1、某浮塢系泊力計算一,風阻力(口本JSDS造船舸裝設(shè)計基準)風吹向側(cè)面Ruy = 9.81 Kuy Auy V g (N)橫向風力系數(shù)Kuy= 0.0735水線以上浮塢側(cè)向受風面積AuV= 1.10 X 204.5 X (16.0-5.4) = 2384.5 (m2)水線以上側(cè)向受風面積(以DWT57000散貨船為例)Auy=4800 (m2)風速 Vu=10 m/s(移船進浮塢中，塢能擺動，則船受風面積不計入)Ruyl = 9.81 X 0.0735 X 2384.5 X 102 = 171.9 (KN)(船3/4移進浮塢及全進塢后)Ruy = 9.81 X 0.0735 X 4800 X 1

2、02 =346.1 (KN)二，潮流阻力(美國石油協(xié)會浮式鉆井裝置輻射狀錨泊系統(tǒng)分析)Fey = 7237 SVC2 (N)(口本JSDS造船舸裝設(shè)計基準)Rw = 9.81X0.1212S ( (Vc+ Vs)2 + 0. 33(Vc+Vs) (N)潮流流向側(cè)面濕表面積 S = 204.5 X 48.6 + (204.5 + 48.6 )X 2X5.4 = 12672.2 ( m2)流速Vc = Vw +Vt (m/s)(本例移塢方向多與流速方向相反，故此處取Vs=o) 風生流速 Vw = 0.01 Vu (m/s) Vw = 0.01 X 10 = 0.1 (m/s)潮生流速 機戔設(shè)計值(

3、m/s) VT =0.8 (m/s)Vc = 0.1+ 0.8 = 0.9 (m/s)Fey = 7237 X 12672.2 X 0.92 = 743.4 (KN)在 Vt =0.8 (m/s)下,因堵塞效應(yīng),取 VT =1.4 (m/s), Vc = 0.1 + 1.4 = 1.5 (m/s)RW = 9.81X 0.1212 X 12672.2 (1.52 + 0.33 X1.5) = 41.4 (KN)(偏小)三，形狀阻力(口本JSDS造船舸裝設(shè)計基準)側(cè)向移塢Rv = 9.81X73.2 As V2s (N)移塢速度 Vs = 10 ( m/min) = 0.167 (m/s)浮塢側(cè)

4、面積 As=L*Tm = 204.5X5.4 = 1104.3 (m2)Rv = 9.81 X 73.2 X 0.167, 11043 = 22.1 (KN)四，移塢總阻力按風力潮流力同方向作用于側(cè)移塢方向(移船進浮塢中，塢能擺動，則船岸成一體，船受風面積不計入)Ri = Ruyl + Fey + Rv=171.9 + 743.4 + 22.1 = 937.4 (KN)(船3/4移進浮塢及全進塢后，船與塢成一體)R = Ruy + Fey + Rv (KN)=346.1 + 743.4 + 22.1 = 1111.6 (KN)五,臘纜絞車拉力(浮塢與碼頭離空)下游艄艇兩纜與浮塢橫向成35方向出

5、纜絞移，則箱纜絞車拉力為Pi = 1/2 *l/cos 35 Ri=1/2 *1/COS 35X 937.4（移船進浮塢及1/4進塢前一下游纜）= 0.610 電=571.8 （KN）P = 1/2 *l/cos 35 R=1/2 *l/cos 35X 1111.6 （船3/4移進浮塢及全進塢后-下游纜）=0.610 R = 678.1 （KN）2009-4-13鐘成就六，首尾纜絞車拉力（浮塢一端頂貼碼頭）設(shè)：浮塢端與碼頭頂貼磨擦系數(shù)f = 0.15浮塢離岸端纜繩拉力N】，纜繩與浮塢橫向角為35或15 （下游纜），12 （上 游纜）浮塢頂岸端纜繩拉力N2纜繩與浮塢橫向角為35岸頂推浮塢力NN】

6、距浮塢中線頂岸點距離 81= 187.5m蜘距浮塢中線頂岸點距離a2 = 33 3mR距浮塢中線頂岸點距離a3=102.2m總阻力R= 1111.6 KN / % = 937.4 KN上游向（Ni成12）則有：N1sinl2 + N=N2sin35NiCOS12 + N2cos35 + fN = R+ N2&2= R解得：0.435R = 483.4 KN= 407.8 KNN2 =0.620R = 689.2 KN= 581.2 KNN = 0.265R = 295.0 KN （碼頭受壓力一未計入移船力）下游向（Ni成15）則有：NSinl5 + N =昭$泊35N1cosl5 + N2co

7、s35 + fN =R+ N2&2= R解得：0.399R = 443.5 KN= 374.0 KNN2 =0.671R = 745.9 KNN2.i= 629.0 KNN =0.282R = 313.3 KN（碼頭受壓力一未計入移船力）下游向（N成35）則有：Nisin35 + N =N2Sin35N1cos35 + N2cos35 + fN =R+ N2&2= R解得：Ni= 0.408R = 453.5 KNNi，i= 382.5 KNN2 =0.774R = 860.4 KNN2.i = 725.5 KNN =0.210R = 233.5 KN（碼頭受壓力一未計入移船力）系泊布置及受力圖如下浮船塢所受約束力入下：6#系駁繩的拉力N6；(計算書中N6改為N2,均為35)1#沉塊的拉力Nl； (N】-12為上游纜，N】下游纜分15或35計算)碼頭的頂推力 N；碼頭與浮船塢接觸處的摩擦力f=uN, u=0. 15；在環(huán)境力及約束力的共同作用下，浮船塢處于平衡