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).

2

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

4

Figure 3 Levels of structural analysis

5

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).

6

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).

7

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).

8

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

10

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.

12

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)

13

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

14

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

15

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.

16

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

18

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).

19

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

20

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.

21

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)

22

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.

23

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.

 

24

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.

25

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