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Plots or Graphs - Generally the most effective format for displaying and conveying the interrelation of experimental variables. Sketches - Quick and informal.

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Presentation on theme: "Plots or Graphs - Generally the most effective format for displaying and conveying the interrelation of experimental variables. Sketches - Quick and informal."— Presentation transcript:

1 Plots or Graphs - Generally the most effective format for displaying and conveying the interrelation of experimental variables. Sketches - Quick and informal method of sharing ideas with others or clarify concepts for yourself. Free body diagrams (FBDs) are an example. Plots, Graphs, and Sketches (1.1)

2 Means of Communicating Ideas Concisely – Axes  X-axis (horizontal, independent variable)  Y-axis (vertical, dependent variable)  Divide major axes into appropriate divisions of 1, 2, 10, 10 X, etc  Label with words, symbols, and units  Minor axes should be distributed evenly Plots and Graphs

3

4 Sketch

5 Sections 1.2 Functions: Independent variables (Cause /Input /Action) Dependent variables (Effect /Output / Reaction)

6 Dependent / Independent Variables Dependent variables (Y-axis) values will depend on the value of independent variables (X-axis). Notation in math and science Notation in math and science - parameter name = f ( independent 1, independent 2, … ) - example : p = f ( , z) Developing the relationship Developing the relationship between the dependent and independent variable Experiment: collect raw data and draw the curve. apply fairing algorithm and interpolate regression analysis. Conservation law or theoretical principles Semi-empirical equation 1) identify the important parameter. 2) develop analytical equations through experiments

7 Sections 1.3 “Area under the curve” “Slope of the curve”

8 Area Under and Instantaneous Slope of a Curve

9 Section 1.4 Unit Systems S.I. Pound force - pound mass Pound - slug: What we will use Carry units through all calculations If the units are wrong, the answer is wrong

10 Units  1 slug is defined as the mass that will be accelerated at a rate of 1 ft/s 2 by a 1 lb force.

11 Unit Analysis If you get unreasonable values: - check the units in the process of calculations To avoid mistakes: - know the units of each parameter - understand the relationship between fundamental dimensions and their units Ruled Lines Method : LT = 4000 ft ³ | 1.99 lb-s ² | ft | 1 LT | ft^4 | s ² | 2240 lb = LT

12 Section 1.5 Significant figures Exact numbers Measurements Addition/subtraction Multiplication/division

13 The number of accurate digits in a number – Example: 2.65 has 3 significant figures – Example: 10 has 1 or 2, 10.0 has 3 Multiplication / Division: Use the same # of significant figures as the number with the least # of significant figures Example: 20 x = 69 Addition / Subtraction: Use the same # of decimal places as the number with the least # of decimal places Example: = 4.8 Significant Figures

14 General Problem Solving Technique Write down applicable reference equation which contains the desired “answer variable”. Solve the reference equation for the “answer variable”. Write down additional reference equations and solve for unknown variables in the “answer variable” equation, if needed. Draw a quick sketch to show what information is given and needed and identify variables, if applicable. Rewrite “answer variable” equation, substituting numeric values with units for variables. Simplify this expanded equation, including units, to arrive at the final answer. Check the answer: Do units match answer? Is the answer on the right order of magnitude?

15 Section 1.6 Linear Interpolation Assume a straight line Use anchor point, output range, and domain fraction from anchor point to input value. Output=anchor+{range*fraction} y3= y1+ {(y2-y1)*(x3-x1)/(x2-x1)} x1x3 y1 y2 y3=? x1 x2 x3 y1=Anchor y2 y3 Input Output

16 Linear Interpolation Example Temperature (°F)Density (lb-s 2 /ft 4 ) Solve for the density at 62.7°F as follows:

17 Section 1.7 Scalars & Vectors Forces & Weight Point forces (centroid of a force distribution) Distributed forces (such as Pressure) Moments & Couples: Force × Distance Static Equilibrium: Hydrostatic Pressure = P hyd =  gz F hyd = P hyd ×Area

18 Rectangular Coordinate System x z y

19 Scalars and Vectors Scalar has only magnitude. - Example : mass, speed, work, energy Vector has its magnitude and direction. - Example : velocity, force, momentum, acceleration - Way of presentation : boldface, symbol with a half/full arrow ( notation should be consistent in the text.) Vectors can be resolved into scalar components in each principal dimension. F FxFx FyFy F

20 Vectors - Vector symbols often have subscripts to better clarify. resultant buoyant force resultant weight of ship = displacement line of action - Naval Engineers uses to represent a change in a property. Ex ; : change in a draft due to parallel sinkage

21 Forces, Moments, and Couples FORCE - a vector quantity with magnitude and a direction. MOMENT - force times a distance with respect to a given origin, i.e. COUPLE - a special case a of moment that results in pure rotation and no translation.

22 Forces F 100 lb/ft 1000 lb 20 ft 10 ft Distributed force and its resultant point force. Point Force

23 Moments d F Example of a moment applied with a wrench Increasing F or d will increase the magnitude of the Moment about the nut Moment = Force x Distance (M = F x d) Units: foot - pounds

24 Couples F F d The magnitude of a couple is calculated by multiplying the magnitude of one force by the distance separating the two forces. The two forces must be equal in magnitude and opposite in direction for pure rotation without translation. Couple = Force  distance C = F  d Units: foot-pounds Two forces creating a couple

25 Static Equilibrium If an object is neither accelerating or decelerating then it is because the following 2 conditions are met: Sum of the forces = 0 Sum of the moments = 0

26 Hydrostatic Pressure “Pressure” is the amount of force applied to a given area (p=F/A) In English units it is pounds/sq. ft. or pounds/sq. in., or “psi” Air pressure is ~ 15 psi. At 440 ft below sea level it is ~ 195 psi!

27 If an object is floating in water and the object and water are both at rest, then the pressure exerted by the water on the object is referred to as hydrostatic pressure and is defined as: P hyd =  gz where  = water density (lb-s 2 /ft 4 ) g = acceleration of gravity (ft/s 2 ) z = depth of object below the water’s surface (ft) Hydrostatic pressure acting over an area (A) results in a Hydrostatic Force where F hyd = P hyd A. Hydrostatic Pressure & Force

28 The Mathematical First, Second and Third Moments These integrals are used in mathematical descriptions of physical problems Where: s = some distance db = some differential property = summation

29 The Mathematical First and Second Moments In Naval Architecture: – “b” could represent length, area, volume, or mass – “s” is a length or distance  First Moment of Mass   Second Moment of Area 

30 Weighted Averages In Naval Architecture, we use the simplified form: Used to find: Longitudinal Center of Flotation (LCF), Longitudinal Center of Buoyancy (LCB) Centers of Gravity (LCG, TCG, VCG)

31 Mathematical Moments and the Parallel Axis Theorem First Moment=  sdm M x =  ydA; M y =  xdA; A T =  dA Centroids: y = M x /A T : x = M y /A T Second Moment=  s²dm I x =  y²dA; I y =  x²dA I x = I xc +Ad 1 2 ; I y = I yc +Ad 2 2 Use Appendix C dA y x x y y x C xcxc ycyc d1d1 d2d2 A

32 Translational vs. Rotational Motion Ship freely floating is subject to 6 degrees of freedom. Three are Translational 1.Heave (z) 2.Sway (y) 3.Surge (x) Three are Rotational 1.Yaw (z) 2.Pitch (y) 3.Roll (x)

33 Ship’s Axes and Degrees of Freedom x y z Heave Surge Sway Pitch/Trim Roll/List/Heel Yaw

34 Section 1.9 Bernoulli’s Equation: Flow Energy + Kinetic Energy + Gravitational Potential Energy=Available Energy (Assumes steady, frictionless, incompressible flow along a streamline) p=hydrostatic pressure at any point in the stream V=speed of the fluid at any point on the stream g=acceleration of gravity z=height of the streamline point above reference  =fluid density Along a line of equal energy (a streamline) in a fluid, the above is a constant.

35 Bernoulli Equation Total pressure is constant in a fluid, if:  inviscid flow (no viscosity)  incompressible flow  steady flow This gives us hydrostatic and hydrodynamic pressure. These are the water loads on the vessel.

36 Pressure Prediction Vertical pressure supports the vessel (lift versus weight) Horizontal pressure is thrust and drag These are the same as an aircraft!

37 Example Problem A 50ft wide, 100ft long barge on an even keel is divided into forward and aft compartments by a transverse bulkhead 60ft aft of the bow. The barge weighs 100LT empty with an even weight distribution. The forward tank is then filled to a height of 5ft and the aft tank to a height of 4ft with oil (  =1.80lb-s²/ft 4 ) What is the total weight of the barge? At what point (fore and aft) would this weight be applied? Which degrees of freedom for the barge were affected by loading the oil and in what direction?

38 Answer: Weight bow compt =  gV= (1.80lb-s²/ft 4 ) (32.17ft/s²)(50ft×60ft×5ft)(1LT/2240lb)= 388LT Weight aft compt =  gV= (1.80lb-s²/ft 4 ) (32.17ft/s²)(50ft×40ft×4ft)(1LT/2240lb) =207LT Weight total =388LT+207LT+100LT=695LT 60ft 40ft 50ft 5ft deep 4ft deep

39 Answer: Application point= = [(100LT×50ft)+(388LT×30ft)+(207LT×80ft)] 695LT = 47.8ft Barge became heavier, draft increased, barge sank in the heave direction. More weight was placed forward, so barge pitched forward. 60 ft 40 ft 50 ft 5ft deep 4ft deep 0ft

40 Example Problem A Landing Craft Medium (LCM8) boat has an empty draft of 1 ft. A 60 LT tank is loaded into the boat. How many 250 lb combat ready Marines can board the LCM and still be able to cross a 4 foot shoal with 1 foot to spare on the way to the beach? Data: LCM dimensions: 75 ft long × 21 ft wide × 4.5 ft high Propulsion: 1300 HP from 2 diesels on separate shafts  T=  w/TPI (T=draft; w=weight; TPI= long tons/inch) 1 LT=2240 lb  g sw =64 lb/ft 3

41 Example Answer 4 foot shoal 1 foot margin 60 LTs + # Marines 1 foot draft empty Max change in draft = 4 foot shoal – 1 foot draft empty – 1 foot margin = 2 feet Vessel Submerged Volume = (75 ft)(21 ft )(1 ft) = 1575 ft 3 (62.4 lb/ ft 3 )(1575 ft 3 )=98280 lb (98280lb)/(2240lb/LT)= LT (Empty) Tons per Inch (TPI) = LT /12 in= LT/ in (24 in)(3.656 LT/ in)= 87.74LT 87.74LT-60LT=27.74 LT is allowable (27.74 LT)(2240 LT/lb)/(250lb/Marine)=248 Marines 4 foot shoal 1 foot margin 60 LT + # Marines 1 foot draft empty Max change in draft = 4’ shoal – 1’draft empty – 1’ margin = 2’ Displacement = Density x Submerged Volume

42 Example Problem A 100 ft Barge with an empty weight distribution of 2LT/ft is loaded with the following additional loads: What is the total weight of the barge and where, along the length, is the center of gravity? Empty Barge=2LT/ft 0100ft X Y 2LT/ft 4LT/ft 1LT/ft

43 Example Answer Weight =(2LT/ft×100ft)+(2LT/ft×30ft)+(4LT/ft×40ft)+(1LT/ft×30ft) = 200LT+60LT+160LT+30LT=450LT Weighted Average of each weight: 200 LT × 50ft= 10,000 LT-ft 60 LT × 15ft= 900 LT-ft 160 LT × 50ft= 8000 LT-ft 30 LT × 85ft= 2550 LT-ft Sum=10, = 21450LT-ft Weighted Average=21450LT-ft/450LT=47.7ft Center of Gravity along X axis=47.7ft Loaded Barge 0 100ft X Y 47.7ft 450 LT


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