Reinforced Concrete Tanks

4th year Civil

Lecture 2
Rectangular Tanks Continued
By: Abdel Hamid Zaghw
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Prof. Abdel Hamid Zaghw

Example
a- Design the open channel tank shown below.
b- Design the top beam and tie

Given:
−fcu = 25 N/mm2 fy = 360 N/mm2
−Spacing between top ties is 5m.
−Spacing between columns in the direction of the open channel is 6m.
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Prof. Abdel Hamid Zaghw

Solution:

Concrete Dimensions: Assume tf = tw = 30cm

Loads:
ww  hw w  5.0 10  50 kN/m2
w f  t f  c  hw w  0.3  25  5.0 10  57.5 kN/m2
Distribution Factors:
0.75  I 5.0
Dba   0.6
0.75  I 5.0  0.5  I 5.0
Dbc  1  Dba  0.4
Fixed End Moments:
ww h 2 50  52
FEM ab     83.33 kN.m
15 15
wf l 2
57.5  52
FEM bc    119.8 kN.m
12 12
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Prof. Abdel Hamid Zaghw

Reactions
50  52 50  52
 5Ra   105.21  0 5Ra   105.21  20.62kN
6 6
50  52
Rb   20.62  104.38kN
2
Zero Shear
10 x 2 2  20.62
 20.62 x  2.03
2 10

Max Positive Moment

10 x 3 10  2.033
M ive  Ra x   20.62  2.03   27.92 kNm
6 6
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Prof. Abdel Hamid Zaghw

Design of Sections
Section 1
Mw = 105.21 kNm, Tw = 143.75 kN (Water Side Section)
Mw 105.21
t 10 4  30  104  30  622.2mm take t = 700 mm
3 3
6 105.21103
f ct M w    1 .288 N/mm 2
f ct  N w  
143.75
 0.205 N/mm 2

7002 700
  f ct ( N w )    .205 
tv  t 1   
   7001     811.6 mm η =1.7
  f ct ( M w )    1.288 
fctr  0.6 fcu  0.6 25  3
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Prof. Abdel Hamid Zaghw

f ct   f ct ( N w )  f ct ( M w )  f ct  .205  1.288  1.493 

f ctr 3
(OK)
 1.7
Reinforcement Calculations
Mu = 105.21 x 1.5 = 157.8 kNm, Tu = 143.75 x 1.5 = 215.6 kN

M u 157.8 1000
e   731.9
Nu 215.6
es = e – t/2 + cover = 731.9 – 350 + 30 = 411.9 mm
Mus = N . es = 88.8 kN.m   0.01
Assume φmax = 18 mm  cr  0.75
f cu Nu 25 215.6 1000 1.15
As  bd   0.011000  670 
 cr f y f 0.75  360 0.75  360
y
 cr
s
= 1538.67 mm2 choose 7φ18 /m
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Prof. Abdel Hamid

Zaghw

Section 2
Mw = 74.48 kNm, Tw = 104.38 kN (Air Side Section, t=300mm)
Mu = 74.48 x 1.5 = 111.72 kNm, Tu = 104.38 x 1.5 = 156.57 kN

M u 111.8 1000
e   714.06
Nu 156.57
es = e – t/2 + cover = 714.06 – 150 + 25 = 589.06 mm
Mus = N . es = 92.23 kN.m   0.06
Assume φmax = 18 mm  cr  0.85
f cu Nu 25 156.57 1000 1.15
As  bd   0.06  1000  275 
 cr f y fy 0.85  360 0.85  360
 cr
s
= 1936.46 mm2 choose 8φ18 /m
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Prof. Abdel Hamid Zaghw

Section 3
Mw = 27.92 kNm, Tw = 58.36 kN (Air Side Section, t=300mm)
Design similar to section 2
Assume φmax = 12 mm  cr  1.0
As = 7φ12 /m

Design of Top beam and Tie

Tie: (axial tension Nw = 103.1 kN)

N w103 f cr 3 1.7 N w103 N w103

  t  0.6 mm
bt  1.7 3 b b
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Prof. Abdel Hamid Zaghw

b=250mm
N w103 103.1103
t  0.6  0.6  247.44mm
b 250
Take tie 250 x 250 mm
Reinforcement Calculations
Nu = 1.5 x 103.1 = 154.65 kN
Assume φmax = 16 mm  cr  0.75
Nu 154.65 103
As    659 mm 2 Choose 4 φ16
fy 360
 cr 0.75
s 1.15

Top Beam Case 1

Section 1 (water side -pure bending Mw = 42.96 kNm)

6M w 106 f cr 3 1.6 6M w Mw
  t  106  107 mm
bt 2  1.6 3 b 3b

Mw 42.96 Take t=800mm

t 107  107  757mm
3b 3  250
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Prof. Abdel Hamid Zaghw

Top Beam Reinforcement Calculations

Case (1)
Section 1
Design as Category 3 R Sec

Section 2
Design as Category 2 L Sec
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Prof. Abdel Hamid Zaghw

Top Beam Reinforcement Calculations Cont.

Case (2)
Section 1
Design as Category 3 L Sec

Section 2
Design as Category 2 R Sec
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Prof. Abdel Hamid Zaghw

Design Wall as Deep Beam

• Load = ow + load from slab

• Load = 0.3 x 5.0 x 25 x1.5 + 557.5 x 1.5 x 2.5=271.875 kN/m

• Get Moment as Continuous beam with span 6m

• M-ive = wL2/12=816 kN.m get T As

• M+ive = wL2/24=408 kN.m get T As

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Prof. Abdel Hamid

Zaghw

Detailing Notes
Note (1)
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Prof. Abdel Hamid Zaghw

Detailing Notes
Note (2)
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Prof. Abdel Hamid Zaghw

Wall Floor Connection

Note (3)

Shallow Cantilever Deep

Shallow and Medium

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Prof. Abdel Hamid Zaghw

Example Detailing
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Prof. Abdel Hamid Zaghw

Deep Tanks

Deflected Load Load in Load in

shape
Distribution Horizontal Vertical
Direction Direction

Deflected
shape in
Deep Tanks
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Prof. Abdel Hamid Zaghw

Design of Deep Tanks

Design in the Vertical Direction

Take α and β in floor loads from Grashoff tables

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Prof. Abdel Hamid Zaghw

Deep Tanks
FEM for wall load in the Vertical Direction

Fixed Free Fixed Hinged Fixed Fixed

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Prof. Abdel Hamid

Zaghw

Design of Deep Tanks Cont.

Design in the horizontal Direction

Distribution Factors:
0.5  I L1 1 L1
Dab  
0.5  I L1  0.5  I L2 1 L1  1 L2
Dad  1  Dab
Fixed End Moments:
0.75 w hL1 0.75 w hL2
2 2
FEM ab  FEM ad 
12 12
Final Moment:

M f  FEM ab  ( FEM ab  FEM ad ) Dab

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Prof. Abdel Hamid Zaghw

Deep Tanks - Design in the horizontal Direction Cont.

Final Bending Moment and Normal Force Diagrams

in the Horizontal Direction
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Prof. Abdel Hamid

Zaghw

Design of Walls as Deep Beam

For Beam L1

Repeat for Beam L2

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Prof. Abdel Hamid Zaghw

Medium Tanks
mL L1 0.76 L1
r 
mh h 0.87h
Take α and β in wall loads form Grashoff tables
if mL L1 mh h PV    wh PH    wh
if mL L1  mh h PV    wh PH    wh assuming in above figure that
mLL1>mhh

Load in Horizontal Direction Load in Vertical Direction

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Prof. Abdel Hamid Zaghw

Design of Medium Tanks

Design in the Vertical Direction

Strip 1 Strip 2
PV 1  1  wh PH 1  1  wh
PV 2   2  wh PH 2   2  wh
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Prof. Abdel Hamid Zaghw

Medium Tanks
FEM for wall load in the Vertical Direction

Fixed Free Fixed Hinged Fixed Fixed

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Prof. Abdel Hamid Zaghw

Design of Medium Tanks Cont.

Design in the horizontal Direction
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Prof. Abdel Hamid Zaghw

Design of Medium Tanks Cont.

Design in the horizontal Direction

Distribution Factors:
0.5  I L1 1 L1
Dab  
0.5  I L1  0.5  I L2 1 L1  1 L2
Dad  1  Dab
Fixed End Moments:
2 2
0.75PH L1 0.75PH L2
FEM ab  FEM ad 
12 12
Final Moment:

M f  FEM ab  ( FEM ab  FEM ad ) Dab

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Prof. Abdel Hamid Zaghw

Medium Tanks - Design in the horizontal Direction Cont.

Final Bending Moment and Normal Force Diagrams

in the Horizontal Direction
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Prof. Abdel Hamid

Zaghw

Design of Walls as Deep Beam

For Beam L1

Repeat for Beam L2

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Prof. Abdel Hamid Zaghw

Corner Effects in Shallow Tanks

Shallow Tank
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Prof. Abdel Hamid

Zaghw

Corner Effects in Shallow Tanks Cont.

Simply Supported Top Hinged Top Fixed Top

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Prof. Abdel Hamid Zaghw

Continuity Factor for Cantilever Walls

All edges
Simply αw
h
supported

(1-α) w

The deflections of strip (1) and strip (2) at point a should be the same
5(1   ) wh 4 5wL4 h4 L4
 (1   )h  L 4 4
  4 4 ,  4 4
384 EI 384 EI L h L h

Grashoff Distribution
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Prof. Abdel Hamid Zaghw

Continuity Factor for Cantilever Walls

Free αw

(1-α) w
Simply Simply
supported supported

Fixed

L
The deflections of strip (1) and strip (2) at point a should be the same
(1   ) wh 4 5wL4 5L4 9.6h 4 (1.76h) 4
 (1   )h 
4
 4  4
8EI 384 EI 48 L  9.6h 4
L  (1.76h) 4

The continuity factor for free edge can be taken equal to 1.76