Principles of Geotechnical Engineering (MindTap Course List)
Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Chapter 8, Problem 8.5P
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Find the hydraulic uplift force at the base of the hydraulic structure per meter length.

Expert Solution & Answer
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Answer to Problem 8.5P

The hydraulic uplift force at the base of the hydraulic structure per meter length is 1,717.5kN/m_.

Explanation of Solution

Given information:

The hydraulic conductivity of the permeable soil layer k is 0.002cm/sec.

The head difference between the upstream and downstream H is 10 m.

The height of the water level H2 is 3.34 m.

The depth of permeable layer up to the tip of the hydraulic structure D is 1.67 m.

The depth of permeable layer D1 is 20 m.

Calculation:

Draw the free body diagram of the flow net for the given values as in Figure 1.

Principles of Geotechnical Engineering (MindTap Course List), Chapter 8, Problem 8.5P , additional homework tip  1

Determine the head loss for each drop using the relation.

ΔH=HNd

Here, Nd is the number of potential drop.

Refer Figure 1.

The number of potential drop Nd is 12.

Substitute 10 m for H and 12 for Nd.

ΔH=1012=0.83m

Determine the pressure head at D using the relation.

(ph)D=(H+H1)flowdis×ΔH

Here, flow dis is the flow net distance.

Substitute 10 m for H, 3.34 m for H1, 2 m for flow dis, and 0.833 for ΔH.

(ph)D=(10+3.34)2×0.833=11.67m

Determine the pressure head at E using the relation.

(ph)E=(H+H1)flowdis×ΔH

Substitute 10 m for H, 3.34 m for H1, 3 m for flow dis, and 0.833 for ΔH.

(ρh)E=(10+3.34)3×0.833=10.84m

Determine the pressure head at F using the relation.

(ph)F=(H+D)flowdis×ΔH

Substitute 10 m for H, 1.67 m for D, 3.5 m for flow dis, and 0.833 for ΔH.

(ph)F=(10+1.67)3.5×0.833=8.75m

Determine the pressure head at G using the relation.

(ph)G=(H+D)flowdis×ΔH

Substitute 10 m for H, 1.67 m for D, 8.5 m for flow dis, and 0.833 for ΔH.

(ph)G=(10+1.67)8.5×0.833=4.586m

Determine the pressure head at H using the relation.

(ph)H=(H+H1)flowdis×ΔH

Substitute 10 m for H, 3.34 m for H1, 9 m for flow dis, and 0.833 for ΔH.

(ph)H=(10+3.34)9×0.833=5.84m

Determine the pressure head at I using the relation.

(ph)I=(H+H1)flowdis×ΔH

Substitute 10 m for H, 3.34 m for H1, 10 m for flow dis, and 0.833 for ΔH.

(ph)I=(10+3.34)10×0.833=5.0m

Determine the hydraulic uplift force at the base of the hydraulic structure per meter length using the relation.

Fuplift=γw[ADE+AEF+AFG+AGH+AHI]=γw{((ph)D+(ph)E2×hDE)+((ph)E+(ph)F2×hEF)+((ph)F+(ph)G2×hFG)+((ph)G+(ph)H2×hGH)+((ph)H+(ph)I2×hHI)} (1)

Here, γw is the unit weight of water, ADE is the area of the pressure head point DE, AEF is the area of the pressure head point EF, AFG is the area of the pressure head point FG, AGH is the area of the pressure head point GH, AHI is the area of the pressure head point HI, hDE is the distance of the point DE, hEF is the distance of the point EF, hFG is the distance of the point FG, hGH is the distance of the point GH, and hHI is the distance of the point HI.

Take unit weight of water γw is 9.81kN/m3.

Substitute 9.81kN/m3 for γw, 11.67 m for (ph)D, 10.84 m for (ph)E, 1.67 m for hDE, 8.75 m for (ph)F, 1.67 m for hEF, 4.586 m for (ph)G, 18.32 m for hFG, 5.84 m for (ph)H, 1.67 m for hGH, 5.0 m for (ph)I, and 1.67 m for (ph)I.

Fuplift=9.81{(11.67+10.842×1.67)+(10.84+8.752×1.67)+(8.75+4.5862×18.32)+(4.586+5.842×1.67)+(5.84+5.02×1.67)}=9.81×(18.796+16.357+122.157+8.706+9.0514)=1,717.5kN/m

Draw the pressure head diagram as in Figure 2.

Principles of Geotechnical Engineering (MindTap Course List), Chapter 8, Problem 8.5P , additional homework tip  2

Therefore, the hydraulic uplift force at the base of the hydraulic structure per meter length is 1,717.5kN/m_.

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