Principles of Foundation Engineering (MindTap Course List)
Principles of Foundation Engineering (MindTap Course List)
9th Edition
ISBN: 9781337705028
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 18, Problem 18.1P

Refer to Figure 18.9. A cantilever sheet pile is driven into a granular soil where the water table is 2 m (L1) below the top of the sand. The properties of the sand are ϕ = 40 ° , γ = 17.5 kN / m 3 , and γ sat = 19 kN / m 3 . It is proposed to excavate to a depth of 6 m (L) below the ground level. Determine the actual depth to which the sheet pile must be driven (L + D), using the net lateral pressure diagram. Note:

D actual = 1.3 ( L 3 + L 4 ) theory

Expert Solution & Answer
Check Mark
To determine

Find the required actual depth of the sheet pile.

Answer to Problem 18.1P

The required actual depth of the sheet pile is 12.25m_.

Explanation of Solution

Given information:

The depth of water level below the sand (L1) is 2 m.

The angle of internal friction for sand (ϕ) is 40°.

The unit weight of the sand (γ) is 17.5kN/m3.

The saturated unit weight of the sand (γsat) is 19kN/m3.

The depth below the ground level (L) is 6 m.

Calculation:

Show the cross section of cantilever sheet pile with dimensions as in Figure (1).

Principles of Foundation Engineering (MindTap Course List), Chapter 18, Problem 18.1P

Refer Figure (1),

Find the depth of water table level (L2):

L2=LL1=62=4m

Find the rankine active pressure coefficient (Ka) using the relation:

(Ka)=tan2(45ϕ2)=tan2(4540°2)=0.217

Find the rankine passive pressure coefficient (Kp) using the relation:

Kp=tan2(45+ϕ2)=tan2(45+40°2)=4.599

Find the difference between the rankine active pressure coefficient and rankine passive pressure coefficient:

KpKa=4.5990.217=4.382

Find the effective unit weight of sand (γ):

γ=γsatγw=199.81=9.19kN/m3

At the water table level:

Find the intensity of the active pressure to the right of the pile (σ1):

(σ1)=KaL1γ=0.217×2×17.5=7.60kN/m2

At the excavation level:

Find the intensity of the active pressure to the right of the pile (σ2):

(σ2)=Ka(L1γ+L2×γ)=0.217×(2×17.5+4×9.19)=19.45kN/m2

Find the depth below the dredge line (L3):

L3=σ2(KpKa)γ=19.454.382×9.19=0.48m

Find the area of the pressure diagram (P):

P=0.5×σ1×L1+σ1×L2+0.5×(σ2σ1)×L2+0.5×σ2×L3=0.5×7.60×2+7.60×4+0.5×(19.457.60)×4+0.5×19.45×0.48=66.37kN/m

Find the area of the pressure diagram into center of pressure (Pz¯):

Pz¯={0.5×σ1×L1(13L1+L2+L3)+σ1×L2(L1+L3)+0.5×(σ2σ1)×L2(13L2+L3)+0.5×σ2×L3(23L3)}={0.5×7.60×2(13×2+4+0.48)+7.60×4(2+0.48)+0.5×(19.457.60)×4(13×4+0.48)+0.5×19.45×0.48(23×0.48)}=158.92kN-m/m

Find the center of pressure to the area z¯:

z¯=Pz¯P=158.9266.37=2.39m

Find the intensity of the passive pressure (σ5):

(σ5)=Kp(L1γ+L2×γ)+(KpKa)γ×L3=4.599×(2×17.5+4×9.19)+4.382×9.19×0.48=349.35kN/m2

Find the area (A1):

A1=σ5γ(KpKa)=349.359.19×4.382=8.68m

Find the area (A2):

A2=8Pγ(KpKa)=8×66.379.19×4.382=13.18m2

Find the area (A3):

A3=6P(2z¯γ(KpKa)+σ5)γ2(KpKa)2=6×66.37[2×2.39×9.19(4.382)+349.35]9.192×4.3822=133.05m3

Find the area (A4):

A4=P(6z¯σ5+4P)γ2(KpKa)2=66.37[6×2.39×349.35+4×66.37]9.192×4.3822=215.89m4

Find the depth below point E (L4):

(L4)4+A1(L4)3A2(L4)2A3(L4)A4=0(L4)4+8.68(L4)313.18(L4)2133.05(L4)215.89=0 (1)

Use trial and error method to calculate depth below point E (L4).

Try (L4)=4.30m:

Substitute 4.30 m for L4 in Equation (4).

(4.30)4+8.68(4.30)313.18(4.30)2133.05(4.30)215.89=0341.88+690.12243.70572.115215.89=00=0

Hence, the assumption is correct.

The depth below point E is 4.30 m.

Find the depth below the dredge line to bottom of the pile (D):

D=L4+L3=4.30+0.48=4.78m

Increase the depth below the dredge line to bottom of the pile by 30%. The depth below the dredge line to bottom of the pile (D) is 6.214 m.

Find required actual depth of the sheet pile (L+D):

(L+D)=6+6.214=12.214m12.25m

Thus, the required actual depth of the sheet pile is 12.25m_.

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