Associate Professor. Tel:

Size: px

Start display at page:

Download "Associate Professor. Tel:"

Gwenda Green
5 years ago
Views:

1 DEPARTMENT OF CIVIL ENGINEERING IIT DELHI Dr. Suresh Bhalla Associate Professor Tel:

2 FOUNDATIONS Geotechnical Engineer Structural Engineer Location and depth criteria Bearing capacity criteria- safety against shear failure of soil Settlement t criteria- i should not settle excessively. Structural drawings follow location and depth criteria Soil pressure does not exceed allowable pressure as per soil report Structurally safe

3 Ultimate bearing capacity, q ult TERMINOLOGIES Total gross pressure at base of foundation which causes shear failure. Ultimate net bearing capacity, q ult, net q ult, net = q ult - q (q = Effective soil pressure at foundation base) ) Safe net bearing capacity, q safe, net q safe, net =(q ult, net )/ Factor of safety Overburden Safe bearing pressure, q safe, pr Maximum net pressure that foundation can transmit without the settlement exceeding the permissible value. Allowable net bearing gpressue, q all, net Lower of (q safe, net and q safe, pr ) WORKING STRESS APPROACH

4 ISOLATED FOOTING 600mm 400mm D = 2m M P B H Natural ground level (NGL) Water table 1m 1m 05m 0.5m Dead + Earthquake P = 1200 kn M = 125 kn-m H = 20 kn γ soil = 18 knm -3 q all, net = 180 knm -2 Design Steps: (1) Size of footing to satisfy base pressure requirements (2) Design of base for bending (3) Check for one-way shear (4) Check for two-wayway shear (5) stability against sliding and overturning

5 (1) SIZE OF FOOTING TO SATISFY BASE PRESSURE REQUIREMENTS D = 2m M 600mm P B 400mm H Natural ground level (NGL) Water table 1m 1m 0.5m Dead + Earthquake P = 1200 kn M = 125 kn-m H = 20 kn γ soil = 18 knm -3 q all, net = 180 knm -2 Under normal conditions q all, gross = q all, net + q = x1 + (18-10)x1= 207 knm -2 Under wind/ earthquake Submerged density Submerged density q all, gross = 1.25 q all, net + q = 1.25 x x1 + (18-10)x1= 251 knm -2 Try 2.5x2.5x0.4m thick footing P (total) = P t = P + Overburden 2.5x2.5x0.4x (25-10) + (2.5x x0.6) (0.6 x x 18) = kn NGL 2m = Footing base Soil M (total) = M t = M + H*(Footing thickness) = x 0.4= 133 kn-m

(1) SIZE OF FOOTING TO SATISFY BASE PRESSURE REQUIREMENTS 600mm P M D =

4 m footing 400mm Natural ground level (NGL) H Water table σ max 1m 1m 0.

-3 q all, net = 180 knm -2 Area A = B 2 = 6.

25 m 3 σ P M t σmax = t + σ A Z min = t A 279.65 kn/m 2 > q 177.

6 (1) SIZE OF FOOTING TO SATISFY BASE PRESSURE REQUIREMENTS 600mm P M D = 2m σ min B For 2.5 x 2.5 x 0.4 m footing 400mm Natural ground level (NGL) H Water table σ max 1m 1m 0.5m Dead + Earthquake P = 1200 kn M = 125 kn-m H = 20kN γ soil = 18 knm -3 q all, net = 180 knm -2 Area A = B 2 = 6.25 m 2 Section modulus Z = (B 3 /6) = 6.25 m 3 σ P M t σmax = t + σ A Z min = t A kn/m 2 > q kn/m 2 all, gross =251 kn/m 2 P t - M t Z Try 2.7 x 2.7 x 0.4 m footing A = 7.29 m 2 Z = 3.28 m 3 P t = kn σ max = kn/m 2 σ min = kn/m 2 min < q all, gross =251 kn/m 2

7 (1) SIZE OF FOOTING TO SATISFY BASE PRESSURE REQUIREMENTS 600mm 400mm D = 2m M P B H Natural ground level (NGL) 1m Water table 1m 0.5m Dead + Earthquake P = 1200 kn M = 125 kn-m H = 20 kn γ soil = 18 knm -3 q = -2 all, net 180 knm P (total) = P t = P + Overburden σ 1,2 = P M t M t P t ± = ± A Z A Z + P overburden A Overburden pressure= 0.4x(25-10) x (18-10) 10) x 18 = 37.8 kn/m 2 σ max = kn/m 2 σ min = kn/m Slightly higher since 2 additional soil considered at col. location σ max = kn/m 2 Exact σ min = kn/m 2

8 (2) DESIGN OF BASE FOR BENDING kn/m kn/m kn/m 2 M25 concrete Fe415 steel d = = 340mm M = knm/m kn/m 2 M u = 1.5 x = knm/m M u / bd 2 = x 106 Nmm 1000mm x (340) 2 Nmm-2 mm2=1.39 A st = 0.417% = 14.2 cm 2 /m. Provide 140 mm c/c = cm 2 /m = 0.42% Dr. Suresh Bhalla, Department of Civil Engineering, IIT Delhi. This material is only for CEP 727 students of IIT Delhi. No part of this

9 (3) CHECK FOR ONE-WAY SHEAR d = 340mm 37.8 kn/m 2 d kn/m kn/m kn/m 2 V = knm/m V u = 1.5 x = kn/m x 10 3 N Nominal τ =0.609 Nmm -2 = shear stress v 1000mm x (340) mm Shear strength of concrete (for 0.42% steel) τ c = Nmm -2 < τ v Pg 73, IS 456 Either increase d or increase reinforcement to 1% (let us increase reinforcement)

10 (4) CHECK FOR TWO-WAY (PUNCHING) SHEAR P d/2 = 170mm 37.8 kn/m d 400+d kn/m kn/m kn/m 2 V = (0.9 x 0.74)( ) = kn V u = 1.5 x = kn τ = x 103 N v = Nmm -2 2( )mm x (340) mm β c = (0.4/0.6) Shear strength of concrete = k s τ c = 1.11 Nmm -2 (Pg. 59, IS 456) < τ v τ c = 0.25 f ck = 1.11 Nmm -2 k s = β c = 1.0 (<=1.0)

11 (4) CHECK FOR TWO-WAY (PUNCHING) SHEAR P d/2 = 170mm 37.8 kn/m d 400+d kn/m kn/m kn/m 2 τ v = Nmm -2 > 1.11 Nmm -2 (k s τ c ) Increase thickness to 500mm => d = 440mm V = (1.04 x 0.84)( ) = kn V u = 1.5 x = kn x 103 N τ = -2 v = Nmm < k s τ c 2( )mm x (440) mm OK

12 *REDESIGN FOOTING FOR BENDINGAND 1-WAY SHEAR REINFORCEMENT DETAILING 37.8 kn/m 2 L d Min 150mm kn/m kn/m kn/m mm c/c (Check) 150 mm c/c (Check)

13 (5) SAFETY AGAINST SLIDING AND OVERTURNING (for a stand alone footing) Sec 20 (p33): IS 456:2000 D = 2m OVERTURNING 600mm P M B 400mm H Natural ground level (NGL) Water table 1m 1m 0.5m Dead + Earthquake P = 1200 kn M = 125 kn-m H = 20 kn γ soil = 18 knm -3 q all, net = 180 knm -2 Restoring moment > 1.2 M o (due to DL) M o (due to IL) Consider a DL = 90% IL = 0 for restoring moment calculation Pg. 33, IS 456 SLIDING Restoring force > 1.4 H Consider a DL = 90% IL = 0 DL = Dead load IL = Imposed load

14 STAND ALONE FOOTING (say for a pipe support) 0.3m P = 50 kn H = 10 kn M = 10 x 2.3 = 23 kn-m 2m 7.04 kn/m 2 q all, net = 180 knm -2 Try 1.2m x1.2m footing A = 1.44 m 2 Z = m 3 M ± + σ Z overburden σ 1,2 = P kn/m 2 A B σ 1,2 = , kn/m 2 Hence, redistribution of base pressure will take place

15 STAND ALONE FOOTING (say for a pipe support) 0.3m e = M/P t 2m e Overburden : = 1.2x1.2x0.3x x0.6x 2.0x25+(1.2x x0.6)x1.7x18 = kn 7.04 kn/m 2 P t = P + Overburden = kn x/ kn/m 2 X=???? e = M/P t = 0.21m B (x/3+e) = B/2 => x = 3(0.5B-e) => x = 1.17m 0.5xB σ 1 = P t =>σ 1 = 156 kn/m 2

16 STAND ALONE FOOTING (say for a pipe support) 03m 0.3m 2m B?? ALTERNATE APPROACH: Increase size such that no point on footing loses contact with soil

17 STAND ALONE FOOTING (say for a pipe support) 0.3m OVERTURNING 2m 38.1 kn/m 2 x/ m 156 kn/m 2 Overturning moment = 1.2 M o (due to DL) M o (due to IL) = x23 = knm Restoring moment = 0.9x59.52x0.6 = knm Marginally unsafe, increase dimension to 1.3m

18 STAND ALONE FOOTING (say for a pipe support) 0.3m SLIDING Coeff. of friction (concrete-soil) Restoring force > 14Si 1.4 Sliding i Force 2m 38.1 kn/m 2 0.9x59.52 x x 10 x/ kn > 14 kn (OK) m 156 kn/m 2

19 ISOLATED FOOTING UNDER BIAXIAL BENDING L y M y x M x x B y σ 4 σ 1,2,3,4(gross) = P A M x Z x M y Z y ± ± + σ overburden σ 1 σ 2 Z x =LB 2 /6 Z y =BL 2 /6 M x,m y : TOTAL MOMENTS ABOUT THE BASE OF FOOTING

20 ISOLATED FOOTING UNDER BIAXIAL BENDING (Reinforcement in x direction) L Plane along xx y M yt σ σ 4 avg x x B σ 1 σ 2 M xt y Along xx σ xx,min = P A M y Z y - + σ σ xx, max = overburden P A M + y + σ overburden Z y

21 ISOLATED FOOTING UNDER BIAXIAL BENDING L y M y σ 1 σ 2 σ 4 σ avg x M x x B y Similarly, Reinforcement in y direction can be determined Design Steps: (1) Size of footing to satisfy base pressure requirements (2) Design of base for bending (3) Check for one-way shear (4) Check for two-way shear (5) stability against sliding and overturning

22 THANK YOU

DESIGN AND DETAILING OF COUNTERFORT RETAINING WALL

DESIGN AND DETAILING OF COUNTERFORT RETAINING WALL When the height of the retaining wall exceeds about 6 m, the thickness of the stem and heel slab works out to be sufficiently large and the design becomes