Design Water Tank

Design Example 1

REF

CALCULATIONS

OUT PUT

This example is to illustrate the design of one way and two way spanning walls in a rectangular water tank. Design an open overhead tank (internal dimensions 3.5 m x 6 m x 2.5 m) supported on a reinforced concrete frame structure as shown below. Use a minimum thickness of 200 mm for walls and floor slab. slab. Over flow pipe 200 mm 2.5 m

3.5 m

Consider the following material properties in the design Density of water = 9.8 kN/m3 Density of concrete = 25 kN/m3  fcu = 35 N/mm2  f y = 460 N/mm2

effective length l x = 6000 6000 + ½ x 2 x 200 = 6200 mm effective height ly = 2500 + ½ x 200 = 2600 mm l  x

6200

l  y



2600



2.4



2

200 mm

2.5 m

6m

200 mm

Hence slab slab will span one way i.e. as a cantilever. cantilever.

Design of water-retaining structures

2

REF

 BS 8007 

2.2.2

CALCULATIONS CALCULATIONS

OUT PUT

Maximum water pressure = 9.8 x 2.5 x 1.4 = 34.3 kN/m2      0      0 Water load F = ½ x 34.3 x 2.5 x 1      5      2 = 42.875 kN/m  BM at base (vertical (vertical bending) = 42.875 x [1/3 [1/3 (2.5) + 0.1] = 40.02 kNm/m

F

     0      0      1

Direct tension (Vertical direction) = 0  BM (Horizontal direction) direction) = 0

2.3

    3      /       )      0      0      5      2       (

The liquid level should be taken to the working top liquid level or overflow level. Maximum water pressure = 9.8 x 1.0 x 2.3 = 22.54 kN/m 2 Water load = ½ x 22.54 x 2.3 = 25.92 kN/m Maximum vertical BM = 25.92 x [1/3 x 2.3 + 0.1] = 22.46 kNm/m Maximum shear force at Base = 25.92 kN/m kN/m

l x

= 3500 + ½ x 2 x 200 = 3700 mm = 2500 + ½ x 200 = 2600 mm

ly l  x

3700

l  y



2600

 1.4 

2

3500 2500

(Two way spanning slab)

Maximum water pressure at ULS

= 9.8 x 2.5 x 1.4 = 34.3 kN/m2

Maximum water pressure at SLS

= 9.8 x 2.3 = 22.5 kN/m 2

 BM at  Base (ULS) = 40.02 kNm/m SF at  Base (ULS) = 42.875 kN/m

Max. BM (SLS) = 22.46 kNm/m Max. SF (SLS) = 25.92 kN/m

REF

Design Example 1

3

CALCULATIONS

OUT PUT

Ultimate limit state analysis of two way spanning walls Method 1 – By elastic analysis with ULS loads (approximate method) Method 2 - Plastic analysis using yield line theory ( exact method) Method 1 To carry out elastic analysis of thin walls any available method can be used. There are design charts (T 53) are available in the Reynolds’s Hand book for different boundary conditions of the wall. The curves given in Table 53 (RHB) can be used to calculate critical service bending moments on vertical and horizontal strips of unit width when the slab is fully fixed or freely supported or unsupported along the top edge. The curves are based on elastic analyses and for a value of Poisson’s ratio of 0.2. Maximum water load (ULS) = f = 34.3 kN/m 2 Using T53/RHB 0.015

For top edge unsupported wall slab l x  = 3700 l z  = 2600 k = l x / l z = 1.4 Mm = 0.015 x 34.3 x 2.6 2 = 3.48 kNm/m Ms = 0.055 x 34.3 x 2.6 2 = 12.75 kNm/m Mhm = 0.008 x 34.3 x 3.7 2 = 3.76 kNm/m Mhs = -0.021 x 34.3 x 3.7 2 = 9.86 kNm/m

0.008 -0.021

-0.021

-0.055

 BM at ULS (method 1) (kNm/m) Mm=3.48 Ms=12.75 Mhm=3.76 Mhs =9.86

Design of water-retaining structures

4

REF

CALCULATIONS

Method 2 – Yield line analysis The ultimate moments in a wall panel subjected to triangular loading can be determined by Yield-line method. A feature of yield line method of designing two-way slabs is that the designer is free to choose the ratio between the moments in each direction and between the  positive and negative units in each direction. However, in case of wall  panel subjected to triangular load distribution results from the storage of liquid, it is more important to ensure that the choice of these ratios in such that the formation of cracks under service loading is minimized. This is achieved if the ratios selected correspond approximately to those given by elastic analysis. The following design procedure is thus suggested. 1. Determine the proportion of the horizontal moments at mid span Mhm and at the supports M hs, and the vertical moment at the base Ms to the vertical moment at mid span M m from the appropriate chart on T53/RHB. 2. Then if i4 = Ms / Mm ,  = Mhm / Mm  and i1 = i3 = Mhs / Mm , determine , i1, i3 and i4 3. Calculate lyr  from l  yr 



2l y (1  i1 )  1  i3

and if the slab is supported at top edge, calculate l xr  from l  xr 



2l x (1  i2 )



1  i4

4. Finally, with f, l  x  (or l  xr ), l  yr  and i , determine M from the scale on Table 58 (RHB) if the slab is supported at top edge and from the chart T 61 otherwise. In the given example. l x  = 3700 l z  = 2600 k = l x / l z = 1.4

From T 53 – top edge unsupported  f = 24.5 kN/m2 Mm = 0.015 x 24.5 x 2.6 2= 2.48 kNm/m Ms = 0.055 x 24.5 x 2.6 2 = 9.12 kNm/m Mhm = 0.008 x 24.5 x 3.7 2 = 2.68 kNm/m Mhs = -0.021 x 24.5 x 3.7 2 = 7.04 kNm/m

OUT PUT

Design Example 1

5

CALCULATIONS REF

OUT PUT

The most suitable values for , i1, i3 and i4 i4



 



i1

9.12 

 M  m  M hm

i

l  yr 

2.48



3.6



1.08

2.68 

 M  m

 3 

l  yr 

l  x

 M   s

2.48

 M hs   M  m



7.04

2l  y



(1  i1 )

1.94 





1.08 x 2.48



1  i3



2.63 2  3.7

1



2.63





2



1.94

0.75

2.6

From T61 /RHB

 M 

 flx

2



0.013

 (for i4 = 3, lyr  / l x = 0.8)

M = M m = 0.013 x 34.3 x 2.6 2 Ms = i4 x Mm = 3.6 x 3.01 Mhm = 1.08 x M m = 1.08 x 3.07 Mhs = 2.63 x 1.08 x 3.01

= 3.01 kNm/m = 10.84 kNm/m = 3.25 kNm/m = 8.55 kNm/m

 BM at ULS (method 2) (kNm/m) Mm=3.01 Ms=10.84 Mhm=3.25 Mhs=8.55

Maximum water pressure = 2.3 x 9.8 = 22.54 kN/m 2 T 53 can be used although there is no water on the top 200 mm. Mm Ms Mhm Mhs

= = = =

0.015 x 22.54 x 2.6 2  = 2.29 kNm/m -0.055 x 22.54 x 2.62 = -8.38 kNm/m 0.008 x 22.54 x 3.7 2  = 2.47 kNm/m -0.021 x 22.54 x 3.7 2  = -6.48 kNm/m

 BM at SLS (kNm/m) Mm= 2.29 Ms = 8.38 Mhm= 2.47 Mhs = 6.48

Design of water-retaining structures

6

REF

CALCULATIONS

Short wall L y  = 3.7 m L  x  = 2.6 m K = ly / l x = 1.42 < 2

OUT PUT

R2

R1

2500

3500

R3

Total water load = ½ x 34.3 x 3.5 x 2.5 R4

 Assuming total load is uniformly distributed Intensity of load (w) = ½ x 34.3 = 17.15 kN/m 2 From T63 /RHB R 1 = R 3 = ½ K (1 - ¼ K) w l x2 = ½ x 1.42 (1 – 1.42/4) x 17.15 x 2.6 2 = 52.3 kN (total load carried by the shorter side) 

Direct tension carried by long wall = 52.3/2.6 = 20.12 kN/m

Maximum water pressure = 22.54 kN/m 2 W = ½ x 22.54 = 11.27 kN/m 2 From T63/RHB R 1 = R 3 = ½ x 1.42 (1 – 1.42/4) x 11.27 x 2.6 2 = 34.9 kN  Direct tension carried by long wall (SLS) = 34.9/2.6 = 13.41 kN/m

Mm Long wall Short 3.01 wall Short 3.48 wall

ULS Mhm Ms (KNm/m) 40.02 10.84

-

Mhs

T

Mrm

KN/m

-

3.25 8.55 Method 2 12.75 3.76 9.86 Method 1

20.12

SLS Mlm Ms (KNm/m)

-

22.46

2.39

8.38

-

Mhs

T KN/m

-

13.41

2.47 6.48

-

Design Example 1

7

CALCULATIONS

OUT PUT

REF

2.7.6

Thickness of the wall = 200 mm Cover = 40

 Assume main bars to be 20 mm and distribution bars to be 12 mm. d = 200 – 40 – 12 – 20/2 = 138 mm M = 40.02 KNm/m K’ = 0.156 No. redistribution  K  

3.4.4.4/  BS8110

 M  2

bd   f  cu



 Z   d 0.5 



40.02  10 10

3



12 Ø 20 Ø

6

 1382  35

0.25 

40

 0.06   K '

Hence no compression r/f is reqd

 K  



200

0.9 

= 0.93 d = 0.93 x 138 = 128.3 mm As



 M  0.87  f    y Z 

6



40.02  10

0.87  460  128.3

T = 20.12 KN/m

 A s





2

779mm / m

20.12  10

3

0.87   460



There are two methods available to calculate r/f under SLS Method 1 – Based on limiting steel stress Method 2 – Based on crack width limitation

2

 50.3mm  /m

Design of water-retaining structures

8

CALCULATIONS

REF

OUT PUT

 BS 8007

Considering the method 1 M (SLS) = 22.46 KNm/m 3

 M  2

bd 

T 3.1



22.46  10 3

10



1.2

 Allowable steel stress f st  = 130 N/mm 2 From design chart T 118 (Reynold’s HB)  = 0.011  f cr   = 6.5 N/mm2 (Maximum compression stress of concrete)  A s bd 



 B2

2

 138



 



0.011

 As = 0.011 x 10 3  x 138 = 1518 mm 2 /m

Ultimate stress of concrete = 0.45 f cu = 0.45 x 35 = 15.8 N/mm2 > f cr   = 6.5 N/mm 2 Hence concrete Stress is not near ultimate. Note: Method of obtaining As by calculating crack width (method 2) will be illustrated later.

T 3.1

 Allowable steel stress = 130 N/mm 2  A st 



T   f   st 



13.41 103 130



103.2mm2 / m

2.6.2.3  A.2 T A.1 Fig. A.1

Minimum steel ratio in each surfaces zone is given by   crit 

  





 f  ct   f   y

 A s bh'





0.0035

0.0035

h ‘ – Depth of surface zone

Design Example 1

9

CALCULATIONS

OUT PUT

REF

 BS8007

h’ = h/2 (Since h  500 mm) = 100 mm  As = 0.0035 x 10 3  x 100 = 350 mm 2 / m  per each layer.

Fig. A.1

R/f in controlling crack spacing has to be provided according to Cl. A.3 (8007). Wmax = Smax R .  . (T 1 + T 2) T 1 = 30o C T 2 = 10o C  = 10 x 10-6 / oC R = 0.5 for rigid end restraints

Considering



W msx S max



0.5  10  10



 A.3 T A.1

S max



 f  ct   f  b



2  10   2  





6

(30  10)

4

Steel ratio based on surface zone.

 f ct / f b  = 0.67

2.2.3.3 Use 12 mm bars and W max  = 0.2 mm S max



and    

0.67 

12



Surface  zone

2   W max

0.2

 1000 mm 2  10 2  10 4 0.67  6  0.004    crit  (0.0035) 1000

S max

h/2 h/2



4





This steel ratio has to be placed in both surface zones.  As = 0.004 x 1000 x 100 = 400 mm2 / m (for each direction) Calculate r/f required for 20 mm bars as well.

200

Design of water-retaining structures

10

EF

CALCULATIONS

OUT PUT

R/f to carry ult. BM (u1) - 779 mm 2 /m (U 1) R/f to carry serv. BM (S1) - 1518 mm 2 /m (S1) R/f to shrinkage and thermal movement - 400 mm2 /m (S3) per each layer.(S3) U1 - 771 S1 - 1518

S3 - 400

( Y20 @ 200 )

S3 - 400

200

R/f to carry ult. tension - 50.3/2 per layer (U 2) R/f to carry serv. tension - 103.2/2 per layer (S 2) R/f to shrinkage and thermal movement - 400 per layer (S 3)

U2 - 50.3/2 S2 - 103.2/2 S3 - 400 ( Y12@275 )

U2 - 50.3/2 S2 - 103.2/2 S3 - 400

200

2.6.2.3

Maximum spacing – lesser of 300 mm and thickness of the member (200mm). Therefore max. spacing = 200 mm

Y12@200

Y20@200

Y12@200

Y12@200

200

Design Example 1

11

REF

OUT PUT

CALCULATIONS

 BS8110

3.5.5

Ultimate Shear force at the base = 42.875 KN/m  



V  bd 

 100  A s

bd 

3.4.5.4

3.12.8.3



10

3

 f  cu  0.8

0.8

3.4.5.2



42.875  10

 

1000  138



 0.31  N  / mm 2

 138 2 35  4.73  N  / mm

 5 or 

100  1570

3

0.8

 f  cu

.

Hence OK

1.14

T3.9 /BS 8110 indicates V c  > v (=0.31) Hence No. shear r/f are required. Note: This is the normal situation.

 Bond stress f b  f bu (design ult anchorage bond stress)  f  b

3.12.8.4



 F  s  .l 

 f  bu



   

0.87 f   y .  . 2 4    .l   f  cu





0.217 f   y  l 



 f  bu

0.5  35  0.7

T 3.28 ( BS 8110) 

2.07  N  / mm 2

The design ultimate anchorage bond stress for horizontal bars in sections under direct tension should not be greater than 0.7 times the values in 3.12.8.4/BS8110.  f  b



 f  bu

0.217  460  12

 2.07

l  l   579mm

 Anchorage length for 12Ø bars = 580 mm

Design of water-retaining structures

12

REF

CALCULATIONS

Use lap length required according to (Cl. 3.12.8.9 / 3.12.8.11 / 3.12.8.13 and T 3.29) Simplified rules for curtailment given in Cl. 3.12.10.3 of BS 8110 do not apply as the wall is not subjected to a UDL.  Bars should extend beyond the point at which it is no longer required for a distance equal to an anchorage length. In this particular example it is not possible to curtail the bars as it violates the maximum bar spacing requirement.

The hand book for the previous code of practical (BS 5337) states that deflection at the top of wall need not be checked for open structures.. But when the reservoir is roofed, or supported at top, the criteria has to be checked. Cl. 3.5.7 / BS 8110 , Cl 2.1/ BS 8007 give the procedure to check deflection. Example 2 explains the applications of these clauses.

This is repetition of step a.1.3 (done by limiting steel stress method) To calculate crack width following procedure is necessary. (a) Calculate serviceability BM (b) Calculate depth of N.A, lever arm and steel stress by elastic theory. (c) Calculate surface strain allowing for stiffening effect of concrete. (d) Calculate the crack width. Consider long wall, h = 200 mm, M = 22.46 kNm/m

d = 138 mm

OUT PUT

Design Example 1

13

CALCULATIONS

OUT PUT

REF

(As) provided = 1570 mm 2 /m ρ



 A s



bd

1570 10

3

138





(Y 20 @ 200)

0.0114

Es  = 200 kN/mm 2 Ec = ½ x static modulus = ½ x 27 (T.7.2 / BS 8110) = 13.5 N/mm2 αe



Es



Ec

200

αeρ 

13.5 

200



13.5 

0.0114  0.169

  2 1 1     d αeρ     2  0.169  1   1  0.436 0.169    X = 0.436 x 138 = 59.4 mm

 x

α

ρ e

Z  d   x

3



138 

 59.4 3



118.2mm

Check stress levels  f s

Eq. (4)



Ms Z.A s



22.46 10 6 2  121 N/mm 118.2  1570 

 f cb

Eq. (5)



2Ms bxZ



0.8  460 N/mm 2

2  22.46 10 6 10 3  59.4  118.2 



6.4 N/mm 2

0.45  35 

Hence O.K. N/mm 2 ε 1



 f s  h   x  E s  d   x 



121 200  10 3

 200  59.4  3    1.08  10  138  59.4 

Design of water-retaining structures

14

REF

CALCULATIONS

b t(h   x)(a 1   x)  3 E s A s(d   x)

ε 2





ε

m

w

1000(200  59.4)(200  59.4) 3  200  10 3  1570(138  59.4) 0.267  10



3

 ε 1  ε 2  0.815   10 3 3 acr  . ε m



 a  C min    2  cr    h   x  Cmin  = 40 + 12 = 52 mm (acr  + 10) 2 = 622 + 100 2 acr   = 107.7 mm 1

 w

3  107.7  0.815   10

3

 107.7  52    200  59.4 

1  2

 0.15mm

 0.2mm Hence O.K.

Note: Since the crack width is not close to the design crack width it is  possible to reduce the reinforcement. Therefore by adopting this method an economical solution can be achieved.

OUT PUT

Design Example 1

15

CALCULATIONS

OUT PUT

REF

For large structures it is possible to save amount of steel required  to control cracking in immature concrete, as well as to obtain a large spacing between bars so that compaction of concrete will be easier by the use of movement joints (Cl. 5.3 / 8007) Cl. 5.3.3 / 8007 gives the options available in providing movement joints. These options are explained by the following examples. Design r/f to control cracking due to shrinkage and thermal movement in immature concrete in a wall 225 mm thick.  f cu  = 35 N/mm2  f y = 460 N/mm2 (10 mm  - type 2 deformed bars) Option (1) - Continuous construction (with no joints)  A2 / 8007

 crit  



 f  ct 

1.6

 f   y





0.0035

460

Wmax = Smax . R  (T 1 + T 2) W max S max

 Also

S max

1





 f  ct   f  b

W max 2  10

4

0.2 2  10

2

4





10  10 6 (30  10)  2  10 

  2  

 0.67 



4

(A.3 / 8007 ) 10 2  

 0.67 

10 2  

     0.00335    crit 

Hence provide crit  = 0.0035 

 A s

 225     2  

1000  

 As  = 394 mm2 /m (each surface zone, each direction) Provide Y 10 @ 175 mm C/C(A s  = 449 mm2)

Design of water-retaining structures

16

REF

CALCULATIONS

crit  = W max S max



 R  (T 1

 Also

 T 2

S max

0.2



 



½ x 10 x 10-6 x (30 + 0) = 1.5 x 10-4

 f  ct   f  b 

1.5  10

0.0035

)=



OUT PUT



  2  



0.67 

10

0.67 

10 2  



2  

4

   

0.0025

provide crit  = 0.0035 i.e. Y10 @ 175 mm c/c .

Minimum steel ration

= 2/3 crit  = 2/3 x 0.0035 = 0.00233

 As given in T 5.1 (Note 2) / 8007, the minimum steel ratio is sufficient in option (3).  As = 0.00233 x 1000 x 225/2 = 263 mm 2 /m  provide Y 10 @ 200 mm c/c (maximum bar spacing )   Joint spacing for option 3 (a) 

4.8 

w



4.8 

 

W max S max S max



 R. .T 1



S max

1 2

(if  Wmax isused) A.3/8007

 10  10

0.2 

1 2

 10  10

6

 

6



30

1333.3mm

30

 Joint spacing =

4.8 + 1.33 = 6.13 m  provide complete joints at 6 m.

  Joint spacing for option 3 (b)  0.5 Smax + 2.4 + W/ = 0.5 Smax + 2.4 + S max = 1.5 Smax + 2.4 = 1.5 x 1.33 + 2.4 = 4.4 m  Joint spacing for option 3 (b)  4.4 m Provide alternate partial and complete joints at 4 m .

REF

Design Example 1

17

CALCULATIONS

OUT PUT

  Joint spacing for option 3 (c) Smax + W/ = Smax + Smax = 2 Smax = 2 x 1.33 = 2.7 m 

 provide partial joints at spacing of 2.5 m.

Control of early thermal contraction and restrained shrinkage in walls and slabs can be effected by a suitable arrangement of r/f with or without joints. At one extreme control can be obtained by  providing substantial amount of r/f in form of small   bars  preferably of high bond type at close spacing and without joints in conc. At the other extreme control can be effected by presence of minimum r/f in form of large  bars together with movement joints at close spacing.

Design of water-retaining structures

18

REF

CALCULATIONS

OUT PUT

REF

Design Example 1

19

CALCULATIONS

OUT PUT

Design of water-retaining structures

20

REF

CALCULATIONS

OUT PUT