aci rectangular tanks2010
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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,
3
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
(From Appendix A) ω = 0.014
'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
Factored tension Nu = 1.65 x 3500 = 5775 lbs/ft width
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
(From Appendix A) ω = 0.039
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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
Factored tension Nu = 1.65 x 2800 = 4620 lbs/ft width
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