FOOTING DESIGN EXAMPLE

County: Any Design: BRG Date: 10/007 Hwy: Any Ck Dsn: BRG Date: 10/007 FOOTING DESIGN EXAMPLE Design: Based on AASHTO LRFD 007 Specifications, TxDOT LRFD Bridge Design Manual, and TxDOT Project 0-4371 17' - 6" x 8' x 6' Footing Rectangular 8' x 5' column, two 4' diameter drilled shafts Limitations: This example will only work when both drilled shafts are in compression (does not encompass drilled shafts in tension). Footing Design- Strut and Tie (AASHTO LRFD 5.6.3) f c the concrete compressive strength, f' c f c 3.6 kip in f y yield strength of the reinforcement f y 60 kip in W col column width W col 8ft D col column depth D col 5ft h depth of footing h 6ft L 1 dimension along the longitudinal axis of the footing L 1 17.5ft L dimension along the transverse axis of the footing L 8ft D DS drilled shaft diameter D DS 4ft EdgeDistance the clear space between the drilled shaft and the end of the footing EdgeDistance 9in s DS the distance between the centroids of the two drilled shafts s DS L 1 EdgeDistance D DS s DS 1 ft 1 of 10

Initially assume that 11 ~ #10 bars will be used for the tension tie reinforcement. Size bar the size of the tension reinforcement bars N b number of the tension reinforcement bars Size bar 10 N b 11 A b area of a single tension reinforcement bar A b 1.7in d b diameter of tension reinforcement bar d b 1.7in d stirrup diameter of #6 stirrup d stirrup 0.75in cc bot clear cover for the transverse reinforcement from the bottom of the footing to the edge of the reinforcement bar cc bot 3in cc side clear cover for the transverse reinforcement from the side of the footing to the edge of the reinforcement bar cc side 3in EndCover concrete cover from the EndCover edge on the footing to the end of the reinforcement bar 3in d c thickness of concrete cover measured from extreme tension fiber to center of the closest tension reinforcement bar d b d c cc bot + + d stirrup d c 4.4 in s bar the distance between the centroids of the tension tie reinforcement bars D DS + d c s bar N b 1 s bar 5.7 in Note: The tension tie reinforcement must be close enough to the drilled shaft to be considered in the truss analysis. Therefore, the tension tie reinforcement must be within a θ t 45 degree distribution angle (i.e. no more than d c away from the drilled shaft on either side) per TxDOT LRFD Bridge Design Manual. of 10

Step One: Draw Idealized Truss Model and Solve for Member Forces Draw Idealized Truss Model: First, Solve for Reactions: The loads transferred from the column to the footing are: (AASHTO LRFD 5.6.3. and TxDOT LRFD Bridge Design Manual) M u P u 3075kip ft 1559kip Note: Design for the worst case created by a positive M u (clockwise) which generates the maximum force in strut BC and node C in the truss model on the following page. Setting the moment about the right column equal to zero: P u R 1 R 1 53.3 kip s DS M u s DS Setting the moment about the left column equal to zero: P u R R 1035.8 kip s DS + M u s DS The following truss model is not applicable for footings with drilled shafts in tension. Check if both drilled shafts are in compression: ( ) ( R > 0) if R 1 > 0, "OK", "NG" "OK" 3 of 10

Then convert M u and P u into two concentrated loads located a distance of W col /4 in from each side of the column. FA M u P u + FB W col M u W col + P u FA 10.8 kip FB 1548.3 kip Both FA and FB are in compression. If FA were in tension, the below truss model would not work. Next place a node at all loads and reactions. Finally draw the most rational truss model. Assume nodes A and B are located at the concentrated loads, FA and FB, at the top of the footing. Assume nodes C and D are located at the center of the drill shafts and a distance of d c above the top of the Drilled Shafts. Note: Nodes represent the center of compression struts or tension ties. Dashed lines represent compression struts and the solid lines represent tension ties, or reinforcement. How to draw a rational truss model: 1)Draw straight lines from node to node to represent the members, struts and ties. )Try to place the tension ties where reinforcement is normally located (i.e. avoid diagonal tension ties). 3)If the angle, θ, between the tension tie and compression strut is too small the compression force will take another route making the current truss model invalid. Therefore, the angle between the compression strut and the tension tie must be greater than 6 degrees per TxDOT LRFD Bridge Design Manual. 4 of 10

In a case where both drilled shafts and column forces are are not in compression see alternate truss models below. One column force in tension One column force and one drilled shaft in tension Solve for Member Forces: Since node C and strut BC have the maximum loads, solve for the forces in BC and CD to design for the worst case. θ atan θ 54.6 deg h d c s DS W col 4 BC BC CD CD R sin( θ) 170 kip BC cos( θ) ( compression) 735 kip ( tension) The figure above shows the boundaries of the compression struts. Strut AB is positioned such that the footing and the column concrete are being utilized. In the case where the column concrete cannot be utilized, nodes A and B should be located at the center of the strut, which will be inside the footing instead of at the top of the footing. Step Two: Choose Tension Tie Reinforcement for Tie CD Assume #10 bars at 5.7in spacing f y 60 ksi ϕ 0.9 (AASHTO LRFD 5.5.4.) A req CD ϕ f y A req 13.6 in A b 1.7in N b 11 A steel A b N b A steel 13.97 in ( ) "good" if A steel A req, "good", "no good, add more steel" 5 of 10

Step Three: Anchorage of Tension Tie (AASHTO LRFD 5.6.3.4. and 5.11.) x the distance from the effective face of the drilled shaft to where the full development of reinforcement is required x x d c tan( θ) 3.1 in The effective face of the drilled shaft is assumed as the side dimension of the equivalent square as defined below. Area Area π*(d DS /) L b ELEVATION NODE C L b π*(d DS /) (TxDOT Project 0-4371 Section 5.5) L b D DS / * π 0.5 L b 0.89*D DS L available L available D DS D DS EdgeDistance + + 0.89 + x EndCover 54.5 in Check if the required development length can be reduced: L db factors: (AASHTO LRFD 5.11..1.3) 1) Check if the clear cover, cc side, is not less than 3in and the reinforcement bar spacing, s bar, is not less than 6in. cc side 3in D DS + d c s bar N b 1 s bar 5.7 in Since s bar is less than 6in, factor 1 1.0. factor 1 1.0 ) Check ratio of A srequired /A sprovided. Since the footing is being designed, the factor of A srequired /A sprovided is assumed to be 1.0. factor 1.0 3) Check if the reinforcement is enclosed in a spiral. Since the footing reinforcement is not enclosed in a spiral, the factor is taken as 1.0. factor 3 1.0 6 of 10

For #11 bars or smaller: L db L db f y in max 1.5 A b, 0.4d b f y, 1in (AASHTO LRFD 5.11..1.1) kip f c kip 50. in ( ) 0.5 L d L db factor 1 factor factor 3 L d 50. in ( ) "good" if L available > L d, "good", "hook required" Step Four: Node Geometry (TxDOT Project 0-4371 Section 5.5) CCT Node - a node which is bounded by two struts (C) and one tie (T) CCC Node - a node which is bounded by three struts (C) w s the width of a strut in plane of the truss model w t the width of the tie, or in a CCC node w t is based on the compression force due to bending of the beam which can be conservatively assumed as the height of the compression block. For simplicity in this design example w t is always taken as the width of a tension tie. Using the width of a tension tie instead of the height of the compression block in most cases should not significantly effect the final result. L b the bearing length corresponding to a node Node A & D Geometry: Since the forces at nodes B and C are larger than those at nodes A and D, the controlling nodal or strut failure will not occur at nodes A and D. Node C Geometry: Assume the bearing length as the side face dimension of the equivalent square for the drilled shaft. L b 0.89 D DS L b 4.7 in w t is the smaller of: w t cc bot 6d b 7.3 in w t cos( θ) + d b 7.3 in 15. in 4. in L b sin( θ) w s w s 39 in 34.8 in w t cos( θ) + L b sin( θ) NODE C GEOMETRY (strut BC and tie CD) 7 of 10

Node B Geometry Due to the ambiguity of a node as complex as node B (two diagonal struts and one horizontal strut), this design example only focuses on strut BC. Node B (strut BC) Geometry: The overall bearing length of node B is W col /, but since there is more than one strut at node B the bearing length will be divided into individual bearing lengths for each strut. The bearing lengths are proportional to the amount of load in each strut. In this example, for simplicity, the bearing length of Node B is divided into two equal sections to accommodate the two diagonal struts. L b W col 4 L b 4.0 in For simplicity w t is taken as the height of the tension tie instead of the height of the compression block. The height of the tension tie at Node B is assumed to be the same as Node C, w t. w t w t w t 7.3 in ELEVATION NODE B (strut BC) w t cos( θ) L b sin( θ) 4. in 19.6 in w s w t cos( θ) + L b sin( θ) w s 3.8 in 8 of 10

Step Five: Check Capacity of Diagonal Struts (TxDOT LRFD Bridge Design Manual and TxDOT Project 0-4371 Section A..1) Strut BC will control, because of the high compressive force. The interface at Node B will control, because it has the smallest cross-sectional area. ν min ν 0.850 kip 0.5 0.85 tan( θ) L b in 0.5 f c ws sin( θ) b BC Effective width of strut BC at node B 0.85 L b,, 0.85 w s sin( θ) (TxDOT Project 0-4371 Section A..1 Eq. A-7 and TxDOT LRFD Bridge Design Manual) W col b BC 4 b BC 4.0 in ϕ 0.70 ϕp n ϕν f c w s b BC (TxDOT Project 0-4371 Section A..1 Eq. A-9 and TxDOT LRFD Bridge Design Manual) ϕp n 1 kip R 1036 kip ( ) "good" if ϕp n > R, "good", "no good" Step Six: Crack Control Reinforcement in Struts (TxDOT LRFD Bridge Design Manual and TxDOT Project 0-4371 Section A..) Strut BC will control, because it deals with the largest applied stresses: L strut length of strut h d c L strut sin( θ) L strut b min b min 8.9 in min( w s, w s ) 4 in b ef maximum effective strut width b ef max L strut, b 3 min + L strut 6 (TxDOT Project 0-4371 Section A.. and TxDOT LRFD Bridge Design Manual) b ef 37.6 in 9 of 10

m slope of the dispersion of compression m b ef (TxDOT Project 0-4371 Section A.. and TxDOT b ef b min LRFD Bridge Design Manual) m 5.44 b the average width of the strut measured perpendicular to the plane of the truss b 0.89 D DS + D col b 51.4 in ρ min BC/(*fy*b*Lstrut*m) but not less than 0.003 BC ρ min max0.003, f y b L strut m ρ min 0.0030 (TxDOT Project 0-4371 Section A.. Eq. A-15 and TxDOT LRFD Bridge Design Manual) Horizontal (Skin) Reinforcement: Assuming size #7 bars and 9" spacing s bar_h 9in A b_h 0.60in ***This check requires that both side faces of the footing be reinforced with the specified bar size at the spacing shown. Vertical Reinforcement (Two Legs - Stirrups): Assuming size #7 bars and 9" spacing s bar_v 9in A b_v 0.60in For the transverse direction of the strut, conservatively use the width of the larger node: ρ ρ 0.0031 A b_h max D col, 0.89 D DS ( ) s bar_h + A b_v max D col, 0.89 D DS ( ) s bar_v 0.5 (TxDOT Project 0-4371 Section A.. Eq. A-14 and TxDOT LRFD Bridge Design Manual) ( ) "Good" if ρ > ρ min, "Good", "NG, add more reinforcement" **Use the same bars and spacing throughout the footing Step Seven: Check Nodal Zone Stress Limits (AASHTO LRFD 5.6.3.5, TxDOT LRFD Bridge Design Manual and TXDOT Project 0-4371 Section A.3.3.) The nodes are all significantly large. By inspection, nodal failure will not occur. 10 of 10