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# Licture_8 - Terzahgi’s bearing capacity equation Terzahgi...

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Unformatted text preview: Terzahgi’s bearing capacity equation Terzahgi Ultimate bearing capacity cohesion term Shallow Foundations (Cont’d) qu = c' Nc + qN q + 1 γBNγ 2 below foundation level above the foundation level c’= effective soil cohesion, γ = soil density, and B = foundation width. Nc , Nq , and Nγ = bearing capacity factors, depends on soil friction angle, φ’ Terzahgi’s bearing capacity equation 1. For continuous (Strip) Foundation In case of local shear failure 1. For continuous (Strip) Foundation 1 qu = c' Nc + qN q + γBNγ 2 qu = 2 1 ' ' ' c' Nc + qNq + γBNγ 3 2 2. For Square Foundation 2. For Square Foundation ' ' ' qu = 0.867 c' Nc + qNq + 0.4 γ BNγ qu = 1.3c' Nc + qN q +0.4 γBNγ 3. For circular Foundation 3. For circular Foundation ' ' ' qu = 0.867 c' Nc + qNq + 0.3γ BNγ qu = 1.3c' Nc + qN q +0.3γBNγ 1 Local shear failure N’c , N’q , and N’γ = modified bearing capacity N’ factors, calculated with: ' N c = cot φ ' N q − 1 _ Bearing capacity factors for local shear failure 1000 Bearing capacity factors Nc, Nq, Nγ ( ) ' Nq = e _ ⎛ 2 ⎜ 3 π / 4 −φ ' / ⎜ ⎝ _ ⎞ 2 ⎟ tan φ ' ⎟ ⎠ 100 ⎛ 1⎜ Kp ' Nγ = ⎜ _ 2⎜ 2' ⎝ cos φ _ ⎛ ⎞ 2 cos ⎜ 45 + φ ' / 2 ⎟ ⎜ ⎟ ⎝ ⎠ ⎞ _ ⎟ − 1 ⎟ tan φ ' 2 10 Nc Nq Nγ 1 ⎟ ⎠ Or from Table 3.2 p.88 For φ = 0 Nc = 5.7 Nq = 1.0 Nγ = 0.0 0.1 and φ' = tan −1 ⎜ tan φ' ⎟ _ ⎛2 ⎝3 ⎞ ⎠ 0.01 0 10 20 30 40 50 Soil friction angle φ (deg.) Example_2.2 Square column foundation is 2 m by 2 m in plan. The foundation depth is 1.5 m, γ = 16.5 kN/m3, φ = 36º and c’ = 0. 36º Assume General shear failure in soil, use Terzaghi’s Terzaghi’ equation and F.S. = 3, calculate the gross allowable load the column can carry. Example_2.4 For the shown continuous strip footing. It is required to determine the ultimate bearing capacity of soil. Assume Local shear failure Use a factor of Safety F.S = 4, calculate the load, qall, the footing can carry. q a ll (Ib/ft) Example_2.3 Resolve Example 2.2 assume Local shear failure in soil. γ = 115 Ib/ft3 φ = 25º 25º ’ = 400 Ib/ft2 c 2 ft 2.5 ft 2 3.4 Factor of Safety Gross allowable- bearing capacity is: allowable- qu = qall F .S 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 ...
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