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Brad Peterson, P.E.Brad Peterson, P.E.
FRIDAYS – 14:00 to 15:40
FRIDAYS – 16:10 to 17:50
Class Website: https://sites.google.com/site/njut2009fall/
M P ’ E il Add Mr. Peterson’s Email Address: [email protected]
Lesson 1, Properties of Fluids, 2009 Sept 04, Rev Sept 18 Lesson 2 Fluid Statics 2009 Sept 11 Rev Sept 18 Lesson 2, Fluid Statics , 2009 Sept 11, Rev Sept 18 Lesson 3, Hydrostatic Force on Surfaces, 2009 Sept 25 Lesson 4, Buoyancy and Flotation, 2009 Sept 25 ?? Lesson 5, Translation and Rotation of Liquid Massesesso 5, a s at o a d otat o o qu d asses Lesson 6, Dimensional Analysis and Hydraulic Similitude Lesson 7, Fundamentals of Fluid Flow Lesson 8, Flow in Closed Conduits Lesson 9, Complex Pipeline Systems Lesson 10, Flow in Open Channels Lesson 11, Flow of Compressible Fluids
L 12 M f Fl f Fl id Lesson 12, Measurement of Flow of Fluids Lesson 13, Forces Developed by Moving Fluids Lesson 14, Fluid Machinery
3.1. Force Exerted by a Liquid on a Plane Surface
Engineers must calculate forces exerted by fluids in order to design constraining structures satisfactorily.
A “plane surface” is a flat surface
F h A specific weight of liquid
cgF h A
depth of the center of gravity
Areacgh
A
AreaA
20 1A m4 0m 0.1A m4.0m
2.0mWater
7.0m
Tank Width 3m
Example
T t l f (F ) th b ttTotal force (FBC) on the bottom of the tank
Total weight (W) of the waterExplain the differenceExplain the difference
20 1A m4 0m 0.1A m4.0m
2.0mWater
7.0m
Tank Width 3.0m
Example
( )BCF pA h A ( )BCF pA h A
3
9.8 (6 ) (7 3 )kN m m mm
m
1235BCF kN
20 1A m4 0m 0.1A m4.0m
2.0mWater
7.0m
Tank Width 3m
Example
( )W Volume ( )W Volume
33
9.8 [(7 2 3 (4 0.1 )]kNW m m m m m 3 ( ( )m
416W kN
20 1A m4 0m 0.1A m4.0m
2.0mWater
7.0m
Tank Width 3m
Example
The Force on the bottom of the tank is:
1235 BCF kN
But the total weight of the water is only:
416W kN
What is the source of the additional force?
20 1A m4 0m 0.1A m4.0m
2.0mWater
7.0m
Tank Width 3m
Example
( )F pA h A( )ADF pA h A
23
9.8 (4 ) (7 3 0.1 )kN m m m mm
m
819ADF kN
20 1A m4 0m 0.1A m4.0m
2.0mWater
7.0m
Tank Width 3m
Example
1235 416 819kN kN Kn 1235 416 819and therefore:
kN kN Kn
BC ADF W F
The large forces that can be developed with a small amount of liquid (the liquid in the tube) acting over a much larger surface.
Hydraulic lift
20 1A m5 0m 0.1A m5.0m
1.0mWater
7.0m
Tank Width 3m
Example – What If???
T t l f (F ) th b ttTotal force (FBC) on the bottom of the tank
Total weight (W) of the waterExplain the differenceExplain the difference
20 1A m5 0m 0.1A m5.0m
1.0mWater
7.0m
Tank Width 3m
Example – What If???
( )BCF pA h A ( )BCF pA h A
3
9.8 (5 1 ) (7 3 )kN m m m mm
m
1235BCF kN
20 1A m5 0m 0.1A m5.0m
1.0mWater
7.0m
Tank Width 3m
Example – What If???
( )W Volume ( )W Volume
33
9.8 [(7 1 3 (5 0.1 )]kNW m m m m m 3 ( ( )m
211W kN
20 1A m5 0m 0.1A m5.0m
1.0mWater
7.0m
Tank Width 3m
Example – What If???
The Force on the bottom of the tank is:
1235 BCF kN
But the total weight of the water is only:
211W kN
What is the source of the additional force?
20 1A m5 0m 0.1A m5.0m
1.0mWater
7.0m
Tank Width 3m
Example – What If???
( )F pA h A( )ADF pA h A
23
9.8 (5 ) (7 3 0.1 )kN m m m mm
m
1024ADF kN
20 1A m5 0m 0.1A m5.0m
1.0mWater
7.0m
Tank Width 3m
Example – What If???
1235 211 1024kN kN Kn 1235 211 1024and therefore:
kN kN Kn
BC ADF W F
Dam is made of Concrete: 323.5 /kN m Foundation soil is impermeable Coefficient of friction between base of dam
d il i 0 48and soil is = 0.48 Find:
a) Factor of safety against slidinga) Factor of safety against slidingb) Factor of safety against overturningc) Pressure intensity on the base of the dam
Foundation◦ That upon which anything stands, and by which it is
supported; the lowest and supporting layer of a structure
Soils◦ The unconsolidated mineral or organic material on
h d f f h hthe immediate surface of the earth Impermeable◦ Not allowing passage especially of liquids;◦ Not allowing passage, especially of liquids;
waterproof
Friction◦ A force that resists the relative motion or tendency
to such motion of two bodies in contact. Coefficient of Friction Coefficient of Friction◦ The ratio of the weight of an object being moved
along a surface and the force that maintains contact b h b d h fbetween the object and the surface
Factor of Safety◦ The ratio of the maximum stress or load which
something can withstand to the stress or load which it was designed to withstand under normal goperation. For example, the common factor of safety used for elevator support cables is 11.
Sliding Sliding◦ Moving in continuous contact with a surface.
Overturning Overturning◦ To turn over, capsize or upset an object
Dam is made of Concrete: 323.5 /kN m Foundation soil is impermeable Coefficient of friction between base of dam
d il i 0 48and soil is = 0.48 Find:
a) Factor of safety against slidinga) Factor of safety against slidingb) Factor of safety against overturning
Vertical Force 0VF Vertical Force 0Remember, pressure forces only act perdicular to the surface.The face of the dam is vertical so the forces on it from the water
VF
The face of the dam is vertical so the forces on it, from the water,are horizontal only.
9.8Horizontal Force H cgkNF h A
m 3 3 (6 1 ) 176.4m m m kN
Calculate the weight of the dam:
1 2 7 23.5P t 1 164 5m m m kN kN 3Part 1 164.5
223 5
kNm
kN
3
23.5Part 2 1 2 7 329.0
T t l i ht f d 164 5 329 0 493 5
kNm m m kNm
kN
Total weight of dam 164.5 329.0 493.5kN
Calculate sliding resistance due to friction:Sliding Resistance = coefficient of friction weightSliding Resistance coefficient of friction weight 0.48 493.5 236.9
Calculate sliding factor of safety FSsliding
sliding resistance 236 9kNsliding resistance 236.9 1.34sliding force 176.4
slidingkNFSkN
Would you feel safe living - downstream of this dam?
AXIS OFAXIS OF ROTATION
Calculate the righting moment
Calculate the overturning moment
See definitions on next page
Moment◦ The turning effect of a force applied to a rotational
system at a distance from the axis of rotation. Coefficient of Friction
Righting Moment◦ The turning force that tends to keep the object
hupright or in its proper position Overturning Moment◦ The turning force that tends tip the object to◦ The turning force that tends tip the object, to
overturn the object or to move it out of its proper position
AXIS OFAXIS OF ROTATION
Calculate the righting moment:
Recall that Part 1 weights 164.5R ll h P 2 i h 329 0
kNkNRecall that Part 2 weights 329.0
Total righting moment 164.5 1.33 329.0 3.00T l i h i 1206
kNkN m kN mkN
Total righting moment 1206kN m
AXIS OF 1 6 4kAXIS OF ROTATION
176.4HF kN
Calculate the overturning moment:
Recall that horizontal force 176.4HF kNThis force is centered 2 from the bottom of the dam. Why???Total overturning moment 176.4 2 352.8
mkN m kN m g
Calculate the overturning factor of safety:
FSoverturning righting momentFSoverturning
1206overturning moment
kN m
1206FSoverturning 3.42352.8
kN mkN m
Lesson 1, Properties of Fluids, 2009 Sept 04, Rev Sept 18 Lesson 2 Fluid Statics 2009 Sept 11 Rev Sept 18 Lesson 2, Fluid Statics , 2009 Sept 11, Rev Sept 18 Lesson 3, Hydrostatic Force on Surfaces, 2009 Sept 25 Lesson 4, Buoyancy and Flotation, 2009 Sept 25 ?? Lesson 5, Translation and Rotation of Liquid Massesesso 5, a s at o a d otat o o qu d asses Lesson 6, Dimensional Analysis and Hydraulic Similitude Lesson 7, Fundamentals of Fluid Flow Lesson 8, Flow in Closed Conduits Lesson 9, Complex Pipeline Systems Lesson 10, Flow in Open Channels Lesson 11, Flow of Compressible Fluids
L 12 M f Fl f Fl id Lesson 12, Measurement of Flow of Fluids Lesson 13, Forces Developed by Moving Fluids Lesson 14, Fluid Machinery