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Lecture 14 - BC14 Eccentric Loading i.e load not at footing...

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BC14 Eccentric Loading: i.e. load not at footing center – overturning moment o Overturning moment equivalent to eccentric load at distance e from footing centerline o Consider footing of width B and length L o Examine pressure distribution under footing o Assume footing is rigid and that soil has linear behavior. Use beam analogy. o Separate flexural and uniform load components BL Q A Q q uniform = = 2 3 flex LB Qe 6 ) 12 / LB ( ) 2 / B )( Qe ( I Mc q = = = o Sum the components: - = + = ± = ± = B e 6 1 BL Q q and B e 6 1 BL Q q LB Qe 6 BL Q q q q min max 2 flex uniform this equation is valid for e B/6 at which point q min = 0 (To prevent tipping, most designers maintain e B/6.) o If (e > B/6) then part of overall pressure diagram is in tension. Since soil doesn’t withstand sustained tension, the pressure distribution becomes triangular and the footing begins to lift at the heel. Since the load Q must act through the centroid of the pressure diagram: ( ) 0 q and L 2e - B 3 4Q L e - 2 B 3 2Q 2 xL Q q e 2 B 3 x and q 2 xL Q min max max = = = = - = = o Can design footing for q max using this approach but the soil does not behave linearly, failure at the toe will redistribute the load, the footing is not rigid, the column connection carrying moment is not rigid, and therefore the actual pressure distribution varies from the above. Q Q M e = B e B/2 x
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BC15 o Meyerhof (1953) proposed an approximation to the above analysis using an “effective area” for the footing based on an “effective width”, B’, or an “effective length”, L’, (depending on the direction of the eccentricity). The analysis shown here is for 1-D eccentricity (e L or e B ). Analysis Procedure: 1. Calculate X = B - 2e B and Y = L – 2e L 2. Assign X and Y to B’ and L’. B’ is the shortest dimension and L’ is the longest. 3. Note that e max = 0.5 because this will give B’ (or L’) = 0. Usual design limit is e B/6 (or e L/6) to keep resultant in center 1/3 of footing (best approximation by this method). 4. Use BC equation as follows: q u = cN c λ cs λ cd λ ci + qN q λ qs λ qd λ qi + ½ γ B’N γ λ γ s λ γ d λ γ i use B = B’ with N γ term use B’ and L’ for calculation of shape factors λ cs , λ qs , λ γ s use actual B for calculation of depth factors λ cd , λ qd , λ γ d 5. Use the effective area to calculate the bearing capacity Q u = q u A’ = q u B’L’ The concept of effective area can be applied to circular footings and 2-D eccentricity. These problems are solved for effective dimensions so that the resultant load acts at the centroid of the effective area. (This is true of the above 1-D problems which are much easier to calculate.) For clays ( φ 0) the bearing capacity decreases linearly with e.
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