Week 4 Challenge
Week 4 Challenge – Explanation
Input Data:
* The Abutment wall is 5m in length and cross-section as per the figure.
* The backfill soil is 7m in height.
* The ground level is 3m above the top of the foundation level.
* Assume water table at GL.
Problem:
Calculate the Earth Pressure acting on the abutment wall and apply it on the staad pro.
Figure:
Solution:
Unit weight of soil:
* Unit weight of soil is defined as the total weight of soil mass (W) per unit of a total volume (V).
ɣ = (W / V)
* Dry Unit weight of soil is defined as the weight of dry soil (Ws) per unit of a total volume (V).
ɣd = (Ws / V)
* Saturated Unit weight of soil is defined as the ratio of the weight of fully saturated soil (Wsat) per volume (V).
ɣsat = (Wsat / v)
* Submerged Unit weight of soil is defined as the saturated unit weight (ɣsat) minus the unit weight of water (ɣw).
ɣ = (ɣsat - ɣw)
Earth Pressure:
Earth Pressure – It is the horizontal thrust exerted by the retained earth on a vertical structural element like water, pier, abutment, etc.
Three conditions exist as follows:
1, Earth Pressure at Rest
When the vertical structural member is rigid and unyielding, the soil mass is in a state of rest and there will be no deformations and displacement. The earth pressure corresponding to this state is earth pressure at rest.
2, Active Earth Pressure
When the vertical structural member moves away from the backfill, the soil mass expands and there is a decrease in pressure. The earth pressure corresponding to this state is active earth pressure.
The structure may fail in overturning.
3, Passive Earth Pressure
When the vertical structural member moves toward the backfill, the soil mass is compressed and the shear resistance developed toward the vertical member and there is an increase in pressure. The earth pressure corresponding to this state is passive earth pressure.
The structure may fail in sliding.
Soil loads on Abutment:
Earth pressure on the bridge,
Case-1
Assume, ka=0.35, ko=0.5, ɣ=18 kN/m^3
k1 = ka x ɣ x h
= 0.35 x 18 x 3
= 18.9 (kN/m^2)
k2 = ko x ɣ x h
= 0.5 x 18 x (3+4)
= 63 (kN/m^2)
Case-2
Assume, ka=0.35, ko=0.5, ɣsat=19 kN/m^3, ɣw=10 kN/m^3
k1 = ka x (ɣsat - ɣw) x h
= 0.35 x (19-10) x 3
= 9.45 (kN/m^2)
k2 = ko x (ɣsat - ɣw) x h
= 0.5 x (19-10) x (3+4)
= 31.5 (kN/m^2)
Soil loads on Foundation:
Soil loads on the foundation of the bridge,
Case-1
Assume, ɣ=18 kN/m^3
P1 = ɣ x h
= 18 x 3
= 54 (kN/m^2)
P2 = ɣ x h
= 18 x (3+4)
= 126 (kN/m^2)
Case-2 (Uplift pressure)
Uplift pressure on the integral abutment bridge,
Assume, ɣw=10 kN/m^3
The “hw” is taken from the water table level to bottom
of foundation.
Pu = ɣw x hw
= 10 x (1 + 3 + 4)
= 80 (kN/m^2)
Case-3 (Inclined backfill and its earth pressure)
Earth pressure for the bridge,
Assume, Angle of internal friction “Φ” is 27 degrees.
The inclination of the backfill “β” is 15 degrees.
ka = cos(β) x (cos(β) – sqrt ((cos(2β) – cos(2Φ))) / (cos(β) + sqrt ((cos(2β) – cos(2Φ)))
= cos(15) x (cos(15) – sqrt ((cos(2x15) – cos(2x27))) / (cos(15) + sqrt ((cos(2x15) – cos(2x27)))
= 0.22
Earth pressure = ka x ɣ x h
= 0.22 x 18 x (1 + 4 + 3)
= 31.68 (kN/m^2)
Staad model – Geometry & Load application:
Geometry
Geometry-Plate thick 300mm
Load Case(1)-Plate load-Hydrostatic-Backfill (7m height)
Load Case(1)-Plate load-Hydrostatic-Water Content (3m height)
Displacement Result-Load Comb(3)
Plate Result-Load Comb(3)-Maximum Bending Moment (Mx-Local)
Plate Result-Load Comb(3)-Maximum Bending Moment (My-Local)
Plate Result-Load Comb(3)-Maximum Shear Force (SQx-Local)
Plate Result-Load Comb(3)-Maximum Shear Force (SQy-Local)
Result Summary:
The retaining wall of 5m length & 8m height is modeled in staad pro. up to 3m height, water content available. up to 7m height, backfill available.
Earth pressure (for backfill and water content) loads are calculated and applied as trapezoidal in hydrostatic load type. Results such as displacement, and plate results (maximum BM, maximum SF) were verified.