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自升式近海移动平台力学设计 PowerPoint PPT Presentation


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自升式近海移动平台力学设计. 材02班 尹熙 002159 材02班 蒋磊 002166 材03班 杨磊 002184 材03班 陈宇 002185 指导老师: 施惠基. 选题讨论时的面红耳赤 卷帙浩繁中的查找搜寻 确定模型的统筹帷幄 分析计算的反复斟酌 分工合作中的争执不下和心照不宣 —— 三个星期以来倾注心血的每个瞬间。。. 仅此纪念. 说明:此 PPT 仅列入整个论文的简略摘要,详细全文参见 WORD 文档. [前言]

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自升式近海移动平台力学设计

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02 002159

02 002166

03 002184

03 002185


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PPTWORD


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5060%


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1976415OCEAN EXPRESS13197974


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a,bAWy,Wz

FxMyMz

A: x(A) = -(Fx/A+My/Wy+Mz/Wz)

B: x(B) = -Fx/A+My/Wy-Mz/Wz

C: x(A) = -Fx/A-My/Wy+Mz/Wz

D: x(A) = -Fx/A+My/Wy+Mz/Wz


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Fx2ABCDFx


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,


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  • P350fig7-6,p3637.1


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2100t=2.058107

60571+1000t=5.936107N(6.916107N)


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W1=5337t=5.230107N

W2=4347t=4.260107N


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70

Vw=Kv=0.0335=1.05m/s

a=0,CD=0.5,CM=2

FP=9.674 104N

x

Ax=34.05.5=187m2

y

Ay=57.65.5=316.8m2


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  • , :

  • M

  • M=W115 21/2 2.774108Nm

  • FPWfoot(x)=(5001039.8/78)x

  • 500t

  • NX =-( Wfoot +W1/D2(1-a)2)=-(8.887x+1850) 103/(1-a2) Pa

  • F1=4.04106 N , Fy=3.94106 N

  • max = -1.85/(1-a2)+104.5/(1-a4)Mpa

  • 2+32 1/2 s/ns ns=2 1.85/(1-a2)+104.5/(1-a4)470/2=235

  • a=0.858 a0.858 d2.574m0.213m


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[cr]= cr /ns=33.26Mpa

[cr]= cr /ns=532.2Mpa

[cr]= cr /ns=133.1Mpa

[cr]= cr /ns=271.5Mpa

cr =47.9Mpa

cr =766.4Mpa

cr =191.6Mpa

cr =391.0Mpa


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  • 412431


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(kl/r)(22EI/s)1/2

cr =s (s2/42E) (kl/r)2

k2k=2.0s =470Mpa(18MnMoNb)

cr =181.8Mpa

ns=1.25+0.133(kl/r) (s /2E)1/2=1.46

[cr]= cr /ns =123.7Mpa


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2[Fpcr]=5.418109 N

W1[Fpcr] 4

2 [Fpcr]= 1.258109 N

W1[Fpcr] 4


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  • hQl


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1


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w

dl

T

dTdl

Tddl


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T+dTcos(+d)-Tcos=0 1

T+dTsin(+d)-Tsin-wdl=0 2

d

cosd1,sindd,

Tsind-cos dT=0 3

Tcosd+sindT- wdl=0 4


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Td= w cosdl 5

dT= w sindl 6

dx= cosdl 7

dy= sindl 8

(9)

8ab

l=( T0/w)(tanb-tana)


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T0T

(9)


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a=0Ta=Tb

TbbTyh

xS


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tan=wl/T0


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4

7wlT0ThS

hw

QT0=Q

l

T

S


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AE

Fndl

Ftdl

dldl


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-T+(T+dT)cosd-wsindl+Ft(1+T/(AE))dl=0

(T+dT)sind-wcosdl-Fn(1+T/(AE))dl=0

cosd1sindddTd0


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FtFn


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ctcn

Vc

w,A,E,c,, ct, cn,V

T11l

Txy


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Tph

1

Tp= T11

n=0

lnl

xnS


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j


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jz

x


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THE END


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