8/10/2019 Coursework (Recovered)

1/14

ap1

N 1 =( L1 0 )( L1 12 )(1 0 )(1

1

2)

= 2 L1( L1 12 )= L1 (2 L1 1 )= 2 L12 L1

d N 1d L1

= 4 L1 1

d N 1d L2

= 0

N 2 =

( L2 0 )( L2 12 )(1 0 )(1 1

2) =

2

L2( L2

1

2 )= L2 (

2

L2 1

)=2

L22

L2

d N 2d L1

= 0

d N 2d L2

= 4 L2 1

N 3 =( L3 0 )( L3

1

2 )(1 0 )(1

1

2)

= 2 L3( L3 12 )= L3 (2 L3 1 )=( 1 L1 L2 )(1 2 L1 2 L2 )

d N 3d L1

= 1 +2 L1 +2 L2 2 +2 L1 +2 L2 = 4 ( L1 + L2 ) 3

d N 3d L2

= 1 +2 L1 +2 L2 2 +2 L1 +2 L2 = 4 ( L1 + L2 ) 3

N 4 =( L1 0 ) ( L2 0 )(

1

2 0 )(

1

2 0 )

= 4 L1 L2

8/10/2019 Coursework (Recovered)

2/14

d N 4d L1

= 4 L2

d N 4d L2

= 4 L1

N 5 =( L3 0 )( L2 0 )(

1

2 0 )(

1

2 0 )

= 4 L3 L2 = 4 L2 (1 L1 L2 )

d N 4d L1

= 4 L2

d N 4

d L2= 4

(1 L1 L2

) 4 L2 = 4(1 L1 2 L2 )

N 6 =( L3 0 )( L1 0 )(

1

2 0 )(

1

2 0 )

= 4 L3 L1 = 4 L1 (1 L1 L2 )

d N 4d L1

= 4(1 L1 L2 ) 4 L1 = 4(1 2 L1 L2 )

d N 4d L2 = 4 L1

ap3.

1.

Half of F 55 acts on line 54-55, and half on 53-54, that is why the force on node 54 will be 5/2 foreach line.

F 55

F 54

F 53

50 55

49

48

54

53

8/10/2019 Coursework (Recovered)

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8/10/2019 Coursework (Recovered)

4/14

A3 = b

b

N 3 P ( y)dy

y+ y2

2 bb

b

b + b

2

2 b+b b

2

2 b

5

P= A3

b

b

N 3 dy

= 2.5

b

b1

2 (1 + yb )dy

= 5

8/10/2019 Coursework (Recovered)

5/14

N 1 = 2 L12 L53

N 6 = 4 L53 L55

N 11 = 2 L552 L55

L55 = y L

L53 = 1 y L

(2 L532 L5 3 )dy= 10

0

L

(2 (1 y L)2

1 + y L)dy=

N 55 P dy= 10 0

L

F 53 =0

L

F 53

F 54

F 55

4"

4#43

55

53

54

8/10/2019 Coursework (Recovered)

6/14

y+2

3 y

3

L2

3

2 y

2

L 0 L= 10 ( L+ 23 L 32 L)= 1.666

(1 +2 y2

L2 3

y L)dy

= 10

10 0

L

y2

2 L y

3

3 L2 0

L=

y L(1 y L)dy = 40

(4 L53 L55 )dy= 4 0 0

L

N 5 4 P dy = 10 0

L

F 54 =0

L

40 ( L2 L3 )= 6.666

(2 L552 L55 )dy= 10

0 L

(2

( y

L)2

y L)

dy=

N 55 P dy = 10 0

L

F 55 = 0

L

2

3 y

3

L y

2

2 L0 L= 10 (23 L2 L2 )= 1.666

10

which doubles because of simmetry

$essel

4.

8/10/2019 Coursework (Recovered)

7/14

b= 2 r

4102= 0.0785

N 1 =1

2 (1 yb )

F 1 = b

b

N 1 P dy

N 1 dy

b

b

12

(1 yb )dyb

b

1

2 (1 yb )dy

b

b

(1 yb )dyb

b

y y

2

2 bb

b

P= F 1

104.7198

0.157= 666.66

8/10/2019 Coursework (Recovered)

8/14

N 53 = 2 L532 L53

N 54 = 4 L53 L55

N 55 = 2 L552 L55

L55 = y L

L53 = 1 y L

(2 L532

L53 )dy= 10 0 L

(2 (1 y L)

2

1 + y L)dy =

N 55 P dy = 10 0

L

F 53 =0

L

y+2

3 y

3

L2

3

2 y

2

L 0 L= 10 ( L+ 23 L 32 L)= 1.666

(1 +2 y

2

L2 3 y L)dy = 10

10 0

L

y2

2 L y

3

3 L2 0

L=

y L(1 y L)dy = 40

(4 L53 L55 )dy= 40 0

L

N 54 P dy = 10 0

L

F 54 =0

L

8/10/2019 Coursework (Recovered)

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40 ( L2 L3 )= 6.666

(2 L552 L55 )dy= 10

0

L

(2 ( y L)

2

y L

)dy=

N 55 P dy = 10 0

L

F 55 = 0

L

2

3 y

3

L y

2

2 L0 L= 10 (23 L2 L2 )= 1.666

10

%1& load a lied in each case'

%2& disc!ssion of the general erfor(ance of each ele(ent'%3& co( are the res!lts of the )4 and *+ ele(ents for each loading case'

%4& co( are of the co( !ted res!lts with theoretical sol!tions obtained fro( +trength of

aterials %deri$ations to be incl!ded&'

%5& o ti(i ation of node n!(bering and half-bandwidth'

% & significance of the res!lts for either bea(1)4 or bea(1*+ in the conte0t of the atch test'

% & !stification of the bo!ndary conditions and loading !sed for the bea( and $essel roble(s'

%#& significance of +aint- enant s rinci le for bea(3)4' and,

%"& a$eraging of nodal stresses.

3. es!lts of the )4 and *+ ele(ents for each loading case

8/10/2019 Coursework (Recovered)

10/14

1 2 3 4 5 6 7 8 9 10 11

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

Beam 1

LST Q4

Fig shows the deflection on 0 a0is for the nodes on the center line of the bea(. he deflections on 0are the sa(e for al 11 col!(ns of rows. he res!lts are the sa(e for the two ty es of ele(ents. 6se0 ected, the deflection is ro ortional with the distance fro( the fi0ed nodes, which ha$e 7dis lace(ent.

1 2 3 4 5 6 7 8 9 100

2

4

6

8

10

12

Beam 2

LST Q4

Fig. dis lays the deflection of the nodes on the center a0is of the bea(. here is a 178 differencebetween the $al!es at the sa(e node. he y dis lace(ents for the other 4 rows of nodes are al(ostthe sa(e, with a less than 18 $ariation fro( these.

8/10/2019 Coursework (Recovered)

11/14

1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

Beam 3

Q4 LST

9n fig. the dis lace(ents on 0 of the nodes laces on the center hori ontal a0is of the bea( areshown. 6gain, a difference of 178 between the sol!tion of )4 and *+ analysis can be noticed. 6s inthe case of bea( 2, all 5 rows of nodes ha$e the sa(e $ertical dis lace(ent %less than 18$ariation&.

1 2 3 4 50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

Vessel

LST Q4

Fig. shows the radial dis lace(ents of the nodes. :ar 1 of the lot re resents the interior nodes, andbar 5 the e0terior nodes. here is a e0tre(ely s(all difference between the sol!tions of the two ty esof ele(ents. he radial dis lace(ent is identical for e$ery radial a0is of the $essel.

9n bea( 1 and $essel, the res!lts are the sa(e for the two ty es of ele(ents, while for bea( 2 and 3the deflection is bigger for *+ than )4.

8/10/2019 Coursework (Recovered)

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8/10/2019 Coursework (Recovered)

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1 2 3 4 5 6 7 8 9 100.00

2.00

4.00

6.00

8.00

10.00

12.00

Beam 2

THEORETICAL LST Q4

y= P x2

6 EI (3 l x)

:ea( 3

1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

Beam 3

THEORETICAL Q4 LST

8/10/2019 Coursework (Recovered)

14/14

y= M x2

2 EI

I = t h2

12

essel

p( i r i

2r )

E (r 02 r i

2 )[(1 )+(1 + )

r i2

r2 ]

ur =

5.

HBW = max ( EHBW n)= 2(6 +1 )= 14

HBW = max ( EHBW n)= 2(12 +1 )= 26

. :ea( 1 res ects the (ain conditions to be a lied the atch test. he bo!ndary conditions are(ini(al in order to restrain rigid body (otion %5 nodes are constrained on 0 a0is and 1 of the( on ya0is, all of the( are e0terior nodes&. he traction is a lied by forces acting on e0terior nodes, b!tnone of the interior nodes is distorted.

Fig. shows that the theoretical res!lts (atch the res!lts of the analysis. he dis lace(ents on 0 arethe sa(e,