Active Wedge Behind A Gravity Retaining Wall Complete 2011

Modeling of the Active Wedge behind a Gravity Retaining Wall

By: Rex Radloff

14.531 Advanced Soil MechanicsUniversity of Massachusetts LowellDepartment of Civil Engineering

University of Massachusetts Lowell 14.531 Advanced Soil Mechanics

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Rankine Active Wedge

• Wall friction is neglected• Active force acts 1/3H from the

base• Front face angle (θ) is 90⁰• Overburden slope (β) is 0⁰• Back face angle (α) is 45 + φ/2

What if the above was not held constant? • How would the magnitude

and location of Pa change?• What are its effects on the

failure criteria?• To what degree?

Rankine wedge behind a gravity retaining wall.

Rankine stress distribution behind a gravity retaining wall.

φw


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Active Wedge with Reactions

Active wedge with consideration towards θ, β, and φw

Note:The weight and centroid of the active wedge is a function of H, Wh, WBs, γ, α , θ, and β

The weight and centroid of the backfill is a function of H, WBs, Wh, γ, θ, and β

W


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Modeling the Active WedgeKnown

Wall dimensions and materials. Soil and wall shear strength

parameters. Grade of overburden soil.

Solve for Angle θ of the front face of the active wedge. Angle α of the back face of the active wedge. Active force, Pa, that is produced by the active wedge. The location of the Pa resultant. FS against overturning and sliding. Eccentricity Maximum stress, qMAX underneath the foundation.


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Example Retaining Wall: Case 1

γ1 = 117 pcfγconc = 150 pcfφ1’ = 34⁰δ’ = 18⁰Ca = 800 psf

Assume firm underlying soil


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θ = 90⁰, φw = 34⁰ Pa = 6.51 Kips, α = 56.8⁰ θ = 90⁰, φw = 0⁰ Pa = 7.12 Kips, α = 62.0⁰ (45 + φw/2)θ = 71⁰, φw = 34⁰ Pa = 11.24 Kips, α =61.2⁰ θ = 71⁰, φw = 0⁰ Pa = 10.84 Kips, α = 71.2⁰

γ = 117 pcf, Cs = 0 psf, Cw = 0 psf, H = 20.75 ft., Wh = 6.00 ft., φ = 34⁰, β = 0⁰


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Failure Analysis: H/Wh = 3.4

Case FS Over Turning

FS Sliding

Eccentricity, ft.

qMAX, ksf

θ = 90⁰, φw = 34⁰,Pa = 6.51 Kips, α = 56.8⁰

3.53 3.47 -1.45 31

θ = 90⁰, φw = 0⁰ , Pa = 7.12 Kips, α = 62.0⁰

3.69 2.46 0.53 18

θ = 71⁰, φw = 34⁰, Pa = 11.24 Kips, α =61.2⁰

0.53 2.68 -0.04 17

θ = 71⁰, φw = 0⁰, Pa = 10.84 Kips, α = 71.2⁰

0.85 1.61 3.48 32The increase in Pa lead to a below acceptable FS for overturning.

The greater distance between the Pa vector and pt.O led to a below acceptable eccentricity.

|e| 2.08 ft


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Example Retaining Wall: Case 2

γ1 = 117 pcfγconc = 150 pcfφ1’ = 34⁰δ’ = 18⁰Ca = 800 psf

Assume firm underlying soil


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θ = 90⁰, φw = 34⁰ Pa = 6.51 Kips, α = 56.8⁰ θ = 90⁰, φw = 0⁰ Pa = 7.12 Kips, α = 62.0⁰ (45 + φw/2)θ = 45⁰, φw = 34⁰ Pa = 29.03 Kips, α =57.2⁰ θ = 45⁰, φw = 0⁰ Pa = 21.41 Kips, α = 84.50⁰

γ = 117 pcf, Cs = 0 psf, Cw = 0 psf, H = 20.75 ft., Wh = 20.75 ft., φ = 34⁰, β = 0⁰


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Failure Analysis: H/Wh = 1.0

Case FS Over Turning

FS Sliding

Eccentricity, ft.

qMAX, ksf

θ = 90⁰, φw = 34⁰,Pa = 6.51 Kips, α = 56.8⁰

7.11 7.07 -1.93 50

θ = 90⁰, φw = 0⁰ , Pa = 7.12 Kips, α = 62.0⁰

15.04 5.19 -0.70 35

θ = 71⁰, φw = 34⁰, Pa = 11.24 Kips, α =61.2⁰

∞ 7.23 -1.90 55

θ = 71⁰, φw = 0⁰, Pa = 10.84 Kips, α = 71.2⁰

∞ 2.36 2.66 42

|e| 4.46 ft

+58%


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Pa vs. α vs. θ

φw = 34⁰

φw = 0⁰ Li

ne o

f in

ters

ectio

n

Produced using Mathematica


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Conclusion The active force (Pa) depends on the angles θ, α, and

φw found among the active wedge.

Any deviation between the calculated active force behind the same retaining wall depends on the combining effects of θ, α, and φw found among the active wedge.

As the angle θ decreases, the Pa will increase as well as the variance between the Pa calculated with and without wall friction.

Hard to openly predict the influence on the failure criteria


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Is it worth consideration?Yes


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References Das, B.M., (2006). “Principles of Geotechnical Engineering – Sixth Edition”

Lambe, T.W., and R.V. Whitman, (1969). “Series in Soil Engineering - Soil Mechanics”

Mangano, Sal, (2010). “Mathematica Cookbook”

AutoCAD 2011®

Wolfram Mathematica®

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Active Wedge Behind A Gravity Retaining Wall Complete 2011