HYDRAULICS
This course deals with the analysis and hydraulic design of by systems such as
reservoirs dams, spillways, gates, open channels, pipe networks, pumps and
turbines; sediment transport in rivers and reservoir; computer hydraulic
modeling.
HOOP TENSION IN
CIRCULAR PIPES
AND TANKS
CIRCUMFERENTIAL
(TANGENTIAL) STRESS
- It is represented by the
forces inside the cylinder
acting
towards
the
circumference perimeter to
the length of the pipe.
A tank or pipe carrying a fluid or
gas under a pressure is
subjected to tensile forces,
which
resist
bursting,
developed across longitudinal
and transverse sections
A.
B.
2 TYPES:
CIRCUMFERENTIAL
STRESS
LONGITUDINAL STRESS
LONGITUDINAL STRESS
EXAMPLES:
PROBLEM #1
A vertical cylindrical tank is 6ft in diameter and 10 ft high. It sides are held in
position by means of two hoops, one at top and one at bottom. The tank is filled
with water upto 9ft.
a.) Compute the hydrostatic force at the side of the tank
b.) Determine the tensile stress at top and bottom
SOLUTION:
𝐚. ) 𝐇𝐲𝐝𝐫𝐨𝐬𝐭𝐚𝐭𝐢𝐜 𝐅𝐨𝐫𝐜𝐞
b.) 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐬𝐭𝐫𝐞𝐬𝐬 𝐚𝐭 𝐭𝐨𝐩 𝐚𝐧𝐝 𝐛𝐨𝐭𝐭𝐨𝐦
For top:
P =𝐅 = γ𝒉𝒃𝒂𝒓 𝑨
𝑴 𝑻𝟐 = 𝟎 clockwise positive
ℎ𝑏𝑎𝑟 = 9 =4.5
2
𝟏
−𝟐𝑻𝟏 𝟏𝟎 + 𝑷( 𝟗) = 𝟎
𝟑
γ = 62.4𝑙𝑏/𝑓𝑡2
A = (9)(6)
P=𝑭 = 𝟔𝟐. 𝟒(𝟒. 𝟓)(𝟗)(𝟔)
P= 𝑭 = 𝟏𝟓𝟏𝟔𝟑. 𝟐𝒍𝒃
𝑻𝟏 = 2274.48lb
For bottom:
𝑭𝑯 = 𝟎 going right as positive
−𝟐𝑻𝟏 −𝟐𝑻𝟐 +𝑷 = 𝟎
𝑻𝟐 = 5307.12lb
PROBLEM #2
A cylindrical container 8ft high and 3ft in diameter is reinforced with two hoops
one foot at each end.
a.) Compute the hydrostatic pressure
b.) What is tension hoop at top and bottom
𝐚. ) 𝐇𝐲𝐝𝐫𝐨𝐬𝐭𝐚𝐭𝐢𝐜 𝐅𝐨𝐫𝐜𝐞
SOLUTION:
𝑭𝑯 = 𝟎 going right as positive
P =𝐅 = 𝜸𝒉𝒃𝒂𝒓 𝑨
𝒉𝒃𝒂𝒓 = 𝟖 =4
𝟏′
𝟐
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟐: −𝟐𝑻𝟏 −𝟐𝑻𝟐 +𝑷 = 𝟎
𝜸 = 𝟔𝟐. 𝟒𝒍𝒃/𝒇𝒕𝟐
𝟒. 𝟑𝟑𝟑′
𝟏′ . 𝟔𝟔𝟕
𝟏′
𝟖
= 𝟐. 𝟔𝟔𝟕’
𝟑
𝑨 = (𝟖)(𝟑)
P=𝑭 = 𝟔𝟐. 𝟒(𝟒)(𝟖)(𝟑)
P= 𝑭 = 𝟓𝟗𝟗𝟎. 𝟒𝒍𝒃
𝑬𝒒𝒖𝒂𝒕𝒆 𝒆𝒒. 𝟏 𝒂𝒏𝒅 𝒆𝒒. 𝟐:
−𝟐𝑻𝟏 𝟒. 𝟑𝟑𝟑 +𝟐𝑻𝟐 𝟏. 𝟔𝟔𝟕 = 𝟎
−𝟐𝑻𝟏 −𝟐𝑻𝟐 +𝑷 = 𝟎
b.) 𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐬𝐭𝐫𝐞𝐬𝐬 𝐚𝐭 𝐭𝐨𝐩 𝐚𝐧𝐝 𝐛𝐨𝐭𝐭𝐨𝐦
For top: 𝑴 𝑨 = 𝟎 clockwise positive
𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝟏:−𝟐𝑻𝟏 𝟒. 𝟑𝟑𝟑 +𝟐𝑻𝟐 𝟏. 𝟔𝟔𝟕 = 𝟎
𝑻𝟏 =832lb
𝑻𝟐 =2163.034lb
DAMS
Dams are built for purpose of
impounding water. It is subjected to
water pressure on the upstream
side are analysed by considering
only one meter strip of the dam.
3 COMMON FAILURES
 FAILURE DUE TO OVERTURNING
 FAILURE DUE TO SLIDING
 FAILURE DUE TO OVER-STRESSING
FBD OF A DAM
P
RM
OM
Credits to Engr. KR
DAMS
FORMULAS:
 FACTOR OF SAFETY AGAINST OVERTURNING
GENERAL NOTES
a) Width of the dam is always
1m
b) Uplift force always acts
upward
c) Uplift force is caused by
penetrating H2O at the
bottom area of the dam
d) Uplift pressure always acts on
the bottom area
e) Toe and Heel may vary. Toe is
always opposite the Head
water
f) Tail water – can be seen if
the dam is operational
g) Tail water is less than the
head water in height/depth
 FACTOR OF SAFETY AGAINST SLIDING
Rx = R.M –O.M
Rx = hydrostatic force
RESISTING MOMENT
R.M= 𝑾𝟏 𝒙𝟏 + 𝑾𝟐 𝒙𝟐
OVERTURNING MOMENT
𝒉
𝟑
O.M= 𝑷( )
DAMS
UPLIFT FORCE
LAW OF MIDDLE THIRD. For a SOUND
STRUCTURAL DESIGN, the resultant force
must cut the base width of the dam
within the middle third.
1. The uplift pressure varies linearly
from full hydrostatic pressure at the
heel to full hydrostatic pressure at the
toe.
PRESSURE DISTRIBUTION AT THE
BASE
 RESULTANT AT THE MIDDLE OF THE BASE:
Ry = P (B) (1)
𝑹𝒚
𝑩
P= ( ) (average)
 RESULTANT AT THE MIDDLE THIRDS NEARER
THE TOE:
2. The uplift pressure varies linearly
from full hydrostatic pressure at the heel
to zero at the toe.
𝑩
𝟑
x= ( )
𝑷𝑩
)
𝟐
Ry= (
DAMS
EXAMPLES:
PROBLEM #1
A concrete dam retaining water is shown in Fig. If
the specific weight of the concrete is 23.5 kN/m3,
find the factor of safety against sliding, the factor
of safety against overturning, and the maximum
and minimum pressure intensity on the base.
Assume there is no hydrostatic uplift and that the
coefficient of friction between dam and foundation
soil is 0.48.
 RESULTANT OUTSIDE MIDDLE THIRDS:
𝑷(𝟑𝒙)
)
𝟐
Ry= (
𝟐𝑹
𝟑𝒙
P= ( )
 RESULTANT WITHIN MIDDLE THIRDS:
M = Rye
e=
Pmax=
Pmin=
𝑩
𝟐
𝑹
𝑩
𝑹
𝑩
−𝒙
(𝟏 +
(𝟏 −
𝟔𝒆
𝑩
)
𝟔𝒆
𝑩
)
PROBLEM #1
A concrete dam retaining water is shown in Fig. If
the specific weight of the concrete is 23.5 kN/m3,
find the factor of safety against sliding, the factor
of safety against overturning, and the maximum
and minimum pressure intensity on the base.
Assume there is no hydrostatic uplift and that the
coefficient of friction between dam and foundation
soil is 0.48.
a. FACTOR OF SAFETY AGAINST SLIDING
(no uplift force)
GIVEN:
Coefficient of friction = 0.48
γc = 23.5kN/m3
SOLVE FOR Ry:
𝟏
𝑾𝟏 = (𝟐𝟑. 𝟓)( 𝟕 𝟐 𝟏 ) =164.5kN
𝟐
𝑾𝟐 = (𝟐𝟑. 𝟓)( 𝟕 𝟐 𝟏 ) = 329 kN
Ry = 𝑾𝟏 + 𝑾𝟐 = 164.5+329 = 493.5kN
SOLVE FOR Rx:
𝑾𝟐
𝑾𝟏
Ry = 𝑾𝟏 + 𝑾𝟐
𝑭
Total weight of the dam
W = γcV
Rx =Fh =F =hydrostatic force = γħA
γc = 23.5kN/m3
Hbar = 3m
Atotal = ( 6)(1)
Rx = γħA = (9.81)(3)(6)(1) = 176.58kN
𝑭𝑺𝒔𝒍𝒊𝒅𝒊𝒏𝒈 = (
𝟎. 𝟒𝟖(𝟒𝟗𝟑. 𝟓)
) = 1.34
𝟏𝟕𝟔. 𝟓𝟖
b.) FACTOR OF SAFETY AGAINST OVERTURNING
(no uplift force)
SOLUTION:
R.M= 𝑾𝟏 𝒙𝟏 + 𝑾𝟐 𝒙𝟐
𝒉
𝟑
O.M= 𝑹𝒙( )
𝑭
Reference: heel to toe
Reference: datum
Solve for R.M
RESISTING MOMENT
R.M= 𝑾𝟏 𝒙𝟏 + 𝑾𝟐 𝒙𝟐
R.M= 164.5 (𝟐 (𝟐)+329(𝟏 + 𝟐))
𝟑
R.M = 1,206.33kN/m
Solve for O.M
OVERTURNING MOMENT
𝒉
𝟑
O.M= 𝑷( )
Rx = 176.58kN
O.M= 𝟏𝟕𝟔. 𝟓𝟖
Toe
𝟔
𝟑
=353.16kN/m
Heel
𝟏𝟐𝟎𝟔. 𝟑𝟑
𝑭𝑺𝒐𝒗𝒆𝒓𝒕𝒖𝒓𝒏𝒊𝒏𝒈 = (
)
𝟑𝟓𝟑. 𝟏𝟔
=3.416
PROBLEM #2
A dam 4m. on top 18 m. at the bottom and 25 m. high has water 20m. deep acting on its
vertical sides. What is the stress at the heel. Wt. of concrete = 2200 kg/m3 .
SOLUTION:
Pmax=
𝑹𝒚
𝑩
(𝟏 +
𝟔𝒆
𝑩
)
SOLVE FOR R.M:
𝟒
𝟐
R.M = 220,000(𝟐+14)+385,000(𝟑 (𝟏𝟒))
SOLVE FOR Ry:
RM = 7,113,333.333kgm
𝑾𝟏 = (𝟐𝟐𝟎𝟎) 𝟒 𝟐𝟓 𝟏 )
SOLVE FOR O.M:
𝟏
𝑾𝟐 = (𝟐𝟐𝟎𝟎)( ) 𝟏𝟒 𝟐𝟓 𝟏 )
𝟐
O.M= 𝑭
𝒉
𝟑
Rx = F = (1000)(10)(20)(1) = 200,000kg
O.M= 𝟐𝟎𝟎, 𝟎𝟎𝟎
Ry = 605,000kg
𝑩
𝟐
e=
𝟑
=1,333,333.333kgm
𝟕,𝟏𝟏𝟑,𝟑𝟑𝟑.𝟑𝟑𝟑−𝟏,𝟑𝟑𝟑,𝟑𝟑𝟑.𝟑𝟑𝟑
𝟔𝟎𝟓,𝟎𝟎𝟎
x=
−𝒙
𝑹𝑴 −𝑶𝑴
𝑹𝒚
x=
Pmax=
𝟐𝟎
e=
𝟔𝟎𝟓,𝟎𝟎𝟎
𝟏𝟖
(𝟏 +
𝟏𝟖
− 𝟗. 𝟓𝟓
𝟐
𝟔(𝟎.𝟓𝟓)
𝟏𝟖
)
=0.55
=39,773.148
𝒌𝒈
𝒎
=9.55
PRACTICE SOLVING
PROBLEM #3 A rectangular dam having a width of 14.8m has a height of 23.2m. The
depth of water on one side of the dam is 20.4m. If the coefficient of friction between the
dam and the foundation is 0.85.
a.) Compute the hydrostatic force acting on the wall.
b.) Compute the factor of safety for sliding if the concrete density is 2310kg/m3.
c.) compute the factor of safety against overturning.
SOLUTION:
a.)F =Rx = γħA = (1000)(10.2)(20.4)(1) == 208080kg
14.8m
P
20.4m
Heel
23.2m
W
Toe
𝟎. 𝟖𝟓(𝟕𝟗𝟑, 𝟏𝟔𝟏. 𝟔𝒌𝒈)
𝒃. )𝑭𝑺𝒔𝒍𝒊𝒅𝒊𝒏𝒈 = (
) = 3.24
𝟐𝟎𝟖, 𝟎𝟖𝟎𝒌𝒈
𝟏𝟒. 𝟖
𝟕𝟗𝟑, 𝟏𝟔𝟏. 𝟔( 𝟐 )
𝒄. )𝑭𝑺𝒐𝒗𝒆𝒓𝒕𝒖𝒓𝒏𝒊𝒏𝒈 = (
)
𝟐𝟎. 𝟒
𝟐𝟎𝟖𝟎𝟖𝟎( 𝟑 )
= 4.148