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The document discusses the design of raft foundations. Raft foundations are used in weak soil conditions, with high structure loads relative to soil bearing capacity, or where isolated footings would interfere. The types of raft foundations include flat plates, plates thickened under columns, waffle slabs, and plates with pedestals. The design process involves dividing the raft into vertical and horizontal strips to calculate soil pressures, bending moments, and shear forces. Column loads are summed and average soil pressure is calculated. A coefficient is applied to column loads to estimate strip forces for reinforcement design. An example shows checking soil pressure limits and calculating member properties.

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22 June 2023 Shallow Foundations, P4, Tarek Nageeb 1 SHALLOW FOUNDATIONS PART 4 ‫الضحلة‬ ‫األساسات‬ ‫الرابع‬ ‫الجزء‬
22 June 2023 Shallow Foundations, P4, Tarek Nageeb 2 DESIGN OF RAFT FOUNDATIONS ‫اللبشة‬ ‫أساسات‬ ‫تصميم‬
22 June 2023 3 Shallow Foundations, P4, Tarek Nageeb FOUNDATION ENGINEERING 1- Site Investigation 2- Design of Shallow Foundations: a- Design of Isolated Footings. b- Design of Isolated Footings under Eccentric Loads. c- Design of Combined Footings. d- Design of Strap Beam. e- Design of Raft Foundations
Raft Foundations are used in: 1- Weak soil conditions (low bearing capacity). 2- High structure loads, relative to soil B/C. 3- Interference between the isolated reinforced concrete footings. 4- Total area of isolated R.C. footings > 2/3 construction site area. 5- Single and multiple basements. 6- The existence of weak soil pockets within the stiff soil area. 7- Isolated footing settlements exceeds the allowable values. Design of Raft Foundations 22 June 2023 4 Shallow Foundations, P4, Tarek Nageeb
‫عندما‬ ‫المسلحة‬ ‫الخرسانة‬ ‫من‬ ‫اللبشة‬ ‫تستخدم‬ : 1 - ‫ضعيفة‬ ‫التربة‬ ‫تحمل‬ ‫قدرة‬ ‫تكون‬ . 2 - ‫التربة‬ ‫تحمل‬ ‫لقدرة‬ ‫بالنسبة‬ ‫كبيرة‬ ‫المنشأ‬ ‫أحمال‬ ‫تكون‬ . 3 - ‫المنفصلة‬ ‫المسلحة‬ ‫القواعد‬ ‫بين‬ ‫تداخل‬ ‫حدوث‬ ‫عند‬ . 4 - ‫ثلثى‬ ‫عن‬ ‫المنفصلة‬ ‫المسلحة‬ ‫القواعد‬ ‫مساحة‬ ‫زيادة‬ ‫عند‬ ‫الموقع‬ ‫مساحة‬ 5 - ‫متعددة‬ ‫بدرومات‬ ‫أو‬ ‫واحد‬ ‫بدروم‬ ‫وجود‬ . 6 - ‫مساحة‬ ‫كامل‬ ‫على‬ ‫وليس‬ ‫التربة‬ ‫من‬ ‫ضعيفة‬ ‫جيوب‬ ‫وجود‬ ‫الموقع‬ 7 - ‫المنف‬ ‫للقواعد‬ ‫بها‬ ‫المسموح‬ ‫القيم‬ ‫عن‬ ‫الكلى‬ ‫الهبوط‬ ‫زيادة‬ ‫صلة‬ Design of Raft Foundations 22 June 2023 5 Shallow Foundations, P4, Tarek Nageeb
Isolated or Strip Footing Raft Foundation Vs. Isolated Footing Raft Foundation Soft Soil 2B B 2B B 22 June 2023 6 Shallow Foundations, P4, Tarek Nageeb To avoid high raft settlement ‫فى‬ ‫كبير‬ ‫هبوط‬ ‫حدوث‬ ‫لتجنب‬ ‫اللبشة‬ ‫استخدام‬ ‫حالة‬
Net and gross allowable soil pressure: qall gross = qall net + gZ qall gross : Gross allowable bearing capacity ‫الكلى‬ ‫التربة‬ ‫جهد‬ qall net : Net allowable bearing capacity ‫الصافى‬ ‫التربة‬ ‫جهد‬ gZ : Weight of soil above F.L. ‫التأسيس‬ ‫منسوب‬ ‫عند‬ ‫التربة‬ ‫عمود‬ ‫وزن‬ Design of Raft Foundations 22 June 2023 7 Shallow Foundations, P4, Tarek Nageeb
Simple settlement calculations: d = mv * H * Ds d : Settlement value ‫الهبوط‬ ‫قيمة‬ mv : Soil compressibility, m2/MN = 1/E Ds : Added soil pressure, kN/m2 Settlements will take place when the applied pressure is higher than g z ‫الحادثة‬ ‫اإلجهادات‬ ‫قيمة‬ ‫تتعدى‬ ‫عندما‬ ‫يحدث‬ ‫سوف‬ ‫الهبوط‬ ‫أن‬ ‫أى‬ ‫التأسيس‬ ‫منسوب‬ ‫عند‬ ‫التربة‬ ‫عمود‬ ‫وزن‬ ‫التربة‬ ‫على‬ Design of Raft Foundations 22 June 2023 8 Shallow Foundations, P4, Tarek Nageeb
Stress Compensation  Total building loads = 40,000.0 kN  Building Area = 20 x 20 m  Determine the depth at which there will be stress compensation between the structure and soil weight  Solution:  Stresses at the foundation level: 22 June 2023 9 Shallow Foundations, P4, Tarek Nageeb 2 kN/m 100.0 20.0 * 0 . 20 40,000.0 Raft of Area Load Building Total    s m 5.56 18.0 100.0 d then, d,     g s g s Equivalent depth, d
Types of Raft or Mat Foundation  a) flat plate  b) Plate thickened under columns 22 June 2023 10 Shallow Foundations, P4, Tarek Nageeb
Types of Raft or Mat Foundation  c) Waffle Slab  d) Plate with Pedestals 22 June 2023 11 Shallow Foundations, P4, Tarek Nageeb
Types of Raft or Mat Foundation Rigid Basement 22 June 2023 12 Shallow Foundations, P4, Tarek Nageeb
Raft [Mat] Foundation Inverted Beam Raft 22 June 2023 13 Shallow Foundations, P4, Tarek Nageeb
Raft [Mat] Foundation Inverted Beam Raft Column Inverted Beam Concrete Plate Section Elevation in the Raft 22 June 2023 14 Shallow Foundations, P4, Tarek Nageeb
22 June 2023 Shallow Foundations, P4, Tarek Nageeb 15 Design of Raft Foundation The resultant is determined from the column loads, and so the eccentricities ex and ey ex R L y x ey B
Design of Raft Foundation 1- Stress Distribution Under the Raft 22 June 2023 16 Shallow Foundations, P4, Tarek Nageeb I x * M I y * M A R f y y x x 2 1    In which; Mx = R * ey and My = R * ex Such that: f1 ≤ qall soil f2 ≥ 0.00 (no tension) 12 B * L I 3 x  12 L * B I 3 y 
f1 f2 f2 f1 f2 f1 22 June 2023 Shallow Foundations, P4, Tarek Nageeb 17 Soil Stresses under the Raft Foundation f1 f2 x y ex R L ey B
22 June 2023 Shallow Foundations, P4, Tarek Nageeb 18 Soil Stresses under the Raft Foundation L B B 1 B 2 B 3 B 4 L1 L2 L3 L4 L5 ‫الش‬ ‫فى‬ ‫موضح‬ ‫هو‬ ‫كما‬ ‫وأفقية‬ ‫رأسية‬ ‫شرائح‬ ‫إلى‬ ‫اللبشة‬ ‫تقسيم‬ ‫يتم‬ ‫المرفق‬ ‫كل‬ . The raft is divided into a number of horizontal and vertical strips, as shown in the figure.
22 June 2023 Shallow Foundations, P4, Tarek Nageeb 19 ‫اإلنحن‬ ‫عزوم‬ ‫حساب‬ ‫ثم‬ ‫شريحه‬ ‫كل‬ ‫أسفل‬ ‫التربة‬ ‫ضغط‬ ‫قيمة‬ ‫تحديد‬ ‫يتم‬ ‫وقوى‬ ‫اء‬ ‫الشريحة‬ ‫تلك‬ ‫على‬ ‫القص‬ The soil pressure under each strip is determined and the bending moments and shear forces are determined L B B 1 B 2 B 3 B 4 L1 L2 L3 L4 L5 P1 P2 P3 P4 P5
22 June 2023 Shallow Foundations, P4, Tarek Nageeb 20 4 5 4 3 2 1 strip n B * L P P P P P A P f        f2 f1 2 f f f 2 1 avg n      P Fstrip  L B 4 P1 P2 P3 P4 P5 4 avg - n strip B * L * f F 
22 June 2023 Shallow Foundations, P4, Tarek Nageeb 21 1 - ‫األعمدة‬ ‫أحمال‬ ‫مجموع‬ ‫حساب‬ ‫يتم‬ 2 - ‫ألعلى‬ ‫أسفل‬ ‫من‬ ‫التربة‬ ‫لضغط‬ ‫متوسط‬ ‫حساب‬ ‫يتم‬ 3 - ‫ومتوسط‬ ‫األعمدة‬ ‫أحمال‬ ‫لمجموع‬ ‫متوسطة‬ ‫قيمة‬ ‫حساب‬ ‫يتم‬ ‫التربة‬ ‫ضغط‬ 4 - ‫معامل‬ ‫حساب‬ ‫يتم‬ () ‫والمتوسط‬ ‫األعمدة‬ ‫حمل‬ ‫بين‬ ‫للفرق‬ ‫الجديد‬ 5 - ‫وقوى‬ ‫العزوم‬ ‫حساب‬ ‫عند‬ ‫األعمدة‬ ‫أحمال‬ ‫فى‬ ‫المعامل‬ ‫يضرب‬ ‫شريحة‬ ‫لكل‬ ‫القص‬ . L B 4 P1 fn-avg P2 P3 P4 P5
Raft Foundation Strip Reinforcement 22 June 2023 22 Shallow Foundations, P4, Tarek Nageeb
Estimation of Soil Reaction under the Raft 22 June 2023 23 Shallow Foundations, P4, Tarek Nageeb An approximate estimate of the coefficient of subgrade reaction is obtained as follows: ks (kN/m3) = (100  120)* soil bearing capacity (kN/m2)
Estimation of Soil Reaction under the Raft 22 June 2023 24 Shallow Foundations, P4, Tarek Nageeb
Estimation of Soil Reaction under the Raft 22 June 2023 25 Shallow Foundations, P4, Tarek Nageeb An approximate estimate of the coefficient of subgrade reaction is obtained as follows: ks (kN/m3) = (80  120)* soil bearing capacity (kN/m2)
A plan of a mat foundation. All columns are 40×40 cm2. qall,net is 125 kPa. The material properties for concrete and steel are: fcu = 25 N/mm2, fy = 360 N/mm2 22 June 2023 26 Shallow Foundations, P4, Tarek Nageeb Example (6): Design of Raft Foundation From Dr. Mashour Ghonaim Book
Solution: Step 1: Check of Soil Pressure: Example (6): Raft Foundation 22 June 2023 27 Shallow Foundations, P4, Tarek Nageeb I x * M I y * M A P f y y x x 2 1    A = Area of the raft = 15.40 * 12.40 = 190.96 m2 4 3 3 x m 3774.0 12 15.40 * 12.40 12 L B I    4 3 3 y m 2446.80 12 12.40 * 15.40 12 B L I   
Total Vertical Unfactored Loads = 440 + (1360*2) + 370 + (1150*2) + (2880*2) + 500 + 1360 + 1440 + 440 = 15330 kN ‫نقطة‬ ‫حول‬ ‫العزوم‬ ‫بأخذ‬ ‫لللبشة‬ ‫األحمال‬ ‫مركز‬ ‫تحديد‬ D 22 June 2023 28 Shallow Foundations, P4, Tarek Nageeb      2 * 2280 2 * 1150 * 2 . 6 370 2 * 1360 440 * 2 . 0 15330 1 X        m 6.282 440 1440 1360 500 * 2 . 12      m 0.082 2 12.40 - 6.282 2 B - X ex    Example (6): Raft Foundation
22 June 2023 29 Shallow Foundations, P4, Tarek Nageeb      1440 2880 1360 * 2 . 5 440 1150 370 * 2 . 0 15330 1 Y           m 7.751 500 1150 440 * 2 . 15 2280 2 * 1360 * 2 . 10       m 0.051 2 15.40 - 7.751 2 L - Y ey    Mx = Ptotal * ey = 15330 * 0.051 = 781.83 kN.m My = Ptotal * ex = 15330 * 0.082 = 1257.06 kN.m Example (6): Raft Foundation
22 June 2023 30 Shallow Foundations, P4, Tarek Nageeb x * 2446 1257.06 y * 3774 781.83 190.96 15330 x * I M y * I M A P f y y x x       The soil pressure can be obtained by applying the following equation: Example (6): Raft Foundation
22 June 2023 31 Shallow Foundations, P4, Tarek Nageeb
22 June 2023 32 Shallow Foundations, P4, Tarek Nageeb Calculation of Bending Moments and Shear Forces for Each Strip: For Strip ADHG (width = 3.20 m) Average Soil Pressure between points (A) and (D) 2 avg n kN/m 77.09 2 75.50 78.69 f     Total Soil Reaction FADHG = fn-avg * B1* L 77.09*3.20*15.40 = 3799 kN Total Column Loads PADHG = 440 + 1360 + 1360 + 370 = 3530 kN Example (6): Raft Foundation
Strip ADHG Average of Soil Pressure and Column Loads 22 June 2023 33 Shallow Foundations, P4, Tarek Nageeb kN 3664.50 2 3530 3799 2 P F F ADHG ADHG avg      ' m / kN 95 . 237 15.40 50 . 3664 L F strip) (per pressure soil Modified avg    0381 . 1 3530.0 50 . 3664 P F ADHG avg    
Strip ADHG Location of Points of Zero Shear: 22 June 2023 34 Shallow Foundations, P4, Tarek Nageeb m 1.92 237.95 77 . 456 x1   m 7.85 237.95 1411.82 77 . 456 x2    on so and m, 79 . 13 237.95 1411.82 1411.82 77 . 456 x3       kN.m 347.1 - 0.20 - 1.92 * 456.77 - 2 92 . 1 * 237.95 M 2 1 x   The maximum negative moment at point (1) (calculated at distance x1): Mx2 = +933.20 kN.m, Mx3 = -778.70 kN.m,
Strip ADHG 22 June 2023 35 Shallow Foundations, P4, Tarek Nageeb
Strip ADHG 22 June 2023 36 Shallow Foundations, P4, Tarek Nageeb
STRIP GHJI (Width = 6.00 m) Average Soil Pressure between points (B) and (E) Total Soil Reaction FGHJI = fn-avg * B1* L = 80.28*6.0*15.40 = 7417.74 kN Total Column Loads PGHJI = 1150 + 2880 + 2880 + 1150 = 8060 kN 22 June 2023 37 Shallow Foundations, P4, Tarek Nageeb 2 avg n kN/m 80.28 2 68 . 78 81.87 f     Average of Soil Pressure and Column Loads kN 87 . 7738 2 0 . 8060 74 . 7417 2 P F F GHJI GHJI avg     
Strip GHJI 22 June 2023 38 Shallow Foundations, P4, Tarek Nageeb ' m / kN 52 . 502 15.40 87 . 7738 L F strip) (per pressure soil Modified avg    9602 . 0 8060.0 87 . 7738 P F GHJIJ avg    
Strip GHJI Location of Points of Zero Shear: 22 June 2023 39 Shallow Foundations, P4, Tarek Nageeb The maximum moments are equal to: Mmax+ve = 1273.20 kN.m Mmax-ve = 992.30 kN.m m 2.20 502.52 18 . 1104 x1   m 7.70 502.52 25 . 2765 18 . 1104 x2    m 20 . 13 502.52 25 . 2765 25 . 2765 18 . 1104 x3    
Strip GHJI 22 June 2023 40 Shallow Foundations, P4, Tarek Nageeb
Strip GHJI 22 June 2023 41 Shallow Foundations, P4, Tarek Nageeb
Strip ACKL, (width=2.7 m) 22 June 2023 42 Shallow Foundations, P4, Tarek Nageeb
Strip ACKL 22 June 2023 43 Shallow Foundations, P4, Tarek Nageeb
Strip KLPR, (width=5.0m) 22 June 2023 44 Shallow Foundations, P4, Tarek Nageeb
Strip KLPR 22 June 2023 45 Shallow Foundations, P4, Tarek Nageeb
22 June 2023 46 Shallow Foundations, P4, Tarek Nageeb Design for flexure: Strip ADHG, with B = 3.20 m Example (6): Raft Foundation Mu = 1.50 * M' = 1.50 * 291.63 = 437.45 kN.m Assume the raft thickness t = 75 cm, then d = t – 7.0 = 75.0 – 7.0 = 68.0 cm kN.m/m' 291.63 3.20 20 . 933 ' M   0.0378 68 * 100 * 250 10 * 745 . 43 d * b * f M R 2 5 2 cu u u    From the R-w chart, R = 0.0378, then w = 0.045
22 June 2023 47 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation Use 7f20/m' (2199 mm2) Bottom reinforcement 2 s y cu s mm 0 . 2172 680 * 1000 * 360 25 * 0.046 d * b * f f * A    w           2 s 2 y min s mm 2823.0 2172 * 1.30 A 3 . 1 mm 1133 680 * 1000 * 360 0.6 d b f 0.6 of smaller A Similarly, the maximum negative moment is equal to 778.70 kN.m kN.m/m' 243.34 3.20 70 . 778 ' M   Design for flexure: Strip ADHG, with B = 3.20 m
22 June 2023 48 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation Mu = 1.50 * M' = 1.50 * 243.34 = 365.01 kN.m From the R-w chart, R = 0.0315, then w = 0.038 0.0315 68 * 100 * 250 10 * 501 . 36 d * b * f M R 2 5 2 cu u u    Design for flexure: Strip ADHG, with B = 3.20 m 2 s y cu s mm 0 . 1794 680 * 1000 * 360 25 * 0.038 d * b * f f * A    w           2 s 2 y min s mm 2332 1794 * 1.30 A 3 . 1 mm 1133 680 * 1000 * 360 0.6 d b f 0.6 of smaller A
22 June 2023 49 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation Use 7f18/m' (1781 mm2) with additional (3.5f16/m' ) (Top Reinforcement) Thus in this direction use a bottom mesh 7f20/m' and a top mesh 7f18/m'
Check of Punching Performed under the column with highest load: Pu = 1.50 Pmax = 1.50 * 2880 = 4320.0 kN 22 June 2023 50 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation d = 68.00 cm ap = c1 + d = 40 + 68 = 108 cm bp = c2 + d = 40 + 68 = 108 cm bpunch = 2*(1.08+1.08) = 2*2.16 = 4.32 m The pressure at point O is equal to 80.8 kN/m2 Thus, the ultimate soil pressure qsu = 1.5 * 80.8 = 121.19 kN/m2
22 June 2023 51 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation The punching load is equal to: Qup = Pu – qsu (a*b) = 4320.0 – 121.19*(1.08*1.08) = 4178.60 kN Check of Punching 2 up up N/mm 1.42 680 * 4320 1000 * 60 . 4178 d * bp Q q    2 c cu cup N/mm 1.6 1.29 5 . 1 25 0.316 f 0.316 q     g c cu p p cup f b a 0.50 0.316 q g           The allowable punching stress is equal to the min. of:
22 June 2023 52 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation Check of Punching 2 cup N/mm 1.94 5 . 1 25 0.40 40 . 0 0.50 0.316 q          50 . 1 f b d * 0.20 0.80 q cu p cup            2 cup N/mm 2.71 50 . 1 25.0 32 . 4 68 . 0 * 4 0.20 0.80 q          qcup = 1.29 N/mm2 > qup = 1.42 N/mm2 Unsafe Therefore, increase the raft depth
Summary for Strips GHJI, ACKL, KLPR 22 June 2023 53 Shallow Foundations, P4, Tarek Nageeb
22 June 2023 54 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation
22 June 2023 55 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation
22 June 2023 56 Shallow Foundations, P4, Tarek Nageeb Analysis and Design of Raft Foundation Using Special Purpose Software Use CSI-SAFE for the analysis and design of raft foundations. Piled Raft Foundation: 1- Use CSI-Safe for the analysis and design of piled-raft foundation. 2- Ignore the soil reactions and use only springs at pile positions, especially in end bearing piles. 3- Rerun the analysis many times to choose the optimum pile positions at which the piles are almost equally loaded to their working loads.

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Foundations-P4-2018 (1).ppt

  • 1. 22 June 2023 Shallow Foundations, P4, Tarek Nageeb 1 SHALLOW FOUNDATIONS PART 4 ‫الضحلة‬ ‫األساسات‬ ‫الرابع‬ ‫الجزء‬
  • 2. 22 June 2023 Shallow Foundations, P4, Tarek Nageeb 2 DESIGN OF RAFT FOUNDATIONS ‫اللبشة‬ ‫أساسات‬ ‫تصميم‬
  • 3. 22 June 2023 3 Shallow Foundations, P4, Tarek Nageeb FOUNDATION ENGINEERING 1- Site Investigation 2- Design of Shallow Foundations: a- Design of Isolated Footings. b- Design of Isolated Footings under Eccentric Loads. c- Design of Combined Footings. d- Design of Strap Beam. e- Design of Raft Foundations
  • 4. Raft Foundations are used in: 1- Weak soil conditions (low bearing capacity). 2- High structure loads, relative to soil B/C. 3- Interference between the isolated reinforced concrete footings. 4- Total area of isolated R.C. footings > 2/3 construction site area. 5- Single and multiple basements. 6- The existence of weak soil pockets within the stiff soil area. 7- Isolated footing settlements exceeds the allowable values. Design of Raft Foundations 22 June 2023 4 Shallow Foundations, P4, Tarek Nageeb
  • 5. ‫عندما‬ ‫المسلحة‬ ‫الخرسانة‬ ‫من‬ ‫اللبشة‬ ‫تستخدم‬ : 1 - ‫ضعيفة‬ ‫التربة‬ ‫تحمل‬ ‫قدرة‬ ‫تكون‬ . 2 - ‫التربة‬ ‫تحمل‬ ‫لقدرة‬ ‫بالنسبة‬ ‫كبيرة‬ ‫المنشأ‬ ‫أحمال‬ ‫تكون‬ . 3 - ‫المنفصلة‬ ‫المسلحة‬ ‫القواعد‬ ‫بين‬ ‫تداخل‬ ‫حدوث‬ ‫عند‬ . 4 - ‫ثلثى‬ ‫عن‬ ‫المنفصلة‬ ‫المسلحة‬ ‫القواعد‬ ‫مساحة‬ ‫زيادة‬ ‫عند‬ ‫الموقع‬ ‫مساحة‬ 5 - ‫متعددة‬ ‫بدرومات‬ ‫أو‬ ‫واحد‬ ‫بدروم‬ ‫وجود‬ . 6 - ‫مساحة‬ ‫كامل‬ ‫على‬ ‫وليس‬ ‫التربة‬ ‫من‬ ‫ضعيفة‬ ‫جيوب‬ ‫وجود‬ ‫الموقع‬ 7 - ‫المنف‬ ‫للقواعد‬ ‫بها‬ ‫المسموح‬ ‫القيم‬ ‫عن‬ ‫الكلى‬ ‫الهبوط‬ ‫زيادة‬ ‫صلة‬ Design of Raft Foundations 22 June 2023 5 Shallow Foundations, P4, Tarek Nageeb
  • 6. Isolated or Strip Footing Raft Foundation Vs. Isolated Footing Raft Foundation Soft Soil 2B B 2B B 22 June 2023 6 Shallow Foundations, P4, Tarek Nageeb To avoid high raft settlement ‫فى‬ ‫كبير‬ ‫هبوط‬ ‫حدوث‬ ‫لتجنب‬ ‫اللبشة‬ ‫استخدام‬ ‫حالة‬
  • 7. Net and gross allowable soil pressure: qall gross = qall net + gZ qall gross : Gross allowable bearing capacity ‫الكلى‬ ‫التربة‬ ‫جهد‬ qall net : Net allowable bearing capacity ‫الصافى‬ ‫التربة‬ ‫جهد‬ gZ : Weight of soil above F.L. ‫التأسيس‬ ‫منسوب‬ ‫عند‬ ‫التربة‬ ‫عمود‬ ‫وزن‬ Design of Raft Foundations 22 June 2023 7 Shallow Foundations, P4, Tarek Nageeb
  • 8. Simple settlement calculations: d = mv * H * Ds d : Settlement value ‫الهبوط‬ ‫قيمة‬ mv : Soil compressibility, m2/MN = 1/E Ds : Added soil pressure, kN/m2 Settlements will take place when the applied pressure is higher than g z ‫الحادثة‬ ‫اإلجهادات‬ ‫قيمة‬ ‫تتعدى‬ ‫عندما‬ ‫يحدث‬ ‫سوف‬ ‫الهبوط‬ ‫أن‬ ‫أى‬ ‫التأسيس‬ ‫منسوب‬ ‫عند‬ ‫التربة‬ ‫عمود‬ ‫وزن‬ ‫التربة‬ ‫على‬ Design of Raft Foundations 22 June 2023 8 Shallow Foundations, P4, Tarek Nageeb
  • 9. Stress Compensation  Total building loads = 40,000.0 kN  Building Area = 20 x 20 m  Determine the depth at which there will be stress compensation between the structure and soil weight  Solution:  Stresses at the foundation level: 22 June 2023 9 Shallow Foundations, P4, Tarek Nageeb 2 kN/m 100.0 20.0 * 0 . 20 40,000.0 Raft of Area Load Building Total    s m 5.56 18.0 100.0 d then, d,     g s g s Equivalent depth, d
  • 10. Types of Raft or Mat Foundation  a) flat plate  b) Plate thickened under columns 22 June 2023 10 Shallow Foundations, P4, Tarek Nageeb
  • 11. Types of Raft or Mat Foundation  c) Waffle Slab  d) Plate with Pedestals 22 June 2023 11 Shallow Foundations, P4, Tarek Nageeb
  • 12. Types of Raft or Mat Foundation Rigid Basement 22 June 2023 12 Shallow Foundations, P4, Tarek Nageeb
  • 13. Raft [Mat] Foundation Inverted Beam Raft 22 June 2023 13 Shallow Foundations, P4, Tarek Nageeb
  • 14. Raft [Mat] Foundation Inverted Beam Raft Column Inverted Beam Concrete Plate Section Elevation in the Raft 22 June 2023 14 Shallow Foundations, P4, Tarek Nageeb
  • 15. 22 June 2023 Shallow Foundations, P4, Tarek Nageeb 15 Design of Raft Foundation The resultant is determined from the column loads, and so the eccentricities ex and ey ex R L y x ey B
  • 16. Design of Raft Foundation 1- Stress Distribution Under the Raft 22 June 2023 16 Shallow Foundations, P4, Tarek Nageeb I x * M I y * M A R f y y x x 2 1    In which; Mx = R * ey and My = R * ex Such that: f1 ≤ qall soil f2 ≥ 0.00 (no tension) 12 B * L I 3 x  12 L * B I 3 y 
  • 17. f1 f2 f2 f1 f2 f1 22 June 2023 Shallow Foundations, P4, Tarek Nageeb 17 Soil Stresses under the Raft Foundation f1 f2 x y ex R L ey B
  • 18. 22 June 2023 Shallow Foundations, P4, Tarek Nageeb 18 Soil Stresses under the Raft Foundation L B B 1 B 2 B 3 B 4 L1 L2 L3 L4 L5 ‫الش‬ ‫فى‬ ‫موضح‬ ‫هو‬ ‫كما‬ ‫وأفقية‬ ‫رأسية‬ ‫شرائح‬ ‫إلى‬ ‫اللبشة‬ ‫تقسيم‬ ‫يتم‬ ‫المرفق‬ ‫كل‬ . The raft is divided into a number of horizontal and vertical strips, as shown in the figure.
  • 19. 22 June 2023 Shallow Foundations, P4, Tarek Nageeb 19 ‫اإلنحن‬ ‫عزوم‬ ‫حساب‬ ‫ثم‬ ‫شريحه‬ ‫كل‬ ‫أسفل‬ ‫التربة‬ ‫ضغط‬ ‫قيمة‬ ‫تحديد‬ ‫يتم‬ ‫وقوى‬ ‫اء‬ ‫الشريحة‬ ‫تلك‬ ‫على‬ ‫القص‬ The soil pressure under each strip is determined and the bending moments and shear forces are determined L B B 1 B 2 B 3 B 4 L1 L2 L3 L4 L5 P1 P2 P3 P4 P5
  • 20. 22 June 2023 Shallow Foundations, P4, Tarek Nageeb 20 4 5 4 3 2 1 strip n B * L P P P P P A P f        f2 f1 2 f f f 2 1 avg n      P Fstrip  L B 4 P1 P2 P3 P4 P5 4 avg - n strip B * L * f F 
  • 21. 22 June 2023 Shallow Foundations, P4, Tarek Nageeb 21 1 - ‫األعمدة‬ ‫أحمال‬ ‫مجموع‬ ‫حساب‬ ‫يتم‬ 2 - ‫ألعلى‬ ‫أسفل‬ ‫من‬ ‫التربة‬ ‫لضغط‬ ‫متوسط‬ ‫حساب‬ ‫يتم‬ 3 - ‫ومتوسط‬ ‫األعمدة‬ ‫أحمال‬ ‫لمجموع‬ ‫متوسطة‬ ‫قيمة‬ ‫حساب‬ ‫يتم‬ ‫التربة‬ ‫ضغط‬ 4 - ‫معامل‬ ‫حساب‬ ‫يتم‬ () ‫والمتوسط‬ ‫األعمدة‬ ‫حمل‬ ‫بين‬ ‫للفرق‬ ‫الجديد‬ 5 - ‫وقوى‬ ‫العزوم‬ ‫حساب‬ ‫عند‬ ‫األعمدة‬ ‫أحمال‬ ‫فى‬ ‫المعامل‬ ‫يضرب‬ ‫شريحة‬ ‫لكل‬ ‫القص‬ . L B 4 P1 fn-avg P2 P3 P4 P5
  • 22. Raft Foundation Strip Reinforcement 22 June 2023 22 Shallow Foundations, P4, Tarek Nageeb
  • 23. Estimation of Soil Reaction under the Raft 22 June 2023 23 Shallow Foundations, P4, Tarek Nageeb An approximate estimate of the coefficient of subgrade reaction is obtained as follows: ks (kN/m3) = (100  120)* soil bearing capacity (kN/m2)
  • 24. Estimation of Soil Reaction under the Raft 22 June 2023 24 Shallow Foundations, P4, Tarek Nageeb
  • 25. Estimation of Soil Reaction under the Raft 22 June 2023 25 Shallow Foundations, P4, Tarek Nageeb An approximate estimate of the coefficient of subgrade reaction is obtained as follows: ks (kN/m3) = (80  120)* soil bearing capacity (kN/m2)
  • 26. A plan of a mat foundation. All columns are 40×40 cm2. qall,net is 125 kPa. The material properties for concrete and steel are: fcu = 25 N/mm2, fy = 360 N/mm2 22 June 2023 26 Shallow Foundations, P4, Tarek Nageeb Example (6): Design of Raft Foundation From Dr. Mashour Ghonaim Book
  • 27. Solution: Step 1: Check of Soil Pressure: Example (6): Raft Foundation 22 June 2023 27 Shallow Foundations, P4, Tarek Nageeb I x * M I y * M A P f y y x x 2 1    A = Area of the raft = 15.40 * 12.40 = 190.96 m2 4 3 3 x m 3774.0 12 15.40 * 12.40 12 L B I    4 3 3 y m 2446.80 12 12.40 * 15.40 12 B L I   
  • 28. Total Vertical Unfactored Loads = 440 + (1360*2) + 370 + (1150*2) + (2880*2) + 500 + 1360 + 1440 + 440 = 15330 kN ‫نقطة‬ ‫حول‬ ‫العزوم‬ ‫بأخذ‬ ‫لللبشة‬ ‫األحمال‬ ‫مركز‬ ‫تحديد‬ D 22 June 2023 28 Shallow Foundations, P4, Tarek Nageeb      2 * 2280 2 * 1150 * 2 . 6 370 2 * 1360 440 * 2 . 0 15330 1 X        m 6.282 440 1440 1360 500 * 2 . 12      m 0.082 2 12.40 - 6.282 2 B - X ex    Example (6): Raft Foundation
  • 29. 22 June 2023 29 Shallow Foundations, P4, Tarek Nageeb      1440 2880 1360 * 2 . 5 440 1150 370 * 2 . 0 15330 1 Y           m 7.751 500 1150 440 * 2 . 15 2280 2 * 1360 * 2 . 10       m 0.051 2 15.40 - 7.751 2 L - Y ey    Mx = Ptotal * ey = 15330 * 0.051 = 781.83 kN.m My = Ptotal * ex = 15330 * 0.082 = 1257.06 kN.m Example (6): Raft Foundation
  • 30. 22 June 2023 30 Shallow Foundations, P4, Tarek Nageeb x * 2446 1257.06 y * 3774 781.83 190.96 15330 x * I M y * I M A P f y y x x       The soil pressure can be obtained by applying the following equation: Example (6): Raft Foundation
  • 31. 22 June 2023 31 Shallow Foundations, P4, Tarek Nageeb
  • 32. 22 June 2023 32 Shallow Foundations, P4, Tarek Nageeb Calculation of Bending Moments and Shear Forces for Each Strip: For Strip ADHG (width = 3.20 m) Average Soil Pressure between points (A) and (D) 2 avg n kN/m 77.09 2 75.50 78.69 f     Total Soil Reaction FADHG = fn-avg * B1* L 77.09*3.20*15.40 = 3799 kN Total Column Loads PADHG = 440 + 1360 + 1360 + 370 = 3530 kN Example (6): Raft Foundation
  • 33. Strip ADHG Average of Soil Pressure and Column Loads 22 June 2023 33 Shallow Foundations, P4, Tarek Nageeb kN 3664.50 2 3530 3799 2 P F F ADHG ADHG avg      ' m / kN 95 . 237 15.40 50 . 3664 L F strip) (per pressure soil Modified avg    0381 . 1 3530.0 50 . 3664 P F ADHG avg    
  • 34. Strip ADHG Location of Points of Zero Shear: 22 June 2023 34 Shallow Foundations, P4, Tarek Nageeb m 1.92 237.95 77 . 456 x1   m 7.85 237.95 1411.82 77 . 456 x2    on so and m, 79 . 13 237.95 1411.82 1411.82 77 . 456 x3       kN.m 347.1 - 0.20 - 1.92 * 456.77 - 2 92 . 1 * 237.95 M 2 1 x   The maximum negative moment at point (1) (calculated at distance x1): Mx2 = +933.20 kN.m, Mx3 = -778.70 kN.m,
  • 35. Strip ADHG 22 June 2023 35 Shallow Foundations, P4, Tarek Nageeb
  • 36. Strip ADHG 22 June 2023 36 Shallow Foundations, P4, Tarek Nageeb
  • 37. STRIP GHJI (Width = 6.00 m) Average Soil Pressure between points (B) and (E) Total Soil Reaction FGHJI = fn-avg * B1* L = 80.28*6.0*15.40 = 7417.74 kN Total Column Loads PGHJI = 1150 + 2880 + 2880 + 1150 = 8060 kN 22 June 2023 37 Shallow Foundations, P4, Tarek Nageeb 2 avg n kN/m 80.28 2 68 . 78 81.87 f     Average of Soil Pressure and Column Loads kN 87 . 7738 2 0 . 8060 74 . 7417 2 P F F GHJI GHJI avg     
  • 38. Strip GHJI 22 June 2023 38 Shallow Foundations, P4, Tarek Nageeb ' m / kN 52 . 502 15.40 87 . 7738 L F strip) (per pressure soil Modified avg    9602 . 0 8060.0 87 . 7738 P F GHJIJ avg    
  • 39. Strip GHJI Location of Points of Zero Shear: 22 June 2023 39 Shallow Foundations, P4, Tarek Nageeb The maximum moments are equal to: Mmax+ve = 1273.20 kN.m Mmax-ve = 992.30 kN.m m 2.20 502.52 18 . 1104 x1   m 7.70 502.52 25 . 2765 18 . 1104 x2    m 20 . 13 502.52 25 . 2765 25 . 2765 18 . 1104 x3    
  • 40. Strip GHJI 22 June 2023 40 Shallow Foundations, P4, Tarek Nageeb
  • 41. Strip GHJI 22 June 2023 41 Shallow Foundations, P4, Tarek Nageeb
  • 42. Strip ACKL, (width=2.7 m) 22 June 2023 42 Shallow Foundations, P4, Tarek Nageeb
  • 43. Strip ACKL 22 June 2023 43 Shallow Foundations, P4, Tarek Nageeb
  • 44. Strip KLPR, (width=5.0m) 22 June 2023 44 Shallow Foundations, P4, Tarek Nageeb
  • 45. Strip KLPR 22 June 2023 45 Shallow Foundations, P4, Tarek Nageeb
  • 46. 22 June 2023 46 Shallow Foundations, P4, Tarek Nageeb Design for flexure: Strip ADHG, with B = 3.20 m Example (6): Raft Foundation Mu = 1.50 * M' = 1.50 * 291.63 = 437.45 kN.m Assume the raft thickness t = 75 cm, then d = t – 7.0 = 75.0 – 7.0 = 68.0 cm kN.m/m' 291.63 3.20 20 . 933 ' M   0.0378 68 * 100 * 250 10 * 745 . 43 d * b * f M R 2 5 2 cu u u    From the R-w chart, R = 0.0378, then w = 0.045
  • 47. 22 June 2023 47 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation Use 7f20/m' (2199 mm2) Bottom reinforcement 2 s y cu s mm 0 . 2172 680 * 1000 * 360 25 * 0.046 d * b * f f * A    w           2 s 2 y min s mm 2823.0 2172 * 1.30 A 3 . 1 mm 1133 680 * 1000 * 360 0.6 d b f 0.6 of smaller A Similarly, the maximum negative moment is equal to 778.70 kN.m kN.m/m' 243.34 3.20 70 . 778 ' M   Design for flexure: Strip ADHG, with B = 3.20 m
  • 48. 22 June 2023 48 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation Mu = 1.50 * M' = 1.50 * 243.34 = 365.01 kN.m From the R-w chart, R = 0.0315, then w = 0.038 0.0315 68 * 100 * 250 10 * 501 . 36 d * b * f M R 2 5 2 cu u u    Design for flexure: Strip ADHG, with B = 3.20 m 2 s y cu s mm 0 . 1794 680 * 1000 * 360 25 * 0.038 d * b * f f * A    w           2 s 2 y min s mm 2332 1794 * 1.30 A 3 . 1 mm 1133 680 * 1000 * 360 0.6 d b f 0.6 of smaller A
  • 49. 22 June 2023 49 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation Use 7f18/m' (1781 mm2) with additional (3.5f16/m' ) (Top Reinforcement) Thus in this direction use a bottom mesh 7f20/m' and a top mesh 7f18/m'
  • 50. Check of Punching Performed under the column with highest load: Pu = 1.50 Pmax = 1.50 * 2880 = 4320.0 kN 22 June 2023 50 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation d = 68.00 cm ap = c1 + d = 40 + 68 = 108 cm bp = c2 + d = 40 + 68 = 108 cm bpunch = 2*(1.08+1.08) = 2*2.16 = 4.32 m The pressure at point O is equal to 80.8 kN/m2 Thus, the ultimate soil pressure qsu = 1.5 * 80.8 = 121.19 kN/m2
  • 51. 22 June 2023 51 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation The punching load is equal to: Qup = Pu – qsu (a*b) = 4320.0 – 121.19*(1.08*1.08) = 4178.60 kN Check of Punching 2 up up N/mm 1.42 680 * 4320 1000 * 60 . 4178 d * bp Q q    2 c cu cup N/mm 1.6 1.29 5 . 1 25 0.316 f 0.316 q     g c cu p p cup f b a 0.50 0.316 q g           The allowable punching stress is equal to the min. of:
  • 52. 22 June 2023 52 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation Check of Punching 2 cup N/mm 1.94 5 . 1 25 0.40 40 . 0 0.50 0.316 q          50 . 1 f b d * 0.20 0.80 q cu p cup            2 cup N/mm 2.71 50 . 1 25.0 32 . 4 68 . 0 * 4 0.20 0.80 q          qcup = 1.29 N/mm2 > qup = 1.42 N/mm2 Unsafe Therefore, increase the raft depth
  • 53. Summary for Strips GHJI, ACKL, KLPR 22 June 2023 53 Shallow Foundations, P4, Tarek Nageeb
  • 54. 22 June 2023 54 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation
  • 55. 22 June 2023 55 Shallow Foundations, P4, Tarek Nageeb Example (6): Raft Foundation
  • 56. 22 June 2023 56 Shallow Foundations, P4, Tarek Nageeb Analysis and Design of Raft Foundation Using Special Purpose Software Use CSI-SAFE for the analysis and design of raft foundations. Piled Raft Foundation: 1- Use CSI-Safe for the analysis and design of piled-raft foundation. 2- Ignore the soil reactions and use only springs at pile positions, especially in end bearing piles. 3- Rerun the analysis many times to choose the optimum pile positions at which the piles are almost equally loaded to their working loads.