Geometric properties of an area PDF

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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): x

 xdA  dA A

,

y

A

 ydA  dA A

(1.83)

A

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:

 x

 xA

,

A

y

  yA A

(1.84)

A

Here x and y represent the coordinates for the centroid of each composite part, and 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: I x   y2 dA A

,

I y   x2 dA A

(1.85)

Polar moment of inertia of an area about the pole O or z axis can be calculate also as (Figure 61):

69

J O  A r dA  I x  I y 2

(1.86)

The relationship between J O and I x , I y is possible since r2  x2  y2 .

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: Ixy   xydA A

(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 I xy for an area will be zero if either the x or y axis is an axis of symmetry for the area (see Figure 63).

70

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 ( I x ' , I y' ), the moment of inertia of the area about a corresponding parallel axis ( I x , I y ) can be determined using the parallel-axis theorem, as (Figure 64): I x  I x'  Ad y2

,

I y  I y'  Ad x2

(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):

J O  J C  Ad 2

(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. 71

The product of inertia of the area with respect to the x - y axis will be (Figure 64):

I xy  I x' y'  Ad x d y

(1.90)

Where, I x' y' is the product of inertia of the area with respect to the centroidal axis. It is important that here the algebraic signs for d x and dy 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 I x' , I y ' and I x' y' for an area with respect to a set of inclined x ' and y ' axes can be calculated when the values for  , I x , I y and I xy 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

x '  x cos   y sin  y '  y cos   x sin 

(1.91)

Using these equations, the moments and product of inertia about the x' and y ' axes become:

72

Ix  I y Ix  I y cos 2  I xy sin 2  2 2 Ix  I y Ix  I y I y'  cos 2  I xy sin 2  2 2 Ix  Iy I x' y'  sin 2  I xy cos 2 2 I x' 

(1.92)

1.20.6. Principal Moments of Inertia

The axes about which the moments of inertia for an area, I x' and Iy ' , 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:  Ix I y   2   sin 2  2I xy cos 2  0 d  2 

dI x '

(1.93)

Therefore, at   p : tan 2 p 

I

 I xy

x

(1.94)

 Iy  2

This equation has two roots,  p1 and  p2 , which are 90o apart and so specify the inclination of each principal axis. The sine and cosine of 2 p1 and 2p 2 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

Imax min

I x I y  I x I y  2      I xy 2 2  

73

(1.95)

Figure 66

If the above trigonometric relations for  p1 and  p2 are substituted into the third of Eqn (1.92) , it will be seen that I x' y' = 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. 74

The two quantities to be calculated are the position of the neutral axis of the section and the moment of inertia INA about the neutral axis. The distance of the neutral axis above the keel is: hNA 

ah a

i i

(1.96)

i

Where, ai is the area and hi 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: 2 I NA  Ixx  AhNA

(1.97)

Where:

I NA = moment of inertia of the entire ship section about its neutral axis Ixx 

i   a h ii

2 i i

= moment of inertia about a horizontal axis x x passing by the keel

iii = moment of inertia of element i about its local neutral axis A   ai = total area

hNA = 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.

75

Ex. 1:

b td

ts D

N

A Y h

B/4 t cg

B/4 to

X

h NA ti

t sg

BL

B/2

Figure 67

Table 9 sectional modulus calculations (1)

Items

(2)

(3)

(4)

ai

Scantling

yi

(5)=(3)×(4)

(6)=(5)×(4)

ai yi

ai yi

bDt d

2

ixi

Deck

b  td

Side

D  ts

Dts

D 2

D ts 2

D ts 4

I. bottom

( B 2)  ti

Bti 2

h

Bhti 2

O. bottom

( B 2)  to

Bto 2

0

S. girder

h  ts g

htsg

C. girder

h  ( tcg 2)

h t cg 2

Sum

bt d

D

a

i

(7)

2

bD t d

2

(9)=(3)×(8)2

xi

ai xi2

( B  b) 2

0

3

(8)

3

(10)

iyi 2

btd ( B  b) 2

4

3

b t d 12

D t s 12

B 2

Dts B

Bh t i 2

0

B 4

B3ti 32

0

0

0

B 4

B to 32

h 2

h t sg 2

h tsg 4

3

h3 tsg 12

B 4

B ht sg 16

0

h 2

h2 t cg 4

h tcg 8

3

h tcg 24

0

0

0

a y

a y

2

i

i

2

i

2 i

3

i

xi

Total area = 2   ai Height of neutral axis above the keel, hNA 

a y a i

i

i

Moment of inertia about neutral axis, I NA  2   i xi   a i y2i  h2NA  a i 

Moment of inertia about center line, I CL  2    i yi   ai x2i  76

4

3

2

a x i

2 i

0 3

B t i 96 B3 to 96

i

yi

Section modulus (deck), ZD 

I NA D  hNA

Section modulus (bottom), Z B 

Section modulus (side), Z S 

I NA h NA

ICL B / 2 

Ex. 2:

Figure 68 example of longitudinally effective material

77

Table 10 sectional modulus calculations

Total area, A 

a

i

 2 0.5706  1.141m2

Height of neutral axis, hNA 

a y a i

i



i

1.9095  3.346 m 0.5706

I xx  2  12.413  0.440   25.706 m4 2 I NA  I xx   ai  hNA  2  12.413  0.440   0.5706 3.3462   12.926 m4  

ZD

I NA 12.926 3   2.286 m D  hNA 9.0  3.346

ZB 

I NA 12.926   3.863m 3 hNA 3.346

78

Ex. 3: 

6m

2

A=80cm

22mm 3m

14mm 2

A=30cm

14mm

16mm

8.5m 18mm

Y 24mm

16mm

20mm

X

1.5m

10m

Figure 69

Table 11 sectional modulus calculations Scantling (m)

ai

yi

aiyi

ai yi

ix i

xi

ai xi

iyi

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

2n d deck plating

Items

2

2

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

1.5×0.012

0.018

0.75

0.0135

0.0101

0.0034

0

0

-

4.2788

42.4512

1.7822

39.348

3.8507

Center girder (1/2) Sum

0.822

Total area = 2  a i  2 0.822  1.644 m 2 Height of neutral axis above the keel, hNA 

a y a i

i

Moment of inertia about neutral axis, 2 I NA  2    ixi   ai yi2  hNA  ai 

 2  1.7822  42.4512  5.205 2  0.822   43.923m 4

79

i



4.2788  5.205 m 0.822

Moment of inertia about center line, I CL  2   i yi   a i x i2   2  3.8507  39.348  86.397 m 4

Section modulus (deck), ZD 

I NA 43.923   5.6349 m3 D  hNA 13  5.205 

Section modulus (bottom), Z B 

Section modulus (side), Z S 

I NA 43.923   8.438 m3 hNA 5.205

ICL 86.397   8.6397 m 3 10  B / 2

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:

 M y I x  M x I xy   M x I y  M y I xy   x    y 2 2 I I I I I I   x y xy x y xy    

   

(1.98)

Where, x and y are the perpendicular distances to the centroidal y -axis and x -axis, respectively. Ix and I y are the centroidal moments of inertia of the beam cross section with 80

respect to the x and y axes, respectively. I xy is the centroidal product of inertia of the beam cross section. M x and M y 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:

 

My Iy

x

Mx y Ix

(1.99)

If additionally the beam is only subjected to bending about one axis, the formula simplifies further:

 

M Mx y Ix 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 MV . For the sagging condition: Maximum tension stress will be,  t 

MV hNA (Bottom) I NA

Maximum compression stress will be,  c  

M V  D  hNA  (Deck) I NA

For the hogging condition: Maximum tension stress will be,  t 

M V D  hNA  (Deck) INA

Maximum compression stress will be,  c  

MV hNA (Bottom) I NA

81

Figure 71 indicates the ship hull under horizontal bending moment M H . 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.

 t   c 

MH B 2  ICL

 Deck Compression (- ve)

c

MV

Deck Tension (+ ve)

c

t

c

t

c

N

t

t

t A

Bottom Tension (+ ve)

MV

Side Tension (+ ve)

Side Tension (+ ve)

c A MV

c

Bottom Compression (- ve)

t

c

c

(a) Sagging

(b) Hogging Figure 70 Vertical bending moment



Deck Tension (+ ve) Deck Compression (- ve)

t c t

C

MH

c

t

c

Side Tension (+ ve)

t

c

t

Side Compression (- ve)

c L MH

Figure 71 Horizontal bending moment

82

t

t

N

t

Side Compression (- ve)

t

MV

c

c

Side Compression (- ve)

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 My 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 My is entirely due to inclination of the vessel, say to an angle  . In this case M y and M x are directly related, being components of the total bending moment MV (which acts in the true vertical plane): M y  M V sin  M x  M V cos 

(1.101)

If x and y are the coordinates of any point in the cross section and, INA and ICL 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:...