This cue card is based on chapter 4 section 3 of Foundation Analysis and Design (5th ed) by Jospeh E. Bowles. To understand the cue card, the user should have the book. Permission to quote from the book has been granted. Comments or questions concerning the cue card should be directed to the developer, Helmut Schmidhofer, on 02 4862 1295 (international replace leading 0 with 61) or email engcomp@gmail.com.

IMPORTANT NOTICE: Terzaghi's, Meyerhof's and Hansen's bearing capacity equations can now be processed ONLINE!

There is currently no method of obtaining the ultimate bearing capacity of a foundation other than an estimate. Keep this statement in mind when using the formulas in table 4.1.

Enter your custom values at lines marked ****, then process all lines. The word "process" means to select (highlight) the relevant text and click the spinning globe in the Engineers' Compendium window. To work with the cue card, you need MATHSERV, which you may download here.

TABLE 4.1 - bearing-capacity equations by the several authors as indicated:

STEP 1 - select the appropriate K_PG by processing only one line of the next eleven lines:

* for PHI is less than 5 degrees, K_PG = 10.8
* for PHI is 5 to 9 degrees, K_PG = 12.2
* for PHI is 10 to 14 degrees, K_PG = 14.7
* for PHI is 15 to 19 degrees, K_PG = 18.6
**** for PHI is 20 to 24 degrees (Example 4-1), K_PG = 25
* for PHI is 25 to 29 degrees, K_PG = 35
* for PHI is 30 to 34 degrees, K_PG = 52
* for PHI is 35 to 39 degrees, K_PG = 82
* for PHI is 40 to 44 degrees, K_PG = 141
* for PHI is 45 to 49 degrees, K_PG = 298
* for PHI is 50 degrees, K_PG = 800

Alternatively, to interpolate the table:
**** angle of internal friction in degrees, PHI = 21
**** lower bound angle in degrees, LBA = 20
**** lower bound K_PG from table, LB = 25
**** upper bound K_PG from table, UB = 35
Interpolated K_PG = LB + (UB-LB)*(PHI-LBA)/5 := 27.

Linear interpolation over-estimates the value, therefore, reduce the result by eye:
**** say K_PG = 26

STEP 2 - select the appropriate S_G by processing only one line of the next three lines:

* for strip footing, S_G = 1.0
* for round footing, S_G = 0.6
**** for square footing, S_G = 0.8

STEP 3 - select the appropriate S_C by processing only one line of the next two lines:

* for strip footing, S_C = 1.0
**** for round or square footing, S_C = 1.3

STEP 4 - given soil properties and footing description (enter and process your values at ****):

**** unit weight of soil in kN/m3, GAMMA = 17.3
**** angle of internal friction in degrees, PHID = 20
**** cohesion of soil in kPa, C = 20
**** depth of base in m, D = 1.2
**** range of sizes in m, B = {1.2,1.5,2.0,2.5,3.0}
**** safety factor, SF = 3

Process the next ten lines:
Soil weight at base, QBAR = GAMMA*D := 20.76
Convert PHI to radians, PHI = PHID*_pi/180 := 0.3491
45 degrees in radians, A45 = 45*_pi/180 := 0.7854
Coefficient A = exp((.75*_pi-PHI/2)*tan(PHI)) := 2.2124
Coefficient NQ = A^2/(2*(cos(A45+PHI/2)^2)) := 7.4387
Coefficient NC = (NQ-1)/tan(PHI) := 17.6903
Coefficient NG = tan(PHI)*(K_PG/(cos(PHI)^2)-1)/2 := 4.9704
Ultimate bearing capacity Q_ULT = C*NC*S_C+QBAR*NQ+.5*GAMMA*B*NG*S_G
Safe bearing capacity in kPa, Q_A = Q_ULT/SF
print using "#.### m square footing, Q__A P(+1) kPa"; B, Q_A

Example 4.1 ends here for Terzaghi's method. In practice, we wish to know the footing size for a range of superimposed loads:

**** range of loads in kN, P = {200,400,600,1000,1500,2000}
**** corresponding size of pedestals in m, BP = {.4,.4,.4,.5,.5,.6}
**** corresponding thickness of pad in m, TP = {.3,.3,.4,.4,.5,.6}
Weight of pedestal in kN, WP = 24*BP^2*(D-TP)

Process from "for I..." to "next I" as one block after editing QA values and test conditions:

for I = 1 to endvalid(P)
  TMP = sqrt(P[I]/200)
  if TMP < 2.0 then
    QA = 220
  elseif TMP < 3.0 then
    QA = 230
  else
    QA = 240
  end if
X[I]=sqrt((P[I]+WP[I]-GAMMA*BP[I]^2*(D-TP[I]))/(QA-24*TP[I]-GAMMA*(D-TP[I])))
next I

print using "for #### kN, base is P(+2) mm square"; P, 1000*X

Explanation of code: the line ...

X[I]=sqrt((P[I]+WP[I]-GAMMA*BP[I]^2*(D-TP[I]))/(QA-24*TP[I]-GAMMA*(D-TP[I])))

is derived from:

QA*X^2 = P + 24*X^2*TP + WP + GAMMA*(X^2-BP^2)*(D-TP)

STEP 1: Enter your values at **** and process all lines (values are from Example 4.2):

**** effective weight of saturated soil in kN/m3, GAMMAD = 9.31
**** angle of internal friction in degrees, PHID = 47
**** cohesion of soil in kPa, C = 0
**** depth of base in m, D = 0.5
**** width in m, B = 0.5
**** length in m, L = 2.0
**** angle in degrees of force v vertical, THETA = 0
**** safety factor, SF = 3

STEP 2: Process the next 21 lines:

overburden pressure in kPa, QBAR = D*GAMMAD := 4.655
45 degrees in radians, A45 = 45*_pi/180 := 0.7854
internal friction in radians, PHI = PHID*_pi/180 := 0.8203
passive pressure coefficient, KP = (tan(A45+PHI/2))^2 := 6.4447
Coefficient NQ = exp(_pi*tan(PHI))*KP := 187.2059
Coefficient NG = (NQ-1)*tan(1.4*PHI) := 414.3268
shape factors, SC = 1 + .2*KP*B/L := 1.3222
SQ = 1 + .1*KP*B/L*(PHID>10) := 1.1611
SG = SQ := 1.1611
depth factors, DC = 1 + .2*sqrt(KP)*D/B := 1.5077
DQ = 1 + (PHID>10)*.1*sqrt(KP)*D/B := 1.2539
DG = DQ := 1.2539
inclination factors, IC = (1 - THETA/90)^2 := 1.
IQ = IC := 1.
if PHID > 0 then
  IG = (1 - THETA/PHID)^2
  NC = (NQ-1)/tan(PHI)
else
  IG = 0
  NC = 0
end if

STEP 3: Ultimate bearing capacities in kPa:

vertical VQULT = C*NC*SC*DC + QBAR*NQ*SQ*DQ + .5*GAMMAD*B*NG*SG*DG := 2672.6984
inclined IQULT = C*NC*DC*IC + QBAR*NQ*DQ*IQ + .5*GAMMAD*B*NG*DG*IG := 2301.8312

These are presented in a separate cue card, see hansen_vesic.doc.