Tractive force (boundary shear) in a flowing sewer: If γw is the unit weight of water, r is the hydraulic mean depth (hydraulic radius), and S is the bed slope, what is the shear (tractive) force per unit area at the invert?

Introduction / Context:
Boundary shear stress (tractive force) governs the ability of flow to move sediments in open channels and partially full pipes. In uniform steady flow, it depends on the weight component of water acting downslope and is a key parameter for self-cleansing sewer design.

Given Data / Assumptions:

• Uniform steady flow in a channel/pipe flowing as an open channel.
• γw = unit weight of water.
• r = hydraulic mean depth (area/wetted perimeter).
• S = slope of energy grade line ≈ bed slope for uniform flow.

Concept / Approach:

The standard relationship for boundary shear stress is τ₀ = γw * r * S. This comes from balancing the downslope component of the water weight with the resisting shear along the wetted perimeter in a control volume under uniform conditions.

Step-by-Step Solution:

Verification / Alternative check:

Derive via force balance on a reach of length L: downslope force = γw * A * L * S; resisting force = τ₀ * P * L; equate and simplify to τ₀ = γw * (A/P) * S = γw * r * S.

Why Other Options Are Wrong:

Exponents 1/2 or 2/3 do not arise in τ₀; S/r is dimensionally inconsistent for stress.

Common Pitfalls:

Confusing τ₀ with Manning/Chézy velocity formulas; using γ (specific weight) units inconsistently; mixing hydraulic radius with diameter.

Final Answer:

γw * r * S