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ANALYSIS OF RESERVOIR

Marco Castelli, matr. 787505

April 12, 2013

Abstract

In reference to the figure 1 we want to define the following quantities:

the stress resultants in the spherical cap, in the conical shell and in the ring beam;

the displacement of point A.

Figure 1: Section of the tank.

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The only loads acting on the structure are due to the fluids pressure (gas and water) intothe tank. Mechanical and geometrical quantities depend from the first letter of name (N)and the third letter of surname (C) of the student: C = 17 e N = 11.

1 Mechanical and geometrical properties

Mechanical properties of the material:

Young Modulus: E= 30 103 N mm2 = 30 106 kN m2;

Poisson Ratio: = 0.2.

Geometry:

Ring beam:

Width: a = 50 cm = 0,5 m;

Height: b = 40 cm = 0,4 m.

Spherical shell: Thickness: ts = 20 cm = 0,2 m;

Angle: 0 =

45 + N10

= 48,4 = 0,8447 rad;

Horizontal radius projection: c =

8 + C5

= 9,1 m.

Conical shell:

Thickness: tc = 30 cm = 0,3 m;

Angle: c = 45 = 0,7854 rad.

Loading:

Spring: k = 50 kNcm2

= 500 000 kN m2

;

Pressure of gas: Pg = 200 kN m2.

We can also calculate the following quantities that are not given:

Radius of spherical shell: Rs = csin 0 =9,1m

sin48,4= 12,17 m;

Height of conical shell: h = c + 2a = 9,1 m + 2 0,5 m = 10,1 m;

Length of conical shell: Lc = hsin c =10,1msin45

= 14,28 m;

Gravity: g = 9,81 m s2;

Density of water: = 1000 kg m3;

Specific weight of water: w = g = 9,81 m s21000 kg m3 = 9,81 kN m3.

2 Analysis of spherical shell

The only loading acting on the shell is the gas under pressure and it is constant along theheight. By convention, as shown in the figure 2, the abscissa z is directed towards the centerof the dome. Pressure is normal to the surface of shell and has the opposite direction to z ofthe shell because the gas under pressure is into the dome. From now on the quantities drawnin the following figures will have sign a positive when directed as z, negative if drawn outsideof the sphere.

rs = Rs sins.

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FTFigure 2: Cross section of the spherical shell.

Ifs = 0, rs = c. The vertical resultant of load applied to the cap is:

V () =

0

Pg cosRs2rs d = PgR

2S

2(1 cos2) . (1)

For = 0:

V (0) = 200 kN m2

(12,17 m)

2

2[1 cos (2 48,4)] = 680,5 kN.

The membrane stress resultants are constant and are given by:

N =PgRs

2=

200 kN m212,17 m

2= 1216,9 kN m1, (2)

N = N Rs(Pg ) =PgRs

2=

200 kN m212,17 m

2= 1216,9 kN m1. (3)

Functions (1), (2) and (3) are represented in figure 3. We can now calculate the deformations.

=1

Ets(N N) =

1216,9 kN m1 0, 2 1216,9 kN m130 106 kN m2 0,2 m = 1,6225 10

4,

=1

Ets(N N) = 1216,9 kN m

1 0, 2 1216,9 kN m130 106 kN m2 0,2 m = 1,6225 10

4.

Strains depend to the membrane stress and N and N are constant therefore also the defor-mations are constant (see also figure 4).

2.1 State of displacement at the edge

We can calculate the radial component s of the displacement and the rotation of the tangentof the meridian. These two quantities come only from the pressure. Since the deformation of the circumference is equal to the deformation of the radius, s is given by:

s = rs.

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5.2E+04 0

unit of measure: kN

(a) Vertical resultant of load applied to the cap:V.

1.22E+031.22E+03

unit of measure: kN/m

(b) N.

1.22E+031.22E+03


(c) N.

Figure 3: Vertical resultant V and membrane stress (N and N).

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0.0001620.000162

unit of measure: m/m

(a) .

0.0001620.000162


(b) .

Figure 4: Deformations and .

Figure 5: Section of the tank.

At the edges

s(0) = Rs sin0 = 1,6225 104(12,17 m)sin48,4 = 1,4765 103 m. (4)To calculate the rotation we use the formula:

s =d()

d ( )cot =

1

Ets

d (N N)

d (1 + ) (N N)cot

Since membrane stress resultants are constant and equal between themselves, the rotation isnull along all the surface. This is understandable because the pressure pushes uniformly onthe whole surface.

2.2 Influence Coefficients at the edges for the flexural solution

The influence coefficients are derivable from the Governing Differential Equation

4w

s4+ 44w = 0

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Figure 6: Conical shell.

where s is the abscissa along the surface and starting from = 0.

Ds =Et3s

12(1 2) =30 106 kN m2 (0,2 m)3

12(1 0.22) = 2,0833 104 kN m

s =4

1

4

Ets

DsRs= 4

1

4 30 10

6 kN m2 0,2 m2,0833 104 kN 12,17 m = 1,5596

s =Ets

R2s=

30 106 kN m2 0,2 m(12,17 m)2

= 4,0517 104

Whit this last three equations we can write the influence coefficient. For the force H0 (figure 5):

H,s =

2ss H0 sin0 =

2

1,5596

4,0517 104H0 sin48,4

= H0 5,7570 105

H,s =2ss

H0 sin2 0 =

2 1,55964,0517 104H0 sin

2 48,4 = H0 4,3051 105

H,s =22ss

H0 sin0 =2 (1,5596)24,0517 104 H0 sin48,4

= H0 8,9787 105

For the moment M0 (figure 5):

M,s =22ss

M0 =2 (1,5596)24,0517 104 M0 = M0 1,2007 10

4

M,s =22ss

M0 sin0 =2 (1,5596)24,0517

104

M0 sin 48,4 = M0 8,9787 105

M,s =43ss

M0 =4 (1,5596)34,0517 104 M0 = M0 3,7452 10

4

3 Analysis of conical shell

The only loading acting on the shell is the pressure of water plus pressure of gas and decreaseslinearly along the abscissa. By convention, as shown in the figure 6, the abscissa z is directedtowards the vertical axis of the cone. Pressure is normal to the surface of shell. From nowon the quantities drawn in the following figures will have sign positive when directed as z,negative if drawn outside of the cone.

rc = x cosc.

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208

307

unit of measure: kN/m2

Figure 7: Pressure profile on the conical shell.

The pressure profile is linear and is represented by the function and in figure 7:

Pc(x) = (PG + w2b) + w (Lc x)sinc.The vertical resultant of load applied to the cone is:

Vx =

0

Pc cosRs2rs d

=

Lc0

Pc()sin0 cos02cos0, d

=x2

6

(2)[w (6b + 3Lc 2x) + 3Pg] (5)

The membrane stress resultants are given by:

Nx =Vx

2rc sinc, (6)

N = Pcx cotc. (7)Functions (5), (6) and (7) are represented in figure 8. For x = Lc:

Vx(Lc) =(14,28 m)2

6

(2)9,81 kN m3

16,68 m + 3 200 kN m2

= 9,0899 102 kN,

Nx(Lc) =9,0899 102 kN210,1 m sin 45

= 1,2855 103 kN m1,

N(Lc) = 2,0785 102 kN m2 14,28 m cot45 = 2,9688 103 kN m1.We can now calculate the deformations.

x =1

Etc(Nx N) ,

=1

Etc(N Nx) .

Strains depend to the membrane stress and N and N are represented in figure 9). Forx = Lc:

x =1,2855 103 kN m1 0, 2 2,9688 103 kN m1

30 106 kN m2 0,3 m = 2,0881 104,

=2,9688 103 kN m1 0, 21,2855 103 kN m1

30 106 kN m2 0,3 m = 3,5843 104.

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5.77E+04

0

unit of measure: kN

(a) Vertical resultant of load applied to the cone:Vx.

0

1.29E+03


(b) Nx.

2.97E+03

0


(c) N.

Figure 8: Vertical resultant Vx and membrane stress (Nx and N).

0

0.000209


(a) x.

0.000358

0


(b) .

Figure 9: Deformations x and .

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3.1 State of displacement at the edge

We can calculate the radial component s of the displacement and the rotation of the tangentof the meridian. These two quantities come only from the pressure of water plus the pressure ofgas. Since the deformation of the circumference is equal to the deformation of the horizontalradius, c is given by:

c = rc.

At the edges

c(Lc) = Lc cosc = 3,5843 104(14,28 m)sin45 = 3,6202 103 m.To calculate the rotation we use the formula:

c =1

tanc

(x) x(x) + x

d(x)

dx

By deriving the (3.1) we obtain:

c =1

2E[Lctc (Pg + 2wb) (Lc + 3+ 3)] (8)

c =1

2E(Lctc) (Pg + 2wb) (Lc + 3+ 3)

=1

2 30 106 kN m2 (14,28 m 0,3 m)

(200 kN m2 + 2 9,81 kN m3 0,4 m) (14,28 m + 3 0,2 + 3)

c = 2,6546 104

3.2 Influence Coefficients at the edges for the flexural solution

The influence coefficients are derivable from the Governing Differential Equation.

4w

s4+ 44w = 0

where s is the abscissa along the surface.

Dc =Et3c

12(1 2) =30 106 kN m2 (0,3 m)3

12(1 0.22) = 7,0312 104 kN m

In order to finding c we need to calculate an equivalent radius for the conical shell:

Rc,eq =h + btanc

sinc=

10,1 m + 0,4mtan45

sin 45= 14,85 m.

s =41

4

Etc

DcRc,eq =

414

30

106 kN m2

0,3 m

7,0312 104 kN 14,85 m = 1,2116

c =Etc

R2c,eq=

30 106 kN m2 0,3 m(14,85 m)2

= 4,0816 104

Whit this last four equations we can write the influence coefficient. For the force H0:

H,c =2cc

H0 sinc =2 1,2116

4,0816 104H0 sin 45 = H0 4,4396 105

H,c =2cc

H0 sin2 c =

2 1,21164,0816 104H0 sin

2 45 = H0 3,3199 105

H,c =22cc

H0 sinc =2 (1,2116)24,0816 104 H0 sin 45

= H0 5,3790 105

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Figure 10: Ring beam and forces H e Mt

For the moment M0:

M,c =22

cc M

0 =2

(1,2116)2

4,0816 104 M0 = M0 7,1931 10

5

M,c =22cc

M0 sinc =2 (1,2116)24,0816 104 M0 sin45

= M0 5,3790 105

M,c =43cc

M0 =4 (1,2116)34,0816 104 M0 = M0 1,7431 10

4

4 Ring beam

In order to analyze the interaction between the ring beam with the shells we have to find theinfluence coefficients of the ring beam. The radius of the middle line of the beam is:

RRB

= c + a = 9,1 m + 0,5 m = 9,6 m.

The area of the cross section of the beam is:

ARB = (2a)(2b) = (2 0,5 m)(2 0,4 m) = 0,8 m2

The inertia modulus is:

IRB =1

122a(2b)3 =

1

122 0,5 m(2 0,4 m)3 = 4,2667 102 m4.

The influence coefficients are: For the force H0:

H,RB =R2RBEARB

H0 =(9,6 m)2

30 106 kN m2 0,8 m2H0 = H0 3,8400 106

For the moment M0:

M,RB =R2RBEIRB

bM0 =(9,6 m)2

30 106 kN m2 4,2667 102 m4 0,4 mM0 = M0 2,8800 105

M,RB =R2RBEIRB

M0 =(9,6 m)2

30 106 kN m2 4,2667 102 m4M0 = M0 7,2000 105

5 Compatibility equations

Solving the problem via flexibility method we write five compatibility equations than corre-spond to the five unknowns (figure 11). The shells are clamped on the ring beam. Compat-ibility equations represent the displacements of the ring beam respect the two shells and thespring and take into account horizontal displacements and rotations.

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FTFigure 11: Scheme of Hyperstatic forces.The vertical forces on the ring from the two shells are given by membrane stress. We canobtain the support reaction on the ring beam by the equilibrium of vertical translation.

Vr = (Nx(Lc)sinc)2h (N(0)sin0)2c 2a(Pg + w2b)2(c + a)2(c + a)

=(908,99 kN sin45)210,1 m

2(9,1 m + 0,5 m)

(680,5 kN sin 48,4)29,1 m

2(9,1 m + 0,5 m)

2 0,5 m(200 kN m2 + 9,81 kN m3 2 0,4 m)

2(9,1 m + 0,5 m)

= 114,12 kN m1

Now we calculate the force on the ring beam (figure 10) that is made than two terms: onehorizontal force H and one moment Mt given by Hmultiplied by the distance from the centerof gravity.

H= X1 X3 X5 + 2b(Pg + wb)

Mt = X2 X4 + a(Vr Nx(Lc)sinc N(0)sin0) + b(X1 X3) +2

3wb

3

The compatibility equations are:

1. Displacement of B = Displacement of B;

2. Rotation of B = Rotation of B;

3. Displacement of C = Displacement of C;

4. Rotation of C = Rotation of C;

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5. Displacement of D = Displacement of D;

Where point B is the edge of the dome and B is the same point but on the ring beam. Thepoint C is the edge of the cone and C is the same point but on the ring beam. Points D andD are referred to the spring. In terms of influence coefficients we can write:

1.H,sX1 + M,sX2 + s = H,RBH+ M,RBMt;

2.H,sX1 + M,sX2 + s = M,RBMt;

3.H,cX3 + M,cX4 + c = H,RBH+ M,RBMt;

4.H,cX3 + M,cX4 + c = M,RBMt;

5.1

kX5 = H,RBH.

By expanding H and Mt and collecting the unknowns X1 X5 we obtain a linear systemKx = b where K is the stiffness matrix, x is the vector of unknowns and b is the vector ofthe knowns. The matrix is:

(H,s + H,RB + bM,RB) (M,s M,RB ) (H,RB bM,RB ) (M,RB ) (H,RB)(H,s bM,RB) (M,s + M,RB ) (bM,RB) (M,RB ) 0

(H,RB bM,RB) (M,RB ) (H,c + H,RB + bM,RB ) (M,c + M,RB) (H,RB)(bM,RB) (M,RB) (H,c + bM,RB) (M,c + M,RB) 0

(H,RB) 0 M,RB 0 (1k

+ H,RB)

And the vector of knowns is:s + H,RB2b(Pg + wb)M,RB a(Vr Nx(Lc)sin c N(0)sin0) + 23wb3

s + M,RBa(Vr Nx(Lc)sin c N(0)sin0) + 23wb3

xic + H,RB2b(Pg + wb) + M,RB

a(Vr Nx(Lc)sinc N(0)sin0) + 23wb3

c + M,RB

a(Vr Nx(Lc)sinc N(0)sin0) + 23wb3

H,RB2b(Pg + wb)

And the vector of unknowns is:

X1X2X3X4X5

The stiffness matrix is symmetric and the terms on the diagonal are positive.

K =

5,8411 105 6,0987 105 7,6800 106 2,8800 105 3,8400 1066,0987 105 4,4652 104 2,8800 105 7,2000 105 0,00007,6800 106 2,8800 105 4,8559 105 8,2590 105 3,8400 1062,8800 105 7,2000 105 8,2590 105 2,4631 104 0,00003,8400 106 0,0000 3,8400 106 0,0000 5,8400 106

b =

4,0715 1031,2304 1027,9153 1031,2569 1026,2645

104

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Kx = b

x =

X1X2X3X4

X5

=

80,539 kN m1

31,079 k N m m1213,935 kN m139,206 k N m m1

194,982 kN m1

We can make a quick check of the stiffness matrix by controlling that all the terms of themain diagonal are positive and that the matrix is symmetric. In this case the matrix could becorrected because the two conditions are satisfied.

During the analysis of the forces acting on the beam we have neglected the thickness ofthe shells in for safety reasons: the redundant forces are increased.

6 Flexural actions

Now we want to compute and graphics the displacements and reactions of the shells. To dothis we calculate the membrane solution at least at the edges of the shells.

In order to do this we can decompose vertical force V in one axial force S e and one

horizontal force H.For the spherical shell:

Ss(0) = N(0) = 1216,9 kN m1,

Hs(0) = Ss(0)cos0 = 1216,9 kN m1 cos48,4 = 807,94 kN m1

For the conical shell:

Sc(Lc) = Nx(Lc) = 1285,5 kN m1,

Hc(Lc) = Sc(Lc)cosc = 1285,5 kN m1 cos45 = 908,99 kN m1

6.1 Solutions of the differential equation

From the theory of shell we can write that:w = Ces

sin(s + ) ,

w = (

2)Ces

sins +

4

,

w = (22)Ces

sins +

2

,

w = (32

2)Ces

sin

s + 3

4

,

w = (44)Ces

sin(s + ) ,With boundary conditions we obtain the two terms C and and those depends to which forcewe are considering. If we consider the case where there is only H, boundary conditions are:

M(s = 0) = 0

Q(s = 0) = Hcos

Since M = D 2w

(s)2 =

Q = D 3w(s)3

we obtain:

D

(22)Csin

2

= 0 =

2

D

(32

2)Csin

pi

2 3

4

= Hcos C= Hcos

D (32

2)sin 4

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If we consider the case where there is only M, boundary conditions are:M(s = 0) = M

Q(s = 0) = 0

Since M

= D2w

(s)2

Q = D 3w(s)3

we obtain:

D

(32

2)Csin

3

4

= 0 = 3

4

D

(22)Csin

3

4

2

= Hcos C= MD

(22)sin

4

Accordingly we can also write:

=w

s

M = D 2

w(s)2

;

M = M;

Q = D 3w

(s)3;

N = Q cot;

N =Q

.

the quantities M, QandN became M,Q

andN

for the dome or M

x , Q

xandN

x for thecone. We proceed with the graphs

For both shells we calculate the horizontal displacement, rotation and internal stresses,

first for the single forces and after by summing of these forces.

6.2 Stress Resultant in the spherical cap

6.2.1 Effects of the gas on the pressure Pg

In this case pressure and normal displacement are constant along the abscissa so the rotationand bending moments are null. Drawings are in figure 12

wPg =Pgs

=200 kN m2

4,0517 104 m = 4,9362 103 m

Pg = 0

M,Pg = 0

M,Pg = 0

6.2.2 Effects of the horizontal force

The horizontal force that insists on the dome is:

Htot = X1 + Hs(0)

The boundary conditions are:

=

2

C=Htot cos

D (32

2)sin 4

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0.00494

unit of measure: m

(a) Normal displacementw.

1.37E+03

0


(b) Shear force in directionQ.

1.22E+03


(c) Axial force in directionN.

1.22E+031.22E+03


(d) Axial force in horizontal di-rectionN.

Figure 12: Effects of the gas on the pressure Pg.

Drawings are in figure 13

6.2.3 Effects of the Moment


Mtot = X2


= 34

C=M

D

(22)sin

4


6.2.4 Sum of the effects on the cap


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0.00373

0.00025

unit of measure: m


0.000252

0.00582

unit of measure: rad

(b) Rotation .

5.27

122

unit of measure: kNm/m

(c) Bending moment in direc-tionM.

1.05

24.4


(d) Bending moment in horizon-tal directionM.

123

590


(e) Shear force in directionQ.

1.63E+03

5.68E+05


(f) Axial force in directionN.

123

1.84E+03


(g) Axial force in horizontal di-rectionN.

Figure 13: Effects of the horizontal force Htot.

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6.37E05

0.000307

unit of measure: m


0.000956

6.41E05


(b) Rotation .

31.1

1.34



6.22

0.269



1.35

31.3



6.54

1.84E+03



151

31.4



Figure 14: Effects of the moment Mtot.

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0.00836

0.00473

unit of measure: m


0.000212

0.00491


(b) Rotation .

31.1

103



6.22

20.6



1.37E+03

590



69.5

5.71E+05



1.11E+03

2.91E+03



Figure 15: Total effects on the cap.

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6.3 Stress Resultant in the conical shell

6.3.1 Effects of the pressure (gas+water)

In this case pressure and normal displacement are linear along the abscissa so the rotation isconstant and bending moments are null. Drawings are in figure 16

wP =Pg + w(Lc s)

c

P =w

c

M,P = 0

M,P = 0

6.3.2 Effects of the horizontal force


Htot = X3 + Hc(x)


=

2

C=Htot cos

D

(32

2)sin

4


6.3.3 Effects of the Moment


Mtot = X4


=3

4

C=M

D

(22)sin

4


6.3.4 Sum of the effects on the cap


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0.0049

0.00833

unit of measure: m


1.29E+03

0


(b) Shear force in directionQx.

0

1.29E+03


(c) Axial force in directionNx.

2.97E+03

0


(d) Axial force in horizontal di-rectionN.

Figure 16: Effects of the gas on the pressure Pg.

7 Displacement of point A

As we know, after a distance from the edge where are applied the disturbances, the theinfluence of those forces is negligible. Point A is on the summit of the cap. If the distance

BA between the edge and the point A is greater than , we can consider only displacementdue to the pressure Pg and neglected the other forces.

=2

s=

2

1,5596= 4,2087

BA = Rs0 = 12,17 m 48,4 = 10,279 m

so BA. The displacement of A is given by the sum of the displacement of the dome andthe displacement of the ring beam due to the rotation around the support. With ( 4) we havethe horizontal displacement of the dome due to the pressure. The orthogonal displacement is:

A =s(0)

sin0=

1,4765 103 m

sin 48,4

= 1,9745

103 m

As in figure 10, the moment Mt is:

Mt = X2 X4 + a(Vr Vs Vc) + b(X1 X3) +2

3gb3 = 356,91 kN m2.

Mt give a negative rotation (clockwise) so the point B goes upward.

B = 2aMtM;RB = 2 0,5 356,91 7,2000 105 = 0,025 m

The sum of the two terms give the total displacement of the point A:

tot = A + B = 1,9745 103 m + 0,0025 m = 0,0046 m

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0.00317

0.000213

unit of measure: m


0.000166

0.00385


(b) Rotation .

9.13

211


(c) Bending moment in direc-tionMx.

1.83

42.3



165

794


(e) Shear force in directionQx.

165

794


(f) Axial force in directionNx.

129

1.92E+03



Figure 17: Effects of the horizontal force Htot.

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3.95E05

0.00019

unit of measure: m


0.00046

3.08E05


(b) Rotation .

39.2

1.69



7.84

0.339



1.32

30.6



1.32

30.6



115

23.9



Figure 18: Effects of the moment Mtot.

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0.0049

0.00808

unit of measure: m


0.000387

0.00316


(b) Rotation .

39.2

187



7.84

37.4



1.29E+03794



2.33

1.29E+03



2.97E+03

1.81E+03



Figure 19: Total effects on the cap.

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8 Comments

The support has the task of pulling downwards the beam and the whole tank. This is unusualin civil engineering but justified because the own weight is neglected. The gas pushes upwardthe dome much more than the water does on the cone. The cone has an average slope of thewalls greater than the sphere and this goes to deflect the push to the sides horizontally.

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