Bundle: Principles Of Geotechnical Engineering, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
Bundle: Principles Of Geotechnical Engineering, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
9th Edition
ISBN: 9781337583817
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
Question
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Chapter 13, Problem 13.18P
To determine

Find the Rankine active force Pa per unit length of the wall and the location z¯ of the resultant force.

Expert Solution & Answer
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Answer to Problem 13.18P

The Rankine active force Pa per unit length of the wall is 287kN/m_.

The location z¯ of the resultant force is 2.49m_.

Explanation of Solution

Given information:

The height (H) of the retaining wall is 8.0 m.

The depth H1 of sand is 3.0 m.

The unit weight γ1 of the sand is 13kN/m3.

The sand friction angle ϕ1 is 30°.

The saturated sand friction angle ϕ2 is 30°.

The cohesion c1  of sand is 0.

The surcharge pressure (q) is 16kN/m2.

The depth H2 of saturated sand is 4.0 ft.

The saturated unit weight γ2 of the sand is 18.8kN/m3.

The cohesion c2 of saturated sand is 0.

Calculation:

Determine the active earth pressure coefficient Ka using the formula.

Ka=tan2(45ϕ2)

Substitute 30° for ϕ.

Ka=tan2(4530°2)=tan2(4515)=tan2(30)=0.333

Determine the total stress σo at 0 m depth using the relation.

σo=q

Substitute 16kN/m2 for q.

σo=16kN/m2

Determine the pore water pressure at 0 m depth using the relation.

u=γw×h

Here, γw is the unit weight of the water.

Take the unit weight of the water as 9.81kN/m3.

Substitute 9.81kN/m3 for γw and 0 m for h.

u=9.81×0=0

Determine the effective active earth pressure σa at 0 m depth using the relation.

σa=σoKa

Substitute 16kN/m2 for σo and 0.39 for Ka.

σa=16(0.333)=5.32kN/m2

Determine the total stress σo at 3 m depth using the relation.

σo=q+γ1×H1

Substitute 16kN/m2 for q, 13kN/m3 γ1, and 3.0 m for H1.

σo=16+13×3.0=55kN/m2

Determine the pore water pressure at 3.0 m depth using the relation.

u=γw×h

Substitute 9.81kN/m3 for γw and 0 m for h.

u=9.81×0=0

Determine the effective active earth pressure σa at 3.0 m depth using the relation.

σa=σoKa

Substitute 55kN/m2 for σo and 0.333 for Ka.

σa=55(0.333)=18.31kN/m2

Determine the total stress σo at 8 m depth using the relation.

σo=q+γ1×H1+(γ2γw)×H2

Substitute 16kN/m2 for q, 13kN/m3 γ1, 3.0 m for H1, 18.8kN/m3 γ2, 9.81kN/m3 for γw, and 5.0 m for H2.

σo=16+13×3+(18.89.81)×5=99.95kN/m2

Determine the pore water pressure at 8 m depth using the relation.

u=γw×h

Substitute 9.81kN/m3 for γw and 5.0 m for h.

u=9.81×5=49.05kN/m2

Determine the effective active earth pressure σa at 8 m depth using the relation.

σa=σoKa

Substitute 99.95kN/m2 for σo and 0.333 for Ka.

σa=99.95(0.333)=33.28kN/m2

Show the variation of effective active earth pressure and pore water pressure for the respective depth as in Figure 1.

Bundle: Principles Of Geotechnical Engineering, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card, Chapter 13, Problem 13.18P

Refer Figure 1.

Determine the active earth pressure per unit length for area 1 using the relation.

A1=bh

Here, b is the width and h is the depth.

Substitute 5.32kN/m2 for b and 8.0 m for h.

A1=(5.32×8.0)=42.56kN/m

Determine the active earth pressure per unit length for area 2 using the relation.

A2=12bh

Substitute (18.315.32)kN/m2 for b and 3.0 m for h.

A2=12×(18.315.32)×3.0=0.5×38.97=19.48kN/m

Determine the active earth pressure per unit length for area 3 using the relation.

A3=bh

Substitute 5.0 m for b and (18.315.32)kN/m2 for h.

A3=5×(18.315.32)=64.95kN/m

Determine the active earth pressure per unit length for area 4 using the relation.

A4=12bh

Substitute 5.0 m for b and (33.2818.31)kN/m2 for h.

A4=12×5.0×(33.2818.31)=0.5×74.85=37.42kN/m

Determine the active earth pressure per unit length for area 5 using the relation.

A5=12bh

Substitute 5.0 m for b and 49.05kN/m2 for h.

A4=12×5.0×49.05=122.62kN/m

Determine the Rankine active force Pa per unit length of the wall using the relation.

Pa=A1+A2+A3+A4

Substitute 42.56kN/m for A1, 19.48kN/m for A2, 64.95kN/m for A3, 37.42kN/m for A4, and 122.62kN/m for A5.

Pa=42.56+19.48+64.95+37.42+122.62=287kN/m

Thus, the Rankine active force Pa per unit length of the wall is 287kN/m_.

Determine the location z¯ of the resultant force by taking the moment about the bottom of the wall.

z¯=A1(H22)+A2(H2+H13)+A3(H22)+A4(H23)+A5(H23)Pa

Substitute 42.56kN/m for A1, 8.0 m ft for H, 5.0 m for H2, 3.0 m for H1, 19.48kN/m for A2, 64.95kN/m for A3, 37.42kN/m for A4, 122.62kN/m for A5, and 287kN/m for Pa.

z¯=42.56(82)+19.48(5+33)+64.98(52)+37.42(53)+122.62(53)287=170.24+116.88+162.45+62.37+204.37287=2.49m

Thus, the location of the resultant force is 2.49m_.

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