· For Assumed Rigid Footing with from 1 To 8 Piers (Load Points) Subjected to Uniaxial or Biaxial Eccentricity %

"FOOTINGS.xls" ProgramVersion 3.7

RECTANGULAR SPREAD FOOTING ANALYSISFor Assumed Rigid Footing with from 1 To 8 Piers (Load Points)

Subjected to Uniaxial or Biaxial EccentricityJob Name: Program Validation Subject: Example Problem #1

Job Number: Originator: A. Tomanovich Checker:From: "Bearing Pressures for Rectangular Footings with Biaxial Uplift"

Input Data: - by Kenneth E. Wilson (Journal of Bridge Engineering, Feb. 1997)

Footing Data:

Footing Length, L = 22.000 ft.

Footing Width, B = 6.000 ft.

Footing Thickness, T = 0.000 ft.

Concrete Unit Wt., c = 0.000 kcf

Soil Depth, D = 0.000 ft.

Soil Unit Wt., s = 0.000 kcf

Pass. Press. Coef., Kp = 3.000Coef. of Base Friction, = 0.400

Uniform Surcharge, Q = 0.000 ksf

Pier/Loading Data:Number of Piers = 1

Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -110.80Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -67.59My (ft-k) = 405.53

FOOTING PLAN(continued)

Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 2:13 PM


Results: Nomenclature for Biaxial Eccentricity:

Total Resultant Load and Eccentricities: Case 1: For 3 Corners in Bearing Pz = -110.80 kips (Dist. x > L and Dist. y > B)

ex = 3.660 ft. (<= L/6)

ey = 0.610 ft. (<= B/6)

Overturning Check:Mrx = 332.40 ft-kips

Mox = -67.59 ft-kips

FS(ot)x = 4.918 (>= 1.5)

Mry = 1218.80 ft-kips

Moy = 405.53 ft-kips

FS(ot)y = 3.005 (>= 1.5)

Sliding Check: Case 2: For 2 Corners in Bearing Pass(x) = 0.00 kips (Dist. x > L and Dist. y <= B)Frict(x) = 44.32 kips

FS(slid)x = N.A.Pass(y) = 0.00 kips

Frict(y) = 44.32 kips

FS(slid)y = N.A.

Uplift Check:Pz(down) = -110.80 kips

Pz(uplift) = 0.00 kips

FS(uplift) = N.A.

Bearing Length and % Bearing Area: Case 3: For 2 Corners in Bearing Dist. x = 28.014 ft. (Dist. x <= L and Dist. y > B)Dist. y = 12.201 ft.

Brg. Lx = 14.238 ft.

Brg. Ly = 2.620 ft.

%Brg. Area = 90.06 %

Biaxial Case = Case 1 6*ex/L + 6*ey/B = 1.608

Gross Soil Bearing Corner Pressures:P1 = 1.151 ksf

P2 = 2.265 ksf

P3 = 0.486 ksf

P4 = 0.000 ksf

Case 4: For 1 Corner in Bearing (Dist. x <= L and Dist. y <= B)

P3=0.486 ksf P2=2.265 ksf

P4=0 ksf P1=1.151 ksf

CORNER PRESSURES

Maximum Net Soil Pressure:Pmax(net) = Pmax(gross)-(D+T)*sPmax(net) = 2.265 ksf

Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly

Line of zeropressure Brg. Lx

Line of zeropressure

Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 2:13 PM




Job Number: Originator: A. Tomanovich Checker:From: "Bearing Pressures for Rectangular Footings with Biaxial Uplift"

Input Data: - by Kenneth E. Wilson (Journal of Bridge Engineering, Feb. 1997)

Footing Data:










Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -925.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -323.75My (ft-k) = 1831.50


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 2:14 PM




ex = 1.980 ft. (> L/6)

ey = 0.350 ft. (<= B/6)



FS(ot)x = 7.500 (>= 1.5)



FS(ot)y = 1.326 (< 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Lx1 = 1.502 ft.

Brg. Lx2 = 2.266 ft.

%Brg. Area = 35.89 %



P2 = 221.902 ksf

P3 = 0.000 ksf

P4 = 0.000 ksf


P3=0 ksf P2=221.902 ksf

P4=0 ksf P1=147.092 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 2:14 PM




Job Number: Originator: A. Tomanovich Checker:From: "How to Calculate Footing Soil Bearing by Computer"

Input Data: - by Eli Czerniak (1963)

Footing Data:










Pier #1Xp (ft.) = -1.000Yp (ft.) = -1.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -250.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = 0.00My (ft-k) = 0.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 2:19 PM




ex = -1.000 ft. (<= L/6)

ey = -1.000 ft. (<= B/6)

Overturning Check:Mrx = N.A. ft-kips

Mox = N.A. ft-kips

FS(ot)x = N.A.Mry = N.A. ft-kips

Moy = N.A. ft-kips

FS(ot)y = N.A.




FS(slid)y = N.A.



FS(uplift) = N.A.

Bearing Length and % Bearing Area: Case 3: For 2 Corners in Bearing Dist. x = N.A. ft. (Dist. x <= L and Dist. y > B)Dist. y = N.A. ft.

Brg. Lx = 15.000 ft.

Brg. Ly = 10.500 ft.

%Brg. Area = 100.00 %

Biaxial Case = N.A. 6*ex/L + 6*ey/B = 0.971


P2 = 0.045 ksf

P3 = 1.315 ksf

P4 = 3.129 ksf


P3=1.315 ksf P2=0.045 ksf

P4=3.129 ksf P1=1.859 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 2:19 PM




Job Number: Originator: A. Tomanovich Checker:From: "How to Calculate Footing Soil Bearing by Computer"

Input Data: - by Eli Czerniak (1963)

Footing Data:










Pier #1Xp (ft.) = -3.750Yp (ft.) = -3.250

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -100.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = 0.00My (ft-k) = 0.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 2:22 PM




ex = -3.750 ft. (> L/6)

ey = -3.250 ft. (> B/6)

Overturning Check:Mrx = N.A. ft-kips

Mox = N.A. ft-kips

FS(ot)x = N.A.Mry = N.A. ft-kips

Moy = N.A. ft-kips

FS(ot)y = N.A.




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Ly1 = 0.000 ft.

Brg. Ly2 = 8.000 ft.

%Brg. Area = 38.10 %



P2 = 0.000 ksf

P3 = 0.000 ksf

P4 = 5.000 ksf


P3=0 ksf P2=0 ksf

P4=5 ksf P1=0 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 2:22 PM

ANALYSIS OF ECCENTRICALLY LOADEDRECTANGULAR 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 andefficient numerical approach satisfying all equilibrium conditions and suitable on computers ispresented in this paper. Microsoft Excel, a cogent tool globally used by structural engineers,under its VBA programming environment is chosen for programming the numerical approachand graphically displaying input and results. A generalized program dealing with anyconditions of eccentricities–zero eccentricity, one-way eccentricity or two-way eccentricity– isdeveloped for analysis of rectangular footings. Several examples, with different eccentricityconditions are chosen to investigate accuracy of results and verify performance of thenumerical approach implemented in the program.

Introduction

The bearing pressure distribution for rigid isolated footing resting on soil subjected to axialload and bending moments can be obtained by,

....................................................……………………...……..(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–axisrespectively and,x, z = Coordinates of point at which bearing pressure is to becalculated. From the above, eccentricity of loading for footing can bederived as,

ex = Eccentricity along X-axis from center of gravity of footing = Mz / Pez = 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 ispositive and the soil below footing is said to be in compression. However, if loading point liesoutside the Kern, magnitude of p at few locations in the footing is negative and that portion offooting is said to be in tension. Since, there exists no mechanism between soil and footing to

Figure 1: Footing Geometry

pP

A

M x

I xz+

M z

I zx+:=

resist the tensile stresses, some portion of footing will remain unstressed and the forceequilibrium will occur in the area of footing which remains in contact with soil. Under thesecircumstances, bearing pressure at different points of footing will be modified and the line ofzero stresses will shift towards loading point. The portion outside line of zero pressure will becompletely unstressed and is called footing uplift area.

Footings with one-way eccentricity, either ex or ez outside kern, solution to the problem issimple. However, for footings with two-way eccentricity ex and ez outside Kern, the solutionis not as simple as that for one-way eccentricity. In the available literature, Teng [1] showsgraphical method, charts and related equations; Roark [2] provides tables and Peck [3]mentions an iterative method for footing with two-way eccentricity. To automate the footingdesign process on computer, tables or charts are cumbersome to implement and the informationis very brief. Hence, for computer implementation of footing design process, a numericalapproach is the best choice. A numerical approach is described in the paper, which solves thisproblem with tangible accuracy. In this approach, it is assumed that pressure varies linearly,the footing is rigid and the effect of soil displacement has no effect on the pressure distribution.

Equilibrium Conditions

In analysis of eccentrically loaded footings, following equilibrium conditions must comply,1. Volume of bearing pressure envelope shall be equal to the applied load P,2. CG of bearing pressure envelope shall coincide with location of applied load P.

For footings having large eccentricities, large area of footing will remain unstressed and hence,the stability of footing demands special attention. Thus, it is imperative to ensure satisfactoryFactor of Safety against overturning. It is also necessary to keep sufficient area of footingremaining in contact with soil and bearing pressure not exceeding the allowable bearingpressure of the soil.

Eccentricity Conditions

For a footing, possible eccentricity conditions can be enumerated as follows:1. ex = 0 and ez = 0 ; or ex, ez within kern area – Compression on entire base of footing2. ex > Lx/6 and ez =0 ; ex outside kern – One-way eccentricity along X axis3. ex = 0 and ez > Lz/6 ; ez outside kern – One-way eccentricity along Z axis4. ex > 0 and ez > 0 ; ex, ez outside kern – Two-way eccentricityIt shall be noted that, conditions 2, 3 and 4 produces tension on some portion of the footing.

Position of Neutral Axis

For footings with loading point outside Kern, the pressure will vary from negative to positivebelow footing base. The points of zero pressure on footing edges can be obtained bysubstituting p = 0 and appropriate coordinate of footing edges in Eq. 1. The initial position ofneutral axis can be obtained by connecting a line between two points having zero stresses onadjacent or opposite edges. However, for static equilibrium to occur, there will be significantshift of initial neutral axis to its final position.

As shown in Figures 1 and 2, following positions of neutral axis can be envisaged.Case 1: No neutral axis – Compression case (Fig. 1)Case 2: One end on BC and other end on CD, Pressure at C = 0Case 3: One end on AB and other end on CD, Pressure at B and C = 0Case 4: One end on BC and other end on AD, Pressure at C and D = 0Case 5: One end on AB and other end on AD, Pressure at B, C and D = 0Case 6: Neutral Axis parallel to Z-axis, Pressure at B & C = 0, Pressure at A= Pressure at DCase 7: Neutral Axis parallel to X-axis, Pressure at C & D = 0, Pressure at A= Pressure at BCases 2 through 5 are for footing with two-way eccentricity and cases 6 and 7 are for footingwith one-way eccentricity.

Numerical Approach

One can imagine that it is almostimpossible to obtain a unifiedmathematical equation that solves all ofthe above-defined cases. Hence, foreffective solution, the numericalapproach is necessary. For a given sizeof footing and loading, the numericalapproach suggested by Peck, et. Al. [3]is adopted and implemented to obtainfaster and accurate solution. Thenumerical procedure essentially worksas follows:

1. Read size of footing and loading.(P, Mx, Mz, Lx, Lz)2. Calculate the geometricalproperties of footing.(A, Ix, Iz, ex, ez)3. Calculate the pressure at corners A,B, C & D.4. Obtain initial position of neutralaxis for problems having tension on thecorners.5. For selected neutral axis, calculate

geometric properties, pressure etc. about neutral axis for portion of footing that remains incontact with soil.

6. Calculate the volume of pressure diagram envelope.7. Calculate the center of gravity of pressure diagram envelope.8. Compare values of P, ex, and ez obtained in step 6 and 7 with input parameters. If

difference is too large, alter the position of neutral axis and repeat step 5 to step 8.

Programming Strategy

The solution methods suggested in the literature are very brief and do not explain a detailedprocedure for implementation of the solution technique on digital computer. A systematic

Figure 2: Positions of Neutral Axis

numerical procedure is described here demonstrating each component of the programmingimplemented for the solution of the problem. Microsoft Excel with its powerful VBA supportis selected for implementing the numerical procedure on computer. The strategy describedhere is for case 2. For other cases, necessary changes are taken care in the generalizedprogram.

1. Read size of footing and applied forces.2. Establish the acceptable numerical error in results and limit of number of iterations.3. Calculate geometrical properties, eccentricities and pressures at each corner of footing.4. Identify the pressure case of footing from Fig. 2 to know initial position of neutral axis.5. For cases 1, 6 and 7, simply solve the problem using known method. For cases 2 to 5, find

out the position of points G and H on appropriate edges of footing where p=0.6. Extend point G on edge AB to locate point E and extend point H on edge AD to locate

point J. Now, the problem is restricted to triangle EAJ, triangle EBG and triangle HDJ.7. In this method, the iterations are performed in two phases. In the first phase, line EJ - the

neutral axis, will be moved, parallel to EJ, towards point K in subsequent iterations. Selectappropriate step for iteration.

8. For each position of neutral axis EJ, calculate distance Z of loading point K, distance b1for corner A, b2 for corner B and b4 for corner D normal to neutral axis EJ.

9. Calculate moment of inertia of polygon ABGHDA about its base GH using, Igh = I(∆EAJ) – I(∆EBG) – I(∆HDJ).10. Calculate pressure at points A, B and D using pi = ( P x Z x bi ) / ( Igh ). In cases 2 to 7,

pressure at C = 0.11. Calculate volume and CG of pressure envelop of polygon ABGHD using properties of

triangle and tetrahedron.12. Compare volume of polygon with applied load P, and center of gravity of pressure envelop

with ex and ez. Calculate percentage error in the achieved solution. If the numerical erroris more than acceptable limit, select another axis EJ at next step and repeat step 8 to 12.

13. Store the positions of neutral axis whenindividual error for P, ex and ez is withinacceptable limit. This results in storage of threepositions of line EJ. This means that, at any ofthese three positions, error for only one of P, ex orez will be within acceptable limits.14. Terminate further iterations when these threepositions of line EJ are traced. Figure 3 shows thelocation of line EJ where individual error for P, exand ez is found within acceptable limits. Thiscompletes the first phase of iterations where line EJis moved parallel to initial neutral axis. It can beinferred that the true solution, the unique positionof line EJ where numerical error for P, ex and ez is

simultaneously within acceptable limits, lies within the band bounded by three positions ofneutral axis. To extract the solution band limits, find out the lower-most and upper-mostposition of EJ. As shown in Fig. 3, the solution band is bounded by a polygon connectedbetween points E1, E2, J1 and J2.

Figure 3: Solution Band

15. It was observed that to achieve a tangible accuracy of 99 percentage or better, a slightlylarger band shall be used than originally extracted. The same is implemented inprogramming by slightly shifting point E1 on left, E2 on right; J1 downward and J2upward before initiating second phase of iterations. In the second phase, the objective is tofind the position of EJ where error for P, ex and ez is within limits simultaneously.

16. The second phase of iterations within the newly formulated solution band is initiated byassuming the neutral axis as a line joining points E1 and J2 (see Fig. 3). Here, the point E1is pivoted first and second point of neutral axis is altered from J2 to J1 with appropriatestep size. At every position of neutral axis during the iterations, all steps to find outvolume of pressure diagram and CG of pressure envelope are repeated as explained earlier.Also, the numerical errors for P, ex and ez are calculated to monitor the convergence andlimit on number of iterations is also verified at each step. If the solution is not convergedwith the selected pivot, then pivot E1 is shifted at the next step towards E2. The entirerange from E1 to E2 will be pivoted during these iterations, with other end from J2 to J1until the true solution is found. While iterating within J2 to J1, if the solution diverges, theprogram abandons further iterations within J2 and J1 and new pivot point within E1 andE2 is selected.

17. It shall be noted that, during iterations, the position of line EJ may get changed from onecase to another. For example, at the beginning of the iterations, the position of line EJmay be representing case 2. However, during subsequent iterations, the position of line EJmay represent case 3, 4 or 5. The program constantly monitors the case of current neutralaxis and calculates required properties accordingly.

18. For true solution to occur, it is imperative that for a particular position of neutral axiswithin solution band, the numerical errors for P, ex and ez, all simultaneously, shall bewithin allowable limits. The very first instance of such convergence is reported andfurther iterations are abandoned. At this point, essential results such as pressures at A, Band D, uplift area, position of final neutral axis are reported by the program.

19. Since, solution search is an iterative process; it is expected that there may be otherpositions of final neutral axis. It is found that the results of other positions do not varymuch for the desired accuracy, and hence, the accuracy of the first instance of solution isacceptable for all practical purposes.

Results and Graphics Interface

After successful execution of the program, the following output is generated:1) The input parameters, 2) position of initial neutral axis, 3) position of final neutral axis, 4)effective compression area, 5) load and loading point coordinates recovered, 6) maximumpressures at corners and 7) numerical difference in recovering P, ex and ez.

Extensive effort is put on the graphical presentation of input and results. Extraordinaryfeatures of Excel chart options are explored and the graphical features of the program includes:1. Footing Geometry: Size of footing, origin, loading point, Kern, initial neutral axis and

final neutral axis.2. Bearing Pressure Diagrams: 2D and 3D presentation of contours showing variation of

pressure, after equilibrium conditions are met, over the footing surface. The footing areais divided into many small parts to produce refined bearing pressure diagram.

Verification Examples

Many practical examples wereselected to validate resultsproduced by the program andmonitor accuracy of the numericalapproach presented here. Theresults were compared with inputdata and not with solutionobtained from any other reference.

Table 1 shows input data and truesolution for selected problems.Note that in all problems atangible accuracy of 99.9% isachieved. The table alsodemonstrates number of iterationsperformed to solve the problemand run time taken on PC with P4-1.5GHz processor and 512MBRAM. Graphical representationof footing geometry and pressuredistribution diagrams forexamples 1, 2, 3 and 5 are shownin Fig. 4.1 to 4.8.

Observations and Conclusions

The numerical approach suggestedin this paper produces impressiveresults having a tangible accuracyof 99.9 percentage or better for allproblems under investigation. Thetime taken for finding the solutionis computationally economical forincredible accuracy achieved.Hence, the numerical approachpresented here can be effectivelyimplemented to automate thefooting analysis and design.

The use of Excel with its VBAenvironment is phenomenally userfriendly and endorses the structural engineers’ acceptance of Excel as a cogent tool forautomating structural design work processes. Even for such a complex problem like footingswith two-way eccentricity, use of Excel is found highly efficient.

Table 1Verification Problems and Comparison of Results

Problem NoItem1 2 3 4 5 6

Geometry and Load Data (Units kN and m)P 278.00 1300.0 1250.0 333.00 2000.0 2000.0Mx 278.00 162.50 2813.0 150.00 1500.0 1500.0Mz 250.00 1800.0 750.00 400.00 4000.0 3000.0Lx 6.00 5.00 6.00 4.00 5.00 5.00Lz 5.00 2.50 5.00 3.00 2.50 2.50ex 0.899 1.385 0.600 1.201 2.000 1.500ez 1.000 0.125 2.250 0.450 0.750 0.750ex/Lx 0.150 0.277 0.100 0.300 0.400 0.300ez/Lz 0.200 0.050 0.450 0.150 0.300 0.300Bearing Pressure at Corners(Before Modification of Pressure)PA 28.72 308.00 179.18 102.75 832.00 736.00PB 12.05 -37.60 129.18 2.75 64.00 160.00PC -10.18 -100.00 -95.85 -47.25 -512.00 -416.00PD 6.48 245.60 -45.85 52.75 256.00 160.00Results obtained by Numerical MethodCase 2 3 4 3 5 5Step 0.0030 0.0020 0.0030 0.0020 0.0020 0.0010P’A 32.41 360.24 749.89 146.10 3000.0 1500.0P’B 11.14 0.00 395.56 0.00 0.00 0.00P’C 0.00 0.00 0.00 0.00 0.00 0.00P’D 4.88 265.47 0.00 50.57 0.00 0.00c 4.624 2.200 4.077 2.923 - -d 2.976 1.200 4.513 0.888 - -

as % of (Lx x Lz)ContactArea 77.07 66.01 14.09 52.36 16.00 32.00Comparison of ResultsPrecovered 277.97 1300.8 1249.8 333.27 2000.0 2000.0exrecovered 0.8984 1.3831 0.5996 1.2000 2.0000 1.5000ezrecovered 1.0008 0.1251 2.2505 0.4503 0.7500 0.7500(%) Error inP 0.0088 0.0652 0.0126 0.0804 0.0000 0.0000ex 0.0647 0.0640 0.0733 0.0838 0.0000 0.0000ez 0.0822 0.0876 0.0219 0.0737 0.0000 0.0000Run Time DataIterations 1245 640 2081 673 946 1067Time(Sec) 7 3 9 3 3 3

Example Problem No. 1 ( Case 2 )

Figure 4.1 : Footing Geometry Figure 4.2: Bearing Presure DiagramExample Problem No. 2 ( Case 3 )

Figure 4.3 : Footing Geometry Figure 4.4: Bearing Presure DiagramExample Problem No. 3 (Case 4 )

Figure 4.5 : Footing Geometry Figure 4.6: Bearing Presure Diagram

Footings with Two-Way Eccentricity

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

-3.00 -2.00 -1.00 0.00 1.00 2.00 3.00

X-Axis: Length of Footing (Lx)

Y-A

xis

: W

idth

of

Foo

tin

g (

Lz)

Footing Load Point Original_NA Final NA

-3.0

0-2

.50

-2.0

0-1

.50

-1.0

0-0

.50

0.00

0.50

1.00

1.50

2.00

2.50

3.00

-2.50-2.00-1.50-1.00-0.500.000.501.001.502.002.50

Points along X Axis

Po

ints

alo

ng

Z A

xis

Base Pressure Distribution Diagram - 2D

-2.000-2.000 2.000-6.000 6.000-10.000 10.000-14.000 14.000-18.000

18.000-22.000 22.000-26.000 26.000-30.000 30.000-34.000 34.000-38.000


-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

-3.00 -2.00 -1.00 0.00 1.00 2.00 3.00


Y-A

xis

: W

idth

of

Foo

tin

g (

Lz)


-2.5

0-2

.25

-2.0

0-1

.75

-1.5

0-1

.25

-1.0

0-0

.75

-0.5

0-0

.25

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

2.25

2.50

-1.25-1.00-0.75-0.50-0.250.000.250.500.751.001.25

Points along X Axis

Po

ints

alo

ng

Z A

xis

Base Pressure Distribution Diagram - 2D

-20.000-20.000 20.000-60.000 60.000-100.000 100.000-140.000140.000-180.000 180.000-220.000 220.000-260.000 260.000-300.000300.000-340.000 340.000-380.000

-3.0

0-2

.50

-2.0

0-1

.50

-1.0

0-0

.50

0.00

0.50

1.00

1.50

2.00

2.50

3.00

-2.50

-1.00

0.50

2.00-2.0

98.0

198.0

298.0

398.0

498.0

598.0

698.0

798.0

Base Pressure Distribution Diagram - 3D 698.000-798.000

598.000-698.000

498.000-598.000

398.000-498.000

298.000-398.000

198.000-298.000

98.000-198.000

-2.000-98.000


-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

-3.00 -2.00 -1.00 0.00 1.00 2.00 3.00


Y-A

xis:

Wid

th o

f Fo

oti

ng

(Lz

)


Example Problem No. 5 (Case 5 )

Figure 4.7 : Footing Geometry Figure 4.8: Bearing Presure Diagram

Acknowledgement

I thank my company M/s. L&T Sargent and Lundy Limited, Vadodara, Gujarat, India, for thesupport, encouragement and providing computational facilities for this programming work.

References

1. Foundation Design, Teng W. C., Prentice-Hall Inc., Englewood cliffs, New Jersey.2. Roark’s Formulas for Stress and Strain, 7th Edition, Young W. C. and Budynas R. G.,

McGraw Hill, Englewood cliffs, New Jersey.3. Foundation Engineering, 2nd Edition, Peck R. B., Hanson W. E., and Thornburn W. H.,

John Wiley and Sons, New York.

-2.5

0-2

.25

-2.0

0-1

.75

-1.5

0-1

.25

-1.0

0-0

.75

-0.5

0-0

.25

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

2.25

2.50

-1.25

-0.50

0.25

1.00

-100.0

300.0

700.0

1100.0

1500.0

1900.0

2300.0

2700.0

3100.0

Base Pressure Distribution Diagram - 3D 2700.000-3100.000

2300.000-2700.000

1900.000-2300.000

1500.000-1900.000

1100.000-1500.000

700.000-1100.000

300.000-700.000

-100.000-300.000


-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00

-3.00 -2.00 -1.00 0.00 1.00 2.00 3.00


Y-A

xis

: W

idth

of

Foo

tin

g (

Lz)





Job Number: Originator: A. Tomanovich Checker:From: "Analysis of Eccentrically Loaded Rectangular Footing Resting on Soil -

Input Data: A Numerical Approach" - by Jignesh V. Chokshi (2004)Note: original verification problems were in metric units,

Footing Data: so neglect units and focus on numerical results.










Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -278.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -278.00My (ft-k) = 250.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 1:48 PM




ex = 0.900 ft. (<= L/6)

ey = 1.000 ft. (> B/6)



FS(ot)x = 2.500 (>= 1.5)



FS(ot)y = 3.336 (>= 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Lx = 1.381 ft.

Brg. Ly = 2.020 ft.

%Brg. Area = 77.06 %



P2 = 32.429 ksf

P3 = 11.119 ksf

P4 = 0.000 ksf


P3=11.119 ksf P2=32.429 ksf

P4=0 ksf P1=4.905 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 1:48 PM
















Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -1300.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -162.50My (ft-k) = 1800.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 1:48 PM




ex = 1.380 ft. (> L/6)

ey = 0.130 ft. (<= B/6)



FS(ot)x = 10.000 (>= 1.5)



FS(ot)y = 1.806 (>= 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Lx1 = 2.786 ft.

Brg. Lx2 = 3.826 ft.

%Brg. Area = 66.12 %



P2 = 361.090 ksf

P3 = 0.000 ksf

P4 = 0.000 ksf


P3=0 ksf P2=361.09 ksf

P4=0 ksf P1=262.938 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 1:48 PM
















Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -1250.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -2813.00My (ft-k) = 750.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 1:49 PM




ex = 0.600 ft. (<= L/6)

ey = 2.250 ft. (> B/6)



FS(ot)x = 1.111 (< 1.5)



FS(ot)y = 5.000 (>= 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Ly1 = 0.488 ft.

Brg. Ly2 = 0.925 ft.

%Brg. Area = 14.13 %



P2 = 748.675 ksf

P3 = 394.703 ksf

P4 = 0.000 ksf


P3=394.703 ksf P2=748.675 ksf

P4=0 ksf P1=0 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 1:49 PM
















Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -333.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -150.00My (ft-k) = 400.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 1:49 PM




ex = 1.200 ft. (> L/6)

ey = 0.450 ft. (<= B/6)



FS(ot)x = 3.330 (>= 1.5)



FS(ot)y = 1.665 (>= 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Lx1 = 1.078 ft.

Brg. Lx2 = 3.112 ft.

%Brg. Area = 52.37 %



P2 = 145.937 ksf

P3 = 0.000 ksf

P4 = 0.000 ksf


P3=0 ksf P2=145.937 ksf

P4=0 ksf P1=50.567 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 1:49 PM
















Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -2000.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -1500.00My (ft-k) = 4000.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 1:49 PM




ex = 2.000 ft. (> L/6)

ey = 0.750 ft. (> B/6)



FS(ot)x = 1.667 (>= 1.5)



FS(ot)y = 1.250 (< 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Lx = 2.000 ft.

Brg. Ly = 2.000 ft.

%Brg. Area = 16.00 %



P2 = 2999.996 ksf

P3 = 0.000 ksf

P4 = 0.000 ksf


P3=0 ksf P2=2999.996 ksf

P4=0 ksf P1=0 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 1:49 PM
















Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -2000.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -1500.00My (ft-k) = 3000.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 1:50 PM




ex = 1.500 ft. (> L/6)

ey = 0.750 ft. (> B/6)



FS(ot)x = 1.667 (>= 1.5)



FS(ot)y = 1.667 (>= 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Lx = 4.000 ft.

Brg. Ly = 2.000 ft.

%Brg. Area = 32.00 %



P2 = 1499.971 ksf

P3 = 0.000 ksf

P4 = 0.000 ksf


P3=0 ksf P2=1499.971 ksf

P4=0 ksf P1=0 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 1:50 PM

Pressures Distribution for Eccentrically Loaded Rectangular Footings on Elastic Soils

IVAN ADRIAN

Department of Steel Structures and Structural Mechanics Civil Engineering Faculty

“POLITEHNICA” University of Timisoara Str. Ioan Curea nr. 1, Timisoara

ROMANIA [email protected]

Abstract: In this paper a rigid foundation of a rectangular shape is analyzed under the action of a vertical load placed anywhere on the foundation area. A general method of analysis for calculating the pressure intensity corresponding to a given settlement for eccentrically loaded square and rectangular footings resting on elastic soils has been found through a comprehensive numerical procedure. The common assumption of linear contact pressure in footing-soil interface is adopted for the solutions. Special attention has been given to the cases where there are inactive parts of foundation, without contact with soil. The solution for these cases is not yet given in any Romanian codes. The procedure has been made clear by giving an illustrative example. Key-Words: foundation, rectangular footing, eccentrically loaded footings, linear surface pressure, contact pressure, active area 1 Introduction The design of a rectangular footing usually starts with an initial assumption of the most suitable dimensions of it. With these dimensions under serviceability conditions normally the bearing capacity of the ground should not be exceeded. Once the suitable base size has been established the footing may be designed for bending, shear and punching shear at ultimate load conditions. Under the action of a vertical load placed anywhere on the foundation area tension between the footing and the elastic soil in most cases may not appear. That’s why sometimes inactive parts of foundation without contact with soil will develop. In such cases the active part of footing will be in full compression, but the maximum pressure should be checked not to exceed the bearing capacity of the ground soil. More than that the active area of the footing is not allowed to be less than 80% of the initial base area. The numerical procedure developed in this paper solves the problem of finding the maximum ground pressure and the size of the active area in the case of eccentrically loaded rectangular footings with linear pressure distribution. It should be noted that the solution for these cases considering a linear pressure distribution is not yet given in any Romanian codes. Footings are commonly assumed to act as rigid structures. As a consequence the vertical settlement of the elastic soil beneath the base must have a planar distribution because a rigid foundation remains plane when it settles. A second assumption is that the ratio of pressure to settlement is constant. So the pressure

distribution below a rigid footing is considered to be linear. Neither of these assumptions is strictly valid, but each is generally considered sufficiently accurate for ordinary problems of design. 2 Pressures distribution computation A vertical load placed anywhere on the foundation area is consider to act on the rectangular shape foundation. As the cases with the entire footing area in compression where treated in detail by Jarquio [1], Vitone and Valsangkar [2], Algin [3] and others in this paper special attention has been given to those cases with inactive parts of the initial base of foundation. The following notations are used: B – width of footing area L – length of footing area ex – eccentricity distance along x direction ey – eccentricity distance along y direction Mx – bending moment with respect to x axis My – bending moment with respect to y axis pi – pressure magnitude at point i of the footing area (compression)bending moment with respect to x axis pmax – maximum compressive pressure V – volume of the pressure distribution diagram x0g – centre of gravity distance along x direction y0g – centre of gravity distance along y direction ariat – initial total area of the footing aria – remaining active area the footing Ixg – moment of inertia with respect to x central axis

Mathematical Models for Engineering Science

ISBN: 978-960-474-252-3 213

(second moment of area) Iyg – moment of inertia with respect to y central axis ixg – radius of gyration with respect to x central axis iyg – radius of gyration with respect to y central axis Ix – moment of inertia with respect to y principal axis Iy – moment of inertia with respect to y principal axis ix – radius of gyration with respect to x principal axis iy – radius of gyration with respect to y principal axis xs – x distance of the corner of central core with respect to x principal axis ys – y distance of the corner of central core with respect to y principal axis Many foundations must resist not only vertical loads but also moments and lateral loads about both axes. When all loads are transferred to the footing we may consider that they are produced by a resultant load F located eccentrically at distances ex and ey from the centroid of the base of the rectangular footing. The method presented here can be used directly to calculate the maximum ground compressive pressure and the size of the active area in the case of eccentrically loaded rectangular footings with linear pressure distribution for all introduced load combinations. There is no need to check if the eccentric load F is inside or outside of the central core (L/6, B/6). The method is based on the bending theory. In the first instance the geometric characteristics of the initial rectangular footing with respect to a central system axes are found. Then it is assumed that the resultant load F is placed in position ex and ey from the centroid of the base. This is equivalent with taking in consideration the effects of all applied loads (including bending moments) in a very simple and effective way. We have

exMyNz

:=

eyMxNz

:= (1)

where Nz is the total applied vertical load. If ex and ey represent a point inside the central core the entire initial footing area will be in compression with the characteristic values of the pressures in the corners given by

p1234Nz

ariat1

xi

ex⋅

iy2

+y

iey⋅

ix2

+⎛⎜⎜⎝

⎞⎟⎟⎠

⋅:=

(2) where xi is the x coordinate of the corresponding corner and yi is the y coordinate of the corresponding corner. When the resultant load F is placed outside the core according to (2) tensions should appear at least at one corner. But tensions cannot develop between the footing and the elastic soil. That’s why a part of the initial rectangular base becomes inactive (without contact with soil). The problem to be solved is to find the edge of the active area diagonal opposite to load F. This edge represents the zero contact pressure line. The active area

is totally in compression. The aim of the proposed iterative numerical computing method is to find the active area such that the total volume of the pressure distribution diagram to equalize the resultant load F and the centroid of this volume to be located along the direction of F. The first assumption for the zero contact pressure line is the neutral axis given by bending theory. The neutral axis is computed using its cutting distances on x and y axes as

xnixg

2 iyg2⋅ ixyg

4−

ext ixg2⋅ eyt ixyg

2⋅−−:=

(3) and

ynixg

2 iyg2⋅ ixyg

4−

eyt iyg2⋅ ext ixyg

2⋅−−:=

(4) Then the active area is considered that part of the initial rectangular base having inside the resultant load F and the neutral axis as one edge. The pressure distribution is computed using the geometric characteristics of this assumed active area using equation (2). If any tension is detected then the actual neutral axis is translated along a direction normal to it in order to reduce the active area, to increase the maximum pressure and in this way to modify the position of the centroid of the total volume of the pressure distribution diagram moving it closer to F direction. This new neutral axis becomes the assumed zero contact pressure line. The changed position of the neutral axis is obtained by modifying the coordinates of its ends as follows

x y( )xb xa−

yb ya−y ya−( )⋅ xa+:=

(5) and

y x( )yb ya−

xb xa−x xa−( )⋅ ya+:=

(6) where (xa, ya) – are the coordinates of one end and (xb, yb) of the other end. The process is repeated until approximately zero contact pressures are obtained along one edge of the active area. The following cases may appear

21

3

45

p3=pmax

B

L

p2

p4

Fig. 1


ISBN: 978-960-474-252-3 214

213

4

p3=pmax

B

L

p2

Fig. 2

2 3

4

p3=pmax

B

L

1 p4

Fig. 3

2 3

p3= pmax

B

L

1

Fig. 4

Due to the applied bending theory relations the total volume of the pressure distribution diagram equalizes the resultant load F. The last check is done to see if the centroid of this volume is along direction of F. Mathematically this means that the load F is positioned in the opposite to zero contact pressures edge corner of the central core. The coordinates of the corner of the core in the principal system of axes of the active area are given by

xsiy

2

xng−:=

(7) and

ysix

2

yng−:=

(8) If xs exα− ε<

(9) and ys eyα− ε<

(10) the final solution of the problem has been reached. The four possible pressure distributions for a rectangular footing under two-way eccentricity with active area where drawn up for initial positive ex and ey values. This numerical procedure is valid for any sign of applied bending moments. The active area and the maximum compression will rotate with respect to the

centroid of the initial footing at bending moment sign changing. 3 Numerical examples The algorithm of the proposed method has been written as an Mathcad application. Example 1. Find the maximum soil pressure and the size of the active area in the case of an eccentrically loaded rectangular rigid foundation with the base footing of L=2,0m and B=1.6m subjected to the action of a centric compressive force Nz=500 kN and two positive bending moments Mx=140 kNm and My=150 kNm with linear pressure distribution. Using the above described procedure we get an active area of aria=2.6168 m2 that is 81.77% of the initial total area ariat=3.2 m2. The shape of the active area corresponds with the first form given by Fig. 1.

Fig. 5

The maximum pressure on the elastic soil is pmax=502.86 m2. Example 2. Modifying in example 1 just the bending moment Mx increasing its value to Mx=200 kNm the shape of the active area corresponds now with the second form given by Fig. 2.

Fig. 6

Using the proposed algorithm we get an active area of aria=2.095 m2 that is just 65.55% of the initial total area ariat=3.2 m2. The maximum pressure on the elastic soil is pmax=657.38 m2.

1− 0.5− 0 0.5 1

0.8 −

0.4 −

0.4

0.8

1− 0.5− 0 0.5 1

0.5 −

0.5


ISBN: 978-960-474-252-3 215

4 Conclusions This paper presents a numerical method to determine the stress in an elastic soil and the size of the active area for a two-way eccentrically loaded rectangular footing, assuming a linear pressure distribution. The method consists in an iterative process which successively adjusts the position of the zero contact pressure edge to reach an active area fully in compression. The process is implemented in a computer program. The program directly evaluates the maximum pressure, the shape and size of the active area. Any load combinations may be considered. It has been found that zero contact pressure edge position of a footing does not depend on the magnitude of the eccentrically applied load, but on its position on the footing. The computer program represents a useful tool for Romanian designers as there is no specified calculus method for the linear pressure distribution in the presence of inactive areas in the Romanian codes. References: [1] R. Jarquio and V. Jarquio, Design of footing area

with biaxial bending, ASCE Journal of Geotechnical Engineering, Vol. 109, No. 10, 1983, pp. 1337-1341

[2] D.M. Vitone and A.J.Valsangkar, Stresses from loads over rectangular areas, ASCE Journal of Geotechnical Engineering Division, Vol. 112, No.10, 1986, pp. 961-964

[3] H.M. Algin, Practical formula for dimensioning a rectangular footing, Engineering Structures, Vol. 29, No.6, 2007, pp. 1128-1134

[4] W.H. Highther and J.C. Anders, Dimensioning footings subjected to eccentric loads, ASCE Journal of Geotechnical Engineering, Vol. 111, No. 5, 1985, pp. 659-665


ISBN: 978-960-474-252-3 216




Job Number: Originator: A. Tomanovich Checker:From: "Pressures Distribution for Eccentrically Loaded Rectangular Footings

Input Data: on Elastic Soils" - by Ivan AdrianNote: original verification problems were in metric units,











Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -500.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -140.00My (ft-k) = 150.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 1:50 PM




ex = 0.300 ft. (<= L/6)

ey = 0.280 ft. (> B/6)



FS(ot)x = 2.857 (>= 1.5)



FS(ot)y = 3.333 (>= 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Lx = 0.705 ft.

Brg. Ly = 0.699 ft.

%Brg. Area = 81.77 %



P2 = 502.858 ksf

P3 = 168.191 ksf

P4 = 0.000 ksf


P3=168.191 ksf P2=502.858 ksf

P4=0 ksf P1=117.98 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 1:50 PM




Job Number: Originator: A. Tomanovich Checker:From: "Pressures Distribution for Eccentrically Loaded Rectangular Footings

Input Data: on Elastic Soils" - by Ivan AdrianNote: original verification problems were in metric units,











Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -500.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -200.00My (ft-k) = 150.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

1 of 2 12/24/2013 1:51 PM




ex = 0.300 ft. (<= L/6)

ey = 0.400 ft. (> B/6)



FS(ot)x = 2.000 (>= 1.5)



FS(ot)y = 3.333 (>= 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Ly1 = 0.539 ft.

Brg. Ly2 = 1.556 ft.

%Brg. Area = 65.47 %



P2 = 657.383 ksf

P3 = 227.783 ksf

P4 = 0.000 ksf


P3=227.783 ksf P2=657.383 ksf

P4=0 ksf P1=0 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

2 of 2 12/24/2013 1:51 PM



Subjected to Uniaxial or Biaxial EccentricityJob Name: Program Validation Subject: Example Problem 9-5, pages 259-263

Job Number: Originator: A. Tomanovich Checker:From: "Foundation Analysis and Design" - Second Edition

Input Data: - by Joseph E. Bowles (McGraw-Hill Book Company, 1968)

Footing Data:










Pier #1Xp (ft.) = 0.000Yp (ft.) = 0.000

Lpx (ft.) = 0.000Lpy (ft.) = 0.000

h (ft.) = 0.000Pz (k) = -60.00Hx (k) = 0.00Hy (k) = 0.00

Mx (ft-k) = -120.00My (ft-k) = 120.00


Resultant of Loads

Kern Area

Y

X

+Pz

+My

+HxQ

Dh

T

L

Nomenclature

Lpx

+Mx

+My

+Hx

+Hy+Pz

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ex = 2.000 ft. (> L/6)

ey = 2.000 ft. (> B/6)



FS(ot)x = 1.500 (>= 1.5)



FS(ot)y = 1.500 (>= 1.5)




FS(slid)y = N.A.



FS(uplift) = N.A.


Brg. Lx = 4.000 ft.

Brg. Ly = 4.000 ft.

%Brg. Area = 22.22 %

Biaxial Case = Case 4 6*ex/L + 6*ey/B = 4


P2 = 22.500 ksf

P3 = 0.000 ksf

P4 = 0.000 ksf


P3=0 ksf P2=22.5 ksf

P4=0 ksf P1=0 ksf

CORNER PRESSURES


Brg. Lx2

Dist. xPmax

Dist. y

Dist. y

Dist. y

Dist. y

Brg. Ly



Brg. Ly1

Dist. xPmax

Brg. Ly2

Dist. xPmax



Brg. Lx1

Dist. xBrg. Lx Pmax

Brg. Ly

B

L

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Documents

· For Assumed Rigid Footing with from 1 To 8 Piers (Load Points) Subjected to Uniaxial or Biaxial Eccentricity %