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Introduction
The design calculation covers the structural design of the swimming pools structures
Length=24 feet
Width=12 feet
height wall = 6 feet Water Depth=6 feet
Geotechnical Design Data
Site Condition and Recommendation
Date
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Since there is no available information about the site condition, the soil bearing capacity shall be
assumed 1,000 psf. The foundation designs will be verified against the data and
recommendations received from the soil investigations
Soil bearing capacity 1,000 psf (pounds per square foot) = lb/ft2
Soil backfills: 1000 lb/ft3
Unit Weight of Soil taken as 120 psf without a soil report
modulus of subgrade reaction 𝐾𝑠 =




𝐼.𝑞𝑎
𝛿
stress/displacement (kip/in2/in)
𝑞𝑎 is the allowable bearing capacity.
𝛿 is the allowable soil settlement
allowable settlement value is assumed (normally 25 mm or 1 inch)
I Safety factor
𝐼. 𝑞𝑎 1 × 1000
𝐾𝑠 =
=
= 12000
1
𝛿
12
𝑞𝑢𝑙𝑡
𝐾𝑠 =
=
∆𝐻
𝑞𝑢𝑙𝑡
𝑞𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 =
→ 𝑞𝑢𝑙𝑡 = 𝑞𝑎 . 𝑆𝐹
𝑆𝐹
𝑞𝑎
1
𝐾𝑠 =
. 𝑆𝐹 ∶ ∆𝐻 = 1 𝑖𝑛( 𝑓𝑡)
∆𝐻
12
𝑞𝑎
𝐾𝑠 =
. 1 = 12000
1
12
𝐾𝑠 = 12. 𝑞𝑎 = 12000 𝐼𝑏⁄𝑓𝑡 2
Materials
All materials shall conform to the applicable standards as stated herein or as specified in the
performance specification
Concrete Masonry Units (CMU)
8-inch-thick concrete masonry units
masonry unit compressive strength f’m ranges between 1,500 and 3,000 psi
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Use f’m = 2,000 psi
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unit weight =135 lb./ft
Em = 900 f 'm for concrete masonry ACI 530 Section 1.8.2.2.1
CMU Size
Nominal Dimensions
Actual Dimensions
DxHxL
DxHxL
8" CMU Full Block
8" x 8" x 16"
7 5/8" x 7 5/8" x 15 5/8"
8" CMU Half-Block
8" x 8" x 8"
7 5/8" x 7 5/8" x 7 5/8"
Mortar Type S
Type S Mortar & masonry block strength of 2000psi, fr Modulus of rapture=163
Grout
Grout, ASTM C-476, Compressive strength f’g = 2500 Psi at 28 days
unit weight 140 psf
Concrete
Concrete
Grade
C40/50
Structural Element Characteristic
Strength
Slab and Wall
Foundations
f′c = 4,000 psi
Normal weight
concrete
density
Modulus of
Elasticity
150 lb/ft3
57,000 √f′c Ib/𝑖𝑛2
Reinforcement Steel
Reinforcements steel to ACI will be specified with the following properties:
Grade
of
Steel
Weight per unit
volume
Yield
Strength
Tensile
Strength Fs
Modulus of
Elasticity
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Grade 40
490
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40 ksi
20 Ksi
29,000,000
psi
Concrete Cover
Element
Foundation Cast against earth
Floor slab
Exposure
Min. Cover
Earth Faces
3 in
Earth Faces Exposed to Weather
Masonry face exposed to earth
Design Loading
Dead Load

Surcharge Load

Service and Floor Finishes =96 psf

Marble tile = 13.34 psf

Water pressure
Water Faces
1½
2 in
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The fluid pressure of water is 62.4 lb/sq.ft. per foot of depth or 62.4 PCF
Assume the density of the fluid in the tank is 63 psf.

Earth Pressure
For the design of earth retaining structural elements, the earth pressure will be determined as
follows:



Ka (active earth pressure) = 1–sin ¢ / 1+sin ¢ = 0.33 (used for check of stability)
Kp (passive earth pressure) = 1+sin ¢ / 1-sin ¢ = 3.0
Ko (at rest pressure) = 1-sin ¢ = 0.50
(used for design of section)
Where ¢ is angle of internal friction, and it is equal to 30º.
Structural Analysis
Load Conditions
Case 1: Leak-Test. For Water Pressure
Analysis for Loading Condition No. 1
The following procedure will be utilized for this loading condition.
• Determine wall forces and bending moments from internal water pressure.
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For a wall with a fixed base and a free top subjected to a triangular load, the wall tension is
calculated by multiplying the coefficients
Check the displacement
Maximum Displacement
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𝜎𝑚𝑎𝑥 = 𝐾𝑠 × 𝛿𝑧,𝑚𝑎𝑥
𝜎𝑧,𝑚𝑎𝑥 = 12000 × 0.043 = 516 𝑝𝑠𝑓
𝜎𝑧,𝑚𝑎𝑥 ≤ 𝜎𝑧,𝑎𝑙𝑙𝑜𝑤. 516 < 1000 𝑜𝑘
The stress under the under the foundation safe
Check Soil Pressure
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𝜎𝑧,𝑚𝑎𝑥 ≤ 𝜎𝑧,𝑎𝑙𝑙𝑜𝑤. 585 < 1000 𝑜𝑘
𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑎𝑐𝑐𝑒𝑝𝑡𝑎𝑏𝑙𝑒 𝑇ℎ𝑒𝑛 𝑡𝑠 = 4′′
12000 × 1
= 1000 > 𝑅𝑧𝑀𝑎𝑥
12
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛
= 534.58 𝐼𝑏 𝑂𝐾
Analysis of long wall
Case 1: Water pressure acting from inside and no earth pressure acting from outside.
During construction, prior to backfilling, the tank will be checked for leaks. This loading condition
represents the situation where the tank is full and the external resistance of the soil is ignored.
An according to ACI-350, resistance provided by the soil is not to be taken into account. This
loading condition also occurs when the tank is leak-tested prior to backfilling.
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Maximum water pressure at base
1
1
𝑃𝑤 = 𝛾𝑤 . ℎ𝑤 . ℎ𝑤 = . 63 × 6 × 6 = 1134 = 𝐼𝑏⁄𝑓𝑡 2
2
2
Wall Bending moment at bottom of wall:
ℎ𝑤
𝑀 = 𝑃𝑤 .
3
6
𝑀 = 1134 × = 2268 𝐼𝑏. 𝑓𝑡
3
Load Combinations
Test Phase – Tank full without backfill



Test 1 = 1.4 DL + 1.4 PwF + 0.9 Pw
Test 2 = 0.9 DL + 0.9 PwF + 1.7 Pw
Test 3 = 1.4 DL + 1.4 PwF + 1.7 Pw
Design of long wall
In Short Direction
Design for vertical bending moments (vertical Steel)
The wall in long direction subject to moment M2-2 and compression axial force F2-2
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Load
+Moment 2-2
Moment 2-2
Axial
Axial Force
Combination
[Ib-ft/ft]
[Ib-ft/ft]
Tension
F2-2 [Ib/ft]
Force
Test 1
1966
-992.3
152.45
-1016.7
Test 2
2013.68
-633.7
727.34
-816.28
Test 3
2500.93
-824.77
547.75
-1047.63
Positive Moment M2-2 = 2500.93 [Ib-ft/ft]
The wall in long direction subject to M2-2 = 2500.93 and F2-2= -1047.63
For a 1-foot-wide strip of wall b=12 inches
Where d = thickness - cover - 𝑑𝑏/2 = 8-2-3/8.1/2= 5.82 in
The required area of steel
𝑀2−2
2500.93 (12)
𝐴𝑠𝑟𝑒𝑞 =
=
= 0.15 𝑖𝑛2 /𝑓𝑡
𝑦
∅𝑓𝑦 (𝑑 − 2) 0.9(40000)0.95(5.82)
Try # 4 bars spaced at 16 incudes, No.4 Diameter 1/2" Area 0.2 (Square inches)
12
𝐴𝑠 = 0.2 (𝑆=16) = 0.15 𝑖𝑛2 ⁄𝑓𝑡
𝑓𝑦 𝐴𝑠
)
1.6𝑓𝑚′ 𝑏
40000(0.15) 1
𝑀𝑛 = 40000(0.15)(5.82 −
)
1.6(2000)12 12
𝑀𝑛 = 𝑓𝑦 𝐴𝑠 (𝑑 −
𝑀𝑛 = 2831 𝐼𝑏 − 𝑓𝑡/𝑓𝑡
∅𝑀𝑛 = 0.9(2831) = 2548.68 > M1 − 1 = 2500.93 Ok
Provide No.4@16 in on inside face (As=0.15 𝒊𝒏𝟐 .)
In Long Direction
Design for horizonal bending moments (horizonal Steel)
The wall in long direction subject to moment M1-1 and Tension axial force F1-1
Load
Combination
+Moment 1-1
- Moment 1-1
Axial
Axial Force
[Ib-ft/ft]
[Ib-ft/ft]
Tension
F1-1 [Ib/ft]
Force
Test 1
1421.12
-397.52
1224.33
-694.312
Date
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Test 2
3182.82
-1117.14
1112.19
-2747
Test 3
2717.7
-850.40
695.47
-2292
Positive Moment M1-1 = 3182 [Ib-ft/ft]
The wall in long direction subject to M1-1 = 3182 and Tension axial force F1-1= 1112.19
For a 1-foot-wide strip of wall b=12 inches
Where d = thickness - cover - 𝑑𝑏/2 = 8-2-3/8.1/2= 5.82 in
The required area of steel
𝐴𝑠𝑟𝑒𝑞 =
𝑀2−2
3182 (12)
=
= 0.191 𝑖𝑛2 /𝑓𝑡
𝑦
0.9(40000)0.95(5.82)
∅𝑓𝑦 (𝑑 − )
2
Try # 4 bars spaced at 12 incudes, No.4 Diameter 1/2" Area 0.2 (Square inches)
12
𝐴𝑠 = 0.2 (𝑆=12) = 0.2 𝑖𝑛2 ⁄𝑓𝑡
𝑓𝑦 𝐴𝑠
)
1.6𝑓𝑚′ 𝑏
40000(0.2) 1
𝑀𝑛 = 40000(0.2)(5.82 −
)
1.6(2000)12 12
𝑀𝑛 = 𝑓𝑦 𝐴𝑠 (𝑑 −
𝑀𝑛 = 3741 𝐼𝑏 − 𝑓𝑡/𝑓𝑡
∅𝑀𝑛 = 0.9(3741) = 3367 > M1 − 1 = 3182
Ok
Provide No.4@12 in on outside face (As=0.2 𝒊𝒏𝟐 .)
Design of Short wall
In long Direction
The wall in long direction subject to moment M1-1 and Tension axial force F1-1
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+Moment 1-1
- Moment 1-1
[Ib-ft/ft]
[Ib-ft/ft]
Combination
Date
Axial
Axial Force
Tension
F1-1 [Ib/ft]
Force
Test 1
1421.12
-397.52
1224.33
-694.312
Test 2
3182.82
-1117.14
1112.19
-2747
Test 3
2717.7
-850.40
695.47
-2292
In short Direction
The wall in long direction subject to moment M2-2 and compression axial force F2-2
Load
+Moment 2-2
- Moment 2-2
Axial
Axial Force
Combination
[Ib-ft/ft]
[Ib-ft/ft]
Tension
F2-2 [Ib/ft]
Force
Test 1
1966
-992.3
152.45
-1016.7
Test 2
2013.68
-633.7
727.34
-816.28
Test 3
2500.93
-824.77
547.75
-1047.63
Calculate the Cracking Moment
𝑀𝑐𝑟
12 8/12
= 𝑆𝑛 𝑓𝑟 =
6
1.3 𝑀𝑐𝑟 = 118.35 𝐼𝑏. 𝑓𝑡 <
Design of walls for Loading Condition case 2.
2
. 163 = 114.8 𝐼𝑏. 𝑓𝑡
M𝑛 𝐿𝑜𝑛𝑔−𝐷𝑖𝑟 3182.82
M𝑛 𝑆ℎ𝑜𝑟𝑡−𝐷𝑖𝑟 2500.93
𝑜𝑘
Date
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This loading condition represents the situation where the tank is empty and the external
pressure of the soil is present. During construction, backfilling and compaction may exert forces
on the structure in considerable excess of the service loading.
Analysis for Loading Condition No. 2
Soil pressure on the wall
𝑃𝑞 = 96 × 0.5 = 48 𝑝𝑠𝑓
𝑃𝑠 = 120 × 6 × 0.33 = 237.6 𝑝𝑠𝑓
Load Combinations
Maintenance Phase – Tank empty with backfill
Maintenanace1 = 0.9 𝐷 + 0.9𝑃𝑞 + 1.7𝑃𝑠
Maintenanace2 = 1.4 𝐷 + 1.4𝑃𝑞 + 1.7𝑃𝑠
Design for vertical and horizontal bending moments
Design of long wall
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Design for vertical bending moments (vertical Steel)
The wall in long direction subject to moment M2-2 and compression axial force F2-2
Load
+Moment 2-2
- Moment 2-2
Axial
Axial Force
Combination
[Ib-ft/ft]
[Ib-ft/ft]
Tension
F2-2 [Ib/ft]
Force
Maintenance1
1813.78
-467.722
534.95
-695
Maintenance2
2424.52
-851.88
465.27
-1043.45
The moments in loading condition case 2 are smaller than the moments loading condition in
case 1, therefore, we consider the same steel
Design for Uplift under loading condition case 3.
Depending on the height of the water, forces may develop underneath the swimming pool that
may be large enough to lift the structure when it is empty. The weight of the slab and walls, as
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well as the weight of the soil resting on the footing projection, must be capable of resisting the
upward force of the water.





Soil weight of 120 pcf
concrete weight of 150 pcf
UCM weight of 135 psf
The thicknesses of the slab 4’’
The thicknesses of the wall 8”
Determine weight of tank:

Walls = height x length x thickness x 135 Ibs/ft3
8
𝑊𝑎𝑙𝑙𝑠 = 6 (24.6 + 12.6 + 24.6 + 12.6) × 12 × 135 = 40176 𝑙𝑏

Bottom slab = length x width x thickness x 150 Ibs/ft3
4
= (24.6) × (12.6) × 12 × 150 = 15498 𝑙𝑏
Weight of tank = 40176 + 15498 = 55674 lb
Determine weight of soil
Soil on footing overhang = soil area x height of soil x 120 pcf
= (30.5 × 18.5) × (24.6 × 12.6) × 6 × 120 = 183,088.8 𝑙𝑏
Total resisting load =55674 + 183,088.8 = 238,762.8 lb
Buoyancy force:
Area of bottom slab = length x width
(24.6 × 12.6) = 37.2𝑓𝑡 2
Water pressure = water head x 120
= (soil height + slab thickness) x 120
4
= ( 6 + 12) × 120 = 760 𝑙𝑏⁄𝑓𝑡 2
Buoyant force = area x pressure
37.2𝑓𝑡 2 × 760 = 28272 𝑙𝑏
𝑆𝑎𝑓𝑒𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟 =
𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝐿𝑜𝑎𝑑 238,762.8
=
= 8.44 > 1.5 𝑂𝐾
𝐵𝑢𝑜𝑦𝑎𝑛𝑡 𝐹𝑜𝑟𝑐𝑒
28272
Design of concrete slab
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One of the criteria for slab design is that it must be able to resist the moment from the bottom of
the wall.
1- Determine the Structural System
𝐿
24
For one way slab 𝐿 𝑙𝑜𝑛𝑔 = 12 = 2 ≥ 2
𝑆ℎ𝑜𝑟𝑡
The slab will be designed as a one-way flexure member spanning in the short direction.
Consider a 1′ wide strip of slab with an ultimate load qu from the soil pressure below the slab.
The soil pressure pushing upward is reduced by the weight of the slab
The soil pressure is multiplied by a factor of 1.7, and the dead weight of the concrete is
multiplied by a factor of 0.9.
Load on the Slab

Self-Weight of Slab = (Unit weight of reinforced concrete X thickness of the slab)
𝑊𝑆 = (150 × 4′′ /12) = 50 𝑝𝑠𝑓

Water Pressure on Base = (Unit Weight of Water X Height of water)
𝑃𝑊𝐹 = 𝛾𝑤 . ℎ𝑤 = 63 × 6 = 378 𝑝𝑠𝑓 = 1 × 63 × 6 = 𝐼𝑏⁄𝑓𝑡 2 (regtancal distribution)

Live Load
𝐿𝐿 = 13.34 𝑝𝑠𝑓

Soil Pressure on Base 𝑃𝑆 = 120 × 12 = 1440 𝑝𝑠𝑓
Load Combination
1.7 Psb + 1.7LL + 1.7PWF − 0.9 Ws
𝑄𝑢 = 1.7 × 120 × 12 × 1 + 1.7(13.34) + 1.7(378) − 0.9 × 150 (
𝑄𝑢 = 3113.27 − 45 = 3068.27 psf
4′′
)×1=
12
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Positive Moment M2-2= 287.139 lb. Ft/ft
= 𝐿/20 𝑥 [0.4 + 𝑓𝑦/100,000] = 12𝑥12/20[0.4 + 40,000/100,000] = 5.76 𝑖𝑛
For a 1-foot-wide strip of wall b=12 inches
Where d = thickness - cover - 𝑑𝑏/2 = 4-1.1/2-3/8.1/2= 2.31 in
The required area of steel
𝐴𝑠𝑏𝑜𝑡𝑡𝑜𝑚 =
𝐴𝑠𝑏𝑜𝑡𝑡𝑜𝑚 =
𝑀2−2 +
287.139 (12)
2
𝑦 = 0.9(40)0.95(2.31 ) = 0.04 𝑖𝑛 /𝑓𝑡
∅𝑓𝑦 (𝑑 − 2)
𝑀2−2 +
287.139 (12)
=
= 0.08 𝑖𝑛2
𝑦
2.29
∅𝑓𝑦 (𝑑 − 2) 0.9 × 40000(3.71 −
2 )
Minimum Reinforcement Ratio for Shrinkage and Temperature
𝐴𝑠𝑚𝑖𝑛 = 0.0020. ts. b = 0.0020 × 4 × 12 = 0.096 𝑖𝑛2
𝐴𝑠𝑏𝑜𝑡𝑡𝑜𝑚 < 𝐴𝑠𝑚𝑖𝑛
Date
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Try # 4 bars spaced at 12 incudes, No.4 Diameter 1/2" Area 0.2 (Square inches)
𝐴𝑠 = 0.2 (
12
) = 0.2 𝑖𝑛2 ⁄𝑓𝑡
12
Note: ACI 350 limits the spacing of shrinkage and temperature reinforcement to 12 in.
Provide No.4@12 in on Top and Bottom (As=0.2 𝒊𝒏𝟐 .)
Note: The Slab thickness increased to (6”) to meet ACI Code, because the steel Grade 40 used
in the design
Moment at the bottom of wall
Note the moment at the bottom of wall is less than the slab of moment
𝑴𝒃𝒐𝒕𝒕𝒐𝒎 𝒘𝒂𝒍𝒍 = 𝟏𝟏𝟓. 𝟔𝟓 < 𝑴𝒔𝒍𝒂𝒃 = 𝟐𝟖𝟕. 𝟏𝟑𝟗 𝑶𝑲
The Slab able to resist the moment from the bottom of the wall.
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Design of wall concrete foundation
ACI 318-14
Discussion
Calculation
Step 1: Foundation type
13.1.1
This strip footing is 7.5 ft below grade level
Therefore, it is considered a shallow foundation
13.3.2.1
The footing will be designed and detailed with the applicable
provisions of Chapter 7,One way slabs,and and Chapter 9,
Beams, of ACI 318-14
Step 2: Material requirements
The comperssive strength of concrete is specified at 28 days
to be 4000 psi.
Density of concrete = 150 lb/ft3
Step 3: Determine footing dimensions
13.3.1.1
Footing width is assumed and then verified through
calculations. Minimum base area of foundation shall be
calculated from unfactored forces
Try B= 8 ft footing width
𝑨𝒓𝒆𝒂 ∶ 𝑨 = 𝟏(𝟖) = 𝟖 𝒇𝒕𝟐 𝒇
Section modulus
The footing thickness must be such that the bottom
13.3.1.2
reinforcement has an effective depth of at least 6 in
Step 4: Footing design
Wall stability
Because there is an out-of-plane (overturning)
lateral force on the stem wall, the overall wall
stability must be checked.
To calculate the stability of a footing, the total
vertical load is calculated and the resisting moment .
(MR) is compared to the resulting overturning
moment (𝑀OTM )
𝑆=
1(8)(8)
= 10.6 𝑓𝑡 3 𝑓
6
Try 12 in. footing thinness
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Step 4: Footing design
To consider a footing stable 𝑀𝑅 ≥ 1.5𝑀OTM
Weights on bearing soil below footing:
(𝑊)𝐹𝑜𝑜𝑡𝑖𝑛𝑔 = 𝑡𝑓 × 𝛾𝑐
Weight of footing:
(𝑊)𝑆𝑜𝑖𝑙
Weight of soil above footing:
Weight of concrete wall pier:
12
𝑘𝑖𝑝
× (0.15
)
12
ft3
= 0.15 𝑘𝑠𝑓
90 − 12
=
0.12 = 0.78 𝑘𝑠𝑓
12 𝑖𝑛/𝑓𝑡
=
(𝑊)𝑐𝑜𝑛𝑐.𝑝𝑖𝑒𝑟 =
90 − 12
0.15
12 𝑖𝑛/𝑓𝑡
= 0.975 𝑘𝑠𝑓
Total vertical dead load:
∑ 𝑷 = (0.15)(8) + (0.78)(8 − 1)
= 7.635
= 7.635 + 0.54 = 8.175 𝑘𝑖𝑝/𝑓𝑡
Vertical distance from the bottom of footing to location of
applied lateral soil & water share.
𝐻 = 7.5 𝑓𝑡
The overturning moment, MOTM, is measured at
base of footing.Total share force at the base
𝑃𝑞 = 96 × 0.5 = 0.048 kip
𝑃𝑠 = 120 × 6 × 0.33 × 3
= 0.7128 kip
𝑃𝑊 = 63 × 6 × 3 = 1.134 𝑘𝑖𝑝
The resisting moment, MR, is calculated as the
𝑉 = 0.048 + 0.7128 − 1.134
= 0.3732 kip
product of vertical load by distance from the
centerline to edge of footing MR=P(B/2).
𝑀OTM = 0.3732 × 7.5
= 2.8 𝑘𝑖𝑝. 𝑓𝑡/𝑓𝑡
To ensure footing stability, the following inequality
𝑀𝑅 ≥ 1.5𝑀OTM
𝑀R =
8.175 × 8
= 32.7 𝑘𝑖𝑝. 𝑓𝑡/𝑓𝑡
2
32.7 ≥ 1.5(2.8) = 4.2 𝑶𝑲
Step 5: Calculate soil pressure
13.3.1.1
Service loads
The maximum soil pressure is calculated from
service forces and moments transmitted by foundation to the
soil.To calculate soil pressure, the location of the vertical
service resultant force is determined.
The distance to the resultant from the front face of stem
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∆M
𝒙 = ∑𝑝
B
𝒆 = 2−𝑥
32.7 − 4.2
= 3.48 𝑓𝑡
8.175
8
𝒆 = − 3.48 = 0.52 𝑓𝑡
2
𝐵 8
= = 1.3 > e = 0.52 𝐎𝐊
6 6
𝒙=
Check if resultant falls within the middle third
of the footing.
Because e<B/6, the footing imposes ompression
to the soil across the entire width.
The resulting soil pressure must be less than the allowable
bearing pressure provided by the geotechnical report.
Maximum and minimum soil pressures are
calculated by:
𝑞1,2 =
∑ 𝑃 𝑃. 𝑒
∓
𝐴
𝑆
8.175 8.175 × 0.52
∓
(8)(1)
10.6
8.175 8.175 × 0.52
𝑞𝑀𝑎𝑥 =
+
(8)(1)
10.6
8.175 8.175 × 0.52
𝑞𝑀𝑖𝑛 =
−
(8)(1)
10.6
𝑞1,2 =
𝑞𝑀𝑎𝑥 = 1.42 𝑘𝑠𝑓 < 1.2 𝐾𝑠𝑓
𝑞𝑀𝑖𝑛 = 0.62 𝑘𝑠𝑓 > 0 𝑂𝐾
𝑂𝐾
Step 6: Factored loads
13.3.2.1
The footing is designed as one-way slab. Calculate the soil
pressures resulting from the applied factored loads
Load Case 1
U1=1.4 DL
5.3.1a
𝑈 = 1.4 × 8.175 = 11.445 kip/ft
Use DL=P=8.175
M OTM=2.8
𝑈 = 1.2 × 8.175 + 1.2 × 1.134
= 11.17 kip/ft
Load Case 2
𝑈2 = 1.2𝐷𝐿 + 1.2 𝑃𝑤
𝑈 = 1.2 × 8.175 + 1.6 × 0.55 +
Load Case 3
U3=1.2 DL+1.6LL+1.2Pw+0.9 Ps+0.9 pq
+1.2 × 8.175 + 0.9(0.048 + 0.7128)
= 21.18 kip/ft
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Step 7: Shear strength
21.2.1(b)
∅𝒔𝒉𝒆𝒂𝒓 = 𝟎. 𝟕𝟓
One-way shear design
Shear strength reduction factor:
∅𝑉𝑐 ≥ 𝑉𝑢
Assume Vs = 0 (no shear
reinforcement)
Vn = Vc
𝑉𝑛 = 𝑉𝑐 + 𝑉𝑢
∅𝑉𝑐 = ∅(2√𝑓𝑐 ′ . 𝑏𝑤 . 𝑑)
Effective depth
𝒅 = 𝒉𝒊𝒆𝒈𝒉𝒕 − 𝒄𝒐𝒗𝒆𝒓 − 𝒅𝒃 ⁄𝟐
Assume Cover =3 in 𝒅𝒃 = 𝟏
𝑑 = 12 − 3 − 1⁄2 = 8.5 𝑖𝑛
∅𝑉𝑐 = ∅(2√𝑓𝑐 ′ . 𝑏𝑤 . 𝑑)
∅𝑉𝑐 = 0.75 2√4000 (12)(8.5)
= 9.67 𝐾𝑖𝑝/𝑓𝑡
* Calculate factored soil pressure at distance d from face of
wall
𝑞𝑢𝑚𝑎𝑥 − 𝑞𝑢𝑚𝑖𝑛 𝐵 𝑡𝑤𝑎𝑙𝑙
𝑞𝑢𝑑 = 𝑞𝑢𝑚𝑖𝑛 + (
)( −
+ 𝑑)
𝐵
2
2
For load case U3
𝑞𝑢𝑚𝑎𝑥 = −520
0.52 − 0.04 8.5 1
𝑞𝑢𝑑 = 0.04 + (
)(
−
8.5
2
2
8.5
+
)
12
= −0.29 kip/ft
𝑞𝑢𝑚𝑖𝑛 = −40
Calculate factored shear force at d from face of
wall: Vu
Vu =
7.4.3.2
Check if : ∅𝑉𝑐
0.52 + 0.29
(3.75 − 0.3 − 0.7)
2
≥ 𝑉𝑢
= 1.11 kip/ft
∅𝑉 = 9.67 𝑐 ≥ 𝑉𝑢
22.5.1.2
= 1.11
kip
𝑶𝑲
ft
Step 8: Flexural strength
Flexure design
5.3.1a
Calculate Mu at face of wall from
U3
𝑞𝑢𝑚𝑎𝑥 − 𝑞𝑢𝑚𝑖𝑛 𝐵 𝑡𝑤𝑎𝑙𝑙
𝑞𝑢𝑤𝑎𝑙𝑙 = 𝑞𝑢𝑚𝑖𝑛 + (
)( +
)
𝐵
2
2
1
𝐵 𝑡𝑤𝑎𝑙𝑙 2
𝑞𝑢𝑤𝑎𝑙𝑙 ( −
) +
2
2
2
𝑞𝑢𝑤𝑎𝑙𝑙
1
𝐵 𝑡𝑤𝑎𝑙𝑙 2
+ (𝑞𝑢𝑚𝑎𝑥 − 𝑞𝑢𝑤𝑎𝑙𝑙 ) ( −
)
3
2
2
𝑀𝑢 =
𝑀𝑢 =
0.52 − 0.04 8.5
)(
8.5
2
8
+ )
2
= 0.29 𝑘𝑠𝑓
𝑞𝑢𝑤𝑎𝑙𝑙 = 0.04 + (
1
0.29(3.75 − 0.3)2 +
2
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Calculate the required area of flexural reinforcement
1
+ (0.52 − 0.29)(3.75 − 0.3)2
3
= 2.54 𝑘𝑖𝑝/𝑓𝑡
0.85𝑓𝑐′ 𝑏 𝑦 = 𝐴𝑠 𝑓𝑦
𝑀𝑢 ≤ ∅𝑀𝑛 = 0.9𝐴𝑠 𝑓𝑦 (𝑑 − 𝑦⁄2)
0.85(4000)12 𝑦 = 𝐴𝑠 (40,000)
𝑦 = 0.98 𝐴𝑠
Check As against the minimum
𝐴𝑠𝑚𝑖𝑛 =
200
𝑏𝑑
𝑓𝑦
∅𝑀𝑛 = 0.9𝐴𝑠 (40,000)(8.5
− 0.98 𝐴𝑠 ⁄2)
≥ 2.82
9.6.1.2
𝐴𝑠𝑚𝑖𝑛 =
200
(12)(8.5)
40,000
= 0.51 𝑖𝑛2 ⁄𝑓𝑡
Use No7 at 12in on center
12
𝐴𝑠 = 0.6 (
) = 0.6 𝑖𝑛2 ⁄𝑓𝑡
𝑆 = 12
∅𝑀𝑛 = 0.9(0.6)(40,000)(8.5
− 0.294)
= 11.012
𝑘𝑖𝑝
𝑓𝑡
> 2.82 𝑂𝐾
Step 9: Footing details
7.6.4.1
Shrinkage and temperature reinforcement:
The area of shrinkage and temperature reinforcement:
𝐴𝑆+𝑇 ≥ 0.0018( 8.5 × 12 )(12) = 0.
13.2.8.3
𝐴𝑆+𝑇 ≥ 0.0018𝐴𝑔
𝐴𝑆+𝑇 ≥= 2.2
Use 8 bars No.5 bottom
(area=2.48 𝑖𝑛2 )
Development length
Placed perpendicular to the flexural
Reinforcement development is calculated at the
reinforcement
maximum factored moment and the code permits
the critical section to be located at the wall face.
Bars must extend at least a tension development
length beyond the critical section.
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𝑙𝑑 =
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3𝑓𝑦 𝜓𝑡 𝜓𝑒 𝜓𝑠
𝑑
𝑐 + 𝑘𝑡𝑟 𝑏
40𝜆√𝑓𝑐′ .
𝑑𝑏
𝑙𝑑 =
3(40.000)
40√4000. 2.5
𝑑𝑏
𝑙𝑑 = 18.9𝑑𝑏
Check if No. 7 can be developed using straight
bars, without hooks
7
𝑙𝑑 = 18.9 ( ) = 16.6 𝑖𝑛
8
𝑙𝑑 = 20 𝑖𝑛
𝑙𝑑 provided perpendicular to
the wall
𝑙𝑑 𝑝𝑟𝑜𝑣 =
8.5 × 12 − 0.3
2
−3
𝑙𝑑 𝑝𝑟𝑜𝑣 = 47.85 𝑖𝑛 > 25 𝑂𝐾
Use straight No.7 bars
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Moment at distance d of the wall face
The program shows the moment at distance d of the wall face =2.6 Kip. ft/ft as calculated
manually.
Also note that the Min Reinforcement steel # 7@12in. o.c. cover the moment 10.4 Kip/ft
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Soil pressure under wall footing
Note: The width of the wall footing increased to (8.5 ft) resist tensile stress under the wall footing
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