@ @ Innovative ship design-MidshipSection Rule Scantling -

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Advanced Ship Design Automation Lab.

http://asdal.snu.ac.kr Seoul

National Univ.

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Innovative ship design – Rule scantling

N av al A rc hite ctu re & O ce an E ngin ee rin g

@ SDAL Advanced Ship Design Automation Lab.

http://asdal.snu.ac.kr Seoul

National Univ.

May 2009

Prof. Kyu-Yeul Lee

Department of Naval Architecture and Ocean Engineering, Seoul National University of College of Engineering

[2009] [13][14]

Innovative ship design

-Midship Section Rule Scantling -

서울대학교 조선해양공학과 학부4학년 “창의적 선박설계” 강의 교재

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Innovative ship design – Rule scantling

N av al A rc hite ctu re & O ce an E ngin ee rin g

@ SDAL Advanced Ship Design Automation Lab.

http://asdal.snu.ac.kr Seoul

National Univ.

Rule Scantling

(선박 구조 설계)

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Innovative ship design – Rule scantling

Ship Structural Design

 Ship Structural Design

what is designer’s major interest?

 Safety :

Won’t ‘IT’ fail under the load?

a ship a stiffener a plate

z

x

L

x

y ^{f x} ( )

global local

a ship

Actual stress on midship section should be less than allowable stress

.

act allow

σ ≤ σ

Allowable stress by Rule : (for example)

2

, σ

_allow

= 175 f

1

[ N mm / ]

, _act . ^mid mid

σ M

= Z

L

x

y ^{f x} ( )

( ) f x

x

y

react

F

( ) V x

( ) M x

( ) y x

Differential equations of the defection curve 4

4

( ) ( ) d y x

EI f x

dx = − what is our interest?

: ( ) : ( ) : ( ) Shear Force V x Bending Moment M x Deflection y x

: ( ) Load f x

cause

( ) ( ) dV x f x

dx = − ( )

, dM x ( ) dx = V x

2

, d y x ( ) ( )

EI M x

dx =

‘relations’ of load, S.F., B.M., and deflection

 Safety :

Won’t it fail under the load?

 Geometry :

How much it would be bent under the load?

,

_act

y i

M M

where

I y Z

σ = =

act allow

σ

≤

σ

Stress should meet :

(4)

Applying beam theory on a ship

z

x

w

( ) V x

( ) M x

배는 부유체로 실제로는 끝단 지지가 없다.

그리고 실제적으로 끝단에서 중량과 부력의 차이에 의한 전단력이 존재하며, 그로 인한 deflection과 slop이 발생한다.

하지만 문제를 풀기 위해서는 처짐의 양이 작다고 가정하고 AP와 FP에 지지점이 있다 고 가정하여 Boundary condition을 적용한다.

R A R _B

( ) y x

y

x w

R B

R A

양 끝단에 지지가 없으면 처짐이 발생 한다.

idealize

( ) M x

x

( ) V x

x ( )

f x

x

L

x y ^f

( )

^x

AP FP

James M. Gere, Mechanics of Materials 6^thEdition, Thomson, Chap.4, p.292

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Innovative ship design – Rule scantling

Ship Structural Design

 Ship Structural Design

what is designer’s major interest?

 Safety :

Won’t ‘IT’ fail under the load?

a ship a stiffener a plate

L

x

y ^{f x} ( )

global local

a ship

Actual stress on midship section should be less than allowable stress

l

act σ

σ _. ≤

Allowable stress by Rule (for example):

2

, σ

_l

= 175 f

1

[ N mm / ]

, _act . ^mid ^S ^W

mid mid

M M

σ M

Z Z

= = +

Hydrostatics

L

x

y ^{f x} ( )

( ) f x

x

y

react

F

( ) V x

( ) M x

( ) y x

Differential equations of the defection curve 4

4

( ) ( ) d y x

EI f x

dx = − what is our interest?

: ( ) : ( ) : ( ) Shear Force V x Bending Moment M x Deflection y x

: ( ) Load f x

cause

( ) ( ) dV x f x

dx = − ( )

, dM x ( ) dx = V x

2

, d y x ( ) ( )

EI M x

dx =

‘relations’ of load, S.F., B.M., and deflection

 Safety :

Won’t it fail under the load?

 Geometry :

How much it would be bent under the load?

,

_act

y i

M M

where

I y Z

σ = =

Stress should meet :

σ

act

: Actual Stress σ

t

: Allowable Stress

act l

σ

≤

σ

M

_S

= Still water bending moment M

_W

= Vertical wave bending moment

Hydrodynamics

z

x

what kinds of load

cause hull girder moment?

f

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Innovative ship design – Rule scantling

Ship Structural Design

 Ship Structural Design

what is designer’s major interest?

 Safety :

Won’t ‘IT’ fail under the load?

a ship a stiffener a plate

global local

a ship

L

x

y ^{f x} ( )

( ) f x

x

y

react

F

( ) V x

( ) M x

( ) y x

Differential equations of the defection curve 4

4

( ) ( ) d y x

EI f x

dx = − what is our interest?

: ( ) : ( ) : ( ) Shear Force V x Bending Moment M x Deflection y x

: ( ) Load f x

cause

( ) ( ) dV x f x

dx = − ( )

, dM x ( ) dx = V x

2

, d y x ( ) ( )

EI M x

dx =

‘relations’ of load, S.F., B.M., and deflection

 Safety :

Won’t it fail under the load?

 Geometry :

How much it would be bent under the load?

,

_act

y i

M M

where

I y Z

σ = =

Stress should meet :

σ

act

: Actual Stress σ

t

: Allowable Stress

act l

σ

≤

σ

S ( ) f x

,

_act . ^S ^W mid

M M

σ Z

= +

l

act σ

σ _. ≤

: load in still water

( )

0^x

( )

S S

V x = ∫ f x dx

S ( ) V x

S ( ) M x

( )

0^x

( )

S S

M x = ∫ V x dx

Hydrostatics Hydrodynamics

. F K diffraction added mass

mass inertia

damping

: still water shear force

: still water bending moment

, M

_S

= Still water bending moment M

_W

= Vertical wave bending moment what kinds of load f cause hull girder moment?

weight

buoyancy

f _S (x) : load in still water

= weight + buoyancy

W ( )

f x : load in wave

( )

0^x

( )

W W

V x = ∫ f x dx

W ( ) V x

W ( )

M x

( )

0^x

( )

W W

M x = ∫ V x dx

: wave shear force

: vertical wave bending moment

f

_W

(x) : load in wave

= added mass + diffraction

+ damping + Froude-Krylov + mass inertia ( )

_S

( )

_W

( ) f x = f x + f x

( )

_S

( )

_W

( ) V x = V x + V x

( )

_S

( )

_W

( )

M x = M x + M x

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Innovative ship design – Rule scantling

Ship Structural Design

 Ship Structural Design

what is designer’s major interest?

 Safety :

Won’t ‘IT’ fail under the load?

a ship a stiffener a plate

z

x

L

x

y ^{f x} ( )

global local

a ship

how we can meet the rule?

.

, .

,

_act ^mid ^s ^w

mid ship N A i

M M M

σ Z I y

= = +

. .L B

. .A N

Deck Upper

σ

<Midship section>

y i

) ( , = y

y y i

, . .

, :vertical wave bending moment , : still water bending moment

, : moment of inertia from N.A. of Midship section

w s ship N A

M M I

Hydrostatics, Hydrodynamics

‘Midship Design’ is to arrange the structural members and fix the thickness of them to

secure enough section modulus to the rule.

L

x

y ^{f x} ( )

( ) f x

x

y

react

F

( ) V x

( ) M x

( ) y x

Differential equations of the defection curve 4

4

( ) ( ) d y x

EI f x

dx = − what is our interest?

: ( ) : ( ) : ( ) Shear Force V x Bending Moment M x Deflection y x

: ( ) Load f x

cause

( ) ( ) dV x f x

dx = − ( )

, dM x ( ) dx = V x

2

, d y x ( ) ( )

EI M x

dx =

‘relations’ of load, S.F., B.M., and deflection

 Safety :

Won’t it fail under the load?

 Geometry :

How much it would be bent under the load?

,

_act

y i

M M

where

I y Z

σ = =

act allow

σ

≤

σ

Stress should meet :

Actual stress on midship section should be less than allowable stress

.

act allow

σ ≤ σ

Allowable stress by Rule : (for example)

2

, σ

_allow

= 175 f

1

[ N mm / ]

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Innovative ship design – Rule scantling

Midship section rule scantling

Main data, geometry and arrangement

 중앙 단면 배치 결정

초기 중앙 단면 계수 가정

Local Scantling

실제 중앙 단면 계수 계산

Required Section Modulus Actual Section Modulus <

Rule Scantling End

실적선 중앙단면 계수

Yes No

Stress Factor 계산

Rule에 의해 결정 (실적선 내용 참조) Design Bending moment 계산

Local scantling

Longitudinal strength check

 Local scantling을 통하여 부재들의 치수가 결정되고 따라서 중앙 단면 계수가 결정 된다 .

 Local scantling 후

Actual section modulus < Required section modulus 인 경우에는, 구조 부재의 배치를 바꾸 거나 , 구조 부재의 치수를 키워주는 방법이 있다.

 I값을 늘리기 위해서는 y값이 큰 deck와 Bottom의 부재를 늘리는 것이 효율적임

 Deck와 Bottom의 Plates, Longitudinals의 치수를 늘림

치수 늘림

. .L B

. .A N

Deck Upper

σ

<Midship section>

y i

) ( , = y

y y i

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Innovative ship design – Rule scantling

1.General 2.Materials

3.Section properties 4.Beam theory

5.Longitudinal Strength 6.Design Load

7.Local Scantling 8.Buckling

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Innovative ship design – Rule scantling

1. General

1.1 선체 중앙단면도

1.2 Stress transmission 1.3 Principal dimensions

1.4 판 두께 선택기준, Grouping of longitudinal stiffener

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Innovative ship design – Rule scantling

1.1 Midship section for 3,700 TEU Containership

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Innovative ship design – Rule scantling

1.2 Stress transmission

Center Girder Side Girder

Longitudinals

Web frame

Inner Bottom Plate

Plate Longi. Web frame Girder

Load

General Materials

Longitudinal Strength Analysis Local Scantling

Buckling and Fatigue

: Stress Transmission

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1) Rule length (L)

: Rule Scantling시 사용하는 선박의 길이

 Greatest moulded breadth in [m], measured at the summer waterline

 Distance on the summer load waterline (LWL) from the fore side of the stem to the axis of the rudder stock

 Not to be taken less than 96%, and need not be taken greater

than 97%, of the extreme length on the summer load waterline (L _WL )

 Starting point of rule length : A.P

LWL L

LWL < < ⋅

⋅ 0 . 97

96 . 0

2) Breadth

ex)

^LBP ^LWL ^0.96·LWL ^{0.97 ·LWL} ^L

250 261 250.56 253.17 250.56

250 258 247.68 250.26 250.00

250 255 244.80 247.35 247.35

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.1 101

DnV rule을 기준으로 구조부재 치수 결정시 다음의 주요치수를 따른다.

1.3 Principal dimensions

*조선소 전문가 조언

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Innovative ship design – Rule scantling

3) Depth (D)

: moulded depth defined as the vertical distance in [m] from baseline to moulded deckline at the uppermost continuous deck measured amidships

 To be calculated based on rule length

1.3 Principal dimensions

4) Draught (T)

: mean moulded summer draught in [m]

5) Brock coefficient (C _B )

1.025 C B

L B T

= ∆

⋅ ⋅ ⋅ ,( moulded displacement in salt water on draught T) ^∆ ^:

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.1 101

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 계산에서 나온 값을 그대로 판의 두께로 사용하는 것이 아니라 생산 되고 있는 판의 두께로 선정한다 .

1) 0.5 mm 간격으로

2) 0.25 mm 이상 : 0.5 mm 3) 0.25 mm 미만 : 0.00 mm

ex) 15.75 mm  16.0 mm 15.74 mm  15.5 mm

1) 판 두께 선택 기준

2) Grouping of longitudinal stiffner

 Average value but not to be taken less than 90% of the largest individual requirement. (DNV)

ex) 100, 90, 80, 70, 60 치수를 갖는 5개의 longi의 평균 치수는 80×5 이다.

그러나 최대 100×90% = 90보다 작으므로 90 ×5 를 배치 하여야 한다.

•Longi 하나마다 각각 요구하는 치수를 만족하는 부재를 배치 하는 것이 아니라 Group으로 묶어서 배치 (생산성 향상을 위함)

•묶어서 배치한 부재는 다음을 만족하여야 한다.

1.4 판 두께 선택기준, Grouping of longitudinal stiffener DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.1 101

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Innovative ship design – Rule scantling

2. Materials

2.1 Material factor 2.2 Material classes

2.3 Corrosion Addition (tk , tc)

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2.1 Material Factors

*Yield Stress(항복응력, σ _y ) [N/mm ² ] or [Mpa]: Hook’s law성립하는 탄 성영역을 넘어서 소성영역이 시작 되는 지점 ²⁾

*NV-NS : Normal Strength Steel( Mild Steel)

1)

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2

 The material factor f ₁ included in the various formulae for scantlings and in expressions giving allowable stresses. ¹⁾

Material Designation

Yield Stress (N/mm

²

)

Material Factor

NV-NS 235 235/235 = 1.00 1.00

NV-27 265 265/235 = 1.13 1.08

NV-32 315 315/235 = 1.34 1.28

NV-36 355 355/235 = 1.51 1.39

NV-40 390 390/235 = 1.65 1.43

NS NV −

σ σ

*NV-XX : High tensile Steel

*High tensile steel

0.2% 정도의 탄소를 함유한 탄소강에 망간 등의 함유량을 증가시켜 성능을 향 상 시킨 강

2)

James M. Gere, Mechanics of Materials 6th Edition,

Thomson, Chap.12, pp.880~094

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Innovative ship design – Rule scantling

2.2 Material Classes

t 20 15

t 25 20

t 30 25

t 35 30

t 40 35

t 50 40

t 15 A/AH A/AH A/AH A/AH A/AH A/AH B/AH

A/AH A/AH A/AH A/AH B/AH B/AH D/DH

A/AH A/AH B/AH D/DH D/DH D/DH E/EH

A/AH B/AH D/DH E/DH E/EH E/EH E/EH

D/DH E/DH E/EH E/EH E/EH E/EH E/EH Thickness

(mm) I II III IV V

Class

1)

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2



In order to distinguish between the material grade requirements for different hull parts, various material classes are applied as defined in Table.

¹⁾

*A: ‘A’ grade ‘Normal strength Steel’

AH: ‘A’ grade ‘High tensile steel’

 Steel Grade 표기가 A,B,D,E 순으로 갈수록 탈산(Deoxidation) 및 열처리과정을 거친다. ²⁾

2)

DSME, “선박구조설계” 2-16 선급강재 steel grade 비교표 , 2005.8

 Steel Grade 에 따라, 첨가되는 화학원소 성분이 다르다. ²⁾

 장점: 저온 취성에 강해짐 (저온에서 취성이 일어남) 단점: 가격이 비쌈

Grade $/ton

A $1,340

AH36 $1,384

… …

E $1,425

EH36 $1,502

Steel Grade별 가격

08.05.30기준

*취성(brittleness): 재료가 급작스래 파괴되는 현상.

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Innovative ship design – Rule scantling

2.2 Material Classes - Typical Example (1)

0.4L

I II I

A2 Deck planting exposed to

weather

A1 Longitudinal bulkhead strake

A3 Side Plate

Structural member 0.4 L Outside 0.4 L A1 Longitudinal bulkhead strakes.

A2 Deck plating exposed to weather.

A3 Side plating. II I

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2 Table B2

A

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Innovative ship design – Rule scantling

2.2 Material Classes - Typical Example (2)

0.4L

I III I

B2 Strength Deck plating

B1 Bottom plating including Keel Plate

B3 Continuous longitudinal members above Strength

Deck

B4 Uppper most strake in LBHD

Structural member 0.4 L Outside 0.4 L

B1 Bottom plating including keel plate.

B2 Strength deck plating.

B3 Continuous longitudinal members above strength deck

B4 Uppermost strake in longitudinal bulkhead.

III I

B

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2 Table B2

Strength Deck

: General defined as the uppermost

continuous deck

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Innovative ship design – Rule scantling

2.2 Material Classes - Typical Example (3)

E/EH When L> 250 m

Structural member 0.4 L Outside 0.4 L C1 Sheer strake at strength deck.

C2 Stringer plate in strength deck

IV

III

(II outside 0.6L amidships)

C2 Stringer Plate in

Strength Deck C1 Shear Strake at Strength Deck

Container Carrier의 경우 0.6L이 Cargo Region 으로 바뀜

0.4L

0.6L

III IV III

II II

C

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2 Table B2

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Innovative ship design – Rule scantling

2.2 Material Classes - Typical Example (4)

C3 Deck Strake at longitudinal bulkhead

Structural member 0.4 L Outside 0.4 L

C3 Deck strake at longitudinal bulkhead.

C6 Bilge strake.

IV

III

(II outside 0.6L amidships)

C6 Bilge Strake 0.4L

0.6L

III IV III

II II

C

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2 Table B2

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Innovative ship design – Rule scantling

2.2 Material Classes - Example ¹⁾

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2 Table B2

(예제) 아래 Table B2에 따라 다음에 보인 VLCC Midship Section에 부합하는 Steel Grade를 기입하시오.

1)

DSME, “선박구조설계” 2-20 예제1) , 2005.8

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Innovative ship design – Rule scantling

2.2 Material Classes - Example ¹⁾

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2 Table B2

(예제) 아래 Table B2에 따라 다음에 보인 VLCC Midship Section에 부합하는 Steel Grade를 기입하시오.

1)

DSME, “선박구조설계” 2-20 예제1) , 2005.8

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Innovative ship design – Rule scantling

2.2 Material Classes - Example ¹⁾

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2 Table B2

(예제) 아래 Table B2에 따라 다음에 보인 VLCC Midship Section에 부합하는 Steel Grade를 기입하시오.

(풀이)

B2 Strength Deck

“III”

‘AH’

‘AH’ ‘AH’ ‘AH’ ‘AH’ ‘AH’ ‘AH’ ‘AH’ ‘AH’

C1 Sheer Strake

‘DH’ “IV”

A3 Side Plating

“II”

‘A’

B1 Keel Plate ‘AH’

“III”

B1 Bottom Plate

“III”

‘AH’ ‘AH’ ‘AH’ ‘AH’ ‘AH’ ‘AH’ ‘AH’ ‘AH’ C6 Bilge Strake

“IV”

‘DH’

1)

DSME, “선박구조설계” 2-20 예제1) , 2005.8

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Innovative ship design – Rule scantling

2.3 Corrosion Addition (tk , tc)

1)

DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2

* 부식(Corrosion):금속재료가 사용 환경 중의 물질과 반응해서

금속이온 또는 비금속 화합물이 되어 소모되어 가는 현상

) 8 (

. 15

1 k s p t mm

t = ^a + _k

σ

ex) Bottom plate thickness ¹⁾

Corrosion Addition을 더해줌

 In tanks for cargo oil and or water ballast the scantlings of the steel structures shall be increased by corrosion additions. ¹⁾

 각 선급에서는 실제 운항하는 선박의 정기 검사를 통해 얻어진 자료를 바탕으로 각 구역별로 Corrosion Margin(0.5~3mm)을 고려한 설계를 규정 함 ²⁾ (DnV Rules, Jan. 2004,Pt.3 Ch.1 Sec.2 Table D1, D2 참조)

2)

DSME, “선박구조설계” 2-10 부식 , 2005.8

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3.1 단면의 중심(Center of area)과 관성모멘트(Moment of inertia)

3.2 단면계수(Section modulus)

3.3 소성모멘트(Plastic moment)와 소성계수(Plastic Modulus)

3.4 유효폭(Effective Breadth)

3. Sectional Properties

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3. Sectional Properties

1) 선체구조물의 특성

: 설계 하중을 견딜수 있도록 ‘판(Plate)’과 ‘보강재(Stiffener)’를 이용하여 설계를 함 .

구조물의 역할 및 위치를 충분히 고려하여 작용하는 하중에 견딜 수 있도록 해야 함 .

2) 하중에 영향을 주는 단면 특성

① 선박에 작용하는 주요 하중의 좌표성분에 따른 분류

: 굽힘하중(Bending Moment), 전단하중(Shear Force), 축하중(Axial Force)

② 하중에 영향을 주는 단면 특성 : 단면적 → 전단하중, 축하중

단면계수→ 굽힘하중

3) 목적

: 단면의 특성을 이해하고, 작용하는 하중을 분석하여, 작용하중에 따른 최적의 부재 배치를 하기 위함

1)

대우조선해양 , 선박구조설계, 3장, 2005.

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3. 1 단면의 중심과 관성 모멘트

: Centroid and Moment of Inertia ¹⁾

dA

x y

y

x . C G

O x

y

A ,

A

xdA x

dA

= ∫

∫ ^,

A

ydA y

yA

= ∫

∫

② If an area is symmetric about an axis, the centroid must lie on that axis because the first moment about an axis of symmetry equals zero.

③ In general, the coordinates may be positive or negative, depending upon the position of the centroid with respect to the reference axes.

2) Moment of Inertia _{(관성모멘트)}

2 ,

x A

I = ∫ y dA I ^y = ∫ A x dA ² ^, I ^xy = ∫ A x y dA ^,

1) Centroid _(도심)

① The Coordinate of the centroid C are equal to the first moments divided by the area

( ^{x y} ^, )

① dA(differential element of area) is multiplied by the square of the distance from the reference axis( Also called “second moments of area”)

② If a different set of axes is selected, the moment of inertia will have different value

• I _x , I _y : x축,y축에 관한 관성모멘트(Moment of Inertia)

• I _xy : x,y축에 관한 관성모멘트적(Product of Inertia)

1) James M. Gere, Mechanics of Materials 6^thEdition, Thomson, Chap.12, pp.880~094

( ^{x y} ^, )

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3. 1 단면의 중심과 관성 모멘트

: Parallel-axis theorem for moments of inertia ¹⁾

2 2

x x , y y

I _′ = + I b A I _′ = I + a A

 Parallel-axis theorem for moments of inertia . (평행축 정리)

x y xy , p x y

I _{′ ′} = I + abA I _′ = I _′ + I _′

[예제] 그림과 같이 직사각형이 넓이 6cm, 높이 12cm의 단면이 있다. 새로운 축 x’ 에 관한 관성 모멘트 (Moment of Inertia) Ix를 구하여라.

y

x . C G

O x′

y′

b a

x 5 cm 12 cm

6 cm

x

[풀이]

I _x _′ = + I _x A b ²

( )

3 2 4

6 12 6 12 11 9, 576

12 cm

×  

= + × =  

① the moment of inertia of an area with respect to any axis in its plane is equal to the moment of inertia with respect to a

parallel centroidal axis plus the product of the area and the square of the distance between the two axes.

② When using the parallel-axis theorem, one of the two paralled axes must be a centroidal axis.

1)James M. Gere, Mechanics of Materials 6^thEdition, Thomson, Chap.12, pp.880~094

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3. 1 단면의 중심과 관성 모멘트 : Rotation of axes ¹⁾



Rotation of axes _{(축의 방향전환)}

① The Moments of inertia of a plane area depend upon the position of the origin and the orientation of the reference axes.

② For a given origin, the moments and product of inertia vary as the axes are rotated about that origin.

③ Below equations give the moment of inertia I

_x1

and the product of inertia I

_x1y1

with respect to the rotated axes in terms of the moments and product of inertia for the

original axes. These equations are called “Transformation equations for moments and products of inertia”

y′

x′

O x

y

y′

x′

x y

θ

( ) ( )

2 2

1 1

cos sin sin 2 cos 2 sin 2

2 2

x x y xy x y x y xy

I _′

=

I θ

+

I θ

−

I θ

=

I

+

I

+

I

−

I θ

−

I θ

( ) ( )

2 2

1 1

sin cos sin 2 cos 2 sin 2

2 2

y x y xy x y x y xy

I _′

=

I θ

+

I θ

+

I θ

=

I

+

I

−

I

−

I θ

+

I θ

( )

^{sin cos}

(

^cos

²

^sin

² )

¹

( )

^{sin 2} ^{sin 2}

x y x y xy

2

x y xy

I _{′ ′}

=

I

−

I θ θ

+

I θ

−

θ

=

I

+

I θ

+

I θ

1 1

x y x y

I

+

I

=

I

+

I

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3. 2 단면계수(Section Modulus) ¹⁾



Section Modulus _{(단면계수)}

① The Maximum tensile and compressive bending stresses acting at any given cross section occur at points located farthest from the neutral axis.

② Corresponding maximum normal stress σ

₁ , σ ₂

are

1 2

x , x

a c

I I

Z Z

y y

= =

1 2

y , y

b d

I I

Z Z

x x

= =

………(x축)

………(y축)

O

x y

. x C G

y d

c a

b y 1

y 2

x 1

x 2

M

x y

σ 2

σ 1

2 dA

y y 1

( y 1

and

y 2 )

1 2

1 , 2

a c

My M My M

I Z I Z

σ = − = − σ = =

③ The quantities Z

_a , Z _c

are known as the section Modulus.

④ Section modulus with respect to X-axis and Y-axis

⑤ Each Section modulus has dimension of length to the third power

⑥ Distance from neutral axis to top and bottom of the beam are always taken as positive quantities.

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3. 3 Plastic Moment & Plastic Modulus ¹⁾



Plastic moment _{(소성모멘트)}

① Beam is in the same condition as a beam in ordinary elastic bending with a linear stress distribution. (a)

② The preceding condition exist until the stress in the beam at the point farthest from the neutral axis reaches the yield stress σ

Y

.(b)

③ If we increase the bending moment above the yield moment, the outer regions of the beam have become fully plastic while a central core (called elastic core) remains linearly elastic. (c)

④ As bending moment increases still further, the beam has reached its ultimate moment resisting capacity, and we can idealize the ultimate stress distribution as consisting of two rectangular parts. The bending moment called

“Plastic moment”(d)

( ) ^a ( ) ^b

( ) ^c ( ) ^d

z

y

z

x

σ Y

c

y

Y y

M I S

c

σ σ

= = ⁽ ^M ^Y : Yield Moment

c : farthest point from the N.A)

M

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④ Therefore, under fully plastic conditions, the neutral axis divides the cross section into two equal areas.

⑤ Since the plastic moment M

_P

is the moment resultant of the stresses acting on the cross section, it can be found by integrating over the cross sectional area A

3. 3 Plastic Moment & Plastic Modulus ¹⁾



Plastic Modulus _{(소성계수)}

① If we consider the cross section under fully plastic condition,

② Every point in the cross section above the neutral axis is subjected to a tensile stress σ

Y ,

and every point below the neutral axis is subjected to a compressive stress σ

Y

. (e)

③ Since the resultant force acting on the cross section is zero, it follows that

( ) ^e ( ) ^f

z

y

1 2

Y A Y A

σ = σ

( T : Tensile force C : Compressive force )

O A 1

A 2

dA y

z

x

M

σ Y

T = C

1 2

A = A

1 2

A

=

A

+

A

1 2

2 A = A = A

( A

₁

: Area above N.A

A

₂

: Area below N.A , A : Total Area )

( ) ( ) ( ) ( 1 2 )

1 1 2 2

1 2 2

Y

P A Y A y Y Y

A y y

M ydA ydA y A y A σ

σ σ σ σ ⁺

= − ∫ − − ∫ = − − =

④ Plastic modulus

( 1 2 )

2 Y P

A y y

M σ +

= M P = σ Y Z ( ₁ ₂ )

2 A y y

Z +

=

C

T

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3. 3 Plastic Moment & Plastic Modulus ¹⁾

3 4

i) Moment of Inertia : I

_x

=12 ×6 / 12=864

cm

y′

x′

x y′

x′

x

30 ^ [예제] 그름과 같은 직사각형의 x, y축에 대한 탄, 소성

단면계수 (Sectional Properties)를 구하여, Shape factor를 구하고, 새로운 x’, y’축에 관한 Moment of Inertia를 구하여라. ¹⁾

y′

y

y 12

6 6 [풀이]

3 4

2

6 12 / 12 216 , 0, 12 6 72

y

xy

I cm

I A cm

= × =

= = × =

ii)Section Modulus : Z

_x

=864 / 6 144=

cm 3

Z

_y

=216 / 3=72

cm 3

( ) ³

iii) Plastic Section Modulus : Z

_px

= 12 6 / 2× × × =3 2 216

cm

( ) ³

Z

_py

= 12 6 / 2× ×1.5 2 108× =

cm

Z 216

iv)Shape factor for x-axis : 1.5 144

px

Z x

= = Z 108

, 1.5

72

py

Z y

= = iv) Moment of Inertia with respect to x' and y' axis :

( ) ( ) ⁴

1 1

864 216 864 216 cos 60 702

2 2

I x _′

= + + −

^

≅

cm

( ) ( ) ⁴

1 1

864 216 864 216 cos 60 378

2 2

I x _′

= + − −

^

≅

cm

( ) ⁴

1 864 216 sin 60 281

x y

2

I _{′ ′}

= −

^

≅

cm

1)대우조선해양, 선박구조설계, 3장, 2005.

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3.4 유효폭(Effective Breadth)



유효폭(Effective Breadth of Attached Plate) ¹⁾

① Lateral Pressure를 받는 Tank Top, Deck, Shell 및 Bulkhead 등 주요 선체 구조 는 하중을 원활이 견딜 수 있도록 보강재 (Stiffener)와 함께 Girder system으로 설계하며 , 대부분 Grillage 구조를 갖는다

② 따라서 , 직접 구조해석 및 보이론에 의한 구조 계산시, 유효폭 (단면계수에 실 질적으로 기여하는 폭 )을 포함하는 단면의 특성치를 적용하고 있음

ex) DNV Rule : effective flange of girder ²⁾

The effective plate flange area is defined as the cross sectional area of plating within the effective flange width. Continuous stiffeners within the effective range may be included. The effective flange width b _e is determined by the following formula.

e ( )

b =  C b m

C = as given in Table for various numbers of evenly spaced point loads(r) on the span b = sum of plate flange width on each side of girder, normally taken to half the

distance from nearest girder or bulkhead 실제폭 유효폭

reduction Factor

1)

대우조선해양 , 선박구조설계, 3.4 선체구조 단면특성, 2005.

2)

DNV rules for ships, Pt.3 Ch.1 Sec.3 C400

b

b e

b

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4.1 Introduction

4.2 Boundary Condition(경계조건)

4.3 Shear Force & Bending Moment Diagram 4.4 Bending Stress

4.5 Deflection of beam

4. Beam Theory

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4. 1 Introduction

2) 가정 1) Beam

① 재료는 Hook’s Law를 따름

① 단면의 치수(폭 혹은 높이)에 비해 길이가 상대적으로 긴 구조물이나 구조부재가 적당한 방법으로 지지되어 있고

② 그 중심 축을 포함하는 평면 내에서 그 면을 굽히려는 힘, 즉 횡하중(길이의 수직 방향으로 가해지는 힘 )을 받는 구조물 또는 구조부재

1)

대우조선해양 , 선박구조설계, 6-1 Beam Theory, 2005.

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4. 2 Boundary Condition(경계조건)

2) 단순 지지 (Simple Support) 1) 자유단(Free)

수평 , 수직 방향의 병진 운동과 회전 운동이 모두 자유로움

한 방향의 병진 운동만 구속함

3) 고정 지지 (Fixed, Embeded)

수직 , 수평 방향의 병진 운동과 회전 운동이 모두 구속됨

0, 0

y ′′ = y ′′′ =

0, 0

y = y′ =

= 0 y

0 EIy′′′ =

= 0 y ′

0 EIy′′ =

No deflection

Slope of deflection curve is 0 Bending moment is 0

Shear force is 0

0, 0

y = y′′ =

경계조건의 종류

Cantilever beam : embedded at the left end, free at the right end

X=0 X=L

ex) 경계조건의 예 (Cantilever)

1)

대우조선해양 , 선박구조설계, 6-2 경계조건, 2005.

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4. 3 Shear force & Bending Moment Diagram

Given) 길이가 l이고 단순 지지된 보에 균일 분포 하중 w가 작용할 때

Find) Shear force & Bending moment 분포도 y w

x L

, V M

w

,

V + dV M + dM

dx

(1) ∑ F = 0

( ) ⁰

V − w dx − V + dV = dV w

dx = − ( ) 1

V = ∫ − w dx = − w x C + ( 0)

2 V x = = wl

2 V = − wx + wl

1 2

C wl

→ =

(2) ∑ M = 0

( ) ⁰

2 M Vdx w dx dx M dM

− − + + + =

미소량인 고차항 무시하면 dM

dx = V 2

M = ∫    − wx + wl    dx

2 2 2 2

w wl

x x C

= − + +

x=0,l 일때 , M=0 이므로, C ₂ =0이다 .

2 2 2

w wl M = − x + x

2 wL

2 − wL

2 8 wL

▶Shear force diagram

▶Bending Moment diagram

• 2009년 창의적선박설계 강의교재, (090530)탄성학1-부호,탄성선

• ref : Gere J.M., Mechanics of Materials, 6th edition, Thomson, 2006, p300~306

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4. 4 Bending Stress

1 d dx

θ ρ

⁼

ε σ = E

•neutral

•surface y

ρ

dx

dθ

: radius of curvature

ρ

① definition

② from the geometry and linearization

, d

dx

κ

=

θ

③ change in length

ds y dx dx ds′

ρ

dx

− = ′ −

x

ε

y

= −

ρ

④ strain or

ε

_x = −

κ

y

x E x

σ

=

ε

⑤ stress ^x y

E or y

σ κ

∴ = −

ρ

−

⑥ moment

dM = − σ x ydA _, ²

I = ∫ A y dA 1

M EI ⁼ ρ

M EI = κ

or

,

κ

1 : curvature

=

ρ

⑦ Flexure formula

x y

( ) f x

x

My σ = I

x

E y

σ

^{= −}

ρ

⑤

Maximum Stress at farthest point.

Let farthest point as c ₁ , c ₂

⑧ Maximum Bending Stress

1 2

max M c , min M c

I I

σ = σ =

1 2

I , I

Z Z

c = c =

section modulus Z ₁ , Z ₂

max min

1 2

M , M

Z Z

σ = σ =

 Bending Stress 유도과정

• 2009년 창의적선박설계 강의교재, (090530)탄성학1-부호,탄성선

• ref : Gere J.M., Mechanics of Materials, 6th edition, Thomson, 2006, p300~306

•neutral

•axis

•<section AA’ >

dA y

y

c x A

A

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4. 5 Deflection of Beam

1 d dx

θ ρ

⁼

ε σ = E

neutral

surface y

ρ

dx

dθ

: radius of curvature

ρ

① definition

② from the geometry and linearization

, d

dx

κ

=

θ

③ change in length

ds y dx dx ds′

ρ

dx

− = ′ −

x

ε

y

= −

ρ

④ strain or

ε

_x = −

κ

y

x E x

σ

=

ε

⑤ stress ^x y

E or y

σ κ

∴ = −

ρ

−

⑥ moment

dM = − σ x ydA _, ²

I = ∫ A y dA 1

M EI ⁼ ρ

M EI = κ

•or

,

κ

1 : curvature

=

ρ

⑦ relationships between loads, shear forces, and bending moments

dM

( )

dx

=

V x

moment equilibrium

dV

( )

dx

= −

f x

force equilibrium

⑧ by the linearization ( )

dx dx dy

ds

≈ ,

θ

≈ tan

θ

=

y

x

, V M

( ) f x

,

V + dV M + dM

2 2

d y

κ = dx ²

2 EI d y M dx =

2 2

d y M dx EI

∴ = d y ³ ₃

EI V

dx = ⁴

4 ( )

EI d y f x dx = −

x y

( ) f x

from 4.4 Bending Stress

• 2009년 창의적선박설계 강의교재, (090530)탄성학1-부호,탄성선

• ref : Gere J.M., Mechanics of Materials, 6th edition, Thomson, 2006, p300~306

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4. 5 Deflection of Beam

1 d

ds

θ

=

ρ

neutral

surface y

ρ

dx

d θ

ds ds′

x

σ x

dV ( )

dx = − f x , dM ( ) dx = V x

2 2

M EI d y

= − dx

4 4 ( )

EI d y f x

∴ dx =

y

y 1

d ds

θ

=

ρ

neutral

surface y

ρ

dx

d θ

ds ds′

x y

σ x

2 2

M EI d y

= dx

4 4 ( )

EI d y f x

∴ dx = −

•4 3/

• 2009년 창의적선박설계 강의교재, (090530)탄성학1-부호,탄성선

• ref : Gere J.M., Mechanics of Materials, 6th edition, Thomson, 2006, p300~306

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Deflection of beam

Example 1 – Cantilever beam

( )

f x

=

w

y

x

4 4 ( )

EI d y f x dx =

recall, differential equations of deflection curves*

Integrate four times Boundary Condition

( ) ( ) ( ) ( )

0 0 0 0

0 0 EIy EIy y L y L

′′′ =

′′ =

′ =

=

distributed load f x ( ) = w

4 4

( )

d y x

EI w

dx

=

4 3 2

1 2 3 4

1 1

( ) 24 6 2

y x w x c x c x c x c

= EI + + + +

3 2

1 2 3

@ @ Innovative ship design-MidshipSection Rule Scantling -

@ SDAL

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N av al A rc hite ctu re & O ce an E ngin ee rin g

@ SDAL Advanced Ship Design Automation Lab.

http://asdal.snu.ac.kr Seoul

National Univ.

May 2009

Prof. Kyu-Yeul Lee

Department of Naval Architecture and Ocean Engineering, Seoul National University of College of Engineering

[2009] [13][14]

Innovative ship design

-Midship Section Rule Scantling -

@ SDAL

Innovative ship design – Rule scantling

N av al A rc hite ctu re & O ce an E ngin ee rin g

@ SDAL Advanced Ship Design Automation Lab.

http://asdal.snu.ac.kr Seoul

National Univ.

Rule Scantling

(선박 구조 설계)

@ SDAL

Innovative ship design – Rule scantling

Ship Structural Design

 Ship Structural Design

what is designer’s major interest?

 Safety :

Won’t ‘IT’ fail under the load?

a ship a stiffener a plate

z

x

L

x

y f x ( )

global local

a ship

Actual stress on midship section should be less than allowable stress

.

act allow

σ ≤ σ

Allowable stress by Rule : (for example)

, σ

= 175 f

[ N mm / ]

, act . mid mid

σ M

= Z

L

x

y f x ( )

( ) f x

x

y

F

( ) V x

( ) M x

( ) y x

( ) ( ) d y x

EI f x

dx = − what is our interest?

: ( ) : ( ) : ( ) Shear Force V x Bending Moment M x Deflection y x

: ( ) Load f x

cause

( ) ( ) dV x f x

dx = − ( )

, dM x ( ) dx = V x

, d y x ( ) ( )

EI M x

dx =

‘relations’ of load, S.F., B.M., and deflection

 Safety :

Won’t it fail under the load?

 Geometry :

How much it would be bent under the load?

,

M M

where

I y Z

σ = =

act allow

y ^{f x} ( )

, _act . ^mid mid

y ^{f x} ( )

R A R _B

x y ^f

^x

y ^{f x} ( )

σ _. ≤

, _act . ^mid ^S ^W

y ^{f x} ( )