ACI Rectangular Tanks2010

UWU/Civil & Environmental Engineering CVNG 3016 – Design of Environmental Engineering Systems 2010

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CVNG 3016 DESIGN OF ENVIRONMENTAL SYSTEMS TOPIC: Design of Reinforced Concrete Liquid Retaining Structures

LECTURER: Dr. William Wilson

DESIGN OF REINFORCED CONCRETE TANKS (ACI 318 / ACI 350) Recommended Reading:

A. Codes 1. ASCE7-05 - Minimum Design Loads for Buildings and other Structures 2. ACI 318-06 – Building Code requirements for Reinforced Concrete 3. ACI 350R – 06 – Environmental Engineering Concrete Structures.

B. Technical Literature 1. Munshi, Javeed A. Rectangular Concrete Tanks (Rev. 5th Ed.), Portland Cement

Association, 1998. 2. Portland Cement Association, 1992. Underground Concrete Tanks 3. Portland Cement Association, 1993. Circular Concrete Tanks without prestressing

RECTANGULAR TANKS

Design Considerations

• Flexure – bending in walls and base

• Shear - wall-to-base, wall-to-wall junctions • Tension - horizontal tension in walls, base • Deflection – vertical/ horizontal deflections of wall • Cracking - thermal, flexural, tension cracks • Flotation - when base is located below water table level • Base Fixity - (i) Fixed (ii) Pinned

Loading Conditions Condition 1 - Internal Water Pressure only (before backfilling, i.e. leakage test) Condition 2 - External Earth Pressure only (before filling tank) Condition 3 - Tank full and Soil backfilled (resistance provided by soil is ignored)

• in reality neither of these conditions may actually exist

• Both may need to be investigated


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M (-)M

M M (-)

Ft

Ft

F t F t

Fv

Fv

Fv

Fv (shear)

+-

(b) Vertical Forces (a) Liquid pressure in Tank

(c) Horizontal Forces

Fv

Fv

Fv Fv

M (+)

Moments pinned base fixed base

FIGURE 1 – FORCES IN RECTANGULAR TANK WALLS


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STRENGTH DESIGN METHOD Basic Requirement: Design Strength ≥ Required Strength

∅(Nominal Strength) ≥ U

nR Uφ ≥

(ACI 318 Sect. 9.2) U = 1.4 (D + F) (2.1)

(ACI 350 cl. 2.6.5) U = 1.7 (D + F) (2.2) D = dead load F = liquid pressure

Sanitary Durability Factors - ACI 350 applies sanitary durability factors (based on crack width calculations) to obtain the Required Strength

Required Strength = Sanitary Coefficient x U

Ur = Cs x U (2.3)

Sanitary Coefficients are: Cs = 1.3 (bending) Cs = 1.65 (direct tension / hoop tension) Cs = 1.3 (shear beyond shear capacity of concrete – stirrup design) Cs = 1.0 (concrete shear)

Strength Design Requirements

(a) Flexural Reinforcement

Design Strength ≥ 1.3U

∅Mn ≥ 1.3(1.4MD+ 1.7 ML+1.7 MF) (2.4)

(b) Direct Tension Reinforcement

Design Strength ≥ 1.65U ∅Nn ≥ 1.65(1.4ND+ 1.7NL+1.7NF) (2.6)


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(c) Stirrup Shear Reinforcement

Design Strength ≥ 1.3 (Vc - ∅Vc)

∅Vs ≥ 1.3 (Vu - ∅Vc) (2.7)

(d) Concrete Shear and Compression Reinforcement

Design Strength ≥ 1.0 U

∅Vn ≥ 1.0Vu (2.8)

(e) Minimum reinforcement (ACI 318-05 cl. 10.5)

3c w

s, min wy y

f b dA b d 200f f

′= ≥

(2.9)

(f) Minimum cover = 2 in. (g) Minimum thickness for walls over 10 ft. high = 12 ins.

Serviceability for Normal Sanitary Exposure (ACI 350, cl. 2.6.6) Crack Control Maximum Design crack width

• Severe exposure = 0.010 in. • Aesthetics = 0.008 in.

Crack width calculation is based on the following equation:

3s cz=f d A (2.10)

Where, Z is a quantity limiting distribution of flexural reinforcement (ACI 350 limits) z ≤ 115 kips/in (crack width = 0.010 in) z ≤ 95 kips/in (crack width = 0.008 in)

• Concrete sections with t ≥ 24″ use minimum temperature and shrinkage reinforcement at each face based on 12″ thickness.

• Size of rebar ≤ #11 • Max. spacing of rebar ≤ 12″ • Minimum cover in tank walls = 2″


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fs = calculated stress in reinforcement at service loads, ksi dc = concrete cover to centroid of closest rebar

A = effective area of concrete surrounding flexural reinforcement with same centroid divided by the number of bars, in2.

The maximum spacing it given by,

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w 2 3c s

Zb2d f

=

(2.11)

t

bw

dc

A= 2dcbw

Fig. 1.0 - Calculation of A


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EXAMPLE 1 Rectangular Tank

Design Data Weight of Liquid, w = 70 lbs/ft3 Weight of Soil, γs = 100 lbs/ft3 ka = 0.3

'cf = 4000 psi

fy = 60,000 psi Wall thickness, t = 18 in. Base slab projection beyond wall = 2.5 ft. Water pressure at base, p = wa = 70 x 10 = 700 lbs/ft2

Longitudinal Section Cross Section

Plan

30'-0″

20'-0″

20'-0″

10'-0 ״

350

2.5'

Fig. E1 – Plan and sections of Tank

30'-0″


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Design for Vertical Bending Moments Wall considered fixed at base and free at top Using PCA Charts to calculate moments (Table 3-29: case 3 – Long side) Ratio of length/height = b/a = 30/10 = 3.0 (long side) Ratio of width/height = c/a = 20/10 = 2.0 (short side) Mx = CMx x pa2/1000 = CMx x 700 x 102 / 1000 = CMx x 70 ft-lbs = CMx x 0.84 in-kips For sanitary structures - Mu = Sanitary coef x 1.7 x M Mux = 1.3 x 1.7 x Mx Mux = 1.3 x 1.7 x CMx x 0.84 in-kips = 1.86 x CMx

Maximum positive moment at 0.7 a (CMx = +10) Mux = +18.6 kips-in Maximum negative moment at bottom (CMx = -129) Mux = -239.9 kips-in

Assuming No. 5 bars at 12in. c/c Cover = 2 in. Wall thickness = 18 in. d = 18 – 2 – 5/16 = 15.7 in. (db = 5/8 in.)

M = + 18.6 kips-in

M = - 239.9 kips-in

Fig E2 – Vertical Moments at mid-length

a = 10'

V = 3500 lbs


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(From Appendix A) ω = 0.023

'2c

sy

f 4A =ωbd 0.023x12x15.7x 0.29inf 60

∴ = =

Check minimum steel (ACI 318-05 cl. 10.5)

' 33c 2

smin wy

f 4000A = b d= x12x15.7=0.595inf 60000 ≥ 200 bwd/fy

2w

y

200b d 200x12x15.7= = =0.628inf 60,000 (governs)

(ACI 318 -05 cl. 10.5.3) Use 4/3 of As required by analysis As = 4/3 x 0.29 = 0.39in2 Provide No. 5 @ 9 in c/c on inside face (As =0.41 in2)

Design for Horizontal Bending Moments

Fig. E3- Horizontal Forces (un-factored) at mid-height of tank) NB: Shear Forces calculations on Page 9 below

+35.93 in-kip

+35.93

1890 lbs

1890 lbs -65.6 in-kip

-65.6 -65.6

-65.6

2590 lbs

2590

18901890

1890

2590

2590

2590

2590

1890

( )22c

M 239.9 =0.0225f bd 0.9x4x12x 15.7φ ′

=


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(a) At Corner - Horizontal Moment Steel

Muy = 1.3 x 1.7 x 65.66 Muy = 145.1 in-kips

u1 2 2c

M 145.1= =0.0136f bd 0.9x4x12x15.7φ


'2c

sy

f 4A =ωbd =0.014x12x15.7x =0.18inf 60

(b) Steel required for Direct Tension in Long Wall

Factored tension Nu = 1.65 x 3213 = 5301 lbs/ft width

2us

y

N 5301A = = =0.1in0.9f 0.9x60,000

Direct tension steel is equally distributed on inside and outside faces of wall.

Total steel required on inside face = 20.10.18+ =0.23in

2

2ws,min

y

200b dA = =0.625inf (governs)

4/3 of As required by analysis = 4/3 x 0.23 = 0.31 in2 Provide No.5 @ 12 in (As = 0.31 in2) horizontal steel on inside face of long walls.


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(c) Horizontal steel near centre of outside face of wall

Design for M = 40.91 in-kips (d) Crack Control - Check Maximum Spacing of bars

Maximum un-factored moment ( )

239.9M= =108.6in-kips1.7x1.3

Stress in steel reinforcement

ss

Mf =A jd

As = 0.41 in2/ft d = 15.7 in

( )29,000n= = 8

57 4000

( )0.41 0.00218

12 x 15.7ρ = =

( )22 0.17k n n nρ ρ ρ= + − = j = 1 – k/3 = 0.94

s108.6f = =17.95ksi

0.41x0.94x15.7∴

( )3

m ax 2 3c s

zs =2xd xf

dc = cover + 2 0.313=2.313in2φ= +

z = 115 kips/in fs = 17.95 ksi

3

max 2 3

115s = =24.6in>9in2x2.313 x17.95

OK


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Design for Shear Forces Using PCA Charts to calculate moments (Table 2-17: CASE 3 – Long side) Ratio of length/height = b/a = 30/10 = 3.0 (long side) Ratio of width/height = c/a = 20/10 = 2.0 (short side) Shear, V = Cs x p x a (a) Check Shear at bottom of Wall

Maximum shear at bottom of long wall, V = 0.50 x 700 x 10 = 3500 lbs

uV =1.7x3500=5950 lbs∴ Since tensile force from adjacent wall is small

'c cV =2 f bd

=2 4000x12x15.7 = 23,831 lbs

V 0.85 x 23,831= 20,256lbs > 5950 lbscφ =

OK (b) Check Shear at side edge of long wall

V = 0.37 x 700 x 10 = 2590 lbs Vu = 1.7 V = 1.7 x 2590 = 4403 lbs Wall subjected to simultaneous tensile force due to shear in short side wall; (ACI 318 cl.11.3.2.3) gives allowable shear as:

'uc c

g

NV =2 1+ f bd500A

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Nu = tension in long wall due to shear in short wall. Shear in short side wall V = 0.27 x 700 x 10 = 1890 lbs.


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Nu = -1.7 x 1890 = -3213 lbs Ag = 18 x 12 = 216 in2

1c c

-3213V =2 1+ f bd500x216

=1.94 4000x12x15.7 = 23,116 lbs

⎛ ⎞⎜ ⎟⎝ ⎠

cV 0.85 x 23,116 = 19,649 lbs > 4403 lbsφ = OK

(e) Shrinkage and Temperature Reinforcement

Assuming the walls will be in one pour of 30 ft long. Minimum Temperature and Shrinkage reinforcement

(Fig 1-2) stA =0.0033bh

2st

1A = x12x18=0.356in2

(No. 5 @ 10”)

< 0.41 in2 (No. 5 @ 9”) OK

Summary of Reinforcement Inside face – vertical No. 5 @ 9 in Outside face – vertical No. 5 @ 10 in (Use No. 5 @ 9 in for consistency) Inside face – horizontal No. 5 @ 12 in Outside face – horizontal No. 5 @ 12 in


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BASE SLAB DESIGN

• Design as a 2-way spanning slab, simply supported at edges.

• Assume that the pressure beneath the slab is uniform and is generated by the weight of the walls spread over entire area.

Unit weight of concrete = 150 lbs/ft3

30'

12″

10'

P

3500 lbs k

3500 lbs

2500 lbs

2500 lbs Mls = 8497 ft-lb

Mss = 15,413 ft-lb

Vls

Vs = 4150 lbs

Fig E4 – Forces on Base Slab


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( ) ( )( )

( )2

Wall Dead load 150 2 x 1.5 x 10 x 30 + 0.75 + 20+0.75 = 231,750 lbs

Factored DL = 1.4(231,750) = 324,450 lbs324,450 = 509 lbs/ft

30.75 x 20.75p

=

=

Using PCA Tables Mss (short span) = 78 x 509 x 202 /1000 = 15,881 ft-lb = 191 in-k Mls (long span) = 43 x 509 x 202 / 1000 = 8755 ft-lb = 105in-k Vss = 0.24 x 509 x 20 = 2443 lbs = 2.4 k Vls = 0.42 x 509 x 20 = 4276 lbs = 4.3 k Ft (long span) = 0.4 x 700 x 10 = 2800 lbs Design of Short Span

Assume slab thickness, h = 12 ins. For #6 bars and 2″ cover d = 12 – 2 – 0.75/2 = 9.625 ins. Short Span (a) Mid-span Moment Steel (NB: ACI 350 is silent on use of sanitary coefficient for slabs, but we will apply

same here)

Muy = 1.3 x 191 Muy = 248.3 in-kips

u' 2 2c

M 248.3= = 0.062f bd 0.9 x 4 x 12 (9.625)φ

(From Appendix A)

3.5 k

191 in-k 2.4k

12″


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ω = 0.064

'2c

sy

f 4A =ωbd =0.064 x 12 x 9.7x = 0.49inf 60

(b) Steel required for Direct Tension in short span


2us

y

N 5775A = = =0.11in0.9f 0.9x60,000

Direct tension steel is equally distributed on inside and outside faces of slab.

Total steel required on top face = 20.110.49+ =0.55 in

2

( ) 2ws,min

y

200 12 x 9.7200b dA = = = 0.39 inf 60000

4/3 of As required by analysis = 4/3 x 0.55 = 0.73 in2 Provide No.6 @ 7 in crs. (As = 0.75 in2) steel on top face of slab in short direction.

(c) Long Span

Muy = 1.3 x 105 in-kips Muy = 136.5 in-kips d = 12 -2 -0.625- 0.625/2 = 9.0625

u1 2 2c

M 136.5= = 0.0385f bd 0.9 x 4 x 12 x 9.0625φ



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'2c

sy

f 4A =ωbd =0.039 x 12 x 9.0625x = 0.283 inf 60

Steel required for Direct Tension in long span


2us

y

N 4620A = = =0.085in0.9f 0.9x60,000

Direct tension steel is equally distributed on inside and outside faces of slab.

Total steel required on top face = 20.0850.283+ =0.326 in

2

( ) 2ws,min

y

200 12 x 9.7200b dA = = = 0.39 inf 60000 (governs)

4/3 of As required by analysis = 4/3 x 0.31 = 0.41 in2

Provide No.5 @ 8 in crs (b) Check Shear at side edge of long span

V = 4276 lbs Vu = 1.0 V = 4276 lbs Wall subjected to simultaneous tensile force (ACI 318 cl.11.3.2.3) gives allowable shear as:

'uc c

g

NV =2 1+ f bd500A

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Nu = 3500 lbs Ag = 12 x 12 = 144 in2


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'c c

-3500V =2 1+ f bd500 x 144

=1.90 4000x12 x 9.7= 14,008 lbs

⎛ ⎞⎜ ⎟⎝ ⎠

cV 0.85 x 14,008 = 11,907 lbs > 4276 lbsφ = OK

(e) Shrinkage and Temperature Reinforcement

Assuming the walls will be in one pour of 30 ft long. Minimum Temperature and Shrinkage reinforcement

(Fig 1-2) stA =0.0033bh

2st

1A = x 0.0033x12x12=0.238 in2

< 0.73 in2

OK Summary of Reinforcement Top face – short direction (mid-span) No. 6 @ 7 in crs Top face – short direction (edges) No. 5 @ 7 in crs (for consistency) Top face - long direction (mid-span) No. 5 @ 7 in (for consistency) Bottom face - both direction (edges) No. 5 @ 7 in Bottom face both direction (mid-span) No. 5 @ 12 in


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Rectangular Tank

Dr. William Wilson 17 March 2010

No. 5 @ 12 in. c/c

No. 5 @ 9 in. c/c No. 5 @ 10 in c/c

No. 5 @ 10 in c/c

Section showing Reinforcement Details for Rectangular Tank

No. 5 @ 12 in. c/c

No. 6 @ 7 in. c/c No. 5 @ 7 in. c/c

No. 5 @ 7 in. c/c both ways

Water stop

Water stop

ACI Rectangular Tanks2010

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