Brad Peterson, P.E.Brad Peterson, P.E. - Weebly · Microsoft PowerPoint - Lesson 03 Fluid Mechanics...

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: bradpeterson@engineer.com

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