Innovative Ship Design

- 3rd Exam -

Friday, June 15th, 2012

14:00-17:00 (3 hours)

Name

SNU ID #

Note: Budget your time wisely. Some parts of this exam could take you much longer

than others. Move on if you are stuck and return to the problem later.

Problem

Number

1 2 3 4 5 6 Total

1 2 3 4 1 2 3 1 2 3 - 1 2 1 2 3

Grader Max 3 2 10 5 3 10 2 5 10 5 5 5 5 5 10 15

/100 Score

[1] A barge ship is floating on the sea as shown in following figure.

L=40m, B=10m, D=10m, T=5m

Density of sea water:� 1Mg/m3

Added mass moment of inertia Iadded is same as the

added mass moment of inertia of the ship.

T

L

B

Dy

z

The equation of roll motion is as follows:

.( ) ( )Gravity Buoyancy F K D Radiation M M M M M

Re ( )storing Exciting M M

( ) ( )I M Body Force Surface Force

( ) ( )Gravity Fluid M M

addedI B

( )Gravity M is the moment due to the gravitational force.

( )Buoyancy M is the moment due to the buoyant force.

.F KM is the moment due to the Froude-Krylov force.

DM is the moment due to the diffraction force.

( , )Radiation M is the moment due to the radiation force.

Two components ( )Gravity M and ( )Buoyancy M produce transverse restoring moment

Re ( )storing M .

Sum of two components . ( )F K M and ( )D M is exciting moment ( )Exciting M .

( , )Radiation M is composed of addedI and B , where addedI is the added mass moment of

inertia and B is roll damping coefficient.

z

y

O

CL

y

z,

FB

FG

Z

B1

B0

K

G

M

y

z( )

i

j

RestoringM

oy'z' : Body fixed frame

oyz : Waterplane fixed frame

G : Center of gravity

K :Keel

B0 : Center of buoyancy

B1 : Center of buoyancy at rotation angle ф

FG : Weight of a ship

FB : Buoyancy force

M : The metacenter is defined as the intersection point of the two lines of action of the buoyant force before and after

inclination. Unless there is an change in the shape of the ship in vicinity of the waterline, point M will remain practically

stationary as the ship is inclined to small angles.

[1.1] Determine the transverse restoring moment in terms of GM with proper assumption. [3 points]

[1.2] Drive the equation of roll motion of the ship. [2 points]

[1.3] Find a general solution of roll motion ( )t assuming there is no exciting moment and damping

moment. [10 points]

[1.4] Drive the relation between GM and natural period of roll motion. [5 points]

[2] Buckling and Euler load

Consider a long slender vertical column of uniform cross-section and length L. The vertical load P

is acting on the top of the vertical column.

[2.1] Derive the differential equation for column buckling based on the differential equation of

the deflection curve of a beam. [3 points]

- Differential Equation of the deflection curve of a beam

2

2

d yM EI

dx

[2.2] Determine Euler load(smallest critical loads). [10 points]

[2.3] To maximize Euler load, where the physical restraint should be put on? [2 points]

x

y

P( ,0,0)P F=

L

[3] Grillage Analysis

Stiffness equations of a beam and shaft, whose length is L, are as follows:

Stiffness equations of a beam:

3 2 3 2

1 1

2 21 1

22

3 2 3 2

22

2 2

12 6 12 6

6 4 6 2

12 6 12 6

6 2 6 4

y y

z z

yy

zz

EI EI EI EI

L L L Lf EI EI EI EI

M L L L L

f EI EI EI EI

L L L LM

EI EI EI EI

L L L L

Stiffness equations of a shaft: 1 1

2 2

x x

x x

GJ GJ

M L L

M GJ GJ

L L

[3.1] Derive the stiffness equations of a grillage structure in the xz-plane for the element 1 and

element 2. [5 points]

xz

y

1yf 2yf

L

,Mz1 Mz2

Mx1 Mx2

1, y 2, y

1, x

1z

2, x

2, z

Element 1 Element 2 Element 1

[3.2] Find displacement and reaction force at each nodes in the following figure by using the stiffness

equations of the grillage structure derived from the problem [3.1] [10 points]

- Moment of inertia I (m4): 16.6×10

-5

- Young’s modulus E (kN/m2): 210×10

6

- Shear modulus G (kN/m2): 84×10

6

- Polar moment of unertia J (m2): 4.6×10

-5

21kN

x

z

y

element 1

element 2

node 1

node 3

node 2

4m

2m

[3.3] The stiffness equation of the grillage structure is derived in the xyz-frame. But the XYZ-frame will

be used in general grillage analysis problem. Perform the coordinate transformation of the stiffness

equation from the xyz-frame to the XYZ-frame. [5 points]

x

z

y

X

Y

Zrotation about

x-axis with -90°

the present coordinate system

xyz-frame

the new coordinate system

XYZ-frame

[4] Write down the name of cargo hold structure members of a tanker. [5 points]

[5] Hull girder strength and Midship section rule scantling

[5.1] Derive the required section modulus(Zreq) for the scantling of the longitudinal stiffener

based on the beam theory. [5 points]

2

3

.

83 kreq

l

l s p wZ cm

p: “pressure” on the load point for the stiffener

l: stiffener span

s: stiffener spacing

l : allowable stress

kw : section modulus corrosion factor in tanks

Unit: 2 2 , , , p kN / m s m l m σ N / mm

[5.2] Derive the required thickness (treq) for the scantling of the plates based on the beam theory.

[5 points]

.

15.8( )a

req k

l

k s pt t mm

p: “pressure” on the load point for the plate

s: span between longitudinals

t: plate thickness

N.A: neutral axis

1: unit length of strip of the plate

l : allowable stress

ak : correction factor for aspect ratio of plate field

kt : corrosion addition

Unit: 2 2, , , t mm s m p kN / m σ N / mm

[6] Rule Scantling for container ship of LBP 316.45m based on the basis ship of 7,000TEU shown in

following table.

Basis Ship(7,000 TEU) Design Ship

Main Dimensions

LOA(m)

LBP(m)

Ls(m)

B mld(m)

D mld(m)

d(design) (m)

d(scant) (m)

Dmld, to second deck (m)

Cb at Td

Cb at Ts

302.0

288.0

284.25

40.0

24.2

12.0

14.5

20.144

0.691

0.721

316.45

302.45

298.4

40.0

24.2

12.0

14.5

20.144

0.725

0.76

Deadweight(t)

(design/scant.)

70,367/96,026

78,900/107,670

Table 1. Principle Particulars

Sectional area of midship for

half part(cm2)

First moment of sectional area

about baseline for half part [cm3]

moment of inertia of

sectional area about

neutral axis [cm4]

Total data 27,160 30.714x106 522.542 x108

Table 2. Sectional Properties of the Basis Ship

Midship Section of the Basis Ship

[6.1] Calculate the rule required minimum design still water bending moment and vertical wave

bending moment of the design ship in accordance with the DNV Rule(Part 3, Chapter 1, Section 5.

longitudinal strength, B. 100, B. 200). [5 points]

[6.2] Calculate the actual hull girder bending stress acting on the bottom and the deck. And

check that the actual bending stress is satisfied with allowable bending stress at midship by the

DNV rule(Part 3, Chapter 1, Section 5. longitudinal strength, C. 300). [10 points]

[6.3] The plate and the longitudinal stiffeners located at the deck indicated in following figure are

most critical members in affecting the actual bending stress. Determine its thickness that satisfy

the allowable bending stress at midship by the DNV rule(Part 3, Chapter 1, Section 5.

longitudinal strength, C. 300). [15 points]

: Plate

: Longitudinal

DNV Rule