 The permissible tractive forces mentioned above refer to straight channels.
 For sinuous channels, the values of permissible tractive force should be lowered to reduce scour.
 A 10% reduction is recommended for slightly sinuous channels,
25% for moderately sinuous, and 40% for very sinuous.
STEPS IN PERMISSIBLE TRACTIVE FORCE METHOD.
 Assume b/y = 6 and z, with Q, S, and n known.
 Assume tractive force on sides is critical (as opposed to tractive force on level ground).
 With b/y and z, enter Fig. 77 left to determine C_{s}
in the acting unit tractive force on the sides T_{s}:
T_{s} = C_{s}γyS
 With d_{25} and grain shape, find angle of repose θ from Fig. 79.
 Calculate φ from:
tanφ = 1/z
 Calculate K from:
K = [1  (sin^{2}φ/sin^{2}θ)]^{1/2}
 Determine permissible unit tractive force on level ground τ_{L} from Fig. 710.
(If the material differs from sides and bottom,
this could be either τ_{Lb} based on the material on the bottom,
or τ_{Ls} based on the material on the sides).
 Calculate the permissible unit tractive force on the sides:
τ_{s} = K τ_{L}
[or: τ_{s} = K τ_{Ls}]
 Set permissible and acting unit tractive forces equal: τ_{s} = T_{s}
τ_{s} = C_{s}γyS
 Solve for flow depth y:
y = τ_{s}/(C_{s}γS)
 With y and b/y, calculate b = y (b/y)
 With Q, b, z, S, and n known, design channel to find y_{n}.
 Test to confirm that: y_{n} ≤ y
If not satisfied, assumed b/y is too small. Assume a greater value and return to step 11.
If satisfied: y = y_{n}
 Once b/y is determined by trial and error, enter Fig. 77 to
determine C_{b}
in the acting unit tractive force on the
bottom T_{L}:
T_{L} = C_{b}γyS
 Calculate T_{L} = C_{b}γy_{n}S
 Compare acting unit tractive force T_{L} with permissible unit tractive force on level ground τ_{L} calculated in step 7. [OR: with τ_{Lb} if different materials]
If T_{L} ≤ τ_{L(b)}, the sides control the design.
The design is OK.
If T_{L} > τ_{L(b)}, the bottom controls the design.
In this case, make T_{L} = τ_{L(b)}.
Then recalculate y_{n} = τ_{L(b)} / (C_{b}γS)
 With new y_{n} and b/y, recalculate b and confirm y_{n} with
ONLINECHANNEL01.
Otherwise, assume new b/y until y_{n} calculated with ONLINECHANNEL01 agrees with y_{n} in previous step.
PERMISSIBLE TRACTIVE FORCE METHOD: EXAMPLE A (sides and bottom are the same)
Given: Q = 600 cfs; b = ?; z = 2; S= 0.001; n = 0.022; sides and bottom: noncohesive material, slightly angular,
d_{25} = 0.7 in.
 Assume b/y = 6.
 Assume tractive force on sides is critical (as opposed to tractive force on level ground).
 With b/y and z, enter Fig. 77 left to determine C_{s} = 0.78.
 With d_{25} and grain shape, find angle of repose θ from Fig. 79:
θ = 34^{o}.
 Calculate φ from:
tanφ = 1/z φ = tan^{1} (1/z) = 26.656^{o}.
 Calculate K from:
K = [1  (sin^{2}φ/sin^{2}θ)]^{1/2} = 0.6
 Determine permissible unit tractive force on level ground τ_{L} from Fig. 710.
τ_{L} = 0.4 × d_{25} (in) = 0.4 × 0.7 = 0.28 psf.
 Calculate the permissible unit tractive force on the sides:
τ_{s} = K τ_{L} = 0.6 × 0.28 = 0.168 psf.
 Set permissible and acting unit tractive forces equal: τ_{s} = T_{s}
τ_{s} = 0.168 = C_{s}γyS = 0.78 × 62.4 × y × 0.001
 Solve for flow depth y:
y = τ_{s}/(C_{s}γS) = 0.168 / (0.78 × 62.4 × 0.001) = 3.45 ft.
 With y and b/y, calculate b = y (b/y) = 3.45 × 6 = 20.7 ft. Assume b = 21 ft.
 With Q = 600, b = 21, z = 2, S = 0.001, and n = 0.022 known, use
ONLINECHANNEL01 to find y_{n} = 4.356 ft.
 Test to confirm that: y_{n} = 4.356 > y = 3.45. Normal depth too high!
If not satisfied, assumed b/y is too small. Assume a greater value and return to step 11.

Assume b/y = 8. Then b = 27.6 ≅ 28 ft.
With Q = 600, b = 28, z = 2, S = 0.001, and n = 0.022 known, use
ONLINECHANNEL01 to find y_{n} = 3.793 > y = 3.45. Normal depth still too high!

Assume b/y = 10. Then b = 34.5 ≅ 35 ft.
With Q = 600, b = 35, z = 2, S = 0.001, and n = 0.022 known, use
ONLINECHANNEL01 to find y_{n} = 3.376 ft.
Test to confirm that: y_{n} = 3.376 < y = 3.45. Normal depth now OK!
 With b/y = 10, enter Fig. 77 to
determine C_{b} = 1.0
 Calculate T_{L} = C_{b}γy_{n}S = 1.0 × 62.4 × 3.376 × 0.001 = 0.211 psf.
 Compare acting unit tractive force T_{L} with permissible unit tractive force on level ground τ_{L} calculated in step 7.
T_{L} = 0.211 < τ_{L} = 0.28. Therefore, the sides control the design.
The design is OK.
PERMISSIBLE TRACTIVE FORCE METHOD: EXAMPLE B (sides and bottom are different)
Given: Q = 600 cfs; b = ?; z = 2; S= 0.001; n = 0.022;
sides: noncohesive material, slightly angular, d_{25} = 0.7 in;
bottom: noncohesive material, with d_{50} = 0.8 mm, with high content of fine sediment in the water.
 Assume b/y = 6.
 Assume tractive force on sides is critical (as opposed to tractive force on level ground).
 With b/y and z, enter Fig. 77 left to determine C_{s} = 0.78.
 With d_{25} and grain shape, find angle of repose θ from Fig. 79:
θ = 34^{o}.
 Calculate φ from:
tanφ = 1/z φ = tan^{1} (1/z) = 26.656^{o}.
 Calculate K from:
K = [1  (sin^{2}φ/sin^{2}θ)]^{1/2} = 0.6
 Determine permissible unit tractive force on level ground τ_{L} from Fig. 710.
τ_{Ls} = 0.4 × d_{25} (in) = 0.4 × 0.7 = 0.28 psf.
τ_{Lb} = 0.09 psf.
 Calculate the permissible unit tractive force on the sides:
τ_{s} = K τ_{Ls} = 0.6 × 0.28 = 0.168 psf.
 Set permissible and acting unit tractive forces equal: τ_{s} = T_{s}
τ_{s} = 0.168 = C_{s}γyS = 0.78 × 62.4 × y × 0.001
 Solve for flow depth y:
y = τ_{s}/(C_{s}γS) = 0.168 / (0.78 × 62.4 × 0.001) = 3.45 ft.
 With y and b/y, calculate b = y (b/y) = 3.45 × 6 = 20.7 ft. Assume b = 21 ft.
 With Q = 600, b = 21, z = 2, S = 0.001, and n = 0.022 known, use
ONLINECHANNEL01 to find y_{n} = 4.356 ft.
 Test to confirm that: y_{n} = 4.356 > y = 3.45. Normal depth too high!
If not satisfied, assumed b/y is too small. Assume a greater value and return to step 11.
 With b/y = 10, enter Fig. 77 to
determine C_{b} = 1.0
 Calculate T_{L} = C_{b}γy_{n}S = 1.0 × 62.4 × 3.376 × 0.001 = 0.211 psf.
 Compare acting unit tractive force T_{L} with permissible unit tractive force on level ground τ_{L} calculated in step 7.
T_{L} = 0.211 > τ_{Lb} = 0.09. Therefore, the bottom controls the design.
The design is not OK.
 Force T_{L} = 0.09. Then: 0.09 = C_{b}γy_{n}S = 1.0 × 62.4 × y_{n} × 0.001
Solve for new y_{n}: y_{n} = 0.09/(1.0 × 62.4 × 0.001) = 1.44 ft.
 Solve for new b by trial and error:
 Assume b/y = 60; b = 86.4; say 87 ft.
With Q = 600, b = 87, z = 2, S = 0.001, and n = 0.022 known, use
ONLINECHANNEL01 to find y_{n} = 2.01 ft. Too high.
 Assume b/y = 100; b = 144 ft.
With Q = 600, b = 144, z = 2, S = 0.001, and n = 0.022 known, use
ONLINECHANNEL01 to find y_{n} = 1.49 ft. Still too high.
 Assume b/y = 106; b = 152 ft.
With Q = 600, b = 152, z = 2, S = 0.001, and n = 0.022 known, use
ONLINECHANNEL01 to find y_{n} = 1.44 ft. OK!
