船舶原理习题答案
a 3 1 b = 4 × ( 2 × 0.2)3 = 7.539 × 10−3 m3 12 12 I 7.539 × 10−3 = = 0.09424m Δ 0.08
BM =
HM = BM − BH = 0.09424 − 0.04714 = 0.0471m
3-5
BM =
I ∇
1 ∇ = π r 2t 3
以上诸式求和: S = 223.35m 2 形心坐标 yB = 0
−lb = −18,
xlb =
L − l 30 − 12 = = 9m 2 2
静矩 lb = −18 × 9 = −162m3 , − l1b1 =-2.4m 2 ,
静矩 l1b1 = −34.4m3 , − l2 b 2 =-2.25m 2 ,
2-1.570795 × 100% = 21.46% 2
π 1
相对误差
(2) 辛浦生法(三坐标)
A=
1π π 1 π 2π [sin 0 + 4sin + sin π ] = × × 4 = 22 2 3 2 3 = 2.09439
相误差 4.72%
(3) 乞贝雪夫法(三坐标)
A= =
2π π 1 π π π 1 π [sin( − ) + sin + sin( + )] 32 2 2 2 2 2 2 2
β = 500 − 36.610 = 13.3930 = 13.390
GZ = GB1 sin β = 2.18sin13.390 = 0.5048m
4-2
GE = 7.1 − 4.6 = 2.5m
则tgα =
GE 2.5 = B1 E 3.4
α = 36.330 ⇒ β = 50 − 36.330 = 13.670
静矩 l2b2 = 32.625m3 ,
xl1b1 =
L 1 30 1 2 − l1 = − × 2 = 15 − = 14.3333m 2 3 2 3 3
xl2b 2 =
L 1 1 − l2 = 15 − × 1.5 = −14.5m 2 3 3
总静矩= − 163.775
X B = −163.775 / 223.35 = −0.7333m
(3)定积分
x2 A = 2 ∫ 4.2[1 − ]dx 900 −30 = 336
2-1 对 x-x 轴惯性矩
30
(相对误差0%)
I = [1575m 2 × 9.4m2 ] × 2 = 27833m 4
10.5
( I xx )1 =
8.3 −8.3
∫
y 2 × 75dy = 14646.1
( I xx ) 2 =
1—1
CB =
∇ 10900 = = 0.55025 LBd 155 × 18 × 7.1
纵向棱形系数
Cp =
∇ 1980 = = 0.6115 AM L 155 × 155 Aw 1980 = = 0.7097 LB 155 × 18 AM 115 = = 0.8998 Bd 18 × 7.1 ∇ 10900 = = 0.7754 Aw d 1980 × 7.1
−10.5
∫
y 2 × 75dy = 14645.95
I xx = 29291.9m 4
对 y-y 轴的惯性矩:
37.5
2
2-2
−37.5
∫
37.5
x 2 dA = 2
−37.5
∫
x 2 2.2dx = 154687.5m 4
b = 1.5m
⎧ L × B = 30 × 8.2 = 246m 2 ⎪ 2 ⎪−lb = −12 × 1.5 = −18m ⎨ 2 ⎪−l1b1 = −2 ×1.2 = −2.4m ⎪−l b = −2.25m 2 ⎩ 2 2
Cwp =
d2 = 0.5 2d ⋅ d
(4) 方形系数: CB = 1-6
π
12
d 3 /[ d
d π π ] ⋅ 2d = d 3 / d 3 = = 0.2618 2 12 12
CB = 0.815 =
则
4400m3 4400m 2 = LBT LB × 2.6
LB = 2076.5m 2
水线面系数
水线面系数:
Cwp =
中横剖面系数:
CM =
垂向棱形系数:
Cvp =
1-2 (1) 中横剖面系数
1 d 2 d2 π π π( ) CM = 2 2 = 8 2 = = 0.7854 d d 4 d 2 2
(2) 纵向棱形系数 C p 每个圆锥体积: V = 浸水圆锥体积: 2[ 纵向棱形系数:
1 1 d π d2 π Ab h = π ( ) 2 d = d = d3 3 3 2 3 4 12
1π 3 π d ] = d3 2 12 12
Cp =
1 d d 3 /[2d π ( ) 2 ] 12 2 2 ⎛ d3 ⎞ 1 π = d 3 / ⎜π ⎟ = = 0.3333 12 ⎝ 4 ⎠ 3
π
(3)水线面系数 C wp 水线面面积
1 S = [ × d × d ]2 = d 2 2
水线面系数
2-7
1 1 V = 6[ × 5.7 + 8.7 + 11.3 + 10.1 + × 8.8] 2 2 3 = 224.1m
该舱载煤吨数
G=
224.1 = 143.654吨 1.56
0
1
2
3
4 8.8 × 24
5.7 × 0 8.7 × 6 11.3 × 12 10.1× 18
对尾体积静矩:
1 1 6[ × 5.7 × 0 + 8.7 × 6 + 11.3 × 12 + 10.1× 18 + × 8.8 × 2.4] 2 2 4 = 2851.2m
GB1 = 2.52 + 3.42 = 6.25 + 11.56 = 4.22 GZ = GB1 sin β = 4.22 × sin13.67 0 = 0.9973
I=
−r
∫y
r
2
1 2 r 2 − y 2 dy = π r 4 4
4
1 2 3 r2 BM = r / π r t = ⋅ 4 3 4 t
3-7
π
KB = 1m I BM = , ∇
I=
d − ( −1.5) 2 d − ( +1.5) 2
KG = 3.875 m ∇ = 2 × 3 × 2 L = 12 L
Cwp = 0.882 =
Aw 2076.5m 2
故 Aw = 1831.43m 2 1-8
∫
π
0
sin xdx = [− cos x]
π
0
= cos 0 − cos π = 2 (精确解)
(1) 梯形法(三坐标)
A=
1 [ y0 + y1 + y2 ] 2 2 2 π 1 π 1 π π = [ sin 0 + sin + sin π ] = × 1 = = 1.570795 2 2 2 2 2 2
π
3
[0.444 + 1 + 0.444] = 1.9771
相对误差 1.1433%
相对误差均取绝对值 1-10 船的水线方程
B x2 y= [1 − ], L = 60m, B = 8.4m 2 (0.5 L)2
一半:
x2 x2 8.4 [1 − ] = 4.2[1 − 2 ] y= (0.5 × 60) 2 30 2 = 4.2[1 − x2 ] 900
体积形心距艉距离 对基平面静矩:
2851.2 / 224.1 = 12.723m
1 1 6[ × 5.7 × 3.7 + 8.7 × 3.5 + 11.3 × 3.3 + 10.1× 3.5 + × 8.8 × 3.6] 2 2 = 776.85
Z B = 776.85 / 224.1 = 3.4665 m
M后 =
X F = 18.182(米)
排水体积
1 A1 = 0.32 sin 600 2 1 1 1 A2 = 1.62 sin 600 , A3 = 4.32 sin 600 , A4 = 5.02 sin 600 2 2 2 1 1 1 A5 = 4.62 sin 600 , A6 = 4.02 sin 600 , A7 = 3.32 sin 600 2 2 2
(1)梯形法(十等分)
1 1 S = 2 × 6 × [ × 0 + 1.512 × 2 + 2.688 × 2 + 3.528 × 2 + 4.032 × 2 + 4.2 + × 0] 2 2 2 = 332.64m
相对误差 1%
辛氏法(十等分)
2 A = × 6 × [0 + 4 × 1.512 + 2.688 × 2 + 3.528 × 4 + 3 4.032 × 2 + 4 × 4.2 + 2 × 4.032 + 4 × 3.528 + 2 × 2.688 + 4 × 1.512 + 0] = 336
3-4
2 0.2 OB = ⋅ = 0.0943 3 2
BM = I ∇
木材总体积=0.2 × 0.2 × 4 = 0.16
1 0.2 BH = ⋅ = 0.04714 3 2
1 ∇ = × 0.16 = 0.08 2
I= =
∫
−
b 2
y 2 dA =
b 2
∫
−