POINT OF VIEW
Analysis of isolated isolated footing subjected 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.
INTRODUCTION When a rectangular/square isolated footing of size L x B is subjected to a set of forces comprising of compressive axial load P and bi-axial moments M x & M y at centre of footing, load P alone may equivalently be considered at eccentricities in X & Y directions from origin at the ; following location from centre of footing : = . =
When pressure at any location under the footing is compressive in nature, Centroid of the effective footing
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The Indian Indian Concrete Concrete Journal June 2014
area coincides with the centre of the footing. The Parameters P, , and dimensions of the foundation along with its sectional properties are known and we get pressure distribution under the foundation at any location (X, Y) with respect to centroidal axis of the footing by using following bending equation :
=
±
∴ =
=
× ±
− ×
×
×
− −
;
− ×
×
... (1)
where L = Length of the footing B = Breadth of the footing X = Perpendicular distance from the Centroidal Y-axis of Y-axis of the point on the Cross-Section at which pressure is to be determined. Y = Perpendicular distance from the Centroidal X-axis of the point on the Cross-Section at which pressure is to be determined. P = Vertical Force Acting on the section at centre.
POINT OF VIEW
ANALYSIS OF THE FOUNDATION
By assuming,
For Case – II to Case – V , analysis of footing has been done and general equations on case to case basis are deduced here. In all the cases, aim is to nd out the effective area, CG of the effective area, sectional properties of the effective area wrt centroidal axes system i.e. through CG of the effective area. As we are not using principle axis to calculate the sectional properties, we are to calculate the product moment of area of the effective footing also and use all such data in General Form of Bending Equation and assembling all the cases into a graph. Assumed that a set of forces P M b M l are acting at the centre of a rectangular isolated footing of size (L x B). P is acting in the vertically downward, M b is acting along B (from down to top of this paper) and M l is acting along L (from left to right of this paper) respectively. When a portion of footing area is lifted, CG of the effective compressed footing area shifts away from the centre and as CG is changed, the equivalent eccentricity of load also changes from say, Therefore, el eb to . . Re vi se d Moments acting at revised CG location are ,
,
,
= ; = . ∴
General Form of Two Dimensional Bending Stress Equation will be :
=
+
+ −
+
+ −
The Indian Concrete Journal June 2014
+ −
= Constant for given section and
loads
=
+ −
= Constant for given section and
loads
=
= Constant for given section and loads
We can write, Am + Bn + C = p. This is the General Form Equation of pressure underneath foundation subjected to high Bi-Axial Eccentricity. This is a straight line equation. Further, it can be observed that at some combinations of m and n, pressure p may remain constant. Now, we know that pressure at neutral axis is equal to 0. Therefore, for any value of Co-Ordinate (m, n) being on the neutral axis, we get, Am + Bn + C = 0. This is nothing but a straight line equation representing the Neutral Axis. Substituting the values of , , for moments in equation (2), we get
...(2)
where, p = Pressure at co-ordinate ( m, n) wrt Centroidal Axis P = Vertical Load at centre of the footing Ar = Revised Effective area of the foundation = Revised Moment at Revised CG location about Revised Centroidal Axis YY = Revised Moment at Revised CG location about Revised Centroidal Axis XX I y = Second Moment of Inertia of the effective area about Revised Centroidal Axis YY I x = Second Moment of Inertia of the effective area about Revised Centroidal Axis XX I xy = Product Moment of Inertia of the effective area m = X-Axis Co-ordinate of the location where pressure is to be found wrt Revised Centroidal Axis n = Y-Axis Co-ordinate of the location where pressure is to be found wrt Revised Centroidal Axis 62
=
∴ =
+
+ + + −
+ + + + = − −
...(3)
General equations for Sectional Properties, Eccentricities & co-ordinate of Maximum Pressure Location wrt Revised Centroidal Axis are found out and used in the above equation on case to case basis for nding out the location of Neutral Axis as well as Maximum Bearing Pressure :
CASE - II When NA cuts AB and AD. Here uplift portion is APQ and effective portion is PQDCBP. This is considered that NA cuts AB at P & AP = yB and NA cuts AD at Q & AQ = xL
POINT OF VIEW
shown later in Figure 6. Following notes may be read in conjunction with the ow 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 ow chart. Different variable names have been used in the Equation Sets with that of the paper for use in VB Macro.
2. As the nal 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, M y, 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 EbB 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 if E 1ByL < E 1/ L
Equal
No Increase y = y + y p=p+1
Decrease y = y - y q=q+1
If p> 0 and q = 0 or If p = 0 and q >0
No
Yes
Yes If x < = 1 and y < = 1
Case = 2 x=x-0 y=y-0
Use Eqn. Set (2) To Calculate Eff. Sectional Properties y = y prev
No Yes If x > 1 and y < = 1
Case = 3 u=2-x v=1-y
Use Eqn. Set (3) To Calculate Eff. Sectional Properties
No Yes Case = 4 u=1-x v=2-y
If x <= 1 and y >1
Use Eqn. Set (4) To Calculate Eff. Sectional Properties
No Yes If x > 1 and y > 1 No
Case = 5 u=2-x v=2-y
Use Eqn. Set (5) To Calculate Eff. Sectional Properties
ERROR Mark -A
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The Indian Concrete Journal June 2014
Calculate E1B L and EbB yB from Eqn. Set (A)
Mark - B
POINT OF VIEW
x prev = x
Mark - B
Mark -A Equal
Yes If Eb By B < Eb/B
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
Yes If x < = 1 and y < = 1
Case = 2 x=x-0 y=y-0
Use Eqn. Set (2) To Calculate Eff. Sectional Properties
No Yes If x > 1 and y < = 1
Case = 3 u=2-x v=1-y
Use Eqn. Set (3) To Calculate Eff. Sectional Properties
No Yes If x <= 1 and y >1
Case = 4 u=1-x v=2-y
Use Eqn. Set (4) To Calculate Eff. Sectional Properties
No Yes If x > 1 and y > 1
Case = 5 u=2-x v=2-y
Use Eqn. Set (5) To Calculate Eff. Sectional Properties
Calculate E1 ByL and EbB yB from Eqn. Set (A)
No
If E1ByL - E 1/L < n and If EbB yB - E b /B < n
Calculate Pmax from eqn. Set (B) Print Case No, x, y, K max, P max
x = x prev
END
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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 rst 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. 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 predened 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. 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 :: 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
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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
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
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
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)
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 Coefcient Kmax = a + ((EbByB + Feccb) * a3 + (ElByL + Feccl) * b3) Max Pressure = Kmax * P / (B * L)
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-efcients CASE - II (Short Side & Long Side Intersected) - A Triangular Part is uplifted
Annexure Table 3. calculation of pressure co-efcients 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 fra ctions ; 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 = K max * P/BL
Example No
y 0.200 0.300 u 0.700 0.600 v 0.800 0.700 a1 0.700 0.600 c1 0.650 0.700 d1 0.500 0.500 a2 0.050 0.050 c2 0.267 0.367 d2 0.667 0.667 ar 0.750 0.650 Fgx 0.624 0.674 Fgy 0.511 0.513 Fix 0.062 0.054 Fiy 0.035 0.023 Fixy 0.003 0.003 Feccl -0.124 -0.174 Feccb -0.011 -0.013 x1 -0.424 -0.374 y1 0.489 0.487 x2 -0.324 -0.274 y2 -0.511 -0.513 x3 0.376 0.326 y3 0.489 0.487 P1 -11.327 -15 .204 Q1 7.267 8.252 R1 2.662 4.084 P2 -9.912 -13.032 Q2 -8.685 -10.138 R2 2.663 3.941 el/L 0.249 0.282 eb/B 0.022 0.026 Kmax 2.840 3.307 Max pressure p = K max * P/BL
Annexure Table 2 . calculation of pressure co-efcients CASE - III (both the short sides intersected by na)
Annexure Table 4. calculation of pressure co-efcients 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
Max pressure p = K max * P/BL
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 rst quadrant. x 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000
Example No
1
2
0.400 0.500 0.600 0.500 0.750 0.500 0.050 0.467 0.667 0.550 0.724 0.515 0.046 0.014 0.002 -0.224 -0.015 -0.324 0.485 -0.224 -0.515 0.276 0.485 -21.46 1 9.535 6.486 -17.884 -12.165 6.013 0.316 0.030 3.956
3
0.500 0.400 0.500 0.400 0.800 0.500 0.050 0.567 0.667 0.450 0.774 0.519 0.037 0.008 0.002 -0.274 -0.019 -0.274 0.481 -0.174 -0.519 0.226 0.481 -32.520 11.267 10.927 -26.016 -15.185 9.634 0.349 0.037 4.918
4
0.600 0.300 0.400 0.300 0.850 0.500 0.050 0.667 0.667 0.350 0.824 0.524 0.029 0.004 0.001 -0.324 -0.024 -0.224 0.476 -0.124 -0.524 0.176 0.476 -54 .857 13.695 20.294 -41.143 -20.139 16.659 0.382 0.047 6.486
5
0.700 0.800 0.900 0.200 0.100 0.000 0.300 0.200 0.100 0.200 0.100 0.000 0.900 0.950 1.000 0.500 0.500 0.500 0.050 0.050 0.050 0.767 0.867 0.967 0.667 0.667 0.667 0.250 0.150 0.050 0.873 0.922 0.967 0.533 0.556 0.667 0.021 0.012 0.003 0.001 0.000 0.000 0.001 0.001 0.000 -0.373 -0.422 -0.467 -0.033 -0.056 -0.167 -0.173 -0.122 -0.067 0.467 0.444 0.333 -0.073 -0.022 0.033 -0.533 -0.556 -0.667 0.127 0.078 0.033 0.467 0.444 0.333 -110.76 9 -320.000 -24 00.000 17.164 20.923 0.000 44.782 140.615 1140.000 -73.846 -160.000 0.000 -29.638 -54.154 -240.000 32.557 77.231 60.000 0.414 0.446 0.475 0.066 0.107 0.250 9.474 17.143 60.000
6
7
8
Assumed x, y are +ve fra ctions ; 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
Max Pressure p = K max * P/BL
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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.
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