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(CIVL151)soln_151_mid03.pdf
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Solution to CIVL 151 midterm exam (26/3/03)
Problem 1 (25%)
a) To find the shaded area, it is first to calculate
Let A be half of the shaded area
then, let the area that is submerged in the water be A
The vertical component of the hydrostatic force is equal to the buoyant force, that is,
Let the line of action of the vertical component of the hydrostatic force acting on the drum gate be Xpv counting from the left of the centerline of the circle.
therefore, the vertical component of the hydrostatic force is 495.37 kN ( ) and the line of action is located 0.89m from the left of the centerline of the circle.
b) The horizontal component of the hydrostatic pressure force acting on the drum gate is distributed as a triangular pressure force diagram.
The horizontal component is calculated as:
the line of action of the horizontal component of the hydrostatic force is acting 2m below from the water surface or 1m above from the base of the drum gate which can calculate through integration as follows,
therefore, the horizontal component of the hydrostatic force is 441kN ( ) and the line of action is located 2m below the water surface or 1m above the bottom of the drum gate.
c) The resultant force F =
= ()
= 663.25 kN
The direction of the resultant force F:
therefore, the resultant force F is 663.25kN and the direction is 48.3 or 0.483 rad.
Problem 3 (10%)
Solution:
The gravity of the rocks and barge: W=1764*1000N
Consider the equilibrium of the floating barge:
The distance between the centers of the gravity and buoyancy (BG) is given by
The distance between the center of buoyancy and the metacentre (BM) is
(Since the second moment of area of the waterline cross
section about y-y is larger than Ixx, the barge is more stable
about the y-y axis)
Stability is defined in terms of the metacentric height (GM)
which is given by
Hence, the barge is unstable and it will tip over.
Problem 4 (15%)
;
; and
;
; and
x
Problem 5 (25%)
a)
Assumptions: Steady and incompressible flow.
Mass flux of water in:
Mass flux of water out:
So net mass flux of water out:
Let Vh be the rate of decreasing of water surface inside the tank.
x cross section area of the tank x Vh = 0.0106
Hence, the rate of water surface decreasing inside the tank is 0.03749 ms-1.
b)
Hence, the maximum flow velocity in inlet pipe 1 is 9 ms-1.