An inclined rectangular sluice gate AB 1.2 m by 5 m size shown in figure is installed to control discharge of water. The end A is hinged. Determine force normal to gate applied at B to open it. [A.M.I.E, Summer,1980]

SOLUTION:

Given,

Size of the gate = 1.2 m ** 5 m

Area of the gate, `A=1.2**5=6 m^2`

Now,

The depth of C.G of the gate from free water surface is,

`bar (h)=5-(BG*cos45)`

`=5-(AB)/2cos45`

`=5-(1.2/2*1/root ()(2))`

`bar (h)=4.576 m`

Now, Total pressure force acting on the gate is,

`P=rho*g*A*bar (h)`

`=(1000**9.81**6**4.576)/1000`

`=269.3 KN`

Let, this force acts at a depth  of `h^**`.

`h^....Show More

SOLUTION:

Given,

Size of the gate = 1.2 m ** 5 m

Area of the gate, `A=1.2**5=6 m^2`

Now,

The depth of C.G of the gate from free water surface is,

`bar (h)=5-(BG*cos45)`

`=5-(AB)/2cos45`

`=5-(1.2/2*1/root ()(2))`

`bar (h)=4.576 m`

Now, Total pressure force acting on the gate is,

`P=rho*g*A*bar (h)`

`=(1000**9.81**6**4.576)/1000`

`=269.3 KN`

Let, this force acts at a depth  of `h^**`.

`h^**=(I_G sin^2theta)/(A*bar (h)+bar (h)`

`I_G=`moment of inertia

`I_G=(b*d^3)/12=(5**1.3^3)/12=0.72 m^4`

Thus,

`h^**=(0.72sin^2theta)/(6**4.576)+4.576`

`=4.589 m`

Now, taking moments about hinge A,we get,

`P**AC-F**1.2=0`

or,`269.3**AC=F**1.2`

From figure,

`AC=AB-BC`

`=AB-(OB-OC)`

`=AB-(5/(sin45)-(h^**)/(sin45))`

`=1.2-(5/(sin45)-4.589/(sin45))`

`=0.618`

Thus,

`F=(269.3**0.618)/1.2`

`F=138.69 KN`