Unformatted text preview: 3.4 Factor of Safety Gross allowable bearing capacity is: Gross allowable qu = qall F .S Shallow Foundations (Cont’d) Net allowable bearing capacity is: allowable qall ( net ) = qall − q q = γD f
Net allowable Column Load: allowableDf Qa ll (DL+LL) Ws /2 Wf Ws /2 Qall = qall ( net ) . A qu 3.5 Water Table effect on Bearing capacity equations If the water table is close to the foundation, bearing capacity equation must be modified. 1. Water table above or at foundation level
The factor q in the equation takes the form: q = (Df – d)γ + d γsub (D Also γ in the last term of the Equation should replaced by: Modification of Bearing capacity equations for water table 2. Water table below the foundation level
Such that 0 < d < B The last term of the bearing capacity equation should replaced by: γ =γ sub + _ d (γ − γ sub ) B In case of d > B γsub = γsat  γw γ =γ _ 1 General bearing capacity equation (Meyerhof 1963) qu = c' N c F csFcd Fci + qN qF qsFqd Fqi +
1. bearing capacity factors Nc , Nq , and Nγ 1 γ BN γ F γsFγd Fγi 2 qu = c' Nc F csFcd Fci + qN q F qsFqd Fqi +
c’= soil cohesion, γ = soil density, and B = foundation width. 1 γBNγ F γsFγd Fγi 2 From Table 3.3 at φ’ 2. Shape factors Nc , Nq , Nγ = bearing capacity factors, Fcs, Fqs, Fγs = shape factor, Fcd, Fqd, Fγd = depth factor, Fci, Fqi, Fγi = load inclination factor, These factor are all empirical factors from experimental work ⎛ B ⎞⎛ N q ⎞ ⎟ Fcs = 1 + ⎜ ⎟⎜ L ⎠⎜ N c ⎟ ⎝ ⎝ ⎠ ⎛ B⎞ Fqs = 1 + ⎜ ⎟ tan φ ' ⎝ L⎠ ⎛ B⎞ Fγ s = 1 − 0.4⎜ ⎟ ⎝ L⎠ qu = c' N c F csFcd Fci + qN qF qsFqd Fqi +
3. Depth factors 1 γ BN γ F γsFγd Fγi 2 Example
A square foundation (B x B) has to be constructed (B as shown in figure. Given that γ = 105 Ib/ft3, γsat = 118 Ib/ft3, Df = 4 ft, D1 = 2 ft, and the gross allowable load Qall with FS = 3 is 150,000 Ib, Ib, calculate the size of the footing.
Qall ⎛D ⎞ Fcd = 1 + 0.4⎜ f ⎟ ⎝B⎠ Fγ d = 1
2 Fqd = 1 + 2 tan φ ' (1 − sin φ ' )
4. inclination factors Df B ⎛ β° ⎞ ⎜1− ⎟ Fci = Fqi = ⎜ 90 ⎟ ⎝ ⎠ 2 ⎛ β° ⎞ Fγ i = ⎜ 1 − ' ⎟ ⎜ φ⎟ ⎝ ⎠ 2 =34º 2 qu = c' N c F csFcd Fci + qN qF qsFqd Fqi +
For c’ = 0.0 c’ 1 γ BN γ F γsFγd Fγi 2 Example
For the column foundation shown below, determine the gross allowable load Qall. Use FS = 4. qu = qN qF qsFqd Fqi +
For β = 0.0 1 γ BN γ F γsFγd Fγi 2 Fqi = Fγi = 1 1 γ BN γ F γsFγd 2 qu = qN qF qsFqd + qu = c' N c F csFcd Fci + qN qF qsFqd Fqi + 1 γ BN γ F γsFγd Fγi 2 3 ...
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 Spring '10
 Magdy
 Geotechnical Engineering, QU, Trigraph, qN qF qsFqd, qF qsFqd Fqi

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