(解)习题一船体形状1.1.根据表1-1七种船型所列数值,试计算表1—2和表1—3空格的主尺度比和船体系数值,并填入表内。最后逐个阅读每种船型的主尺度比和船体系数值,并比较各种船型数值的差别,建立每种船型主尺度比和船体系数的数值概念。表1-1船长型宽型深型吃水船中剖面面积水线面积排水体积船型LBDdAmAwVmmmmm2m2m310000t级远洋杂货舱14720.4012.408.20164.602381164757500t排水量海洋客货船12417.6010.905.5091.771702685420000t级远洋油船17025.0012.609.50235.603621313316141kw港作拖船27.008.003.802.80*18.14159.42871441kw长江拖船43.6010.003.603.00*27.30345.77503700t排水量长江客货船10516.404.703.6057.30135236892400t长江油船93.5513.804.803.40*45.9811253380*:这三种船型均为设计尾倾,其首和尾吃水依次为dF=2.25m,dA=3.35m;dF=3.2m,dA=2.8m;dF=3.2m,dA=3.6m。解:答在表格的的方框内。表1—2船型L/BB/dD/dB/DL/D10000t级远洋杂货船7.22.491.511.6511.857500t排水量海洋客货船7.053.201.981.6211.3820000t级远洋油船6.82.631.321.9813.49614lkW港作拖船3.382.861.362.117.111441kW长江拖船4.363.331.22.7812.113700t排水量长江客货船6.404.561.313.4922.342400t长江油船6.784.061.412.8819.49表1—3船型CbCwCmCpCvp10000t级远洋杂货船0.670.7940.9840.6810.8447500t排水量海洋客货船0.5710.7800.9480.6010.73220000t级远洋油船0.7760.8520.9920.7820.911614lkw港作拖船0.4750.7380.8100.5860.6441441kw长江拖船0.5730.7930.9100.6300.7233700t排水量长江客货船0.5950.7850.9710.6130.7582400t长江油船0.7700.8720.9800.7860.8831.2.已知某船:L=120mld=5.8m;V=7350m3,Cb=0.62,Cw=0.75;试求其Aw船舶原理习题集(解)2解:AW=CWLBCb=LBdVLB=dCVbAW=bWCCdV=8.5*62.07350*75.0=1532.95=1533m21.3.名词解释:船中,舷弧,梁拱,型表面,型线图,型线图的三个基准面,平行中体,船长,型宽,型深,型吃水,船体系数答:略。习题二船体的近似计算2.1.已知某船吃水为4.2m的水线分为10个等分,其横向坐标间距l=3.5m,自首向尾的横向坐标值(半宽,m)分别为:0,3.30,5.30,5.90,5.90,5.90,5.90,5.85,5.22,3.66,1.03。试分别用梯形法则和辛氏法则求其水线面积。解:梯形法计算:通用公式:A=niiyl0(-20nyy)niiy0=0+3.3+5.3+4*5.90+5.85+5.22+3.66+1.03=47.96ε=20nyy=203.10=0.515因水线面面积是对称的,故AW=2AAW=2niiyl0(-ε)=2*3.5*(47.96-0.515)=332.08m2辛氏一法计算:公式A=31l(y0+4y1+2y2+4y3+……4yn-3+2yn-2+4yn-1+yn)=31*3.5*(0+4*3.3+2*5.3+4*5.9+2*5.9+4*5.9*+2*2.9+4*5.85+2*5.22+4*3.66+1.03)=168.13AW=2A=2*168.13=336.26m22.2.某船中横剖面的半宽坐标,自基线起向上分别为0,2.43,5.23,6.28,6.60,6.75,6.80m,两半宽坐标间的垂向间距为1m。试用辛氏法则求船中横剖面的面积。解:辛氏一法:船舶原理习题集(解)3Am=2*31(1*0+4*2.43+2*5.23+4*6.28+2*6.6+4*6.75+6.8)=2*30.77=61.54m2辛氏二法:Am=2*83(1*0+3*2.43+3*5.23+2*6.28+3*6.6+3*6.75+6.8)=2*30.91=61.8m22.3.某5000t货船各水线面积如表2—1所示。水线(m)基线0.40.81.21.62.02.42.83.2水线面积Aw(m2)09801123116512001225124012581268水线(m)3.64.04.44.85.25.66.06.46.8水线面积Aw(m2)128012931305132013351350136013801400试用梯形法求6.8m水线以下的排水体积。解::梯形法:I=niiy00+980+1123+1165+1200+1225+1240+1258+1268+1280+1293+1305+1320+1335+1350+1360+1380+1400=21482ε=214000=700排水体积V=l(I-ε)=0.4*(21482-700)=8312.8m3辛氏法联合应用:V=31*0.4*(0+4*980+2*1123+4*1165+2*1200+4*1225+2*1240+4*1258+2*1268+4*1268+2*1293+4*1305+2*1320+4*1335+1350)+83*0.4*(1350+3*1360+3*1380+1400)=8363.1m3习题三浮性3.1.已知某船重量分布如表3—9所示,试求船舶的重量和重心坐标。表3-9项目重量(t)重心距基线高(m)垂向力矩(t.m)重心距船中(m)纵向力矩(t.m)第一货舱12839.88+54.54第二货舱29897.52+34.89船舶原理习题集(解)4第三货舱33657.23+11.89第四货舱30797.80-33.45第五货舱17649.94-56.52空船52158.95-4.99燃油10453.01+2.72轻柴油2652.01-13.80机油1810.70-2.18滑油4310.70-7.80淡水3795.07-49.8解::列表计算:重量重心距基线高垂向力矩重心距船中纵向力矩(t)(m)(t.m)(m)(t.m)第一货舱1283.0009.88012676.04054.54069974.820第二货舱2989.0007.52022477.28034.890104286.210第三货舱3365.0007.23024328.95011.89040009.850第四货舱3079.0007.80024016.200-33.450-102992.550第五货舱1764.0009.94017534.160-56.520-99701.280空船5215.0008.95046674.250-4.990-26022.850船员,行李……76.00012.400942.400-5.500-418.000燃油1045.0003.0103145.4502.7202842.400轻柴油265.0002.010532.650-13.800-3657.000机油18.00010.700192.600-2.180-39.240滑油43.00010.700460.100-7.800-335.400淡水379.0005.0701921.530-49.800-18874.200合计19521.0007.935154901.610-1.789-34927.240项目按图表3-9所示数值计算(上表),将计算结果iigZPZW00=154901.61、W=W0+iP=19521.00代入公式,得Zg=mPWZPZWiiig935.71952161.154901000,将计算结果iigXPXW00=-34927.24、W=W0+iP=19521.00代入公式,得Xg=mPWXPXWiiig789.100.1952124.34927000.3.2.若题1船在航行中途消耗油和谈水如表3—10,求此时船舶的重心坐标及移动方向。表3-10项目重量(t)重心距基线高(m)垂向力矩(t.m)重心距船中(m)纵向力矩(t.m)燃油8382.18+3.82轻柴油236.50.96-15.16滑油4310.70-7.8机油1810.70-2.18船舶原理习题集(解)5淡水3144.47-55.80总计解::由上题已知:W0=19521t,Zg0=7.935m,Xg0=-1.789m重量重心距基线高垂向力矩重心距船中纵向力矩(t)(m)(t.m)(m)(t.m)燃油838.0002.1801826.8403.8203201.160轻柴油236.5000.960227.040-15.160-3585.340滑油43.00010.700460.100-7.800-335.400机油18.00010.700192.600-2.180-39.240淡水314.0004.4701403.580-55.800-17521.200总计1449.5002.8364110.160-12.611-18280.020项目由题中给的表计算得iP=1449.5t,Zpi=2.836,Xp=-12.611W1=W0+iP=19521-1449.5=18071.5Zg1=ipigPWZPZWi000=5.1807116.4110154899=8.344,上移Xg1=iiigPWXPXW000=5.1807178.1828074.34927=-0.921,前移。3.3.某船某航次离港时船舶重量为w=19503t,重心距基线高为zg=7.93m,到港时油和淡水共消耗1526.6t,其重心距基线高为3.13m,试求到港时船舶重心距基线高。又以上算出的船舶重心偏高,为保证船舶在到港前的安全,在第三和第四压载舱内加装压载水,其重量和重心见表3—11,试求压载到港后船舶重心距基线高。表3-11项目重量(t)重心距基线高(m)垂向力矩(t.m)第三压载舱(左.右)3360.72第四压载舱600.76总计已知:W=19503t,Zg=7.93,P=1526.6,Zg=3.13,求Zg1=?解::列表计算:船舶原理习题集(解)6重量重心距基线高垂向力矩(t)(m)(t.m)第三压载舱(左.右)336.0000.720241.920第四压载舱60.0000.76045.600总计396.0000.726287.520项目Zg1=PWPZWZpg=m338.86.15261950313.3*6.152693.7*19503=Zg2=3233221)()(PPPWZPZPZPWPPg=603366.15261950376.0*6072.0*336338.8*)6.152619503(=8.174m3.4.已知船舶重量为16700t,今有船内重量10t自底舱上移14m后,又水平右移9m,试求船舶重心移距及其方向。已知:W0=16700tP=10tlzp=14mlpy=9m求lgz,lgy解::lzg2=WPlzp=1670014*10=0.0084m(上移)lyg2=WPlyp=167009*10=0.0054m(右移)3.5.已知船舶重量为16700t,今有船内重量50t自首部水平后移60m,试求船舶重心移距及其方向。已知W=16700tlx=60mP=50t求G’G=?解:G’G=WxPl=17006)60(*50=
本文标题:船舶习题解1-5
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