design of water tanks - • home | cairo · pdf file · 2012-12-12design of...

Design of Water Tanks:

Part (1)

Prof. Dr. Hamed Hadhoud

Cairo University Prof. Dr. Hamed Hadhoud

1


Types of Tanks

Elevated Tanks

Resting on Soil & Underground Tanks 2

Tank Walls

Walls

Shallow Medium Deep

L

H

L

H

L

H

L/H ≥ 2 H/L ≥ 2 L/H <2 & L/H >0.5


3

Shallow Walls

L

H

• L/H ≥ 2 (for same continuity conditions, or for different continuity conditions) where m is the continuity factor • Loads are transmitted in the vertical direction

gwH

Total Pressure

gwH

Vertical Strip

0.0

Horizontal Strip


1 m

4

2/ HmLm HL

Deep Walls

• H/L ≥ 2 (for same continuity conditions, or for different continuity cond.) where m is the continuity factor

• Loads are transmitted in the horizontal direction • Due to the fixation of the wall to the floor, the

lower portion of the wall doesn’t deflect horizontally resulting in transmitting part of the load in the vertical direction

gwH

Total Pressure

gwH

Vertical Strip Horizontal Strip

L

H H/4

gwH

H/4

Hwg4

3

H/4


5

2/ LmHm LH

Medium Walls

• L/H <2 & L/H >0.5 (for same continuity conditions, or for different continuity cond.)

• Loads are transmitted in both horizontal and vertical directions

• Due to the fixation of the wall to the floor, the lower portion of the wall doesn’t deflect horizontally resulting in transmitting part of the horizontal load in the vertical direction

gwH

Total Pressure Vertical Strip Horizontal Strip

H/4 hP

4

3

H/4

L

H

Pv Ph Pv Ph Ph

Grashoff Tables a and b

or use equations

Pv= a gw H (or b gw H) Ph= b gw H (or a gw H)


44

4

44

4

,

,

LmHm

HmdirLfactorload

LmHm

LmdirHfactorload

LH

H

LH

L

6

HmLmr HL / LmHmror LH /

5.0/&2/ HmLmHmLm HLHL

Tank Floors

gwH

tf

DL = gw tf + floor finishes LL = gw H wf= DL + LL

L1

L2

Grashoff Tables a and b

wa= a wf (in the short direction) wb= b wf (in the long direction)

Or use equations


4

22

4

11

4

112

4

22

4

11

4

221

LmLm

LmdirLinfactorload

LmLm

LmdirLinfactorload

7

2211 / LmLmr 1122 / LmLmror

Continuity Factors

8


End 1 End 2 Factor (m)

Continuous Continuous 0.76

Continuous Hinged 0.87

Continuous Free 1.76

Horizontal Beam

Roof Slab

Free

Statical System of Tank Walls

9


Cantilever H ≤ 3 m

Hinged @ top H ≤ 4 m

Fixed @ top H ≤ 5 m

Horizontal Beam

Roof Slab

tie

tie

tie

Column

Analysis & Design of Elevated

Rectangular Tanks: Vertical Sections

10


1.0 m 1.0 m

w2 w3

w1

w2 w3

w1

OR Top horizontal beam

Case of Cantilever Wall

11


w2 w3

w1

Statically Determinate

BMD

NFD

+

+ +

Case of Wall Hinged @ Top

(Analysis using 3M Equation)

12


w2 w3

w1

A

B C

D Twice Statically Indeterminate From Symmetry: MB = MC

Applying 3M equation @B

)(6)()(20 21 rrLMLHM BB H

L

35045

3

3

3

2

1

HwHwr

24

3

1

2

Lwr

Get MB

HR

Hw

HwM AB

6

4

6

2

32

2Get RA

RA

BMD

MB


(Analysis using Moment Distribution Method)

13


w2 w3

w1

A

B C

D

H

L

HR

Hw

HwM AB

6

4

6

2

32

2Get RA

RA

BMD

MB

Joint B

Member BA BC

K 0.75 I/L 0.5 I/L

D.F.

F.E.M. MBA MBC

Balance M. -(MBA+MBC)*DBA -(MBA+MBC)*DBC

Carry Over M. 0 0

Moment FM1+ Balance M + 0 FM2+ Balance M + 0

BCBA

BABA

KK

KD

BCBA

BCBC

KK

KD

Case K Carry over factor

Fixed-Fixed

K=I/L 0.5

Fixed-Hinged

K=0.75 I/L 0.0

Symmetrical K=0.5 I/L 0.0



(Analysis using Moment Distribution Method)

Fixed-end Moments

14 w= g h 15

3hg

20

3hg

117

3hg

124

3hg


15


w2 w3

w1

A

B C

D Maximum Positive BM in the wall @ point of zero shear

XH

wXR

Q

A2

2

1

0H

L

Get X

Get Mm

RA

X

RA

X

H

w2

32

1 2 XX

H

wXXRM Am

Maximum Positive BM in the floor

BMD

MB

Mm

Mf

Bf MLw

M 8

2

1


16


w2 w3

w1

A

B C

D Tension Force @ floor

Af R

Hw

HwT

2

4

2

3

2H

L

Get Tf

Get T,wallmax

RA

2, 1

max

LwwallT

Maximum Tension Force @ wall bottom

NFD

T,wallmax

Tf

+ + +

Get Tm

H

XwallTTm max,

Tension Force @ wall section with max. +ve BM


17


A

B C

D

Critical sections

1

2

3 4

Section Type M N

1 Air-Side Section Mm Tm

2 Water-Side Section MB T,wall max

3 Water-Side Section MB Tf

4 Air-Side Section Mf Tf

Analysis & Design of Elevated

Rectangular Tanks: Horizontal Sections

18


Consider a strip @ H/4 from wall bottom From symmetry once-statically indeterminate

p1

p1

p2 p2

L1

L2

H/4 p

24246)()(2

3

22

3

112211

LpLpLMLLMML

Get M

2&

2

222

111

LpT

LpT

B.M.D.

M M

M M

1 3

2

4

Design critical sections; 2&3 Water-side 1&4 Air-side

1T 1T

2T

2T

Design of walls as deep beams

(in-plane action)


L

H

• The Tank walls will act as beams carrying the tank floor gravity load

• Those beams are usually deep beams • Check the condition for the deep beam

beamcontinuousforL

H

beamsimpleforL

H

4.0

8.0

19

L

H

L


(in-plane action)


min,

87.0

@37.0

@43.0

86.0

S

s

y

US

CT

UU

CT

CT

CT

CT

Af

TA

y

MT

caseanyinHywhere

momentnegativebeamcontinuousforLy

momentpositivebeamcontinuousforLy

beamssimpleforLy

g

20


(in-plane action)


21

Detailing


22

Vertical Section

Deep Beam RFt

Water stop (@ construction joint)

Detailing


23

Vertical Section

Deep Beam RFt

Water stop (@ construction joint)

Detailing


24

Vertical Section

Deep Beam RFt

Detailing


25

Vertical Section

Deep Beam RFt

Detailing


26

Vertical Section

Deep Beam RFt

Detailing


27

Vertical Section

Deep Beam RFt

Detailing


28

Horizontal Section

Detailing


29

Horizontal Section

Example (1)


30

Design and give full details for the conduit shown below

Loads


31

Straining Actions


32

Vertical Strip:

Floor Longitudinal Strip:

Vertical Strip

Floor Longitudinal Strip

Design of Critical Sections


33

Design of section (1):



34


34



35


35



36


1.5*18.5= 27.8

7.8

421 27.8



37

Design of beam action of wall:

Own weight= 25*0.3*2= 15 kN/m’

Floor load=24.5*2*0.79= 38.7 kN/m’

Wu= 1.5*(15+38.7)= 80.6 kN/m’

beamDeepL

H 4.0

5

2

my

mm

Lymomentnegative

my

mm

Lymomentpositive

mHy

CT

CT

CT

CT

CT

74.1

74.185.1537.0

37.0@

74.1

74.115.2543.0

43.0@

74.187.0

mkNLw

M

mkNLw

M

UveU

UveU

.9.16712

56.80

12

.5.20110

56.80

1022

max,

22

max,

momentnegativeforSame

usemmA

mmf

TA

kNy

MT

S

s

y

US

CT

U

veU

184900

308

15.1

360

10005.96

5.9674.1

9.167

2

min,

2

g

Reinforcement Details


38

418 418

Example (2)


39

Calculate bending moments and normal forces due to the shown water pressures using both 3M equation and moment distribution methods (tf=tw=0.3 m)

5 m

5 m

50kN/m2

60 kN/m2

Using 3M equation


40

5m

5m

50kN/m2

60 kN/m2

A D

B C

Twice Statically Indeterminate From Symmetry: MB = MC

Applying 3M equation @B

mkNM

MM

B

BB

.33.108

)24

560

45

550(6)5()55(20

33

kNR

RM

A

AB

20

56

550 2

Using moment distribution method


41

Joint B

Member BA BC

K 0.75 I/L 0.5 I/L

D.F.

F.E.M.

Balance M. -(83.33-125)*0.6 = 25

-(83.33-125)*0.4= 16,67

Carry Over M. 0 0

Moment 108.33 -108.33

6.0

5.075.0

75.0

BCBA

BABA

KK

KD

4.0

5.075.0

5.0

BCBA

BCBC

KK

KD

33.83

15

550 2

BAM

mkN

M BC

.125

12

560 2

kNRRM AAB 2056

550 2

5m

5m

50kN/m2

60 kN/m2

A D

B C

0.0 BABC MM

0.0 BABC MM

design of water tanks - • home | cairo · pdf file · 2012-12-12design of...

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