ANALYSIS OF ECCENTRICALLY LOADED RECTANGULAR FOOTING RESTING ON SOIL – A NUMERICAL APPROACHJignesh V Chokshi, L&T Sargent & Lundy Limited, Vadodara, India For analysis of isolated rectangular footings with large bi-axial eccentricity, an accurate and efficient numerical approach satisfying all equilibrium conditions and suitable on computers is presented in this paper. Microsoft Excel, a cogent tool globally used by structural engineers, under its VBA programming environment is chosen for programming the numerical approach and graphically displaying input and results. A generalized program dealing with any conditions of eccentricities–zero eccentricity, one-way eccentricity or two-way eccentricity– is developed for analysis of rectangular footings. Several examples, with different eccentricity conditions are chosen to investigate accuracy of results and verify performance of the numerical approach implemented in the program. Introduction The bearing pressure distribution for rigid isolated footing resting on soil subjected to axial load and bending moments can be obtained by, p := P A + Mx Ix z+ Mz Iz x ....................................................……………………...……..(1) In the equation 1, p = Bearing pressure under footing base at point (x, z), P = Axial Load; A = Area of Footing, Mx, Mz = Moment about X–axis and Z–Axis respectively, Ix, Iz = Moment of Inertia of footing about X–axis and Z–axis respectively and, x, z = Coordinates of point at which bearing pressure is to be calculated. From the above, eccentricity of loading for footing can be derived as, Figure 1: Footing Geometry ex = Eccentricity along X-axis from center of gravity of footing = Mz / P ez = Eccentricity along Z-axis from center of gravity of footing = Mx / P For isolated rectangular footings, called footings now onwards, when the loading point k(ex, ez) lies in middle third of the footing, called Kern (shaded area in Fig. 1), magnitude of p is positive and the soil below footing is said to be in compression. However, if loading point lies outside the Kern, magnitude of p at few locations in the footing is negative and that portion of footing is said to be in tension. Since, there exists no mechanism between soil and footing to CG of bearing pressure envelope shall coincide with location of applied load P. bearing pressure at different points of footing will be modified and the line of zero stresses will shift towards loading point. following equilibrium conditions must comply. ex outside kern – One-way eccentricity along X axis 3. It is also necessary to keep sufficient area of footing remaining in contact with soil and bearing pressure not exceeding the allowable bearing pressure of the soil. 1. A numerical approach is described in the paper. ex > Lx/6 and ez =0 . However. 2. Thus. tables or charts are cumbersome to implement and the information is very brief. ez outside kern – Two-way eccentricity It shall be noted that. Roark [2] provides tables and Peck [3] mentions an iterative method for footing with two-way eccentricity. The initial position of neutral axis can be obtained by connecting a line between two points having zero stresses on adjacent or opposite edges. The portion outside line of zero pressure will be completely unstressed and is called footing uplift area. some portion of footing will remain unstressed and the force equilibrium will occur in the area of footing which remains in contact with soil. for computer implementation of footing design process. ex. possible eccentricity conditions can be enumerated as follows: 1. the stability of footing demands special attention. ex = 0 and ez = 0 . a numerical approach is the best choice. ez within kern area – Compression on entire base of footing 2. solution to the problem is simple. or ex. it is imperative to ensure satisfactory Factor of Safety against overturning. charts and related equations. Volume of bearing pressure envelope shall be equal to the applied load P. either ex or ez outside kern. Equilibrium Conditions In analysis of eccentrically loaded footings. To automate the footing design process on computer. conditions 2. In the available literature. Eccentricity Conditions For a footing. it is assumed that pressure varies linearly. the pressure will vary from negative to positive below footing base. there will be significant shift of initial neutral axis to its final position. However. ez outside kern – One-way eccentricity along Z axis 4. Footings with one-way eccentricity. The points of zero pressure on footing edges can be obtained by substituting p = 0 and appropriate coordinate of footing edges in Eq. which solves this problem with tangible accuracy. 3 and 4 produces tension on some portion of the footing. large area of footing will remain unstressed and hence. for static equilibrium to occur. Position of Neutral Axis For footings with loading point outside Kern. the footing is rigid and the effect of soil displacement has no effect on the pressure distribution. the solution is not as simple as that for one-way eccentricity. Hence. In this approach. Teng [1] shows graphical method.resist the tensile stresses. 1. ex = 0 and ez > Lz/6 . for footings with two-way eccentricity ex and ez outside Kern. . For footings having large eccentricities. Under these circumstances. ex > 0 and ez > 0 . C & D. alter the position of neutral axis and repeat step 5 to step 8. Read size of footing and loading. following positions of neutral axis can be envisaged. Programming Strategy The solution methods suggested in the literature are very brief and do not explain a detailed procedure for implementation of the solution technique on digital computer. the numerical approach is necessary. Hence. 4. calculate geometric properties. Pressure at B and C = 0 Case 4: One end on BC and other end on AD. 8. Mx. et. Lz) 2. Calculate the volume of pressure diagram envelope. B. For a given size of footing and loading.As shown in Figures 1 and 2.(A. Ix. Pressure at C and D = 0 Case 5: One end on AB and other end on AD. Calculate the geometrical properties of footing. pressure etc. Mz. Lx. For selected neutral axis. Pressure at B & C = 0. Pressure at C = 0 Case 3: One end on AB and other end on CD. Pressure at A= Pressure at D Case 7: Neutral Axis parallel to X-axis. [3] is adopted and implemented to obtain faster and accurate solution. Calculate the center of gravity of pressure diagram envelope. If difference is too large. 7. A systematic . Pressure at B. ex. and ez obtained in step 6 and 7 with input parameters. Compare values of P. about neutral axis for portion of footing that remains in contact with soil. C and D = 0 Case 6: Neutral Axis parallel to Z-axis. Numerical Approach One can imagine that it is almost impossible to obtain a unified mathematical equation that solves all of the above-defined cases. Calculate the pressure at corners A. 6. Figure 2: Positions of Neutral Axis 5. ez) 3. Iz. Pressure at A= Pressure at B Cases 2 through 5 are for footing with two-way eccentricity and cases 6 and 7 are for footing with one-way eccentricity. ex. Al. (P. Pressure at C & D = 0. Case 1: No neutral axis – Compression case (Fig. 1) Case 2: One end on BC and other end on CD. the numerical approach suggested by Peck. for effective solution. Obtain initial position of neutral axis for problems having tension on the corners. The numerical procedure essentially works as follows: 1. This results in storage of three positions of line EJ. In cases 2 to 7. E2. As shown in Fig. Figure 3 shows the location of line EJ where individual error for P.the neutral axis. at any of these three positions. 7. Calculate geometrical properties. the problem is restricted to triangle EAJ. 3. Establish the acceptable numerical error in results and limit of number of iterations. ex or ez will be within acceptable limits. parallel to EJ. 9. Igh = I(∆EAJ) – I(∆EBG) – I(∆HDJ). line EJ . 1. the iterations are performed in two phases. select another axis EJ at next step and repeat step 8 to 12. error for only one of P. 5. eccentricities and pressures at each corner of footing. triangle EBG and triangle HDJ. This means that. For cases 1. 10. . 3. In this method. towards point K in subsequent iterations. In the first phase. For other cases. Calculate pressure at points A. 13. Terminate further iterations when these three positions of line EJ are traced. 12. It can be inferred that the true solution. For cases 2 to 5. pressure at C = 0. 8. necessary changes are taken care in the generalized program. 14. This completes the first phase of iterations where line EJ is moved parallel to initial neutral axis.numerical procedure is described here demonstrating each component of the programming implemented for the solution of the problem. Extend point G on edge AB to locate point E and extend point H on edge AD to locate point J. ex and ez is found within acceptable limits. 4. The strategy described here is for case 2. Read size of footing and applied forces. 6 and 7. Now. the solution band is bounded by a polygon connected between points E1. Store the positions of neutral axis when individual error for P. will be moved. B and D using pi = ( P x Z x bi ) / ( Igh ). 6. Calculate volume and CG of pressure envelop of polygon ABGHD using properties of triangle and tetrahedron. b2 for corner B and b4 for corner D normal to neutral axis EJ. Identify the pressure case of footing from Fig. Select appropriate step for iteration. Compare volume of polygon with applied load P. Microsoft Excel with its powerful VBA support is selected for implementing the numerical procedure on computer. lies within the band bounded by three positions of neutral axis. find out the position of points G and H on appropriate edges of footing where p=0. For each position of neutral axis EJ. 2. the unique position Figure 3: Solution Band of line EJ where numerical error for P. ex and ez is within acceptable limit. 2 to know initial position of neutral axis. simply solve the problem using known method. ex and ez is simultaneously within acceptable limits. Calculate moment of inertia of polygon ABGHDA about its base GH using. J1 and J2. To extract the solution band limits. calculate distance Z of loading point K. distance b1 for corner A. 11. Calculate percentage error in the achieved solution. find out the lower-most and upper-most position of EJ. If the numerical error is more than acceptable limit. and center of gravity of pressure envelop with ex and ez. it is expected that there may be other positions of final neutral axis. Extraordinary features of Excel chart options are explored and the graphical features of the program includes: 1. It was observed that to achieve a tangible accuracy of 99 percentage or better. 19. The program constantly monitors the case of current neutral axis and calculates required properties accordingly. 16. For true solution to occur. At this point. during iterations. ex and ez are calculated to monitor the convergence and limit on number of iterations is also verified at each step. ex and ez. all steps to find out volume of pressure diagram and CG of pressure envelope are repeated as explained earlier. Here. The second phase of iterations within the newly formulated solution band is initiated by assuming the neutral axis as a line joining points E1 and J2 (see Fig. the position of line EJ may represent case 3. the accuracy of the first instance of solution is acceptable for all practical purposes. The same is implemented in programming by slightly shifting point E1 on left. 5) load and loading point coordinates recovered. 2) position of initial neutral axis. initial neutral axis and final neutral axis. 4) effective compression area. 2. If the solution is not converged with the selected pivot. origin. all simultaneously. E2 on right. The very first instance of such convergence is reported and further iterations are abandoned. solution search is an iterative process. 3). the objective is to find the position of EJ where error for P. if the solution diverges. after equilibrium conditions are met. with other end from J2 to J1 until the true solution is found. position of final neutral axis are reported by the program. 6) maximum pressures at corners and 7) numerical difference in recovering P. Results and Graphics Interface After successful execution of the program. The entire range from E1 to E2 will be pivoted during these iterations. during subsequent iterations. the program abandons further iterations within J2 and J1 and new pivot point within E1 and E2 is selected. the position of line EJ may be representing case 2. 4 or 5. 18. However. the position of line EJ may get changed from one case to another. J1 downward and J2 upward before initiating second phase of iterations. Extensive effort is put on the graphical presentation of input and results. Bearing Pressure Diagrams: 2D and 3D presentation of contours showing variation of pressure. loading point. B and D. It shall be noted that. It is found that the results of other positions do not vary much for the desired accuracy. .15. Kern. Since. 3) position of final neutral axis. At every position of neutral axis during the iterations. it is imperative that for a particular position of neutral axis within solution band. at the beginning of the iterations. over the footing surface. shall be within allowable limits. While iterating within J2 to J1. ex and ez. ex and ez is within limits simultaneously. 17. and hence. a slightly larger band shall be used than originally extracted. the numerical errors for P. the following output is generated: 1) The input parameters. the numerical errors for P. then pivot E1 is shifted at the next step towards E2. essential results such as pressures at A. Footing Geometry: Size of footing. uplift area. the point E1 is pivoted first and second point of neutral axis is altered from J2 to J1 with appropriate step size. Also. The footing area is divided into many small parts to produce refined bearing pressure diagram. In the second phase. For example. The results were compared with input data and not with solution obtained from any other reference.976 1.8 1249. The time taken for finding the solution is computationally economical for incredible accuracy achieved.56 0.00 4.050 0.0000 0.0000 0.00 Mx 278.0 333.09 52.47 0.00 0.923 d 2.9 percentage or better for all problems under investigation.00 160.0088 0.9% is achieved.00 1800.41 360.00 ex 0.00 32.05 -37. Graphical representation of footing geometry and pressure distribution diagrams for examples 1.85 52.01 14. The table also demonstrates number of iterations performed to solve the problem and run time taken on PC with P4 -1.0737 Run Time Data Iterations 1245 640 2081 673 Time 7 3 9 3 (Sec) (Units kN and m) 2000.4503 (%) Error in P 0.85 -47.0 4000.00 0.899 1.8.00 5.0 2.00 4.00 179.00 1300.00 3.00 400.0020 0.0000 0.00 736.00 0.385 0.00 Lx 6.0000 0.0 3000.0 2000.00 6.00 P’C 0.00 0. Hence.0 1500. .00 50.00 Mz 250.0804 ex 0.077 2.0020 0.75 PB 12.0030 0.888 as % of (Lx x Lz) Contact Area 77.300 0.000 1.00 256.0 1500.00 2.50 2813.00 -512.00 16.750 0.48 245.89 146.00 2000.3831 0.00 0.1251 2.277 0.0000 946 3 1067 3 The use of Excel with its VBA environment is phenomenally user friendly and endorses the structural engineers’ acceptance of Excel as a cogent tool for automating structural design work processes.200 0. Table 1 shows input data and true solution for selected problems.00 0.25 PD 6.60 129.14 0.0126 0.0010 3000.200 4.1 to 4.50 2.18 2. use of Excel is found highly efficient. Note that in all problems a tangible accuracy of 99.200 4.750 0.0 5.0000 0.18 -100. 2.300 ez/Lz 0.00 162.0652 0.0 1250.250 0.00 2.600 1.8984 1.0733 0.2505 0.00 395.50 5.450 ex/Lx 0.201 ez 1. Table 1 Verification Problems and Comparison of Results Problem No Item 1 2 3 4 5 6 Geometry and Load Data P 278.8 333.36 Comparison of Results Precovered 277.7500 0.0000 1.125 2.300 0.50 2.18 102.450 0.57 P’D c 4.00 -95.0647 0.5GHz processor and 512MB RAM.624 2.400 0.0 2000.0822 0.0876 0.150 Bearing Pressure at Corners (Before Modification of Pressure) PA 28.00 0.5996 1. Even for such a complex problem like footings with two-way eccentricity.5000 0.10 P’B 11.0008 0.72 308.300 832. Observations and Conclusions The numerical approach suggested in this paper produces impressive results having a tangible accuracy of 99. the numerical approach presented here can be effectively implemented to automate the footing analysis and design.75 PC -10.0640 0. 3 and 5 are shown in Fig.2000 ezrecovered 1.0219 0.88 265.00 -416.0838 ez 0.0020 P’A 32.0 0.Verification Examples Many practical examples were selected to validate results produced by the program and monitor accuracy of the numerical approach presented here.07 66.27 exrecovered 0.00 5.00 160.150 0.100 0.0 1500.00 64.60 -45.000 0.0 150.513 0.00 5 5 0.0030 0.00 Lz 5.97 1300.500 0.0 750. 4.00 0.7500 0.24 749.75 Results obtained by Numerical Method Case 2 3 4 3 Step 0. 00 -1. 3 (Case 4 ) Footings with Two-Way Eccentricit y 3.75 -1. 2 ( Case 3 ) Footings with Two-Way Eccentricity 3.75 -0.00 0.00 0.50 -2.0 -2.50 3.00 -1.000 -3.00098.00 -3.00 2.00 -3.000-22.00 -3.000 198.50 -0.000 X-Axis: Length of Footing (Lx) Load Point Original_NA Final NA Figure 4.000-220.00 -2.00 -2.000 140.50 0.00 -3.50 0.000 100.00 -1.00 -2.50 2. 1 ( Case 2 ) Footings with Two-Way Eccentricity 3.00 -1.00 -2.000-18.000 180.000-380.00 2.00 0.00 3.00 Footing -2.75 -1.00 -1.00 0.000 598.000 18.50 1.00 -3.50 -1.000-6.50 2.50 -1.000198.00 Final NA X-Axis: Length of Footing (Lx) Load Point Original_NA Figure 4.000 26.000 6.000 30.000-100.25 1.00 1.50 1.50 -1.00 -2.0 298.00 1.50 2.1 : Footing Geometry Example Problem No.25 0.75 0.000-260.000-180.00 1.00 -1.00 1.6: Bearing Presure Diagram Points along Z Axis .00 698.00 2.00 1.000 Figure 4.000-300.00 -1.000698.50 -2.00 0.50 -2.2D Y-Axis: Width of Footing (Lz) 2.00 -0.000 98.50- 1.00 -1.25 Points along Z Axis Points along X Axis 20.000 22.3 : Footing Geometry Example Problem No.00 -3.000-2.00 -1.00 0.50 -1.50 -1.00 -0.000-140.50 0.0 -3.00 0.000 300.25 0.4: Bearing Presure Diagram Base Pressure Distribution Diagram .00 0.Example Problem No.50 -2.00 Footing -2.00 1.00 1.000 Figure 4.000 260.00 0.00 798.00 0.50 3.000298.000 398.0 598.00 X-Axis: Length of Footing (Lx) Load Point Original_NA Final NA -20.00 -1.0 98.50 Points along X Axis 2.000-38.00 2.0 198.000-34.00 3.50 1.00 1.00- 2.000-30.000 10.00 498.50 0.00 -0.50 -0.000 298.5 : Footing Geometry Figure 4.00 2.0 398.000498.25 -2.000 60.000 340.000-10.000-340.00 -2.000-60.25 1.25 -1.000 -2.000 Y-Axis: Width of Footing (Lz) 2.00 Footing -2.75 2.000-14.00 -2.00 -0.2: Bearing Presure Diagram Base Pressure Distribution Diagram .75 1.000-26.00 -0.3D 698.50 0.000 14.000598.25 0.00 2.00 -2.00 1.000798.0 498.00 1.000398.25 2.000 0.2D Y-Axis: Width of Footing (Lz) 2.00 2.000 34.000 220.00 Figure 4.25 -0.000-20.50 1.00 2.00 Base Pressure Distribution Diagram .0 1.50 0.00 3. B.00 Footing 700.7 : Footing Geometry Figure 4.00 -0. India.0 -100. 5 (Case 5 ) Footings with Two-Way Eccentricit y 3. Englewood cliffs.75 -0.0 1.75 -1.75 1.000 700.00 2700.50 -1.000 -2.75 2.. E.Example Problem No.25 0.50 -0. Roark’s Formulas for Stress and Strain.00 1. Englewood cliffs.0003100.00 -1.8: Bearing Presure Diagram Acknowledgement I thank my company M/s.000 -100.50 1900. Hanson W.000 1.50 -1. and Budynas R.000-0. New Jersey. References 1. Peck R.00 1. 3. for the support.0002300.00 -1.00 -3.00 0. 2.00 Base Pressure Distribution Diagram .25 0.00 3100. and Thornburn W.25 300.00 3..25 -2. 2nd Edition. Vadodara.50 1. G.0 1500. L&T Sargent and Lundy Limited. C.25 -1.000 2300.50 0. H.00 -2. C.0 1900.. John Wiley and Sons.0001900. New York.00 -3.0 1100. Teng W. encouragement and providing computational facilities for this programming work.0 -2. Young W.000 Y-Axis: Width of Footing (Lz) 2. .000 1500.0 700.0001500. Foundation Engineering..00 2.0 2300.00 2.0002700. New Jersey. Prentice-Hall Inc...50 -2.0001100.25 2.25 0. Foundation Design.00 0. McGraw Hill.25 1.3D 2700. Gujarat.00 -1.000300.000 1100. 7th Edition.00 0.0 300.00 Final NA X-Axis: Length of Footing (Lx) Load Point Original_NA Figure 4.