Fundamentals of Geotechnical Engineering (MindTap Course List)
Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
Question
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Chapter 14, Problem 14.8P

(a)

To determine

Find the Rankine’s active force per unit length of the wall.

Find the location of the resultant.

(a)

Expert Solution
Check Mark

Answer to Problem 14.8P

The Rankine’s active force per unit length of the wall is 34.31kN/m_.

The location of the resultant is 0.89m from bottom_.

Explanation of Solution

Given information:

The height (H) of the retaining wall is 3.05 m.

The depth (H1) of water table is 1.52 m.

The unit weight (γ1) of the soil is 16.51kN/m3.

The saturated unit weight (γ2) of the soil is 19.18kN/m3.

The angle (ϕ1) of internal friction above water table is 30°.

The angle (ϕ2) of internal friction below water table is 30°.

The surcharge (q) on the soil is 0.

Calculation:

Find the submerged unit weight (γ) of the soil using the equation:

γ=γ2γw

Here, γw is the unit weight of water.

The unit weight (γw) of the water is 9.81kN/m3.

Substitute 19.18kN/m3 for γ2 and 9.81kN/m3 for γw.

γ=19.189.81=9.37kN/m2

The angle of internal friction above and below the water table is same. Therefore,

ϕ1=ϕ2=ϕ=30°.

Find the coefficient of Rankine’s active earth pressure for frictionless wall (Ka) using the equation:

Ka=tan2(45°ϕ2)

Substitute 30° for ϕ.

Ka=tan2(45°30°2)=13

Find the active earth pressure (σa) at 0 m depth using the equation:

σa0=γ1zKa

Substitute 16.51kN/m3 for γ1, 0 m for z, and 13 for Ka.

σa0=16.51×0×13=0kN/m2

Find the active earth pressure (σa1) at water table (1.52 m depth from top) using the equation:

σa1=γ1H1Ka

Substitute 16.51kN/m3 for γ1, 1.52 m for H1, and 13 for Ka.

σa1=16.51×1.52×13=8.37kN/m2

Find the active earth pressure (σa2) at bottom (3.05 m depth) using the equation:

σa2=[γ1H1+γ(HH1)]Ka

Substitute 16.51kN/m3 for γ1, 9.37kN/m2 for γ, 1.52 m for H1, 3.05m for H, and 13 for Ka

σa2=[(16.51)(1.52)+(9.37)(3.051.52)](13)=13.15kN/m2

Find the pore water pressure (u) using the equation:

u=γw(HH1)

Substitute 9.81kN/m3 for γw, 3.05m for H, and 1.52 m for H1.

u=9.81×(3.051.52)=15kN/m2

Sketch the active earth pressure diagram as shown in Figure 1.

Fundamentals of Geotechnical Engineering (MindTap Course List), Chapter 14, Problem 14.8P , additional homework tip  1

Find the active earth pressure per unit length at component 1 (Pa1) using the equation:

Pa1=12σa1×H1

Substitute 8.37kN/m2 for σa1 and 1.52 m for H1.

Pa1=12×8.37×1.52=6.36kN/m

Find the active earth pressure per unit length at component 2 (Pa2) using the equation:

Pa2=σa1×(HH1)

Substitute 8.37kN/m2 for σa1, 3.05m for H, and 1.52 m for H1.

Pa2=8.37×1.53=12.81kN/m

Find the active earth pressure per unit length at component 3 (Pa3) using the equation:

Pa3=12(σa2σa1)×(HH1)

Substitute 13.15kN/m2 for σa2, 8.37kN/m2 for σa1, 3.05m for H, and 1.52 m for H1.

Pa3=12(13.158.37)×(3.051.52)=3.66kN/m

Find the active earth pressure per unit length at component 4 (Pa4) using the equation:

Pa4=12×u×(HH1)

Substitute 15.0kN/m2 for u, 3.05m for H, and 1.52 m for H1.

Pa4=12×15×(3.051.52)=11.48kN/m

Find the total active earth pressure per unit length (Pa) using the equation:

Pa=Pa1+Pa2+Pa3+Pa4

Substitute 6.36kN/m for Pa1, 12.81kN/m for Pa2, 3.66kN/m for Pa3, and 11.48kN/m for Pa4.

Pa=6.36+12.81+3.66+11.48=34.31kN/m

Therefore, the Rankine’s active force per unit length of the wall is 34.31kN/m_.

Refer Figure 1.

Find the location (z1) of active earth pressure per unit length at component 1 from bottom of the wall using the equation:

z1=(HH1)+H13

Substitute 3.05m for H and 1.52 m for H1.

z1=(3.051.52)+1.523=2.04m

Find the location (z2) of active earth pressure per unit length at component 2 from bottom of the wall using the equation:

z2=(HH1)2

Substitute 3.05m for H, and 1.52 m for H1.

z2=(3.051.52)2=0.765m

Find the location (z3) of active earth pressure per unit length at component 3 from bottom of the wall using the equation:

z3=(HH1)3

Substitute 3.05m for H and 1.52 m for H1.

z3=(3.051.52)3=0.51m

Find the location (z4) of active earth pressure per unit length at component 4 from bottom of the wall using the equation:

z4=(HH1)3

Substitute 3.05m for H and 1.52 m for H1.

z4=(3.051.52)3=0.51m

Find the location of resultant force (z¯):

Take moment about the bottom of the wall.

Pa×z¯=(Pa1×z1)+(Pa2×z2)+(Pa3×z3)+(Pa4×z4)

Substitute 34.31kN/m for Pa, 6.36kN/m for Pa1, 2.04m for z1, 12.81kN/m for Pa2, 0.765m for z2, 3.66kN/m for Pa3, 0.51m for z3, 11.48kN/m for Pa4, and 0.51m for z4.

34.31×z¯=(6.36×2.04)+(12.81×0.765)+(3.66×0.51)+(11.48×0.51)34.31z¯=12.974+9.799+1.867+5.855z¯=30.49534.31=0.89m

Therefore, the location of the resultant is 0.89m from bottom_.

(b)

To determine

Find the Rankine’s active force per unit length of the wall.

Find the location of the resultant.

(b)

Expert Solution
Check Mark

Answer to Problem 14.8P

The Rankine’s active force per unit length of the wall is 141.13kN/m_.

The location of the resultant is 2.04m from bottom_.

Explanation of Solution

Given information:

The height (H) of the retaining wall is 6 m.

The depth (H1) of water table is 3 m.

The unit weight (γ1) of the soil is 15.5kN/m2.

The saturated unit weight (γ2) of the soil is 19.0kN/m2.

The angle (ϕ1) of internal friction above water table is 30°.

The angle (ϕ2) of internal friction below water table is 36°.

The surcharge (q) on the soil is 15kN/m2.

Calculation:

Find the submerged unit weight (γ) of the soil using the equation:

γ=γ2γw

The unit weight (γw) of the water is 9.81kN/m3.

Substitute 19.0kN/m2 for γ2 and 9.81kN/m3 for γw.

γ=199.81=9.19kN/m2

Find the coefficient of Rankine’s active earth pressure for frictionless wall (Ka1) above water table using the equation:

Ka1=tan2(45ϕ12)

Substitute 30° for ϕ1.

Ka=tan2(45°30°2)=13

Find the coefficient of Rankine’s active earth pressure for frictionless wall (Ka1) below water table using the equation:

Ka2=tan2(45ϕ22)

Substitute 36° for ϕ.

Ka2=tan2(45362)=0.26

Find the active earth pressure (σa) at 0 m depth using the equation:

σa0=γ1zKa1+qKa2

Substitute 15.50kN/m3 for γ1, 0 m for z, 13 for Ka1, and 15kN/m2 for q.

σa0=(15.50×0×13)+(15×13)=5kN/m2

Find the active earth pressure (σa1) at water table (3.0 m depth from top) using the equation:

σa1=γ1H1Ka1+qKa1

Substitute 15.50kN/m3 for γ1, 3.0 m for H1, 13 for Ka1, and 15kN/m2 for q.

σa1=(15.5×3×13)+(15×13)=20.5kN/m2

Find the active earth pressure (σa3) at bottom (3.0 m depth) using the equation:

σa3=γ1H1Ka2+qKa2

Substitute 15.50kN/m3 for γ1, 3.0 m for z1, 0.26 for Ka2, and 15kN/m2 for q.

σa3=(15.5×3×0.26)+(15×0.26)=15.99kN/m2

Find the active earth pressure (σa2) at bottom (6.0 m depth) using the equation:

σa4=[γ1H1+γ(HH1)]Ka2+qKa2

Substitute 15.5kN/m3 for γ1, 19kN/m3 for γ2, 9.19kN/m3 for γ, 3 m for H1, 6m for H, and 0.26 for Ka2.

σa4=[(15.5)(3)+(9.19)(63)]0.26+[15×0.26]=19.26+3.9=23.16kN/m2

Find the pore water pressure (u) using the equation:

u=γw(HH1)

Substitute 9.81kN/m3 for γw and 6m for H, and 3 m for H1.

u=9.81×(63)=29.43kN/m2

Sketch the active earth pressure diagram as shown in Figure 2.

Fundamentals of Geotechnical Engineering (MindTap Course List), Chapter 14, Problem 14.8P , additional homework tip  2

Find the active earth pressure of component 1 per unit length due to surcharge load (Pa1) using the equation:

Pa1=σa0×H

Substitute 5.0kN/m2 for σa0 and 6.0 m for H.

Pa1=6×5=30kN/m

Find the active earth pressure per unit length at component 2 (Pa2) using the equation:

Pa2=12(σa1σa0)×H1

Substitute 20.5kN/m2 for σa1, 5kN/m2 for σa0, and 3.0 m for H1.

Pa2=12×(20.55)×3=23.25kN/m

Find the active earth pressure per unit length at component 3 (Pa3) using the equation:

Pa3=(σa3σa0)(HH1)

Substitute 15.99kN/m2 for σa3, 5kN/m2 for σa0, 6.0 m for H, and 3.0 m for H1.

Pa3=(15.995)×(63)=32.97kN/m

Find the active earth pressure per unit length at component 4 (Pa4) using the equation:

Pa4=12(σa4σa3)×(HH1)

Substitute 23.16kN/m2 for σa2, 15.99kN/m2 for σa1, 6.0 m for H, and 3.0 m for H1.

Pa4=12×(23.1615.99)×(63)=10.76kN/m

Find the active earth pressure per unit length at component 5 (Pa5) using the equation:

Pa5=12×u×(HH1)

Substitute 29.43kN/m2 for u, 6.0 m for H, and 3.0 m for H1.

Pa5=12×29.43×(63)=44.15kN/m

Find the total active earth pressure per unit length (Pa) using the equation:

Pa=Pa1+Pa2+Pa3+Pa4+Pa5

Substitute 30kN/m for Pa1, 23.25kN/m for Pa2, 32.97kN/m for Pa3, 10.76kN/m for Pa4, and 44.15kN/m for Pa5.

Pa=30+23.25+32.97+10.76+44.15=141.13kN/m

Therefore, the Rankine’s active force per unit length of the wall is 141.13kN/m_.

Refer Figure 1.

Find the location (z1) of active earth pressure per unit length at component 1 from bottom of the wall using the equation:

z1=H2

Substitute 6m for H.

z1=62=3m

Find the location (z2) of active earth pressure per unit length at component 2 from bottom of the wall using the equation:

z2=(HH1)+H13

Substitute 6m for H and 3 m for H1.

z2=(63)+33=4m

Find the location (z3) of active earth pressure per unit length at component 3 from bottom of the wall using the equation:

z3=(HH1)2

Substitute 6m for H and 3 m for H1.

z3=(63)2=1.5m

Find the location (z4) of active earth pressure per unit length at component 4 from bottom of the wall using the equation:

z4=(HH1)3

Substitute 6m for H and 3 m for H1.

z4=(63)3=1m

Find the location (z5) of active earth pressure per unit length at component 5 from bottom of the wall using the equation:

z5=(HH1)3

Substitute 6m for H and 3 m for H1.

z5=(63)3=1m

Find the location of resultant force (z¯).

Take moment about the bottom of the wall.

Pa×z¯=(Pa1×H1)+(Pa2×H2)+(Pa3×H3)+(Pa4×H4)+(Pa5×H5)

Substitute 141.13kN/m for Pa, 30kN/m for Pa1, 3m for z1, 23.25kN/m for Pa2, 4m for z2, 32.97kN/m for Pa3, 1.5m for z3, 10.76kN/m for Pa4, 1m for z4, 44.15kN/m for Pa5, and 1m for z5.

141.13×z¯=[(30)×3+(23.25)×4+(32.97)×(1.5)+(10.76)×(1)+(44.15)(1)]z¯=90+93+49.46+10.76+44.15141.13=2.04m

Therefore, the location of the resultant is 2.04m from bottom_.

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