lecture26 - Lecture 26 – Footings August 8, 2003 CVEN444...

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Unformatted text preview: Lecture 26 – Footings August 8, 2003 CVEN444 Lecture Goals Footing Classification Footing Design Footing Examples Footings Definition Footings are structural members used to support columns and walls and to transmit and distribute their loads to the soil in such a way that the load bearing capacity of the soil is not exceeded, excessive settlement, differential settlement,or rotation are prevented and adequate safety against overturning or sliding is maintained. Types of Footings Wall footings are used to support structural walls that carry loads for other floors or to support nonstructural walls. Types of Footings Isolated or single footings are used to support single columns. This is one of the most economical types of footings and is used when columns are spaced at relatively long distances. Types of Footings Combined footings usually support two columns, or three columns not in a row. Combined footings are used when tow columns are so close that single footings cannot be used or when one column is located at or near a property line. Types of Footings Cantilever or strap footings consist of two single footings connected with a beam or a strap and support two single columns. This type replaces a combined footing and is more economical. Types of Footings Continuous footings support a row of three or more columns. They have limited width and continue under all columns. Types of Footings Rafted or mat foundation consists of one footing usually placed under the entire building area. They are used, when soil bearing capacity is low, column loads are heavy single footings cannot be used, piles are not used and differential settlement must be reduced. Types of Footings Pile caps are thick slabs used to tie a group of piles together to support and transmit column loads to the piles. Distribution of Soil Pressure When the column load P is applied on the centroid of the footing, a uniform pressure is assumed to develop on the soil surface below the footing area. However the actual distribution of the soil is not uniform, but depends on may factors especially the composition of the soil and degree of flexibility of the footing. Distribution of Soil Pressure Soil pressure distribution in cohesive soil. Soil pressure distribution in cohesionless soil. Design Considerations Footings must be designed to carry the column loads and transmit them to the soil safely while satisfying code limitations. 1. The area of the footing based on the allowable bearing soil capacity 2. Two-way shear or punch out shear. 3. One-way bearing 4. Bending moment and steel reinforcement required Design Considerations Footings must be designed to carry the column loads and transmit them to the soil safely while satisfying code limitations. 1. Bearing capacity of columns at their base 2. Dowel requirements 3. Development length of bars 4. Differential settlement Size of Footings The area of footing can be determined from the actual external loads such that the allowable soil pressure is not exceeded. Area of footing = Total load ( including self - weight ) allowable soil pressure Strength design requirements qu = Pu area of footing Two­Way Shear (Punching Shear) For two-way shear in slabs (& footings) Vc is smallest of 2 + 4 f b d Vc = c0 βc where, β c = b0 = ACI 11-35 long side/short side of column concentrated load or reaction area < 2 length of critical perimeter around the column When β > 2 the allowable Vc is reduced. Design of two­way shear 1. Assume d. 2. Determine b0. b0 = 4(c+d) for square columns where one side = c b0 = 2(c1+d) +2(c2+d) for rectangular columns of sides c1 and c2. Design of two­way shear 3. The shear force Vu acts at a section that has a length b0 = 4(c+d) or 2(c1+d) +2(c2+d) and a depth d; the section is subjected to a vertical downward load Pu and vertical upward pressure qu. Vu = Pu − qu ( c + d ) for square columns Vu = Pu − qu ( c1 + d ) ( c2 + d ) for rectangular columns 2 Design of two­way shear 4. Allowable φVc = 4φ Let Vu=φ Vc d= f c b0 d Vu 4φ f c b0 If d is not close to the assumed d, revise your assumptions Design of one­way shear For footings with bending action in one direction the critical section is located a distance d from face of column φVc = 2φ f c b0 d Design of one­way shear The ultimate shearing force at section m-m can be calculated L c Vu = qu b − − d 2 2 If no shear reinforcement is to be used, then d can be checked Design of one­way shear If no shear reinforcement is to be used, then d can be checked, assuming Vu = φ Vc d= Vu 2φ f c b Flexural Strength and Footing reinforcement The bending moment in each direction of the footing must be checked and the appropriate reinforcement must be provided. As = Mu a φf y d − 2 Flexural Strength and Footing reinforcement Another approach is to calculated Ru = Mu / bd2 and determine the steel percentage required ρ . Determine As then check if assumed a is close to calculated a a= f y As 0. 8 5 f c b Flexural Strength and Footing reinforcement The minimum steel percentage required in flexural members is 200/fy with minimum area and maximum spacing of steel bars in the direction of bending shall be as required for shrinkage temperature reinforcement. Flexural Strength and Footing reinforcement The reinforcement in one-way footings and two-way footings must be distributed across the entire width of the footing. Reinforcement in band width Total reinforcement in short direction where β= = 2 β +1 long side of footing short side of footing Bearing Capacity of Column at Base The loads from the column act on the footing at the base of the column, on an area equal to area of the column cross-section. Compressive forces are transferred to the footing directly by bearing on the concrete. Tensile forces must be resisted by reinforcement, neglecting any contribution by concrete. Bearing Capacity of Column at Base Force acting on the concrete at the base of the column must not exceed the bearing strength of the concrete N1 = φ ( 0.85 f c A1 ) where φ = 0.65 and A1 =bearing area of column Bearing Capacity of Column at Base The value of the bearing strength may be multiplied by a factor A2 / A1 ≤ 2.0 for bearing on footing when the supporting surface is wider on all sides than the loaded area. The modified bearing strength N 2 ≤ φ ( 0.85 f c A1 ) A2 / A1 N 2 ≤ 2φ ( 0.85 f c A1 ) Dowels in Footings A minimum steel ratio ρ = 0.005 of the column section as compared to ρ = 0.01 as minimum reinforcement for the column itself. The number of dowel bars needed is four these may be placed at the four corners of the column. The dowel bars are usually extended into the footing, bent at the ends, and tied to the main footing reinforcement. The dowel diameter shall not exceed the diameter of the longitudinal bars in the column by more than 0.15 in. Development length of the Reinforcing Bars The development length for compression bars was given ld = 0.02 f y d b / fc but not less than 0.003 f y d b ≥ 8 in. Dowel bars must be checked for proper development length. Differential Settlement Footing usually support the following loads: 1. Dead loads from the substructure and superstructure 2. Live load resulting from material or occupancy 3. Weight of material used in back filling 4. Wind loads General Requirements for Footing Design 1. A site investigation is required to determine the chemical and physical properties of the soil. 2. Determine the magnitude and distribution of loads form the superstructure. 3. Establish the criteria and the tolerance for the total and differential settlements of the structure. General Requirements for Footing Design 4. Determine the most suitable and economic type of foundation. 5. Determine the depth of the footings below the ground level and the method of excavation. 6. Establish the allowable bearing pressure to be used in design. General Requirements for Footing Design 7. Determine the pressure distribution beneath the footing based on its width 8. Perform a settlement analysis. Example – Wall Design a plain concrete footing to support a 16 in. thick concrete wall. The load on the wall consist of 16 k/ft dead load (including the self-weight of wall) and a 10 k/ft live load the base of the footing is 4 ft below final grade. fc = 3 ksi and the allowable soil pressure = 5 k/ft2 Example ­ Wall Assume a depth of footing. (1.5 ft or 18 in.) The weight of concrete and the soil are: 1 ft. 3 Wc = γ d = 150 lb/ft *18 in. * 12 in. = 225 lb/ft 2 1 ft. Ws = γ s ds = 100 lb/ft * 4 ft − 18 in. * 12 in. = 250 lb/ft 2 3 Example ­ Wall The effective soil pressure is given as: qeff = qs − Wc − Ws = 5000 lb/ft − 225 lb/ft − 250 lb/ft 2 2 = 4525 lb/ft 2 ⇒ 4.525 k/ft 2 2 Example ­ Wall Calculate the size of the footing for 1-ft of wall: Actual Loads = DL + LL = 16 k/ft + 10 k/ft = 26 k/ft 26 k/ft Width of footing = = 5.75 ft 2 4.525 k/ft ⇒ Use 6 ft Example – Wall Calculate net upward pressure: Actual Loads = 1.2 DL + 1.6 LL = 1.2 ( 16 k/ft ) + 1.6 ( 10 k/ft ) = 35.2 k/ft 35.2 k/ft Net upward pressure qn = = 5.87 k / ft 2 6 ft Example – Wall Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering. d = h − cover − 1.5d b d = 18 in. − 3 in − 1.5 ( 1.0 in ) = 13.5 in. Example – Wall The depth of the footing can be calculated by using one-way shear 1 ft 6 ft 16 in 12 in L c − 13.5 in 1 ft − − −d = 22 2 2 12 in = 1.21 ft L c Vu = qn ( l2 ) − − d 2 2 = 5.87 k/ft 2 ( 1 ft ) ( 1.21 ft ) = 7.1 k Example – Wall The depth of the footing can be calculated by using one-way shear 1000 lb 7.1 k Vu 1k d= = φ 2 fc b 12 in 0.75 2 3000 1 ft 1 ft = 7.2 in. The footing is 13.5 in. > 7.2 in. so it will work. Example – Wall Calculate the bending moment of the footing at the edge of the wall 1 ft 6 ft 16 in 12 in L c = 2.33 ft − = − 2 2 2 2 L c − ( 2.33 ft ) 1 ft L c 2 2 2 M u = qn − b = 5.87 k/ft ( 2.33 ft ) () 2 2 2 2 = 15.98 k-ft Example – Wall Calculate Ru for the footing to find ρ of the footing. 12 in. 15.98 k-ft * Mu 1 ft Ru = 2 = = 0.0877 ksi 2 bd ( 12 in.) * ( 13.5 in ) Example – Wall From Ru for the footing the ρ value can be found. 1.7 Ru Ru = ω f c ( 1 − 0.59ω ) ⇒ ω − 1.7ω + =0 φ fc 2 0.0877 ksi 1.7 − ( 1.7 ) − 4 1.7 0.9 ( 3 ksi ) ω= = 0.03312 2 ρ fy 0.03312 ( 3 ksi ) = 0.03312 ⇒ ρ = = 0.00166 fc 60 ksi 2 Example – Wall Compute the area of steel needed 12 in. As = ρ bd = 0.00166 1 ft 13.5 in.) = 0.27 in 2 ( 1 ft The minimum amount of steel for shrinkage is As = 0.0018 bh = 0.0018 ( 12 in.) ( 18 in.) = 0.389 in 2 The minimum amount of steel for flexure is 200 200 As = bd = 12 in.) ( 13.5 in.) = 0.54 in 2 ⇐ Use ( fy 60000 Example – Wall Use a #7 bar (0.60 in2) Compute the number of bars need As 0.54 in 2 n= = = 0.9 ⇒ Use 1 bars/ ft 2 Ab 0.60 in Example – Wall Check the bearing stress. The bearing strength N1, at the base of the wall, 16 in x 12 in., φ = 0.65 N1 = φ ( 0.85 f c A1 ) = 0.65 ( 0.85 ( 3 ksi ) ( 16 in ) ( 12 in ) ) = 318.2 k The bearing strength, N2, at the top of the footing is N 2 = N1 A2 ≤ 2 N1 A1 Example – Wall A2 = ( 6 ft ) ( 1 ft ) = 6 ft 2 1 ft A1 = 16 in ( 1 ft ) = 1.33 ft 2 12 in. The bearing strength, N2, at the top of the footing is A2 6 ft 2 = = 4.5 2 A1 1.33 ft > 2 ⇒ N 2 = 2 N1 = 2 ( 318.2 k ) = 636.4 k Example – Wall Pu =35.2 k < N1, bearing stress is adequate. The minimum area of dowels is required. 0.005 A1 = 0.005* ( 16 in ) ( 12 in ) = 0.92 in 2 Use minimum number of bars is 2, so use 4 # 8 bars placed at the four corners of the column. Example – Wall The development length of the dowels in compression from ACI Code 12.3.2 for compression. ld = 0.02d b f y fc = 0.02 ( 0.875 in ) ( 60000 psi ) 3000 psi = 19.17 in ⇒ Use 20 in The minimum ld , which has to be greater than 8 in., is ld = 0.0003d b f y = 0.0003 ( 0.875 in.) ( 60000 psi ) = 15.75 in. ≥ 8 in. Example – Wall Therefore, use 4#7 dowels in the corners of the column extending 20 in. into the column and the footing. Note that ld is less than the given d = 15.75 in., which is sufficient development length. Example – Wall The development length, ld for the #7 bars for the reinforcement of the footing. fy f ydb ld = ⇒ ld = d b 20 f c 20 f c ( 60000 psi ) ( 0.875 in ) = 20 3000 psi = 47.9 in There is not adequate development length provided need to design a hook. L c 72 in. 16 in. ld = − cover − = − 3 in. − = 25 in. 2 2 2 2 Example – Square Footing Design a square footing to support a 18 in. square column tied interior column reinforced with 8 #9 bars. The column carries an unfactored axial dead load of 245 k and an axial live load of 200 k. The base of the footing is 4 ft. below final grade and allowable soil pressure is 5 k/ft2 Use fc = 4 ksi and fy = 60 ksi Example – Square Footing Assume a depth of footing. (2 ft or 24 in.) The weight of concrete and the soil are: Wc = γd = 150 lb/ft * 24 in. * 3 1 ft. 12 in. = 300 lb/ft 2 1 ft. = 200 lb/ft 2 Ws = γ s d s = 100 lb/ft * 4 ft − 24 in. * 12 in. 3 Example – Square Footing The effective soil pressure is given as: qeff = qs − Wc − Ws = 5000 lb/ft 2 − 300 lb/ft 2 − 200 lb/ft 2 = 4500 lb/ft 2 ⇒ 4.5 k/ft 2 Example – Square Footing Calculate the size of the footing: Actual Loads = DL + LL = 245 k + 200 k = 445 k Area of footing = 445 k 4.5 k/ft = 98.9 ft 2 2 Side of footing = 9.94 ft ⇒ Use 10 ft Example – Square Footing Calculate net upward pressure: Actual Loads = 1.2 DL + 1.6 LL = 1.2( 245 k ) + 1.6( 200 k ) = 614 k 614 k Net upward pressure qn = = 6.14 k / ft 2 100 ft 2 Example – Square Footing Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering. d = h − cover − 1.5d b d = 24 in. − 3 in − 1.5(1.0 in ) = 19.5 in. Example – Square Footing Calculate perimeter for two-way shear or punch out shear. The column is 18 in. square. bo = 4( c + d ) = 4(18 in. + 19.5 in.) = 150 in. 1 ft = 3.125 ft c + d = (18 in. + 19.5 in.) 12 in Example –Square Footing Calculate the shear Vu 2 Vu = Pu − qn ( c + d ) = 614 k − 6.14 k/ft ( 3.125 ft ) = 554 k 2 The shape parameter β= 10 ft 10 ft =1 2 Example – Square Footing Calculate d value from the shear capacity according to 11.12.2.1 chose the largest value of d 4 Vc = 2 + f c b0 d βc α d Vc = s + 2 f c b0 d bo αs is 40 for interior, 30 for edge and 20 for corner column Vc = 4 f c b0 d Example – Square Footing The depth of the footing can be calculated by using two way shear 1000 lb 554 k Vu 1k d= = φ 4 f c b0 0.75 4 4000 (150 in ) ( = 19.47 in. ) Example – Square Footing The second equation bo is dependent on d so use the assumed values and you will find that d is smaller and α = 40 Vu d= 40d φ b + 2 f c b0 o 1000 lb 554 k 1k = = 10.81 in. 40(19.5 in ) 0.75 + 2 4000 (150 in ) 150 in ( ) Example – Square Footing The depth of the footing can be calculated by using one-way shear 1 ft 18 in L c 10 ft 12 in − 19.5 in 1 ft − −d = − 2 2 2 2 12 in = 2.625 ft L c Vu = qn ( l2 ) − − d 2 2 = 6.14 k/ft 2 (10 ft ) ( 2.625 ft ) = 161.2 k Example – Square Footing The depth of the footing can be calculated by using one-way shear 1000 lb 161.2 k Vu 1k d= = φ 2 fc b 12 in 0.75 2 4000 10 ft 1 ft = 14.2 in. The footing is 19.5 in. > 14.2 in. so it will work. Example –Square Footing Calculate the bending moment of the footing at the edge of the column 1 ft 18 in L c 10 ft 12 in = 4.25 ft − = − 2 2 2 2 L c − ( 4.25 ft ) (10 ft ) L c 2 2 M u = qn − b = 6.14 k/ft ( 4.25 ft ) 2 2 2 2 = 554.5 k - ft Example –Square Footing Calculate Ru for the footing to find ρ of the footing. 12 in. 554.5 k - ft * Mu 1 ft Ru = 2 = bd (120 in ) * (19.5 in ) 2 = 0.1458 ksi Example – Square Footing From Ru for the footing the ρ value can be found. 1.7 Ru Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω + =0 φf c 2 0.1458 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) ω= = 0.04152 2 ρ fy 0.04152( 4 ksi ) = 0.04152 ⇒ ρ = = 0.00277 fc 60 ksi 2 Example – Square Footing Compute the area of steel needed 12 in. As = ρ bd = 0.0027710 ft (19.5 in.) = 6.48 in 2 1 ft The minimum amount of steel for shrinkage is As = 0.0018 bh = 0.0018(120 in.) ( 24 in.) = 5.18 in 2 The minimum amount of steel for flexure is 200 200 (120 in.) (19.5 in.) = 7.8 in 2 As = bd = fy 60000 ⇐ Use Example – Square Footing Use a #7 bar (0.60 in2) Compute the number of bars need As 7.8 in 2 n= = = 13 ⇒ Use 13 bars Ab 0.60 in 2 Determine the spacing between bars s= L − 2 * cover ( n − 1) = 120 in - 2( 3 in ) 12 = 9.5 in Example – Square Footing Check the bearing stress. The bearing strength N1, at the base of the column, 18 in x 18 in., φ = 0.65 ( ) N1 = φ ( 0.85 f c A1 ) = 0.65 0.85( 4 ksi ) (18 in ) = 716 k 2 The bearing strength, N2, at the top of the footing is N 2 = N1 A2 ≤ 2 N1 A1 Example – Square Footing A2 = (10 ft ) = 100 ft 2 2 2 1 ft = 2.25 ft 2 A1 = 18 in 12 in. The bearing strength, N2, at the top of the footing is A2 100 ft 2 = = 6.67 > 2 ⇒ N 2 = 2 N1 = 2( 716 k ) = 1432 k 2 A1 2.25 ft Example – Square Footing Pu =614 k < N1, bearing stress is adequate. The minimum area of dowels is required. 0.005 A1 = 0.005 * (18 in ) = 1.62 in 2 2 Use minimum number of bars is 4, so use 4 # 8 bars placed at the four corners of the column. Example – Square Footing The development length of the dowels in compression from ACI Code 12.3.2 for compression. ld = 0.02d b f y fc = 0.02(1 in ) ( 60000 psi ) 4000 psi = 18.97 in ⇒ Use 19 in The minimum ld , which has to be greater than 8 in., is ld = 0.0003d b f y = 0.0003(1 in ) ( 60000 psi ) = 18 in ≥ 8 in Example – Square Footing Therefore, use 4#8 dowels in the corners of the column extending 19 in. into the column and the footing. Note that ld is less than the given d = 19.5 in., which is sufficient development length. Example – Square Footing The development length, ld for the #7 bars for the reinforcement of the footing. fy f ydb ld ( 60000 psi ) ( 0.875 in ) = 41.5 in = ⇒ ld = = d b 20 f c 20 f c 20 4000 psi There is adequate development length provided. ld = L 2 − cover − c 2 = 120 in 2 − 3 in − 18 in 2 = 48 i n Example – Square Footing Final Design Example – Restricted Footing Design a footing to support a 18 in. square column tied interior column reinforced with 8 #9 bars. The column carries an unfactored axial dead load of 245 k and an axial live load of 200 k. The base of the footing is 4 ft. below final grade and allowable soil pressure is 5 k/ft2 Use fc = 3 ksi and fy = 60 ksi. Limit one side of the footing to 8.5 ft. Example – Restricted Footing Assume a depth of footing. (2 ft or 24 in.) The weight of concrete and the soil are: Wc = γd = 150 lb/ft * 24 in. * 3 1 ft. 12 in. = 300 lb/ft 2 1 ft. = 200 lb/ft 2 Ws = γ s d s = 100 lb/ft * 4 ft − 24 in. * 12 in. 3 Example – Restricted Footing The effective soil pressure is given as: qeff = qs − Wc − Ws = 5000 lb/ft 2 − 300 lb/ft 2 − 200 lb/ft 2 = 4500 lb/ft 2 ⇒ 4.5 k/ft 2 Example – Restricted Footing Calculate the size of the footing: Actual Loads = DL + LL = 245 k + 200 k = 445 k 445 k Area of footing = = 98.9 ft 2 4.5 k/ft 2 98.9 ft 2 Side of footing = = 11.64 ft ⇒ Use 12 ft 8.5 ft Example – Restricted Footing Calculate net upward pressure: Actual Loads = 1.2 DL + 1.6 LL = 1.2( 245 k ) + 1.6( 200 k ) = 614 k 614 k Net upward pressure qn = ( 8.5 ft ) (12 ft ) = 6.02 k / ft 2 Example – Restricted Footing Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering. d = h − cover − 1.5d b d = 24 in. − 3 in − 1.5(1.0 in ) = 19.5 in. Example – Restricted Footing The depth of the footing can be calculated by using the one-way shear (long direction) 1 ft 18 in L c 12 ft 12 in − 19.5 in 1 ft − −d = − 2 2 2 2 12 in = 3.625 ft Vu =135.5 k in short direction L c Vu = qn ( l2 ) − − d 2 2 = 6.02 k/ft 2 ( 8.5 ft ) ( 3.625 ft ) = 185.5 k Example – Restricted Footing The depth of the footing can be calculated by using one-way shear design 1000 lb 185.5 k Vu 1k d= = φ 2 fc b 12 in 0.75 2 4000 8.5 ft 1 ft = 19.2 in. The footing is 19.5 in. > 19.2 in. so it will work. Example – Restricted Footing Calculate perimeter for two-way shear or punch out shear. The column is 18 in. square. bo = 4( c + d ) = 4(18 in. + 19.5 in.) = 150 in. 1 ft = 3.125 ft c + d = (18 in. + 19.5 in.) 12 in Example – Restricted Footing Calculate the shear Vu Vu = Pu − qn ( c + d ) 2 = 614 k − 6.02 k/ft ( 3.125 ft ) = 555.2 k 2 2 The shape parameter β= 12 ft 8.5 ft = 1.41 Example – Restricted Footing Calculate d from the shear capacity according to 11.12.2.1 chose the largest value of d. 4 Vc = 2 + f c b0 d βc α d Vc = s + 2 f c b0 d bo αs is 40 for interior, 30 for edge and 20 for corner column Vc = 4 f c b0 d Example – Restricted Footing The depth of the footing can be calculated for the two way shear 1000 lb 555.2 k Vu 1k d= = 4 4 φ 2 + f c b0 0.75 2 + 4000 (150 in ) β 1.41 = 16.13 in. Example – Restricted Footing The third equation bo is dependent on d so use the assumed values and you will find that d is smaller and α = 40 V d= u 40d φ b + 2 f c b0 o 1000 lb 555.2 k 1k = 40(19.5 in ) 0.75 + 2 4000 (150 in ) 150 in = 10.84 in. ( ) Example – Restricted Footing The depth of the footing can be calculated by using the two way shear 1000 lb 555.2 k Vu 1 k = 19.5 in. d= = φ 4 f c b0 0.75 4 4000 (150 in ) ( ) Example – Restricted Footing Calculate the bending moment of the footing at the edge of the column (long direction) 1 ft 18 in L c 12 ft 12 in = 5.25 ft − = − 2 2 2 2 L c − ( 5.25 ft ) ( 8.5 ft ) L c 2 2 M u = qn − b = 6.02 k/ft ( 5.25 ft ) 2 2 2 2 = 705.2 k - ft Example – Restricted Footing Calculate Ru for the footing to find ρ of the footing. 12 in. 705.2 k - ft * Mu 1 ft Ru = 2 = bd 12 in 2 8.5 ft * (19.5 in ) 1 ft = 0.2182 ksi Example – Restricted Footing Use the Ru for the footing to find ρ . 1.7 Ru Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω + =0 φf c 2 0.2182 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) ω= = 0.06294 2 ρ fy 0.06294( 4 ksi ) = 0.06294 ⇒ ρ = = 0.004196 fc 60 ksi 2 Example – Restricted Footing Compute the amount of steel needed 12 in. As = ρ bd = 0.004196 8.5 ft (19.5 in.) = 8.35 in 2 1 ft The minimum amount of steel for shrinkage is As = 0.0018 bh = 0.0018(102 in.) ( 24 in.) = 4.41 in 2 The minimum amount of steel for flexure is 200 200 (102 in.) (19.5 in.) = 6.63 in 2 As = bd = fy 60000 Example – Restricted Footing Use As =8.36 in2 with #8 bars (0.79 in2). Compute the number of bars need As 9.33 in 2 n= = = 11.8 ⇒ Use 12 bars Ab 0.79 in 2 Determine the spacing between bars s= L − 2 * cover ( n − 1) = 102 in - 2( 3 in ) 11 = 8.73 in Example – Restricted Footing Calculate the bending moment of the footing at the edge of the column for short length 1 ft 18 in L c 8.5 ft 12 in = 3.5 ft − = − 2 2 2 2 L c − ( 3.5 ft ) (12 ft ) L c 2 2 M u = qn − b = 6.02 k/ft ( 3.5 ft ) 2 2 2 2 = 442.5 k - ft Example – Restricted Footing Calculate Ru for the footing to find ρ of the footing. 12 in. 442.5 k - ft * Mu 1 ft = 0.0970 ksi Ru = 2 = bd 12 in 2 12 ft * (19.5 in ) 1 ft Example – Restricted Footing Use Ru for the footing to find ρ . 1.7 Ru Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω + =0 φf c 2 0.0970 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) ω= = 0.0274 2 ρ fy 0.0274( 4 ksi ) = 0.0274 ⇒ ρ = = 0.00183 fc 60 ksi 2 Example – Restricted Footing Compute the amount of steel needed 12 in. As = ρ bd = 0.0018312 ft (19.5 in.) = 5.12 in 2 1 ft The minimum amount of steel for shrinkage is As = 0.0018 bh = 0.0018(144 in.) ( 24 in.) = 6.22 in 2 The minimum amount of steel for flexure is 200 200 (144 in.) (19.5 in.) = 9.36 in 2 As = bd = fy 60000 Example – Restricted Footing Use As =9.36 in2 with #6 bar (0.44 in2) Compute the number of bars need As 9.36 in 2 n= = = 21.3 ⇒ Use 22 bars Ab 0.44 in 2 Calculate the reinforcement bandwidth Reinforcement in bandwidth = 2 = 2 = 0.83 Total reinforcement β + 1 1.41 + 1 Example – Restricted Footing The number of bars in the 8.5 ft band is 0.83(22)=19 bars . Total # bars - band bars outside # bar = 2 22 − 19 = = 1.5 ⇒ Use 2 bars 2 So place 19 bars in 8.5 ft section and 2 bars in each in (12ft -8.5ft)/2 =1.75 ft of the band. Example – Restricted Footing Determine the spacing between bars for the band of 8.5 ft s= L ( n − 1) = 102 in 18 = 5.67 in Determine the spacing between bars outside the band s= L − cover n = 21 in - 3in 2 = 9 in Example – Restricted Footing Check the bearing stress. The bearing strength N1, at the base of the column, 18 in x 18 in., φ = 0.65 ( ) N1 = φ ( 0.85 f c A1 ) = 0.65 0.85( 4 ksi ) (18 in ) = 716 k 2 The bearing strength, N2, at the top of the footing is N 2 = N1 A2 ≤ 2 N1 A1 Example – Restricted Footing A2 = ( 8.5 ft ) (12 ft ) = 102 ft 2 2 1 ft 2 A1 = 18 in = 2.25 ft 12 in. The bearing strength, N2, at the top of the footing is A2 102 ft 2 = = 6.74 > 2 ⇒ N 2 = 2 N1 = 2( 716 k ) = 1432 k 2 A1 2.25 ft Example – Restricted Footing Pu =614 k < N1, bearing stress is adequate. The minimum area of dowels is required. 0.005 A1 = 0.005 * (18 in ) = 1.62 in 2 2 Use minimum number of bars is 4, so use 4 # 8 bars placed at the four corners of the column. Example – Restricted Footing The development length of the dowels in compression from ACI Code 12.3.2 for compression. ld = 0.02d b f y fc = 0.02(1 in ) ( 60000 psi ) 4000 psi = 18.97 in ⇒ Use 19 in The minimum ld , which has to be greater than 8 in., is ld = 0.0003d b f y = 0.0003(1 in ) ( 60000 psi ) = 18 in ≥ 8 in Example – Restricted Footing Therefore, use 4#8 dowels in the corners of the column extending 19 in. into the column and the footing. Note that ld is less than the given d = 19.5 in., which is sufficient development length. Example – Restricted Footing The development length, ld for the #8 bars ld db = fy 20 f c ⇒ ld = f ydb 20 f c ( 60000 psi ) (1.0 in ) = 47.4 in = 20 4000 psi There is adequate development length provided. ld = L 2 − cover − c 2 = 144 in 2 − 3 in − 18 in 2 = 60 in Example – Restricted Footing The development length, ld for the #6 bars ld db = fy 25 f c ⇒ ld = f ydb 25 f c ( 60000 psi ) ( 0.75 in ) = 28.5 in = 25 4000 psi There is adequate development length provided. ld = L 2 − cover − c 2 = 102 in 2 − 3 in − 18 in 2 = 39 in Example – Restricted Footing ­ Final design 23 #6 12 #8 Example – Multi­Column Footing Design a rectangular footing to support two square columns. The exterior column (I) has a section 16 x 16 in., which carries DL of 180 k and a LL of 120 k. The interior column (II) has a section of 20 x 20 in., which carries a DL of 250 k and a LL of 140 k. The base of the footing is 5 ft. below final grade and allowable soil pressure is 5 k/ft2 Use fc = 4 ksi and fy = 60 ksi The external column is located 2 ft from the property line. Example – Multi­Column Footing Determine the location of an equivalent point and its location select the datum at column I ∑x F x= ∑F i i i = 16 ft ( 250 k + 140 k ) + 0 ft (180 k + 120 k ) ( 250 k + 140 k ) + (180 k + 120 k ) = 9.04 ft. ⇒ Use 9 ft. Extend the footing up to the property line, so the length is l = 9 ft + 2 ft. = 11 ft. So the length of the footing is 2(11 ft.) = 22 ft. Example – Multi­Column Footing Assume a depth of footing. (36 in.) The weight of concrete and the soil are: Wc = γd = 150 lb/ft * 36 in. * 3 1 ft. 12 in. = 450 lb/ft 2 1 ft. 5 ft − 36 in. * = 200 lb/ft 2 Ws = γ s d s = 100 lb/ft * 12 in. 3 Example – Multi­Column Footing The effective soil pressure is given as: qeff = qs − Wc − Ws = 5000 lb/ft 2 − 450 lb/ft 2 − 200 lb/ft 2 = 4350 lb/ft 2 ⇒ 4.35 k/ft 2 Example – Multi­Column Footing Calculate the size of the footing: Actual Loads = DL + LL = 250 k + 140 k = 390 k Actual Loads = DL + LL = 180 k + 120 k = 300 k Total Loads = AL1 + AL2 = 390 k + 300 k = 690 k Area of footing = Side of footing = 690 k 4.35 k/ft 2 158.6 ft 2 22 ft = 158.6 ft 2 = 7.21 ft ⇒ Use 7.5 ft Example – Multi­Column Footing Calculate net upward pressure: Actual Loads = 1.4 DL + 1.7 LL = 1.4(180 k ) + 1.7(120 k ) + 1.4( 250 k ) + 1.7(140 k ) = 456 k + 588 k = 1044 k Net upward pressure qn = 1044 k ( 22 ft ) ( 7.5 ft ) = 6.33 k / ft 2 Example – Multi­Column Footing Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering. d = h − cover − 1.5d b d = 36 in. − 3 in − 1.5(1.0 in ) = 31.5 in. Example – Multi­Column Footing Compute the shear and bending moment diagrams. Shear Forces V ( x ) = qb ( x ) − w 400 = 6.32 k/ft 2 ( 7.5 ft ) x − w 358.7 k 300 = 47.454 k/ft ( x ) − w Force (kips) 200 63.3 k 100 0 -100 0 2 4 6 8 10 12 16 18 -150.3 k -200 -300 14 -329.5 k -400 location (ft) 20 22 The columns are considered point loads but shear values are taken at each side of the column. Example – Multi­Column Footing The location of the maximum moment is 1 ft 1 ft − 10 in = 14.5 ft 16 ft − 8 in 12 in 12 in x= 329.5 k 329.5 k + 358.7 k (14.5 ft ) = 6.9 ft Example – Multi­Column Footing Compute the shear and bending moment diagrams. x2 M ( x ) = qb − wi ( x − xi ) 2 Bending Moment 400 Bending Moment (k-ft) 200 42.2 k-ft 0 -200 0 2 4 6 8 10 12 14 16 18 20 -400 -600 -800 -1000 -1200 -1278.9 k-ft @ 9.61 ft -1400 Location (ft) x2 = 6.32 k/ft ( 7.5 ft ) − wi ( x − xi ) 2 x2 = 47.454 k/ft − wi ( x − xi ) 2 2 249.9 k-ft 22 The columns are considered point loads but moments are taken at each side of the column. It will not balance because center is at 9.04 ft Example – Multi­Column Footing The maximum shear force occurs at the edge of the 20 in. column. So maximum shear is measured at distance d from the column. 1 ft Vmax − q ( d ) = 358.7 k − 47.454 k/ft 31.5 in 12 in = 234.1 k Example – Multi­Column Footing The depth of the footing can be calculated by using one-way shear 1000 lb 234.1 k Vu 1k d= = = 24.2 in. 12 in φ 2 fc b 0.85 2 4000 7.5 ft 1 ft The footing is 31.5 in. > 24.2 in. so it will work. Example – Multi­Column Footing Calculate perimeter for two-way shear or punch out shear. The column is 20 in. square. bo = 4( c + d ) = 4( 20 in. + 31.5 in.) = 206 in. 1 ft = 4.292 ft c + d = ( 20 in. + 31.5 in.) 12 in Example – Multi­Column Footing Calculate the shear Vu Vu = Pu − qn ( c + d ) 2 = 588 k − 6.70 k/ft ( 4.292 ft ) = 464.6 k 2 2 The other column will not be critical, Pu = 456 k for the 16 in. column Example – Multi­Column Footing The depth of the footing can be calculated by using two way shear d= Vu φ 4 f c b0 = 1000 lb 464.6 k 1k ( 0.85 4 4000 ( 206 in ) ) = 10.5 in. Example – Multi­Column Footing Calculate Ru for the footing to find ρ of the footing. 12 in. 1278.9 k - ft * Mu 1 ft = 0.1719 ksi Ru = = ( 90 in ) * ( 31.5 in ) 2 bd 2 Example – Multi­Column Footing From Ru for the footing the ρ value can be found. Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω + 2 1.7 Ru φf c 0.1719 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) =0 2 ω= ρ fy fc 2 = 0.04917 ⇒ ρ = 0.04917( 4 ksi ) 60 ksi = 0.04917 = 0.00328 Example – Multi­Column Footing Compute the area of steel needed 12 in. ( 31.5 in.) = 9.29 in 2 As = ρ bd = 0.00277 7.5 ft 1 ft The minimum amount of steel for shrinkage is As = 0.0018 bh = 0.0018( 90 in.) ( 36 in.) = 5.8 3 in 2 The minimum amount of steel for flexure is 200 200 ( 90 in.) ( 31.5 in.) = 9.45 in 2 As = bd = fy 60000 ⇐ Use Example – Multi­Column Footing Use a #9 bar (1.00 in2) Compute the number of bars needed A 9.45 in 2 n= s = = 9.45 ⇒ Use 10 bars Ab 1.0 in 2 Determine the spacing between bars s= L − 2 * cover ( n − 1) = 90 in - 2( 3 in ) 9 = 9.33 in Example – Multi­Column Footing The minimum amount is steel is going to be due to the flexural restrictions. So below the columns with positive moment, the reinforcement will be 10 # 9 bars running longitudinally. The development length will have to be calculated. Example – Multi­Column Footing The development length, ld for the #7 bars for the reinforcement of the footing. fy f ydb ld ( 60000 psi ) (1.128 in ) = 53.5 in = ⇒ ld = = d b 20 f c 20 f c 20 4000 psi The bars have more than 12-in. of concrete below them, therefore ld = 1.3 ld. ld = 1.3( 53.5 in ) = 69.6 in ⇒ Use 70 in. Example – Multi­Column Footing To determine the reinforcement in the short direction. The bandwidth of the two columns must be determined for the 16 in. column. 2 ft − 16 in 1 ft + 31.5 in 1 ft = 5.3 ft ⇒ Use 5.5 ft Band = 16 in + 2 12 in. 12 in. Compute the moment at the edge qnet = 456 k 7.5 ft = 60.8 k/ft L= 7.5 ft 2 1 ft = 3.08 ft − 8 in 12 in Example – Multi­Column Footing The bending moment will be M u = qnet l2 2 ( 3.08 ft ) 2 = ( 60.8 k/ft ) 2 Compute the Ru = 289.0 k - ft 12 in. 289 k - ft * Mu 1 ft Ru = = = 0.053 ksi 12 in bd 2 5.5 ft * ( 31.5 in ) 2 1 ft Example – Multi­Column Footing From Ru for the footing the ρ value can be found. Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω + 2 1.7 Ru φf c 0.053 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) =0 2 ω= ρ fy fc 2 = 0.01484 ⇒ ρ = 0.04917( 4 ksi ) 60 ksi = 0.01484 = 0.001 Example – Multi­Column Footing Compute the area of steel needed 12 in. ( 31.5 in.) = 2.08 in 2 As = ρ bd = 0.001 5.5 ft 1 ft The minimum amount of steel for shrinkage is As = 0.0018 bh = 0.0018( 66 in.) ( 36 in.) = 4.28 in 2 The minimum amount of steel for flexure is 200 200 ( 66 in.) ( 31.5 in.) = 6.93 in 2 As = bd = fy 60000 ⇐ Use Example – Multi­Column Footing Use a #9 bar (1.00 in2) Compute the number of bars needed A 6.93 in 2 n= s = = 6.93 ⇒ Use 7 bars Ab 1.0 in 2 Determine the spacing between bars s= L − cover ( n − 1) = 66 in - ( 3 in ) 6 = 10.5 in Example – Multi­Column Footing To determine the reinforcement in the short direction. The 20-in. interior column extends beyond 4 ft from the center therefore the band is 7.5 ft x 7.5 ft. Compute the moment at the edge qnet = 588 k 7.5 ft = 78.4 k/ft L= 7.5 ft 2 1 ft = 2.92 ft − 10 in 12 in Example – Multi­Column Footing The bending moment will be M u = qnet l2 2 ( 2.92 ft ) 2 = ( 78.4 k/ft ) 2 Compute the Ru = 334.3 k - ft 12 in. 334.3 k - ft * Mu 1 ft Ru = = = 0.045 ksi 12 in bd 2 7.5 ft * ( 31.5 in ) 2 1 ft Example – Multi­Column Footing From Ru for the footing the ρ value can be found. Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω + 2 1.7 Ru φf c 0.045 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) =0 2 ω= ρ fy fc 2 = 0.01257 ⇒ ρ = 0.01257( 4 ksi ) 60 ksi = 0.01257 = 0.00084 Example – Multi­Column Footing Compute the area of steel needed 12 in. ( 31.5 in.) = 2.38 in 2 As = ρ bd = 0.00084 7.5 ft 1 ft The minimum amount of steel for shrinkage is As = 0.0018 bh = 0.0018( 90 in.) ( 36 in.) = 5.83 in 2 The minimum amount of steel for flexure is 200 200 ( 90 in.) ( 31.5 in.) = 9.45 in 2 As = bd = fy 60000 ⇐ Use Example – Multi­Column Footing Check the bearing stress. The bearing strength N1, at the base of the column, 16 in x 16 in., φ = 0.7 ( ) N1 = φ ( 0.85 f c A1 ) = 0.7 0.85( 4 ksi ) (16 in ) = 609 k 2 The bearing strength, N2, at the top of the footing is N 2 = N1 A2 ≤ 2 N1 A1 Example – Multi­Column Footing 2 A2 = ( 5.5 ft ) = 30.25 ft 2 2 1 ft = 1.78 ft 2 A1 = 16 in 12 in. The bearing strength, N2, at the top of the footing is A2 = A1 30.25 ft 2 1.78 ft 2 = 4.125 > 2 ⇒ N 2 = 2 N1 = 2( 609 k ) = 1218 k Example – Multi­Column Footing Pu =456 k < N1, bearing stress is adequate. The minimum area of dowels is required. 0.005 A1 = 0.005 * (16 in ) = 1.28 in 2 2 Use minimum number of bars is 4, so use 4 # 7 bars placed at the four corners of the column. Note if the Pu > N1 the area of steel will be As ( Pu − N1 ) = fy As long as the area of steel is greater than the minimum amount. Example – Multi­Column Footing The development length of the dowels in compression from ACI Code 12.3.2 for compression. ld = 0.02d b f y fc = 0.02( 0.875 in ) ( 60000 psi ) 4000 psi = 16.6 in ⇒ Use 17 in The minimum ld , which has to be greater than 8 in., is ld = 0.0003d b f y = 0.0003( 0.875 in ) ( 60000 psi ) = 15.75 in ≥ 8 in Example – Multi­Column Footing Therefore, use 4#7 dowels in the corners of the column extending 17 in. into the column and the footing. Note that ld is less than the given d = 31.5 in., which is sufficient development length. Example – Multi­Column Footing Use a #9 bar (1.00 in2) Compute the number of bars need As 9.45 in 2 n= = = 9.45 ⇒ Use 10 bars Ab 1.0 in 2 Determine the spacing between bars s= L − cover ( n − 1) = 90 in - ( 3 in ) 9 = 9.67 in Example – Multi­Column Footing Check the bearing stress. The bearing strength N1, at the base of the column, 20 in x 20 in., φ = 0.7 ( ) N1 = φ ( 0.85 f c A1 ) = 0.7 0.85( 4 ksi ) ( 20 in ) = 952 k 2 The bearing strength, N2, at the top of the footing is N 2 = N1 A2 ≤ 2 N1 A1 Example – Multi­Column Footing 2 A2 = ( 7.5 ft ) = 56.25 ft 2 2 1 ft = 2.78 ft 2 A1 = 20 in 12 in. The bearing strength, N2, at the top of the footing is A2 A1 = 56.25 ft 2 2.78 ft 2 = 4.5 > 2 ⇒ N 2 = 2 N1 = 2( 952 k ) = 1904 k Example – Multi­Column Footing Pu =588 k < N1, bearing stress is adequate. The minimum area of dowels is required. 0.005 A1 = 0.005 * ( 20 in ) = 2.0 in 2 2 Use minimum number of bars is 4, so use 4 # 8 bars placed at the four corners of the column. Example – Multi­Column Footing The development length of the dowels in compression from ACI Code 12.3.2 for compression. ld = 0.02d b f y fc = 0.02(1 in ) ( 60000 psi ) 4000 psi = 18.7 in ⇒ Use 19 in The minimum ld , which has to be greater than 8 in., is ld = 0.0003d b f y = 0.0003(1 in ) ( 60000 psi ) = 18 in ≥ 8 in Example – Multi­Column Footing Therefore, use 4#8 dowels in the corners of the column extending 19 in. into the column and the footing. Note that ld is less than the given d = 31.5 in., which is sufficient development length. Example – Settlement Determine the footing areas required for equal settlement (balanced footing design) if the usual live load is 25% for all footings. The footings are subjected to dead loads and live loads as indicated by the table. The net soil pressure is 6 ksf. Footing Number 1 Dead Load (kips) Live Load (kips) 2 3 4 5 120 180 140 190 210 150 220 200 170 240 Example – Settlement Find the ratio of the live load to dead load, the largest ratio will control the settlement. Column 3 has the largest ratio. Example – Settlement Compute the usual loading for the footing, DL + 0.25LL Column 3 has the largest ratio. Example – Settlement Determine the need area for the footing with the largest LL/DL ratio. A= DL + LL qnet = 140 k + 200 k 6 k/ft 2 = 56.67 ft 2 The usual net soil pressure acting on the footing is qnet = DL + ( % live load ) ( LL ) A = 140 k + 0.25( 200 k ) 56.67 ft 2 = 3.353 k/ft 2 Example – Settlement Use the qnet (3.353 k/ft2) to determine need area for each of the other footings to have the same settlement. Usual Loading 157.5 k Area = = = 46.97 ft 2 q net 3.353 k/ft 2 Compute the qnet for each of the footings q net Total Load (120 k ) + (150 k ) = = = 5.75 k/ft 2 New Area 46.97 ft 2 Example – Settlement Use the qnet (3.353 k/ft2) to determine need area for each of the other footings to have the same settlement. Footing Number 1 3 4 5 Dead Load (kips) Live Load (kips) 120 150 180 220 140 200 190 170 210 240 Ratio (LL/DL) 1.25 1.22 1.43 0.89 1.14 Usual Loading (kips) 157.5 235.0 190.0 232.5 270.0 Standard Area (ft 2) 45.00 66.67 56.67 60.00 75.00 New Area(ft2) 2 46.97 70.09 56.67 69.34 80.53 New qnet 5.75 5.71 6.00 5.19 5.59 Example – Combined Loading A 12-in. x 24 in. column of an unsymmetrical shed is subjected to an axial load PD of 220 k and MD = 180 k-ft due to dead load and an an axial load PL = 165 k and a moment ML= 140 k-ft due to live load. The base of the footing is 5 ft below final grade, and the allowable soil bearing pressure is 5 k/ft2. Design the footing using fc = 4 ksi and fy = 60 ksi Example – Combined Loading Find the combined actual loads, P0 and M0 P0 = PDL + PLL = 220 k + 165 k = 385 k M 0 = M DL + M LL = 180 k - ft + 140 k - ft = 320 k - ft Determine the eccentricity of the footing 12 in 320 k - ft M0 1 ft = 9.97 in ⇒ Use 10 in. e= = P0 385 k Example – Combined Loading Assume a depth of footing, 24 in. The weight of concrete and the soil are: Wc = γd = 150 lb/ft * 24 in. * 3 1 ft. 12 in. = 300 lb/ft 2 1 ft. 5 ft − 24 in. * = 300 lb/ft 2 Ws = γ s d s = 100 lb/ft * 12 in. 3 Example – Combined Loading The effective soil pressure is given as: qeff = qs − Wc − Ws = 5000 lb/ft 2 − 300 lb/ft 2 − 300 lb/ft 2 = 4400 lb/ft 2 ⇒ 4.4 k/ft 2 Example – Combined Loading Calculate the size of the footing: Actual Loads = DL + LL = 385 k Area of footing = 385 k 4.4 k/ft 2 = 87.5 ft 2 Compute the sizes of the footing if width is 9 ft. Side of footing = 87.5 ft 2 9 ft = 9.72 ft ⇒ Use 10 ft Example – Combined Loading Use the long section and place the column 10 in. off-center for the 10 ft segment Example – Combined Loading Calculate net upward pressure: Actual Loads = 1.2 DL + 1.6 LL = 1.2( 220 k ) + 1.6(165 k ) = 528.0 k 528.0 k Net upward pressure qn = ( 9 ft ) (10 ft ) = 5.87 k / ft 2 Example – Combined Loading Calculate the depth of the reinforcement use # 8 bars with a crisscrossing layering. d = h − cover − 1.5d b d = 24 in. − 3 in − 1.5(1.0 in ) = 19.5 in. Example – Combined Loading The depth of the footing can be calculated by using the one-way shear (long direction) 1 ft 24 in 10 ft L c 12 in − 19.5 in 1 ft + 10 in 1 ft − −d+e = − 2 2 2 2 12 in 12 in = 3.208 ft Vu =139.4 k in short direction L c Vu = qn ( l2 ) − − d + e 2 2 = 5.87 k/ft 2 ( 9 ft ) ( 3.208 ft ) = 169.4 k Example – Combined Loading The depth of the footing can be calculated by using one-way shear design 1000 lb 169.4 k Vu 1k d= = = 16.53 in. φ 2 fc b 12 in 0.75 2 4000 9 ft 1 ft The footing is 19.5 in. > 16.53 in. so it will work. Example – Combined Loading Calculate perimeter for two-way shear or punch out shear. The column is 12 in. x 24 in. bo = 2( c1 + d ) + 2( c2 + d ) = 2(12 in. + 19.5 in.) + 2( 24 in. + 19.5 in.) = 150 in. 1 ft = 2.625 ft c1 + d = (12 in. + 19.5 in.) 12 in 1 ft = 3.625 ft c2 + d = ( 24 in. + 19.5 in.) 12 in Example – Combined Loading Calculate the shear Vu (c + d )2 Vu = Pu − qn = 528.0 k − 5.87 k/ft 2 ( 2.625 ft ) ( 3.625 ft ) = 472.2 k The shape parameter β= 10 ft 9 ft = 1.11 Example – Combined Loading Calculate d from the shear capacity according to 11.12.2.1 chose the largest value of d. 4 Vc = 2 + f c b0 d βc α d Vc = s + 2 f c b0 d bo αs is 40 for interior, 30 for edge and 20 for corner column Vc = 4 f c b0 d Example – Combined Loading The depth of the footing can be calculated for the two way shear 1000 lb 472.2 k Vu 1k d= = 4 4 φ 2 + f c b0 0.75 2 + 4000 (150 in ) β 1.11 = 11.84 in. Example – Combined Loading The third equation bo is dependent on d so use the assumed values and you will find that d is smaller and α = 40 d= Vu 40d φ + 2 f c b0 b o 1000 lb 472.2 k 1k = = 9.22 in. 40(19.5 in ) 0.75 + 2 4000 (150 in ) 150 in ( ) Example – Combined Loading The depth of the footing can be calculated by using the two way shear 1000 lb 472.2 k Vu 1k d= = φ 4 f c b0 0.75 4 4000 (150 in ) ( = 16.59 in. ) Example – Combined Loading Calculate the bending moment of the footing at the edge of the column (long direction) 1 ft 24 in 10 ft L c 12 in + 10 in 1 ft = 4.83 ft − +e = − 2 2 2 2 12 in L c − + e L c 2 2 M u = qn − + e b 2 2 2 ( 4.83 ft ) ( 9 ft ) = 616.2 k - ft = 5.87 k/ft ( 4.83 ft ) 2 Example – Combined Loading Calculate Ru for the footing to find ρ of the footing. 12 in. 616.2 k - ft * Mu 1 ft = 0.1801 ksi Ru = 2 = bd 12 in 2 9 ft * (19.5 in ) 1 ft Example – Combined Loading Use the Ru for the footing to find ρ . 1.7 Ru Ru = ωf c (1 − 0.59ω ) ⇒ ω − 1.7ω + =0 φf c 2 0.1801 ksi 1.7 − (1.7 ) − 41.7 0.9( 4 ksi ) ω= = 0.05158 2 ρ fy 0.05158( 4 ksi ) = 0.05158 ⇒ ρ = = 0.00344 fc 60 ksi 2 Example – Combined Loading Compute the amount of steel needed 12 in. As = ρ bd = 0.00344 9 ft (19.5 in.) = 7.24 in 2 1 ft The minimum amount of steel for shrinkage is As = 0.0018 bh = 0.0018(108 in.) ( 24 in.) = 4.67 in 2 The minimum amount of steel for flexure is 200 200 (108 in.) (19.5 in.) = 7.02 in 2 As = bd = fy 60000 Example – Combined Loading Use As =8.36 in2 with #8 bars (0.79 in2). Compute the number of bars need As 8.1 in 2 n= = = 10.25 ⇒ Use 11 bars Ab 0.79 in 2 Determine the spacing between bars s= L − 2 * cover ( n − 1) = 108 in - 2( 3 in ) 10 = 10.2 in Example – Combined Loading Calculate the bending moment of the footing at the edge of the column for short length 1 ft 12 in L c 9 ft 12 in = 4 ft − = − 2 2 2 2 L c − ( 4 ft ) (10 ft ) L c 2 2 M u = qn − b = 5.87 k/ft ( 4 ft ) 2 2 2 2 = 469.6 k - ft Example – Combined Loading Calculate Ru for the footing to find ρ of the footing. 12 in. 469.6 k - ft * Mu 1 ft Ru = 2 = bd 12 in 2 10 ft * (19.5 in ) 1 ft = 0.1235 ksi Example – Combined Loading Use Ru for the footing to find ρ . Ru = ωf c (1 − 0.59ω ) ⇒ ω 2 − 1.7ω + 1.7 − 1.7 Ru =0 φf c (1.7 ) 2 − 41.7 0.1235 ksi 4 ksi ω= = 0.03503 2 ρ fy 0.03503( 4 ksi ) = 0.03503 ⇒ ρ = = 0.00234 fc 60 ksi Example – Combined Loading Compute the amount of steel needed 12 in. As = ρ bd = 0.0023410 ft (19.5 in.) = 5.46 in 2 1 ft The minimum amount of steel for shrinkage is As = 0.0018 bh = 0.0018(120 in.) ( 24 in.) = 5.18 in 2 The minimum amount of steel for flexure is 200 200 (120 in.) (19.5 in.) = 7.80 in 2 As = bd = fy 60000 Example – Combined Loading Use As =9.36 in2 with #6 bar (0.44 in2) Compute the number of bars need As 7.80 in 2 n= = = 17.7 ⇒ Use 18 bars Ab 0.44 in 2 Calculate the reinforcement bandwidth Reinforcement in bandwidth = 2 = 2 = 0.947 Total reinforcement β + 1 1.11 + 1 Example – Combined Loading The number of bars in the 9 ft band is 0.947(18)=17 bars . Total # bars - band bars outside # bar = 2 18 − 17 = = 0.5 ⇒ Use 1 bars 2 So place 17 bars in 9 ft section and 1 bars in each in (10ft - 9ft)/2 =0.5 ft of the band. Example – Combined Loading Determine the spacing between bars for the band of 9 ft s= L ( n − 1) = 108 in 16 = 6.75 in Determine the spacing between bars outside the band s= L − cover n = 6 in - 3in 1 = 3 in Example – Combined Loading Check the bearing stress. The bearing strength N1, at the base of the column, 12 in x 24 in., φ = 0.65 N1 = φ ( 0.85 f c A1 ) = 0.65( 0.85( 4 ksi ) (12 in ) ( 24 in ) ) = 636.5 k The bearing strength, N2, at the top of the footing is N 2 = N1 A2 ≤ 2 N1 A1 Example – Combined Loading A2 = 9 ft (10 ft ) = 90 ft 2 1 ft 1 ft 24 in = 2 ft 2 A1 = 12 in 12 in. 12 in. The bearing strength, N2, at the top of the footing is A2 90 ft 2 = = 6.71 > 2 ⇒ N 2 = 2 N1 = 2( 636.5 k ) = 1273 k 2 A1 2 ft Example – Combined Loading Pu =628 k < N1, bearing stress is adequate. The minimum area of dowels is required. 0.005 A1 = 0.005 * (12 in ) ( 24 in ) = 1.44 in 2 Use minimum number of bars is 4, so use 4 # 8 bars placed at the four corners of the column. Example – Combined Loading The development length of the dowels in compression from ACI Code 12.3.2 for compression. ld = 0.02d b f y fc = 0.02(1 in ) ( 60000 psi ) 4000 psi = 18.97 in ⇒ Use 19 in The minimum ld , which has to be greater than 8 in., is ld = 0.0003d b f y = 0.0003(1 in ) ( 60000 psi ) = 18 in ≥ 8 in Example – Combined Loading Therefore, use 4#8 dowels in the corners of the column extending 19 in. into the column and the footing. Note that ld is less than the given d = 19.5 in., which is sufficient development length. Example – Combined Loading The development length, ld for the #8 bars ld db = fy 20 f c ⇒ ld = f ydb 20 f c ( 60000 psi ) (1.0 in ) = 47.4 in = 20 4000 psi There is adequate development length provided. ld = L 2 − cover − c 2 = 144 in 2 − 3 in − 18 in 2 = 60 in Example – Combined Loading The development length, ld for the #6 bars ld db = fy 25 f c ⇒ ld = f ydb 25 f c ( 60000 psi ) ( 0.75 in ) = 28.5 in = 25 4000 psi There is adequate development length provided. ld = L 2 − cover − c 2 = 102 in 2 − 3 in − 18 in 2 = 3 9 in ...
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