UWU/Civil & Environmental Engineering
CVNG 3016 – Design of Environmental Engineering Systems
2010
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
-
• Flotation
thermal, flexural, tension cracks
- when base is located below water table level
• Base Fixity - (i) Fixed (ii) Pinned
Loading Conditions
Condition 1 Condition 2 Condition 3 -
• in reality neither of these
conditions may actually exist
• Both may need to be investigated
Internal Water Pressure only (before backfilling, i.e. leakage test)
External Earth Pressure only (before filling tank)
Tank full and Soil backfilled (resistance provided by soil is ignored)
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UWU/Civil & Environmental Engineering
CVNG 3016 – Design of Environmental Engineering Systems
2010
Moments
pinned base
fixed base
+
Fv
(a) Liquid pressure in Tank
Fv
M
Ft
M (-)
Fv
Ft
Ft
M
(b) Vertical Forces
Fv
M (+)
Fv
(shear)
Ft
Fv
M (-)
Fv
Fv
(c) Horizontal Forces
FIGURE 1 – FORCES IN RECTANGULAR TANK WALLS
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CVNG 3016 – Design of Environmental Engineering Systems
2010
STRENGTH DESIGN METHOD
Basic Requirement:
Design Strength ≥ Required Strength
∅(Nominal Strength) ≥ U
φ Rn ≥ U
(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:
(bending)
Cs = 1.3
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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CVNG 3016 – Design of Environmental Engineering Systems
2010
(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)
As, min =
3
f c′
fy
b w d ≥ 200
(f) Minimum cover = 2 in.
bw d
fy
• Concrete sections with t ≥ 24″ use
minimum
temperature
and
(2.9)
shrinkage reinforcement at each
face based on 12″ thickness.
• Size of rebar ≤ #11
• Max. spacing of rebar ≤ 12″
• Minimum cover in tank walls = 2″
(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
Aesthetics
= 0.010 in.
= 0.008 in.
Crack width calculation is based on the following equation:
z=f s 3 d c 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)
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CVNG 3016 – Design of Environmental Engineering Systems
2010
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,
Z3
bw = 2 3
2d c f s
(2.11)
A= 2dcbw
t
dc
bw
Fig. 1.0 - Calculation of A
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EXAMPLE 1
CVNG 3016 – Design of Environmental Engineering Systems
2010
Rectangular Tank
20'-0″
10'-0‫״‬
30'-0″
2.5'
Cross Section
350
20'-0″
Longitudinal Section
30'-0″
Plan
Fig. E1 – Plan and sections of Tank
Design Data
Weight of Liquid, w =
70 lbs/ft3
Weight of Soil, γs =
100 lbs/ft3
=
0.3
ka
'
=
4000 psi
fc
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
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CVNG 3016 – Design of Environmental Engineering Systems
2010
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
a = 10'
M = + 18.6 kips-in
V = 3500 lbs
M = - 239.9 kips-in
Fig E2 – Vertical Moments at mid-length
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.)
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UWU/Civil & Environmental Engineering
M
φ f c bd
′
2
=
239.9
0.9x4x12x (15.7 )
(From Appendix A)
∴ A s =ωbd
2
CVNG 3016 – Design of Environmental Engineering Systems
2010
=0.0225
ω = 0.023
f c'
4
= 0.023x12x15.7x
= 0.29in 2
fy
60
Check minimum steel (ACI 318-05 cl. 10.5)
3
fc'
3
4000
x12x15.7=0.595in 2 ≥ 200 bwd/fy
fy
60000
200b w d 200x12x15.7
=
=
=0.628in 2 (governs)
fy
60,000
Asmin =
bw d=
(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
2590 lbs
2590
-65.6 in-kip
1890 lbs
-65.6
1890
1890
2590
2590
+35.93 in-kip
+35.93
1890
-65.6
1890
-65.6
1890 lbs
2590
2590
Fig. E3- Horizontal Forces (un-factored) at mid-height of tank)
NB: Shear Forces calculations on Page 9 below
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UWU/Civil & Environmental Engineering
(a)
CVNG 3016 – Design of Environmental Engineering Systems
2010
At Corner - Horizontal Moment Steel
Muy = 1.3 x 1.7 x 65.66
Muy = 145.1 in-kips
Mu
145.1
=
=0.0136
1
2
φ f c bd 0.9x4x12x15.7 2
(From Appendix A)
ω = 0.014
A s =ωbd
(b)
f c'
4
=0.014x12x15.7x
=0.18in 2
fy
60
Steel required for Direct Tension in Long Wall
Factored tension Nu = 1.65 x 3213 = 5301 lbs/ft width
As=
Nu
5301
=
=0.1in 2
0.9f y 0.9x60,000
Direct tension steel is equally distributed on inside and outside faces of wall.
Total steel required on inside face =
As,min =
0.18+
0.1
=0.23in 2
2
200b w d
=0.625in 2 (governs)
fy
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)
CVNG 3016 – Design of Environmental Engineering Systems
2010
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
M=
239.9
=108.6in-kips
(1.7x1.3)
Stress in steel reinforcement
M
A s jd
As = 0.41 in2/ft
fs =
n=
29,000
57
ρ=
(
4000
)
d = 15.7 in
=8
0.41
= 0.00218
(12 x 15.7 )
k = 2 ρ n + ( ρ n ) − ρ n = 0.17
2
j = 1 – k/3 = 0.94
∴ fs =
108.6
=17.95ksi
0.41x0.94x15.7
z3
s m ax =
( 2 x d c2 x f s3 )
dc = cover +
φ
2
z = 115 kips/in
= 2 + 0.313=2.313in
fs = 17.95 ksi
115 3
s max =
=24.6in>9in
2x2.313 2 x17.95 3
OK
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UWU/Civil & Environmental Engineering
CVNG 3016 – Design of Environmental Engineering Systems
2010
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
∴ Vu =1.7x3500=5950 lbs
Since tensile force from adjacent wall is small
Vc =2 fc' bd
=2 4000x12x15.7 = 23,831 lbs
φ Vc = 0.85 x 23,831= 20,256lbs > 5950 lbs
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:
⎛
Nu ⎞
V c = 2 ⎜ 1+
⎟
⎜
500A g ⎟⎠
⎝
f c' b d
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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CVNG 3016 – Design of Environmental Engineering Systems
2010
Nu = -1.7 x 1890 = -3213 lbs
Ag = 18 x 12 = 216 in2
-3213 ⎞ 1
⎛
Vc =2 ⎜1+
⎟ f c bd
⎝ 500x216 ⎠
=1.94 4000x12x15.7 = 23,116 lbs
φVc = 0.85 x 23,116 = 19,649 lbs > 4403 lbs
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)
A st
=0.0033
bh
1
A st = x12x18=0.356in 2 (No. 5 @ 10”)
2
< 0.41 in2
(No. 5 @ 9”)
OK
Summary of Reinforcement
Inside face – vertical
Outside face – vertical
Inside face – horizontal
Outside face – horizontal
No. 5 @ 9 in
No. 5 @ 10 in (Use No. 5 @ 9 in for consistency)
No. 5 @ 12 in
No. 5 @ 12 in
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CVNG 3016 – Design of Environmental Engineering Systems
2010
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.
12″
10'
•
P
30'
Vs = 4150 lbs
3500 lbs
Vls
Mls = 8497 ft-lb
2500 lbs
Mss = 15,413 ft-lb
2500 lbs
3500 lbs
k
Fig E4 – Forces on Base Slab
Unit weight of concrete = 150 lbs/ft3
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UWU/Civil & Environmental Engineering
CVNG 3016 – Design of Environmental Engineering Systems
2010
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 lbs
324,450
p=
= 509 lbs/ft 2
( 30.75 x 20.75)
Using PCA Tables
Mss (short span)
Mls (long span)
Vss
Vls
Ft (long span)
=
=
=
=
=
78 x 509 x 202 /1000
43 x 509 x 202 / 1000
0.24 x 509 x 20
0.42 x 509 x 20
0.4 x 700 x 10
= 15,881 ft-lb = 191 in-k
= 8755 ft-lb = 105in-k
= 2443 lbs = 2.4 k
= 4276 lbs = 4.3 k
= 2800 lbs
Design of Short Span
2.4k
12″
191 in-k
3.5 k
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
Mu
248.3
=
= 0.062
'
2
φfcbd 0.9 x 4 x 12 (9.625)2
(From Appendix A)
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UWU/Civil & Environmental Engineering
CVNG 3016 – Design of Environmental Engineering Systems
2010
ω = 0.064
f c'
4
A s =ωbd =0.064 x 12 x 9.7x
= 0.49in 2
fy
60
(b)
Steel required for Direct Tension in short span
Factored tension Nu = 1.65 x 3500 = 5775 lbs/ft width
As=
Nu
5775
=
=0.11in 2
0.9f y 0.9x60,000
Direct tension steel is equally distributed on inside and outside faces of slab.
Total steel required on top face =
As,min =
0.49+
0.11
=0.55 in 2
2
200b w d 200 (12 x 9.7 )
=
= 0.39 in 2
fy
60000
4/3 of As required by analysis = 4/3 x 0.55 = 0.73 in2
Provide No.6 @ 7 in crs.
direction.
(As = 0.75 in2) steel on top face of slab in short
(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
Mu
136.5
=
= 0.0385
1
2
φfc bd 0.9 x 4 x 12 x 9.06252
(From Appendix A)
ω = 0.039
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UWU/Civil & Environmental Engineering
CVNG 3016 – Design of Environmental Engineering Systems
2010
f c'
4
A s =ωbd =0.039 x 12 x 9.0625x
= 0.283 in 2
fy
60
Steel required for Direct Tension in long span
Factored tension Nu = 1.65 x 2800 = 4620 lbs/ft width
As=
Nu
4620
=
=0.085in 2
0.9f y 0.9x60,000
Direct tension steel is equally distributed on inside and outside faces of slab.
Total steel required on top face =
As,min =
0.283+
0.085
=0.326 in 2
2
200b w d 200 (12 x 9.7 )
=
= 0.39 in 2 (governs)
fy
60000
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:
⎛
Nu ⎞
V c = 2 ⎜ 1+
⎟⎟
⎜
500A
g
⎝
⎠
f c' b d
Nu = 3500 lbs
Ag = 12 x 12 = 144 in2
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UWU/Civil & Environmental Engineering
CVNG 3016 – Design of Environmental Engineering Systems
2010
-3500 ⎞ '
⎛
Vc =2 ⎜1+
⎟ fc bd
⎝ 500 x 144 ⎠
=1.90 4000x12 x 9.7= 14,008 lbs
φVc = 0.85 x 14,008 = 11,907 lbs > 4276 lbs
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)
A st
=0.0033
bh
1
A st = x 0.0033x12x12=0.238 in 2
2
< 0.73 in2
OK
Summary of Reinforcement
Top face – short direction (mid-span)
Top face – short direction (edges)
Top face - long direction (mid-span)
Bottom face - both direction (edges)
Bottom face both direction (mid-span)
No. 6 @ 7 in crs
No. 5 @ 7 in crs (for consistency)
No. 5 @ 7 in (for consistency)
No. 5 @ 7 in
No. 5 @ 12 in
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UWU/Civil & Environmental Engineering
CVNG 3016 – Design of Environmental Engineering Systems
2010
Rectangular Tank
No. 5 @ 10 in c/c
No. 5 @ 12 in. c/c
No. 5 @ 12 in. c/c
No. 5 @ 10 in c/c
No. 5 @ 9 in. c/c
Water
stop
No. 6 @ 7 in. c/c
No. 5 @ 7 in. c/c
Water stop
No. 5 @ 7 in. c/c both ways
Section showing Reinforcement Details for Rectangular Tank
Dr. William Wilson
17 March 2010
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