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ENGR5002 / Week 1
Week 1 Exercise 4Exercise sheetCurrent spec2:118 min

Week 1 Exercises, Exercise 4 (Slanted Gate)

A rectangular gate of width W=1.5W = 1.5 m (out of plane) is installed in a vertical water reservoir wall. The gate has a slanted length L=1.5L = 1.5 m and is hinged at the bottom point BB. The gate is inclined at an angle θ=30∘\theta = 30^\circ with respect to the horizontal. The total water depth is H=3.0H = 3.0 m.

Data: ρw=1000 kg/m3\rho_w = 1000\ \mathrm{kg/m^3}; g=9.81 m/s2g = 9.81\ \mathrm{m/s^2}; pa=101.3 kPap_a = 101.3\ \mathrm{kPa}.

Gate AB, 1.5 m long at 30 deg to the horizontal, hinged at B on the reservoir bed (depth H = 3.0 m) and resting at A against the bottom of the wall; water above the gate, air at p_a below it.
Gate AB, 1.5 m long at 30 deg to the horizontal, hinged at B on the reservoir bed (depth H = 3.0 m) and resting at A against the bottom of the wall; water above the gate, air at p_a below it.
Formulas you may need
  • Hydrostatic force on a plane surface: F=ρgzCGAF = \rho g z_{CG} A (on the formula sheet)
  • Centre of pressure along the plane: lCP=lCG+IGA lCGl_{CP} = l_{CG} + \dfrac{I_G}{A\, l_{CG}} (on the formula sheet)
  • Rectangle: A=bhA = bh, IG=bh312I_G = \dfrac{b h^3}{12} (on the formula sheet, shape table)
  • Components on an inclined plate: FH=Fsin⁡θF_H = F\sin\theta, FV=Fcos⁡θF_V = F\cos\theta (equivalently FH=ρgzCGAprojF_H = \rho g z_{CG} A_{proj}, FVF_V = weight of fluid above) (learn this)
  • Moment equilibrium about the hinge: ∑MB=0\sum M_B = 0 (learn this)
  1. (a)
    Calculate the resultant hydrostatic force acting on the gate (magnitude and direction).
  2. (b)
    Calculate the horizontal and vertical components of the resultant force.
  3. (c)
    Find the centre of pressure depth zCPz_{CP}.
  4. (d)
    Determine the horizontal force RR exerted by the wall at point AA.