advanced structure analysis
TRANSCRIPT
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Arab academy for Science, Technology and Maritime Transport
Submitted to:Dr. El Sayed Hegazy
Prepared by:Eng. Mohamed Abdel Aziz Fahmy
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Advanced Structure Analysis
Introduction
1- Longitudinal strength calculation2- Local strength calculation3- Transverse strength calculation4- Docking strength calculation5- Building berth calculation6- Launching calculation7- Strength calculation in damaged condition8- Strength after repair
1- Longitudinal strength calculation
CH.1: Loads acting on ships during service
There are two types of loading (stresses) acting on a ship duringservice:
a) General structure stresses: which appear at any point along theships length (with different values)
b) Local stresses: which appear in certain places due to local loadsacting on it
a) Generalstructure stresses
Water pressure Racking stressesLongitudinal
bending of shipshull
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10 2 3 4 5
Stepped Weight Curve (WC)
Bouyancy Curve (BC)
LoadCurve
Shear Force Diagram (SFD)
Still Bending Moment Diagram (M still )
Zero Shear
Qmax
Mmax
B
G
6 7 8 9 10
W
L
FPAP
BonjeanCurves
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To be noticed that:
The calculating components (MStill, MHog , MSag ) are called static components of bending moment There is another component; dynamic component of bending moment(MDynamic ) to be added to the above mentioned components to get thetotal bending moment
MTotal = MStatic + MDynamic
There are imperial formula to calculate MDynamic
Stress Distribution on Mid-Ship SectionFrom simple beam theory we get:
I y M .
where,
M: Bending moment acting on the sectionI : Moment of inertia about the neutral axis of the sectiony: Distance of the point in the section at which a stress to be calculated from
the neutral axis
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Calculation of the moment of inertia of a mid-shipsection about the neutral axis of the section
Steps:
1. Draw the mid-ship section of the ship showing only all longitudinalstiffeners with its scantling
2. Numbering all these longitudinal members
No Item A i(Scantling Area)
Xi (Distancefrom base
line)
M (Moment
about baseline)
Yi (xi e)
**
ii***
I i (ii + A i * (x i e) 2
i (M/ I * yi)
123...
A M
** e = M / A
*** ii = I + ( Ai*L2
)
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Approximate formula for calculating maximum staticbending moment (M max ) at mid-ship section
Mmax = L / k , where k = 29 32 passengerin tons 22 30 cargo
Qmax = / k L : in meters 40 50 tanker35 40 Bulk carrier
Notice:
Shearing force at any section is equal to the sum of all forces locatedon the right or the left of the section
Bending moment at any section is equal to the sum of all momentslocated on the right or the left of the section
e.g.:A box barge has 5 compartments 200 m long. The weight of the barge is
uniformly distributed along the length. 500 tons of cargo is added at each ofthe end compartments.
Draw the shearing force diagram (S.F.D) and bending moment diagram(B.M.D) and find the value of maximum bending moment.
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W.C.
500 ton 500 ton
500/40 12.5 t/m
1000/200 5 t/m
7.5 t/m 12.5 - 5
A B
Q A =Q max =7.5*40 =300
Zero Shear
MA = (7.5*40)*20 = 6000
B.C.
Load Curve
S.F.D.
B.M.D.
MB = [(7.5*40)*80 ]-[(5*60)*30 ] = 24000 - 9000 = 15000
QA = 7.5 * 40 = 300 . . . . . . . . . . . . Q maxQB = 0 . . . . . . . . . . . . Zero shear
MA = 7.5 * 40 * 20 = 6000 MB = 300 * 80 300 * 30
= 24000 9000= 15000 . . . . . . . . . . . . M max
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Solution of statically undetermined beams
In these problems the number of unknowns more than three
qMA MB
YA YB
Unknowns = 4
Equilibrium Equations X = 0 Y = 0 M = 0 Eq. 4
Compatibility Equation:
1. Super position method
M A M BP
Y A Y B
M = 0 M A = MB Y = 0 Y A = YB = P/2 X = 0 = = 0
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EI PL
EI L M
EI L M
B
A
16
6
3
2
3
2
1
A =
2. Three Moments Equations
M A M D
Y A Y D
B C
M B M Cq(x)
A D
6 unknowns (M A, MB, MC, MD, YA, YD), So 6 equations are needed:
A = 0 BA = BC Y = 0
D = 0 CB = CD M = 0
From 3:
)()(6336
X q
BC C B
X q
BA B A
EI L M
EI L M
EI L M
EI L M
From 4:
)()( 6336 X qCD
DC X q
CBC B
EI L M
EI L M
EI L M
EI L M
M A
M B
P
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e.g.:Draw the Bending Moment Diagram for the following continuous beam.
4 unknowns (M 0, M1, M2 and M 3), So 4 equations are needed:
0 = 0 10 = 12 3 = 0 21 = 23
06324
)()()(
103
0
1000000
EI L M
EI L M
EI Lq
M M q
03645
)()()(
323
332333
EI L M
EI L M
EI qL
M M q
82
201
0
Lq M M
1152
22
3
qL M M
2
EI L M
EI L M
EI Lq
M M q
3624
)()()(
103
0
11001001010 3
EI L M
EI L M
EI PL
M M P
6316
)()()(
212
2121121212 4
3=4 PL Lq
M M 375.04
20
20 5
EI L M
EI L M
EI PL
M M P
3616
)()()(
212
2211212121
6
EI L M
EI L M
EI qL
M M q
633607
)()()(
323
3232232323
7
6=7 qLPL M M 467.0375.031 8
From 1, 2, 5, 8 the unknown moments M0, M1, M2 and M 3 can be determined
M0 = ]375.3636)3.04
([ 0 Pqq
L L M1 = ]18625.17074.0[ PqL L
M2 = ]36363.0[ PqL L M3 = ]181812[ PqL
L
qPq (t/m)
0 1 2 3
o
M max =0.1283qLat 0.5774L
M max =PL/4at point of load
M max =q L/8at 0.5L
o
M 0 M 1 M 2 M 3
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EX.:Draw the Bending Moment Diagram for the following continuous beam.
2 unknowns ( M 1 and M 2), So 2 equations are need
10 = 12 21 = 23
EI
L M
EI
qL
M M q
345
)()()(
13
1100101010 1
EI
L M
EI
L M
LEI
b LPab
M M P
636
)(
)()()(
21
2121121212 2
1=2
2
621
31
3
636*645
345 M
EI L M
EI L M
LEI PL
EI L M
EI qL L
EI
3
EI
L M
EI
L M
LEI a LPab
M M P
366)(
)()()(
21
2211212121 4
EI L M
EI Lq
M M q
324
)()()(
23
0
32322302323 5
4=5 EI
L M EI Lq
3242
30 6
From 3, 6 the unknown moments M1, M2 can be determined
q P q
0 1 2 3
o
M max =0.1283qLat 0.5774L
M max =3PL/16at point of loadM max =q L/8at 0.5L
o
M 0 M 1 M 2 M 3
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i j
M i,i-1 M i,j
j+1i-1
Q
i j
M ijQ
M ji
L ij
Slop Deflection Method
It has advantage to the other methods that it can be used in case of springtype support
Sign convention:CW CCW
For Moment: negative positive +ve -ve
B.M.D +ve -ve
Slop : positive negative
: Rigid Body Rotation
L A A B B
tan
Note:In case of rigid body support = 0
L I stiffness
Slop Deflection equation:
Assume the moment direction in +ve
ijijiij
ij ji
ij
ijiji Q EI
L M
EI
L M )(
63 . . . . . . .1
ijij jij
ij ji
ij
ijij j Q EI
L M
EI
L M )(
36 . . . . . . .2
Solving 1 & 2 for M ij & M ji . . . . . .
=+ve =-ve
AB
P
A' B'
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]32[2
)]()(2[2
ij jiij
ijij jiji
ij
ijij L
EI QQ
L
EI M . . . . . . .3
]32[2
)]()(2[2
iji jij
ijijiij j
ij
ij ji L
EI QQ
L
EI M . . . . . . .4
Assume supports i & j are fixed ,so
)]()(2[2
ij jijiij
ijij QQ L
EI M M ij . . . . . . .5
)]()(2[2
ijiij jij
ij ji QQ L
EI M M ji . . . . . . .6
Equation 5 & 6 are obtained on the assumption that the endsi & j are fixed.
Therefore the moments M ij & M ji are called fixed endsmoments and they can be obtained directly from tables ofbending of simple beams.
For example:
P
i j
M ij M ji
La b
q
i j
M ij M ji
L
q
i j
M ij M ji
L
M ij 22
LPab
12
2qL 30
2qL
M ji 22
LPba
12
2qL 20
2qL
So equations 3 & 4 become
]32[2 ij jiij
ijijij L
EI M M . . . . . . .7
]32[2
iji jij
ij ji ji L
EI M M . . . . . . .8
By applying equilibrium equations at each support we get:
01, ijii M M
0]32[2]32[2 1,11,
1,1, ij ji
ij
ijijiiii
ii
iiii L
EI M
L EI M . . . . . . 9
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Equation 9 contains three unknowns jii ,, 1 is called ThreeSlops EquationApplying equilibrium equations at any support, we get thenecessary number of equations to find out the unknownslops at all supportsSubstitution the slops in equations 7 & 8, we get theunknown moments at each support
Note:When drawing the B.M.D. draw the arrow direction andneglect the signDetermine the moments value
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Moment Distribution Method
This method is useful in case of more than two members meetingin one joint
Sign convention:CW CCW
For Moment: negative positive +ve -ve
B.M.D +ve -ve
Slop : positive negative
m i = Applied external moment
From slop deflection method (previous lecture)
]32[2
444
44 iiij
iii L
EI M M . . . . . . .1
But, 04 04i 04i M (No load)
444
44
4iii
i
ii K L
EI M . . . . . . .2
Where, 44
44 4
4i
i
ii Ek L
EI K ,
4
44
i
ii L
I k =Stiffness Factor
Similarly we can get
111 iii K M . . .3 222 iii K M . . .4 333 iii K M . . .5
In General ijijij K M . . . . . . .6
=+ve =-ve
1
3
2 4
m ii
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Balance of joint i
)( 4321
4321
iiiii
iiiii
K K K K
M M M M m
. . . . . . .7
4321 iiii
i
i K K K K m
. . . . . . .8
Substitute 8 in 2
miiiii
ii K K K K
K M .
4321
44 , 44 4 ii Ek K
miiiii
ii k k k k
k M .
4321
44 . . . . . . .9
Similarly
miiiii
ii k k k k
k M .
4321
11 . . . . . . .10
miiiii
ii k k k k
k M .
4321
22 . . . . . . .11
miiiii
ii k k k k
k M .
4321
33 . . . . . . .12
In General ijijn j jij
k
k
1 ,where n=number in joint i . . . . . . .13
Result 1:The external moment effecting on joint i m i is distributed to othermember with a suitable ratio to the stiffness factor of all unitedmember in joint
Moment at the far end (at point 1, 2, 3 and 4)1. consider the case when the other end is fixed
036 4
44
4
444
i
ii
i
ii
EI L M
EI L M
44 21
ii M M 21 carry over factor
Result 2:The moment value at the far end for any member is equal to thehalf of moment that appears at the near end when the far end isfixed
4i
Mi4 M4i=?
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consider the case when the other end is simply supported
In this case 04i M ,where this end is simply supported
Means that carry over factor = 0
Result 3:The moment at the far end in the case of simply supported end,the carry over factor is equal to zero (C.O.F = 0), hence themoment is zero.
)32(2 44444 iiiii k M M
0 0 002 4 i
24i . . . . . . .14
Substitute 14 in 1
)32(2 44444 iiiii k M M
0 0
)43(4)
23(2
)2
2(2
422
4
44
ii
x
ii
iiii
Ek Ek
k M
. . . . . . .15
Compare between 15 & 2
44
3ik
= Reduced stiffness factor = ij
Rij k k 4
3
Result 4:When the far end is simply supported, so we deal with the reducedstiffness factor
2. Consider the case when the member intersect with the axis ofsymmetry of the system
jiij M M
In this case jiij M M
Hence carry over factor = 1
4i
Mi4 M4i=?
ji
Mij Mji
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Result 5:The moment at the far end in the case of symmetric axis, the carryover factor is equal to one (C.O.F = 1)
ij
ij ji
ij
ijiji EI
L M EI
L M 63
ij
ijiji EI
L M 21
)2
1(4)(2
2
2
2
iij
x
iij
iij
ijij
Ek Ek
L
I E M
. . . . . . .16
Compare between 16 & 2
ijk 21
= Modified stiffness factor = ij M ij k k 2
1
Result 6:When the member at the axis of symmetry, so we deal with thereduced stiffness factor
Summary
Far End Carry OverFactorStiffness
FactorMoment at Far
End
21
ij
ijij L
I k ij M 2
1
0 ij Rij k k 43 0
1 ij M ij k k 21 jiij M M
iijij m M . 1ij at one joint (sum of distribution factor at one joint
should be equal to one)
Mij
Mij
ji
Mij Mji
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Grillage in Ship Structures
Grillage is a system of inter connecting beams.
There are two types of grillages:
Open Grillage Closed Grillage
Ships hull consists o f number of grillages ( Deck grillage,Side grillage, Bottom grillage and Bulkhead grillage )
Solution Of Open Grillage
Assumptions:
1. The beams intersect at right angle, which is true for shipstructure.
2. The external loads are carried only by members of highernumber.
3. The other members are loaded by reactions at connectingNodes.
4. Neglecting all reactions between members at the nodesexcept vertical reactions.
The purpose of grillage solution is to find out the unknownvertical reactions at nodes.
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Example: Solve the following grillage and then draw the B.M.D. foreach member. The grillage is subjected to uniform pressure (p)(Kg/cm 2)
To get the unknowns R 1 & R2 , fromthe slop and deflection tables atpoints 1, 2 for both transverse
members ab & cd and longitudinalmember ef
W1 (long) = W 1 (trans ab) . . . . . 1
W2 (long) = W 2 (trans cd) . . . . . 2
221212111
2121111...
)()()()(
R f R f R f
RW RW RW longW . . . . . 3
222222121
2222122
...
)()()()(
R f R f R f
RW RW RW longW
. . . . . 4
1
34
1111
0052.0384
)()()(
R Ei B
q Ei
B
RW qW transW . . . . . 5
2
34
2222
0052.0384
)()()(
R Ei B
q Ei B
RW qW transW . . . . . 6
R2 R1 R2'
12 2'
ac
bd
c'
d'
e f1
a
b
R1q=p.s(t/m)
1
c
d
R2q=p.s(t/m)
R2=R 2'L=2B
R2
R1
R2'
f11=0.0052 EIL
f12=0.0026 EIL
f12'=0.0026 EIL
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Coefficient f 11 is presenting the deflection value at point 1 asa result of a unit force at the same point.Coefficient f 12 is presenting the deflection value at point 1 as
a result of a unit force at point 2.In General: Coefficient f ij is presenting the deflection value atpoint i as a result of a unit force at point j
Notice that: f ij = f ji Coefficient f ij is called influence coefficient and can be directlycalculated from the beam formulas table
By applying equations 1 & 2, So 3 = 5 & 4 = 6
From the formulas table:
EI B
EI pL
f 33
11 0416.0192
EI B
EI L L L L
EI L
L L L L L
L EIL
L L
axbxbL EIL
x pa f
33222
3
22
3
22
12
0208.0384
)88
9
8
18(
6416
)2424
33
4
33(
6
)2
()4
(1)33(
6
212112 f f f
EI B
EI L
EIL
L L
EILb pa
f 33
3
33
3
33
22 0176.064643
27
3
)4
3()4(1
3
EI B
EI L L L L
EI L
L L L L L
L EIL
L L
bxaxaL EIL
x pb f
33222
3
22
3
22
22
0084.012288
13)
1616
9
16
36(
61616
)4444
33
4
33(
6
)4
()4
(1)33(
6
So
1
34
2
3
2
3
1
3
1
34
221212111
0052.02384
0208.00208.00416.0
0052.0384...
R Ei B pB
Ei B
R EI B
R EI B
R EI B
R Ei B
q Ei B
R f R f R f
p B R R 221 0013.00416.00364.0
2
34
2
3
2
3
1
3
2
34
222222121
0052.02384
0084.00176.00208.0
0052.0384
...
R Ei B pB
Ei B
R EI B
R EI B
R EI B
R Ei B
q Ei
B R f R f R f
p B R R 221 0013.00208.00208.0
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R 2 R 1 R 2'
cd
R
q=p.s(t/m)
e f
Bending Moment Diagram
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Effective Breadth of Plating
beff
max max
beff
Broad Flanged BeamFor simple beam theory we know that
I y M .
where,
:stress at any point in the section, y:distance of the pointfrom N.A of the section, I:moment of inertia of the section.
This theory is based on the assumption that the stressdistribution is uniform across the breadth of the flange.
This is not true in the case of broad flanged beam, where thestress is not uniformly distributed (see the figure).
To calculate the stresses in broad flanged beam by usingsimple beam theory, we can use the concept of effectivebreadth of plating.
Thus the effective breadth of plating (b eff) is defined as thatpart of the plate which if computed as uniformly stressedequal to the value of maximum stress would be able towithstand the same load acting on the original flange.
The value of effective breadth (b eff) depends on:Boundary conditions of the beam (fixed, simple, cantilever)Type of loadingScantling of the beam
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There are many methods to calculate the effective breadth
ratio
b
beff :
o According to some classification societies(beff)=40t 60t , where t: flange thickness.
o Simple formulas:
Simple supported beam Lbeff 31
Fixed end beam Lbeff 61
Using the concept of effective breadth, any cross stiffenedplate (closed grillage) can be converted to
open grillage and solve it as before.
Notice that: if calculated b eff is bigger than the spacing between girders (s),
then b eff must be taken equal to (s)If beff(calculated)>s, then b eff = s
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Transverse Strength of Ships
i1 i2
i3
i4
i5q = p.sp = T t/mq = Ts t/m
Transverse Strength of Cargo Ships
I
2I
3I
6I
: wave length h: wave height L: ship lengthL = h = /20
Hog
Sag
0.5ha b
cd
e
f
2I
ec
Tanker
Bracket floor Solid floor
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Why stiffeners 2 is lighter than 1?(Consider that they have the samestiffness)
Because its obvious that stiffeners 1 is approximatelydoubled weight stiffeners 2 by the number of stiffeners
Determine the neutral axis of the given section.
)(00078.0
])73.0(001.08.0[12
)1(001.0])23.0(1001.0[])27.0(0015.02[
),(
)(73.00048.00035.0
)001.08.0()1001.0()0015.02()5.01001.0()10015.02(
)(
4
23
22
3132332
2221
211
m
iineglect i y Ai y Ai y A I
m
A y A
A M
e ii
I
5I
ab c
def
2IIdealized Mid-Ship Section
I
2I
5I
(T+0.5h-D)
1 2
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Example: Draw the B.M.D for the following section.
a b
d c
ef
i
i
i
2i
i
) / (111
.
) / (12
2
mt
s pq
mt D
p
)(1
)(20
ms
m D B
Stiffness factor
ab iik Rab 43
43
bc ik bc
cd iik Rcd 43
43
ce ik ce ef ik ef 2
Distribution factor
73
7443i
ik k
k
bcab
abba = 0.43
74
74
ii
k k k
bcab
bcbc = 0.57
114
114
ii
k k k k
cecd bc
bccb = 0.36
113
11443
ii
k k k k
cecd bc
cd cd = 0.27
114
114
ii
k k k k
cecd bc
cece
= 0.36
31
3ii
k k k
ef ce
ceec
= 0.33
32
32ii
k k
k
ef ce
ef ef
= 0.67
Fixed End MomentsMab = Mba = Mbc = Mcb = Mcd = Mdc = 0
Mce =30
10030
)10(130
22ql = +3.33
Mec =20
10020
)10(120
22ql = -5
Mef =12
10012
)10(112
22ql = +8.33
Mfe =12
100
12
)10(1
12
22ql = -8.33
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Advanced Structure Analysis Mohammad Abdel Aziz Fahmy 28
Joint a b c d e fMember ab ba bc cb cd ce dc ec ef fe
- 0.43 0.57 0.36 0.27 0.36 - 0.33 0.67 -F.E.M. 0 0 0 0 0 +3.33 0 -5 +8.33 -8.33
Balance -1.2 -0.9 -1.2 -1.1 -2.2
C.O. -0.6 -0.55 -0.6 -1.1Sum 0 0 -0.6 -1.2 -0.9 +1.58 0 -6.7 +6.13 -9.43Balance +0.2 +0.14 +0.2 +0.2 +0.4
C.O. +0.1 +0.1 +0.1 +0.2Sum 0 0 -0.5 -1 -0.76 +1.88 0 -6.4 +6.43 -9.23
Balance -0.04 -0.03 -0.04 -0.01 -0.02C.O. -0.02 -0.005 -0.02 -0.01Sum 0 0 -0.52 -1.04 -0.79 +1.83 0 -6.43 +6.41 -9.24
Balance +0.22 +0.3 +0.007 +0.01C.O. +0.15 +0.004 +0.005Sum 0 +0.22 -0.22 -0.89 -0.79 +1.834 0 -6.423 +6.42 -9.235
Balance -0.055 -0.042 -0.055 +0.001 +0.002C.O. -0.027 +0.000 -0.027 +0.001Sum 0 +0.22 -0.247 -.0.945 -0.832 +1.78 0 -6.45 +6.422 -9.234
Balance +0.012 +0.015 +0.009 +0.019C.O. +0.008 +0.005 +0.009Sum 0 +0.232 -0.232 -0.937 -0.832 +1.785 0 -6.441 +6.441 -9.225
a0.232
d
f
0.232
0.832
1.785
0.937
6.441
6.441
b
c
e
0
0
9.225
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Local Strength of Ships
1. Local strength of double bottom:
a) Idealization of D.B.:The bottom structure is idealized by an open grillage (b eff):
The longitudinal members are keels and side girdersThe transverse members are solid floors
The section of the member is I-beam
b) Boundary conditions:If not specified, we can apply the following simple rule:
- When a weak member connected to a strongmember, the end of the weak member is consideredfixed, while the end of the strong member isconsidered simply supported.
c) Effect of longitudinals:the thickness of plating is increased by:
s
f t t o where,
t0: original thickness, f: area of each longitudinal, s:longitudinal spacing
d) Load on D.B.:
B LwhT p
p p p
hT p
ocw
w
)5.0(
)5.0(
arg
w = cargo
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Advanced Structure Analysis Mohammad Abdel Aziz Fahmy 30
( T+0.5 h - y )
I1
I2
2. Local strength of side assembly:
a) Transverse system of framing:
b) Longitudinal system of framing:
Sidelongitudinals
A
BC
D
E
T+0.5h-y
T+0.5h
BHD
Web frames
A B C D E
q = (T+0.5h-y).s
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Advanced Structure Analysis Mohammad Abdel Aziz Fahmy 31
3. Local strength of deck structure:
a) Transverse system of framing:
H.O.
A
B
C
D
E
A
B
C
D
E
A
E
BHD BHDDeck Beams
B
C
D
Loads on deck consists of:
- Cargo pressure = w/A- Shipped water: by equation from ship building
specifications( related to the class)
b) Longitudinal system of framing:
A
B C D E
F
BHD
Deck centergirder
Deck side girder Deck transverse
Deck longitudinals
p = p cargo + p water
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Example: Given the following data of a single deck dry cargo ship withlongitudinal system of framing;L =124 m B =16 m D =9.5 m T =7.2 mDistance between transverse bulkheads 14 mDistance between solid floors 3.5 m
One center girder (keel)Specific volume of cargo 1.8 m 3 /tonDistance between longitudinals (s) 0.7 mArea of inner bottom longitudinal 26.8 cm 2 Area of outer bottom longitudinal 27.36 cm 2 Scantling of vertical keel 1200x15 mmThickness of inner bottom plating 10 mmThickness of outer bottom plating 13 mmThickness of horizontal keel 16 mmThickness of solid floor 13 mm
Normal stress in inner bottom plating due to hogging -920 Kg/cm 2 Normal stress in outer bottom plating due to hogging -1260 Kg/cm 2 In sagging condition, these values are respectively 520, 710 Kg/cm 2
Find the value of maximum resultant stress in bottom plating
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Advanced Structure Analysis Mohammad Abdel Aziz Fahmy 33
Finite Element Method1-Node ElementF = k u where, F = Force
k = stiffness of springu = Displacement
, when u = 1 k F
2-Nodes Element
2221212
2121111
uk uk F
uk uk F
2
1
2221
1211
2
1 .uu
k k k k
F F
k ij = stiffness at i as a result of unit displacement at j
3-Nodes Element
3
2
1
3
2
1
333231
232221
131211
333231
232221
131211
3
2
1
3
2
1
.
000
000
000
000
000
000
v
v
v
u
u
u
k k k
k k k
k k k
k k k
k k k
k k k
F
F
F
F
F
F
y y y
y y y
y y y
x x x
x x x
x x x
y
y
y
x
x
x
4
3
2
1
4
3
2
1
44434241
34333231
24232221
14131211
44434241
34333231
24232221
14131211
4
3
2
1
4
3
2
1
.
0000
0000
0000
0000
0000
0000
00000000
v
v
v
v
u
u
uu
k k k k
k k k k
k k k k
k k k k
k k k k
k k k k
k k k k k k k k
F
F
F
F
F
F
F F
y y y y
y y y y
y y y y
y y y y
x x x x
x x x x
x x x x
x x x x
y
y
y
y
x
x
x
x
F
1 2
F 1 F 2
k u
F 1 F 2
F 1 F 211 1
k u12 2
k u21 1
k u22 2
=
+
Stiffness Matrixof
2-Nodes Member
1
2
F x 1, u 1
3
F x 2, u 2
F x 3, u 3
F y 2, v 2
F y 3, v 3
F y 1, v 1
Stiffness Matrixof
Triangular Plate Element
1 2F x 1, u 1
3
F x 2, u 2
F x 3, u 3
F y 2, v 2
F y 3, v 3
F y 1, v 1
4
F x 2, u 2F y 2, v 2
Stiffness Matrixof
Rectangle Plate Element
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Advanced Structure Analysis Mohammad Abdel Aziz Fahmy 34
Finite Element Method Procedures
Solution of a structure by FEM is summarised as follows:1. Divide the structure into a number of finite elements (the
element could be a beam, a rectangle plate or a triangleone), then numbering the nodes
2. Assume displacement function at any point
yC xC C v
yC xC C u
654
321 where, C 1 - C 6 are unknown constants
6
5
4
3
2
1
.1000
0001),(
C
C
C C
C
C
y x y x
vu y x
C y x H y x ).,(),( . . . . . . 1
From this equation we can get the vertical and horizontal displacement at any point (x,y)
3. Displacement at Nodes
C y x
y x y x
v
u.
111000
000111)1,1(1
1
11
C y x
y x y x
v
u.
221000
000221)2,2(2
2
22
C y x
y x y x
v
u.
331000
000331)3,3(3
3
33
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Advanced Structure Analysis Mohammad Abdel Aziz Fahmy 35
6
5
4
3
2
1
3
2
1
3
2
1
3
2
1
.
331000
221000
111000
000331
000221
000111
C
C
C
C
C
C
y x
y x
y x
y x
y x
y x
v
v
v
u
u
u
Or C A. From this equation we can get the displacement at nodes
.1AC . . . . . . 2
By substitution 2 in 1
.),(),( 1A y x H y x . . . . . . 3
From this equation we can get the relation between displacement at any point inside theelement and the displacement at nodes
4. Express the relation between strain and displacement atnodes
Strain x, y, xy
xy
y
x
y x
),( xv
yu
yv
xu
xy y x ,,
By using the previous equation at step 2 we get:
53654321
66654
22321
)()(
00)(
00)(
C C yC xC C x
yC xC C y x
v yu
C C yC xC C y y
v
C C yC xC C x x
u
xy
y
x
6
5
4
3
2
1
.
010100
100000
000010
),(
C
C
C
C
C
C
y x
xy
y
x
GC y x ),(
By substitution 2 in it
1
),( GA y x or
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Advanced Structure Analysis Mohammad Abdel Aziz Fahmy 36
B y x ),( . . . . . . 4
From this equation we can get the strain at any point as a function of the nodal displacements
The Matrix 1GA B is called Strain Matrix
5. Stress calculation at any point inside element
xy
y
x
but
E G
E
E
xy
x y y
y x x
)1(2
)(1
)(1
xy
y
x
xy
y
x
E
.
)1(200
01011 or
xy
y
x
xy
y
x E
y x
.
2
100
01
01
1),(
2 . . . . . . 5
Or
),(),( y x D y x . . . . . . 6
By substitution 4 in 6 DB y x ),(
S y x ),( . . . . . . 7
From this equation we can get the stress at any point
The Matrix S is called Stress Matrix
6. Determine stiffness matrix for each element
2
1
2
1
2
1
2
1
.
v
v
u
u
K
F
F
F
F
y
y
x
x
111 )()( k F , 222 )()( k F
nnn k F )()( where, n = number of elements
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7. Stiffness matrix for whole structure
nnn k
k
k
k
F
F
F
F
)(
.
.
.
.)(
)(
)(
.
0000000
0.000000
00.00000
000.0000
0000.0000000000
0000000
0000000
)(
.
.
.
.)(
)(
)(
3
2
1
3
2
1
3
2
1
so we can get
F = K . . . . . . 8
Where, F : Force matrix at nodes for whole structure
K : Stiffness matrix for whole structure
: Displacement matrix for all nodes in the structure
8. When knowing the boundary conditions, applyingcompatibility and equilibrium modes, we can solve equation8 to get all the nodal displacements.