Transcript Document

Plots, Graphs, and Sketches (1.1)
• 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 and Graphs
• 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,
10X, etc
 Label with words, symbols, and units
 Minor axes should be distributed evenly
Plots and Graphs
Sketch
Sections 1.2
 Functions:
Independent variables (Cause /Input /Action)
Dependent variables (Effect /Output / Reaction)
Dependent / Independent Variables
Dependent variables (Y-axis) values will depend on the value
of independent variables (X-axis).
Notation in math and science
- parameter name = f ( independent 1, independent 2,…)
- example :
p = f (, z)
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
Sections 1.3
“Area under the curve”
“Slope of the curve”
Area Under and Instantaneous Slope of a Curve
Slope
Area Under
Curve
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
Units
System
Length
Time
Force
Mass
pound –slug
(BG)
meter
(m)
foot
(ft)
second
(s)
second
(s)
pound force –
pound mass
foot
(ft)
second
(s)
newton
(N)
pound
(lb)
pound
force
(lbf)
kilogram
(kg)
slug
(slug)
pound
mass
(lbm)
SI
gc
1 kg  m
N  s2
1 slug  ft
lb  s 2
32.2 lb m  ft
lb f  s 2
 1 slug is defined as the mass that will
be accelerated at a rate of 1 ft/s2 by a
1 lb force.
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 ft³ | 1.99 lb-s² | 32.17 ft | 1 LT
|
= 114.32 LT
ft^4
|
s²
| 2240 lb
Section 1.5
 Significant figures
Exact numbers
Measurements
Addition/subtraction
Multiplication/division
Significant Figures

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 3.444 = 69
 Addition / Subtraction: Use the same # of
decimal places as the number with the least # of decimal
places
Example: 3.6 + 1.212 = 4.8
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?
Section 1.6
Output
 Linear Interpolation
y2
Assume a straight line
y1
Use anchor point, output range, and
domain fraction
x1
x3
Input
from anchor point to input value.
y3=?
 Output=anchor+{range*fraction}
 y3= y1+ {(y2-y1)*(x3-x1)/(x2-x1)}
Output
y2
y3
y1=Anchor
x1
x3
x2
Input
Linear Interpolation Example
Temperature (°F)
Density (lb-s2/ft4)
61
1.9381
62
1.9379
63
1.9377
Solve for the density at 62.7°F as follows:
62.7F  62F 62.7 F  62 F

63F  62F
63 F  62 F
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:  F  0  M
0
Hydrostatic Pressure = Phyd = gz
Fhyd = Phyd×Area
Rectangular Coordinate System
z
y
x
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.)

F F

F F
Fy
Vectors can be resolved into scalar components in
each principal dimension.
F

Fx
Fx  F cos
Fy  F sin
Vectors
- Vector symbols often have subscripts to better clarify.

F
resultant buoyant force
B

resultant weight of ship = displacement
s

line of action
s

F
B
- Naval Engineers uses  to represent a change in a property.
Ex ;
: change in a draft due to parallel sinkage
T
PS
Forces, Moments, and Couples
• FORCE - a vector quantity with magnitude
and a direction. F  m  a
• MOMENT - force times a distance with
respect to a given origin, i.e.
M  F d
• COUPLE - a special case a of moment that
results in pure rotation and no
translation.
Forces
1000 lb
F
20 ft
100 lb/ft
10 ft
Point Force
Distributed force and its
resultant point force.
Moments
F
d
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
Couples
Two forces creating a couple
F
 Couple = Force  distance
C=Fd
Units: foot-pounds
d
F
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.
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
F  0
M  0
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!
Hydrostatic Pressure & Force
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:
Phyd = gz where  = water density (lb-s2/ft4)
g = acceleration of gravity (ft/s2)
z = depth of object below the
water’s surface (ft)
Hydrostatic pressure acting over an area (A) results in
a Hydrostatic Force where Fhyd = Phyd A.
The Mathematical First, Second
and Third Moments
• These integrals are used in mathematical
descriptions of physical problems
 s db
2
s
 db
s
3
db
Where:
s = some distance
db = some differential property
 
= summation
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 
x
dm

y
dA

2
Weighted Averages
X ave 

 ( X )(WeightingFactor )
i 1
i
In Naval Architecture, we use the simplified form:
X F 
X1 F1  X 2 F2  X 3 F3  

x 
X ave 
F1  F2  F3  
 F

i 1
ave
i i

i 1 i
Used to find:
Longitudinal Center of Flotation (LCF),
Longitudinal Center of Buoyancy (LCB)
Centers of Gravity (LCG, TCG, VCG)
Mathematical Moments and
the Parallel Axis Theorem
y
 First Moment=sdm
 Mx=ydA; My=xdA; AT=dA
dA
x
 Centroids: y = Mx/AT : x = My/AT
y
 Second Moment=s²dm
 Ix=y²dA; Iy=x²dA
x
y
yc
 Ix= Ixc+Ad12; Iy= Iyc+Ad22
d2
 Use Appendix C
xc
C
A
d1
x
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)
Ship’s Axes and Degrees of Freedom
z
Pitch/Trim
Surge
x
Heave
Roll/List/Heel
y
Sway
Yaw
Section 1.9
Bernoulli’s Equation:
V12
p2 V22

 gz1 

 gz 2  constant

2

2
p1
Along a line of equal energy (a streamline)
in a fluid, the above is a constant.
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
Bernoulli Equation
1
1
2
2
p1  V1  gz1  p2  V2  gz2
2
2
• 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.
Pressure Prediction
 Vertical pressure supports the vessel (lift
versus weight)
 Horizontal pressure is thrust and drag
These are the same as an aircraft!
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/ft4)
 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?
Answer:
60ft
50ft
5ft 4ft
deep deep
Weightbow compt=gV=
(1.80lb-s²/ft4) (32.17ft/s²)(50ft×60ft×5ft)(1LT/2240lb)=
388LT
Weightaft compt=gV=
(1.80lb-s²/ft4) (32.17ft/s²)(50ft×40ft×4ft)(1LT/2240lb)
=207LT
 Weighttotal=388LT+207LT+100LT=695LT
40ft
40 ft
5ft
deep
4ft
deep
388LT@30ft
0ft
50 ft
Answer:
60 ft
207LT
@80ft
100LT@50ft
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.
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
gsw=64 lb/ft3
Example Answer
 Displacement = Density x Submerged Volume
Vessel Submerged Volume = (75 ft)(21 ft )(1 ft) = 1575 ft3
(62.4 lb/ ft3)(1575 ft3)=98280 lb
(98280lb)/(2240lb/LT)= 43.875 LT (Empty)
4 foot
shoal
Tons per Inch (TPI) = 43.875 LT /12 in= 3.656 LT/ in
4 foot
(24 in)(3.656 LT/ in)= 87.74LT
87.74LT-60LT=27.74 LT is allowable
Max change in draft =
4 foot shoal –
60 LT
+ draft empty –
1 foot
# Marines
1 foot margin
= 2 feet
1 foot draft empty
1 foot
margin
1 foot
margin
1 foot draft empty
shoal
60 LTs + #
Marines
(27.74 LT)(2240 LT/lb)/(250lb/Marine)=248 Marines
Max change in draft
= 4’ shoal – 1’draft empty – 1’ margin
= 2’
Example Problem
A 100 ft Barge with an empty weight
distribution of 2LT/ft is loaded with the
following additional loads:
Y
<-30ft->
2LT/ft
<-40ft->
4LT/ft
<-30ft->
1LT/ft
Empty Barge=2LT/ft
0
100ft
X
What is the total weight of the barge and where,
along the length, is the center of gravity?
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=10,000+900+8000+2550 = 21450LT-ft
Weighted Average=21450LT-ft/450LT=47.7ft
 Center of Gravity along X axis=47.7ft
450 LT
Y
47.7ft
Loaded Barge
0
100ft
X