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Lecture #07The Design of Spread Footings- Design Criteria- Design Procedure- Example Problem for a Square Footing- Example Problem for a Rectangular Footing-Example Problem for a Continuous Footing- Complex Cases

The Design Procedure.1. Determine the structural loads and member sizes at the foundation level;2. Collect all the geotechnical data; set the proposed footings on the geotechnical profile;3. Determine the depth and location of all foundation elements,4. Determine the bearing capacity,5. Determine possible total and differential settlements; check effects at 2B depths;6. Select the concrete strength (and possibly the mix),7. Select the steel grade,8. Determine the required footing dimensions,9. Determine the footing thickness, T (or D in some textbooks),10. Determine the size, number and spacing of the reinforcing bars,11. Design the connection between the superstructure and the foundation, and12. Check uplift and stability against sliding and overturning of the structure-soil system.The first studies performed on foundation structural failures were done by Professor Talbot atthe University of Illinois in 1913. Advances in the next 50 years include Prof. F.E. Richart’stests at the University of Michigan. His results were synthesized into the methodology usedtoday by a committee sponsored by the ACI and ASCE and published in 1962.Spread footings is still the most popular foundation around the world because they are moreeconomical than piles, adding weight to them does not affect any other member, and theirperformance has been excellent.

Selection of materials.Spread footings are usually designed to use 3 ksi f'c 4 ksi, whereas modern structuralmembers frequently use concrete between the range of 4 ksi f'c 8 ksi. A higher concretestrength helps reduction the member’s size. However, the footing’s design is govern by thebearing capacity and settlement. That means that the strength of the soil might be limitingfactor, and a higher concrete strength would not be relevant.Where a footing must carry a load greater than 500 kips, an f'c 5 ksi might be justified.Since flexural stresses are usually small, a grade 40 steel would usually be adequate, althoughit is currently unavailable in the US. The most common grade used for construction is Grade60 steel, which is almost universally used in the world today.

The typical details of a spread footing, as sketched for drafting. The standard thicknesses T aregiven in English Units as multiples of 3": 12", 15", 18"., etc. A high precision in specifyingthe depth of excavation Df is unnecessary. ACI code specifies that at least 3 inches of concretecover must be included from ground contact, which takes into consideration irregularities in theexcavation and corrosion factors.

The same footing as built. Note that forming is required in this site due to the presence of a cleangravelly sand, that would not stand vertically at the sides of the excavation.

Design Criteria.1) The qall and Q control the footing dimensions B x L (footing area A);2) The designer controls the depth Df (embedment of the foundation);3) Shear (v) controls the footing’s thickness T (d 3” the diameter of rebars)a) Diagonal tension (punching shear) for square footings, andb) Wide beam shear for rectangular footings (that is, when L / B 1.2).

Analysis.The analysis of a square or rectangular footing may first be performed by assuming there is nosteel in the member. The depth d from the top of the footing to the tension axis is,S Fy 0Qu 2d vc (b d) 2d vc (c d) (c d)(b d) qo(shear on 4 faces)(bottom face)Set Qu BL qo d2(4 vc qo) d (2 vc qo)(b c) - (BL - cb) qo 0For the special case of a square column, where c b w,d2 (vc qo/4) d (vc qo/2)w - (B2 - w2) qo/4 0For the case of a round column, with a diameter,d2 (vc qo/4) d (vc qo/2) a - ( BL - Acol ) qo/4 0

Design Steps.Step 1. Compute the footing area via B x L,for a square footing BxB SQRT(Q / qall)for a rectangular footing BxL Q / qallwhere Q is the critical load combination (not Qu).Step 2. Find the soil reaction under ultimate structural loads to check bearing capacity.find the "ultimate" contact bearing, qo Qu / BLand check that qo quStep 3. Compute the shear in the concrete vc.Case (a) for a square footing, check for diagonal tension (punching shear),vall 4 SQRT (f'c) where 0.85 for shear.For example, for f'c 3000 psi, vall 186 psi

Case (b) for a rectangular footing, check for wide beam shear,vall 2 SQRT(f'c)where 0.85 for shear.For example, for f'c 3000 psi, vall 93.1 psiStep 4. Find the effective footing depth d.(Note that use of d via this method eliminates the need to use steel for shear, whichis used only for flexure. Use the appropriate equation from the Analysis Section.Step 5. Compute the required area of steel As (each way) for bending (flexure).Bending moment /unit widthM q L2 / 2 (for a cantilever beam)Mu qult L2 / 2 As fy (d – a /2)Check p, so that the maximum allowed percentage of steel is not exceeded.Step 6. Compute bond length, column bearing, and the steel area required for dowels.Use as a minimum an As 0.005 Acol (usually with 4 equal bars).Step 7. Draft the above information into a complete drawing showing all the details.

Example #1.Design a square reinforced concrete footing for the following conditions:- The column has a DL 100 kips, a LL 120 kips, and is a 15” x 15” with 4 #8 bars;- The footing is upon a soil with qall 4 ksf with a FS 2.5; use f’c 3000 psi and ƒy 50 ksi.Solution.Step 1. Find the footing dimensions (for service loads).qall QB2 B Q220 7.42 ftqall4Use B 7.5 ft.Step 2. Check the ultimate parameters (that is, the actual soil pressure qo under Qu)The ultimate contact pressure,Qu 1.4 DL 1.7 LL 1.4 (100 ) 1.7 (120 ) 140 204 344 kips qo Qu344 6.1 ksf 10 ksf for qu22B(7.5)OK

Step 3. Compute the allowable shear stress in the concrete.vall 4ϕ f 'c 4(0.85) 3000 186 psi 26.8 ksfStep 4. Find the effective footing depth d (in this case punching shear governs),q q q d 2 4vc o d vc o w ( B2 w2 ) o 04 2 4 2 6.16.16.11515 22d 4)(26.8) d 26.8 2 12 7.5 12 4 04 which yields two solutions: d1 1.20 ft and d2 - 2.46 ft. Choose d 1.20 ft15”B 7.5’15” d

As a check, use the modified equation for d, which is,4d 2 2 ( b c ) d B 2qo 0vowhich yields two solutions: d1 1.28 ft and d2 - 2.52 ft. Choose d 1.28 fta difference of only 7%. Therefore use the largest d 1.28’ 15.4 in. Use d 16 in.It is not necessary to check for wide-beam shear.B 7.5’Step 5. Compute the required flexural steel area As.The unit strip of the cantilever arm L’ is,L' ( B w) 2[7.5' 215]12 3.13 ft15”1’L’

The cantilever moment M is,qo L '2 (6.1)(3.13) 2 (12)M 359 in kips22a where M u ϕ As f y d with ϕ 0.9 for flexure2 As f yAs ( 50 )but a 1.63 As0.85 f 'c b 0.85 ( 3)(12 in.)which yields 16 As 0.81 As 7.97 0 whence As 0.50 in 2 / ftTotal As 7.5 ft (0.5 in 2/ft) 3.75 in 2Check ρ As / bd 0.5 / (12) (16) 0.0026 0.002 (minimum) 0.021 (maximum)In B 90 in (7.5’)use 12 #5 bars (As 3.72 in 2) @ 7.5” ccor 7 #7 bars (As 4.20 in 2) @ 12” ccor 5 #8 bars (As 3.85 in 2) @ 18” ccOK

Step 6. Check the development length for bond Ld (ACI 12.2, 12.6).If Ab the area of each individual bar, and db the diameter of barLd0.04 A f ) [0.04(0.60)(50000)]( 21.9 in 12 in (minimum)byf 'c3000also check with Ld 0.0004 d b fy 0.0004 (0.875 in.) (50,000 psi) 17.5 inches.Step 7. Check the column bearing to determine need of dowels.Area of the column Ag w2 (15 in.)2 225 in 2Effective area of footing A2 (b 4d)2 [15 in 4(16 in)]2 6,240 in 2Checking the contact stress between column and footing,f c ϕ f 'cA2Agwhere ϕ 0.70 andA2 2Agf c (0.7)(3000)(2) 3750 psi 3000 psiNeed dowels.

The actual ultimate contact pressure at ultimate loads is,fc Qu / Ag 344 kips / 225 in 2 1.52 ksi 3.0 ksiOKHowever, dowels are always required, with at least As 0.005 AgAs required 0.005 (225 in 2) 1.125 in 2The diameter φ of the dowels are φ column bars 0.15” (maximum diameter difference)Use 4 #7 bars or (4 x 0.60 in 2) 2.40 in 2 As dowels, whereas column has #8’sCheck diameter difference 1.00” – 0.875” 0.125 0.15”OKStep 8. Check the embedment length Ld for the dowels.Ld 0.2 f y dbf 'cor 0.0004 f y d bor 8 incheschoose the largest of the three0.2(50,000)(0.875) 16 inches 8 inchesOK3000Ld 0.0004(50,000)(0.875) 18 inches 8 inches OKLd

Note however that d 16 in., need a 2” hook to reach 18”, but minimum hook is 6”.Step 9. Sketch the results.N 220 kipsQu 344 kips4 #7 dowels19”3” clearanceB 7.5’7 #7 @ 12” c-c each way

Example #2.Design a rectangular footing to carry a moment induced by wind, with the following data,DL 800 kN, LL 800 kN, M 800 kN-m and qall 200 kN/m2 with FS 1.5. Squarecolumns with c 500 mm, f'c 21 MPa and fy 415 MPa.Step 1. Find the footing dimensions.This time, perform a trial and error selection of B x L.Set BxL B2 and check the increasein the soil pressure due to wind load.B2 Q / qall 1600 kN / 200 kN/m2 8 m2 B 2.82 me M / Q 800/1600 0.5 L 6(0.5) 3 m from (6e/L)If L 3 m try a footing 2.5 m x 4 m.

qavg Q / A 1600 kN / 10 m2 160 kPaqmax Q / BL [ 1 6e/L] (1600 /10)[1 6(0.5)/4] 280 kPaNote that qmax exceeds qavg by 33% , increase size to 2.75 m x 4.5 m.qavg Q / A 1600 / (2.75)(4.5) 130 kPaqmax 130[1 6(0.5)] 217 kPa 200 x 1.33 266 kPaIterate one again and settle with B 3 m and L 5 m.Step 2. Check the ultimate parameters (that is, the actual soil pressure qo under Qu).Pu 1.4DL 1.7LL 1.4(800) 1.7(800) 2480 kNMu 1.4DL 1.7LL 1.4(300) 1.7(500) 1270 kN-me M / Pu 1270 / 2480 0.152qmax (Pu / A){1 6e/L] (2480/15)[1 6(0.152)/5] 266 kPa qu 300 kPa OKqmin (Pu /A)[1 - 6e/L] 2480/15[1 - 6(0.152)/5] 64.5 kPaqavg 2480 / 15 165 kPa qall 200 kPaOK

Step 3. Compute the allowable shear stress in the concrete.The diagonal tension for f'c 21 MPa, vc 1.29 MPaStep 4. Find the effective footing depth d. Using the simplified equation,4d2 2(b c) d – BL qo / vc 04d2 2(0.5m 0.5m) d - (15)(165) / 1290 0 which yields d 0.50 mNow find the depth d for wide beam,from x 0 to x 2.25 – ddv q dxV ? qdx (266 - 40.2x)dx {266x – (40.2)(2)/2} 598 - 266d - 20.1(2.25 - d)2Vc vc /2d 1290 / 2d d 0.60 m use the highest, d 0.60 m.

Step 5. Compute the required longitudinal flexural steel area As.As f y415 Asa 23.3 As0.85 f 'c b 0.85(21)(1)Mu 2.25 0Vdx 2.25 [266 x (40.2)(2)/2]dx 597 kN m0a)2from whence As 28.2 cm 2 / m M u 597 kN m ϕ As f y ( d As 2 0.0515 As 0.0137 0 use 1 7 # 2 5 mm rebarsCheck ? As / bd (0.00282)/(1)(0.6) 0.0047 0.002 (minimum) 0.021 (maximum) OKStep 6. Compute the required transverse flexural steel area As.Using a high average q v(qavg. qmax) v(165 266) 215 kPaMu wl2/2 (q[(b-0.5)/2]2)/2 215(215[(3-0.5)/2]2)/2 168 m-KNand again As(d - a/2) Mu/ fyAs2 - 0.0515As 0.000386 0 As 7.61 cm2/mcheck ? As / bd 7.61 x 10-4 m2 / (1)(0.6) 0.00126 0.002 less than minimum use minimum As 0.002bd 0.002(1)(0.6) 0.0012 m2/m 12cm2/m

Therefore, the longitudinal steel, 28.2 cm2/m x 3m 84.6 cm2 x 1 m2 /10,000cm2 8.46 x 10-3 m2 17 # 25 mm bars placed at 17.6 cm c-c.The transverse steel, 12 cm2/m x 5m 6.0 x 10-3 m2 13 # 25mm bars placed at 38 cm c-c.Step 7. Draft a sketch of the results.

Example #3.Design continuous wall footing for a warehouse building, given DL 3 k/ft, LL 1.2 k/ft, theqall 2 ksf (with FS 2), ?soil 110 pcf, and f'c 3000 psi, fy 60 ksi, ?concrete 150 pcf.Step 1. Assume a footing thickness D 1 ft.By ACI 7.7.3.1, the minimum coverIn soils is 3”. Also assume using#4 bars ( 1/2") d 12" - 3" - 0.5"/2 8.75“

Continuous, or strip, or wall footings, are reinforced transversely to the direction of the wall.The footing acts, in essence, as two small cantilever beams projecting out from under the walland perpendicular to it, in both directions. The reinforcement is placed at the bottom, where thestresses are in tension (flexure). Although the main reinforcement is transverse to the axis ofthe wall, there is also a requirement for longitudinal reinforcement to control temperatureshrinkage and concrete creep.

Step 2. Find the soil pressure qo under ultimate loads.Estimate the footing width B Q / qall 4.2 k / 2 ksf 2.1 ft, assume B 3 feetACI 9.2 required strength U 1.4D 1.7LU 1.4(3) 1.7(1.2) 4.2 2.0 6.2 kipsThe soil pressure at ultimate loads,qo U / (B)(1) 6.2 kips /(3ft)(1ft) 2.1ksf qu 4 ksf

Step 3. Check the shear strength of the concrete.The critical section for shear occurs at a distance d from the face of the wall (ACI 15.5 and11.11.1.2).

the ultimate shear Vu (12" - 8.75") (1ft) ft / 12 in x 2.1 k/ft2 0.57 kips /ft of wallCheck the concrete strength,vu vc 2 v(f'c) bd (0.85)(2)v(3000)(12”)(8.75) 9.8 kips/ft of wall OKSince vu vc can reduce D thickness of the footing, to say 0.85 ft 10” d 10" - 3" - 0.5/2 6.5" 6" (dmin ACI 15.7)rechecking:Vu Vc 2 v(f'c) bd 2(0.85)v(3000) (12in)(6.50in) 7.3 kips/ft of wallOKStep 4. Compute the required transverse flexural steel area As.From ACI 15.4Mu qo l2 / 2 where l 12” (2.1 ksf/2)(1ft2) 1.05 k-ft /ft of wall

but a As fy / 0.85 f'c b (60 ksi) As / (0.85)(3 ksi)(12 in) 1.96 As (inches)b 12" of wall, and Mn As fy (d - a/2) As (60 k/in2)(6.5" - a/2)but Mu Mn 0.9 Mn 1.05 kips-in / in 0.9(60 k/in 2) As (6.5" - 0.98As)53As2 - 351As - 1.05 0 which yields As1 6.6 in 2 per ft. of wallAs2 0.003