Internal Erosion Suite

Appendix A - Derivation of Hydraulic Shear Stress Equations

Each of the following four derivations of hydraulic shear stress on the surface of a continuous flaw (crack, gap, or pipe) from flow of water in the flaw is based on these simplifying assumptions:

• Cross-section of the flaw is uniform from upstream to downstream (waterside to landside).

• Steady uniform flow occurs through the flaw.

• Head loss is linear from upstream to downstream (waterside to landside).

• Frictional resistance is uniform along the surface of the flaw.

• Frictional resistance is equal to the driving force.

Hydraulic Shear Stress on the Surface of a Cylindrical Pipe

Figure: Flow through a cylindrical pipe.

Equation approximates the average hydraulic shear stress acting on the pipe by assuming static equilibrium.

$$\sum F = \gamma_{W}H_{1}\frac{\pi D^2}{4}-\tau(\pi D)L-\gamma_{W}H_{2}\frac{\pi D^2}{4} = 0$$

where:

F = force
D = diameter of the pipe
L = length of the pipe
H₁ = hydraulic head at upstream end of pipe (headwater elevation minus base of pipe elevation)
H₂ = hydraulic head at downstream end of pipe (tailwater elevation minus base of pipe elevation)
γ_w = unit weight of water
τ = average hydraulic shear stress

Figure: Free-body diagram for flow through a cylindrical pipe.

Solving for the average hydraulic shear stress results in Equation and Equation.

$$\tau(\pi D)L = \gamma_{W}(H_{1}-H_{2})\frac{\pi D^2}{4}$$

$$\tau = \frac{\gamma_{W}(H_{1}-H_{2})D}{4L}$$

Equation calculates the average hydraulic gradient across the flow path.

$$i=\frac{H_{1}-H_2}{L}$$

Therefore, Equation can be rewritten as Equation.

$$\tau = \frac{\gamma_{W}iD}{4}$$

Alternatively, the average hydraulic shear stress acting on the walls of the pipe can be approximated by a more general formula for the hydraulic shear stress along a pipe as shown in Equation.

$$\tau = \gamma_{W}i\frac{A}{P_w}$$

where:

A = average cross-sectional area of the flow

$$A=\frac{\pi D^2}{4}$$

where:

P_w = average wetted perimeter

$$P_{w} = \pi D$$

Substituting Equation, Equation, and Equation into Equation results in Equation, Equation, and Equation.

$$\tau = \gamma_{W}\left(\frac{H_{1}-H_{2}}{L}\right)\left(\frac{\left(\frac{\pi D^2}{4}\right)}{\pi D}\right)$$

$$\tau = \frac{\gamma_{W}(H_{1}-H_{2})D}{4L}$$

$$\tau = \frac{\gamma_{W}iD}{4}$$

Solving Equation for the pipe diameter and setting τ = τ_c (critical shear stress) results in the critical pipe diameter for initiation of concentrated leak erosion (D_cr) as shown in Equation, Equation, and Equation.

$$\gamma_{W}(H_{1}-H_{2})D = 4L\tau_{c}$$

$$D_{cr} = \frac{4L\tau_c}{\gamma_{W}(H_{1}-H_2)}$$

$$D_{cr} = \frac{4L\tau_c}{\gamma_{W}(H_{1}-H_2)}$$

Hydraulic Shear Stress on the Surface of a Horizontal Crack

Figure: Flow through a horizontal crack.

Equation approximates the average hydraulic shear stress acting on the crack by assuming static equilibrium.

$$\sum F = \gamma_{W}H_{1}WX-\tau(2WL+2XL)-\gamma_{W}H_{2}WX = 0$$

where:

F = force
W = vertical crack dimension (crack width)
X = horizontal crack dimension (crack width parallel to the embankment centerline)
L = length of the crack
H₁ = hydraulic head at upstream end of crack (headwater elevation minus base of crack elevation)
H₂ = hydraulic head at downstream end of crack (tailwater elevation minus base of crack elevation)
γ_w = unit weight of water
τ = average hydraulic shear stress

Figure: Free-body diagram for flow through a horizontal crack.

Solving for the average hydraulic shear stress results in Equation and Equation.

$$\tau(2WL+2XL) = \gamma_{W}(H_{1}-H_{2})WX$$

$$\tau = \frac{\gamma_{W}(H_{1}-H_2)WX}{L(2W+2X)}$$

Equation calculates the average hydraulic gradient across the flow path.

$$i=\frac{H_{1}-H_2}{L}$$

Therefore, Equation can be rewritten as Equation.

$$\tau=\frac{\gamma_{W}iWX}{2W+2X}$$

Alternatively, the average hydraulic shear stress acting on the walls of the crack can be approximated by a more general formula for the hydraulic shear stress along a crack as shown in Equation.

$$\tau =\gamma_{W}i\frac{A}{P_w}$$

where:

A = average cross-sectional area of the flow

$$A=WX$$

P_w = average wetted perimeter

$$P_{w} = 2W+2X$$

Substituting Equation, Equation, and Equation into Equation results in Equation, Equation, and Equation.

$$\tau = \gamma_{W}\left(\frac{H_{1}-H_2}{L}\right)\left(\frac{WX}{2W+2X}\right)$$

$$\tau = \frac{\gamma_{W}(H_{1}-H_{2})WX}{L(2W+2X)}$$

$$\tau = \frac{\gamma_{W}iWX}{2W+2X}$$

Solving Equation for the crack width (vertical crack dimension) and setting τ = τ_c (critical shear stress) results in the critical crack width for initiation of concentrated leak erosion (W_cr) as shown in Equation to Equation.

$$L\tau_c(2W+2X) = \gamma_{W}(H_{1}-H_{2})WX$$

$$L\tau_c(2W+2X) - \gamma_{W}(H_{1}-H_{2})WX = 0$$

$$2L\tau_{c}W+2L\tau_{c}X- \gamma_{W}(H_{1}-H_{2})WX = 0$$

$$W[2L\tau_{c}-\gamma_{W}(H_{1}-H_2)X] = -2L\tau_{c}X$$

$$W_{cr} = \frac{-2L\tau_{c}X}{2L\tau_{c}-\gamma_{W}(H_{1}-H_2)X}$$

$$W_{cr} = \frac{-2\tau_{c}X}{2L\tau_{c}-\gamma_{W}iX}$$

Because W << X, the average hydraulic shear stress in Equation can be approximated as shown in Equation and Equation.

$$\tau = \frac{\gamma_{W}iWX}{2(0)+2X}$$

$$\tau \approx \frac{\gamma_{W}iW}{2}$$

Solving Equation for the crack width and setting τ = τ_c (critical shear stress) results in the critical crack width for initiation of concentrated leak erosion as shown in Equation.

$$W_{cr} \approx \frac{2\tau_c}{\gamma_{W}i}$$

Hydraulic Shear Stress on the Surface of a Vertical Rectangular Crack

Figure: Flow through a vertical rectangular crack.

Equation approximates the average hydraulic shear stress acting on the crack by assuming static equilibrium.

$$\sum F = \frac{1}{2}\gamma_{W}H_{1}^{2}W-\tau(2L\left(\frac{H_{1}+H_2}{2}\right)+WL)-\frac{1}{2}\gamma_{W}H_{2}^{2}W=0$$

where:

F = force
D = depth of crack from top of core
H₁ = hydraulic head at upstream end of crack (headwater elevation minus base of crack elevation)
H₂ = hydraulic head at downstream end of crack (tailwater elevation minus base of crack elevation)
L = equivalent length of crack for net hydraulic head
W = width of crack at top of core
γ_w = unit weight of water
τ = average hydraulic shear stress

Figure: Free-body diagram for flow through a vertical rectangular crack.

Solving for the average hydraulic shear stress results in Equation and Equation.

$$\tau L(H_{1}+H_{2}+W) = \frac{1}{2}\gamma_{W}W(H_{1}^{2}-H_{2}^2)$$

$$\tau = \frac{\gamma_{W}W(H_{1}^{2}-H_{2}^2)}{2L(H_{1}+H_{2}+W)}$$

Alternatively, the average hydraulic shear stress acting on the walls of the crack can be approximated by a more general formula for the hydraulic shear stress along a crack as shown in Equation.

$$\tau=\gamma_{W}i\frac{A}{P_w}$$

where:

i = average hydraulic gradient across the flow path

$$i = \frac{H_{1}-H_{2}}{L}$$

A = average cross-sectional area of the flow

$$A=\frac{1}{2}(WH_{1}+WH_{2})$$

$$A=\frac{W}{2}(H_{1}+H_2)$$

P_w = average wetted perimeter

$$P_{w} = \frac{1}{2}((2H_{1}+W)+(2H_{2}+W))$$

$$P_{w}=H_{1}+H_{2}+W$$

Substituting Equation, Equation, and Equation into Equation results in Equation and Equation.

$$\tau = \gamma_{W}\left(\frac{H_{1}-H_2}{L}\right)\frac{\frac{W}{2}(H_{1}+H_2)}{H_{1}+H_{2}+W}$$

$$\tau = \frac{\gamma_{W}W(H_{1}^{2}-H_{2}^2)}{2L(H_{1}+H_{2}+W)}$$

Solving Equation for the crack width at the top of the core and setting τ = τ_c (critical shear stress) results in the critical crack width for initiation of concentrated leak erosion (W_cr) as shown in Equation to Equation.

$$2L\tau_{c}(H_{1}+H_{2}+W)=\gamma_{W}W(H_{1}^{2}-H_{2}^2)$$

$$2L\tau_{c}(H_{1}+H_{2})+2L\tau_{c}W = \gamma_{W}W(H_{1}^{2}-H_{2}^2)$$

$$\gamma_{W}W(H_{1}^{2}-H_{2}^2)-2L\tau_{c}W=2L\tau_{c}(H_{1}+H_2)$$

$$W(\gamma_{W}(H_{1}^{2}-H_{2}^2)-2L\tau_c) = 2L\tau_{c}(H_{1}+H_2)$$

$$W_{cr} = \frac{2L\tau_{c}(H_{1}+H_2)}{\gamma_{W}(H_{1}^{2}-H_{2}^2)-2L\tau_c}$$

Because H₂ is often zero (i.e., no tailwater) and W << H₁, the average hydraulic shear stress in Equation can be approximated as shown in Equation, Equation, and Equation.

$$\tau \approx \frac{\gamma_{W}W(H_{1}^{2}-0)}{2L(H_{1}+0+0)}$$

$$\tau \approx \frac{\gamma_{W}WH_1}{2L}$$

$$\tau \approx \frac{\gamma_{W}iW}{2}$$

Solving Equation for the crack width and setting τ = τ_c (critical shear stress) results in the critical crack width for initiation of concentrated leak erosion as shown in Equation.

$$W_{cr} \approx \frac{2\tau_{c}}{\gamma_{W}i}$$

Hydraulic Shear Stress on the Surface of a Vertical Triangular Crack

Figure: Flow through a vertical triangular crack.

Equation approximates the average hydraulic shear stress acting on the crack by assuming static equilibrium.

$$\sum F = \frac{1}{6}\gamma_{W}H_{1}^{2}W_{1}-\tau L \left(\sqrt{H_{1}^{2}+\left(\frac{W_1}{2}\right)^2} + \sqrt{H_{2}^{2}+\left(\frac{W_2}{2}\right)^2}\right) - \frac{1}{6}\gamma_{W}H_{2}^{2}W_{2} = 0$$

where:

F = force
D = depth of crack from top of core
H₁ = hydraulic head at upstream end of crack (headwater elevation minus base of crack elevation)
H₂ = hydraulic head at downstream end of crack (tailwater elevation minus base of crack elevation)
H_z = height of water in crack at a distance z from upstream end of crack
L = equivalent length of crack for net hydraulic head
W = width of crack at top of core
W₁ = width of crack at upstream end of crack
W₂ = width of crack at downstream end of crack
W_z = width of crack at a distance z from upstream end of crack
Z = distance measured along base of crack from upstream end of crack
γ_W = unit weight of water
τ = average hydraulic shear stress

Figure: Free-body diagram for flow through a vertical triangular crack.

Solving for the average hydraulic shear stress results in Equation and Equation.

$$\tau L \left(\sqrt{H_{1}^{2}+\left(\frac{W_1}{2}\right)^2} + \sqrt{H_{2}^{2}+\left(\frac{W_2}{2}\right)^2}\right) = \frac{1}{6}\gamma_{W}(H_{1}^{2}W_{1}-H_{2}^{2}W_2)$$

$$\tau = \frac{\gamma_{W}}{6L}\frac{(H_{1}^{2}W_{1}-H_{2}^{2}W_2)}{\sqrt{H_{1}^{2} + \left(\frac{W_1}{2}\right)^{2}} + \sqrt{H_{2}^{2} + \left(\frac{W_2}{2}\right)^{2}}}$$

Solving for crack width at H₁ and H₂ using similar triangles results in Equation, Equation, and Equation.

$$\frac{W}{D} = \frac{W_1}{H_1} = \frac{W_2}{H_2}$$

$$W_{1} = \frac{WH_1}{D}$$

$$W_{2} = \frac{WH_2}{D}$$

Substituting these crack widths into Equation results in Equation, Equation, and Equation.

$$\tau = \frac{\gamma_{W}}{6L}\frac{\left(H_{1}^{2}\left(\frac{WH_{1}}{D}\right) - H_{2}^{2}\left(\frac{WH_{2}}{D}\right)\right)}{\sqrt{H_{1}^{2}+\left(\frac{WH_1}{2D}\right)^{2}}+ \sqrt{H_{2}^{2}+\left(\frac{WH_2}{2D}\right)^{2}} }$$

$$\tau = \frac{\gamma_{W}W}{6DL}\frac{(H_{1}^{3}-H_{2}^3)}{H_{1}\sqrt{1+\frac{W^2}{4D^2}} + H_{2}\sqrt{1+\frac{W^2}{4D^2}}}$$

$$\tau = \frac{\gamma_{W}W}{6DL}\frac{(H_{1}^{3}-H_{2}^3)}{(H_{1}+H_{2})\sqrt{1+\frac{W^2}{4D^2}}}$$

The derivation of the equation used to calculate the hydrostatic forces is shown in Figure A-9.

Figure: Hydrostatic force on a triangular face.

Alternatively, the average hydraulic shear stress acting on the walls of the crack can be approximated by a more general formula for the hydraulic shear stress along a crack as shown in Equation.

$$\tau = \gamma_{W}i\frac{A}{P_w}$$

where:

i = average hydraulic gradient across the flow path

$$i = \frac{H_{1}-H_{2}}{L}$$

A = average cross-sectional area of the flow

$$A = \frac{1}{L}\int_0^L\frac{1}{2}H_{z}W_zdz$$

where:

H_z = height of water in crack at a distance z from upstream end of crack
W_z = width of crack at a distance z from upstream end of crack

$$H_{z} = H_{1}- \frac{(H_{1}-H_2)z}{L}$$

$$W_{z} = \frac{WH_{z}}{D} = \left[H_{1} - \frac{(H_{1}-H_{2})z}{L}\right]\left(\frac{W}{D}\right)$$

$$A = \int_0^L\frac{1}{2}\left[H_{1}-\frac{(H_{1}-H_2)z}{L}\right]\left[H_{1}-\frac{(H_{1}-H_2)z}{L}\right]\left(\frac{W}{D}\right)dz$$

$$A=\frac{W}{2DL}\int_0^L\left(H_{1}-\frac{H_{1}z}{L}+\frac{H_{2}z}{L}\right)\left(H_{1}-\frac{H_{1}z}{L}+\frac{H_{2}z}{L}\right)dz$$

$$A = \frac{W}{2DL}\int_0^L\left(H_{1}^{2}-\frac{H_{1}^{2}z}{L}+\frac{H_{1}H_{2}z}{L}-\frac{H_{1}^{2}z}{L}+\frac{H_{1}^{2}z^{2}}{L^2}-\frac{H_{1}H_{2}z^2}{L^2}+\frac{H_{1}H_{2}z}{L}-\frac{H_{1}H_{2}z^2}{L^2}+\frac{H_{2}^{2}z^2}{L^2}\right)dz$$

$$A = \frac{W}{2DL}\left(H_{1}^{2}z - \frac{H_{1}^{2}z^2}{2L}+\frac{H_{1}H_{2}z^2}{2L}-\frac{H_{1}^{2}z^2}{2L}+\frac{H_{1}^{2}z^3}{3L^2}-\frac{H_{1}H_{2}z^3}{3L^2}+\frac{H_{1}H_{2}z^2}{2L}-\frac{H_{1}H_{2}z^3}{3L^2}+\frac{H_{2}^{2}z^3}{3L^2}\right)\bigg|_0^L$$

$$A = \frac{W}{2DL}\left(H_{1}^{2}L-\frac{H_{1}^{2}L^2}{2L}+\frac{H_{1}H_{2}L^2}{2L}-\frac{H_{1}^{2}L^{2}}{2L}+\frac{H_{1}^{2}L^{3}}{3L^2}-\frac{H_{1}H_{2}L^3}{3L^2}+\frac{H_{1}H_{2}L^2}{2L}-\frac{H_{1}H_{2}L^3}{3L^2}+\frac{H_{2}^{2}L^3}{3L^2}\right)$$

$$A=\frac{W}{12DL}\left(6H_{1}^{2}L-3H_{1}^{2}L + 3H_{1}H_{2}L-3H_{1}^{2}L + 2H_{1}^{2}L - 2H_{1}H_{2}L + 3H_{1}H_{2}L - 2H_{1}H_{2}L + 2H_{2}^{2}L\right)$$

$$A = \frac{W}{12D}(2H_{1}^2+2H_{1}H_{2}+2H_{2}^2)$$

$$A = \frac{W}{6D}(H_{1}^{2}+H_{1}H_{2}+H_{2}^2)$$

where:

P_w = average wetted perimeter

$$P_{w} = \frac{1}{2}\left(2\sqrt{H_{1}^{2}+\left(\frac{W_1}{2}\right)^2} + 2\sqrt{H_{2}^{2}+\left(\frac{W_2}{2}\right)^2}\right)$$

$$P_{w} = \frac{1}{2}\left(2\sqrt{H_{1}^{2}+\left(\frac{WH_{1}}{2D}\right)^2} + 2\sqrt{H_{2}^{2}+\left(\frac{WH_2}{2D}\right)^2}\right)$$

$$P_{w} = \sqrt{H_{1}^{2}+\left(\frac{WH_{1}}{2D}\right)^2} + \sqrt{H_{2}^{2}+\left(\frac{WH_2}{2D}\right)^2}$$

$$P_{w} = H_{1}\sqrt{1+\frac{W^2}{4D^2}} + H_{2}\sqrt{1+\frac{W^2}{4D^2}}$$

$$P_{w}= (H_{1}+H_{2})\sqrt{1+\frac{W^2}{4D^2}}$$

Substituting Equation, Equation, and Equation into Equation results in Equation, Equation, and Equation.

$$\tau = \gamma_{w}\left(\frac{H_{1}-H_2}{L}\right)\frac{\frac{W}{6D}(H_{1}^{2}+H_{1}H_{2}+H_{2}^2)}{(H_{1}+H_2)\sqrt{1+\frac{W^2}{4D^2}}}$$

$$\tau = \frac{\gamma_{w}W}{6DL}\frac{(H_{1}^{3}+H_{1}^{2}H_{2}+H_{1}H_{2}^{2}-H_{1}^{2}H_{2}-H_{1}H_{2}^{2}-H_{2}^{3})}{(H_{1}+H_{2})\sqrt{1+\frac{W^2}{4D^2}}}$$

$$\tau = \frac{\gamma_{w}W}{6DL}\frac{(H_{1}^{3}-H_{2}^3)}{(H_{1}+H_{2})\sqrt{1+\frac{W^2}{4D^2}}}$$

Solving Equation for the crack width and setting τ = τ_c (critical shear stress) results in the critical crack width for initiation of concentrated leak erosion (W_cr) as shown in Equation to Equation.

$$\tau_{c}(H_{1}+H_{2})\sqrt{1+\frac{W^2}{4D^2}}=\frac{\gamma_{w}W(H_{1}^{3}-H_{2}^{3})}{6DL}$$

$$\tau_{c}^{2}(H_{1}+H_{2})^{2}\left(1+\frac{W^2}{4D^2}\right) = \frac{\gamma_{w}^{2}W^{2}(H_{1}^{3}-H_{2}^{3})^2}{36D^{2}L^2}$$

$$\tau_{c}^{2}(H_{1}+H_{2})^{2}+\frac{\tau_{c}^{2}(H_{1}+H_2)^{2}W^2}{4D^2} = \frac{\gamma_{w}^{2}W^{2}(H_{1}^{3}-H_{2}^{3})^2}{36D^{2}L^2}$$

$$\frac{\gamma_{w}^{2}W^{2}(H_{1}^{3}-H_{2}^3)^2}{36D^{2}L^2} - \frac{\tau_{c}^{2}(H_{1}+H_2)^{2}W^2}{4D^2} = \tau_{c}^{2}(H_{1}+H_{2})^{2}$$

$$W^{2}\left(\frac{\gamma_{w}^{2}(H_{1}^{3}-H_{2}^3)^2}{36D^{2}L^2} - \frac{\tau_{c}^{2}(H_{1}+H_2)^{2}}{4D^2}\right) = \tau_{c}^{2}(H_{1}+H_{2})^{2}$$

$$W^{2} = \frac{\tau_{c}^{2}(H_{1}+H_{2})^{2}}{\left(\frac{\gamma_{w}^{2}(H_{1}^{3}-H_{2}^3)^2}{36D^{2}L^2} - \frac{\tau_{c}^{2}(H_{1}+H_2)^{2}}{4D^2}\right)}$$

$$W_{cr} = \frac{\tau_{c}(H_{1}+H_{2})}{\sqrt{\frac{\gamma_{w}^{2}(H_{1}^{3}-H_{2}^3)^2}{36D^{2}L^2} - \frac{\tau_{c}^{2}(H_{1}+H_2)^{2}}{4D^2}}}$$

Because H₂ is often zero (i.e., no tailwater) and W << D, the average hydraulic shear stress in Equation can be approximated as shown in Equation, Equation, and Equation.

$$\tau \approx \frac{\gamma_{w}W}{6DL}\frac{(H_{1}^{3}-0)}{(H_{1}+0)\sqrt{1+0}}$$

$$\tau \approx \frac{\gamma_{w}WH_{1}^2}{6DL}$$

$$\tau \approx \frac{\gamma_{w}iWH_1}{6D}$$

Solving Equation for the crack width and setting τ = τ_c (critical shear stress) results in the critical crack width for initiation of concentrated leak erosion as shown in Equation.

$$W_{cr} \approx \frac{6D\tau_{c}}{\gamma_{w}iH_{1}}$$