Chapter 8, Problem 8.5P

To determine

Find the hydraulic uplift force at the base of the hydraulic structure per meter length.

Answer to Problem 8.5P

The hydraulic uplift force at the base of the hydraulic structure per meter length is $\underset{\_}{1,717.5\text{kN/m}}$.

Explanation of Solution

Given information:

The hydraulic conductivity of the permeable soil layer k is $0.002\text{cm/sec}$.

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

The height of the water level $H_2$ 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 $D_1$ is 20 m.

Calculation:

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

Determine the head loss for each drop using the relation.

$\Delta H=\frac{H}{N_d}$

Here, $N_d$ is the number of potential drop.

Refer Figure 1.

The number of potential drop $N_d$ is 12.

Substitute 10 m for H and 12 for $N_d$.

$\begin{array}{c}\Delta H=\frac{10}{12}\\ =0.83\text{m}\end{array}$

Determine the pressure head at D using the relation.

${\left(p_h\right)}_D=\left(H+H_1\right)-\text{flow}\text{dis}\times \Delta H$

Here, flow dis is the flow net distance.

Substitute 10 m for H, 3.34 m for $H_1$, 2 m for flow dis, and 0.833 for $\Delta H$.

$\begin{array}{c}{\left(p_h\right)}_D=\left(10+3.34\right)-\text{2}\times 0.833\\ =11.67\text{m}\end{array}$

Determine the pressure head at E using the relation.

${\left(p_h\right)}_E=\left(H+H_1\right)-\text{flow}\text{dis}\times \Delta H$

Substitute 10 m for H, 3.34 m for $H_1$, 3 m for flow dis, and 0.833 for $\Delta H$.

$\begin{array}{c}{\left({\rho }_h\right)}_E=\left(10+3.34\right)-\text{3}\times 0.833\\ =10.84\text{m}\end{array}$

Determine the pressure head at F using the relation.

${\left(p_h\right)}_F=\left(H+D\right)-\text{flow}\text{dis}\times \Delta H$

Substitute 10 m for H, 1.67 m for D, 3.5 m for flow dis, and 0.833 for $\Delta H$.

$\begin{array}{c}{\left(p_h\right)}_F=\left(10+1.67\right)-\text{3}\text{.5}\times 0.833\\ =8.75\text{m}\end{array}$

Determine the pressure head at G using the relation.

${\left(p_h\right)}_G=\left(H+D\right)-\text{flow}\text{dis}\times \Delta H$

Substitute 10 m for H, 1.67 m for D, 8.5 m for flow dis, and 0.833 for $\Delta H$.

$\begin{array}{c}{\left(p_h\right)}_G=\left(10+1.67\right)-\text{8}\text{.5}\times 0.833\\ =4.586\text{m}\end{array}$

Determine the pressure head at H using the relation.

${\left(p_h\right)}_H=\left(H+H_1\right)-\text{flow}\text{dis}\times \Delta H$

Substitute 10 m for H, 3.34 m for $H_1$, 9 m for flow dis, and 0.833 for $\Delta H$.

$\begin{array}{c}{\left(p_h\right)}_H=\left(10+3.34\right)-\text{9}\times 0.833\\ =5.84\text{m}\end{array}$

Determine the pressure head at I using the relation.

${\left(p_h\right)}_I=\left(H+H_1\right)-\text{flow}\text{dis}\times \Delta H$

Substitute 10 m for H, 3.34 m for $H_1$, 10 m for flow dis, and 0.833 for $\Delta H$.

$\begin{array}{c}{\left(p_h\right)}_I=\left(10+3.34\right)-\text{10}\times 0.833\\ =5.0\text{m}\end{array}$

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

$\begin{array}{c}{F}_{\text{uplift}}={\gamma }_w\left[A_{DE}+A_{EF}+A_{FG}+A_{GH}+A_{HI}\right]\\ ={\gamma }_w\left\{\begin{array}{l}\left(\frac{{\left(p_h\right)}_D+{\left(p_h\right)}_E}{2}\times h_{DE}\right)+\left(\frac{{\left(p_h\right)}_E+{\left(p_h\right)}_F}{2}\times h_{EF}\right)+\\ \left(\frac{{\left(p_h\right)}_F+{\left(p_h\right)}_G}{2}\times h_{FG}\right)+\left(\frac{{\left(p_h\right)}_G+{\left(p_h\right)}_H}{2}\times h_{GH}\right)+\\ \left(\frac{{\left(p_h\right)}_H+{\left(p_h\right)}_I}{2}\times h_{HI}\right)\end{array}\right\}\end{array}$ (1)

Here, ${\gamma }_w$ is the unit weight of water, $A_{DE}$ is the area of the pressure head point DE, $A_{EF}$ is the area of the pressure head point EF, $A_{FG}$ is the area of the pressure head point FG, $A_{GH}$ is the area of the pressure head point GH, $A_{HI}$ is the area of the pressure head point HI, $h_{DE}$ is the distance of the point DE, $h_{EF}$ is the distance of the point EF, $h_{FG}$ is the distance of the point FG, $h_{GH}$ is the distance of the point GH, and $h_{HI}$ is the distance of the point HI.

Take unit weight of water ${\gamma }_w$ is $9.81{\text{kN/m}}^{\text{3}}$.

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, 11.67 m for ${\left(p_h\right)}_D$, 10.84 m for ${\left(p_h\right)}_E$, 1.67 m for $h_{DE}$, 8.75 m for ${\left(p_h\right)}_F$, 1.67 m for $h_{EF}$, 4.586 m for ${\left(p_h\right)}_G$, 18.32 m for $h_{FG}$, 5.84 m for ${\left(p_h\right)}_H$, 1.67 m for $h_{GH}$, 5.0 m for ${\left(p_h\right)}_I$, and 1.67 m for ${\left(p_h\right)}_I$.

$\begin{array}{c}{F}_{\text{uplift}}=9.81\left\{\begin{array}{l}\left(\frac{11.67+10.84}{2}\times 1.67\right)+\left(\frac{10.84+8.75}{2}\times 1.67\right)+\\ \left(\frac{8.75+4.586}{2}\times 18.32\right)+\left(\frac{4.586+5.84}{2}\times 1.67\right)+\\ \left(\frac{5.84+5.0}{2}\times 1.67\right)\end{array}\right\}\\ =9.81\times \left(18.796+16.357+122.157+8.706+9.0514\right)\\ =1,717.5\text{kN/m}\end{array}$

Draw the pressure head diagram as in Figure 2.

Therefore, the hydraulic uplift force at the base of the hydraulic structure per meter length is $\underset{\_}{1,717.5\text{kN/m}}$.