Advanced Structure Analysis

8/7/2019 Advanced Structure Analysis

1/37

Arab academy for Science, Technology and Maritime Transport

Submitted to:Dr. El Sayed Hegazy

Prepared by:Eng. Mohamed Abdel Aziz Fahmy


2/37

Advanced Structure Analysis Mohammad Abdel Aziz Fahmy 2

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


3/37


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


4/37


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


5/37


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

)


6/37


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.


7/37


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


8/37


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


9/37


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


10/37


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


11/37


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


12/37


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'


13/37


]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


14/37


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


15/37


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


16/37


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=?


17/37


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


18/37


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


19/37


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.


20/37


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


21/37


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


22/37


R 2 R 1 R 2'

cd

R

q=p.s(t/m)

e f

Bending Moment Diagram


23/37


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


24/37


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


25/37


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


26/37


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


27/37


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


28/37


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


29/37


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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( 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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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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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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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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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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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.

Advanced Structure Analysis

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