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Footing Analysis: Axial Load & Biaxial Moments - Numerical Approach
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POINT OF VIEW Analysis of isolated footing subjected to axial load and high biaxial moments and numerical approach for its solution Bijay Sarkar & & In this paper a rigid isolated foundation of square or rectangular shape is analyzed on the action of a vertical load and high biaxial moments at centre. The pressure intensity corresponding to any given set of above loads on footings resting on elastic soils has been found through a general method of analysis and solution is made 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 where there are inactive parts of foundation, without contact with soil and necessary equations on case to case basis are deduced. Flow Chart for computer procedure is also provided at end. The solution for these cases is not yet given in any Indian Standards. && area coincides with the footing. The the centre of & ; = . = ; = . = Parameters P, , and dimensions of the foundation along = . properties are known and we get == with ;its ; sectional = . ; = . the foundation at any location = pressure , distribution under , , , (X, Y) with respect to centroidal axis of the footing by ,, , , using bending equation : , ,following = ± = ×± ± × ×± × ; == ±± ××±± ×× = ± × ± × ∴ = − × × ∴ = × − ×× − × × − × ∴ × ∴ ×× ×× −− ×× == −− × ∴ = − × × − × × −= −− −…………….….…….. ∴ =∴ (1) …………….….…….. ( −− −− ∴∴== …………….….…….. (1) …………….….…….. (1) = − − ... (1) ∴ …………….….…….. (1) × × Introduction ×× When a rectangular/square isolated footing of size L x B × where × × is subjected to a set of forces comprising of compressive L= of the footing ××Length axial load P and bi-axial moments Mx & My at centre of × of the footing B = Breadth footing, load P alone may equivalently beconsidered & = Perpendicular X distance from the Centroidal at eccentricities in X & Y directions from origin at the point & the of on the Cross-Section at which pressure ; Y-axis following location from centre of footing : to . be determined. ; = = is = distance from the Centroidal X-axis of = Y = Perpendicular ; = . = on the Cross-Section at which pressure is to be the =point = , , = = = determined. When pressure at any location under the footing is , P ==Vertical Force Acting on the section at centre. = compressive in nature, Centroid of the effective footing = × × = ± × ± × × × ± × ± × × 60 The Indian Concrete Journal June 2014 + <= −− >= − , or,>= , or, + <= − − × × ∴ = − × × ± × = ± × × = ± × ± = ± × ± × ×= ±× −× × × × ∴ −× = ± ± ∴×× ±= −×= × × ± − × × × × ∴ = − × × × × − ± POINT + = , assuming = ∴ = − × × − × × OF VIEW ∴ = − × × − × × ∴ = − × × − × × = − ∴−× ∴× …………….….…….. − × × −(1) × × = ∴− −× −−− =× …………….….…….. (1) = × × × × A = Compressive area of the section = BL Due to uplift of the footing area from soil, Neutral Axis −×−× × × × × × ∴ = − − …………….….…….. (1) and yaxis at . M y = Bending moment remain at the about∴the (1) equation (1), do not = − Y-axis −= Centroidal …………….….…….. as calculated (1)from ∴ = − − …………….….…….. ∴ = − − …………….….…….. (1) × calculated location, however, it gets shifted to other × P e −(1) − …………….….…….. ∴ (1) (1) −x …………….….…….. …………….….…….. = − − = − …………….….…….. location to adjust the phenomena of uplift (1) − × redistributing …………….….…….. − − (1) × on xaxisan and − × − P × e moment about the Centroidal X-axis = the base pressure under the footing for obtaining x×Bending M y × × moment of area Xequilibrium. × about Centroidal × Ix = Second × × axis = × × First we divide the whole footing into four equal quadrants. moment ofarea Centroidal Y-axis = Iy = Second × about × Assumed, biaxial moments are such that vertical load P e = alone may be considered acting at in the upper x = Eccentricity of Load along X-Axis = = rightmost quadrant wrt centre. Thereby, the lower left most corner of the footing will experience the least Eccentricity of Load along Y-Axis e = = y = = = = = + > , the lower left most pressure. When = = = = × × = = corner of the footing starts to uplift i.e. the neutral axis = = However, above equation remains valid till × whole starts to enter into the footing area from lower left most × , − − >= × corner intersecting the footing sides. Depending on the + × +<= <= − − >= , or, or, area of footing remains × × cross-sectional of axis with the sides of the footing, intersection neutral + − >= , or, − <= underneath foundation is the , or, scenario + bearing − − >= <= pressure under compression i.e. , or, − + <= , or, different cases : >= <= −− − >= , or, + divided − − >= into five or, − − >= , or, + <= , assuming , , + <= + = = = >= , or, + <= , assuming , = , or, + = which can = be represented as a general >= >= ,or, + +<=<= + = , assuming When = , NA = lies outside the footing area. Effective 1. , = = + = , assuming = , = line + = , assuming assuming form = , = shape remains rectangular and equation (1) is of straight equation .+ = , ,assuming yaxis and at and yaxis at . , assuming , + = = = + = , assuming assuming = , = = , = Issue is already discussed above applicable. ng = , = ssuming = , = i.e. intersecting the x-axis at and yaxis at . (Case – I). at . and yaxis at at . . and and yaxis yaxis − on xaxis and on yaxis. yaxis at Considering opposite quadrant, the . yaxis. atand . and on . the adjacent two sides of xaxis −yaxis − −and axis y-axis .ataton 2. When NA intersects . xis at . and − lower most corner of the footing. Contact area xaxis on yaxis. − and on left x-axis we can get the co-ordinates onxaxis − on yaxis. − on and =Issue is. discussed = ;shape. reduces to apentagonal and − − on xaxis yaxis. on xaxis on yaxis. and − −on on xaxis and on yaxis. − − on yaxis. we −axisand − Connecting on −xaxis and onyaxis. under Case – II. − on these co-ordinates, on y-axis. nd yaxis. xis and − on yaxis. geta diamond + shorter + shaped bounded zone around the centre the 3. When NA intersects + two opposite + = of the footing which is called “KERN” or “Central Core” Contact sides of the footing. area − reduces to a non − of the footing. So as the load is located rectangular shape. Issue is discussed under within this zone, whole cross-sectional area is effective in transferring Case – III. loads to soil as compression. Problem Definition When the eccentricities are such that the load location crosses the boundary of the “KERN”, equation (1) shows a negative pressure i.e. tensile pressure at some zone under the footing. As underneath soil can not resist such tensile forces, the footing area in that zone gets detached & uplifted from soil and thus the above equations for base pressure calculation do not remain anymore valid. 4. When NA intersects the two opposite longer sides Contact area reduces to a non-rectof the footing. angular shape. Issue is discussed under Case – IV. 5. When NA intersects the adjacent two sides of up per right mostcorner of the footing. Contact area reduces to a triangular shape. Issue is discussed under Case – V. The Indian Concrete Journal June 2014 61 POINT OF VIEW By assuming, Analysis of the foundation + For Case – II to Case – V, analysis of footing has been = + = − done and general equations on case to case basis are = Constant for given section and +− deduced here. In all the cases, aim is to find out the = loads − effective area, CG of the effective area, sectional properties + > , of the effective + area wrt centroidal axes system i.e. through + of the = CG effective area. As we are not using principle = − = Constant for given section and + > , + > , +− we are to axis +to calculate > , the sectional properties, = the product moment of area of the effective loads − calculate = footing also and use all such data in General Form of = Equation Bending and assembling all the cases into a = = Constant for given section and loads P, M , M are acting graph. Assumed that a set of forces b l i s o l a t e d at the centre of a rectangular , Am + Bn + C = p. This is the General Form , We can write, footing of size (L x B). P is acting in the vertically + + + Equation of =pressure underneath foundation subjected to = M downward, is acting along B (from down to top of = , b + + − − + high Bi-Axial This is a straight line equation. − Eccentricity. + + this paper) and M l is acting along L (from left to ∴ = + + at some combinations of Further, it can be observed that ∴ = + − right respectively. When a portion of of this paper) + + + and n,pressure − constant. remain m pmay + footing area is lifted, CG of the effective compressed ∴ = ++ + − = = centre footing and as CG is .area shifts away from the + + = know ………………………..(3) axis is equal to0. − + we ∴ = Now, +−− at+neutral + that pressure changed, the equivalent eccentricity of load also changes ………………………..(3) ∴ = Therefore, + + for −any Co-Ordinate +− (m, n) being on valueof +− from say, Therefore, . . − = . el , e;b to = . ∴ = the +neutral axis,weget, + Am + Bn + C = ………………………..(3) is nothing . Moments 0. This Revised acting at revised CG location are − − = = but a straight = line equation representing the Neutral = ; = . = ; = . Axis. = ; = . + Bending Stress +of Dimensional General Form Two ∴ = + + ……………………..(2) = ( − ) × = where = − Equation will : be − = ( −Substituting = − , , for moments in ) × thewhere valuesof= − ,, + + + + equation (2), we get ……………………..(2) = + + + = + + = (−+ ) × = ……………………..(2) where = − − + + + ……………………..(2) = + − + − −...(2) + + ∴ + − − = =∴ where = ++ = + + = = =∴ =where = −− where, = = where = p = Pressure Centroidal ) wrt at co-ordinate (m, n + + = + × = where = + + + + Axis + = ∴ = ∴ =× = +where + …………………… + + = + − + = − + ……………………… P ∴+ = Vertical Load at centre of the footing ...(3) − where = × = = − − Ar = Revised Effective area of the foundation = − × equations = where = − for Sectional Eccentricities = Revised Moment at Revised CG location about = −General × = where = Properties, − Revised Centroidal Axis YY & co-ordinate of Maximum Pressure Location wrt Revised = − × = where = − = Revised Moment at Revised CG location about Centroidal Axis are found out and used in the above +equation = +=finding (−− =)on )×× where where (= − − )× = where Revised Centroidal Axis XX case==to case = − out the location = + = (× basis = + for where I y where = × = = as wellas Maximum Bearing Pressure : = Second Moment of Inertia of the effective area +of Neutral Axis + = × = where = about Revised Centroidal Axis YY I x = Second Moment of Inertia of the effective area =× ==-II where = = = where = where = CASE = = where = = × = where = about Revised Centroidal Axis XX NA cuts AB and AD. I xy When Here uplift portion is APQ = × = where = = Product Moment of Inertia of the effective area effective + portion is PQDCBP. considered that m and +++is where = + ×× == where ==This × = where = = X-Axis Co-ordinate of the location where pressure = = NA cuts AB at P & AP = yB and NA cuts AD at Q & AQ = × = where = is to be found wrt Revised Centroidal Axis = × = where = = xL n = Y-Axis Co-ordinate of the location where pressure = × = where = is to be found wrt Revised Centroidal Axis = − × = where = =− − = where = − × = where = −× 62 The Indian Concrete Journal June 2014 + ++ = + − − ++ = == − − − + − = + + = = = +− VIEW POINT OF − − = + + = = − + == fraction − in side Therefore, uplift AB is y & in AD is x and = − − = = ineffective neglecting the triangular uplifted portion APQ, , = PQ is the location of NA properties of effective portion PQDCBP all the sectional = AP = yB ; PB = vB = (1-y)B ; v = 1-y , foundation = of the calculated as follows : = area are AQ = xL ; QD = uL = (1-x) L ; u = 1-x + + + , ∴ O is the origin of Centroidal Axes XX & YY = + , , Revised + and + + of the − 1. Effective Area Section Centre of ∴ + ,= Y + + , : Gravity − + ++ + + + B + ,,∴ = + C − ++ ∴∴ = = ++ +axis at +system ………………………..(3) +origin orthogonal Considering an located ∴ = + + − + + + − ∴ = + CG area +of the foundation + + O −effective of O+ + at − X ∴A,∴ co-ordinate = = + − + + ………………………..(3) + + + + ∴ = R ∴ = + − PQDCBP follows calculated is + : +− S as − ………………………..(3) P − + + ∴ = + + + + − ………………………..(3) ………………………..(3) ∴∴ = = + + +− + + +− ………………………..(3) − − − − (i)∴ Effective Area PBCS) : = +(Rectangular + + + + − + ………………………..(3) ∴ = + − + + + + Y A Q D ∴ = + −………………………..(3) − = ………………………..(3) ∴ = = (+− ) ×− + where = −− − − = (− ) × = where = − = ( ) (( )) ( ) ( () ) ( ) X Centre from : = of −Gravity × = AB where = − −= ×× = where= = − = = − = − Figure 1. Effective area of the section and revised centre of gravity = wherewhere = = − × = where = − = = − = − ×= = where where = × = where = = = = − × where = − = − = − × = where + = = +where = where = × = = where = = where = ofGravity := = from AD Centre (ii) × = where = =& (iii), From (i), + + where = × = = where = × = = = where = = × = where = == = +where = + ==+ where = = = × = where = + × = == where = + = where + TotalEffective Area, × ==×where where = where = = × = == + = where = = − × = where = × + = =− × = += × (Rectangular = where where = + =: QRSD) = + + = − = × Where = − − ++ + = ×= =× where = (ii) = Effective ×− Area where = = + = where = == ×× = = × = where = where == where = = − +× = where = − where = × = = = = where + − −× × where = = +− − + = − = = where × Where = − = = × = = + + = − = × Where = − = where = = +− × = where +=− + + where = × = = where = × × = = = Centre − = where = − − + + = − = AB: = = − where = = ××=Where =+ −− == from × from −−Gravity +×× = == of =+ where Therefore, Distance of revised AB + Where = = × = − − Centre of Gravity where + × = + =+ = + − where = = + ×=× =×==where = = += where : + + − + + = − = × = − × where += = =× × = = where + = = × = Where + −= + = = − = × = × = +× = where = + = + − + −− − == +=×××===where == =+ where where + + = − × = × + Centreof Gravity + where from ==AD: where === − × = × Where, == = − = − × = × ××= = = × × = =where = − − where = = = − + + − where = × = + = − + − where where + − − = + = = ===××× = = = × = AD × distance ×: Similarly, of revised Centre of Gravity from = where = = = × = = = = × = × = − where = − − − + − = × = + − where = × = = where =(Triangular (iii)Effective : × Area + PQR) = = = × = + = − − × × = × = − = × = = == where − = where = × = − = × = = where + + − − = × = where = Where, = = × = × = × − = + − × − × − − − Centre of Gravity from AB: = + − − × − × − × = + × − − × − × − = + × = where = − − − − Moment 2. Second of Inertia wrt revised Centroidal Axis = + × − − × − × − = = × += × where − = − × − × −: XX from AD: = + Centre of Gravity − − − − − − − = + the value − where = ×= = + and Area × −= − × − × Putting − about –MI simplifying, I = MI of XX-axis of Area APQ ofABCD x = = × = where = − − + − − XX-axis = − − Putting + − − about the (and value ofand simplifying, of − simplifying, ) the value Putting = + + = − = × Where = − = + − − − − − Indian − Journal June 2014 = + 63 The Concrete = + + = − = × Where = − Putting the value of and simplifying, and simplifying, ( − ) Putting thevalue of = Where = − − = + + = − = ×−− − = × Where = − + + − − × = × = − ++− = −− + == +++ − = −−− =+ + + × ===−− = ×× = −×× − = = × == =+ + + − − = × = × = = × = × = VIEW − OF −−= + +− POINT − − −−−= × = × = − − − = = × = × = − − − − − − × = × = − Moment of Inertia (PMI) of the effective − 4. − Product + + = − == × = × = wrt revised Centroidal Axes : + × − × − section × − − − − − − + + × = × = If NA lies outside the section, Centroidal Axes are = − − = × = = − − × − + ××− ×× −− ××× ×−−× − − = = − ×= − − − =−+ symmetrical at the centre of the footing. As in such + − − × − × − = + − − − − + × − − × − × − = + × − × − condition, Centroidal axes are the principle axes, Product − − × Moment of Inertia of the section becomes zero. But, when NA lies within the footing area, the Centroidal Axes may of − −simplifying, −and − × == Putting × − + the − − − − ×++value −− −− =+ =− −− − − + − − − = + − − not be symmetrical or in other words, the Vertical and − + = − − − Putting Horizontal are no more the principle axes and simplifying, the valueaxes of atCentroid = + × − − ( ×− )− × − of the section. Principle axes may be in other orientation the value of value ×Putting − and simplifying, +simplifying, =×and − of he of−Putting simplifying, Putting the value of simplifying, the value of simplifying, and Putting value and simplifying, Putting the and − − − = the the +value of − thanthe vertical and horizontal axes. − and simplifying, Putting value of Therefore, when we of and − ( ) Putting the value simplifying, Putting value of and simplifying, are + the = considering − Vertical ×and Horizontal axes at centroid ( ) − ) ( −− ) − ( − − = + − − − + ( ) − = × of the effective section, we must consider the Product ( ) − −− == + = −−− −+ and −(− ) + − Putting + simplifying, the value of − the effective area which Inertia − + −− −− of of ( − ) contact = ( − ) Moment − = − += = − + × Putting the value of and simplifying, − + × × = × Putting willsimplifying, take if pressures calculated with Putting value of − of of thecare as and simplifying, are − being and − thevalue of and simplifying, thevalue Putting == × ( ) − respect to Principle Axes. × = = × mplifying, = −×× = × + − − ) × (−−−)) − Putting − ( ( − the value of and simplifying, = × = − + = − + × − = × (+ − − Where, + ×× of the Revised ) =+ =×−− = − = =Moment × Effective −− of Inertia +× ×−− = − +Product + (PMI) +− × − −− + =− − − − = + × ) = Where, − − Footing Area, × − − = − − − ( − ) = Where, = − + × Where, − + × = × − = −= + = =PMI ×of ABCD × × −Ixy − − wrt revised CG PMI of APQ wrt = − + × = =× + × − − × − × × − − ABCD wrt revised CG PMI of APQ wrt revised CG CG Ixy revised = PMI Inertia of 3. Second Moment of wrt Centroidal Axis YY: − − − = − − + × = × − + × Where, = − = − Where, + × = − + × − Where, × − − − ××− ×× ofrevised ×× − of −++of −−− −− APQ Whole CG − Ixy ABCD PMI = − wrt − CG − −PMI × wrt+revised of ABCD × CG× =PMI revised of APQ =MI − CG ×× + −I×= × × − − × Area about YY-axis – MI Ixy = ABCD wrt PMI of wrt revised y== = + × − − × − × × − == + + − × × − − × − − − × × −− + × APQ − YY-axis − × × −about Triangular Area Where, = + × − − = − − − − × +wrt × ×C revised in above ×and wrt − −Putting −Ixy + × revised ×, =PMI − CGPMI = simplifying, − − of APQ of ABCD ofAPQ === −+ ++ − − −− −− Ixy ABCD wrt PMI − ABCD × wrt of Ixy = of revised CG PMI wrt − revised wrt CG ×− − ×PMI CG − revised of APQ = PMI revised CG + = − − −− −×−− = + − −−− ++ − = − − − simplifying, in above and , Putting = + × − − × − × × − G Putting , in above and simplifying, G = − − − − × + × Putting , in above and simplifying, = − − + − − − x y Ixy = PMI of ABCD wrt revised CG PMI of APQ wrt revised CG = − − − − × + × = − − − − × + × × − − − × − × × − = − − + − − = ×− −×simplifying, + + −× ×− − − − − + − =Putting −−in − and and , above = − = − − − Putting , in above simplifying, Putting , in above and simplifying, − − = + − − − − − andsimplifying, simplifying, Puttingthe thevalue valueofofand Putting , = × − − Putting in above and simplifying, = − − + − = − − + − − Putting the value of F and simplifying, = × gx − = − = − + − + −− − − ) ( − − + ×( − ) − = == + × = − = + − − − −− − − + − − + − − − −− = × − − − == × − = ; = × and − − − − Where, + − − = −× = − − − = = and ; + −− − − − += −− = = − and = ; − − − − − and along B and L respectively was calculated based on Centroid b − − + × − Where, Where, == − − Where, − + × = − − + − − − to uplift of some of the − − isrelocated − −− =− ; portion = due = footing area, eccentricit on the revised = ; the and and = ;and = new Centroid tofind out moments : − − − L was − − and along − B and respectively calculated based on Centroid and along B and L respectively was calculated based on Centroid being at centre of the footing. As Cent (i) Revised Eccentricity along B : − − is relocated due to uplift of some portion of the footing area, eccentric is relocated due to uplift are required to be recalculated ba = = area,eccentricities ofthe; footing andof some portion : on the new Centroid tofind out the revised moments on the new Centroid to find out the revised moments : − − + − = + − = + − × Ixy = PMI of ABCD wrt revised CG PMI of APQ wrt revised CG = Ixy = PMI of ABCD wrtJune revised The Indian Concrete Journal 2014 CG PMI of APQ wrt revisedCG based on Cen along B: calculated and Eccentricity was along Band L respectively was (i) Revised Eccentricity B : (i) Revised Bcalculated and L respectively was calculated based on of Centr and along based on Centroid being at centre the and along along B and L respectively == ×× 64 − of− APQ − −revised + × − − CG ×CG × wrt = PMI ABCD=wrt = = = ofPutting − + +− −×× ,revised PMI and in above == + +− − simplifying, = = Putting − − − × + × × − − − = +Eccentricity , in and simplifying, (ii) −along =L Revised := − ; above − Eccentricity Where, = = Revised : − L + + Eccentricity ==+ + (ii) =LRevised ++=−Eccentricity −Eccentricity ×× ;= +−along = (ii) × −− = = Where, − − − (ii) Revised along along L : − (ii) Revised along L : ++ ==== == Where, == −−−− ;;; ; − −: ++ Where, Where, = == Where, = Where, − = −−−−− − = ; − = × × − simplifying, × + − + Putting − −and ++Where, in above === Where, − = − ; = , = −=−; − = − − + − − POINT OF VIEW +− = += =− Where, , in above and −− = − =Where, + ng + =− − − + + − ×× − == =× − = simplifying, − − + − =− − − −+ = + −++ −− +− = + − = = + + − − = = + − × = = + − × == L: == −−; ; ++ along = (ii) Where, Eccentricity Where, Revised Co-ordinate Centroidal System : ng , in above and simplifying, = − − + − − − = × −− = =− ×− + − − − − − = = + Where, ; = − − ;the footing = = − Co-ordinate = point Where, intersecting and; = + − × = (i) and whereNA = ; + =is== Where, Where, − = = − +of ; =−++ PWhere, − + − − = − + = − : − = −= × − − side − = where − = − × =AB + − − −− where = − × − + − − − =−− Revised = − of ==−=−− = where = − × − = ×5. Load wrt. revised Centroidal = where = − × = Eccentricities where = − = − × = where − × = − − + − − − −− = − = where = = = − × − −−−−− − = where = − = − × = where = − = − × Axes : where − Where, = − = = + ; − = − ×− −− where − − = =− = = × = − −= − + = − × − where − ( ) = calculated −B and − where along Lrespectively − = × − −− and was based on Centroid being at centre of the footing. As Centroid == − where where = = − − = = − − × × ( ) = where = − = × − = = −eccentricities − −− e +and ; e − =Earlier and − − where −−− − B along and L respectively )where == ==based where == −− ==required ××××((( −−−− b uplift l of ) where −− −− = − some eccentricities to be recalculated ) − − to = due − portion area, == where = and is relocated of the footing ; are = − ( ) = where = − = × − − − − − was based o n Centroid being at centre of calculated − − − ( ) = where = − = × − on the new Centroid to find out the revised moments : ( ) = where = − = × − − − − = − − − + − ( −−− − ) where where − =− − ) = Eccentricity where ; along −where ×=×(= ==−− ==−−××= = = the footing. is relocated due to uplift of some As − and − Centroid =B = : −−where −(i) Revised = − = − × − − wher = = − − = = − − × × − = where ×× ( − ( ) ) = = where where = = − − = = − − = = ; footing portion of the area, eccentricities are required to − − − − − and B and L respectively was being at centre ofQthe footing. As Centroid the footing = outbased + on− Centroid of point + based = new + Centroid − × calculated be to find the (ii) re-calculated − = along −on − the Co-ordinate where NA is intersecting −− =is Bto =: and revised ;due relocated and moments along L respectively was calculated based on Centroid being at centre of the footing. As Centroid uplift of some portion of the footing area, eccentricities are required to be recalculated based side AD : ( ) = − where = − = × − − = × ( − ) where = − − − − − = × ( ) = − )where where = − is due to uplift the footing area, eccentricities are required to be recalculated based = == of=some −the = revised − × of moments ( ( ) wher = − − = × × − − − onrelocated the new Centroid to find out : where ( ) = where = − = × − portion − − ( ) where = and along B and L respectively was calculated based on Centroid being at centre of the footing. As Centroid = − = × − − − = − − = × (( − )) where Centroid to find out the revised moments : − − on the new = − where = = − = × − − ( ) where = (i) Revised Eccentricity along B : = − = × − where = = ) where (i)isBRevised Eccentricity along B : portion basedof where = −−− =of the− −are required = × × −be d along and L respectively was calculated on Centroid being at centre footing. As(Centroid −= relocated due to uplift of some to− recalculated −= −eccentricities (i) Revised Eccentricity the footing area, along B : ) where based = −− = − = −×( )− where ( = = × − tofind revised moments : = are new = of Centroid + Where, =area, − eccentricities ; = − − ocated due on to the uplift some portion of out the the footing based required where = −×(− − ))where − to be recalculated = − = × ( − = −Centroid + )where −being B = +respectively out −the +=calculated centre − −− − = −B = where = − = − × Lfind nd along was at of the footing. As Centroid = along −:×=based where e new Centroid to revised moments : ×(on (i) and Revised Eccentricity − ( ( ) ) where where = = where −− =−−−− = − ==− − ×− ==××where = + − = + − × = + − − ==== − ==−−− be =recalculated −− ×××× based ocated Eccentricity due to uplift of some of the footing area, eccentricities are to where −− − =−− = − along where evised B : portion == required where = − = where = − = − × − − −− − he new Centroid the revised moments : = where = − = − × = + − =to find + out − = + − × = where = − = − × − −− − −× = where −) − 7. − −location × where = −− of =maximum ××((−− = = = = : × − where where = == ) = + − = =Revised + Eccentricity − = along + −B Co-ordinate pressure C : − −) whe − where ( ( ) = = − − = = × × − − − ( ) where = = − = × − − = = where where = = − − = = − − × × = − ; = − = + Where, − −− = − ; = − = − = ( − ( ) ) Where, = − = − = + − = + − = = ++ − × ( ) = − = − = −− − == ) ) (−− = ====−−−− ((× −− −− = = ) = where − − ( ) = + Where, ; = − ; = = − = − = − − ( ) = where = − = − × ( ) = − = − = − = ×−( − ) where where wher where = = − − = = −− × − − =−−× = −= −= −= × ( ) − − = − = ( = − ( )) = = = + Where, = − ; = − ( ==== (−−−− )) ==− −−− ) − = ( = ( =)=) − = − === ((−−) ) = (−)−− = − =− = ( − − = + Where, = − ; = ) = ( − ) = = − − − = ( − = = − = − × − where RELATIONSHIP = ) KNOWN =−− ((−−) = == )BETWEEN − − − = = and = ((−− ) = = ( − ) ) − −AND (FOOTING − ( ) ECCENTRICITIES = − = − and = = & and and and and (ii) Revised Eccentricity along L : (− : = FRACTIONS (ii) Revised Revised Eccentricity Eccentricity along L :: = )y (ii) −and = & x& LIFTED and − and = ( − ) ( −)) = − − ( − & = −& = (− ) = = = and and & & & & and along = −L: = ( − (ii) Revised Eccentricity points and lie on NA, Pressure +) − As Revised Eccentricity and ) Revised Eccentricity L :+along & = For (ii) −− L××: & = = + − = ++ −−along == + (i) = = coordinate as ≠ , substitut form & for locations. at these Now general of NA for NA & and p is(i)0For in = two = coordinate and as , subs ≠ substitu (i)(i)For For ==== bending ====& & for NA coordinate and asas ≠ ≠≠ ,,, ,substitu For for NA coordinate and substi pressure (i) for NA coordinate and as ( ) = − = − (i) For for NA coordinate and as substitu dimensional equation (3), we put the ≠ = =for NA coordinate and ≠ , substitu + − = + − × = + − (i)For as and and = (i) = NA NA coordinate ≠ by, substitu for = =of for coordinate asand , substituting a − × ≠ one Forco-ordinates + − = =+along and and &and as + − (i) For = and points +−L :− = = − =×+ = + NA evised − an ( += = −= (i) For ; = −− )for = forNA NA coordinate a (i) coordinate Eccentricity For == ++ Where, = − Where, ( = − ; = − ∴ − +of( − )+) area effective sectional one and other properties the of + + + − + − & For = =+++− for for NA coordinate as (−−−coordinate − ( − ∴ + + +as ≠ )))+++and = +For −−− NA (i) ≠ and (i) +and & the + = ++zero. & (& −− ∴ ∴ &equate − same to − + + + + ) ∴ + + footing − + ∴ + + + + ( − − ( ) ( + − ∴ + + + + − − − + − == ∴ ∴ === + Where, ;; ; + −−= −+ × Where, + ;==− − + ++ + +− −+ + + and Where, == − (substit (− − += − −− + + −− − ∴ + + − (i) For = = for NA coordinate and as , ≠ ∴ + + − = for and coordinate (i) For = −−≠ NA as ,)substitu −− =as NA For NA co-ordinate and ( ( ) + + (i) (i) For For = = = for for NA coordin coord (i) For =∴ ∴ + =+ for NA and ≠−,− as coordinate ++in + + + (3), we simplify substitu substituting all the terms eqn to get, & −− − = + Where, ; = − −++( ( −−)) + − +− − − (g + + ∴ + + ( ) − + − +simplify + ≠ +, substituting ∴ as all we to + − (i) For = = for NA coordinate and the terms in eqn (3), ∴ ∴ + + + + − ∴ + + + + −− where == where == −− == −−×× − − − − − 6. Co-ordinate of points P & Q of NA intersecting the + ( − ) footing side AB & AD wrt revised − ==− −− where = = − = − × ( − ) − where − = − × + = + −− where × ∴ = − = − + + − = − = == −− == ××((−−))where − − where = − −− − − ∴ × + × + = ......…….……(4) ... (4) −− = ( ) where = − = × − = where − =− − × = where − = × − ( ) = where = = × − − − − − − The Indian Concrete Journal June 2014 − + ( − ) ( − 65 )− POINT OF VIEW ∴ × + + = ......…….……(4) × ( − ) − − + ( − = + ( − ) − − − +− ∴ + + − − − ... (6) ∴ × + × + = ......…….……(4) = ∴ × + ;× + = = …….…….…(5) ……………….(6 × + = ......…….……(4) − + ( − ) ( − ) − − − Where ; ; = Where = − − − ×+ ) ( − ) − Where = ( − ) − −= ......…….……(4) × − + ∴ ×∴ + (......…….……(4) ; = + × + = ( − ) ;(6), , seen that all right hand side = =it(may − )be =( −......…….……(4) =) = Where − × ++ ( − );− In equation − − − ; = are ) − ( − ) − terms − − + ( − in x & y only. = + + )−− ) − = = where, = −+−+ − ) ) −) ( ( −((−)− −− − ( CASE – III ( −......…….……(4) + ( − ) − − + = ; = Where Where ∴ × + × + = ......…….……(4) = ; = ( −− ) ;+ − ( − ) − = = + − − )−+ + ∴−= (= ......…….……(4) (× − +) × + − + − When NA cuts AB and DC. Here uplift portion is APQDA − − + − − + − = = where, = ; = − = wh and effective − Similarly, at−= = portion is PQCBP. This is considered that for NA −location ( − )∴(ii) ( ) − × + × + = ......…….……(4) (−−()−) ) (−((−)−−))−−NA − − cuts AB at P where, AP = yB and NA cuts DC at Q + + ==−=......…….……(4) + × Where + ; ( + ; = ( ) ( ) − + − − − ( ) ( ) − +(ii) − − − where, − − − +DQ NA =;+ + ==xB − = = Where = Similarly, for location at + − = = = ; ……… + + − − − − − ) − − − ( − − +=(−− ) for NA at = = − − location − − where, = + = = ∴ + − + ( − ) + + ( − ) − Therefore, considering effective fraction in side AB is v − = ......…….……(4) + − × + − uplift part in AB is y & in CD is x, (= −(−) ) − ( − ; )−= + & in DC is u, and − )Where ; = ( ) ( ) − − − + − − − ( ) ( ) − + − − − = for and (ii) Similarly, location +Similarly, +=at + for NA += (ii) Similarly NA location we have − at at (ii) for location = = (−+ ) ; = = ( − ) , where x, u and y, v are ∴ + ) − NA =−+ − = + + ) ( ) − − for at = = ( ) ( − + − − NA (location − − −+ (3), −eqn = − parts of B & B respectively. We will now involve =× the substituting all we +get fraction ×+terms + +in = …….…….…(5) − + − ( + − ∴ − ) − ( − )− onlythe effective + − fractions u & v to ease formation of the = ; = − + − ) ( ( ) − − equations ( ) ( )which − − − + − = = can = in terms of − − as follows transferred ( − )( − )− (−− ( − − bewhere, × + )+− += + + − (+−+() − ) ( ) ) ∴ − − nat = = + ∴ + + + = ∴ × + = …….…….…(5) + − ( ) − − + + + = +location (ii) Similarly, NA at =; == + −+ +− x=& = for above. − + …….…….…(5) y from × +Where =− − − − for NA = − − location at = Similarly, (ii) − − = = where, = ; = − = − Effective − + ( −()− ) − ( Area of the Section and Revised Centre of − ) − − + ( 1. ) − +−−+ ( ) Where − ; = = + ( ) + − ) )−(− (ii) Similarly, location = + = − Gravity ∴ =+)−at …….…….…(5) + ( :− ) ∴ +NA ×− = = − + for ( + ++× × ; =−= × location − + +∴ − )−…….…….…(5) arly, for NA at = …….…….…(5) −− + = + + = ( ) + − ( −× − −∴ ( − − + − + + ( − )− + ) − − + = axis system origin located at + Considering anorthogonal = + − − − − − + A, co-ordinate ) − = area of=the effective where, )− − CG atO ( foundation (− ) =of )−− ( ( − + − ( ) − + − ( − Where ( − ) − ) Where ; = − + (= ;−= )− (− −)−PQCBP )(is−calculated NA at − )= − : − ∴(location ( =rly, for …….…….…(5) − − + + − +)= ;= + − + as=follows + ...(5) − (−∴−)−=×+ × +−+−=−++…….…….…(5) +( − − ) + −+= × −++ × = …….…….…(5) + +− − ∴ − − − − ……………….(6) = = ; ( ) − + − − ) − −((×−−(+)−)−−=)−…….…….…(5) (−++(−)((−)− ) ) +− ( ; − ; − − =+ − + ∴ × + + Where = = +( − =) + + − ( ) − − ( ) ( ) + − × −+= − + −+ − = − + PQ is the location of NA = …….…….…(5) + Where − −− ; Where ; − = − − = + + − AP = yB ; PB = vB = (1-y)B ; v = 1-y ……………….(6) = =− ; − − − − − − − ; =− ……………….(6) DQ = xB ; QC = uB = (1-x) B ; u = 1-x , ( − + − ) = ( − ) ; (= (−−−) − O is the origin of Centroidal Axes XX & YY = + ( − −) ; + (= − ) − +)−−= ×( Where += …….…….…(5) ( − ; = − ) ( ) − − − + − + − ++ −−− = =(− − = − ( ) = −− ) +……………….(6) ( − )+ − − ……………….(6) , ( − ; =) ; = = ; Y ; = = + − ……………….(6) == = where, − − , − − + − − −= ( (− −) ( ) − + − − − ) − ; = = C − + ( − ) B ( − ) − = where, = ) ++ − (− =− + ) ( − − − = where, ; where, = − = = − = = X − O ;+− ()−, − ), = − + − =) = ……………….(6) X = = ( ( )=where, −=)(;−= , = (− = = ; ……………….(6) −− − − − S = − = − ……………….(6) P ; = −) = −= = + ; ( − )− where, = (−− − where, = − where, =; Q = − = = − = where, = = where, = + + −−= = where, = − − where, = A = D = where, = −= ……………….(6) = − − = =− ( − );, ……………….(6) );− Y (;− − == − − ), − = ( − ); =( = − =− = where, = ; = ;=−− =where, == =where, where, −&(5), = = Solving (4) − where, = ; = = where, = = −where, = − = = − ere, = − = == = = where, − where, Figure 2. Effective area of the section and revised centre of gravity = (; − ) ; == (− ) , ……………….(6) = ) −; =( − ) , == − where, =−=− − = − = where, = − −= − where, e, = ; = where, ; where, = − = = = = =− where, = = where, = = The Indian Concrete Journal June 2014 66 == == where, == ; = − = where, = − where, ; ==(− =) ) ,where, = − − − − + = + = + = = − − − − + ……………….(6) + = = = + ……………….(6) == = == where, ; ;; ……………….(6) = where, + = − − − − = = = where, − − = − =+ + (−++ ) + ( ) + − = − + = = = + OF − ……………….(6) VIEW − + = where, =POINT ; = where, = + =+ ++= = − = = ; ……………….(6) ==(( = = ) ; ( ) , − = − −), ) , = = − − = where, = (− − );) ; =−=((− ( − ) = +) + + ++( − +where, + = (i) Effective Area (Rectangular PBCS) : == = == = + − where, + + + ( ) ( = + − + + + − ) = , = where, = = − = =− where, where, ( ) , = == == ) =; === −where, where, = ( = = + ( − ) + + ( − ) = + + =++− == ) = ( + − =)+ + + (=− = where, where, + −= where, where, where, == where, = ; ;; = where, where, where, = == = ===−−− + = − ) ====where, = =−−− = = = = = where, +=( − ) (= = = + ( − ) + where, ( )( ) ( )( ) = − − − + − − + − ( ) ( ) = + − + + − = − = Moment where, 4. − of+the Inertia (PMI) ) ( ) Product == ++ (( ++ ) effective −−section == −−of here, = ; = − = where, = − ) ( ) + + ( − = − − = ==where, = −−−= = where, = ;−− (revised − + ( − )( − ) = )( Centroidal − ) − Axes : − == − where, wrt = where, = ( ) where, = + − + + ( − ) = = = = = where, = −Revised =B + ( − )+(i) Eccentricity along : ( ) = + − − (ii) Effective Area (Triangular PQS) ) = = + ( − ) + + ((− − ) = (− − − )(− )−−− + (− )( − ) − where, = − )) + = = Revised +; (Eccentricity −== ) +++(along == ( −−=B+:))+++( +)(( −− = − = == where, = = where, (i) = + − = + − − × − + ( = = + ( − ) = = where, = ; ( − − ) + ) = where, == ;; == =−==( − where, )( + ( = −=)− ;− − − )(− ) − where, = ; − = = = where, ) ++ − along +×(B:=−) + =+Revised == + −= = = where, = −( −= = where, (i) Eccentricity ; = −−)) ( − − − − − ( = + ( − + −−+)()((−− −))−− ) =++ ==)(( ( ( −−=)()(−−+) −− −− −− =−− where, = =5.Revised ; = + Where, = − Eccentricities of Load with respect to revised − − − − − − − − == = ;== where, == = − ) −( − ) + ( − )( − = = where, where, = along where, = where, where, Revised − − Eccentricity =− = ; (i) ( )( = − == − == B : = + Axes Centroidal = − : = + − × = where, = = + + − + − ( ) = + = = = ( +− Where, − ) − (; − =)( − ) = = −+ )( = ( − ) (i) Revised Eccentricity along B : Revised − − − (i) Eccentricity + (B:− )( − ) = − where, − ++ = − = −+ − (i) ) − + along −== −)(Revised × − Eccentricity =(i) (+ − = From & − = = = + = = + = (ii), where, + + = + = = = = along = = =+ + = + (i) Revised Eccentricity B : = + = = + + = = + = =where, ; = = + Where, = − Eccentricity == ++ −−along =B= : ++ −−×× == ++ −− Total Effective Area (i) Revised + where, = + + + == + + + where, where, (i) Revised Eccentricity along B : = + − = + − = + − × =+ + + =where, where, = + + = = = = + = = + Where, where, = = = = − ; = = + − × = + − = + − + + ++ ++ + = = ++ − + + = += = + −+= ×Where, = −−− + + + + == = Where, = = = + ;; = = = = , = where, = = + =+ = = = + + = = == where, = = + Where, ; = − + + ++ where, = + Where, = − ; = = + + + ++ =+ == += where, = = + where, + where, = = =+ = where, = = = + where, − ; = Where, = = + + ++ = += + += + + + = == = = + = = == = + == + + + , = where, = = = where, = (ii) Revised Eccentricity along L : = (ii) Revised Revised Eccentricity along (ii) along LL :: + + + = + ++ where, (ii) Revised Eccentricity along L : + = == += where, = + where, where, where, = = = (ii) Revised Eccentricity = + −along =L : + − × = + − + = 2. − revised CentroidalAxis Second Inertia = + − × = + − Moment = of where, wrt + ( − ) + + (= + − −) = = XX : (ii) Revised = along + L−: = + − × = + − − Eccentricity + + =−− + ( − )+ = ++−( −= ) + − =× + − = −− )) − ++ (( , = where, == ++ =(( −− )) ++ − = = ( ) ( ) ( ) ( ) = + − + + − = = + − + + − = − + + ( − + ( = = + Where, = − ; = + ) + − ) + = = + − − = − = + − × + Where, = − ; = = + ( − ) + + ( −= ) = + − = + −− ) = − ) + = Where, ; + = − Where, − ( ( + + − ) + −−= ++ (( −−) )+ ==+ = (+−) Where, + ( = = = − ; = )) −+ == −−)) ++ (( −− ++ (( = − + ( ) ( ) ( ) ( ) = + − + + − = = + − + + − = − + ( − ) = = + (− ) + =Axis (+ =Where, − = − ; = + 3. Second Moment of Inertia wrt revised Centroidal − ) = − = ( − )( − ) − YY + ( − )( − ) = : − − = ( − ) = − ( −−)() −+ )(+ −=(()+−− +=((−− )() − ) = + (=−) ++ ( ) ( ) − = −=− ) )=+−() − )( − ) = ( =−(==− == ( −−)()( −− )) −− ( − )++ (( −− )()( −− )) ( )( ) = − − − )( ) ( )( ) =(( − − − + − − − ( ) + ( − )( − ) = − ) − = (− )( = ( − − ) (i) Revised Eccentricity =along − B=: ( − − ) 67 The Indian Concrete Journal June 2014 − ) (i) Revised ( ( − ) Eccentricity B=: = ( − ) along )( ) ( )( ) − Revised − − + − − ( )( ) ( )( ) ( (i) = − − − + − − =− ) Eccentricity along B = ( (i) Revised Eccentricity along B :: (ii) Revised Eccentricity along L : (ii) Revised Eccentricity along L : (ii) Revised Eccentricity along L : (ii) Revised Eccentricity along L : Revised Eccentricityalong alongL L: : (ii)(ii) Revised Eccentricity ∴ + + − + ( − − ) + + − = + − = + − × = + − = + − = + − × = + − = (ii)+ Revised − of = + P & − Q ×NA Wherefrom, −the points we get = Eccentricity 6. Co-ordinate Eccentricity along :−intersecting ××+ (ii) along L (ii) Revised along L= :=of = Eccentricity (ii) − +L:Eccentricity Revised +Eccentricity =along :+ −− = = + + − Revised Revised − along (ii) L:L+ wrt revised Centroidal footing side AB & DC Co-ordinate + …(7) System × + × + = ......…….……(7) =+ ∴ = − ; Where, =: = + +− = −+ +× − + ++ −− + = − = −= ×× = == + = +− Where, ;=+ = = −− + = −+ ++ −Where, −××+= + −; − = == ++=−− =+= = = ; the Co-ordinate = + − ++ P where L : (i) − + ( − − ) footing Where, is=intersecting of point NA = Where, Where, == −− == ++ =Where ; ; − ;+ ( − − ) = − Where side AB : ( − + −= − ) ; ; − − Where Where = + ( + + + = − ; = + Where, × = = −+ − − + + Where, Where = = ; = − = ; + = = −==+ = − Where, = = − = Where, = =) ; ; − Where, − ( − −+ = =+ + − −(− − ; )− = − Where = = ( − − =−)− = ( − − ) − − = = ( − − ) = ( ) − − − − ( ) = − − = ( − − ) + − ( − − ) − = ( ) − ; ===(−(− − + − − ( ) − + − − ) ) − − = − =−− Co-ordinate of NA is intersecting the ; = = = point Where = −Where Q=where (ii) ; = =− − − − ( ) − − − footing − side CD : ( ) − + − − = ( − ) = Where =−−++((−−−−)) ; ( ( − )− ) − ) − − ) − − ) == ((−−==−(()−− − −(−−− ) = −))) = − = + +; −(−−= (− Where ( = − Where = ( − ) = (= − ; + − ( ) ) + − − − − ) − − − ( ) − + − (− = ( − − − ) ) = =()( −− − = − = + Where ; = ( − + ( ) − − − − − ) − = ( − − − = = = ( − − ) = ( − − ) − + ( − − −) − ) − (− − ) − − (−)−−−−)) ==( ( = + = (− ==(()−−) = ( − −+ ) − = (= −( −) ) ( ) − − − − − − ) − ( ) ( − + − − − 7. − Co-ordinate of maximum pressure location C : == = = = + ( − − )= − = − − − − ( ) −== −)− + ( − − ) = = = −− = =((− −( ) ( ) )− − − − −−− −) = = ( − = ( − ) − location + + = (ii) Similarly, at = ) = and = (NA ( = for ( − ) ) + ( − − ) − − ( ) ( − + − − + + ∴ + ( substituting − terms +∴ in( eqn − − −)− the ) = ( − (3), we get ==(( −− − − ++− all ) )) ( ) ( ) − − − = = ) − = + ( ) = − − ( ( − + − − − ) + − +)− = = − ( − + ∴ + ( ) = − ( − −)+ + = + − =+ = ( − ) − − ∴ + ( ) ( − − − − − ((−−) )) − ( − − ==(()−=−= = +(− ( − ) + ) ( ) ) ( ) − − + + − − = (− ( ) ( −− − = (= ) ) −(−DIMENSIONLESS RELATIONSHIP BETWEEN − − ∴ + ∴ + + − + + + = = …….…….…(8) × + − − ∴ × + AND ECCENTRICITY RATIOS = = ( ) ∴ × + × + = …….…….… ( ) = − ( = − ) − FOOTING LIFTED = (= DIMENLESS ∴= × =+ × + = …….…….…(8) −(−FRACTIONS ) ) x & y : and (− ( − − ) − ) + ( − ) ( − ) + and = and + × + = = = = +∴ =× ..(10) + + ∴ + ( ) ( ) − + − − and × + × +− ∴ ));+ ( − − ) − on = = ( −= ) +…….…….…(8) Where and As and and points & lie =…….…….…(8) − (− − NA, Pressure ( ∴ × + × + = & Where = )+)−+ −− ( − − ) ∴ + +− + (( Now in general & form of two ( ) p is 0 at these locations. − − ( ( − + − − Where ; = + +− ( − ∴) + + +) ( − & ) ∴ ) + ( − − Wherefrom, we get dimensional bending − and &&pressure equation (3), we put the Where = Where ; = ( ) ( ) − + − − + − =and for NA and as ≠ , − coordinate ==and −+ + ∴ = NA coordinate − ( ) ( ) − + − − and and for and as , ≠ co-ordinates of NA points & one by one ( ≠ Where = for NA coordinate and as , ) + − (− −) ∴ × + × + == …….…….…(8) ; = − =allother = coordinate ≠ , areaof and sectional ofand theaseffective for NA properties ( ) ( ) − − + − − = forforNA NAcoordinate coordinate andasas ≠≠ , , ( − − − and =&=& & == )+(==−(…….…….…(8) −) ) to zero. & footing and equate the same ( − ∴ × + − × )++ & = = −( − ...(8) − ) + ∴ × + × + = …….…….…(8) ) ( ) (−− + − − ( ) ( ) + − − − − − ( − − ×+ × + − − ) ∴ = …….…….…(8) = − + ( ) − − − = ; = ∴ Where + + + + = ) and and = =NA for NAcoordinate coordinate as ,+ ≠− ( +) + ( −−× − )+ ∴ =+ for − , ) …….…….…(8) − and −(=− (i) for NA co-ordinate as For = + coordinate and as ,+≠ − ≠as coordinate + ∴ ( − − ( − − ) − −× == NA − ) ( + − − − = = = for for NA coordinate and and as as , , ≠ ≠ (−−− )) (+ ∴ + = −−− )) + ++ =( in ((−(−−−−−)−+)+)++) −−− −=)(−(− to ∴ + all + the + substituting eqn (3), simplify − (−+get, terms Where ; ( − − − =−( we − − − ( ) ( − + − − ∴ ∴ + + + + + + + + = = () −+) +; − −))+) (−− ) + = − ((− − Where (−−= − = + = Where = − − − − ( ) ( ) ( − + − − −; ) +− (− − − )− − ( − − = − Where ; = + Where ; ( ) − ) ( ) + − (− −− ( ( ) + − +− −− )−−)) (−− − ) −− − ( ) − + − − + − ( ( = nate and as∴≠× , − − ∴ = +++ × +++ = ......…….……(7) − + − −==− −(+ +− − ) ) − + ( − )−)− +++ × ∴ =(( + + ∴ + = ......…….……(7) ×+∴ ( − − ) + ( ( − − − ∴ ∴ + + + + + + + + = = ) ( ) ) + − − − −)+ ( − ) −−+ × × + =......…….……(7) ( − − − − ∴ − = + + − − − − ∴ × + × + = ......…….……(7) = (− )−)−+−(− ) − − ) + ( −−− × ++ == ......…….……(7) ......…….……(7) − ∴∴× × × ++ ; −= = = = − )− ……………….(9) − − ;) + (=−− ……………… − = −( − ( − − ) − − + ( − − ) = = = ; − − − − ( ) ( ) ( ) ( ) − + − − − − + − + = ∴ =+× + …………… − + −; −= = − = + ×= ......…….……(7) =......…….……(7) ......…….……(7) ++ ×+ + − ∴ ×∴ − − × × +× − ∴ ∴ × + + × × + + = = ......…….……(7) ......…….……(7) = ; − − ( − − ) ( − ) + ( − − + − − − ; , ) ( ) = (=− = − − )+ (−( − ( − −(−)− = = ( − )+ −……………….(9) ; = + = − − +− ( − ), ) = ( − ) ; = +− ( − − ) ( − − ( ) ; ( ) , = − = − = + .....…….……(7) , − = ( − ) ; = ( − ) − − = ( ) = where, = = ( − ) ; = ( − ) , The Indian 68 =Concrete Journal June 2014 = ; ……………….(9) − − POINT = + OF − VIEW = + − × = + − = ..(10) + + + ( − ) − = − = where, = − ; = = wh − − = where, = where, = − = ==where, − POINT OF VIEW = − = = − =− ; = = w − + ( − − ) ( − − ) − = + + × + = …….…….…(8) ∴ × + where, − − ( −−− ) + ( − ) − + ( − − ) where, =− ; = = w where, − ===− = = ==where, ; = − = = + + ( − − )+ ( − −) − − − where, = ) − − − ( − = ( − ) +(− − − where, = = ) + ) + = where, + − CG : = ( − of = Area == where, A2 Where = − − ; − = = − −− − − − Solving (7) & (8), −−where, where, =− == = where, == where, = =− − = = = −===;; = where, = = = where, − = ; = = where, − ( − ) + ( − − ) ( − − − ) ) ( + − = + (……………….(9) = where, − ) + + = ( −= − − ) + − ∴= − + = ..(10) − − = ; ……………….(9) − − ... (9) From (i) & (ii) above, ; = = = where, = − − − − + − − −+ + where −= −=where, == = where = where, === = −+= + = − = =where, = ; = Effective = = where, = Area x In equation (9), see that all right hand side terms are Total in & y. , ) + ( − − ) − − (− =−−+)+−= +(+− )= + ∴ × + × +( − = …….…….…(8) + where =++ + + ; = = = = where, = = = ( − ) , where, = ( −) ; = = ; = − = where, = = = = = w − = = − − = where, CASE – IV + + = where = = + = ( − ) + ( − − ) CG of total Effective Area + : + When NA cuts BC and AD. Here uplift portion ;is ABPQA = = = = where, = ere, =Where + − This is considered + + + + where and effective portionis PQDCP. −that − +== =+ = ( ) = where, = = = = + where, = + = = + ; = = + = + where = = = = where, NA cuts BC at P where, BP = yL and NA cuts AD at Q = == where, == = = − − = ; ……………….(9) where, AQ==xL − ( − − ) + ( − ) − ++ + + where, + == = + = == +where, = === where, = = = + Therefore, considering fraction in side v = − effective where =BC is + = = + − + ) + =+ − − ) = = = = + ( = − & in AD is u, and uplift part in BC is y & in AD is x, = + ( − ) + (where, + ( − we have = ( − ) ; = ( − ),, where x, u and y, v are + + + ( − − ) + ( fraction parts of(L− &L )respectively. ) now ) =)where, will + + ( − −We −+ +− + == =involve +==( + = + where, (=−+ + where, = = + = = = − ) = = only the effective fractions u & v to ease formation of the − in − −( − ) ( −) equations as follows which can be easily transferred (=− ) ++) + ( − +)+ (+ −− )+) = = = + +(− (= = ( ) = where, = − terms of x & y from above. + + + − ( − ) ++=((− − ) ) = where, = == ) +++−((−= − + = = + ) )( ( ) = + − + − = 1. Effective Area of the Section and Revised Centre of − − = = ; ……………….(9) Gravity : ( − ) ( −() + − − − (− −) )((− = = + ((−−=)()+(−− ) − )(+ ) − ) += − − Considering an orthogonal axis system origin located at ( − ()− ) )= ( − ) = = + ( − (+−−PQ (is+)the ( − + )==+) +( −+location −() of + )NA A, co-ordinate of CG at O of the effective foundation )( − ( ) = area + − − += BP =yL ; PC = vL ( − )( − ) ;v =1-y = (1-y)L PQDCP is calculated as follows : B : = 1-x) = ( − ) ; = ( − ) , AQ =BxL: ; QD = uL = (1-x)L( ; u− = ( − − ) − )( ( )( + − − Axes XX & YY O is the origin of Centroidal ) (i) Effective Area (Rectangular QSCD) : ( − ) = −− =)×= ( )−Y−( = + ( − B: )+ = +(+−− − =−−) − ×+ )+ ) = + = + − ( )( ( )( = − − + )− = ( − )( − + ( − )( − ) = = ( ) = where, = B : S C P = + B− = + − × = + − CG of Area A1 : = − = where, = − ; = = where, = ( − ) X B : O X B=: ( − )(− = ) −+ − += (+− − )( × − = ) + = − = − ; = = where, = = where, = where, = − ; = = where, = − = + −+ × − = + − − + −− = = + + − − = = = × where, = = = Q D A B : PQS) : (ii) Effective Area (Triangular ∴ +− − ( − − ) + ( − ) Y − − − where, = = where, = = = + − = + − × = + − = revised centre and = − − = where, Figure 3. Effective area of the section of gravity − − − − − = where, = == where, ; = = where, = The Indian Concrete Journal June 2014 69 = = where, = = + = = + Where, = = where =− = −; = ; = = = where, = − + Where, − − == − where, = − ; = = where, = − where, + + = = = = + = where = + = + − − − + +− = + Where, = − ; = =where, = =where, where, = = = = = OF = + = where, = + ; VIEW POINT = where, L=:L :+ − = = − =+ + = + − − =+ = where = where, + + = = = where, = where, 2. Second Moment of Inertia wrt revised Centroidal Axis = = = = where, = + − × = + − = = + − L : = : − + + + − + − = + − × = −XX + =+ + where, where = = ==− += = = + = − where, + + = =−where, where, − (−= −= = + (; − =)= = = + + ; ) = = = = +where, + + = where, − × = + − +++ = =+ − − += = + −=where Where, ; ; == = + = + == ++Where, = − == −;− Where, where, = = = = ; + = = = where, = = + = − + + ) (− = ) ++ =+ + (= − where ++ + = = + + = + = ( − ) = where, == +where + + = + = − ; = + ( − ) Where, = + ( where, − ) + = = = = = − + −)− ) Where, + + − ; = + ( − ) + =−( =+ ( − ) ==(L−+:= + where = = ; = = += = ++ Where, = + = − + + + + = Moment of =Inertia = = =Centroidal where, = = where, Where, = − Second 3. wrt revised + Axis + + = + = ) ( ) − = = − ( Where, ; = + = = intersecting = − where, ( ) = − = Co-ordinate 6. of points P & Q of NA YY : ) + − + ( − ) + ( − + ( ) − − − = + − = + − × = + the = = + + + (+− ) += + Where, ( (− = AD Centroidal −)where, )++(==−)( − = )footing ) wrt+revised − )BC −& = − ; = (LL−:: =−(side Co-ordinate = ()=−−=) = = = − = )(where, ( − : = + ( − ) + ( − ) (= System + + + − L : + ( − )= =− + ) + = − L = = where, ) + == + ( − = = − : + + = ( − − + = − = Where, where, = + ; ) ( − ) = = = = −ofpoint the Pwhere NA is intersecting footing = + − = + = + − × (i) Co-ordinate L : = + − = + − = + − × ) ( )( ) ( − )( − − + − − = ( − ) ( − ) ) + + + + + − B : ( + − = side = + − = + − × = )= + + =( BC : = = where, ) − = ( − ) + ( − )( =−− ) = = + − = + − × = + − = ( − )( − ()−− ) = ( − ()−− − )= )+ − = + − × = +++ − = = ( + ( −− )) ==− ) == ++ (( − + ; = = + = − = Where, ) Where, − = ( −+ = − = + − )− + Moment = ; = =× + + = (− + =(() −+) = − 4. Product the (PMI) effective of − of Inertia Where, = + = section (− −Axes ) (( − )() − ) = = ( − )= + Where, Centroidal = = − ; (+−)()(wrt + : ) −− − + ) = += = = ( − ) )(+ − + − ( ) − B : (−−−−=)) + Where, = − ; = = = ( − ( + − = + − × = ( − (+)− )− ) = NA is intersecting the ( = − − of)point (ii) Co-ordinate Q where )(=(−− )footing ) + ) − ++( (−−) == (+− (−)( = = ( − − ) − ) AD : = + ( − − = ) + = −(side )− + =)()− − − × = = + + − = + ( = − and ( − ) + ( − ) and = ( − ) = = ) + ( −( )− ) (+ ( = + − − )( − ) = − (( −− −−=(−))− −)=()− ) = = ( − ) B : & & = ( − − ) = + − = = + − = + − × and ( )( ) ( )( ) = − − − + − − ( − )===+−−(−(= )(−(−−)(−)−) ) = ( )( ) = − − − = = respect − =−) Revised Eccentricities of Load with to revised = ==( − =)(− 5. − ( ) = & = + − + − = + − × ) =7.Co-ordinate ( − )(Axes = Centroidal − )− + ( − )( − − maximum : pressure location C : = of = − , B : = =≠ , =≠ ( − along B: ) (i) Revised Eccentricity ( ) = − = + − × = + − ) = ( − ) = (( − − B : = ) ) ( − ) = ( − = + − = + − × = + − ≠, and = +=−((−− ))−+ = + − = + − × = (= )(+ ( − − − − ()− ) ( − ∴ + +++ ∴ + + KNOWN BETWEEN ) − = (− = + − = + − × = + − − & RELATIONSHIP ECCENTRICITIES AND FOOTING ) + x &( −y : ) − − (FRACTIONS ( − ) + ∴ + LIFTED + + and = = and + − and = + Where, = − ; = ......…….……(10) ∴ × + × + = As + lie on NA, Pressure p is 0 & ∴ points × ;+ and ==......…….……(10) + +× == − + ; =Where, − = + Where, = & and , ≠ at these Now and in general & of two dimensional form locations. + equation & (3), we put the co-ordinates of + L : Where, = − ; = ∴ × + bending ×= = +pressure = = = ......…….……(10) NA points & = one = by one and all other L : Revised = = (ii) Eccentricity along LL:: sectional properties of the effective ( − offooting ) and ≠ , ( − − ) + area equate , ≠ ∴ + + + = the same to zero. = = + − = + − × = + − − , ≠ + , − + =− + − × = ≠ = + − = + − × = =+ − ≠ , + ) + ( − ) + − × (( − − − − + − = += − + ) + ( − ) + + = − ; = Where, = ∴ + ∴ + + ( − − + × + ) ( + − ......…….……(10) ∴ × + = + + − − − ( − ) = − ; =∴+( ) + =Where, = + Where, == − + ; − The Indian Concrete Journal June 2014 70 ∴ + − + + ) + ( − + ( − −−) = ( − − ) − + − (+− ) ( − −∴ ) −+ − − − ++ & + + ∴ + − + = = ∴ − × + × + = …….…….…(11) − ( − − ) + ( − ) POINT VIEW = = = = OF Where = = = −− + ( − ; − ) & ( − − ) − and ∴ + + ∴ × + × + ++ = …….…….…(11) − − ( ) − − − − ∴ × + × + = …….…….…(11) ; = + ≠ , ≠ , = = for NA co-ordinate and as ( ) (i) For − − − − Where = & (− − + ) + ∴ + (−−) + substituting all the terms in eqn (3), we simplify to get, = − ) − − ( − − − + = =≠ , Where ; = − ( =− ( ) ( ) ) + ( − − − − + − ∴ × + × + = …….…….…(11) ) ( ) ( ) − − + − ( ) ( ) ( − − + − + − = ; Where = − ∴ + + + ∴ −+=+ ( − − ) −= − − +−(− + − − × + × ≠ , ∴ …….…….…(11) )+ ( − ) −(− +(−−=− = − )+ ( − −+ ) ( − − ) + ( − ) (− − )− (− ) + ( − − − + ( − ∴ + + + = − (;− − = = ( − ) + ( − −− Where ) ) − + ) −− ) = + ( ) ( − − − + − − + = − + + − +( − −)− + − − + × + × (+ ∴ =)+ ......…….……(10) + −=−......…….……(10) − )+ ( − =) − × + × ∴ ( − − − Where ; = = ∴ + + + + = = − − − = − = ……………… ; − = − − ( − − ) − − + ( − − ) ∴ we × get, + × + = ......…….……(10) Wherefrom = + ( − − ) − + …(10) − − − − = + + ∴ + = ……………… − (− ; ) − − + (− = = − − − ……………….(12) − ...(10) = ; − ∴ × + × + = ......…….……(10) + − − −) ,− =( − ); −=( Solving (10) & (11), ( − − ) + ( − ) ) )+ − Where = ( − − )+(−(−− ; ( −− ) Where = Where = ; ; ; = × + × +……………….(12) ∴ = − ( − ) , = …….…….…(11) ; = = ( −) − − − − = = where, = = ( − ) ; = ( − ) , − − = ( − ) = ; ……………… ( ) − − + ...(12) ) ( − ) + ( − − ) − ) + ( − − Where ; = ((−− −) + ) ( − = (;− (− ) + ( − − ) − − ) + ( − − ) − ) = − = = − = = where, = − Where ; = ; = Where − ==( − ) , = = − = where, ) ; = = (− where, − − − ) ( − ) + ( − − ( − − ) In equation = (see − that ) , all right hand side terms are in = ( − ) ;(12), = ) ( ) ( ( ) ( − − + − − + − − ) ( ) ( ) − + − − ) y. ( − − x & ) + ) (+−( −−where, − = + = where, = ( − − )+(−(−−))+ ) ) − − (= −−) + = = )(− (−− − − + ( = = ; + + ( ) ( ) where, − − + − = − = − − = + + − = + − − ; − − − Where = V CASE − = – = where, = ( − ) + ( − − ) (− ) ) + )( =− )= ( − ) + (−= − )( − − ) +When − NA cutsBC and DC. Here uplift portion is ABPQDA + + ( − − ) +( − ) + ( − +(=− = ) ) − + where, = − − − and + is PQC. This isconsidered that NA effective portion − = ) + = ( −−−) + − = = ( = = P where, DC − cuts − BC yL and NAcuts at−Qwhere, at − − BP − ( ) − + − = = ; ……… where, = − = ( − − ) − +=xL( − − DQ ( ) ) − − − − (− − +)− ∴ + + − = ) − − (+− − + ∴ Similarly, + for + = + (ii) location and =+ − − + = NA at =+ + + +in −(− − − )= =all∴ ) −terms −we (the − ( − substituting ) equ (3), get considering effective fraction in side BC is v + ( − − ) + + ( − ) +Therefore, ( − ) ( − − ) − + in DC is u, and part in DC is x, we − BC isy+&in = + uplift ) − ( − − ) − ) − − − ) ( − − & −− +∴(+ + + + = − + + + = have where x, u and y, v are − ), ; )= ( ) ×( − −= ) − −− + (= −(−− − ∴ × + + …….…….…(11) ∴ + = × + + − − ∴ ×+ + = × …….…….…(11) fraction parts of B & L +respectively. We will now involve + × = +…….…….…(11) − ∴ − & and only the effective fractions u & v to ease formation of the = = can be easilytransferred in − =as )follows = which + equations where, = ( ) ( − − − − − …….…….…(11) ( −∴−);×− ) += −× ++ =terms − +) − y above. = (− −=…….…….…(11) ) from of−x+& Where − ( − − ) ( −−− + = = − −(+− + ( − ; + = Where ) ( ) − − − ∴ ; = = = − × + …….…….…(11) ×++ Where − − ∴ + Section = and Revised Centre of − + + 1.− Area − ofthe Effective − − ( − − )Gravity − : = − =−where, + ( − − ) ( ) ) − Wherefrom − + − − we get, Where ; = = ; =−−− ) )−− (+−−−−(− − ))− ) − −+ − ( (((− ) − − −(− −= + − = + + …….…….…(11) ) ( ) − − − + − − ; = Considering an orthogonal axis system origin located at − Where + = − + + − + −= − …(11) ∴ × + ×= + …….…….…(11) − − the effective foundation area of CG at Oof − A,co-ordinate − PQC is calculated as follows : ( − − ) − − + ( − − ) − + ( ) ) − − + −(−− − )= + + ( − − ) − + − + ; + = ( +− −(−)− ) 2014 − − The Indian Concrete Journal June ( ) − − − − − − = =− − + + − Where ; = = − − = ;−− ……………….(12) − − 71 = = + Where, = − ; + B : = + Where, = − ; = = ( − ) ; = (− ) , L : POINT OF VIEW B : L : = ( − ) ; = (− ) , = + − = + − × = + − − = + − = + − × = + Total Effective Area : L : where, = = = += − + = − + =− + − = × += − + − = × = = where, = = + =−− = + − == + − × Where, ; = + + + = + Where, = − ; = + = − = where, Where, = − ; += = =+ + Where, = − ; = = − = where, L : Where, = − ; = + = − ; = − = where, = − = + =( − − ) = ( L− : − ) = − ; = − = where, = − ==( − + )− = + − × = + − − 6. Co-ordinate P & Q of NA intersecting 2. Second Moment ofInertia Centroidal of points ; = − wrt = revised where, =Axis the = − = ( −−=(−) ) side XX : footing BC & DC=wrt+revised Centroidal Co-ordinate = + − = + − − × = = where, = System: ; where, = − = − = = − = (− ) = − ; = − = where, = − + where, = − = =where, = ; = − = = ( − )+ =− Where, = − ; = = = (i) Co-ordinate of point the footing = P where NA is intersecting = ( − ) = where, + − ; where, = − = − = = of Inertia wrt revised side BC : + Where, 3. Second Moment ; = = = − Centroidal ( ) = − − = −= where, where, =; YY = == − = = −Axis where, = : = = =−(− )( ) − = ; where, = === − − − ; where, = − = = − = where, ===− == ( − − ) = = where, = where, ; = where, = − − = − = = = where, = ( − − )= ( − − ) where, = = where, = ( − ) = = == = where, = 4. = where, where, == = = = =the effective Moment of Product of Inertia = (PMI) where, ==( − where, == = = where, = = = ) = : = where, = where, = = = (ii) Co-ordinate section wrt revised Centroidal Axes of point Q where NA is intersecting the where, = = footing side AD : = = = ( − ) where, = where, = = B: = where, = = = where, = = = = = where, = : = where, B = == = where, ===((−− ) : Eccentricities BRevised with respect to revised of Load − ) 5. = = = where, :== where, Axes − = = Centroidal = + = + − + − × where, = = + − = ( − − ) = + = + − × − == where, where, BB: :== + = =+ = − = − = along + B−:: ×= Eccentricity (i) Revised where, = = B: = + − = + − − + × = + PQ is the locationof NA + + − =× + = −; == + + − =Where, B : ; += −=+ − × = BP+ =yL ;−PC=vL= (1-y)L ; v = 1-y = =− − = + Where, B : DQ = xB ; QC = uB= (1-x)B ; u = 1-x Where, B: = − ; = + = + O is the origin of Centroidal Axes XX & YY = + − × = + − = + − + = + − = + − × = + − =+ = =− ;+ − L:== + −+ = Where, + − × = + − = + − × = + − + = L: = + Where, = − ; = − ; = = + Where, + − L : = + − × = + − + P = = − = Where, C ;+ +−= L: ==++− + −× = + − + − = + − × = = =+ Where, ; = − B + + − = ; + =− := + − + L + = Where, L : ×= =− Where, = − ; = = + O X + + Where, = − ; = + − = + − × = + − L:= = + − = + − ×= +=− + = −+ −+× = + − Q L : = + − Where, = = + + ; L : = + LL::= =− − ; ; = = Where, = Eccentricity + (ii) Revised along D Where, A + − − = + − × = = + + = + − = + − × = + − ; = = Where, = − ===(+− −+ = + − + − × == = = + −+ × = + − + +− − ; = +)Where, = −4. Figure Effective area = + Where, ; =of the section and revised centre of gravity + − ===(−+−−− ×) = + − + = ( − ) Where, = − ; + = = − −+ ) Where, = =+=(− ) = − ; = + + − )Journal ( − Concrete Where, = The Indian June 2014 72 == ; + = − ( ) = − − Where, = − ; = = + + − ( ( ) ) − = ( − ) ( − − ) + ( − ) ( − − () − = ( − − ) ( ( −(−) +)+ ((− −(=−) −)+ (( −)+− − − ) )+ ; Where = = ( ( − ) ) ( − − ) + (−( − ) ) ( ) − − − − + − Where + = ; = −+ ∴ =× × ++ ......…….……(13) = ∴ +× ++ × + = ......…….……(13) − + Where = ∴ ; + − − − −− = − OF VIEW POINT − ( ( ) ( (( )( ( ) ()) ) ()( ( ) ()) ) ) (( ( )) +−− − − − + − −−−− + −+ − − − − − + − Where = and = − + = +; = −Where ; 7. Co-ordinate of maximum pressure location C : = = − + − − + − − = (ii) for NA location at Similarly, + −− ∴ × + × + =+− ......…….……(13) + − = == = − substituting all the terms in eqn (3), we get − − = − = − ( ) ) ( ( ) ( ) ( ) = − ( ) = − ()+−( −()− − ) ) (− () −+ = ( − ) = ( − ) − ()+− )+ ((−− −− ))++ ((−− ) ( −− ( = ( − ) = = + = = ; = ( − ) + (−−= )()− Where ∴ (−++− ++) ) =( −= ) −− ∴ == ( = += + − − + + − = = = ( − ) ) = ( − KNOWN RELATIONSHIP = ( − )BETWEEN − AND FOOTING ECCENTRICITIES ( − − ) + ( − ) ( − ()−+ )(+ − −(− )− ) y : LIFTED FRACTIONS x & ( − − ) +) +∴ ((−×− − +)+ ) × + + + = += + ( − = = = + −= ) + ( − − )∴ ( ∴∴ +× ++ × + = −…….…….…(14) and + + − and − ∴ + + + + = − − − and on NA, Pressure p is lie As points and 0 =(−−and ) & and = & at these locations. Now in general form of two dimensional and ( − ) + (−×+−+ )×+( −= ) +…….…….…(14) (−−−−))+(+−(+−) + ∴ ( − ∴ + == (− ∴ + + + + & & (3), ) = = bending pressure equation we put the co-ordinates of ( ) ( − + − − Where ) = …(14) ( − ∴ + = + × −× = − …….…….…(14) − ; − = & = ∴ × = + × + = …….…….…(14) Where = & = by NA points one andall other & one NA for − = = for NA = = for NA sectional properties of the effective area of footing and NA = (− =)for − )(− ) + (− (− the = same = to zero. ) ) +(−(−)(+(−−) NA (− =) +…….…….…(14) = for NA equate (i) For for NA = × += ×for − − −) − ∴ (×− + =(…….…….…(14) ∴ + ( ) ,∴ × −−+ + ++ =≠(−, ) ≠ ; Where + + + − − =(=− − = ( ) ( ) ( ) ( ) − + − − − − + − = Where ; − = − = + − ; , = , ≠ co-ordinate and≠Where as , substituting all the terms in − − and − ≠ ,and − − , ≠ ≠ , to get, (3), we simplify eqn ( − ) + ( − − ) ( − − ) + ( − ) ( − (((−(−)−−+−−−−−)+()−)))−( − )) ((=− (− −− − +) & ( − − −) +( ))+(−−()−+) + − (− )++ (− − + ∴ + + ( ( ) ( ) − − + − − & ( ( ) ( ( ) ( ) − − − + − − ( ) ∴ × + × + = …….…….…(14) − + Where = = ;+ = Where ; = = ( ) ( ) ( ) ( + + +=+ + +))= − + (−− += ∴ + − ++ + + ∴ +− ( ) − − ++) −−−− ( − − − − ) ∴ (−+ )+ ) ( )+ − −) − − − = + ( ( ) + =( − − − ( ) ( ( ) ( − − + − − − − + − + ∴ + and+ − −− =− +− ∴ + + + + − − ∴ for + NA + = for − − + + == == NA − − − − ∴ × + × + ((=−−......…….……(13) ((−−−−)) ( − () +− (−−(−−) +− )+()−+(− )+) ( − − ))++ = = & + ; = + = =− ......…….……(13) × ∴ +× ×− +Where ≠≠×,, ∴ )+×=+(......…….……(13) ( − ∴ − − ) −− −− ) ++(= ......…….……(13) ......…….……(10) ∴ + × = + × × = + + ∴ ∴ × ......…….……(13) + × + = ......…….……(13) × + × + = ( − ) + ( − − ) = = for −(− − ) + ( − ) NA Where = ; = )()−+ )( − ;− (−()−−)(+−)++()−+(−−−(−−−)−)+ ) ( − ) ( − − (−)+− ) −+ − (−− ()−++ ( − ( −−)+=−− )( ) Where Where =( )− − ) (− ) + ( − − ) (− ; =− − =) + ( ( − ∴ ) (−=−− ) == ( − − =−) + − + =++ + ++)+ ∴ ++ , +)+=− ( (− Where ;− ( − + ≠ = = − = Where ;−−− − ; − − ……(13) Where − − Wherefrom, we get, )− −− (− − ) − −) ( ) ( ( − + + − − (=− ( −)+− ()−+ )( − ) (+− )(+− ()−+ − + ( ) ) − − − ( −(−− ) + (− ( − (−(−− )−( − ) ( −+ ) + − −)+−)…(13) )+ +) (×−) )+−+−− + )++ −)+)( + − ) ( (−) − ∴ ++ − −( + ( ) = ∴ += = ......…….……(13) & − (−) ( ∴ = ××+ ......…….……(13) ( ( ) )+ = (− − ) − + − = × = ( ) ) ( ) ( − + − − + − Solving (13) (14), + + + + + − − + −+ −= − = + + ; − = − − − − − − − − − − − ...(15) = = ; ……………… + − ) −− −− ) (−− −−=) + (−− ) ++ = (− ( − − Where ; = = Where ; = = ) × (−+( ) ......…….……(13) )×+) +=(=−=−−−=......…….……(13) = ) ( − ) + ( − − − = −∴ −= + + (−+ ) − = = += + = = = − − (− − )+ ( − − ) (+− )+− (− − ( − − ) + ( − ) ) ) + ( − ) − − − + − − ( ) ( ) ( ) ( ) − − + − − + − − ( ) ( ) ( ) ( ) ∴ + + + + − − + − − + − − In equation (15), see that side)( ( − − in = terms are (−−)−) (++− (all + −right )+−hand )(+ = − −+ = +− ; ( () − ) )−+ = ; +∴ Where ∴ − = = + + + + ( + ) + − ) ( −=− )+= ( − − −(− − ) +−( −− −− ( ( x)−− & +y −− − ∴ + + ( − + ())+ ( − −+−− − − + − ) + =(−− ) ( ) + ) ∴ − = ......…….……(13) +− + = + + ∴ − + + + + − − − − Solution Problem − = − = ; Lor B. and −(∴ = (…….…….…(14) − − × + × + ) ( ) − + − − ) ( ) ) ( ) + − − + − − − ) + ( − ) ( − = = ; …………… ) ( ) ∴ ) + × (× −+ += + …….…….…(14) − = −+…….…….…(14) (==− )+==∴ + + × + + × for − generating (A)Method thewhole graph ( − − ∴ ) + ; ×=+(−−=− …….…….…(14) +( − ) − for all cases −) × × = + = −…….…….…(14) …….…….…(14) + + ∴ × + ∴ × + × + = − − : − ( − ) + ( − − ) ( − − ) + ( − ) ; ( −− )=++ −− −− ) −− =−− + − ( ( Where (+ ( ( ( +−( −) +)−+ )− ) −−)++− ++) ( − − (− )+)( −) − − − ( −= ++)Where ) + ∴ −()−−+ ∴ () ==)of x and y which = ++ ( − all functions Where =( = ( ) are ; − − −− − ) (+− − (=; − + = ) + −) ) (− + ( − − ) ( + − − − ) +−(−=−;−) = ( −−;−) +=(−− ) ) + ( − ) − …….…(14)Where = ( − − ; = −Where = − Where turn are Lor or B. B. and in the are known −− ) (−− −fractions −− )of+ L )− ) ( ( ( + − − ( − ) +∴∴ (××− ++−××) ++( (−==−=)…….…….…(14) ( ) ( ) − − + − L or B. and + + ((−−)−+−)(+)−+()(− − ) ) +)(+− ( ) ( (−)+−+) + − − = − …….…….…(14) − + =)+ ∴ + + = = + − − + + ( ( ) ( ) + − + − − − − ) ) − ( − & ( ()− −− ) )+) −) − (− − )++ ( − ) ( ) ( + − − ( ) ( ( − + − − − − − = + + − − − − ; + + = + The Indian Concrete Journal − 73 June 2014 − = = + − − − − − −……………….(15) − = ; = − − − − − − − ; = = − ……………….(15) − ……………….(15) = = − ……………….(15) = = = − ; ; − ……………….(15) ; − = − − − − − and − ……………….(15) L or B. = = ; − POINT OF −− ∴ VIEW −− × = = = and ; L or B. ……………….(15) − − known parameters. Equations involving on (the right )+ ( − ) − ( − ) + ( − ) − Where − L or B. and = + + + + = side and = y on ; ……………….(15) terms involving hand x, the right hand − − − L or B. and − ……………….(15) LB.or B.and and are all the cases deduced for L or(Case – IItoL Case side or B.– V). and , and Right hand side parameters shall on be calculated L or B. and Equations involving known on the left hand side case to case basis depending on the NA location. After , , and and L or B. and generating numerous sets of dimensionless parameters terms and involving x, y on the right hand side are also & in Xaxis and two of x and y, and , and against different sets deduced for all the cases (Case – II toCase –V).These , and & equations with two x& y for case unknowns and one particular inXaxis in Xaxis and considering of in X-axis and in Y-axis, we can L or B. and and Kmax , non-linear are simultaneous equations and can be used & , and corresponding and Kmax values. plot yand values in Xaxis and x values, , & Kmax in Xaxis and for generating the graph as a whole by assuming various & K K max max thex-curves, get After plotting manyof y such sets, we & finding & , and in Xaxis and combination of x & y and the corresponding & Kmax & , Kmax & K -curves max plotted and in K Xaxis , . 8 (Eight) curves againstKmax and max & and in Xaxis & . However, for a particular and known , K example hand calculations for each case max K max using the above be adopted Kmax & efficient trial and error & method &shall values, , , in Xaxis and Kmax , & ina tabular form & K basis equations on casetoKcase are placed theand max the number of decrease to iterations. From equations, , max and , & 2 for , (Annexure Table Kmax 1 for Case – II, Table and , , Case – III, can see we that though it is an iterative process to find K max K max Table & Table 3 for Case – IV, V) atAnnexure-A. & , and , and ,, 4 for Case – and , , and and , whole graph showing for out values of Here, the curves NA locations the values of x and y for known , and , K max and , , contours and x and y are known , are drawn , however, think and in ,respect of in reverse, if we that Pressure & and , , and , and and , and unknowns parameters , the problem x, y, , and by , preparing a computer are Knowing program and is, presented in and , , and , and , Knowing x, y, and , gets much easier. For several such known values of x and Figure. 5 above. and and ,, and and by using y,we canfind out corresponding and , intensity and (B) To (for ( ) ( find − out + the maximum − ) pressure − a ) + , and (6) for Case-II, (9) for Case-III,∴(12) , = Case+ + Knowing equation for x, y,+ and given set of : ) ( − )+ (− −) − ( − )+ and − ( and , Knowing x, y, and , ∴ x,y,Knowing = + + , + x, y, , IV, (15) for we can + and Case-V. Knowing Knowing x, y, , y, and , : Onceafull Knowing x, and , − − 1. MANNUAL METHOD graph is with calculate x, corresponding maximum pressure on the base Knowing y, and , ready, design use this graph for our future − can ( − ) , (we footing by ) + ( − ) of , putting all known parameters corner C the ∴ = + + + of Knowing and , ∴ = × values off and x, y, purpose by reading maximum ) with with ( − ()−+pressure ()quantity ) + + − )+) (− of C in pressure along )+) ( with coordinates − equation (3) ,K and ( − −) ( ( − coefficient − NA (− ( ) ( ( +− )).(1+ − − and uplift ∴ ∴= ∴= +×= + + max ( ) ( ) − + + + − + , + ( − simplifying : ∴ = + + + − ∴ = + ( − + ,) − ( − − −− with ) ( − −+ ) + ( ) , .−( against − ) + with given set of for design purpose ( ) ( ) − + − − .(16) ∴ = x, y, and + , + + − Knowing )+ + Where = + + ( ) ( ) ( ) ( − + − − + − − − to + and this with (get − −) .(16) −the maximum −) with ,)+ (− + =+(×+− ∴ = ∴ ,K + (multiply ) max . , . = + + + + with − − Where pressure under footing. with − − ∴ = ∴× = × ∴ = ∴ × = × , . , . ) + ( − ) (− (−)+PROCEDURE ) − (−) +: A)computer ( − + ( 2. COMPUTER with ( − == × program .(16) + + + ( − )∴∴ ( ) ( ) − + − Where = + + + , . ...(16) − + .(16) ( − − ∴+ += may created the non-linear ( of ( ) ( ( − )+ − + − )+) (− −)for solution )+)be −−, × ( ( − . ( ) ( ) ( ) ( +− ))+ − + − − + ) − Where = + + ) + = (+ + . − Where Pmax ( − +. ( − Where = + + + , equations for finding out the uplift fractions of the − − Where ) = + + (− )+ − ( − ) + − ( − ) + ( − − − − Where = + + + + & ∆y Once footing against the known values of ∆x , . P . P . max ( ) ( ) ( ) ( ) max − + − − + − − − ∴ = × + Where + + + = the uplift fractions i.e. the neutral axis location is − − ∆x∆x && ∆y∆y Pmax. Pmaximum max =. a ∆n known, we can calculate the Kmax and Chart for ∆x − + P−footing. . A Flow − + − pressure the & ∆y under max ∆x && ∆yyprev + + Where = + + − + − xprev − P−max . ∆n ∆n aany a computer = framinga program in= language is − + + + . ∆x &Pmax ∆y − ∆x &xprev ∆y && yprev yprev xprev ∆n = a ∆n = a. Pmax ∆x & ∆y ( ) 74 ( ) ( ) ( ( ) The Indian Concrete Journal June 2014 ) ( ) ( ) xprev & yprev ∆n = a ∆n = a &∆y yprev x∆x prev& POINT OF VIEW Location of NA and maximum pressure chart of highly eccentric foundation 0.5000 0.4500 0.4000 0.3500 Eh / B 0.3000 0.2500 0.2000 0.1500 0.1000 0.0500 0.0000 0.0000 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 EL / L Figure 5. Location of NA and maximum pressure chart of highly eccentric foundation shown later in Figure 6. Following notes may be read in conjunction with the flow chart : 1. Sets of Equations used in the Flow Chart are all taken from the above discussion in the paper and are provided at the end of the flow chart. Different variable names have been used in the Equation Sets with that of the paper for use in VB Macro. 2. As the final equations are dimensionless (as discussed above), L or B of the footing are not necessary to use in a program and hence, equations in the given Equation Sets are provided as independent of L or B except for calculation of the Maximum Pressure Pmax. 3. For each of the Equation Sets, different function may be created and called as necessary by passing The Indian Concrete Journal June 2014 75 POINT OF VIEW FLOW CHART START Case 2 : Calculate Eff. Sectional Properties from Eqn Set (2) Read Mx , My, P, B, L Calculate Eb/B = Mx /(P x B) Calculate E /L = My/ (P x L) 1 Assume x = 1 ; Y=0.1 Calculate E bB y B and E1B y L from Eqn. Set (A) Initialise p=0, q=0, r=0, s=0, x & Y=a small value Consider a less value of x & y for each new iteration Y prev = Y Yes Equal if E 1B yL < E 1/ L No Increase y = y + y p=p+1 Decrease y = y q=q+1 y If p> 0 and q = 0 or If p = 0 and q >0 No Yes If x < = 1 and y < = 1 Yes Case = 2 x=x-0 y=y-0 Use Eqn. Set (2) To Calculate Eff. Sectional Properties y = y prev No If x > 1 and y < = 1 Yes Case = 3 u=2-x v=1-y Use Eqn. Set (3) To Calculate Eff. Sectional Properties No If x <= 1 and y >1 Yes Case = 4 u=1-x v=2-y Use Eqn. Set (4) To Calculate Eff. Sectional Properties No If x > 1 and y > 1 No Yes Case = 5 u=2-x v=2-y Use Eqn. Set (5) To Calculate Eff. Sectional Properties ERROR Mark -A 76 The Indian Concrete Journal June 2014 Calculate E 1B L and EbB yB from Eqn. Set (A) Mark - B POINT OF VIEW x prev = x Mark - B Mark -A Yes If Eb By B < Eb /B Equal No Increase x = x + x r=r+1 Decrease x = x - x s=s+1 If r > 0 and s = 0 or If r = 0 and s > 0 No Yes If x < = 1 and y < = 1 Yes Case = 2 x=x-0 y=y-0 Yes Case = 3 u=2-x v=1-y Use Eqn. Set (2) To Calculate Eff. Sectional Properties No If x > 1 and y < = 1 Use Eqn. Set (3) To Calculate Eff. Sectional Properties No If x <= 1 and y >1 Yes Case = 4 u=1-x v=2-y No Yes If x > 1 and y > 1 Case = 5 u=2-x v=2-y Use Eqn. Set (4) To Calculate Eff. Sectional Properties Use Eqn. Set (5) To Calculate Eff. Sectional Properties Calculate E1B yL and EbB yB from Eqn. Set (A) No If E1ByL - E 1/L < n and If EbByB - E b /B < n Calculate Pmax from eqn. Set (B) Print Case No, x, y, K max, P max x = x prev END The Indian Concrete Journal June 2014 77 POINT OF VIEW parameters for necessary calculations and getting the result. 4. The values of fractional part x & y have been varied from 0 to 2 (considering x= 0 to 1 for edge AD & x= 1 to 2 for edge DC and y= 0 to 1 for edge AB & y= 1 to 2 for edge BC). 5. A check is to be kept on x & y so that calculated eccentricity always falls on the first quadrant (+ve) i.e. upper rightmost quadrant of the footing. 6. Different values of ∆x & ∆y may be chosen by calling a function for the same. For each new iteration, ∆x & ∆y shall be less than the ∆x & ∆y values of the previous iteration to gradually converge into the solution. 7. Mark-A and Mark-B of one page of the Flow Chart shall be considered merged with the same Mark of other page to study the Flow Chart. Case – IV : Equation Set (4) : Effective Sectional Properties Area1 = u ; cgx1 = 1 - u / 2 ; cgy1 = 1 / 2 ; Area2 = 1 / 2 * (v - u) ; cgx2 = (3 - 2 * u - v) / 3 ; cgy2 = 2 / 3 Area = Area1 + Area2 CGX = (Area1 * cgx1 + Area2 * cgx2) / Area ; CGY = (Area1 * cgy1 + Area2 * cgy2) / Area Icgx = u / 12 + Area1 * (CGY - cgy1) ^ 2 + (v - u) / 36 + Area2 * (CGY - cgy2) ^ 2 Icgy = u ^ 3 / 12 + Area1 * (CGX - cgx1) ^ 2 + (v - u) ^ 3 / 36 + Area2 * (CGX – cgx2) ^ 2 Icgxy = Abs(Area1 * (CGX - cgx1) * (CGY - cgy1) - (v - u) ^ 2 / 72 + Area2 * (CGX - cgx2) * (CGY - cgy2)) X1 = 1 - v – CGX ; Y1 = 1 – CGY ; X2 = 1 - u – CGX ; Y2 = -CGY 8. Abbreviations Used in Flow Chart : P = Axial force, Mx = Moment about x-axis, My = Moment about y-axis, Eb/B = (Actual ecc. / fdn. Dimn.) along y-axis, El/L = (Actual ecc. / fdn. Dimn.) along x-axis, ElByL and EbByB = Ecc./fdn dimn based on assumed x & y i.e. NA location to compare the same with the actual ones to reach at a solution for x & y. ∆n = a negligible predefined number to verify whether a solution is reached or not. p, q, r, s = variables keeping record whether y or x is increasing or decreasing. xprev & yprev are keeping the values of x & y which are being changed in the later part of the program. Case – V : Equation Set (5) : Effective Sectional Properties Area = v * u / 2 ; CGX = 1 - v / 3 ; CGY = 1 - u / 3 Icgx = v * u ^ 3 / 36 ; Icgy = v ^ 3 * u / 36 ; Icgxy = Abs(v ^ 2 * u ^ 2 / 72) X1 = 1 - v – CGX ; Y1 = 1 – CGY ; X2 = 1 – CGX ; Y2 = 1 - u - CGY Set Of Equations for Use in Flow Chart :: Equations given below may be directly used in computer program in excel sheet (VB macro) with the same variable names and suitable changes of operators are required for other languages :: Equation Set (B) : Max Pressure Equation at Corner of footing : X3 = 1 – CGX ; Y3 = 1 - CGY a3 = (Icgy * Y3 + Icgxy * X3) / (Icgx * Icgy - Icgxy ^ 2) ; b3 = (Icgx * X3 + Icgxy * Y3) / (Icgx * Icgy - Icgxy ^ 2) Max Pressure Coefficient Kmax = a + ((EbByB + Feccb) * a3 + (ElByL + Feccl) * b3) Max Pressure = Kmax * P / (B * L) Case – II : Equation Set (2) : Effective Sectional Properties Area = 1 - x * y / 2 ; CGX = (3 - x ^ 2 * y) / (6 - 3 * x * y) ; CGY = (3 - x * y ^ 2) / (6 - 3 * x * y) Icgx = 1 / 12 * (1 - (x * y ^ 3) / 3 + x * y / 3 * (3 - 2 * y) ^ 2 / (x * y - 2)) Icgy = 1 / 12 * (1 - (x ^ 3 * y) / 3 + x * y / 3 * (3 - 2 * x) ^ 2 / (x * y - 2)) Icgxy = Abs((CGX - 0.5) * (CGY - 0.5) + ((x ^ 2 * y ^ 2 / 72) - (x * y / 2 * (CGX - x / 3) * (CGY - y / 3)))) X1 = -CGX ; Y1 = (y - CGY) ; X2 = (x - CGX) ; Y2 = -CGY 78 Case – III : Equation Set (3) : Effective Sectional Properties Area1 = v ; cgx1 = 1 / 2 ; cgy1 = 1 - v / 2 ; Area2 = (u - v) / 2 ; cgx2 = 2 / 3 ; cgy2 = (3 - 2 * v - u) / 3 ; Area = Area1 + Area2 ; CGX = (Area1 * cgx1 + Area2 * cgx2) / Area ; CGY = (Area1 * cgy1 + Area2 * cgy2) / Area Icgx = v ^ 3 / 12 + Area1 * (CGY - cgy1) ^ 2 + (u - v) ^ 3 / 36 + Area2 * (CGY – cgy2) ^ 2 Icgy = v / 12 + Area1 * (CGX - cgx1) ^ 2 + (u - v) / 36 + Area2 * (CGX – cgx2) ^ 2 Icgxy = Abs(Area1 * (CGX - cgx1) * (CGY - cgy1) - (u - v) ^ 2 / 72 + Area2 * (CGX - cgx2) * (CGY - cgy2)) X1 = -CGX Y1 = 1 - v - CGY X2 = 1 - CGX Y2 = 1 - u - CGY The Indian Concrete Journal June 2014 Equation Set (A) : Stress Equation on NA for calculating EbByB and ElByL : Feccb = 0.5 – CGY ; Feccl = 0.5 – CGX ; a = 1 / Area P1 = (Icgx * X1 + Icgxy * Y1) / (Icgx * Icgy - Icgxy ^ 2) ; Q1 = (Icgy * Y1 + Icgxy * X1) / (Icgx * Icgy - Icgxy ^ 2) R1 = (a + Feccl * P1 + Feccb * Q1) P2 = (Icgx * X2 + Icgxy * Y2) / (Icgx * Icgy - Icgxy ^ 2) ; Q2 = (Icgy * Y2 + Icgxy * X2) / (Icgx * Icgy - Icgxy ^ 2) R2 = (a + Feccl * P2 + Feccb * Q2) ElByL = (Q2 * R1 - Q1 * R2) / (P2 * Q1 - Q2 * P1) EbByB = (P2 * R1 - P1 * R2) / (P1 * Q2 - Q1 * P2) References 1. “Foundation Design” by Wayne C. Teng published by Prentice Hall of India Private Limited, New Delhi-110001 Published in : 1979 pp 130-133 POINT OF VIEW Annexure Table 1. Calculation of pressure co-efficients CASE - II (Short Side & Long Side Intersected) - A Triangular Part is uplifted Annexure Table 3. calculation of pressure co-efficients CASE - IV (both the long sides intersected by na) Example No → 1 2 3 4 5 6 7 8 Notations used Assumed x, y are +ve fractions ; x<1 in AD portion and y<1 in AB portion in paper ↓ x 0.000 0.000 1.000 1.000 0.900 0.800 0.700 0.300 y 0.000 1.000 0.000 1.000 0.700 0.600 0.500 0.400 a1 1.000 0.000 1.000 0.000 0.300 0.400 0.500 0.600 c1 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 d1 0.500 1.000 0.500 1.000 0.850 0.800 0.750 0.700 a2 0.000 1.000 0.000 0.000 0.070 0.120 0.150 0.280 c2 0.500 0.500 1.000 1.000 0.950 0.900 0.850 0.650 d2 0.000 0.500 0.000 0.500 0.350 0.300 0.250 0.200 a3 0.000 0.000 0.000 0.500 0.315 0.240 0.175 0.060 c3 0.000 0.000 0.667 0.667 0.600 0.533 0.467 0.200 d3 0.000 0.667 0.000 0.667 0.467 0.400 0.333 0.267 ar 1.000 1.000 1.000 0.500 0.685 0.760 0.825 0.940 Fgx 0.500 0.500 0.500 0.667 0.592 0.574 0.557 0.526 Fgy 0.500 0.500 0.500 0.667 0.623 0.595 0.571 0.523 Fix 0.083 0.083 0.083 0.028 0.042 0.050 0.057 0.074 Fiy 0.083 0.083 0.083 0.028 0.051 0.058 0.063 0.073 Fixy 0.000 0.000 0.000 0.014 0.019 0.019 0.017 0.009 Feccl 0.000 0.000 0.000 -0.167 -0.092 -0.074 -0.057 -0.026 Feccb 0.000 0.000 0.000 -0.167 -0.123 -0.095 -0.071 -0.023 x1 -0.500 -0.500 -0.500 -0.667 -0.592 -0.574 -0.557 -0.526 y1 -0.500 0.500 -0.500 0.333 0.077 0.005 -0.071 -0.123 x2 -0.500 -0.500 0.500 0.333 0.308 0.226 0.143 -0.226 y2 -0.500 -0.500 -0.500 -0.667 -0.623 -0.595 -0.571 -0.523 x3 0.500 0.500 0.500 0.333 0.408 0.426 0.443 0.474 y3 0.500 0.500 0.500 0.333 0.377 0.405 0.429 0.477 P1 -6.000 -6.000 -6.000 -24.000 -13.209 -11.326 -9.900 -7.543 Q1 -6.000 6.000 -6.000 0.000 -4.132 -4.168 -4.195 -2.594 R1 1.000 1.000 1.000 6.000 3.181 2.545 2.069 1.317 P2 -6.000 -6.000 6.000 0.000 0.630 0.039 -0.468 -4.047 Q2 -6.000 -6.000 -6.000 -24.000 -14.519 -11.854 -10.094 -7.552 R2 1.000 1.000 1.000 6.000 3.182 2.436 1.952 1.344 el/L 0.083 0.167 0.000 0.250 0.170 0.149 0.130 0.139 eb/B 0.083 0.000 0.167 0.250 0.227 0.206 0.187 0.103 Kmax 2.000 2.000 2.000 6.000 4.108 3.560 3.143 2.496 Max Pressure p = Kmax * P/BL Example No 1 2 3 4 5 6 7 8 Assumed x, y are +ve fractions ; x<1 in AD portion, y<1 in BC portion, x>y to keep the eccentricity in first quadrant. x 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 y 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 u 0.700 0.600 0.500 0.400 0.300 0.200 0.100 0.000 v 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 a1 0.700 0.600 0.500 0.400 0.300 0.200 0.100 0.000 c1 0.650 0.700 0.750 0.800 0.850 0.900 0.950 1.000 d1 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 a2 0.050 0.050 0.050 0.050 0.050 0.050 0.050 0.050 c2 0.267 0.367 0.467 0.567 0.667 0.767 0.867 0.967 d2 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 ar 0.750 0.650 0.550 0.450 0.350 0.250 0.150 0.050 Fgx 0.624 0.674 0.724 0.774 0.824 0.873 0.922 0.967 Fgy 0.511 0.513 0.515 0.519 0.524 0.533 0.556 0.667 Fix 0.062 0.054 0.046 0.037 0.029 0.021 0.012 0.003 Fiy 0.035 0.023 0.014 0.008 0.004 0.001 0.000 0.000 Fixy 0.003 0.003 0.002 0.002 0.001 0.001 0.001 0.000 Feccl -0.124 -0.174 -0.224 -0.274 -0.324 -0.373 -0.422 -0.467 Feccb -0.011 -0.013 -0.015 -0.019 -0.024 -0.033 -0.056 -0.167 x1 -0.424 -0.374 -0.324 -0.274 -0.224 -0.173 -0.122 -0.067 y1 0.489 0.487 0.485 0.481 0.476 0.467 0.444 0.333 x2 -0.324 -0.274 -0.224 -0.174 -0.124 -0.073 -0.022 0.033 y2 -0.511 -0.513 -0.515 -0.519 -0.524 -0.533 -0.556 -0.667 x3 0.376 0.326 0.276 0.226 0.176 0.127 0.078 0.033 y3 0.489 0.487 0.485 0.481 0.476 0.467 0.444 0.333 P1 -11.327 -15.204 -21.461 -32.520 -54.857 -110.769 -320.000 -2400.000 Q1 7.267 8.252 9.535 11.267 13.695 17.164 20.923 0.000 R1 2.662 4.084 6.486 10.927 20.294 44.782 140.615 1140.000 P2 -9.912 -13.032 -17.884 -26.016 -41.143 -73.846 -160.000 0.000 Q2 -8.685 -10.138 -12.165 -15.185 -20.139 -29.638 -54.154 -240.000 R2 2.663 3.941 6.013 9.634 16.659 32.557 77.231 60.000 el/L 0.249 0.282 0.316 0.349 0.382 0.414 0.446 0.475 eb/B 0.022 0.026 0.030 0.037 0.047 0.066 0.107 0.250 Kmax 2.840 3.307 3.956 4.918 6.486 9.474 17.143 60.000 Max pressure p = Kmax * P/BL Annexure Table 2 . calculation of pressure co-efficients CASE - III (both the short sides intersected by na) Annexure Table 4. calculation of pressure co-efficients CASE - V (Long Side & Short Side Intersected) - A Triangular Part remains effective Example No 1 2 3 4 5 6 7 8 Assumed x, y are +ve fractions and x<1 for DC portion, y<1 in AB portion, x<y to keep the eccentricity in first quadrant x 0.000 0.200 0.300 0.400 0.500 0.600 0.700 0.800 y 1.000 0.300 0.400 0.500 0.600 0.700 0.800 0.900 u 1.000 0.800 0.700 0.600 0.500 0.400 0.300 0.200 v 0.000 0.700 0.600 0.500 0.400 0.300 0.200 0.100 a1 0.000 0.700 0.600 0.500 0.400 0.300 0.200 0.100 c1 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 d1 1.000 0.650 0.700 0.750 0.800 0.850 0.900 0.950 a2 0.500 0.050 0.050 0.050 0.050 0.050 0.050 0.050 c2 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.667 d2 0.667 0.267 0.367 0.467 0.567 0.667 0.767 0.867 ar 0.500 0.750 0.650 0.550 0.450 0.350 0.250 0.150 Fgx 0.667 0.511 0.513 0.515 0.519 0.524 0.533 0.556 Fgy 0.667 0.624 0.674 0.724 0.774 0.824 0.873 0.922 Fix 0.028 0.035 0.023 0.014 0.008 0.004 0.001 0.0003426 Fiy 0.028 0.062 0.054 0.046 0.037 0.029 0.021 0.012 Fixy 0.014 0.003 0.003 0.002 0.002 0.001 0.001 0.001 Feccl -0.167 -0.011 -0.013 -0.015 -0.019 -0.024 -0.033 -0.056 Feccb -0.167 -0.124 -0.174 -0.224 -0.274 -0.324 -0.373 -0.422 x1 -0.667 -0.511 -0.513 -0.515 -0.519 -0.524 -0.533 -0.556 y1 0.333 -0.324 -0.274 -0.224 -0.174 -0.124 -0.073 -0.022 x2 0.333 0.489 0.487 0.485 0.481 0.476 0.467 0.444 y2 -0.667 -0.424 -0.374 -0.324 -0.274 -0.224 -0.173 -0.122 x3 0.333 0.489 0.487 0.485 0.481 0.476 0.467 0.444 y3 0.333 0.376 0.326 0.276 0.226 0.176 0.127 0.078 P1 -24.000 -8.685 -10.138 -12.165 -15.185 -20.139 -29.638 -54.154 Q1 0.000 -9.912 -13.032 -17.884 -26.016 -41.143 -73.846 -160.000 R1 6.000 2.663 3.941 6.013 9.634 16.659 32.557 77.231 P2 0.000 7.267 8.252 9.535 11.267 13.695 17.164 20.923 Q2 -24.000 -11.327 -15.204 -21.461 -32.520 -54.857 -110.769 -320.000 R2 6.000 2.662 4.084 6.486 10.927 20.294 44.782 140.615 el/L 0.250 0.022 0.026 0.030 0.037 0.047 0.066 0.10714 eb/B 0.250 0.249 0.282 0.316 0.349 0.382 0.414 0.44643 Kmax 6.000 2.840 3.307 3.956 4.918 6.486 9.474 17.143 Example No Max pressure p = Kmax * P/BL Max Pressure p = Kmax * P/BL 1 2 3 4 5 6 7 8 Assumed x, y are +ve fractions ; x<1 in DC portion, y<1 in BC portion x y u v a1 c1 d1 a2 c2 d2 ar Fgx Fgy Fix Fiy Fixy Feccl Feccb x1 y1 x2 y2 x3 y3 P1 Q1 R1 P2 Q2 R2 el/L eb/B Kmax 0.300 0.200 0.700 0.800 0.280 0.733 0.767 0.000 0.000 0.000 0.280 0.733 0.767 0.008 0.010 0.004 -0.233 -0.267 -0.533 0.233 0.267 -0.467 0.267 0.233 -53.571 0.000 16.071 0.000 -61.224 19.898 0.300 0.325 10.714 0.400 0.300 0.600 0.700 0.210 0.767 0.800 0.000 0.000 0.000 0.210 0.767 0.800 0.004 0.006 0.002 -0.267 -0.300 -0.467 0.200 0.233 -0.400 0.233 0.200 -81.633 0.000 26.531 0.000 -95.238 33.333 0.325 0.350 14.286 0.500 0.600 0.700 0.800 0.400 0.500 0.600 0.700 0.500 0.400 0.300 0.200 0.600 0.500 0.400 0.300 0.150 0.100 0.060 0.030 0.800 0.833 0.867 0.900 0.833 0.867 0.900 0.933 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.150 0.100 0.060 0.030 0.800 0.833 0.867 0.900 0.833 0.867 0.900 0.933 0.002 0.001 0.000 0.000 0.003 0.001 0.001 0.000 0.001 0.001 0.000 0.000 -0.300 -0.333 -0.367 -0.400 -0.333 -0.367 -0.400 -0.433 -0.400 -0.333 -0.267 -0.200 0.167 0.133 0.100 0.067 0.200 0.167 0.133 0.100 -0.333 -0.267 -0.200 -0.133 0.200 0.167 0.133 0.100 0.167 0.133 0.100 0.067 -133.333 -240 -500 -1333.333 0.000 0.000 0.000 0.000 46.667 90.000 200.000 566.667 0.000 0.000 0.000 0.000 -160.000 -300.000 -666.667 -2000 60.000 120.000 283.333 900 0.350 0.375 0.400 0.425 0.375 0.400 0.425 0.450 20.000 30.000 50.000 100.000 0.900 0.800 0.100 0.200 0.010 0.933 0.967 0.000 0.000 0.000 0.010 0.933 0.967 0.000 0.000 0.000 -0.433 -0.467 -0.133 0.033 0.067 -0.067 0.067 0.033 -6000 0.000 2700 0.000 -12000 5700 0.450 0.475 300 0.800 0.900 0.200 0.100 0.010 0.967 0.933 0.000 0.000 0.000 0.010 0.967 0.933 0.000 0.000 0.000 -0.467 -0.433 -0.067 0.067 0.033 -0.133 0.033 0.067 -12000 0.000 5700 0.000 -6000 2700 0.475 0.450 300 The Indian Concrete Journal June 2014 79 POINT OF VIEW Bijay Sarkar holds a degree in Civil Engineering from Jadavpur University, Kolkata. He is a Superintending Engineer (Civil) in Engineering & Planning Cell of Project Department at DVC Head Quarters, Kolkata. He has a long experience in civil construction and design for more than last 25 years in Damodar Valley Corporation (DVC), a PSU under Ministry of Power, Govt of India. He is experienced in structural design works for power house & boiler building structures, mill & bunker structures, coal conveying structures of power plants for 500MW capacity & above owned by DVC. His keen interest is on preparing software modules in respect of civil engineering aspects. 80 The Indian Concrete Journal June 2014
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