Transcript Eng. 6002 Ship Structures 1 Hull Girder Response Analysis

Eng. 6002 Ship Structures 1
Hull Girder Response Analysis
Lecture 3: Estimation of weight
distribution and still water
bending moment
Overview



A method for determining the distribution of
buoyant forces was described in the previous
lecture.
Today we will look at a method for estimating the
distribution of weight along the ship.
We will also discuss still water bending moments
Estimation of Weight Distribution
For ships with parallel middle body (cargo ships)
 If the weight distribution is unknown and we
need to estimate the distribution, the Prohaska
method can be used

The weight distribution is a trapezoid on top of a
uniform distribution
Estimation of Weight Distribution cont.
The weights are distributed as shown, with:
Whull
W 
L
Estimation of Weight Distribution cont.

Note that since a and b average to W we can
say:
b
a
 1.5 
W
2W
Estimation of Weight Distribution cont.

To move the lcg, the fore and aft ends of the
load diagram must be adjusted by equal and
opposite amounts
Longitudinal Strength of Ships:
Murray’s Method for SWBM
Although computer methods have emerged as a
practical way of calculating the still water
bending moment, it makes sense to discuss
Murray’s Method for hand calculations
 Based on the idea that forces and moments in a
ship are self balancing (no net forces transferred
to world)
Longitudinal Strength of Ships:
Murray’s Method for SWBM

He proposed that any set of weight and
buoyancy forces are in balance
Longitudinal Strength of Ships:
Murray’s Method for SWBM

Furthermore, for any cut at x, the moment at
the cut can be determined by:
BM ( x)  y1L1  y2 L2  y5 L5  y3 L3  y4 L4
Longitudinal Strength of Ships:
Murray’s Method for SWBM

Applying this idea to a ship
Longitudinal Strength of Ships:
Murray’s Method for SWBM

The bending moment at amidships is:

The estimate of max bending moment can be
improved by averaging these
Longitudinal Strength of Ships:
Murray’s Method for SWBM

The bending moment at amidships is:

We can simplify the buoyancy part by:
Longitudinal Strength of Ships:
Murray’s Method for SWBM

Murray suggested a set of values for the average
moment arm, as a function of ship length,CB,
and the ratio of draft to length
x  La  CB  b
T/L
0.03
0.04
0.05
0.06
a
0.209
0.199
0.189
0.179
b
0.03
0.041
0.052
0.063
Example using Murray’s Method

Tanker with L = 278m, B=37m, CB=0.8
Example using Murray’s Method

To find BMB we need the
draft
W    CB  L  B  T  

T 
CB  L  B  
140690

0.8  278  37 1.025
 16.68 m

So T/L = 16.68/278 = 0.06
Example using Murray’s Method

Murray’s table gives
a=0.179 and b=0.063

And the buoyancy
bending moment is
x  278.179  0.8  0.063
 57.32 m
1
BM B    x
2
1
 140690  57.32
2
 4,032,428 t  m
Example using Murray’s Method


The still water bending
moment is then
Assuming the wave
bending moments are:
WBMsag=583800 t-m and
WBMhog=520440 t-m
SWBM  BM W  BM B
hog
sag
 3,129,220  4,032,428
 903,145 t  m
sagging
Example using Murray’s Method

The total bending
moment is
Total BM  903,145  583,800
 1,486,945 ( sag )