1
1. Hull Girder Response Analysis – Prismatic Beam
1.1. Requirements of structural design
Structural design is an iterative process through which the layout and scantlings for a
structure are determined, such that it meets all the requirements of structural adequacy. The
overall configuration, which are in general dictated by non-structural consideration, such as
volume and space requirements, global stability, safety, etc. are required to be achieved in the
design.
In general terms the major steps that are involved can be summarized as follows:
a) Identify load and load combinations acting on the structure as a whole, or on its main sub-
components.
b) Select initial structural layout and scantlings. In general this is based on past experience
with similar structures.
c) Identify structure’s main components, and determine through structural analysis the loads
and load combinations acting on each component.
d) Identify relevant limit states and associated factors of safety.
e) Check structural adequacy. If any limit state is violated, adjust scantlings and repeat the
analysis and the structural checks. Perform the iterations required to converge to a
structurally adequate design.
f) Check other limit state, such as fatigue, which requires the selection of main structural
detail configurations. Also check the adequacy of the design against accidental loads. If
the structure is found to be inadequate, then new design iterations have to be conducted.
g) ‘Optimize’ structural design. Once an adequate design has been achieved it is in general
possible to ‘optimize’ it for a given objective. The objective depends on the structure’s
intended use, and can be, for example, the structural weight or the cost of fabrication and
installation. Thus, once a new configuration and set of scantlings are derived, structural
adequacy (Step d) has to be checked again, in an iterative fashion (Figure 1).
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Figure 1 Strength based design procedure
1.2. Classification of structural analysis
The ship structure can be classified into primary, secondary and tertiary elements as shown
in Figure 2. The first level of structure usually considered is the complete hull as a beam; this
is called the primary structure or hull girder as shown in Figure 3. In this level the ship is
idealized as a simple beam – a floating box girder that is internally stiffened and subdivided –
and in which the decks and bottom structure are flanges and the side shell and any
longitudinal bulkheads are the webs. Superstructure may also be considered depending on its
effectiveness. A part of the overall structure is cut out to show the different forces and
moments to be dealt with in beam theory. That part is called a hull module as shown in Figure
3
3. Secondary structure consists of stiffened panels and grillages bounded by the decks,
bulkheads and the shell. Tertiary structure may be panels of plates bounded by stiffeners or
elements of stiffeners themselves (Figure 2).
The forces and moments to be considered are:
1- Vertical longitudinal bending moment (most significant).
2- Horizontal longitudinal bending moment .
3- Longitudinal twisting moment .
4- Vertical shear force .
Stress can be classified in a similar way to the structure in which the stresses are occurring
and to the loads, which cause the stresses:
Primary – stresses due to bending, shear and torsion in the main hull girder.
Secondary – stresses in a stiffened grillage due to bending and membrane effects.
Tertiary – membrane stresses in panels between stiffeners.
The following assumptions must be taken:
a) Plane cross sections remain plane.
b) Prismatic beam (no openings or discontinuities).
c) Deflection and distortion caused by shear and torsion do not affect hull girder
bending.
d) Material is homogeneous and elastic.
Figure 2 Primary, Secondary and Tertiary structure
Tertiary
Secondary
Primary
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Figure 3 Levels of structural analysis
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1.3. Ship loads and Stresses
Loads can also be classified according to how they vary with time. They are either static,
slowly-varying or rapidly-varying. The principal loads on ships are:
1.3.1. Static loads
Vertical shear and longitudinal bending in still water: A ship floating in still water
has unevenly distributed weight owing to both cargo distribution and structural
distribution. The buoyancy distribution is also non-uniform since the underwater
sectional area is not constant along the length. Total weight and total buoyancy are of
course balanced. But at each section there will be a resultant force or load, either an
excess of buoyancy or excess of load. Since the vessel remains intact there are vertical
upward and downward forces tending to distort the vessel, which are referred to as
vertical shearing forces. The variation in the vertical loading will tend to bend the
vessel either to sagging or to hogging condition depending on the relative weight and
buoyancy forces (Figure 4).
Figure 4 Vertical shear and longitudinal bending in still water
Longitudinal shear in still water: When the vessel hogs and sags in still water and at
sea, shear forces similar to the vertical shear forces will be present in the longitudinal
plane. Vertical and longitudinal shear stresses are complementary and exist in
conjunction with a change of bending moment between adjacent sections of the hull
girder. The magnitude of the longitudinal shear force is greater at the neutral axis and
decreases towards the top and bottom of the hull girder (Figure 5).
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Figure 5 longitudinal shear forces
Dry docking loads: docking a ship on blocks imposes very high vertical loads on
ship’s bottom. As all ships may be expected to be docked at some time, it is necessary
to design for the docking condition.
Thermal loads: stresses in the ship structure can be caused by different temperatures
in one part with respect to another part, e.g. high air temperature in addition to solar
radiation may lead to the upper deck being as much as 40oC hotter than the hull below
water causing thermal stresses in the ship’s hull girder.
Grounding loads: for most ships, grounding is an accident condition and does not
directly affect the design of the structure. For vessels expected to ground, such as
landing craft and small boats, bending stresses on the hull resulting from grounding
must be calculated as well as local loads.
Lifting loads: some small vessels may be designed to be lifted in slings or from lifting
eyes. This kind of loading can be calculated using simple beam theory for the light
displacement condition.
1.3.2. Slowly varying loads
Vertical shear, longitudinal shear and longitudinal bending in seaway: When a ship is
in a seaway the waves with their troughs and crests produce a greater variation in the
buoyancy forces and therefore can increase the bending moment, vertical and
longitudinal shear forces. Classically the extreme effects can be illustrated with the
vessel balanced on a wave of length equal to that of the ship. If the crest of the wave
is amidships the buoyancy forces will tend to hog the vessel. If the trough is
amidships the buoyancy forces will tend to sag the vessel (Figure 6).
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Figure 6 Wave bending moments
Horizontal bending and torsion: these are caused by wave action. A ship heading
obliquely (45o) to a wave will be subjected to righting moments of opposite direction
at its ends twisting the hull and putting it in torsion. In most ships, horizontal bending
and torsional moments are much lower than bending in the vertical plane and can
usually be ignored. However, torsional moments in ship with extremely wide and long
deck openings (such as container ships) are significant (Figure 7).
Figure 7 Torsion
Racking: When a ship is rolling, the deck tends to move laterally relative to the
bottom structure and the shell on one side to move vertically relative to the other side.
This type of deformation is referred to as ‘racking’. Transverse bulkheads primarily
resist such transverse deformation (Figure 8).
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Figure 8 Racking
Sloshing of liquid cargo: specially significant in tankers.
Shipping of green water on deck: if a ship proceeds at speed even into moderate seas,
green water is thrown onto the deck. The forecastle and bridge front will be the worst
affected parts.
Wave slap on sides and foredecks: this is due to the action of the waves as they hit the
ship.
Panting: Panting refers to tendency for the shell plating to work in and out in a bellow
like fashion, and is caused by the fluctuating pressures on the hull at the ends when
the ship is amongst waves. These forces are most severe when the vessel is running
into waves and is pitching heavily.
Inertial loads: they are caused by the motion of heavy masses such as masts,
containers and other heavy objects following the motion of the hull. Provided that the
local accelerators are known, the estimation of inertial loads is straightforward
( ).
Berthing loads: they are extremely variable. They depend on the officer’s skill, the
weather conditions and the structure to which the ship is berthed.
Launching loads: these loads should be checked by the shipbuilder. The bending
stresses in the hull girder are moderate. The fore poppet should be carefully designed
and the fore end of the hull structure may be temporarily stiffened if necessary.
Ice loads: these are localized loads. Some ships are strengthened to have ice breaking
capability.
Wheel loads: they result from vehicles, for example on RO/RO vessels. They consist
of dead weight and inertia loads.
Distortion of structure
Rolling of shipacceleratesstructure, tendingto distort it
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1.3.3. Rapidly varying loads
Slamming: this is impact between the ship’s hull and the water surface. It occurs when
the vessel is driven into head seas where some part of the bottom of the forward end
of the ship comes out of the water and then re-enters. These slamming stresses are
likely to be most severe in a lightly ballasted condition, and occur over an area of the
bottom shell aft of the collision bulkhead (Figure 9).
Figure 9 Slamming
Vibration: there are particular locations on a ship where vibration response may be
important. For example, in way of weapons, machinery, and in the stern region.
Collision loads: this is an accident condition. It is not usual to design a structure to
withstand collisions. However, there are some exceptions such as nuclear powered
ships and tankers.
Loads due to underwater explosion: these are taken into consideration only for navy
ships.
Impact loads: due to weapons effects in naval ships.
Springing loads: are sea excitation forces. Springing is a continuous and steady
vibration, it occurs when the natural frequency of the hull and the wave frequency
coincide (resonant response).
0.25L or 0.30L
0.05LSlamming region
Pitching
Summer load
waterlineShip forced downon to waterwave profile
Heaving
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1.4. Failure modes
During structural design and analysis, care must be taken to ensure that all possible failure
modes are considered. The possible failure modes are:
1.4.1. Fatigue
The majority of loads on ships are cyclic. Indeed, most structural failures that occur in service
are the result of fatigue damage. Generally, fatigue is not included in the main design process,
but later in detail design after the final construction drawings are being produced.
1.4.2. Brittle fracture
It depends on the material from which the ship is constructed. The risk is increased by the
presence of stress concentrations, notch like defects, exposure to low temperature, impact or
high loading rates. Welding procedures are an important factor too.
1.4.3. Yielding or plastic collapse
When yielding occurs, very small increases in load cause large increases in deformation. The
two most critical types of load causing yielding in ship structures are lateral load and inplane
load. Under lateral loading, a mechanism is formed by the formation of plastic hinges. Under
inplane loads, yielding occurs when the combined stresses reach the yield point of the
material. Usually, buckling will occur before pure yielding in the case of inplane compressive
loads.
1.4.4. Buckling
According to the structural configuration and the loading conditions, it takes many forms. It
can occur in the plating between stiffeners, in the stiffeners’ webs or flanges (by tripping or
flexure) or in an entire stiffened panel or grillage. All form of buckling may result in
complete collapse of the structure. The initial deformations and residual stresses that occur
during fabrication almost always lead to some loss of buckling strength.
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1.4.5. Excessive deflection
In way of machinery, there are limits on allowable deflections. Large deflections may
interfere with the performance of the equipment nearby.
1.5. Function of ship’s structure
A ship is capable of bending in a longitudinal vertical plane and hence there must be material
in its structure which will resist this bending.
Any material distributed over a considerable portion of the length of the ship will contribute
to its longitudinal strength. Example of such items:
1- Side and bottom shell plating
2- Inner bottom plating
3- Decks
4- Deck and bottom longitudinals
5- Side longitudinals
Items which contribute to transverse strength:
1- Floors
2- Side frames
3- Beams
4- Transverse watertight bulkheads
No. 1,2 and 3 form transverse rings.
1.6. Review of load, shear and moment relationships
Consider the beam shown in Figure 10 is subjected to a distributed transverse load of varying
intensity .
Consider an element of the beam ∆ in length.
Let = average value of and the resultant load . ∆ acts at a distance of . ∆ from the
right-hand face of the element.
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Figure 10 simple beam subjected to an arbitrary distributed load
f f x Q Q x M M x (1.1)
From vertical equilibrium:
. 0a a
QQ f x Q Q f
x
(1.2)
The slope of the shear force diagram at any point is equal to the load intensity at that point:
0
limx
dQ Qf
dx x
(1.3)
Taking moments about :
. . . . 0 . .a a
MM Q x f x x M M Q f x
x
(1.4)
The slope of the moment diagram at any point is equal to the shear force at that point:
2
20limx
dM M d MQ f
dx x dx
(1.5)
Hence, the relationships between the load intensity and the shear force and between the shear
force and the bending moment for the beam will be given by:
2
1
1 2 1. . .x
x
dQ f dx Q f dx C Q Q f dx (1.6)
L
X1
X2
x x1 2
f(x)
A B
xx
O
M M+
Q
Q+Q
f f+f
F .xa x (01)
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The change of shear force between two points is equal to the area under the load intensity
diagram between these two points.
2
1
2 2 1. . .x
x
dM Q dx M Q dx C M M Q dx (1.7)
The change in bending moment between two points is equal to the area under the shear force
diagram between these two points.
1.7. Basic relationships for ship hull girder
Overall static equilibrium requires that the total upwards buoyancy force equals the weight of
the ship and that these two vertical forces coincides; that is, the longitudinal center of
buoyancy (LCB) must coincide with the longitudinal center of gravity (LCG).
The first requirement is:
0 0
L Lg a x dx g m x dx g (1.8)
Or B W
Where:
a x = immersed cross-sectional area
m x = mass distribution (mass per unit length)
= mass density of sea water (or fresh water if appropriate)
g = gravitational acceleration
= displacement
W = weight
B = buoyancy
Similarly, equilibrium of moments requires that:
0 0
L L
Gg a x xdx g m x xdx g l (1.9)
Or . .B LCB W LCG LCB LCG
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Where, Gl = distance from origin to LCG.
However, over any given unit length of the hull the forces will not balance out. At any point
x :
Buoyancy per unit length = ( )b x ga x
The weight per unit length = ( )w x m x g
Hence, the net force (load) per unit length = ( ) ( ) ( )f x b x w x
If this net loading is integrated along the length there will be, for any point, a force tending to
shear the structure such that:
0 0
,x x
Shear force Q x f x dx ga x m x g dx (1.10)
the integration being from one end to the point concerned.
Integrating a second time gives the longitudinal bending moment. That is:
0 0 0
,x x x
Bending moment M x Q x dx ga x m x g dxdx (1.11)
Put the other way, load per unit length f x = d dQ x = 2 2d dM x .
0 0
andx L
L L
M Mx dx dx
EI EI (1.12)
0
and .x
Lv x x dx x v x x (1.13)
Where:
= slope
v = vertical displacement
= deflection
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Figure 11 summary of hull girder bending
1.8. Characteristics of shear force and bending moment curves
For any given loading of the ship, the draughts at which it floats can be calculated. Knowing
the weight distribution, and finding the buoyancy distribution from the Bonjean curves, gives
the net load per unit length.
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Certain approximations are needed to deal with distributed loads such as shell plating. Also
the point at which the net force acts may not be in the center of the length of the increment
used. However, these approximations are not usually of great significance and certain checks
can be placed upon the results (Figure 12):
First the shear force and bending moment must be zero at the ends of the ship. If after
integration there is a residual force or moment this is usually corrected arbitrarily by
assuming the difference can be spread along the ship length.
The shearing force is approximately asymmetric, has a maximum or minimum value at
points about a quarter of the length from the ends and is zero near amidships.
From the relationships deduced above when the net load is zero the shear force will
have a local maximum or minimum value and the moment curve will show a point of
inflection.
Where net load is a maximum the shear force curve has a point of inflection.
Where shear force is zero, the bending moment is a local maximum or minimum.
Bending moment will have zero slopes at both ends with small values forward and aft
of the quarter points.
Figure 12 shearing force and bending moment
When the ship is distorted so as to be concave up it is said to sag. The deck is then in
compression with the keel in tension. When the ship is convex up it is said to hog. The deck
is then in tension and the keel in compression.
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1.9. Weight distribution
The calculation of the longitudinal distribution of weight or mass m x is a difficult process,
partly because m x is made up of discrete items rather than being a continuous and regular
curve, and partly because at the design stage many of the individual weights are known only
approximately.
The weights in a ship fall into two main categories: those which are relatively unchanged,
such as the ship’s own structural weight; and those which do change, such as cargo, fuel,
stores, and ballast. The first group constitutes the “lightweights” of a ship, that is, the weight
when it is without cargo, fuel, and so on (this condition is referred to as the “lightship”
condition). The second group is called the “deadweight“. The deadweight changes with each
different cargo loading, and hence there are usually several loading conditions which need to
be investigated. The two most common conditions are “full load” and “ballast”.
In most cases the following information should be specified for each distinct weight item:
1. Total weight.
2. Vertical and longitudinal center of gravity lcg.
3. Longitudinal extend.
4. The type of distribution over this extend.
In specifying the extend and distribution of individual weights, it is helpful and even
necessary to use some approximations and idealizations. Nearly all items can be represented
in terms of one or more of three basic types of distribution: point, uniform distribution, and
trapezoidal distribution.
Typical examples of point loads are machinery (one point load at each foundation point),
masts, winches, and transverse bulkheads.
Examples of uniform loads are: hull steel within the parallel midbody, and cargo, fuel,
ballast, and other homogenous weights within prismatic spaces. Outside the parallel midbody
and particularly toward the ends of the ship a trapezoidal distribution is appropriate, although
even here some items can be accurately represented as uniform loads, such as superstructure.
For a trapezoid, say of length l , the relevant information may be specified in two different
ways: either as total mass, oM , with a specified position of center of gravity within this
length (say a distance x from the center; see Figure 13) or in terms of mass per unit length at
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the forward and after ends: fm and am . The formulas for converting from one form to
another are:
Figure 13 trapezoidal representation of a weight
6
2
f a
f a
f a
o
m mlx
m m
l m mM
(1.14)
and
2
2
6
6
o oa
o of
M M xm
l lM M x
ml l
(1.15)
1.9.1. Hull weight distribution
Hull weight is traditionally defined as lightship minus the weight of the anchor, chain, anchor
handling gear, steering gear and main propulsion machinery. Numerous approximation methods
for distributing hull weight have been proposed in the past. These approximations are general and
appropriate only for initial stage design due to their low fidelity.
A useful first approximation to the hull weight distribution is obtained by assuming that two-
thirds of its weight follows the still water buoyancy curve and the remaining one-third is
distributed in the form of a trapezoid, with end ordinates such that the center of gravity of the
entire hull is in the desired position (Figure 14).
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Figure 14 approximation for hull weight distribution
Trapezoidal approximation is useful for ships with parallel midbody. This approximation uses a
uniform weight distribution over the parallel midbody portion and two trapezoids for the end
portions, with end ordinates again chosen such that the LCG of the hull is in the desired
position as shown in Figure 15. The ordinates indicated in the figure are given by:
Figure 15 Trapezoidal approximation
Hull Weight
Ordinate .Length
HWCoeff
L (1.16)
Where the coefficient is as indicated in Table 1:
Table 1 coefficients of the trapezoidal approximation
1 2 3
K 0.333 0.333 0.250
a 0.567 0.596 0.572
b 1.195 1.174 1.125
c 0.653 0.706 0.676
C.G. Aft 0.0052L 0.0017L 0.0054L
1 Fine ships – Merchant type 2 Full ships – Merchant type 3 Great lakes Bulk freighters
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Biles presented the hull weight by a trapezoid, frequently called the ‘coffin diagram’, and
gave the following values of the ordinates for passenger and cargo ships (Figure 16):
At F.P. 0.566
over 3amidships 1.195
at A.P. 0.653
H
H
H
W L
L W L
W L
(1.17)
Where HW = hull weight, L = length of ship
Figure 16 coffin diagram
The centroid of the diagram as given is at 0.0056L abaft amidships. It is permissible to make
small adjustments to the end ordinates in order to ensure that the centroid of the diagram
corresponds to the longitudinal center of gravity of the hull.
The desired shift of the centroid can be secured by transferring a triangle from one trapezium
to the other as indicated by the dotted lines. The shift of centroid of the triangle is 7 9 L .
Thus, if x is the end ordinate of the triangle to be shifted:
Area of triangle = 13
2x L
Moment of shift = 27 7
6 9 54
Lx L xL
Shift of centroid = 2754 HxL W
Thus 54 shift of centroid
7HW
xL L
Prohaska has given detailed consideration to the diagram (Figure 17) and suggested values
for a number of different types of ships as given in Table 2.
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Figure 17 Prohaska’s approximation
Table 2 Prohaska’s values
Comstock representation is typically used to approximate the hull weight. In this
approximation, 50% of the hull weight is distributed as a rectangular in the middle 0.4 length,
and 50% in two trapezoids so as to give the required LCG. If HW is the total weight to be
distributed and d is the LCG of the weight from amidships, then (Figure 18):
Figure 18 Comstock’s representation
1.25
1 203
1 203
HWh
Lh d
xL
h dy
L
(1.18)
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Cole has proposed a parabolic rule. This method is useful in ships without parallel middle
body. The hull weight is presented by a rectangle and a superimposed parabola (Figure 19).
Obviously the centroid of this diagram is at amidships. The centroid can be shifted to a
desired position by swinging the parabola as follow (Figure 19, Figure 20):
1) Through the centroid of the parabola draw a line parallel to the base and in length equal
to twice the shift desired forward or aft.
2) Through the point so obtained draw a line to the base of the parabola at amidships.
3) The intersection of this line with the horizontal drawn from the intersection of the
midship ordinate with the original parabola determines the position of one point on the
new curve.
4) Parallel lines are drawn at other ordinates as shown and the new curve determined.
Figure 19 Cole’s proposal
Figure 20
Small errors in the area and the centroid of the diagram can be corrected by adjustments of
the base line. An error in weight can be corrected by raising or lowering the base line. The
position of the centroid can be adjusted by tilting the base line as shown in the Biles method.
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1.9.2. Total lightweight distribution
When the hull weight distribution has been obtained, the other items of the lightweight (weight of
the anchor, chain, anchor handling gear, steering gear and main propulsion machinery) can be
added at their centers of gravity. The resulting curve for the lightship weight can be obtained as
shown in Figure 21.
Figure 21 lightship weight distribution
1.9.3. Deadweight distribution
For cargo and ballast, the weight per unit length is related to the cross-sectional area of the
relevant cargo or ballast space, and their weight distribution may be taken as the product of
the sectional area curve of the relevant space times the mass density of the cargo or ballast. If
the total volume of cargo spaces and the cargo deadweight of the ship being known, then:
3volumeof cargospacesstowage ratein
cargodeadweightft ton (1.19)
Because the cargo is the largest item of weight and because there are so many possible
variations in its distribution, there are often some distributions and combinations that would
cause excessive values of bending moment and that therefore must be avoided. It is more
efficient to have the cargo holds or tanks either completely full or completely empty. Given
such extreme differences it is important that they be spread out, rather than grouped together,
because the latter would give excessive shear force and/or bending moment as shown in
Figure 22. Figure 23 shows a typical curve of buoyancy, weight, load, shear force and
bending moment for a 30 000 T.D.W. bulk carrier with ore in holds No. 1,3,5 and 7 only.
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Figure 22 effect of cargo distribution
Figure 23 30 000 T.D.W. bulk carrier with ore in holds No. 1,3,5 and 7 only
When the weights per unit length for the deadweight items have been obtained, they are
added to the curve of the total lightweight, giving the total weight curve as shown in Figure
24. After the curve is plotted, it should be checked for the total area, giving the weight of the
ship in that particular loading condition. Its centroid will give the longitudinal center of
gravity of the ship. A sample weight curve is given in Figure 25.
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Figure 24 Total weight curve
Figure 25 typical weight distribution
1.9.4. Modified weight curve
The described weight curve shows many discontinuities. The sudden changes that occur in
the weight curve are not at regular intervals in the length direction. This makes some
difficulties during integration, particularly by a tabular method. To overcome this difficulty,
the length of the ship is divided into a number of equal parts and we assume that the weight
per unit length is constant over each division. In this way a stepped weight curve is produced
as shown in Figure 26.
Figure 26 Stepped weight curve
Indicates deadweight items
26
To produce this stepped weight curve, the total weight in each division is calculated and then
is divided by the length of the division. This will give the mean weight per unit length for that
division. Having obtained the stepped weight curve in this way, the total area and position of
its centroid should be checked so as to give the correct weight and center of gravity of the
ship.
The steel weight of the far portions in the forward and aft must also be included in the weight
curve, thus, the weight curve must be corrected to enclose these weights between the
perpendiculars. This can be done through transferring the end weights to the nearest two
intervals to compensate for the moment of shifted weight, as shown in Figure 27, such that:
1 2
1
2
3
2
1
2
P P P
XP P
S
XP P
S
(1.20)
Figure 27 inclusion of end weight
1.10. Buoyancy distribution in still water
The still water buoyancy is a static quantity and depends on the geometry of the underwater
portion of the hull. The buoyancy due to waves is both dynamic and probabilistic. It is
assumed that all the usual hydrostatic information is available for the ship and that Bonjean
curves of area are also available. The problem is then to find the distribution of buoyancy that
will give these values of displacement and center of buoyancy so that the ship shall be in
static equilibrium either in still water or on a wave.
P
P
PX
S S
0 1 2
1
2
27
For the still water condition, the mean draft mT is determined from the hydrostatic curves
according to the loading condition, i.e. at the magnitude of displacement, as shown in Figure
28. If LCB (corresponding to mT ) and LCG are not equal, then the total trim tT caused by this
difference is determined according to the following equation:
1t
cm
LCB LCGT
MCT
(1.21)
Figure 28 Hydrostatic curves
The magnitudes of forward and aft trims are determined based on the location of the center of
flotation LCF (+ve Fwd) as shown in Figure 29, such that:
0.5
0.5
F t
A t
L LCFt T
LL LCF
t TL
(1.22)
Figure 29 Trim calculation
The end drafts are then determined by adding trim to, or subtracting trim from, the mean draft
according to the condition of trim. Afterwards, the end drafts should be drawn on a profile of
TPC
MCT
LCB LCF
Tm
T
Fwd +veAft -ve
L
LCF tF
t A
Tt
Tm
28
the ship in the normal way to obtain the waterline at which the ship floats as shown in Figure
30.
Figure 30 Bonjean curves
If the Bonjean curves of area are also shown on this profile, it is a simple matter to lift off the
immersed areas where the waterline intersects the various sections. The buoyancy per unit
length at any section is then simply the area of the section multiplied by the density of water.
It must be checked that the areas lifted from the Bonjean curves for the obtained trim line
give the correct displacement and LCB. If the trim is large, some discrepancy may exist so
that:
'
'LCB LCB
(1.23)
The position of the waterline must be corrected by moving it a distance ( ' ) TPC and
tilting it an amount 1'( ' ) cmLCB LCB MCT , where:
& LCB are the required displacement and LCB,
'& 'LCB are those obtained from Bonjean curves’ calculations.
Next areas are lifted and displacement and center of buoyancy calculations are repeated. This
second approximation is usually sufficient.
1.11. Buoyancy distribution in seaway
The mass distribution is the same in waves as in still water assuming the same loading
condition. The differences in the forces acting are the buoyancy forces and the inertia forces
on the masses arising from the motion accelerations, mainly those due to pitch and heave. For
the present the latter are ignored and the problem is treated as a quasi-static one by
considering the ship balanced on a wave.
Waterline
Bonjean curve of area
29
The buoyancy forces vary from those in still water by virtue of the different draughts at each
point along the length due to the wave profile and the pressure changes with depth due to the
orbital motion of the wave particles. This latter, the Smith effect, is usually ignored in the
standard calculation to be described next. Ignoring the dynamic forces and the Smith effect
does not matter as the results are used for comparison.
The concept of considering a ship balanced on the crest, or on the trough, of a wave is clearly
an artificial approach although one which has served the naval architect well over many
years. Nowadays the naval architect can extend the programs for predicting ship motion to
give the forces acting on the ship. Such calculations have been compared with data from
model experiments and full scale trials and found to correlate quite well.
The strip theory is commonly used for calculating ship motions. The ship is divided into a
number of transverse sections, or strips, and the wave, buoyancy and inertia forces acting on
each section are assessed allowing for added mass and damping. From the equations so
derived the motions of the ship, as a rigid body, can be determined. The same process can be
extended to deduce the bending moments and shear forces acting on the ship at any point
along its length. This provides the basis for modern treatments of longitudinal strength.
1.11.1. The trochoidal wave
Figure 31 shows the profile of a regular wave which may be considered to be a deep sea
wave. Wave of this type is oscillating waves in which the water particles move in closed
paths without bodily movement of fluid. The wave form moves over the surface and energy is
transmitted. The distance between successive crests is the wave length L . The distance from
the trough to the crest is the wave height h .
Figure 31 profile of a regular wave
Observations on ocean waves have shown that the crests are sharper than the troughs, which
assumes that the wave profile is a trochoid. The trochoidal theory shows that the paths of the
30
water particles are circles, whereas the classical theory shows that the paths are ellipses
which tend to circles as depth of water increases.
Both theories show that for a deep sea wave, if conditions are considered some distance
below the surface, then the radius of the orbit circles of the particles diminishes. If 2or h is
the radius of the surface particles and r is their radius at some sub-surface distance y below
the free surface then:
2
exp expo o
y yr r r
R L
(1.24)
Where y is considered positive downwards.
It will be seen then that the disturbance below the surface diminishes rapidly with depth, the
situation being as shown in Figure 32, where the sub-surface trochoids as they may be called
rapidly flatten out.
In shallow water where the influence of depth is important, the elliptical orbits of the particles
flatten out with depth below the free surface and at the bottom the vertical movement is
prevented altogether and the particles move horizontally only.
Figure 32
A trochoid is a curve produced by a point at radius r within a circle of radius R rolling on a
flat base. The equation to a trochoid with respect to the axes shown in Figure 33, is:
31
Figure 33 construction of a trochoid
sin
1 cos
x R r
z r
(1.25)
One accepted standard wave is that having a height from trough to crest of one twentieth of
its length from crest to crest. In this case, 2L R and 2 40r h L and the equation to
the wave is:
sin2 40
1 cos40
L Lx
Lz
(1.26)
Research has shown that the 20L wave is somewhat optimistic for wave lengths from 90m
up to about 150m in length. Above 150m, the 20L wave becomes progressively more
unsatisfactory and at 300m is probably so exaggerated in height that it is no longer a
satisfactory criterion of comparison. This has resulted in the adoption of a trochoidal wave of
height 0.607 L as a standard wave in the comparative longitudinal strength calculation.
This wave has the equation:
0.607sin
2 2 , and in meters0.607
1 cos2
L Lx
x z LL
z
(1.27)
The 0.607 L wave has the slight disadvantage that it is not non-dimensional, and units must
be checked with care when using this wave and the formulae derived from it.
32
The wave height 1.1 L is another approximation suggested by Lloyd’s Register.
To draw a trochoidal wave surface:
1 – Divide selected wave length ( WL L ) by a convenient number of equally spaced points (
S = spacing).
2 – With each point as a center, draw a circle of diameter equal to the selected wave height
(e.g. 20h L ).
3 – In each of the circles, draw a radius at an angle increase of the fraction of 360o as the
spacing of the circles to wave length.
4 – Connect the ends of the radii
2 2W
S S SS R
L R R
(1.28)
1.11.2. Pressure in waves (Smith effect)
In still water the pressure at any point is proportional to the distance below the free surface,
but it is not so in the case of a wave. It can be shown that the still water level corresponding
to any point in a trochoidal wave lies at a distance 2 22r R r L below the line drawn
through the orbit center of the particles corresponding to that point. Thus at the surface the
still water level would be 2or L below the orbit centers; and for sub-trochoid of orbit
center at distance y below the orbit center of the surface trochoid, the still water level would
be 2r L below the orbit centers.
The distance between these two still water levels is:
2 2
or ry
L L
(1.29)
and the pressure in the wave can be shown to be proportional to this distance. Hence, pressure
in wave = 2 2og y L r r . This is shown in Figure 34. Its effect is that pressure is
reduced below the static value at the crest of a wave and is increased beyond the static value
at the trough, which influences the buoyancy of ship amongst waves (Figure 35). The
33
influence of dynamic considerations on the pressure in a wave is often referred to as the
‘Smith effect’.
Figure 34
Figure 35 Buoyancy distribution of ship in waves (a) wave crests at ends. (b) troughs at ends
1.11.3. The standard static longitudinal strength approach
Too long a wave length, several times the length of the ship, makes the profile look like a
horizontal line on the ship and the buoyancy distribution would be essentially as that in still
water. On the other hand, a very short wave length, i.e., small fraction of the length of the
ship, makes small undulations in the still water buoyancy curve that would have little effect
on the bending moment produced on the ship.
It would appear that there would be some length of wave in between these two extremes that
would have the maximum effect on the buoyancy distribution. The standard that has usually
been accepted is that the ship is assumed to be poised, in a state of equilibrium, on a
trochoidal wave of length equal to that of the ship. This is not easy and can involve a number
34
of successive approximations to the ship's attitude before the buoyancy force equals the
weight and the center of buoyancy is in line with the center of gravity. Clearly this is a
situation that can never occur in practice but the results can be used to indicate the maximum
bending moments the ship is likely to experience in waves.
Two conditions are considered, one with a wave crest amidships and the other with wave
crests at the ends of the ship. In the former the ship will hog and in the latter it will sag
(Figure 36). Figure 37 shows the resulting buoyancy distribution. The bending moments
obtained include the still water moments. It is useful to separate the two. Whilst the still water
bending moment depends upon the mass distribution besides the buoyancy distribution, the
bending moment due to the waves themselves depends only on the geometry of the ship and
wave.
The influence of the still water bending moment on the total moment is shown in Figure 38.
Whether the greater bending moment occurs in sagging or hogging depends on the type of
ship depending, mainly, upon the block coefficient. At low block coefficients the sagging
bending moment is likely to be greater, the difference reducing as block coefficient increases.
Figure 36 ship on wave
Figure 37 Shape of buoyancy distribution
Wave profile
Wave profile
Sagging
Hogging
Buoyancy instill water
Buoyancy (Wavecrests at amidships)Buoyancy (Wave crests
at perpendiculars)
35
Figure 38 still water and wave bending moments
1.11.4. W. Muckle’s Method
The position of the still-water level corresponding to the wave form is first calculated. This is
a distance 2 2r R below the center of the wave, where 2r h and 2R L , so that:
Still-water level below wave center = 2 4h L
The still-water level of the wave is now placed on the waterline at which the ship would float
in still water for the condition of loading for which the calculation is being carried out.
If the sagging condition is being considered, then the wave must be raised a certain amount
and tilted. Thus, the amount by which the wave must be raised at any position in the length of
the ship can be written:
y a bx L (1.30)
where a and b are constants yet to be determined and x is measured from, say, the after
perpendicular.
Figure 39 shows the Bonjean curve of area for a section in the length of the ship. The point C
is where the wave cuts the section before it has been adjusted by the amount y . The
corresponding sectional area is oA .
36
Figure 39
The buoyancy curve should satisfy the following two equilibrium conditions.
The total volume after the wave has been shifted must be equal to the required volume for the
condition of loading, hence:
0
L
x
WA dx
(1.31)
The moment of the total displacement after the wave has been shifted must be equal to the
moment of the ship weight, hence:
0
L
G B B x G B
WX X X A xdx X X
(1.32)
Where, GX and BX are the LCG and the LCB, respectively, measured from the after
perpendicular. W is the total weight of the ship.
Let the area at a position m m above the position C be mA . An approximation of the Bonjean
curves will show that the curve between these two positions could be very closely represented
by a straight line. If yA is the area at any height y above C, then:
y om o m oy o
A AA A A AA A y
m y m
(1.33)
From Eqn (1.30):
m o m o m oy o o
A A A A A Ax xA A a b A a b
L m m m L
(1.34)
From Eqn (1.31):
37
0 0 0 0
L L L Lm o m o
y o
A A A A x WA dx A dx a dx b dx
m m L
(1.35)
From Eqn (1.32):
2
0 0 0 0
L L L Lm o m o
y o
GB
A A A Ax x x xA dx A dx a dx b dx
L L m L m L
XX W
L L
(1.36)
We thus have two equations (1.35) and (1.36) from which the two unknowns a and b can be
calculated.
1.11.4.1. Example
A ship 460 ft in length and of 8996 tons displacement with the center of gravity 6.32 ft aft of
amidships.
For a wave having a length equal to the length of the ship and a height equal to 20L the
still-water level is
2 211.50.905
2 2 460 2
r
R
ft
Below the center of the wave. This still-water level was put on the water line at which the
vessel would float in still water, and ordinates were read from the Bonjean curves where the
wave cut the various sections.
As the sagging condition was being considered, the final position of the wave would be
higher than this, so that ordinates of area at positions 4 ft above these intersections were also
lifted as described in the foregoing.
Table 3 shows the values of oA and 4A read from the curves, and also shows how the various
integrations were made by using Simpson’s first rule.
38
Table 3
Section oA S.M.
Area function Lever
Moment function 4
A 4 o
A A Area
function Moment function
Moment of inertia function
0 125 12 63 0 - 210 85 43 - -
12 465 2 930 1
2 465 615 150 300 150 75
1 770 1 770 1 770 948 178 178 178 178 121 960 2 1920 1
21 2880 1175 215 430 645 968
2 1005 121 1507 2 3014 1235 230 345 690 1380
3 790 4 3160 3 9480 1025 235 940 2820 8460
4 475 2 950 4 3800 718 243 486 1944 7776
5 310 4 1240 5 6200 555 245 980 4900 24500
6 412 2 824 6 4944 655 243 486 2916 17496
7 705 4 2820 7 19740 948 243 972 6804 47628
8 900 121 1350 8 10800 1115 215 323 2584 20672
128 838 2 1676 1
28 14246 1025 187 374 3180 27030
9 625 1 625 9 5625 770 145 145 1305 11745 129 300 2 600 1
29 5700 375 75 150 1425 13538
10 - 12 - 10 - - - - - -
18435 87664 6152 29541 181446
From Table 3 the values of the various summations are calculated and these are as follows:
3
0
4618435 282670
3
L
oA dx ft
3
0
87664 46134418
10 3
L
o
xA dx ft
L
34
0
6152 4623583
4 4 3
LoA A
dx ft
34
0
29541 4611324
4 4 10 3
LoA A x
dx ftL
234
0
181446 466956
4 4 10 10 3
LoA A x
dx ftL
The required volume of displacement is 38996 35 314860 ft
3230 6.32314860 153104
460G GX XW
ftL L
39
Equations (1.35) and (1.36) may now be written as:
282670 23583 11324 314860a b
134418 11324 6956 153104a b
That is,
0.4802 1.365a b
0.6143 1.6501a b
From these we obtain 0.3441a and 2.126b .
The area to be added to each section is now = 4 0.3441 2.1264
oA Ax L
Table 4 shows these areas together with the calculation for the additional displacement and its
centroid.
Table 4
Section Additional area S.M. Area function Lever Moment function
0 7.3 12 3.6 5 18.0
12 17.0 2 34.0 1
24 153.0
1 24.8 1 24.8 4 99.2 121 35.6 2 71.2 1
23 249.2
2 44.2 121 66.3 3 198.9
3 57.7 4 230.8 2 461.6
4 72.6 2 145.2 1 145.2
5 86.2 4 344.8 0 1325.1
6 98.4 2 196.8 1 196.8
7 111.3 4 445.2 2 890.4
8 109.9 121 164.8 3 494.4
128 100.6 2 201.2 1
23 704.2
9 81.8 1 81.8 4 327.2 129 44.3 2 88.6 1
24 398.7
10 - 12 - 5 -
2099.1 3011.7
Additional displacement = 2099.1 46
91935 3
tons
40
Centroid from amidships = 3011.7 1325.1
46 36.96 ford.2099.1
ft
Original displacement obtained from 0
2826708076
35
L
oA dx tons
Corresponding LCB = 134418
0.5 460 11.27 aft.282670
ft
Final displacement = Original displacement + Additional displacement = 8076 + 919 = 8995
tons
Final LCB = 8076 11.27 919 36.96 57050
6.348076 919 8995
ft
aft of amidships
These results compare very favorably with the required values of 8996 tons displacement
with a position of center of buoyancy of 6.32 ft aft of amidships.
It remains now to add the areas in the second column of Table 4 to the values oA obtained
originally from the Bonjean curves. The resulting areas, when divided by 35, give the
ordinates of the buoyancy curve in tons/ft.
Should the hogging condition be considered instead of the sagging condition, the wave would
be required to be lowered instead of raised. It would be necessary, therefore, to take areas
from the Bonjean curves at positions 4 ft below those at which the areas oA are read.
Otherwise, the calculation would be exactly the same as outline here.
1.12. Load distribution
The load on the structure at any point in the length of the ship is the difference between the
weight per unit length and the buoyancy per unit length:
p b w (1.37)
The total area enclosed by the load curve should be zero, the area underneath the base being
considered negative, i.e. total vertical force = 0 or static equilibrium.
If the load curve is integrated, then the shearing force on the structure is obtained, so that:
Q p dx (1.38)
41
By integrating the shearing force curve, the bending moment on the structure can be
obtained:
M Q dx (1.39)
The shearing force and bending moment curves are shown in Figure 40. Both of these curves
should be zero at x L , which is the consequence of the equilibrium condition. The load
distribution is what mainly affects shear force and bending moment (maximum values).
Figure 40 Load, shearing force and bending moment curves
1.13. Integration procedure
The integration is performed using one of the following two methods:
1.13.1. First method:
In this method, the integration of the weight and buoyancy curves is performed separately.
The first integral of these two curves gives the curves shown in Figure 41 and the shearing
force is the difference between the two curves. If these two first integral curves are
integrated, then curves of the type shown in Figure 42 are obtained and one again the
difference between these two curves gives the bending moment.
Figure 41 Integration of weight and buoyancy curves
Shearing force
Load
Bending moment
Shearingforce
Integral of buoyancyInteg
ral o
f weig
ht
42
Figure 42 Double integration of weight and buoyancy curves
1.13.2. Second method:
The tabular method of integration is one used very frequently and is particularly suited where
the stepped type of weight curve has been employed. Suppose that the length of the ship is
divided into a number of equal intervals of length s (usually 40 intervals) and that mb and
mw are the mean ordinates of the buoyancy and weight curves for an interval. The integral of
the load curve for this division is then given by m mb w s . Then the shearing force at any
point will be:
m mQ s b w (1.40)
If mQ is the mean value of the shearing force for an interval, then the bending moment is
given by:
mM s Q (1.41)
Table 5 shows how the calculation can be performed using the second method.
Note that if there is an error in shearing force or bending moment, it is corrected as shown in
Figure 43 and Table 5. Since 40Q and 40M must be equal to zero, the base lines are corrected.
Bendingmoment
b dxdx
w dxdx
43
Figure 43 Base line correction for shear force and bending moment
Bending moment
Shear
ing
forc
e
B
A
O Original base lineCorrected base line for bending moment
Corrected base line for shearing force
44
Table 5 Tabular integration of shearing force and bending moment
1 2 3 4 5 4 2 6 7 8 9 6 8 10 11 12 13 11 12
Stn w m
w b m
b m m
b w Q s Lever
x L Correction
40.x L Q s
Corrected
Q s mQ 2M s
Correction
40
2.x L M s
Corrected2M s
0 0
w 0
b 0 0 0 0 0 0 0
0 12w w 0 1
2b b 1 1m m
b w 1
2Q s
1 1
w 1
b 1Q s 1 40 40
1 40. Q s 21M s 40
21 40. M s
1 22w w 1 2
2b b 2 2m m
b w 1 22Q Q s
2 2
w 2
b 2Q s 2 40 40
2 40. Q s 22M s 40
22 40. M s
2 32w w 2 3
2b b 3 3m m
b w 2 32Q Q s
3 3
w 3
b 3Q s 3 40 40
3 40. Q s 23M s 40
23 40. M s
3 42w w 3 4
2b b 4 4m m
b w 3 42Q Q s
4 4
w 4
b 4Q s 4 40 404 40. Q s 2
4M s 40
24 40. M s
. . . .
. . . . . . . .
. . . .
39 39
w 39
b 39Q s 39 40 4039 40. Q s 2
39M s 40
239 40. M s
39 402w w 39 40
2b b 40 40m m
b w 39 402Q Q s
40 40
w 40
b 40Q s 1 40Q s 240M s 2
40M s
+
=
=
+
+
+
=
=
+
+
+
+
=
=
=
=
45
1.14. Wave bending moment for a box-shaped vessel on a sine wave
Figure 44
The wave surface equation is given by:
2
cos2
h xy
L
(1.42)
Where, h = wave height ( 20h L or 1.1h L ), and L = length of the vessel.
The wave bending moment is the bending moment resulting from the difference in the
buoyancy between the wave surface and the still water surface.
2
( ) cos2
h xLoad p x By B
L
(1.43)
Where, is the specific weight of water and B is the breadth of the vessel
0 0
2 2cos sin
2 2 2
x x h x h L xShear force Q x p x dx B dx B
L L
(1.44)
The maximum shear force, maxQ occurs when ( ) 0dQ dx p x , at 4x L and 3 4x L
max
max
42 2 4
3 44
h L BhLQ x L B
BhLQ x L
(1.45)
0 0
2 2
2 20
2sin
2 2
2 2cos 1 cos
8 8
x x
x
h L xBending moment M x Q x dx B dx
L
BhL x BhL xM x
L L
(1.46)
L
0 X
YB
h
46
The maximum bending moment, maxM occurs when 0dM dx Q , hence, sin 2 0x L
, at 0x , 2x L , and x L
2 2
max 2 22 2
8 4
BhL BhLM x L
(1.47)
1.15. Wave bending moment for a diamond-shaped vessel on a sine wave
The wave surface equation is given by:
2
cos2
h xy
L
(1.48)
The breadth is a function of x as follows:
Figure 45
20
22
22
Bx LB x x
LBx L
B x B x LL
(1.49)
2. . cos
2
h xLoad p x B x y B x
L
(1.50)
L
0 X
YB
h
B
L/2 L/2
47
0 0
2
2 2cos 0
2 22 2
2 cos2 2 2
x x
x
L
B h x LShear force Q x p x dx x dx x
L LL Bx h x L
Q x Q B dx x LL L
(1.51)
Finally, we obtain:
2
max 212
BhLM
(1.52)
max max
1
3M Diamond M Box (1.53)
1.16. Wave bending moment and shear force for a ship on a sine wave
1.16.1. Wave bending moment
The magnitude of the wave bending moment for a ship shape will be between the wave
bending moments of diamond and box shapes, such that:
Diamond Ship BoxM M M (1.54)
Let, . 1 .Ship Box DiamondM M M , where = factor determined by the following:
0.5BC Diamond & 1BC Box
1. 0.5 1BC Ship , then, 2 0.5BC , 0.5BC
For diamond shape: 0.5BC , and 0
For box shape: 1.0BC , and 1
Therefore,
2 0.5 1 2 1 .
2 0.5 1
Ship B Box B Diamond
Box B B Diamond
M C M C M
M C C M
(1.55)
and
ax 4 1M B MaxM Ship C M Diamond (1.56)
48
1.16.2. Shear force
For a box shape:
max
max
21 cos
2
2sin
M xM x
L
dM x xQ x M
dx L L
(1.57)
The maximum shear force, maxQ occurs when 0dQ dx p x , at 4x L and 3 4x L
max max
max max
Q M for box shapeLC
Q M for ship shapeL
(1.58)
For passenger ship: C = 3.5
For cargo ship: C = 3.75
For tankers: C = 4.3
1.17. Examples
Ex. 1:
Consider a vessel of constant rectangular cross section, 140m long, 20m beam and 13m deep,
with total mass 25 830 tones, 20 830 of which is uniformly spread over the length and the rest
distributed uniformly over the central 10m. Calculate the bending moments and shearing
forces.
Figure 46
49
Solution:
The mass distribution will be constant at 20830
148.8140
t/m over the entire length; plus
5000500
10 t/m over the central 10m.
The still water buoyancy distribution will be constant at 25830
184.5140
t/m
The maximum bending moment (BM) will be at amidships.
The BM amidships due to buoyancy = 25830 140 9.81
44342 4 1000
MNm
The BM amidships due to weight = 20830 140 5000 5 9.81
36372 4 2 2 1000
MNm
The net BM amidships = 4434 – 3637 = 797 MNm sagging moment
Suppose now that the vessel is poised on a sinusoidal wave equal to its own length and of
height 0.50.607( )L , that is a wave 140m long and 7.2 m high. The wave height at any point
above the still water level, when the wave crests are at the ends, is:
Figure 47 wave load
7.2 2
cos2 140
xh
(1.59)
The wave buoyancy distribution = 3 2 21.025 20 3.6 9.81 10 cos 0.724cos
140 140
x x
MN/m
By integration, the wave shearing force is:
0
2 140 20.724cos d 0.724 sin
140 2 140
x x xQ x
Wave crests at the ends (Sagging) Wave crest amidships (Hogging)
50
Integrating again the bending moment is:
2 2
2 20
140 2 140 2 140 20.724 sin d 0.724 cos 0.724 359 1 cos
2 140 4 140 4 140
x x x xM x
Putting x = 70 the wave bending moment at amidships is found to be 718MNm sagging.
With the wave crest amidships the wave moment would be of the same magnitude but
hogging. The total moments are obtained by adding the still water and wave moments,
giving:
Sagging = 797 + 718 = 1515MNm
Hogging = 797 – 718 = 79MNm
Had the mass of 5000 tones been distributed uniformly over the whole ship length the still
water bending moment would have been zero, giving equal sagging and hogging wave
bending moments of 718 MNm.
51
Figure 48
2321.4 tones2321.4 tones
4642.9 tones
148.8 t/m
500 t/m
184.5 t/m
35.7 t/m35.7 t/m
464.3 t/m
2321.4 tones
-2321.4 tones
+
-
75446.4 t.m
+
81250 t.m
Mass distribution
Still water buoyancy distribution
Net load distribution
S F D
B M D
2nd degree parabola2nd degree parabola
52
Ex. 2:
The barge shown in Figure 49 floats at a uniform draught of 1 m in sea water when empty. A
heavy weight, uniformly distributed over the middle 5 m of the barge, increases the draught
to 2 m. It may be assumed that the buoyancy curves for the barge (loaded and unloaded) and
the weight distribution of the unloaded barge are constant over the parallel length of the
barge, decreasing linearly to zero at the ends. Draw the curves of weight, buoyancy, load,
shear force and bending moment for the barge loaded and at rest in salt water. Find the
locations and maximum values of shear force and bending moment.
Figure 49
Solution:
Weight of empty barge emptyW = 5 3
2 14 5 1 1.025 87.1252
tons
32 14 87.125 5.125
2empty
empty empty
ww w
t/m
Weight of loaded barge loadedW = 5 3
2 14 5 2 1.025 2 87.125 174.252
tons
174.25 87.12517.425
5heavy weightw
t/m
314 2 174.25
2
bb
buoyancy load b = 10.25 t/m
Maximum shear force maxQ = 30.75 tons, at 7.5m from ends.
Maximum bending moment maxM = 7.6875 8 23.0625 4.75 30.75 1.25 132.609 t.m
(sagging), at amidships.
3.0 m 7.0 m 3.0 m7.0 m 5.0 m
53
Figure 50
5.125 t/m
22.55 t/m
10.25 t/m
Weight diagram
Buoyancy diagram
5.125 t/m
-12.3 t/m30.75 t
23.0625 t7.6875 t4.8 m
1.3 m
8.0 m
Net load diagram
30.75 t
7.6875 t +
--7.6875 t
-30.75 t
2 nd
Shear force diagram
7.6875 t.m
94.1719 t.m
132.609 t.m
+ +
3 rd
2 nd
2 nd
Bending moment diagram
3.0 m 7.0 m 7.0 m 3.0 m
5.0 m
54
Ex. 3:
The sectional area curve of a box-shaped barge of length 60m and breadth 12m shows that
the area of cross section over 30m at mid-length is constant and then diminishes uniformly to
zero at each end. The barge carries a uniformly distributed weight (including its own) of 7
t/m over its entire length. The barge may carry the following loads:
(a) Two uniform loads of 16 t/m each 15m long and located at the vessel’s ends.
(b) A uniform load of 16 t/m over the middle half of the vessel’s length.
Find the magnitude of bending moment amidships for each of the above two loading
conditions. Also sketch the curves of weight, buoyancy, load, shearing force and bending
moment.
Solution:
Figure 51
Total weight totalW = 2 15 23 30 7 = 900 tons
152 30 900
2
bb
Buoyancy load b = 20 t/m
(a) Figure 52
Bending moment amidships maxM = 13 15 7.5 3 15 (30 7.5) 0.5 15 20 25 3300
t.m (Hogging)
(b) Figure 53
Bending moment amidships maxM = 63.375 (15 9.75 3) 18.375 (30 5.25 3) 45 7.5
= 300 t.m (Sagging)
15.0 m 15.0 m 15.0 m15.0 m 12.0 m
55
Figure 52
Weight diagram
15.0 m
23 t/m23 t/m
7 t/m
30.0 m 15.0 m
20 t/m
Buoyancy diagram
Net load diagram
13 t/m
23 t/m
3 t/m3 t/m
23 t/m
150 t
750 t
900 t
-345 t
-555 t
-900 t
+
-
345 t
22.5 m
210 t
22.5 m
600 t150 t
20.0 m
2 nd
2 nd
345 t
150 t
-2587.5 t.m
20.0 m
-27000 t.m
-16087.5 t.m
-7762.5 t.m
-20250 t.m
27000 t.m
750 t.m
14250 t.m
3000 t.m
18000 t.m
-
Shear force diagram
Bending moment diagram
2 nd
2 nd
3 rd
3 rd
2 nd
2 nd
56
Figure 53
Weight diagram
15.0 m
23 t/m
7 t/m
30.0 m 15.0 m
20 t/m
Buoyancy diagram
Net load diagram
13 t/m
7 t/m3 t/m
105 t
22.5 m
690 t
22.5 m
600 t150 t
20.0 m
105 t
150 t
20.0 m
Shear force diagram
Bending moment diagram
7 t/m
13 t/m
7 t/m
5.25 m
9.75 m
18.375 t
63.375 t
90 t
18.375 t
-45 t
10.5 m
45 t
-18.375 t
2 nd2 nd
+
-
+
--
+
300 t.m
-64.313 t.m
-37.5 t.m
-183.75 t.m
-128.625 t.m
5.63 m
79.28 t.m
637.5 t.m
+
--
+
18.375 t
63.375 t
2 nd2 nd
3 rd3 rd
57
Ex. 4:
The ordinates for the curve of buoyancy for the fore body of a ship at intervals of 35 ft
commencing from the F.P. are 0, 6, 14, 24, 34, 41 and 43 tons/ft. The weight distribution
from amidships to 70 ft forward is 25 tons/ft and from that point to the F.P. is 20 tons/ft.
The shearing force (S.F.) and the bending moment (B.M.) at amidships are desired.
Solution:
The maximum shearing force will be at the point marked A in Figure 54, where there is
equality of weight and buoyancy.
Figure 54
Buoyancy (tons/ft) S.M. Function Lever Function (F.P.) 0 1 - 6 -
6 4 24 5 120 14 2 28 4 112 24 4 96 3 288 34 2 68 2 136 41 4 164 1 164
(Amidships) 43 1 43 0 - 423 820
Buoyancy (F.P. to amidships) = 1
423 353
= 4935 tons
L.C.B. = 820
35423
= 67.8 ft forward amidships
Moment of buoyancy about amidships = 4935 67.8 = 334590 ft.tons
58
Weight (tons) Lever (ft) Moment (ft.tons) 25 × 70 = 1750 35 61250 20 × 140 = 2800 140 392000
4550 453250
Shearing force at amidships = 4550 – 4935 = –385 tons
Bending moment at amidships = 334590 – 453250 = –118660 ft.tons (hogging)
The shear force and bending moment at any section can be obtained in a similar manner.
1.18. Correction for changes in weight (Influence lines)
It is important to be able to calculate the effect of the addition or the removal of weight on the
hull girder bending moment. A useful technique for this is to construct an influence line
diagram. An influence line shows the effect on the maximum bending moment of the addition
of a unit weight at any position x along the ship’s length. Two influence lines are normally
drawn, one for the maximum hogging and one for the maximum sagging conditions.
The weight P is added at a distance Px forward of amidships (see Figure 55). As a results,
the ship will undergo a parallel sinkage v and a non-dimensional trim t .
P F
W L
P x xP total trimv and t
gA L gI
(1.60)
Where:
WA = area of waterplane
LI = longitudinal moment of inertia of waterplane about CF
P FP x x = moment causing trim
LI
L
= moment causing unit trim
59
Figure 55 Influence line for changes in max
M due to added weight
x is positive forward of amidships and the change in waterplane is assumed small.
Let R denote the position of maximum bending moment, maxM , located at a distance Rx
from amidships. The total change in maxM can be determined by taking moments about R
(from the right hand side):
1. Moment of added weight:
P P RM P x x (1.61)
2. Moment of buoyancy of parallel sinkage v :
Rv R
W
PM gv
A
MM (1.62)
Where:
60
02
FP
R z d M = first moment of RA about R
RA = area of waterplane forward of R
3. Moment of buoyancy of wedge forward of R :
0
2FP
t R F R R R F
P FR R R F
L
M g z x x t d g I x x t
P x xI x x
I
M
M
(1.63)
Where:
2
02
FP
RI z d = second moment of area RA about R
Then,
max ,R P FR R R F P R
W L
x xM P I x x x x Sagging positive
A I
MM (1.64)
The term between pointed brackets exists only if P is forward of R .
The influence lines are straight lines which cross the axis at approximately the quarter points
of the vessel. Therefore, a weight added within this length causes an increased sagging
moment and an added weight outside this length causes an increased hogging moment. To
construct an influence line diagram, maxM should be evaluated at 2L , Rx , and 2L . If a
weight is removed; then P replaces P in all of the above (see Figure 56).
Figure 56 Influence of adding and removing weight
-L/2 +L/2
X P
R
P removedP added
M /Pmax
61
By making suitable approximations, it is possible to simplify equation (1.64). If both LCF
and R are taken as being at amidships, then 0R Fx x , 2R LI I , and (0.5)0.5PR W WA xM ;
where (0.5) PWx is the distance from amidships to the centroid of the forward half waterplane.
Then, equation (1.64) becomes:
max (0.5)0.5 0.5PW P PM P x x x (1.65)
Thus, if the weight is added forward of amidships the results is:
max (0.5)2 PW P
PM x x (1.66)
This has a direct and relatively simple physical interpretation. It is 0.5 times the moment of
P about the centroid of the half waterplane area.
In addition to their use in design, influence lines are a helpful tool for the ship operator and
are sometimes provided as a part of the loading manual. However, it should be noted that
they are intended for small weight changes only; certainly not more than 10% of the
displacement. If the change exceeds this amount, a new bending moment calculation should
be performed.
In some cases, it is desired to find the change in bending moment ( )M x (due to addition of
weight P at Px ) at some arbitrary distance x along the vessel, or perhaps along the entire
length, instead of only the change in maxM . The change in bending moment at any section is
calculated by taking the moment of buoyancy loads, due to parallel sinkage and due to trim,
about that section. The graphical illustration for the shear force and bending moment due to
added weight P is shown in Figure 57.
Figure 57 change in shear force and bending moment distribution
62
1.19. Deflection of ship hull girder
1.19.1. Types of deflection
The ship hull girder deflection can be:
a – Built-in: due to fabrication
b – Thermal: due to temperature differences
c – Loading: consists of bending deflection and shear deflection
1.19.2. Review of flexural (bending) deformations
Figure 58 Bending of beam element
For a beam element of length dx subjected to a positive moment M :
, straindl
Edx
(1.67)
Where, dl is shortening of top fiber.
dx dl dl c
dc dx
(1.68)
At top fiber:
d
d
dx
dlc
M M
O
d
63
compression,c
E E ve
(1.69)
Also,
1M c M
I EI
(1.70)
Based on Eqns (1.68) and (1.70)
2
2
1dv d d v M
dx dx dx EI
(1.71)
The slope can be written as:
M M
d dx dx AEI EI
(1.72)
The displacement v :
dv dx v dx B (1.73)
The total angle change L :
0 0
L L
L
Md dx
EI (1.74)
If p x is the load per unit length, then:
4
4
d vEI p x
dx (1.75)
1.19.3. Calculation of the bending deflection in still water
From the theory of bending, it was seen that M EI . The curvature of a beam can be
expressed in terms of the coordinates of a point on the beam such that if is the deflection
of the beam at a distance x from one end, it can be shown that:
2
2
1 d M
dx EI
(1.76)
The deflection of a beam is obtained from this formula by simply integrating twice, hence:
64
M xddx A
dx EI x
(1.77)
2M xdx Ax B
EI x (1.78)
Where A and B are any arbitrary constants and their values will depend on the end
conditions of the beam. For simple cases of loading, the value of M is given by a simple
mathematical expression, and if the beam is of uniform section, then I is constant and the
integration can be readily carried out mathematically.
For a ship hull girder, the bending moment cannot be represented by any simple
mathematical expression. In addition, the magnitude of I is not constant along the length of
the ship. Thus, it is necessary to calculate the value of the moment of inertia at a number of
sections along the ship’s length and then plot a curve of M I . This curve may have abrupt
changes in it if there are sudden changes in the section of the ship. For the ship hull girder,
the following applies:
2 2
2 2
d v d d
dx dx dx
(1.79)
Where,
v = vertical displacement of hull girder
= slope of hull girder
= deflection of hull girder
The deflection of a ship hull girder is obtained from Eqn (1.78). The integration of the M I
curve can be done graphically or, alternatively, it may be done in tabular form by dividing the
length of the ship into a sufficient large number of sections and taking the mean ordinate for
each of these divisions. The resulting curve is shown in Figure 59. The second integration
will then be carried out in the same way and the curve is also shown in Figure 59. To obtain
the deflection of the ship hull girder from this curve, it is only necessary to join the ends with
a straight line and use this line as the base from which the deflection is measured.
The constants of the integration can be evaluated as follows:
At 0, 0 0x B
65
At , 0x L
2
0 0
0L L M x
dx ALEI x
(1.80)
Then
2
0 0
1 L L M xA dx
L EI x (1.81)
0 0
L L
L
Mdx dx
EI
L
= slope of new base line for deflection = L ordinate of
0
x Mdx
EI curve,
since 0
L
L
Mdx
EI and
0
L
L Ldx L
Therefore, the deflection at any distance x along the ship’s length will be evaluated from the
following equation:
2 2
0 0 0 0
x x L L
L
M x M xxx dx dx v x x
EI x L EI x (1.82)
Figure 59 Determination of ship hull girder bending deflection
1.19.4. Examples
Ex. 1:
M I values are given for a vessel as follows ( 4MN.m/m ):
MI
dx dx
Deflection
MI
dx
Base line for deflectionMI
66
Stn 0 1 2 3 4 5 6 7 8 9 10
M I 0 0.6 1.8 2.6 2.7 3.4 4.0 2.6 1.1 0.3 0
Calculate the deflection of the keel at amidships.
L = 130 m, E = 200 2GN/m , S = 130/10 = 13 m
Solve using tabular method (trapezoidal rule)
Solution:
2 2
0 0 0 0
1 x x L LM x M xxx dx dx
E I x L I x
Table 6 Tabular integration of deflection
Stn M I Mean 1 M
dxS I Mean 2
2
1 Mdx
S I
0 0 0 0 0.3 0.15 1 0.6 0.3 0.15 1.2 0.9 2 1.8 1.5 1.05 2.2 2.6 3 2.6 3.7 3.65 2.65 5.025 4 2.7 6.35 8.675 3.05 7.875 5 3.4 9.4 16.55 3.7 11.25 6 4.0 13.1 27.8 3.3 14.75 7 2.6 16.4 42.55 1.85 17.325 8 1.1 18.25 59.875 0.7 18.6 9 0.3 18.95 78.475 0.15 19.025
10 0 19.1 97.5
at amidships =
6
2 22 2 39
1 1 10 116.55 97.5 16.55 13 97.5 13 27.2 10 m
2 200 10 2S S
E
+
=
=
+
=+
=
=
=
=
=
=
=
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+=
=
=
=
=
=
=
=
=
=
67
Ex. 2:
The mean immersed cross-sectional areas between ordinates of a ship, 300m in length,
balanced on a hogging wave, as read from the Bonjean curves are given in Table 7, together
with the mass distribution. The second moment of area of the midship section is 752 4m .
Making the assumption that the second moment of area is constant along the length, estimate
the difference in slope between ordinates number 4 and 10 and the deflection of the middle
relative to the ends.
Table 7
Ordinate 0 1 2 3 4 5 6 7 8 9 10 Area ( ) 0.5 2.5 71 302 565 633 490 243 49.5 10.5 Mass (ton) 2690 4920 6400 9260 7830 9370 9900 10670 6790 5030
Solution:
Difference in slope between ordinates number 4 and 10 =
310 4
0 0
49.252 12.263 3010.006
752 208 1000M x dx M x dx
EI
radians
Deflection of the middle relative to the ends =
45 5 10 102 2
0 0 0 0
30.295 0.5 237.12 301 10.457
2 752 208 1000M x dx M x dx
EI
m
68
ab
cd
ef
(e-c
)g
hi
j (
g-i)
kl
mn
(l-
m)
op
qr
Ord
inat
eA
rea
Buo
yanc
yM
ass
Wei
ght
Loa
dSh
ear
(Q/s
)L
ever
(x/
L)
(x/L
).(Q
10/s
)C
orre
cted
(Q
/s)
Qm
/sM
omen
t (M
/s2 )
(x/L
).(M
10/s
2 )C
orre
cted
(M
/s2 )
Mm
/s2M
dx/s
3 (
Mdx
) m/s
3
Mdx
dx/s
4
m2
MN
/mto
nM
N/m
MN
/mM
N/m
MN
/mM
N/m
MN
/mM
N/m
MN
/mM
N/m
MN
/mM
N/m
MN
/mM
N/m
00
00
00
00
00
0.5
0.00
526
900.
880
0.87
50.
437
0.21
40.
107
10.
875
0.1
0.00
20.
872
0.43
70.
009
0.42
80.
214
0.10
72.
50.
025
4920
1.60
91.
584
1.66
61.
257
0.84
32
2.45
80.
20.
005
2.45
32.
104
0.01
82.
086
1.47
10.
950
710.
714
6400
2.09
31.
379
3.14
83.
655
3.29
93
3.83
70.
30.
007
3.83
05.
251
0.02
75.
224
5.12
64.
248
302
3.03
792
603.
028
-0.0
093.
833
7.13
68.
694
43.
829
0.4
0.01
03.
819
9.08
40.
036
9.04
812
.263
12.9
4356
55.
681
7830
2.56
0-3
.121
2.26
810
.178
17.3
525
0.70
80.
50.
012
0.69
511
.352
0.04
511
.307
22.4
4130
.295
633
6.36
593
703.
064
-3.3
01-0
.943
10.8
3227
.856
6-2
.593
0.6
0.01
5-2
.608
10.4
100.
054
10.3
5633
.272
58.1
5149
04.
927
9900
3.23
7-1
.690
-3.4
388.
632
37.5
887
-4.2
830.
70.
017
-4.3
006.
972
0.06
36.
908
41.9
0495
.739
243
2.44
310
670
3.48
91.
046
-3.7
605.
024
44.4
168
-3.2
370.
80.
020
-3.2
573.
211
0.07
23.
139
46.9
2814
0.15
549
.50.
498
6790
2.22
01.
723
-2.3
761.
947
47.9
019
-1.5
150.
90.
022
-1.5
370.
835
0.08
10.
754
48.8
7518
8.05
710
.50.
106
5030
1.64
51.
539
-0.7
450.
377
49.0
6310
0.02
41
0.02
40
0.09
00.
090
049
.252
237.
120
Tab
le 8
69
1.20. Geometric properties of an area
1.20.1. Centroid of an area
The x and y coordinates defining the location of the centroid C of an area are determined
using the formulas (see Figure 60):
,A A
A A
xdA ydAx y
dA dA
(1.83)
Figure 60
Often an area can be sectioned or divided into several parts having simpler shapes. Provided
the area and location of the centroid of each of these “composite shapes” are known, then
Eqn (1.83) can be written as:
,xA yA
x yA A
(1.84)
Here x and y represent the coordinates for the centroid of each composite part, and A
represents the sum of the areas of the composite parts or simply the total area.
1.20.2. Moment of inertia for an area
Moment of inertia for an area about the x and y axes shown in Figure 61, is defined as:
2 2,x yA AI y dA I x dA (1.85)
Polar moment of inertia of an area about the pole O or z axis can be calculate also as (Figure
61):
70
2O x yA
J r dA I I (1.86)
The relationship between OJ and xI , yI is possible since 2 2 2r x y .
Figure 61
1.20.3. Product of Inertia for an Area
The product of inertia for the area A shown in Figure 62 is defined as:
xy AI xydA (1.87)
Figure 62
The product of inertia may be positive, negative, or zero, depending on the location and
orientation of the coordinate axes. For example, the product of inertia xyI for an area will be
zero if either the x or y axis is an axis of symmetry for the area (see Figure 63).
71
Figure 63
1.20.4. Parallel-Axis Theorem for an Area
If the moment of inertia for an area is known about a centroidal axis ( 'xI , 'y
I ), the moment of
inertia of the area about a corresponding parallel axis ( xI , yI ) can be determined using the
parallel-axis theorem, as (Figure 64):
' '
2 2,x y y xx yI I Ad I I Ad (1.88)
The polar moment of inertia about an axis perpendicular to the x - y plane and passing
through the pole O ( z axis) will be (Figure 64):
2O CJ J Ad (1.89)
Figure 64
The form of each of the above equations states that the moment of inertia of an area about an
axis is equal to the area’s moment of inertia about a parallel axis passing through the
“centroid” plus the product of the area and the square of the perpendicular distance between
the axes.
72
The product of inertia of the area with respect to the x - y axis will be (Figure 64):
' 'xy x yx yI I Ad d (1.90)
Where, ' 'x yI is the product of inertia of the area with respect to the centroidal axis. It is
important that here the algebraic signs for xd and yd be maintained when applying Eqn
(1.90).
1.20.5. Moments of Inertia for an Area about Inclined Axes
The moments and product of inertia 'xI , 'y
I and ' 'x yI for an area with respect to a set of
inclined 'x and 'y axes can be calculated when the values for , xI , yI and xyI are known.
As shown in Figure 65, the coordinates to the area element dA from each of the two
coordinate systems are related by the transformation equations:
Figure 65
'
'
cos sin
cos sin
x x y
y y x
(1.91)
Using these equations, the moments and product of inertia about the 'x and 'y axes become:
73
'
'
' '
cos 2 sin 22 2
cos 2 sin 22 2
sin 2 cos 22
x y x yxyx
x y x yxyy
x yxyx y
I I I II I
I I I II I
I II I
(1.92)
1.20.6. Principal Moments of Inertia
The axes about which the moments of inertia for an area, 'xI and 'y
I , are maximum and
minimum is called the principal axes of inertia for the area, and the corresponding moments
of inertia with respect to these axes are called the principal moments of inertia.
The angle p , which defines the orientation of the principal axes for the area, can be
found by differentiating the first of Eqn (1.92) with respect to and setting the result equal
to zero. Thus:
'
2 sin 2 2 cos 2 02
x yxxy
dI I II
d
(1.93)
Therefore, at p :
tan 2
2xy
p
x y
I
I I
(1.94)
This equation has two roots, 1p and
2p , which are 90o apart and so specify the inclination
of each principal axis.
The sine and cosine of 1
2 p and 2
2 p can be obtained from the triangles shown in Figure 66,
which are based on Eqn (1.94). If these trigonometric relations are substituted into the first or
second of Eqn (1.92) and simplified, the result is:
2
2maxmin 2 2
x y x yxy
I I I II I
(1.95)
74
Figure 66
If the above trigonometric relations for 1p and
2p are substituted into the third of Eqn (1.92)
, it will be seen that ' 'x yI = 0; that is, the product of inertia with respect to the principal axes is
zero. Since it was indicated in Section 1.20.3 that the product of inertia is zero with respect to
any symmetrical axis, it therefore follows that any symmetrical axis and the one
perpendicular to it represent principal axes of inertia for the area.
1.20.7. Composite Areas
Many cross-sectional areas consist of a series of connected simpler shapes, such as
rectangles, triangles, and semicircles. In order to properly determine the moment of inertia of
such an area about a specified axis, it is first necessary to divide the area into its composite
parts and indicate the perpendicular distance from the axis to the parallel centroidal axis for
each part.
1.21. Section modulus calculation
In most cases the critical hull girder cross section will be that section which contains the least
amount of effective material, that is, the section containing the largest hatches or other
openings. It depends also, on the distance of these from the neutral axis.
In general, the net sectional areas of longitudinal members are to be used in the section
modulus calculation. Small isolated openings need not be deducted provided the openings
and the shadow area breadths of other openings in any one transverse section do not reduce
the section modulus by more than a few percent.
75
The two quantities to be calculated are the position of the neutral axis of the section and the
moment of inertia NAI about the neutral axis. The distance of the neutral axis above the keel
is:
i iNA
i
a hh
a
(1.96)
Where, ia is the area and ih the height of neutral axis above the keel for element i .
From the parallel axis theorem, the moment of inertia about the neutral axis is:
2NA xx NAI I Ah (1.97)
Where:
NAI = moment of inertia of the entire ship section about its neutral axis
2xx ii i iI i a h = moment of inertia about a horizontal axis xx passing by the keel
iii = moment of inertia of element i about its local neutral axis
iA a = total area
NAh = distance from xx to global neutral axis
It will be noted that the moment of inertia of what is called horizontal material about its own
neutral axis is sufficiently small to be negligible. The calculation is usually carried out for
one side of the ship only, and therefore the results have to be multiplied by two, as will be
illustrated in the following examples.
76
Ex. 1:
Figure 67
Table 9 sectional modulus calculations
(1) (2) (3) (4) (5)=(3)×(4) (6)=(5)×(4) (7) (8) (9)=(3)×(8)2 (10)
Items Scantling i
a i
y i i
a y 2i ia y xi
i i
x 2i ia x yi
i
Deck d
b t d
bt D dbDt 2
dbD t 0 ( ) 2B b 2( ) 4
dbt B b 3 12
db t
Side s
D t s
Dt 2D 2 2s
D t 3 4s
D t 3 12s
D t 2B 2 4s
Dt B 0
I. bottom ( 2)i
B t 2i
Bt h 2i
Bht 2 2i
Bh t 0 4B 3 32i
B t 3 96i
B t
O. bottom ( 2)o
B t 2o
Bt 0 0 0 0 4B 3 32o
B t 3 96o
B t
S. girder sgh t
sght 2h
2 2sg
h t 3 4sg
h t 3 12sg
h t 4B 2 16
sgB ht 0
C. girder ( 2)cg
h t 2cg
h t 2h 2 4
cgh t 3 8
cgh t 3 24
cgh t 0 0 0
Sum i
a i i
a y 2i ia y xi
i 2
i ia x yi
i
Total area = 2 ia
Height of neutral axis above the keel, i iNA
i
a yh
a
Moment of inertia about neutral axis, 2 22NA xi i i NA iI i a y h a
Moment of inertia about center line, 22CL yi i iI i a x
b
hNA
DN A
B/2
h
B/4 B/4
X
Y
BL
td
t s
t it o t sgt cg
77
Section modulus (deck), NAD
NA
IZ
D h
Section modulus (bottom), NAB
NA
IZ
h
Section modulus (side), / 2
CLS
IZ
B
Ex. 2:
Figure 68 example of longitudinally effective material
78
Table 10 sectional modulus calculations
Total area, 22 0.5706 1.141miA a
Height of neutral axis, 1.9095
3.3460.5706
i iNA
i
a yh
a
m
42 12.413 0.440 25.706 mxxI
2 2 42 12.413 0.440 0.5706 3.346 12.926 mNA xx i NAI I a h
312.9262.286m
9.0 3.346NA
DNA
IZ
D h
312.9263.863m
3.346NA
BNA
IZ
h
79
Ex. 3:
Figure 69
Table 11 sectional modulus calculations
Items Scantling
(m) ia
iy
i ia y 2
i ia y xii
ix 2
i ia x yii
Upper hatch side girder - 0.008 13 0.104 1.352 - 4 0.128 -
Lower hatch side girder - 0.003 10 0.03 0.3 - 4 0.048 -
Strength deck plating 6×0.022 0.132 13 1.716 22.308 - 7 6.468 0.396
2nd deck plating 6×0.016 0.096 10 0.96 9.6 - 7 4.704 0.288
Side plating 11.5×0.014 0.161 7.25 1.1673 8.4626 1.77 10 16.1 -
Bilge 1.5×0.016 0.024 0.75 0.018 0.0135 0.0045 10 2.4 -
Inner bottom plating 10×0.018 0.18 1.5 0.27 0.405 - 5 4.5 1.5
Bottom plating 10×0.02 0.2 0 0 0 - 5 5 1.6667
Center girder (1/2) 1.5×0.012 0.018 0.75 0.0135 0.0101 0.0034 0 0 -
Sum 0.822 4.2788 42.4512 1.7822 39.348 3.8507
Total area = 22 2 0.822 1.644mia
Height of neutral axis above the keel, 4.2788
5.205m0.822
i iNA
i
a yh
a
Moment of inertia about neutral axis,
2 2
2
4
2
2 1.7822 42.4512 5.205 0.822
43.923m
NA xi i i NA iI i a y h a
6m
10mX
Y
22mm
14mm
18mm
20mm24mm
16mm
A=80cm2
A=30cm2
14mm3m
8.5m
1.5m16mm
80
Moment of inertia about center line,
2 42 2 3.8507 39.348 86.397 mCL yi i iI i a x
Section modulus (deck),
343.9235.6349m
13 5.205NA
DNA
IZ
D h
Section modulus (bottom), 343.9238.438m
5.205NA
BNA
IZ
h
Section modulus (side),
386.3978.6397 m
/ 2 10CL
S
IZ
B
1.22. Horizontal bending
Unless the ship is moving head on into long-crested seas, longitudinal bending in a horizontal
plane can arise. Horizontal bending will arise when a ship is moving obliquely across waves.
Under this circumstance, horizontal forces are generated which can result in horizontal
acceleration of the masses making up the total mass of the ship. There will be no gravitational
components of force in this case. Horizontal forces can only be evaluated by a detailed study
of the hydrodynamic forces and the motions such as yawing and swaying which generate
acceleration. In general the horizontal bending moments created are of much less magnitude
than the vertical bending moments.
1.23. Response of the structure
Having determined the shear forces and bending moments it is necessary to determine the
response of the structure to these forces and moments, which simply means the calculation of
the stresses in the structure and if required the overall deflection.
The normal stress distribution on a given cross section of a beam subjected to unsymmetrical
bending is in the form:
2 2
y x x xy x y y xy
x y xy x y xy
M I M I M I M Ix y
I I I I I I
(1.98)
Where, x and y are the perpendicular distances to the centroidal y -axis and x -axis,
respectively. xI and yI are the centroidal moments of inertia of the beam cross section with
81
respect to the x and y axes, respectively. xyI is the centroidal product of inertia of the beam
cross section. xM and yM are the bending moments about the x and y axes, respectively.
is the normal stress in the beam due to bending.
If the coordinate system is chosen to give a product moment of area equal to zero, the
previous formula simplifies to:
y x
y x
M Mx y
I I (1.99)
If additionally the beam is only subjected to bending about one axis, the formula simplifies
further:
x
x
M My
I Z (1.100)
Where, /Z I y is the section modulus. The maximum stresses will occur when y is a
maximum that is at the top and bottom of the section.
This relationship was derived for beams subject to pure bending and in which plane sections
remained plane. Although a ship's structure is much more complex, applying the simple
formula has been found to give reasonable results.
Figure 70 indicates the ship hull under vertical bending moment VM .
For the sagging condition:
Maximum tension stress will be, V NAt
NA
M h
I (Bottom)
Maximum compression stress will be, V NA
cNA
M D h
I
(Deck)
For the hogging condition:
Maximum tension stress will be, V NA
tNA
M D h
I
(Deck)
Maximum compression stress will be, V NAc
NA
M h
I (Bottom)
82
Figure 71 indicates the ship hull under horizontal bending moment HM . One side will have a
maximum compression and the other will have a maximum tension. Since the ship is
symmetric about the center line, then the magnitude of both of the maximum compression
and the maximum tension stresses are equal.
2H
t cCL
M B
I
(a) Sagging (b) Hogging
Figure 70 Vertical bending moment
Figure 71 Horizontal bending moment
MV
MV
N
A
Compression(- ve)
Tension(+ ve)
c
c
t
t
c
c
Deck
Bottom
Compression(- ve)
Side
Tension(+ ve)
Sidet
tt
c
MV
MV
N
A
Tension(+ ve)
t
t
Deck
t
t
Compression(- ve)
SideTension(+ ve)
Side
c
c
Compression(- ve) c
c
Bottom
t
c
Compression(- ve)
Deck
Tension(+ ve)
Deck
Compression(- ve)
Side
Tension(+ ve)
Side
t
t
t
t
t
c
c
c
c
c
C
L
MH
MH
83
1.24. Combined vertical and horizontal bending
Vertical bending assumes that the ship is upright and that the bending moment is in the ship’s
vertical plane. If the ship is at an angle of heel due to rolling, it will also be subjected to
horizontal bending, that is, a bending moment yM acting in the ship’s horizontal plane (see
Figure 72). For this bending moment, the neutral axis is the ship’s vertical centerline.
Figure 72 Neutral axis with simultaneous horizontal and vertical bending
Let us first take the case in which yM is entirely due to inclination of the vessel, say to an
angle . In this case yM and xM are directly related, being components of the total bending
moment VM (which acts in the true vertical plane):
sin
cos
y V
x V
M M
M M
(1.101)
If x and y are the coordinates of any point in the cross section and, NAI and CLI are the
moments of inertia about the horizontal axis in the upright condition and about the centerline,
respectively, then the stress at ,x y is:
cos sinV V
x yNA CL
M My x
I I
(1.102)
When = 0 it follows that
cos sin
0NA CL
y x
I I
(1.103)
84
or
tanNA
CL
Iy x
I (1.104)
This gives the equation of the neutral axis in the inclined condition. It will be seen that this is
inclined to the neutral axis in the upright condition at an angle given by:
tan tanNA
CL
Iy
x I (1.105)
If the vessel were such that NA CLI I , then tan tan , and the neutral axis would remain
horizontal. In general this is not so, CLI being larger than NAI , and so the neutral axis is
inclined to the horizontal.
Referring to Eqn (1.102), the angles of heel at which the greatest and least stresses occur may
be obtained by putting 0d d , hence:
sin cos
0V V
NA CL
M Mdy x
d I I
(1.106)
or
tan NA NA
CL CL
I Zx
y I Z (1.107)
The greatest and least stresses will also be associated with the maximum values of y and x
and this means that these stresses will occur at the corners of the section.
In practice the horizontal and vertical bending moments are not so directly coupled and do
not necessarily occur simultaneously. Their relationship varies with different sea conditions
and depends mainly on ship heading.
85
1.24.1. Examples:
Ex.1:
For the given mid ship section indicated in Figure 73:
Vertical bending moment = VM
Horizontal bending moment = HM
Angle of heel 0
Figure 73
Solution:
V Hx y
NA CL
M y M x
I I
1
2V NA H
NA CL
M D h M B
I I
,
2
2V NA H
NA CL
M D h M B
I I
3
2HV NA
NA CL
M BM h
I I ,
4
2HV NA
NA CL
M BM h
I I
5 0 ,
6V NA
NA
M D h
I
,
7
2H
CL
M B
I
B/2B/2
N A
h NA
D
MH
MV
c
c
t
t
12
3 4
5
6
7
C
L MH
MV
y
x
86
Ex.2:
For the given mid ship section indicated in Figure 74:
Vertical bending moment = VM
Horizontal bending moment = HM
Angle of heel =
Figure 74
Solution:
cos sinx V HM M M , cos siny H VM M M
yxx y
NA CL
M xM y
I I
1
2yx NA
NA CL
M BM D h
I I
,
2
2yx NA
NA CL
M BM D h
I I
3
2yx NA
NA CL
M BM h
I I ,
4
2yx NA
NA CL
M BM h
I I
5 0 ,
6x NA
NA
M D h
I
,
7
2y
CL
M B
I
B/2
B/2
N
A
hNA
D
1
2
3
4
5
6
7
C
L
y
x
MV
MHX
Y
M cosV
M sinV
M cosH
M sinH
MV
MH
=f(M )c
V
=f(M )c
H
=f(M )t
V
=f(M )t
H
=f(M )c V
=f(M )t H=f(M )t V
=f(M )c H
87
Ex.3:
For the ship indicated in example 3, page 96:
L = 130 m SWM = 200 MN.m (sagging)
B = 20 m WM = 100 MN.m
D = 13 m E = 206.9E9 pa
= 15o
Two strain gauges of base length of 25.4 cm are located on the extreme outer points of the
stringer plates (port and starboard sides). Calculate the extension in each strain gauge.
Solution:
Figure 75
Total moment, 200 100 300SW WM M M MN.m (sagging)
1
cos sin 242.4383 MPaNA
NA CL
M D h M B
I I
2
cos sin 260.4124 MPaNA
NA CL
M D h M B
I I
,l
El E
11 0.052 mm
l
E
, 22 0.074 mm
l
E
N
A
C
L
MM
cos
M sin
Strain gauge [1]
Strain gauge [2]
88
1.25. Hull girder shear stress
1.25.1. Shear stress in open sections
Figure 76 shows a thin-walled symmetric box girder subjected to a vertical shear force Q .
From elementary beam theory it is known that:
dM Qdx (1.108)
Due to this change in the bending moment, the bending stresses A and B on the two faces
of the differential segment dx are not equal.
If we isolate a portion of the differential segment by making two cuts, one at the centerline
and the other at an arc length s from the centerline, the imbalance in the longitudinal normal
stress forces must be counterbalanced by longitudinal shear stress forces across the cut
sections.
However, because of symmetry, there can be no shear stress in the center plane cut and hence
the balancing force must come entirely from the shear stress at the other cut.
Figure 76 Free body diagram for transverse shear
89
Longitudinal equilibrium therefore states that:
0 0
s s
B At dx t ds t ds (1.109)
Substituting My I on both faces:
0 0
s sB AM M dM
t dx yt ds yt dsI I
(1.110)
Substituting dM Qdx gives:
0
sQt yt ds
I (1.111)
For convenience, let the integral on the right hand side be assigned as m , so:
0
sm yt ds (1.112)
Note that m is the first moment about the neutral axis of the cumulative section area starting
from the open end (shear-stress-free end) of the section.
Substituting for m in Eqn (1.111) and solving for :
Qm
tI (1.113)
The product t has special significance in the torsion of thin-walled sections, and has some
analogies to the flow of an ideal fluid within a closed pipe. This product is therefore referred
to as the shear flow and is assigned the symbol q :
Qm
q tI
(1.114)
Since Q and I are constants for the entire section, the shear flow is directly proportional to
m . In fact, the ratio Q I may be regarded as simply a scaling factor, and once the
distribution of m has been calculated the shear flow distribution is identical to it but with
different units. Still another advantage of q is that its value does not vary abruptly with local
thickness changes, as does .
The calculation of m is illustrated in Figure 77 for an idealized hull girder. For horizontal
portions the moment arm y is constant and m therefore increases linearly with arc length.
90
This occurs in the deck and bottom if there is no camber or deadrise. For instance, in the
deck:
Figure 77 calculation of moment term m by integrating along branches
1 1 andD A Dm s gt s m m b gt b
In the side shell m is parabolic:
2 22 2 2 20
1
2
s
A s A sm s m yt ds m gs s t
The integration is always commenced at the open end of any branch because that simplifies
the computation. It is best to stop at the neutral axis and to finish that branch by starting from
the other end. If this is not possible then the integration can proceed across the neutral axis,
provided that a negative moment arm is used for all points on the other side.
As shown in Figure 78 this need not be on the centerline; it may be at the edge of a hatch or
other opening. If an imaginary cut where made at point C, the shear force at that point would
have to balance the net imbalance in bending stress forces in the second deck and all plating
above it; then:
C A Bm m m (1.115)
Since q is directly proportional to m :
C A Bq q q (1.116)
91
Figure 78 conservation of shear flow at corners and branch points
This illustrates one of the reasons for the use of the term shear flow. At any junction or
branch point, the increment in the shear flow is equal to the flow contributed or taken away
by the branch, as shown in Figure 79 and Figure 80.
Figure 79 sample diagrams indicating shear flow through some beam sections
Figure 80 sample diagram indicating shear flow through an idealized mid ship section
92
It should be noted that since deck and side plating may differ, this rule of continuity of shear
flow does not hold for . Figure 81 illustrates how changes with changes in thickness.
Figure 81 change of due to change of thickness
1.25.2. Examples:
Ex. 1:
Determine the distribution of the shear stress over a rectangular cross section of a beam.
Solution:
Figure 82
221
2 2 2 2 4
d d b dm A y b y y y
3
12
bdI
N A
d/2
d/2 y
A
ymax
y
b
93
22
22
3 3
2 4 6
412
b dQ y
Qm Q dy
bdtI bdb
This result indicates that the shear-stress distribution over the cross section is parabolic
(Figure 82).
at top and bottom 02
dy
max
3 30 neutral axis
2 2 avg
Qy
bd
where, avg
Q
bd is the average shear stress over the entire rectangular cross section.
It is important to realize that max also acts in the longitudinal direction of the beam (Figure
83).
Figure 83
Ex. 2:
A steel wide-flange beam has the dimensions shown in Figure 84. If it is subjected to a shear
of Q = 80 kN, plot the shear-stress distribution acting over the beam’s cross-sectional area.
Figure 84
94
Solution:
Since the flange and web are rectangular elements, then like the previous example, the shear-
stress distribution will be parabolic and in this case it will vary in the manner shown in Figure
85. Due to symmetry, only the shear stresses at points B, B´ and C have to be determined.
Figure 85
3 3 2 6 41 10.015 0.2 2 0.3 0.02 0.3 0.02 0.11 155.6 10 m
12 12I
For point B´ , 'Bt = 0.3 m, thus (Figure 86):
'
' ' 3 30.11 0.3 0.02 0.66 10 mB
m y A
So,
'
'
'
3 3
6
80 10 0.66 101.13 MPa
155.6 10 0.3B
BB
Qm
It
For point B , Bt =0.015 m and 'B Bm m , hence (Figure 86):
3 3
6
80 10 0.66 1022.6 MPa
155.6 10 0.015B
BB
Qm
It
Figure 86
95
For point C, Ct =0.015m and 'A is composed of two rectangles (Figure 87):
' ' 3 30.11 0.3 0.02 0.05 0.015 0.1 0.735 10 mCm y A
Thus,
3 3
max 6
80 10 0.735 1025.2 MPa
155.6 10 0.015C
CC
Qm
It
Figure 87
Ex. 3:
The following beam’s cross-sectional area is subjected to a shear Q of 1000 Kg. Plot the
shear flow distribution acting over
Figure 88
Solution:
Table 12
Item Scantling
(Cm) ia
iy
i ia y 2
i ia y xii
Upper flange 20×1 20 32.5 650 21125 1.667 Web 30×1.5 45 17 765 13005 3375 Attached plate 40×2 80 1 80 80 26.667
145 1495 34210 3403.333
20
40
30
96
Figure 89 shear flow distribution
149510.31
145NAh Cm
2 2 2 43403.333 34210 145 10.31 22199.37 CmNA xi i i NA iI i a y h a
1 2 5 6 0q q q q
31 32
100010 1 32.5 10.31 10 Kg/Cm
22199.37
Q Qq q m AY
I I
3 31 32 2 10 20 Kg/Cmq q q
7 3
1000 32 10.3132 10.31 1.5 35.885 Kg/Cm
22199.37 2q q
45 46
100020 2 10.31 1 16.776 Kg/Cm
22199.37q q
4 45 46 2 16.776 33.552 Kg/Cmq q q
7 4
1000 10.31 210.31 2 1.5 35.885 Kg/Cm
22199.37 2q q
(Check)
23 3 20 20 20 1 Kg/CmF q
23 3 1.5 20 1.5 13.33 Kg/CmW q
24 4 40 33.552 40 0.839 Kg/CmP q
Q=1000 Kg
123
4
56B L
N A
q31q32
q3
q4
q45q46
7
23
456
7
1
10 kg/cm
16.776 kg/cm
20 kg/cm
35.885 kg/cm
33.552 kg/cm
97
24 4 1.5 33.552 1.5 22.368 Kg/CmW q
27 7 1.5 35.885 1.5 23.923 Kg/Cmq
Figure 90 shear stress distribution
Ex. 4:
The following mid ship section is subjected to a shear Q of 1000 t. Plot the shear flow
distribution acting over
Figure 91
Solution:
Table 13
Item Scantling (m) i
a i
y i i
a y 2i ia y xi
i
Strength deck plating 4×0.018 0.072 12 0.864 10.368 - 2nd deck plating 4×0.014 0.056 9 0.504 4.536 - Side plating 12×0.015 0.18 6 1.08 6.48 2.16 Inner bottom plating 10×0.016 0.16 1.5 0.24 0.36 - Bottom plating 10×0.022 0.22 0 0 0 - 0.688 2.688 21.744 2.16
3F
3W
4W
4P
13.33 kg/cm2
23.923 kg/cm2
22.368 kg/cm2
1.0 kg/cm2
0.839 kg/cm2
23
4 56
7
1
4
4
10
1.5
12
18 mm
14 mm
15 mm
3
16 mm
22 mm
98
Total cross sectional area A = 1.376 2m
Height of neutral axis above the keel, 2.688
3.907 m0.688
i iNA
i
a yh
a
Moment of inertia about neutral axis,
2 2 2 42 2 2.16 21.744 3.907 0.688 26.804 mNA xi i i NA iI i a y h a
Figure 92
Strength deck:
1 0q
1 1 1sd sd NA
Qq s q t s D h
I
10004 0.018 4 12 3.907 21.739 t/m
26.804sdq
Second deck:
4 0q
4 4 4nd nd nd NA
Qq s q t s h h
I
1 2
43
8 6
9 7
N A
s1
s4
s8
s9
s2
s3
s6
s7hdb
hnd
hNA
tsd
tnd
tss
tds
tb
tib
tg
tns
D
5
99
10004 0.014 4 9 3.907 10.641 t/m
26.804ndq
Side (below strength deck):
22 24
2ss sd ss NA
sQq s q t s D h
I
1000 33 21.739 0.015 3 12 3.907 32.808 t/m
26.804 2ssq
Side (below 2nd deck):
33 33 4
2ds ss nd ds nd NA
sQq s q q t s h h
I
0 3 4 32.808 10.641 43.448 t/mds ss ndq q q
1000 5.0935.093 32.808 10.641 0.015 5.093 9 3.907 50.706 t/m
26.804 2dsq
(This is the maximum value which is located at the neutral axis)
Bottom:
9 0q
9 9 9b b NA
Qq s q t s h
I
100010 0.022 10 3.907 32.067 t/m
26.804bq
Inner bottom:
8 0q
8 8 8ib ib NA db
Qq s q t s h h
I
100010 0.016 10 3.907-1.5 14.368 t/m
26.804ibq
Bilge:
77 710
2g b g NA
sQq s q t s h
I
100
1000 1.51.5 32.067 0.015 1.5 3.907 34.717 t/m
26.804 2gq
Side (below neutral axis):
66 61.5 10
2ns g ib ns NA db
sQq s q q t s h h
I
0 1.5 10 34.717 14.368 49.085 t/mns g ibq q q
1000 2.4072.407 34.717 14.368 0.015 2.407 3.907 1.5 50.706 t/m
26.804 2nsq
(This is the maximum value which is located at the neutral axis) – (check)
Figure 93 shear flow distribution
1.26. Ship section idealization
The idealized ship section is based on using effective thickness concept for the decks, sides,
and bottom structure. The effective thickness takes account of longitudinal stiffeners and
girders and is calculated as follows:
21.739 t/m
10.641 t/m
21.739 t/m
32.808 t/m43.448 t/m
50.706 t/m
32.067 t/m
14.368 t/m
32.067 t/m
34.717 t/m49.085 t/m
101
1
1
n
ii
e
at t
ll
ns
(1.117)
Where:
t = plate thickness
et = effective (idealized) plate thickness
ia = sectional area of longitudinal member
l = stiffened panel width through which n longitudinal members are distributed
s = distance between each two consecutive longitudinal members
Figure 94 A ship section of a container ship and corresponding idealization
102
1.27. Changes to section modulus
The provision of the required section modulus is necessarily an iterative process. As the
design progress it will be necessary to add or subtract material in the hull girder cross section.
The calculation of the moment of inertia of the ship sections I is a lengthy computation and
it would not be desirable to have to repeat it for every change of area.
A typical situation is shown in Figure 95, in which:
a = area added at height y above the neutral axis
I = moment of inertia
A = total cross sectional area of the ship
Dy = distance of the deck from the original neutral axis
Ky = distance of the keel from the original neutral axis
Figure 95 effect of adding area
The effect of the addition is to raise the neutral axis a distance h and to increase the
moment of inertia to a value I I (about the new neutral axis). The net effect on the deck
and bottom can vary, depending on the location of a .
For example, although the addition shown would reduce the deck stress (because it increases
I and decreases Dy ), it might increase the keel stress because Ky is increased and this might
out-weight the increase in I .
For a given bending moment, the stress in the keel will not be increased if the section
modulus KI y is not reduced; that is, if:
103
0K K
I I I
y h y
(1.118)
or
0K K KIy y I Iy I h (1.119)
or
K
I h
I y
(1.120)
With the addition of the area a , the rise of the neutral axis is:
ay
hA a
(1.121)
and the additional moment of inertia is:
22I ay i A a h (1.122)
If the material is added below the original neutral axis the value of y is negative. If the
material is removed then the value of a is negative and also i is negative in the foregoing
equation.
If the local moment of inertia of the added area i is small and assumed negligible; then,
substitution for h from (1.121) gives:
2 2 2
2 a y AayI ay
A a A a
(1.123)
There are two cases which require separate consideration:
1) The material is added within the exiting cross section, that is, Dy y (or y < Ky if
the material is added below the neutral axis).
2) The material is added above the existing section, Dy y , such that the maximum stress
now occurs in the added material. This would happen if, for example, a superstructure
deck were to become longitudinally effective.
104
1.27.1. Material added within the section ( y < Dy )
In this case, both h and I would act to decrease the deck stress. At the keel, the condition
expressed in (1.120) becomes:
2
k
Aay ay
I A a y A a
(1.124)
From Eqns (1.121) and (1.123), the new section modulus at the deck 'DZ is:
2
'D
DD
AayII I A aZ
ayy h yA a
(1.125)
Which may be written as:
'
1D D
DD
Z AyrZ
r
(1.126)
Where D
D
ayr
A a y
The corresponding expressions for the new section modulus at the keel are:
2
'K
KK
AayII I A aZ
ayy h yA a
(1.127)
and its alternative form
'
1K K
KK
Z AyrZ
r
(1.128)
With K
K
ayr
A a y
1.27.2. Material added above the strength deck ( y > Dy )
If the material is added above the strength deck, the maximum stress now occurs in the added
material. The distance of this material from the new position of the neutral axis is:
105
ay Ay
y h yA a A a
(1.129)
and the condition that the section modulus to the new material should be not less than DZ is:
0D
I I I
y h y
(1.130)
Which becomes, after substitution and rearrangement:
2
1
1
D
yA
ya
AyI
(1.131)
For the keel the new section modulus is as given in Eqns (1.127) and (1.128). The required
condition for the keel stress not to increase, is usually satisfied because y is large.
1.27.3. Examples:
Addition of a steel deck
For the following mid ship section, calculate the area of the added deck a at a height y
above the neutral axis, such that 1 and 1M M .
Figure 96
h
h
yK
yD
original N.A.
new N.A.
added deck a
y
y
B/2
106
Solution:
ayh
A a
1
ay Ayy y h y
A a A a
221I I ay A a h
1 1 11
1 1
D DM y My y y
I I I I
22
1 DyAy
A a II ay A a h
2
1
1
D
yA
ya
AyI
For a transverse system of framing,
aa Bt t
B
Where, t is the thickness of the deck plating.
For a longitudinal system of framing,
1
n
l ii
a Bt a
, 1B
ns
Where, s is the distance between longitudinals, n is the number of longitudinals and la is
the cross sectional area of a longitudinal.
If all the longitudinals have the same cross sectional area la then,
1
nl
l l lii
a naa na a Bt na t
B
If is the ratio of the total cross sectional area of the longitudinals,
1at
B
107
Removal of a steel deck
For the following mid ship section, calculate the area of the removed deck a at a height y
above the neutral axis, such that 1 and 1M M .
Figure 97
Solution:
ayh
A a
1 D D
ayy y h y
A a
221
2 22
I I ay A a h
ay aAyI ay A a I
A a A a
1 1 11
1 1
D DM y My y y
I I I I
2
1D
DD D
y A
ay
y y y AyI
h
h
yK
yD
original N.A.
new N.A.
removed deck a
y
y
108
Addition of a hatch coaming
For the following mid ship section, calculate the thickness t and the height H of an added
hatch coaming, such that 1 and 1M M .
Figure 98
Solution:
ayh
A a
1 2 2 2
H ay H Ay Hy y h y
A a A a
221I I i ay A a h
1 1 11
1 1
D DM y My y y
I I I I
where,
2D
Hy y , a tH ,
3
12
tHi
It is difficult to solve the previous equation analytically. Then, by assuming a suitable design
value for the hatch coaming height H and plotting the new sectional modulus 1 1I y as a
function of the hatch coaming thickness t , the range of the acceptable thickness will be as
indicated in Figure 99.
h
h
yK
yD
original N.A.
new N.A.
y
H
added hatch coamingthickness = t
109
Figure 99
Other cases of study
Addition of a tank top
Addition of a deck
Addition of a tween deck
Replacement of the upper deck
Effect of corrosion
Addition of a longitudinal bulkhead
High tensile steel
I1/y1
t
()(I/yD)
Rejected Accepted
110
1.28. Buckling of a simply supported plate
Figure 100 shows simply supported panel plating subjected to a uniform in-plane
compressive stress a in the x-direction. The buckling or critical stress is:
Figure 100 buckled shape of a long plate
22 22 2
2 2
2
3
212 1
a cr
m na D
a b
tm
EtD
v
(1.132)
Where,
D = Plate flexural rigidity
v = Poisson’s ratio
t = Plate thickness
a = Plate length
b = Plate width
The parameters m and n indicate the number of half-waves in each direction in the buckled
shape. Both must be integers, and it can be seen that the value of n that gives the smallest
value of a cr is 1n . Hence the plate will buckle into only one half-wave transversely,
and the buckling or critical stress is:
111
222
2
1a cr
D am
a t m b
(1.133)
This equation is usually written in a more general form in terms of a buckling coefficient k
and the plate width b :
2
2
2
a cr
Dk
b t
mb ak
a mb
(1.134)
The expression for the buckling coefficient k depends on the type of boundary support.
For design applications, Eqn (1.134) is usually written in the alternative form:
2
2
212 1
a cr
tKE
b
kK
v
(1.135)
In Figure 101 the coefficient k is plotted against aspect ratio /a b for various values of m .
By setting 0dk dm , from which is obtained m a b ; that is, the stress is lowest (and
therefore truly critical) when the number of half-waves in the longitudinal direction is equal
to the aspect ratio. Under these conditions, 4k .
Figure 101
112
If the length a is not an exact multiple of the width b , then the panel will buckle into the
nearest whole number of half-waves which will make the critical stress a minimum. The
value of k is therefore given by the solid (and discontinuous) curve in Figure 101 from
which it may be seen that, although the value of k is somewhat greater for non-integer aspect
ratios, this effect is slight and diminishes as aspect ratio increases. Hence, for long simply
supported plates it is usually assumed that 4k .
For a wide plate, in which the aspect ratio is less than 1.0, m will be equal to unity.
Therefore, Eqn (1.133) becomes:
222
21a cr
D a
a t b
(1.136)
This gives the critical stress for a simply supported wide plate. For design purposes Eqn
(1.136) may be written as:
2
222
21
12 1
a cr
tKE
a
aK
bv
(1.137)
The question of wide plates versus narrow plates leads to the consideration of the relative
merits of stiffening a large sheet of plating in the longitudinal or in the transverse direction.
Let the plating be of length L and breadth B and subjected to a uniform compressive stress
a .
If longitudinal stiffeners are fitted at a spacing s as in Figure 102-(a) then the buckling stress
is found from Eqn (1.135) with b s :
2
2
212 1
a cr
tKE
s
kK
v
(1.138)
On the other hand, if the stiffeners are fitted transversely at the same spacing, as in Figure
102-(b), a cr is obtained from Eqn (1.137) with a s and b B :
113
2
222
21
12 1
a cr
tKE
s
sK
Bv
(1.139)
The term s B is generally quite small for a ship (seldom greater than 1 6 ) and so the term in
square brackets is approximately unity. Therefore, the buckling strength of longitudinally
stiffened plating is nearly four times as great as that of transversely stiffened plating. This
shows the great advantage of longitudinal over transverse stiffening in ship structures, and the
former is used wherever possible.
Figure 102
1.28.1. Example
A superstructure deck is added above the main deck of a ship. If the main characteristics of
the original mid ship section and the added superstructure deck are as follow:
Original mid ship section Superstructure deck Sectional area, A 2.26 2m Height above main deck, y 2.8 m
Moment of inertia, I 27 4m Breadth 9.15 m Main deck height above neutral axis,
Dy 8.2 m Frame spacing or distance
between longitudinals 0.84 m
Sectional area of a longitudinal 420 10 2m
If the total vertical bending moment is the same before and after the modifications M =
30500 . .t m , modulus of elasticity of steel E = 72.1 10 2/t m and Poisson’s ratio v = 0.3.
For both of the transverse and the longitudinal system of framing:
a
L
B
s
(a)
aL
B s
(b)
114
Calculate the thickness of the superstructure deck such that bending stress in the
superstructure deck after modification is the same as that in the main deck before
modification.
Check buckling of a deck plate between stiffeners, if occurred; calculate the necessary
thickness of the superstructure deck to avoid buckling with a factor of safety = 1.5.
Solution:
As the stress in the added superstructure deck after modification equals the stress in the main
deck before modification, 1 and the bending moment is the same before and after the
modification, 1M M ; then, from the example in page (105):
1 , 1
22 2
111 2.26 18.2
0.069352.26 11
1 127
D
yA
ya a m
AyI
21
30500 8.29262.963 /
27DMy
t mI
For transverse system of framing:
30.069357.58 10
9.15a B t t m
2 22 22 2
2 2
22 37 2
0.841 1 0.9191
9.1512 1 12 1 0.3
7.58 100.9191 2.1 10 1571.248 /
0.84a cr
sK
Bv
tKE t m
s
1 a cr Buckling occurs
Considering a factor of safety = 1.5:
2
1 11 22
1a acr cr
M y Ay M tKE
I A a sI ay A a h
By substitution with:
115
ayh
A a
and a Bt
2
2 2 3 319.13 10s
IAt B I Ay t AyM t mKE
For longitudinal system of framing:
9.151 1 9.89 10
0.84
Bn
s
430.06935 10 20 10
5.393 109.15
ll
a naa Bt na t m
B
2 2
2 2
22 37 2
43.6152
12 1 12 1 0.3
5.393 103.6152 2.1 10 3129.557 /
0.84a cr
kK
v
tKE t m
s
1 a cr Buckling occurs
Considering a factor of safety = 1.5:
2
1 11 22
1a acr cr
M y Ay M tKE
I A a sI ay A a h
By substitution with:
ayh
A a
and la Bt na
2
2 2 2 3 310.5 10l
sIA na I Ay t B I Ay t AyM t m
KE