PE-19-Combined-Footings

PE-19-Combined-Footings - 19 - Combined Footings Problem...

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19 - Combined Footings Problem #01: A simple combined footing. Problem #02a: Rectangular footing with an eccentric column. Problem #02b: Rectangular footing with an eccentric column. Problem #03: Strap footing (grade beam) for two columns. Problem #04: Design of a combined footing. Problem #05: Optimization via trapezoidal footing. 410
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**Combined Footing-01: A simple combined footing. (Revision: Jun-09) Design a combined footing for the following conditions: f c = 3,750 psi, f y = 20,000 psi and q all = 8.5 ksf. Dead and Live Loads DL = 1050 k + DL = 1210 k = 2260 k LL = 325 + LL = 615 = 940 k Total load = 1375 k + TL= 1825 k = 3200 k Solution: 1825 (24') 13.5' 2.5 16.0' 3200 X ⎛⎞ == ± ⎜⎟ ⎝⎠ = Locate centroid: Length: () ( ) 2 16.0' = 32.0' Clearance: 32.0' - 26.5' = 5.5' Width = all Combined Load 3200 12.0' Length x q 21 8.5 =≈ Soil pressure : ( ) kips ft p = 8.33 kSF 12 ft = 100 Required depth: ( ) 6018 d = = 42" 0.298 12 (3” added for cover + 1” margin for error) d = 46” 411
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() 1175000 = 175 psi 7 12 12 54 8 ⎛⎞ ⎜⎟ ⎝⎠ V = Longitudinal Steel: Bottom steel + A s = 2 6018 90.85 in (1.44)(54) = Σ O = 2 1175000 = 104.26 in 7 280 54 8 Top steel -A s = 2 1010 = 12.3 in 1.44 (57) = Use 52 - #11 bars in 2 layers 412
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9 - #11 bars at column, extend 7 - #11 bars full length Transverse steel: M = (5')(32')(8.33 )(2.5') 3330 ksf k ft = A s = 22 3330 41.6 in x = 23.8 in (1.44)(55) = Σ O = () (5)(32)(8330) = 100 in 7 280 55 8 ⎛⎞ ⎜⎟ ⎝⎠ Use 15- #11 bars in left column at 5.5” c/c. 12- #11 bars in right column at 10” c/c. Web reinforcing: V c =(0.75)(12)(12)(55) 515 k = V’ = V – V c = 1175 – 515 = 660 k 6 sets of #6 stirrups A v = 5.28 in 2 Spacing = s = ' V jd f A u v = 7 (5.28)(16)( )(55) 8 6 660 in = For 12” spacing: V’ = 330 + 515 = 845 kips For 24” spacing: V’ = 165 + 515 = 680 kips 413
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**Combined Footing-02a: Rectangular footing with an eccentric column. (Revision: Jun-09) Find a suitable rectangular combined footing for the conditions shown below. Given '3 6 0 2 . 5 cy a l l f ksi f ksi q ksf == = ( q ult = 10 ksf ) and use ACI 318-08. Column #1 Column #2 12 in x 12 in 16 in x 16 in = 1.33 ft x 1.33 ft 4 # 7 bars 6 # 8 bars D L = 80 kips D L = 120 kips L L = 60 kips L L = 110 kips 140 kips 230 kips Edge of column #1 is at the Property Line (PL); the spacing between the columns is 20 ft on- center. STEP 1: Convert column loads to ultimate; then, convert q all to q ult . ( ) ( ) () () 11 12 140 1.2 80 1.6 60 192 230 1.2 120 1.6 110 320 370 512 u u PP kips + = + = k i p s k i p s Ratio: 1.38 Ultimate U Service S = () 1.38 2.5 3.45 oa l l U qq k S = = s f 415
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STEP 2: Determine footing dimensions L and B 11 2 2 0 o M Rx Px == + where the moments are taken about the left corner. ( ) ( ) 512 192 0.5 320 20.5 12.98 , : 13 , 2 26 x x ft say Thus L x =+ = The rounding to 26 ft will prevent complete closure of the moment diagram, but negligibly. () 12 5.70 26 3.45 uu o PP B Lq + === Use 5 ft 9 in. For construction; B = 5.75 feet. ' 3.45 5.70 19.66 kip q ft STEP 3: Draw shear ( V ) and moment ( M ) diagrams. The column loads are treated as concentrated loads acting at the centers of the columns. The shear and moments diagrams are shown below. 416
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STEP 4: Determine the footing depth T based on one-way (wide-beam) and two-way (punching) shear, as shown in figure below. Note that the punching analysis reflects on a three-sided section for column 1 and four-sided section for column 2. We shall first determine d via a wide-beam analysis, and then check for punching shear. From the shear diagram, the maximum shear is near column 2 at a distance d from the face of column 2.
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This note was uploaded on 09/07/2011 for the course CEG 4012 taught by Professor Staff during the Spring '10 term at FIU.

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PE-19-Combined-Footings - 19 - Combined Footings Problem...

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