# Home Stretch Tips Extra Innings Hydrostatic Pressure By Tamer

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Hydrostatic Pressure
This is
the one
a ls
Fin

By:
Tamer E. El-Diraby
Assoc. Prof. & Director, i2c
Water                                           Water
d1                                                                     P=ρd1g                                d1                    P=ρd1g

P=ρd1g                                                                                                                                air
P=ρd2g                                       air
d2                                                                                                                                      b
d2-d1
P=ρd2g                                                                                     d2
a                            P=ρd2g
air                                                              P=ρd2g                                       ground
ground                                                ground

If a vertical structural                    If an inclined structural
Water pressure at any point   If a horizontal structural         member is holding water,                    member is holding water,
is constant and is equal to    member is holding water,         then water pressure                         then water pressure
P=ρd1g [N/m2]                 then water pressure               is linearly variable and is                 is linearly variable and is
is constant and is equal to       P=ρd1g [N/m2]                               P=ρd1g [N/m2]
P=ρdg [N/m2]                      at the top                                  at the top
and P=ρd2g [N/m2]                           and P=ρd2g [N/m2]
at the bottom                               at the bottom
Depth of water is what matters
The Force for each
running meter of the                               In this case, we use b instead of (d2-d1).
No area, No Force          The Force for each            structure is the area                               The first, is the area of the rectangle
running meter of the         under the pressure block.                           whose depth is P=ρd1g and height is (b).
structure is the area         In this case the area of the trapezoid.             This is equivalent to
under the pressure block.     We normally divide that into two forces.             P=ρd1g x (b) [n/m] acting at mid height.
In this case, F=Pxa [N/m]     The first, is the area of the rectangle whose       The second is the area of the remaining triangle,
depth is P=ρd1g and height is (d2-d1).              whose base is P=ρd1g – ρd2g the height of this
This is equivalent to                               triangle is again (b). This is equivalent to
P=ρd1g x (d2-d1) [n/m] acting at mid height.       F= (ρd1g – ρd2g) x (b)/2 [N/m]
The second is the area of the remaining triangle,
whose base is P=ρd1g – ρd2g the height of this      acting at one third [b/3] from the bottom.
triangle is again (d2-d1). This is equivalent to
F= (ρd1g – ρd2g) x (d2-d1)/2 [N/m]
acting at one third [(d2-d1)/3] from the bottom.
The wall that separates mud (density=1760 kg/m3)
and fresh water is supported by a continuous hinge
at the base A and anchor bolts BC which are
embedded in the bedrock at C. The attachments of

2.25
the anchor bolts at the wall and the bedrock are                                  Mud
equivalent to pin connections. What is the

3.5
maximum safe uniform spacing of anchor bolts if
the failure load for each bolt is 120KN and the load                                             Water
factor is 3?                                                       C
2m

1.25
A

Step-1: Calculate pressure values

P1 = (1760)(9.81)(3.5) = 60429 N/m2
= 60.429 KN/m2
P2 = (1000)(9.81)(2) = 19620 N/m2
= 19.62 KN/m2
Note:The width of the water or the mud behind the
wall does not affect the value of the pressure at the
bottom.                                                 T

Step-2: Calculate forces                                                F1
1.167

In this case we will calculate the forces on a length                                                    F2
of the b meters of the wall where b is the spacing
P1                  Ax       P2

0.667
between anchors
F1 = (60.429)(3.5)(1/2)b = 105.75 KN/m
F2 = (19.62)(2) (1/2)b = 19.62 KN/m

Step-3: Stability of the wall
ΣMA=0                                                                         b
T x 1.25 + 19.62 x 0.667 – 105.75 x 1.167=0
T=88.26 KN/m                                                       b
Spacing= [120/3]/88.25 = 0.453 m
in other words, if each anchor can (safely) carry 40
kN, and the structural system requires 88.25 kN for
each meter, how many anchors do we need?                Anchors
Alternative solution
Wall                      b
Instead of finding the force for each meter, find the
force for each spacing. Let us assume that the
spacing is b.
F1= 105.75 x b ; F2= 19.62 x b ; T = 120/3 = 40
KN
ΣMA=0= -105.75 x b(1.167) + 19.62 x b(0.667) + 40(1.25) = 0
b = 0.453m
The 10m long walkway is constructed on the
edge of a fresh water reservoir. Determine the                        walkway
reaction components at the fixed support A due
to the fluid pressure on the walkway.
4
3

1
A
Step-1: Calculate pressure values                                           B

water

1.5
P1 = (1000)(9.81)(1) = 9810 N/m2
= 9.81 KN/m2

2
Step-2: Calculate forces
In this case we will calculate the forces on a length of the 10 meters of the walkway
F1 = (9.81)(2)(10) = 196.2 KN

F2 = (1/2)(9.81)(1.667)(10) = 81.75 KN

67
1 .6
Ay                                     0.6F2
B
Ax
MA
P1              F2
F1
0.8F2

Step-3: Stability of the wall
ΣMA=0
MA – 196.2(1) – 0.8(81.75)(2.444) – 0.6(81.75)(0.333) = 0

MA = 372.41 KN.m

ΣFx=0
AX = 0.6F2 = 49.05 kN

ΣFy=0
Ay + 196.2 + 0.8(81.75) = 0
AY = - 261.6 kN
The gate AB is 3 m wide and is hinged        Water level
at B and separates water as shown.
Find the weight of the gate such that it
does not rotate.
6m
Step 1: Find the Pressure
a. Pressure at B due to the 2m water

Note, that it is the depth of the water at                                                   Water level
point B which counts (not the length of
EB which is submerged in water)                                                                                         2m
42°

The Pressure P1 at B is:                            Water level
P1=1000x2x9.81=19.62 kN/m2

b. Pressure at A                                     4.66m
The pressure at A is caused by the                                                                                      6m
water head to the left. The depth of that
A                                  2 .9
water at A is (8- 5 sin 42 =4.666).                                                                      88
m
P2=1000x4.666x9.81= 45.71 kN/m2                     45.71kN/m2
c. Pressure at B due to the 8 meter
E
P3=1000x8x9.81=78.48 kN/m2                                                                     D
2m
Note: to simplify the force                                                                 42°                  19.62kN/m2
calculations, we will split the                                                                    C

trapezoidal water pressure on AB into

/m 2 /m 2
a rectangle (depth=45.71 kN/m2 ) and a

kN
.71
triangle (depth=78.48-45.71=32.77
45
kN
kN/m2)
.48
/m 2

78
kN
.73
32
Step 2: Find the Forces                    Water level

The resultant of the pressure from the
2m water head is calculated as:
F1=19.62x[2.988/2]x3=87.92 kN.
This force is acting at the centroid of                                                                6m
the triangle (2.988/3=.9963m from B).                    A
1.857

Note: the pressure at B is a function of
the water depth (in this case 2m). This
pressure, however, is acting on the                                 E                         F1
D                2m
inclined surface (EB), which has a                                         W
F2                                    B
length of 2.988 (2/sin 42). As such, the                                                  C
F3
force is calculated using the 2.988m.

6
99
0.
Note also that the width of the gate is                      2. 5
1.
66
3m, as such F1 represents the total                                         7

force acting on the gate (not the
force/m)

F2=45.71x5x3=685.65 kN acting at
2.5m from B (on the incline).
F3=32.77x[5/2]x3=245.65 kN acting at
5/3m (=1.666m) from B (on the
incline).
Finally, the weight is acting at 2.5m
from B on the incline which is
equivalent to a moment arm of 2.5
cos42=1.857m (the perpendicular
distance between W and B).

Note, that for all other forces, the
distances calculated are perpendicular
to the force as such they represent the
moment arm for these forces too.

Step 3: equilibrium
ΣMB=0=Wx1.857+87.92x.996-
685.65x2.5-245.65x1.667=0

W=1096.28 kN

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