Examine the gate arrangement shown. The gate extends from B up to C above the wa

Question

Examine the gate arrangement shown. The gate extends from B up to C above the waterline. The gate is rectangular with a width (out-of-page) of b = 10 m. There is a stop in the water at the base of the gate (B) and a pivot (A) located partway along the gate at a distance of h_AB = 2 m along the gate up from the base. The height of the fluid above the base of the gate is H = 10 m. Assume that alpha = 30degree, rho = 1000 kg/m^3, g = 9.81 m/s^2, and ignore the weight of the gate itself. For the given arrangement, calculate the force exerted by the stop and by the hinge on the gate. Clearly indicate the direction of the forces. If the water level H decreases, the force at the stop will decrease as well until it is eventually zero, at which point the gate will open. At what value of H will this occur for the above gate system?

Explanation / Answer

solution:

for part a

1)here value of H=10 m,hence inclined depth of plate of gate is d and given by

d=H/sin30

so we get that

d=20 m

here C.G. of plate is midpoint of d

d'=d/2 and d'=10 m

x'=d'/sin30

x'=5 m

here area of plate is

A=b*d=10*20=200 m2

moment of area of plate is

Ig=bd^3/12=6666.667 m4

hence force exerted at center of pressure h' is

F=density*g*A*x'

F=9810 KN

where position of C.P. is

h'=x'+Ig*sin30^2/A*x'

on putting value we get that

h'=6.666 m

dh'=13.332 mhence for hinge to be stable moment at honge A should be zero

Ma=0

F*(dha)=Fs*2

Fs=force at stopper,Hm=2 m

F=9810 KN,dha=20-13.332=6.668=6.668-2=4.668

hence we get

Fs=9810*4.668/2

fs=22896.54 KN

for part b

when force acted as pressure force is act below point A of hinge gate will open and we get that

for that condition,

for hinge pointA

dh'=d-2

h'=d-2/2

wher position by formula for C.P.

h'=x'+Ig*sin30^2/A*x'

on putting value we get that

x'=d/4

Ig=bd^3/12

so we get

h'=d/4+d/12

h'=d/3

h'=d-2/2

on equating we get

d=6 m

H=3 m,when water level is H=3 m then gate will open

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