wri 03 forcesstaticfluids

03 Forces Due To Static Fluids

Water Resources

Dave Morgan

Fall 2007

Forces Due To Static Fluids WRI 1/48

Readings for “Forces Due To Static Fluids”

All readings for this topic are from Applied Fluid Mechanics by Mott:

Read sections:4.4 - 4.8Study Example Problems: 4.2 - 4.7


Forces on Horizontal Plane Areas

Pressure on a horizontal flat plane area due to a static fluid isuniform over the plane area since the whole plane area is at thesame depth

The force on the horizontal flat plane area is given by F = pA, wherep is the uniform pressure and A is the area of the plane area (thisfollows from p = F/A, the definition for pressure).


Forces on Horizontal Plane Areas (Example)

1.3 m

1.0 m

0.50 mOil (sg=0.93)

Water

Example:

Determine the force exerted by the oil and water upon the bottom planearea of the barrel



1.3 m

1.0 m

0.50 mOil (sg=0.93)

Water

Solution:Pressure at oil-water boundary:

PO−W = γ ·h= (0.93)(9.81 kN/m3)(0.50 m)

= 4.5617 kPa

Pressure at bottom of the barrel:

PBB = PO−W +(9.81 kN/m3)(1.0 m)

= 4.5617 kPa+9.81 kPa= 14.372 kPa



1.3 m

1.0 m

0.50 mOil (sg=0.93)

Water

Solution:Force on the bottom of the barrel:

PBB = 14.372 kPa

F = pA

= p · πd2

4

= (14.372 kN/m2) · π(1.3 m)2

4= 19.076 kN

F ≈ 19.1 kN


Forces on Horizontal Plane Areas

1.3 m

1.0 m

0.50 mOil (sg=0.93)

Water

Question:In the previous example, we took thepressure at the surface as 0, assuming thatit was gauge pressure.

But, in reality, there is an atmosphericpressure of around 101.3 kPa at the surface.

Should we add atmospheric pressure to thegauge pressure at the bottom of the barreland increase the force accordingly?

If so, what is the force on the bottom of thebarrel? If not, why not?


Forces on Horizontal Plane Areas (Exercise)

A

B

133 kPa

3.0 m

3.37 msg = 1.59

Exercise:A pressurized tank contains liquid with a specific gravity of 1.59. Theinspection hatch at A has dimensions 400 mm x 250 mm. The accesshatch at B has dimensions 500 mm x 750 mm.

Determine the force exerted by the fluid on the hatch at A and on thehatch at B



A

B

133 kPa

3.0 m

3.37 msg = 1.59

Solution:

PA = 133 kPa+ (1.59)(9.81 kN/m3)(3.0 m)

= 133 kPa+46.794 kPa= 179.79 kPa

FA = PA ·AA

= (179.79 kN/m2)(0.4 m×0.25 m)

= 17.979 kN

FA ≈ 18.0 kN



A

B

133 kPa

3.0 m

3.37 msg = 1.59

Solution:

PB = 133 kPa+ (1.59)(9.81 kN/m3)(6.37 m)

= 133 kPa+99.358 kPa= 232.36 kPa

FB = PB ·AB

= (232.36 kN/m2)(0.5 m×0.75 m)

= 87.134 kN

FB ≈ 87.1 kN


Forces on Vertical Rectangular Plane Areas

Consider a static volume of waterretained by a vertical wall or dam

Water pressure on the wall is 0 at thesurface

Water pressure is at a maximum at thebottom of the volume of water and canbe calculated from ∆p = γ ·h

Water pressure is proportional to thedepth (since ∆p = γ ·h, it follows that∆p ∝ h)



0Consider a static volume of waterretained by a vertical wall or dam






0

h

γh

Consider a static volume of waterretained by a vertical wall or dam






0

h

γh

pavg

=γh/2

The average pressure is at half-depth

Pavg = γ · h2

The magnitude of the resultant forceexerted on the wall by the water is

FR = Pavg ·A

= γ · h2·A

where A is the area of the rectangularplane area


Forces on Inclined Rectangular Plane Areas

0

h

γh

pavg

=γh/2

The same argument used for vertical rectangular plane areas applies toinclined regular plane areas and, again,

pavg = γ · h2, FR = γ · h

2·A

The direction of the pressure remains perpendicular to the plane area


Forces on Rectangular Plane Areas (Exercise)

Water

60O

5.25 m

Exercise:The wall has a rectangular plane area in contact with the water, isinclined at 60◦ to the horizontal and is 17 m long.Determine the force exerted on the dam plane area by the water.


Forces on Rectangular Plane Areas (Exercise)

Water

60O

5.25 m

Solution:The average pressure is at half-depth:

Pavg = γ · h2

= (9.81 kN/m3) · (2.625 m)

= 25.751 kN/m2

The area A of the dam wall:

A =

(5.25sin60◦

m)· (17 m)

= 103.06 m2

The resultant force on the wall is:FR = pavg ·A

= (25.751 kN/m2) · (103.06 m2)

= 2653.8 kN

FR ≈ 2650 kN


Forces on Submerged Rectangles

Find the force on a submergedrectangular gate

The pressure triangle is as before,except the pressure isn’t 0 at thetop of the rectangle so our previousreasoning no longer holds (andneither does FR = γ · h

2 ·A)

We are only interested in thepressure that the fluid exerts on thegate

The average pressure on the gate isat the gate’s mid-height, which canbe easily calculated if the gatedimensions and location are known



0



2 ·A)







2 ·A)





pavg



2 ·A)




Forces on Submerged Rectangle (Exercise)

Exercise:A vertical retaining wall supports water to a depth of 4.75 m. There aretwo rectangular hatches in the wall.

1 The top of the first hatch is at a depth of 1.25 m; the hatch is 2.25m wide × 1.5 m high. What is the magnitude of the force exertedupon the hatch by the water?

2 A second hatch has dimensions 3.75 m wide × 1.6 m high. At whatdepth can the top of this hatch be placed below the surface if themaximum allowable force for the hatch is 128 kN?



1.25 m

1.5 m

Solution (Part 1):

Exercise sketch....

The average pressure on the hatch is atmid-height of the hatch(h = 1.25 m+0.75 m = 2.0 m)

pavg = γ ·h= (9.81 kN/m3)(2.0 m)

= 19.62 kN/m2

The force on the hatch is:F = pavg ·A

= (19.62 kN/m2)(3.375 m2)

= 66.218 kN



1.25 m

1.5 m

2.0 m

pavg

Solution (Part 1):

Exercise sketch....


pavg = γ ·h= (9.81 kN/m3)(2.0 m)

= 19.62 kN/m2


= (19.62 kN/m2)(3.375 m2)

= 66.218 kN



2.0 m

pavg

Solution (Part 1):

Exercise sketch....


pavg = γ ·h= (9.81 kN/m3)(2.0 m)

= 19.62 kN/m2


= (19.62 kN/m2)(3.375 m2)

= 66.218 kN



F

Solution (Part 1):

Exercise sketch....


pavg = γ ·h= (9.81 kN/m3)(2.0 m)

= 19.62 kN/m2


= (19.62 kN/m2)(3.375 m2)

= 66.218 kN



d

1.6 m

Solution (Part 2):

Let d be the depth of the top of thehatch

Average pressure on the hatch is ath = (d+0.8 m)


128 = (9.81)(d+0.8)(1.6×3.75)

d+0.8 =128

(9.81)(1.6×3.75)

= 2.1747d = 1.3747

d ≈ 1.37 m



d

1.6 m

d+0.8 m

pavg

Solution (Part 2):




128 = (9.81)(d+0.8)(1.6×3.75)

d+0.8 =128

(9.81)(1.6×3.75)

= 2.1747d = 1.3747

d ≈ 1.37 m



d

Solution (Part 2):




128 = (9.81)(d+0.8)(1.6×3.75)

d+0.8 =128

(9.81)(1.6×3.75)

= 2.1747d = 1.3747

d ≈ 1.37 m


Centre of Pressure

h

The liquid exerts a force all overthe plane area; the force increaseswith depth.

The resultant of the force, FR , actsthrough the centroid of the pressuretriangle at a depth of 2h/3.(Like the resultant of uniformlyvarying loads in statics....)

For most problems we will consider,the total force can be thought of asacting at this point, called thecentre of pressure

Use pavg to find the averagepressure and magnitude of theresultant force, FR = pavgA, andthe centre of pressure to find theposition and direction of FR


Centre of Pressure

2h/3

h/3

FR






Centre of Pressure

2h/3

h/3

FR

pavg

=γh/2






Centre of Pressure

pavg

=γh/2F

R2h/3

h/3

The result is identical for an inclined rectangular plane area


Flumes

A flume is a man-madechannel used to transportliquid, often in an elevatedwooden box-like structure.This one was built in 1909near the Sandy River,Portland, Oregon

(http://www.portlandgeneral.com/community_and_env/hydropower_and_fish/phototour/sandy/little_sandy_flume.asp,accessed 11th September, 2007)


Flumes

This flume is 35 miles long.

It was completed in 1888 to supply water to San Diego.(http://www.sandiegohistory.org/timeline/timeline2.htm,

accessed 11th September, 2007)


Flumes

(http://www.co.yamhill.or.us/pics/Sheridan/, accessed 11th September, 2007)


Flumes

Hanging Flume, Montrose County, Colorado, US

This flume is on the World Monuments Foundation’s list of the 100 MostEndangered Sites 2006. Built during the Gold Rush, this 21 kilometreflume transported more than 30 million litres of water a day for use in

hydraulic gold mining.(http://wmf.org/resources/sitepages/united_states_hanging_flume.html, accessed 11th September, 2007)


Centre of Pressure (Example)

1.6 m

0.35 m

2.05 mPosts, spaced 1.75 m on centre

Tierods

Example:

Water flows slowly through a flume, assumed to have a pinnedconnection at the bottom of the sidewalls and held in place against waterpressure by tie-rods through posts every 1.75 m.Determine the tension in the tie rods.



1.6 m

0.35 m

2.05 m

Solution:

pavg = γ · h2

= (9.81 kN/m3)(1.025 m)

= 10.055 kPa

FR = pavg ·A= (10.055 kN/m2)

×(2.05 m×1.75 m)

= 36.072 kN

36.072 kN is the force exerted onthe sidewall for each 1.75 m span ofthe flume; this is the force restrainedby the tie-rods.



0.35 m

2.05 m

T

FR

A

36.072 kN

T

0.68333 m

Solution (cont’d):

Draw an FBD, showing the forces acting on one side of the flume.Take moments about A



0.35 m

2.05 m

A

36.072 kN

T

0.68333 m

Solution (cont’d):

ΣMA = T · (2.05 m+0.35 m)

−(36.072 kN×0.68333 m)

= 0

T =(36.072 kN×0.68333 m)

(2.05 m+0.35 m)

= 10.270 kN

The tension in the tie-rods is 10.3 kN.


Centre of Pressure, Submerged Rectangular Plane Area

Where is the centre of pressure for asubmerged rectangular plane area?

The pressure area is a trapezoid, not atriangle, so the centroid is not atone-third height

In statics, this problem was handled byseparating the trapezoid into a triangleand a rectangle

The centres of pressure for thesesimple shapes are known

The centres of pressure and themagnitudes of the forces can becombined to solve a variety of problems


Centre of Pressure, Submerged Plane Area

With the method described, theresultant force can be calculated tofind the tension in the bolts holdingthe inspection hatch onto the tank

There is a formula that locates thecentre of pressure, Lp more readily...Find Lc , the depth of the centroid ofthe plane areaFind A, the area of the surface, and IC ,the moment of inertia of the plane areaabout its horizontal centroidal axisThen,

Lp = Lc +Ic

Lc ·A



FR

Lp

With the method described, theresultant force can be calculated tofind the tension in the bolts holdingthe inspection hatch onto the tankThere is a formula that locates thecentre of pressure, Lp more readily...

Find Lc , the depth of the centroid ofthe plane areaFind A, the area of the surface, and IC ,the moment of inertia of the plane areaabout its horizontal centroidal axisThen,

Lp = Lc +Ic

Lc ·A



FR

Lp

Lc

With the method described, theresultant force can be calculated tofind the tension in the bolts holdingthe inspection hatch onto the tankThere is a formula that locates thecentre of pressure, Lp more readily...Find Lc , the depth of the centroid ofthe plane area

Find A, the area of the surface, and IC ,the moment of inertia of the plane areaabout its horizontal centroidal axisThen,

Lp = Lc +Ic

Lc ·A



FR

Lp

Lc

With the method described, theresultant force can be calculated tofind the tension in the bolts holdingthe inspection hatch onto the tankThere is a formula that locates thecentre of pressure, Lp more readily...Find Lc , the depth of the centroid ofthe plane areaFind A, the area of the surface, and IC ,the moment of inertia of the plane areaabout its horizontal centroidal axis

Then,

Lp = Lc +Ic

Lc ·A



FR

Lp

Lc

With the method described, theresultant force can be calculated tofind the tension in the bolts holdingthe inspection hatch onto the tankThere is a formula that locates thecentre of pressure, Lp more readily...Find Lc , the depth of the centroid ofthe plane areaFind A, the area of the surface, and IC ,the moment of inertia of the plane areaabout its horizontal centroidal axisThen,

Lp = Lc +Ic

Lc ·A


Submerged Plane Area (Example)

100 mm 600 mm

Castor Oilsg = 0.96

2.5 m

Example:

A tank containing castor oil has a1.0 m wide × 600 mm high inspectionhatch.

The top of the hatch is at 2.5 m below thesurface of the castor oil. The hatch cover isattached to the tank by eight bolts, four atthe top of the hatch and four at the bottomof the hatch.

The bolts are offset from the hatch openingby 100 mm, as shown.

Calculate the tension in each of the top andin each of the bottom bolts.

(Assume that all of the top bolts have thesame tension and that all of the bottombolts have the same tension.)



100 mm 600 mm

Castor Oilsg = 0.96

2.5 m

Solution:Calculate the area of the hatch:

A = 1.0 m×0.6 m = 0.6 m2

Calculate the moment of inertia of thehatch:

Ic =bh3

12=

(1.0 m)(0.6 m)3

12= 0.018 m4

Find Lc , the location (depth) of thecentroid of the hatch area:

Lc = 2.5 m+0.6 m2

= 2.8 m



100 mm

Castor Oilsg = 0.96

2.8 m

Lc

Solution:Calculate the area of the hatch:

A = 1.0 m×0.6 m = 0.6 m2

Calculate the moment of inertia of thehatch:

Ic =bh3

12=

(1.0 m)(0.6 m)3

12= 0.018 m4

Find Lc , the location (depth) of thecentroid of the hatch area:

Lc = 2.5 m+0.6 m2

= 2.8 m



100 mm

Castor Oilsg = 0.96

2.8 m

Lc

Solution:

A = 0.6 m2, Ic= 0.018 m4, Lc = 2.8 m

Lp = Lc +Ic

Lc ·A

= 2.8 m+0.018 m4

2.8 m×0.6 m2

= 2.8107 m



100 mm

Castor Oilsg = 0.96

2.8 m

LcL

p

2.8107 m

Solution:

A = 0.6 m2, Ic= 0.018 m4, Lc = 2.8 m

Lp = Lc +Ic

Lc ·A

= 2.8 m+0.018 m4

2.8 m×0.6 m2

= 2.8107 m



Castor Oilsg = 0.96

2.8 m

pavg

Solution:Find the average pressure on the hatch:

pavg = γ ·Lc

= (0.96)(9.81 kN/m3)(2.8 m)

= 26.369 kN/m2

Find the magnitude of the resultant force:

FR = pavg ·A= (26.369 kN/m2)(1.0 m×0.6 m)

= 15.822 kN



Castor Oilsg = 0.96

pavg

FR

2.8107 m

2.8 m

Solution:Find the average pressure on the hatch:

pavg = γ ·Lc

= (0.96)(9.81 kN/m3)(2.8 m)

= 26.369 kN/m2

Find the magnitude of the resultant force:

FR = pavg ·A= (26.369 kN/m2)(1.0 m×0.6 m)

= 15.822 kN



Castor Oilsg = 0.96

pavg

FR

0.4 m

0.4 m

2.4 m

Solution:The upper row of bolts is at a depth of2.4 m and the lower row is at 3.2 m

The upper row of bolts is 0.4107 m abovethe centre of pressure

The lower row of bolts is 0.3893 m belowthe centre of pressure

We now have the values needed to calculatethe tensions in the bolts



Castor Oilsg = 0.96

pavg

FR

0.4 m

0.4 m

2.4 m

0.4107 m







Castor Oilsg = 0.96

pavg

FR

0.4 m

0.4 m

2.4 m

0.4107 m

0.3893 m







Castor Oilsg = 0.96

FR

0.4107 m

0.3893 m







FR

S

T

15.822 kN

B

A

0.4107 m

0.3893 m

Solution:Let S be the sum of the tensions inthe upper bolts and T be the sum ofthe tensions in the lower bolts.

ΣMB = S× (0.8 m)

−(0.3892 m)× (15.822 kN)

= 0S = 7.6974 kN

ΣMA = −T × (0.8 m)

+(0.4107 m)× (15.822 kN)

= 0T = 8.1226 kN

The tension in each upper bolt is7.6974/4≈ 1.92 kN

The tension in each lower bolt is8.1226/4≈ 2.03 kN


Submerged Plane Area (Exercise)

10 kN

0.75 m

1.0 m

0.9 m

dThis is an example of a“self-levelling” gate. It is hingedalong its top edge.

When the water exceeds a certainspecified height, the hydrostaticforce on the gate is just sufficient toopen the gate. Water drains until itis again at the specified height andthe gate closes.

Find the value d for which the gateopens.



10 kN

0.75 m

1.0 m

0.9 m

d

A

B



10 kN

0.75 m

1.0 m

0.9 m

d

A

B

Solution:

A = 0.9 m×1.0 m = 0.9 m2

Ic = (0.9 m)(1.0 m)3

12 = 0.075 m4

Lp−Lc = ICLc ·A = 0.075 m4

Lc ·(0.9 m2)

= 0.083333Lc

Cont’d

pavg = γ ·h = 9.81 ·Lc(kN/m2)

FR = (9.81Lc)(0.9) = 8.829Lc(kN)

Lp−Lc = ICLc ·A = 0.075

Lc ·(0.9)

= 0.083333Lc

(m)



10 kN

0.75 m

1.0 m

0.9 m

d

A

B

Cont’d (2)

When the gate opens, the reactionat B is 0 so the only two momentsto consider are the moment due toFR and the moment due to the10 kN weight

Cont’d (3)

ΣMA = FR(0.5+ (Lp−Lc))−10(0.75)

= 8.829Lc(0.5+0.08333/LC )−7.5

= 4.4145Lc +0.73572−7.5

= 0

Lc = 1.5322 m

The centroid of the gate is 1.5322 m fromthe surface when the gate opens.Therefore

d = 1.5322 m−0.5 m

d ≈ 1.03 m


Inclined vs. Vertical Plane Areas

FR

Lp

Lc

With a vertical submerged planearea, we have

Lp = Lc +Ic

Lc ·A

What is the situation when theplane area under investigation is notvertical?


Inclined vs. Vertical Plane Areas

FR

Lp

Lc

Lp = Lc +Ic

Lc ·A

The formula still applies if Lc and Lp aretaken parallel to the plane area underinvestigation. (A vertical plane area is justa special case of the general formula.)

Lc is the distance along the slope of the plane area from thecentroid of the plane area to the surface of the liquid

Lp is the distance along the slope of the plane area from thecentre of pressure to the surface of the liquid

(The formula doesn’t hold for horizontal plane areas. Why not?)


Submerged Inclined Plane Area (Example)

1.0 m

1.0 m

1.0 m

50O

sg = 0.823

2.05 m

Example:

Find the magnitude, direction and centre of pressure (point ofapplication) of the resultant force on the triangular hatch in theV-shaped vessel illustrated.



1.0 m

1.0 m

1.0 m

50O

sg = 0.823

2.05 m

B

A

Solution:

The length AB is the “height” of the hatch:

AB1.0

= sin60◦

AB = 0.86603 m

The area of the hatch:

A =bh2

=(1.0 m)(0.86603 m)

2= 0.43302 m2

The moment of inertia of the hatch:

Ic =bh3

36=

(1.0 m)(0.86603 m)3

36= 0.18042 m4

The centroidal axis of the triangle is at one-thirdheight from the base (two-thirds down from thetop):

BC =23×AB= 0.57735 m



1.0 m

1.0 m

1.0 m

50O

sg = 0.823

2.05 m

B

AC

Solution:

The length AB is the “height” of the hatch:

AB1.0

= sin60◦

AB = 0.86603 m

The area of the hatch:

A =bh2

=(1.0 m)(0.86603 m)

2= 0.43302 m2

The moment of inertia of the hatch:

Ic =bh3

36=

(1.0 m)(0.86603 m)3

36= 0.18042 m4

The centroidal axis of the triangle is at one-thirdheight from the base (two-thirds down from thetop):

BC =23×AB= 0.57735 m



1.0 m

1.0 m

1.0 m

50O

B

AC

D

50O

Lc

Lc

Lc

Lc

Lc

Lc

Lc

Lc2.05 m

Cont’d:

Lc , the distance from the surface to thecentroidal axis of the hatch along the slope ofthe hatch:

Lc = DC

= DB +BC

=2.05sin50◦

m+0.57735 m

= 3.2534 m

The depth at the centroid of the triangularhatch:

hc = Lc · cos40◦

= 2.4922 m



1.0 m

1.0 m

1.0 m

50O

B

AC

D

50O

Lc

Lc

Lc

Lc

Lc

Lc

Lc

Lc

Frame Submerged Inclined Surface (Example)

L_{c}, the distance from the surface to the centroidal axis of the hatch along the slope of the hatch is:L_{c}

_ _ _ _ _ _ _ _ _ _ _ _ _ _

The depth of L_{c} has the average pressure for the hatch, h_{C}

_ _ _ _ _ _ _ _ _ _ _ _ _ _

p_{avg}

2.4922 m

Cont’d:

Lc , the distance from the surface to thecentroidal axis of the hatch along the slope ofthe hatch:

Lc = DC

= DB +BC

=2.05sin50◦

m+0.57735 m

= 3.2534 m

The depth at the centroid of the triangularhatch:

hc = Lc · cos40◦

= 2.4922 m



50O

C

D

50O

Frame Submerged Inclined Surface (Example)

L_{c}, the distance from the surface to the centroidal axis of the hatch along the slope of the hatch is:L_{c}

_ _ _ _ _ _ _ _ _ _ _ _ _ _

The depth of L_{c} has the average pressure for the hatch, h_{C}

_ _ _ _ _ _ _ _ _ _ _ _ _ _

p_{avg}

2.4922 m

Cont’d:

The average pressure on the hatch is thepressure at the centroid of the hatch:

pavg = γ ·hc

= (0.823)(9.81 kN/m3)(2.4922 m)

= 20.121 kPa

The resultant force on the hatch is:

FR = pavg ·A= (20.121 kN/m2)(0.43302 m2)

= 8.7128 kN



50O

sg = 0.823

2.05 m

C

D

50O

FR

Cont’d:

The average pressure on the hatch is thepressure at the centroid of the hatch:

pavg = γ ·hc

= (0.823)(9.81 kN/m3)(2.4922 m)

= 20.121 kPa

The resultant force on the hatch is:

FR = pavg ·A= (20.121 kN/m2)(0.43302 m2)

= 8.7128 kN



50O

sg = 0.823

C

D

50O

Lp

FR

Cont’d:

The resultant force, FR , acts at the centre ofpressure which is at a distance Lp , along theslope of the hatch, from the surface where

Lp = Lc +IC

Lc ·A

= 3.2534 m+0.18042 m4

(3.2534 m)(0.43301 m2)

= 3.3815 m

The depth of the centre of pressure is

Lph = 3.3815cos40◦ = 2.5904 m

FR has direction 320◦ (go figure... ;-)



50O

sg = 0.823

50O

Lp

2.5904 m

FR

Cont’d:

The resultant force, FR , acts at the centre ofpressure which is at a distance Lp , along theslope of the hatch, from the surface where

Lp = Lc +IC

Lc ·A

= 3.2534 m+0.18042 m4

(3.2534 m)(0.43301 m2)

= 3.3815 m

The depth of the centre of pressure is

Lph = 3.3815cos40◦ = 2.5904 m

FR has direction 320◦ (go figure... ;-)



2.59 m

FR

Solution:

FR is 8.712 kN at 320◦, acting on the hatch at a depth of 2.59 m


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