Question

Determine the shear stress required to move the single grain on the side slopes, if the critical shear stress required moving the similar grain on the horizontal bed is 2.91 N/m^2. Consider the angle of the side slope with the horizontal as 30° and the angle of repose of soil as 37°.

(a) 1.61 N/m^2

(b) 2.61 N/m^2

(c) 2.91 N/m^2

(d) 1.00 N/m^2

The question was posed to me in a national level competition.

The question is from Irrigation Channels Design in division Sediment Transport and Irrigation Channels Design of Irrigation Engineering

Answer 1

Correct answer is (a) 1.61 N/m^2

The explanation: The equation required is Tc’ = (1 – Sin^2Q/Sin^2R)1/2.Tc where Q = 30°, R = 37° and Tc = 2.91 N/m^2.

Tc’ = (1 – Sin^230°/Sin^237°)1/2x 2.91 = 1.61 N/m^2.