Question

2.123 The concrete (specific weight 150 lb/ft3 ) seawall of Fig. P2.123 has a curved surface...

2.123 The concrete (specific weight 150 lb/ft3 ) seawall of Fig. P2.123 has a curved surface and restrains seawater at a depth of 24 ft. The trace of the surface is a parabola as illustrated. Determine the moment of the fluid force (per unit length) with respect to an axis through the toe (point A).

2.123 The concrete (specific weight 150 lb/ft3 ) s

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Answer #1
Concepts and reason

When surface is submerged in a fluid, the forces developed on surface due to fluid.

Calculation of these forces is important role in design of tanks, dams and hydraulic structures.

Force must be perpendicular to the surface for static fluids since no shear stresses are involved.

Fundamentals

Write the formula for hydrostatic force.

FR= pÅ
=rhA

Here, the specific weight of fluid is , area of the dam is A and centroidal distance of dam is hc.

Write the formula for weight of the dam.

W yV

Here, the volume of the fluid is .

Write the formula for moment.

Μ-Σa

Draw the free body diagram of the system when it has turned through an angle.

D
С
XC
W
У1
А
В
15 ft
24 ft

Compute the horizontal component of fluid force acting on the wall.

Fyh,A

Here, is the weight density of the fluid, У
is the height at which the horizontal component acts and is the projected vertical area on which the force acts h the height of the sea wall and l the length of the sea wall.

Substitute 64 lb/ft
3
for, 24/2 ft
for, 24 ftx1 ft
X
for (since unit length).

24
64x-
x(24x1)
2
18432 lb

Draw the curved section of concrete wall.

dx
С
dA
х
24 y
24 ft
- y=0.2x2
В
Хо

Calculate the value of хо
using the curve equation.

y 0.2x2
…… (1)

Substitute, 24 ft for y.

24 0.2(x
xV120 ft

Determine the area of BCD.

[dA=
(24-y)dx
0
…… (2)

Substitute equation (1) in equation (2).

fan- (24-02x)d
0.2
A 24x
3

Substitute V120 ft
for хо
.

A 24(120
3
= 175.27 ft
0.2
V120

Calculate the vertical force acting on the concrete wall.

W = yxV
=rx Axl

Substitute for, 175.27 ft
2
for A, and ft
for l (since unit length).

W yx Axl
=64x175.27x1
=11217.28 lb

Calculate the centroid of the area.

Хо
х, хА-[хал
хdA
о

Here, the distance of the centroid from y-axis is х.
and the area of the section is A.

Substitute (24-y)xd
for dA
and 0.2x3
for У
.

x,x A = fx
cx(24-y)xcdx
0
-xx(24-0.2x )x dx
0
1(24
х.
A 2
0.2
2
4

Substitute for хо
and 175.27 ft
2
for A.

24
1
17527 п (2(120 п) -02(120 п))
х.
=4.11 ft

Calculate the moment of the fluid about the point A.

М 3 Fу -W(15-х.)

Here, the distance of centroid of the parabolic section from x-axis is У

Substitute 18432 lb
for , 24
3
forУ
, 11217.28 lb
for W and 4.11 ft
for х.
.

24
-11217.28 х(15-4.11)
м - 18432х.
=25299.82 lb-ft

Ans:

Therefore, the moment about the point A is 25299.82 lb-ft
.

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