Ship Hydrostatics and Stability 2010 Part 4 ppt

Basic ship hydrostatics 65 B/2 (a) T m (b) Figure 2.24 Monohull versus catamaran We can obtain the same displacement volume with two hulls of breadth B/2, the same length, L, and the same draught, T. Assuming that the distance between the centrelines of the two hulls is 3B/2, the resulting metacentric radius is -=-_ 2 \(B/1) Z B T (3B\ 2 BM =LBT[U + ^ L (^) 19B 2 48T The first term between parantheses represents the sum of the moments of inertia of the waterlines about their own centrelines. The second term accounts for the parallel translation of the hulls from the plane of symmetry of the catamaran. The second term is visibly the greater. The ratio of the catamaran BM to that of the monohul is — The improvement in stability is remarkable. Catamarans offer also the advantage of larger deck areas and, under certain conditions, improved hydrodynamic performances. On the other hand, the weight of structures increases and the overall performance in waves must be carefully checked. It may be worth mentioning that also many vessels with three hulls, that is trimarans, have been built. Moreover, a company in Southampton developed a remarkable concept of a large ship with a main, slender hull, and four side hulls; that is a pentamaran. Example 2.8 - Submerged bodies Submerged bodies have no waterplane; therefore, their metacentric radii are equal to zero. Then the condition of initial stability is reduced to GM = KB - KG > 0 In simple words, the centre of gravity, G, must be situated under the centre of buoyancy, B. We invite the reader to draw a sketch showing the two mentioned points and derive the condition of stability by simple mechanical considerations. Submerged bodies do not develop hydrostatic moments that oppose inclinations, as they do not develop hydrostatic forces that oppose changes of depth. 66 Ship Hydrostatics and Stability Example 2.9 - An offshore platform Figure 2.25 is a sketch of an offshore platform of the semi-submersible type. The buoyancy is provided by four pontoons, each of diameter b and length L The platform deck is supported by four columns. The depth of the platform is H and the draught, T. Our problem is to find a condition for the height of the centre of gravity, KG, for given platform dimensions. To do this, we calculate the limit value of KG for which the metacentric height, GM, is zero. The metacentric radius is given by BM = - where the moment of inertia of the waterplane, /, and the volume of displace- ment, V, are 7=4 TTb^ 64 7T& 2 , ~ In calculating the volume of displacement, V, we did not take into account the overlapping between column and pontoon ends. In conclusion '-26) (2.79) 4(7 + T) 4(^ + T) where we neglected the term in 6 2 , usually small in comparison with other terms. Figure 2.25 A semisubmersible platform Basic ship hydrostatics 67 Table 2.6 Calculation of KB Pontoons Columns Total Volume 7Tb 2 l -rrtfiT b/9} 7TO (t + Of ) Vertical arm Moment 6/2 Z£L (T \ M /r> 7rb ( 2T + bT ~ b ) (J. \ O)/* 4 2T 2 +bT+4b£ -rrb 2 (2b£+2T 2 +bT — b 2 ) 4(l+T) 4 4(1 + T) The height of the metacentre above the baseline is given by _ _ 2T 2 4- bT 4- / 2 Z1 The condition for initial stability is The height of the centre of buoyancy above the base-line is calculated in Table 2.6. Neglecting the term in — b 2 we obtain (2 ' 80) (2.81) KG < 4 (£ + T) (2.82) To rewrite Eq. (2.82) in non-dimensional form we define a = b/l, /3 = T/l and obtain T/r S~1 f) /O2 I /Q I "I ¥ - ^iirrlr (183) 2.14 Exercises Exercise 2.1 - Melting icebergs In Example 2.1 we learnt that if an ice cube melts in a glass of water, the level of water does not change. Then, why do people fear that the meltdown of all icebergs would cause a water-level rise and therefore the flooding of lower coasts? Show that they are right. Hint: Icebergs are formed on the continent and they are made of fresh water, while oceans consist of salt water. The density of salt water is greater than that of fresh water. 68 Ship Hydrostatics and Stability Exercise 2.2 - The tip of the iceberg Icebergs are formed from compressed snow; their average density is 0.89 tm~ 3 . The density of ocean water can be assumed equal to 1.025 tm~ 3 . Calculate what part of an iceberg's volume can be seen above the water and explain the meaning of the expression The tip of the iceberg'. Hint: See Exercise 2.1. Exercise 2.3 - Draughts of a parallelepipedic barge Consider a parallelepipedic (or, with another term, a box-shaped) barge charac- terized by the following data: Length, L 10m Breadth, B 3m Mass, A/0 30 t Find the draught, TI, in fresh water, and the draught, T 2 , in ocean water. See the Appendix of this chapter for various water densities. Exercise 2.4 - Whisky on the rocks Instead of considering a cube of ice floating in a glass of water, as in Example 2.1, let us think of a cube of ice floating in a glass of whisky. What happens when the cube melts? Exercise 2.5 -A lemma about moving masses in three-dimensional Prove the lemma in Section 2.7 for a three-dimensional system of masses and a three-dimensional displacement of one of the masses. Exercise 2.6 -A wooden parallelepiped The floating condition of a wooden, homogeneous block of square cross-section depends on its specific gravity. Three possible positions are shown in Figure 2.26. (a) (b) (c) Figure 2.26 Different floating conditions of a wooden, parallelepipedic block Basic ship hydrostatics 69 1. Find the ranges of specific gravity enabling each position. 2. For each range find a suitable kind of wood. To do this look through tables of wood properties. 3. Can you imagine other floating positions? In the affirmative, calculate the corresponding specific-gravity range. Hint: A floating position is possible if the corresponding metacentric height is positive. Exercise 2.7 - B and Mcurves - variable heel Table 2.7 contains the same data items as Table 2.4, but calculated at 5-degree intervals. With this 'resolution' it is possible to plot smooth B and M curves. First, write the data on a file lido9a similar to file lido9. Next, modify the programme cited in Example 2.6 to plot only the B and M curves of the vessel whose data are called from the keyboard. Run the program with the data given at 5-degree intervals and print a hardcopy of the resulting plot. Exercise 2.8 -B and M curves - variable trim Table 2.8 contains data of the vessel Lido 9 for constant volume of displacement equal to 44.16 m 3 , upright condition, and trim varying between -0.3 and l.lm. Table 2.7 Data of vessel Lido 9 at 44.16 m 3 volume of displacement and 5-degree heel intervals, trim = -0.325 m Heel angle (°) 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 Draught (m) 1.729 1.711 1.659 1.575 1.462 1.324 1.163 0.985 0.796 0.600 0.402 0.198 -0.012 -0.235 -0.464 -0.693 -0.919 -1.140 -1.354 KN (m) 0 0.399 0.776 1.122 1.432 1.716 1.979 2.215 2.419 2.595 2.749 2.870 2.945 2.960 2.931 2.874 2.788 2.678 2.539 NB (m) 1.272 1.255 1.204 1.121 1.009 0.872 0.711 0.528 0.326 0.107 -0.126 -0.372 -0.625 -0.883 -1.140 -1.393 -1.640 -1.878 -2.108 NM (m) 4.596 4.438 4.119 3.711 3.341 3.073 2.857 2.464 2.105 1.830 1.537 1.082 0.479 -0.185 -0.543 -0.869 -1.171 -1.446 -13.314 LCB (m) -1.735 -1.740 -1.761 -1.799 -1.841 -1.887 -1.932 -1.971 -2.002 -2.047 -2.106 -2.113 -2.072 -2.041 -2.025 -2.008 -1.994 -1.981 -1.970 NM L (m) 23.371 23.693 24.008 23.730 23.813 23.464 23.154 22.822 22.810 23.133 21.837 19.757 17.473 16.162 15.117 14.298 13.633 13.121 12.792 70 Ship Hydrostatics and Stability Table 2.8 Data of vessel Lido 9 at 44.16 m 3 volume of displacement, 0.1 m trim intervals, upright condition Trim (m) -1.000 -0.900 -0.800 -0.700 -0.600 -0.500 -0.400 -0.300 -0.200 -0.100 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.100 Draught (m) 1.673 1.653 1.668 1.683 1.697 1.709 1.721 1.732 1.742 1.750 1.758 1.765 1.772 1.777 .782 .786 .789 .792 .793 .794 .795 1.795 NB (m) 1.174 1.192 1.208 1.224 1.238 1.251 1.24 1.276 1.286 1.295 1.304 1.311 1.319 1.324 1.329 1.333 1.336 1.338 1.339 1.340 1.340 1.338 NM (m) 4.536 4.550 4.564 4.577 4.585 4.589 4.592 4.598 4.604 4.610 4.614 4.615 4.614 4.612 4.610 4.606 4.603 4.599 4.599 4.597 4.594 4.590 LCB (m) -2.777 -2.623 -2.468 -2.313 -2.157 -2.001 -1.848 -1.697 -1.548 -1.401 -1.257 -1.114 -0.971 -0.829 -0.690 -0.546 -0.407 -0.270 -0.135 0.000 0.135 0.269 NM L 23.681 23.904 24.069 24.163 24.145 23.954 23.584 23.293 22.951 22.556 22.108 22.137 22.115 22.046 21.910 21.707 21.431 21.116 20.895 20.871 20.834 20.829 The LCB values in column 5 are equivalent to the KN values in Example 2.6, Figure 2.22. Write the data on an M-file, Iido9b. m, and use the program b_curve to plot the B- and M-curves. Here the M-curve is the locus of the longitudinal metacentre. 2.15 Appendix -Water densities Density (tm~ 3 ) Fresh water Eastern Baltic Sea Western Baltic Sea Black Sea Oceans Red Sea Caspian Sea Dead Sea 1.000 1.003 1.015 1.018 1.025 1.044 1.060 1.278 Numerical integration in naval architecture 3.1 Introduction In Chapter 2, we have learnt that the evaluation of ship properties, such as dis- placement and stability, requires the calculation of areas, centroids and moments of plane figures, and of volumes and centres of volumes. Such properties are cal- culated by integration. In the absence of an explicit definition of the hull surface, in terms of calculable mathematical functions, the integrations cannot be car- ried out by analytic methods. The established practice has been to describe the hull surface by tabulated data, as shown in Chapter 1, and to use these data in numerical calculations. Two methods for numerical integration are described in this chapter: the trape- zoidal and Simpson's rules. The treatment is based on Biran and Breiner (2002). The rules are exemplified on integrands defined by explicit mathematical expres- sions; this is done to convince the reader that the two methods of numerical integration are efficient, and to allow an evaluation of errors. The first examples are followed in Chapter 4 by Naval-Architectural applications to real ship data presented in tabular form. Many Naval-Architectural problems require the calculation of the definite integral •& f(x) dx of a function bounded in the finite interval [a, b]. We approximate the definite inte- gral by the weighted sum of a set of function values, /(zi), /(x 2 ), , /(x n ), evaluated, or measured, at n points Xi G [a, 6], i = 1,2, , n, i.e. fb I -* a (3.1) In Naval Architecture, the coefficients a^ are called multipliers; in some books on Numerical Methods they are called weights. 72 Ship Hydrostatics and Stability There are several ways of deriving formulae for numerical integration - also called quadrature formulae - of the form shown in Eq. (3.1); three of them are mentioned below: 1. By geometrical reasoning, considering f a f(x) dx as the area under the curve /(x), between x = a and x = b. 2. By approximating the function f(x) by an interpolating polynomial, P(x), and integrating the latter instead of the given function, so that •b P(x) dx 3. By developing the given function into a Taylor or MacLaurin series and integrating the first terms of the series. The first approach yields a simple intuitive interpretation of the rules for numer- ical quadrature and of the errors involved. This interpretation enables the user to derive the rules whenever required, and to adapt them to particular situations, for instance, when changing the subintervals of integration. On the other hand, each rule must be derived separately. The advantages of the other approaches are: • The derivation is common to a group of rules which thus appears as particular cases of a more general method. • The derivation yields an expression of the error involved. In the next two sections, we shall use the geometrical approach to derive the two most popular rules, namely the trapezoidal and Simpson's rules. These two methods are sufficient for solving most problems encountered in Naval Architec- ture. The error terms will be given without derivation; however, interpretations of the error expressions will follow their presentation. 3.2 The trapezoidal rule Let us consider the function f(x) represented in Figure 3.1. We assume that we know the values /(:TI), /(x2), . . . , f(x$) and we want to calculate the definite integral fXS 1= f(x)dx (3.2) The integral in Eq. (3.2) represents the area under the curve f(x). Let us connect the points /(xi), /(#2)> . . • , f(x$) by straight line segments (the dashed-dotted Numerical integration in naval architecture 73 Given integrand Trapezoidal approximation X 3 Figure 3.1 The derivation of the trapezoidal rule lines in the figure). We approximate the area under the curve by the sum of the areas of four trapezoids, i.e. the area of the trapezoid with base #i£ 2 and heights /(#i), /(x 2 ), plus the area of the trapezoid with base x 2 x 3 and heights /(x 2 ), /(^s)* and so on. We obtain (X2 - (X3 _ (3 .3) For constant x-spacing, # 2 — x\ = £3 — x 2 = • • • = h, Eq. (3.3) can be reduced to a simpler form: f(x n -i) + f(x n (3.4) We call the intervals [x\, x 2 ], [0^2, ^3], and so on, subintervals. As an example, let us calculate •90° sin 74 Ship Hydrostatics and Stability The calculation presented in tabular form is as follows: Angle (°) 0 15 30 45 60 75 90 Sum sin x 0 0.2588 0.5000 0.7071 0.8660 0.9659 1.0000 - Multiplier 1/2 1 1 1 1 1 1/2 - Product 0 0.2588 0.5000 0.7071 0.8660 0.9659 0.5000 3.7979 The calculations were performed with MATLAB and the precision of the display in the short format, i.e. four decimal digits, was retained. To obtain the approximation of the integral, we multiply the sum in column 4 by the constant subinterval, h: TTX — ) x 3.7979-0.9943 180 J Above we measured the interval in radians, as we should do in such calculations. Equation (3.3) in matrix form yields (x 2 - xi)(x 3 - x 2 )(x 4 - - x 4 ) 2/1 + 2/2 2/2 + 2/3 V3 + 2/4 (3.5) The generalized form of Eq. (3.5) is implemented in MATLAB by the trapz function that can be called with two arguments: 1. the column vector x, 2. the column vector y, of the same length as x, or a matrix y, with the same number of rows as x. If the points xi, x 2 , , x n are equally spaced, i.e., if the trapz function can be called with one argument, namely the column vec- tor (or matrix) y. In this case, the result must be multiplied by the common x-interval, h. [...]... Stations 0 and 1 and appropriately call it Station \ We introduce another intermediate station between Stations 9 and 10 and call it 9| We invite the reader to check that the corresponding sequence of trapezoidal multipliers is now 1 /4, 2 /4, 3 /4, 4/ 4, ! , , ! , 4/ 4,3 /4, 2 /4, 1 /4 = 1 /4, 1/2, 3 /4, !, ,!, 3 /4, 1/2, 1 /4 84 Ship Hydrostatics and Stability 0 1 / 2 1 2 3 4 5 6 7 8 Figure 3 .4 Intermediate... 1 1 1/2 — X (3) 0.00 5.63 11.25 16.88 22.50 28.13 33.75 39.38 45 .00 — /(*) (4) 0.00 177.98 142 3.83 48 05 .42 11390.63 22 247 .31 3 844 3.36 61 046 .63 91125.00 — (45 /8)Sum/3 = Products (5 = 2 x 4 ) 0.00 177.98 142 3.83 48 05 .42 11390.63 22 247 .31 3 844 3.36 61 046 .63 45 562.50 185097.66 1 041 1 74. 32 The error is E = 1025 156.25 - 1 041 1 74. 32 - -16 018.07 and the relative error is E ET = 100 x = -1.56% 1 025 156.25 (d)... 2 3 4 5 6 7 8 9 Sum Integral Simpson's multiplier (2) 1 4 2 4 2 4 2 4 1 - X (3) 0.00 5.63 11.25 16.88 22.50 28.13 33.75 39.38 45 .00 — /(*) (4) 0.00 177.98 142 3.83 48 05 .42 11390.63 22 247 .31 3 844 3.36 61 046 .63 91125.00 — (45 /8)Sum/3 = Products (5 = 2 x 4 ) 0.00 711.91 2 847 .66 19221.68 22781.25 88989.26 76886.72 244 186.52 91125.00 546 750.00 1025156.25 The error is E = 1025 156.25 - 1025 156.25 = 0 and. .. 11.25 22.50 33.75 45 .00 - (4) 0.00 142 3.83 11390.63 3 844 3.36 91125.00 (45 /4) Sum/3 = The error is E = 1 025 156.25 - 1 089 228.52 - - 64 072.27 Products (5-2x4) 0.00 142 3.83 11390.63 3 844 3.36 45 562.50 96820.31 1089228.52 88 Ship Hydrostatics and Stability and the relative error is E Er = 100 x = -6.25% 1025156.25 (c) The following values were calculated in MS Excel: No of ordinate (1) 1 2 3 4 5 6 7 8 9 Sum... format long a = 45 ~4/ 4 = 1.025156250000000e+006 (b) Trapezoidal rule, five ordinates: x = 0: 4 5 / 4 : 45 ; Y = X.~3; b = trapz(x, y) = 1.089228515625000e+006 error = a - b = -6 .40 72e+0 04 percent_error = 100*(a - b)/a = -6.2500 ° 90 Ship Hydrostatics and Stability (c) Trapezoidal rule, nine ordinates: X = 0: 45 /8: y = X.~3; 45 ; c = trapz(x, y) = 1. 041 1 743 1 640 6250e+006 error = a - c = -1.6018e+0 04 percent_error... (the trapezoids) and the solid line (the given 76 Ship Hydrostatics and Stability Table 3.1 Results by trapezoidal and by Simpson's rule Subinterval Integral Trapezoidal rule 7T /4 TV/8 7T/16 7T/32 7T/ 64 7T/128 Simpson's rule 2.51885577576 342 2.55791212776767 2.56758 149 868107 2.56999300727997 2.5705955211 149 2 2.570 746 12688700 2.5730762 042 8711 2.57093091176909 2.5708 046 2231886 2.570796 843 47960 2.57079635905990... in MS Excel: No of ordinate (1) 1 2 3 4 5 Sum Integral Simpson's multiplier (2) 1 4 2 4 1 _ x (3) 0.00 11.25 22.50 33.75 45 .00 - f(x) (4) 0.00 142 3.83 11390.63 3 844 3.36 91125.00 (45 /4) Sum/3 = The error is E = 1025 156.25 - 1025 156.25 = 0 Products (5-2x4) 0.00 5695.31 22781.25 153773 .44 91125.00 273375.00 1025156.25 Numerical integration in naval architecture 89 and the relative error is Er = 100 x... J\^3) 4 /(*„) x /(2 i) ' i Jf \*.r 2 ) + ^ ( Numerical integration in naval architecture 83 Table 3 .4 Integral with variable upper limit - comparing the analytic result with that obtained in MATLAB Angle (°) 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 Analytic result Numerical result 0 0 0.0152 0.0603 0.1 340 0.2 340 0.3572 0.5000 0.6580 0.82 64 1.0000 1.1736 1. 342 0 1.5000 1. 642 8 1.7660... 1.9397 1.9 848 2.0000 0.0152 0.0602 0.1336 0.23 34 0.3563 0 .49 87 0.6563 0.8 243 0.9975 1.1707 1.3386 1 .49 62 1.6386 1.7616 1.8613 1.9 348 1.9798 1.9 949 In our experience, the MATLAB procedure is slightly more exact than the Excel spreadsheet Table 3 .4 compares the result yielded by Eq (3.15) with those obtained with the MATLAB function The agreement between the results obtained, analytically in Excel, and in... shown in column 2 For columns 3 and 4, the algorithm is Write 0 in column 4, line 1 For i = 1 : (n — 1) • • • • Pick up the value in column 4, line / Go left and add the value in column 3, line i Go down and add the value in column 3, line / +1 Write the result in column 4, line / -j-1 End In column 5, line i, we write the result of the product of the content of column 4, line z, by half of the subinterval . condition Trim (m) -1.000 -0.900 -0.800 -0.700 -0.600 -0.500 -0 .40 0 -0.300 -0.200 -0.100 0.000 0.100 0.200 0.300 0 .40 0 0.500 0.600 0.700 0.800 0.900 1.000 1.100 Draught (m) 1.673 1.653 1.668 1.683 1.697 1.709 1.721 1.732 1. 742 1.750 1.758 1.765 1.772 1.777 .782 .786 .789 .792 .793 .7 94 .795 1.795 NB (m) 1.1 74 1.192 1.208 1.2 24 1.238 1.251 1. 24 1.276 1.286 1.295 1.3 04 1.311 1.319 1.3 24 1.329 1.333 1.336 1.338 1.339 1. 340 1. 340 1.338 NM (m) 4. 536 4. 550 4. 5 64 4.577 4. 585 4. 589 4. 592 4. 598 4. 6 04 4.610 4. 6 14 4.615 4. 6 14 4.612 4. 610 4. 606 4. 603 4. 599 4. 599 4. 597 4. 5 94 4.590 LCB (m) -2.777 -2.623 -2 .46 8 -2.313 -2.157 -2.001 -1. 848 -1.697 -1. 548 -1 .40 1 -1.257 -1.1 14 -0.971 -0.829 -0.690 -0. 546 -0 .40 7 -0.270 -0.135 0.000 0.135 0.269 NM L 23.681 23.9 04 24. 069 24. 163 24. 145 23.9 54 23.5 84 23.293 22.951 22.556 22.108 22.137 22.115 22. 046 21.910 21.707 21 .43 1 21.116 20.895 20.871 20.8 34 20.829 The . multipliers is now 1 /4, 2 /4, 3 /4, 4/ 4, !, ,!, 4/ 4,3 /4, 2 /4, 1 /4 = 1 /4, 1/2, 3 /4, !, ,!, 3 /4, 1/2, 1 /4 84 Ship Hydrostatics and Stability 01/21 2 3 4 5 6 7 8 Figure 3 .4 Intermediate ordinates. angle (°) 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 Draught (m) 1.729 1.711 1.659 1.575 1 .46 2 1.3 24 1.163 0.985 0.796 0.600 0 .40 2 0.198 -0.012 -0.235 -0 .46 4 -0.693 -0.919 -1. 140 -1.3 54 KN (m) 0 0.399 0.776 1.122 1 .43 2 1.716 1.979 2.215 2 .41 9 2.595 2. 749 2.870 2. 945 2.960 2.931 2.8 74 2.788 2.678 2.539 NB (m) 1.272 1.255 1.2 04 1.121 1.009 0.872 0.711 0.528 0.326 0.107 -0.126 -0.372 -0.625 -0.883 -1. 140 -1.393 -1. 640 -1.878 -2.108 NM (m) 4. 596 4. 438 4. 119 3.711 3. 341 3.073 2.857 2 .46 4 2.105 1.830 1.537 1.082 0 .47 9 -0.185 -0. 543 -0.869 -1.171 -1 .44 6 -13.3 14 LCB (m) -1.735 -1. 740 -1.761 -1.799 -1. 841 -1.887 -1.932 -1.971 -2.002 -2. 047 -2.106 -2.113 -2.072 -2. 041 -2.025 -2.008 -1.9 94 -1.981 -1.970 NM L (m) 23.371 23.693 24. 008 23.730 23.813 23 .46 4 23.1 54 22.822 22.810 23.133 21.837 19.757 17 .47 3 16.162 15.117 14. 298 13.633 13.121 12.792 70