Question 3 of 13: Counterweight Mass for a Hinged Dam Gate

Question 3: Counterweight Mass for a Hinged Dam Gate (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The mass of the counterweight at which the gate is on the verge of opening.

Approach. Compute the hydrostatic resultant force and its centre of pressure on the gate, then take moments about the top hinge: the counterweight's moment (weight × 1.2 m arm) must equal the water force's moment (F × distance from hinge down to the centre of pressure) at the threshold of opening.

1. Resultant hydrostatic force on the gate. Depth to centroid, yc = 3.0 + 2.0/2 = 4.0 m; area A = 1.5(2.0) = 3.0 m²: $$ F = \rho g y_c A = (1000)(9.81)(4.0)(3.0) = 117\,720\ \text{N} $$
2. Depth to the centre of pressure. Using I_c = bh^3/12 = 1.5(2.0)^3/12 = 1.0 m⁴: $$ y_p = y_c + \frac{I_c}{y_c A} = 4.0 + \frac{1.0}{(4.0)(3.0)} = 4.0833\ \text{m} $$
3. Lever arm from the hinge (top of gate, depth 3.0 m) to the centre of pressure. $$ d = y_p - 3.0 = 1.0833\ \text{m} $$
4. Moment balance about the hinge. Setting the weight's moment equal to the water force's moment at the threshold of opening: $$ mg(1.2) = F\,d \quad\Rightarrow\quad m = \frac{Fd}{g(1.2)} = \frac{(117\,720)(1.0833)}{(9.81)(1.2)} $$ $$ \boxed{m \approx 10\,833\ \text{kg} \approx 10.8\ \text{tonnes}} $$