ch. 5: distributed forces

Ch. 5: Distributed Forces

5.0 Outline 234 Introduction 235 Center of Mass 236 Centroids (Line, Area, Volume) 241 Composite Bodies and Figures 257 Theorem of Pappus 265 Fluid Statics 272

5.0 Outline

234


5.0 Introduction

5.0 Introduction

When forces are applied over a region whosedimensions are not negligible compared with otherpertinent dimensions, we must account for the actualmanner in which the force is distributed by summing upthe effects of the distributed force over the entire region.For this purpose, we need to know the intensity of theforce at any location and we will use the integrationto find their total effects.

235


5.1 Center of Mass

5.1 Center of Mass

Location of the CG and CM are found by the principle of moment

CG: point where the resultant gravitational force W actsCM: point at which the total mass is visually concentrated

sum of the moments = moment of the sum

236


Center of Gravitiy

5.1 Center of Mass

Assumption1) uniform intensity of the earth’s force field over the body2) parallel field of force due to the gravitational attraction

the moment of the resultant gravitational force Wabout any axis = the sum of the moments aboutthe same axis of the gravitational force dW acting onall particles of the body

x y z x y z

dW

dW

=

∫∫

CG

r i + j+ k r = i + j+ k

rr =

237


Center of Mass

5.1 Center of Mass

( )

W mg & dW gdm

dm

dm

dm dV, e.g. x, y, z , dV dxdydz

dV

dV

ρ ρ ρ

ρ

ρ

= =

= = =

∫∫

∫∫

CM

CM

rr =

rr =

Assumption1) uniform intensity of the earth’s force field over the body2) parallel field of force due to the gravitational attraction

CM = CG if the gravity field is treated as uniform and parallel

CM is independent of gravitational effect and hence isa unique property of the body

238


Center of Volume

5.1 Center of Mass

Assumption1) constant intensity of the earth’s force field over the body2) parallel field of force due to the gravitational attraction

CV depends only on the shape of the body,Independent of the material

dV

dV∫∫

CV

rr =

CV = CM if the mass intensity (density) is uniformover the volume

239


5.1 Center of Mass

Look for symmetry to simplify the calculation

rectangular body rectangular coordinates

circular body polar coordinates

For a homogeneous body,CM will lie on a lineor plane of symmetry

240


5.3 Centroids (Line, Area, Volume)is purely the geometrical (shape) propertyof the body since any reference to its mass properties has disappeared.

5.3 Centroids

Line:dL

, not always on the linedL

∫∫

CC

rr =

assuming the cross sectional areais constant over the length

241

Ch. 5: Distributed ForcesdA

, not always on the surfacedA

∫∫

CC

rr =

5.3 Centroids

Area:

assuming the thicknessis constant over the entire area

dV, not always inside the volume

dV∫∫

CC

rr =

dA first moment of area=∫ Cr

Volume:

242


5.3 Centroids

Guidelines in choosing the differential element &setting up the integralsa) Order of element: prefer 1st order diff. element to

higher order, so only one integration will be required

( )( )2

A l y dy dxdy

V r y dy dxdydzπ

= =

= =

∫ ∫∫∫ ∫∫∫

243


5.3 Centroids

b) Continuity: choose an element which can beintegrated in one continuous operation to coverthe whole object

( ) ( )2 2

1

x x

1 20 x

A h x dx h x dx= −∫ ∫ ( )2y

0

A l y dy= ∫

h1(x)

h2(x)

(x2, y2)

244


dxdy∫

5.3 Centroids

c) Neglect high order terms: higher order terms maybe dropped compared with lower order terms

1 1A ydx dxdy y dy dx ydx2 2

note: dxdy dxdy

= + = + =

≠

∫ ∫ ∫ ∫

∫ ∫∫

dxdy∫∫

245


( )O Ox , y

5.3 Centroids

d) Choice of coordinates: to appropriately describethe boundaries

Oy

O O0

A x y xdy= − ∫O

2

0

1A r d2

θ

θ= ∫

246


5.3 Centroids

e) Centroidal coordinate of the element:use the coordinate of the centroid of the elementfor the moment arm in setting up the momentof the differential element

( ) ( )C Cfirst moment x dA x y l y dy= =∫ ∫( ) ( )2

C Cfirst moment x dV x z r z dzπ= =∫ ∫

247


5.3 Centroids

P. 5/1 Locate the centroid of the area of a circularsector w.r.t. its vertex

248


5.3 Centroids

P. 5/1

o

oc o o c

r ro

o o o o0 0

by symmetry, Y 0select differential element to be strip of partial circular ringwith r varying from 0 to r

r sinXA x dA dA 2 r dr , x

r sin 2rsinX 2 r dr 2 r dr , X3

ααα

α αα αα α

=

= = =

= =

∫

∫ ∫

249


5.3 Centroids

P. 5/1

( )c c

2 2

by symmetry, Y 0select differential element to be strip of sectorwith varying from to

1 2XA x dA dA r rd , x r cos2 3

1 2 1 2rsinX r d rcos r d , X2 3 2 3

α α

α α

θ α α

θ θ

αθ θ θα− −

=

−

= = =

= × =

∫

∫ ∫

250


5.3 Centroids

P. 5/2 Locate the centroid of the area shownin the figure by direct integration

251


5.3 Centroids

P. 5/2

( ) ( )

( )c

2

c1 c2

by symmetry, the centroid lies on the line y a xdifferential area horizontal strip strip of sectorwith varying from 0 to /2

1XA x dA dy ad sin , dA ady a ad2

dA a sin 1/ 2 d2x a/2, x a3

θ π

θ θ θ

θ θ

= −= −

= = = −

= −

= =

∫

( )

( )( )( )

/ 2 /2 /22 2 2

0 0 0

sin

2 1X a sin 1/ 2 d a/2 a sin d asin a d3 2

10 3 a2aX , Y a X3 4 3 4

π π π

θ

θ θ θ θ θ θ

ππ π

− = × − ×

−= = − =

− −

∫ ∫ ∫

θ

xc1

2/3asinθ

252


5.3 Centroids

P. 5/3 Determine the coordinates of the centroidof the volume obtained by revolving the shadedarea about the z-axis through 90°angle

253


5.3 Centroids

P. 5/3

( )

( )22 2 2 2

2 2

c

by symmetry, the centroid lies on the vertical plane

making 45 with the x-axis X Ydifferential volume as shown in the figurewith y varying from 0 to a

y z a a , z a a y2 ydV zdy a a y ydy

4 2

x

π π

° ∴ =

+ − = = − −

= × = − −

( ) ( )

( ) ( )

( ) ( )

2 2

c c

a a2 2 2 2

c0 0

a a 2 22 2 2 2

c0 0

ysin /4 a a ycos /4 2y/ y , z z/2

/4 2

XV x dV X a a y ydy 2y/ a a y ydy2 2

4 3X a Y4

a a yZV z dV Z a a y ydy a a y ydy

2 2 2

Z a/4

ππ π

ππ ππ

π

π π

− −= = = = =

= − − = × − −

= − =

− − = − − = × − −

=

∫ ∫ ∫

∫ ∫ ∫

4y/(π√2)z/2

ydy

254


5.3 Centroids

P. 5/4 Locate the center of mass G of the steelhalf ring.

255


5.3 Centroids

P. 5/4by symmetry, the centroid lies on the intersection lineof the zero horizontal plane and the middle vertical plane,

only r needed to be specifieddifferential volume as shown in the figurewith x varying

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

2 2 2 2 2

2 2

c

a a2 2 2 2

ca a

2 2

from a to a

x y a , y a x

dV 2 a x dx R x

R x sin / 2 2r R x/ 2

2rV r dV r 2 R x a x dx R x 2 R x a x dx

a 4Rr2 R

π

ππ π

π ππ

π

− −

−

+ = = −

= − × −

−= = −

= − − = − × − −

+=

∫ ∫ ∫

x

y

x

dx

R-x

256


5.3 Composite Bodies and Figures


Discrete version in finding the centroidCM may be easily known for subparts of the object

mx mx mzX Y Z

m m m= = =∑ ∑ ∑∑ ∑ ∑

257



Depending on the geometry and the density ofthe object, mass m may be replaced by l, A, or V.

Addition vs. Subtraction

Table consultation

Systematic tabulation of i ci i i cim , r , m , m r∑ ∑

258



P. 5/5 Determine the coordinates of the centroidof the shaded area

259



P. 5/5

i Ai xci yci Aixci Aiyci

1 0.1 0.2 0.125 0.02 0.0125

2 0.01875 0.45 0.083 8.4375e-3 1.5625e-3

3 -0.0113 0.2 0.125 -2.262e-3 -1.4137e-3

i ci

i

i ci

i

A xX 243.6 mm

A

A yY 117.7 mm

A

= =

= =

∑∑∑∑

13

2

260



P. 5/6 A cylindrical container with an extendedrectangular back and semicircular ends is allfabricated from the same sheet-metal stock.Calculate the angle α made by the backwith the vertical when the container rests inan equilibrium position on a horizontal surface.

261



P. 5/6i xi yi zi Ai Aixi Aiyi Aizi

1 200 0 100 8e4 16e6 0 8e6

2 200 -150 -300/π 6πe4 12πe6 -9πe6 -18e6

3 0 -150 -200/π 1.125πe4 0 -1.6875πe6 -2.25e6

4 400 -150 -200/π 1.125πe4 4.5πe6 -1.6875πe6 -2.25e6

262



P. 5/7

i ci

i

i ci

i

i ci

i

A x 16E6 16.5 E6X 200 mmA 8E4 8.25 E4

A y 12.375 E6Y 114.62 mmA 8E4 8.25 E4

A z 8E6 22.5E6Z 42.75 mmA 8E4 8.25 E4

ππ

ππ

π

+= = =

+

−= = = −

+

−= = = −

+

∑∑∑∑∑∑

1

2

3

4

x

y

z

263



P. 5/7

1150 Y

tan 39.6Z

α − − = = ° α

y

z

mg

YZ

N

264


5.4 Theorem of Pappusis about the surface or the volume created byrevolving the planar curve or the planar area aboutthe nonintersecting line in its plane.

Revolved Surface

area is the same as the lateral area of a cylinderof length L and radius

5.4 Pappus Theorem

A ydL yLθ θ= =∫

y

265


Revolved Volume

volume is obtained by multiplying the planar area bythe circumference of the circular path of the centroidabout the revolving axis

5.4 Pappus Theorem

V ydA yAθ θ= =∫

266


Usage1. determining the area or volume of revolution2. find the centroid of planar curve or planar area

when the corresponding area or volume are known

5.4 Pappus Theorem

267


5.4 Pappus Theorem

P. 5/8 A hand-operated control wheel made of aluminum hasthe proportions shown in the cross-sectional view.The area of the total section shown is 15,200 mm2,and the wheel has a mass of 10 kg. Calculate thedistance to the centroid of the half-section. Thealuminum has a density of 2.69 Mg/m3.

r

268


5.4 Pappus Theorem

P. 5/8

10 / 2690 2 7600 6

77.85 mm

M V E rV

r

ρ π = = = × − × =

269


5.4 Pappus Theorem

P. 5/9 Calculate the mass m of concrete required to constructthe arched dam shown. Concrete has a density of2.40 Mg/m3.

270


5.4 Pappus Theorem

P. 5/9

( ) 2

2

find the centroid w.r.t. the rotation axis first4 7080 80 40 120 70 120

4 3 80 80 70

4 175.52 mcalculate the revolved volume from Pap

i i

i

y Ay y

A

ππ

π

× × × + − × × + = = × − ×

=

∑∑

[ ]

[ ]

2

pus to find the mass

/ 3 175.52 80 80 704

2400 1.126 9 kg 1.126 6 Mg

V yA V

m V m V E E

πθ π

ρ

= = × × × − ×

= = × = =

271


5.5 Fluid Staticsconsider the equilibrium of bodies subjected toforces due to fluid pressure.

Fluid force on a surface – normal compressive force– shear force ( = 0, static)

In equilibrium, fluid pressure at any given point isthe same in any direction. The pressure is a functionof the vertical dimension.

5.5 Fluid Statics

dp gdhρ=

pressure increases with depth

272


5.5 Fluid Statics

o o

o

dp gdhWith constant, we may integrate and getp p gh absolute pressure, where p is the surface pressure at h 0.

If p atmospheric pressure and is used as the reference value,p gh gage pressure

Uni

ρρ

ρ

ρ

=

= + = =

== =

[ ] [ ] ( )3 2 2 2 2t: Pa kg/m m/s m kg-m/s / m N/m = = =

273


5.5 Fluid Statics

Resultant force due to pressure distribution on surface

1. Flat surface

274


5.5 Fluid Statics

Variation of p over the depth is linear (ρgh) sectionof the pressure distribution is trapezoidal

Pressure distribution 12345678

An element of plate area over which p acts is dA = xdy An increment of the resultant force is dR = pdA

275


5.5 Fluid Statics

( ) ( )

av

R dR pdA gh y x y dy

volume of pressure distribution

or R g hdA ghA p A

average altitude h is the depth corresponding

to the centroid O of the flat surface area, A,exposed to the pressure. At

ρ

ρ ρ

∴ = = =

=

= = =

∫ ∫ ∫

∫

av

that depth, the pressure

gh is the average pressure, p ,of the pressure distribution acting on the surface area.ρ

Magnitude

276


5.5 Fluid Statics

ypxdyRY ydR, Y

pxdy

centroidal (C) y-coord of the volume of pressure distribution through which the resultant actually passes

center of pressure, P projection of centroidal volume Calong the pre

= =

=

=

∫∫ ∫

'

'

'

ssured force direction onto the platepiercing point of the line of action R to the platecentroid of the plate area, A

ydAif x is constant, Y

dA

centroidal y-coord of the area A (not A) of the pr

=≠

=

=

∫∫

essure profile

Direction perpendicular to the flat surface

Line of action using principle of moment

277


5.5 Fluid Statics

Resultant force due to pressure distribution on surface

2. Cylindrical surface with constant width, b

dR continuously changesdirection

278


5.5 Fluid Statics

( ) ( )x yx y

R volume of pressure distribution ghA

h depth corresponding to the centroid O of the curve AB

R b pdL b pdy, R b pdL b pdx

ρ= =

=

= = = =∫ ∫ ∫ ∫

Magnitude

Direction( )1

y xtan R / Rθ −=

Line of action

y x

y x

xdR xpdx ydR ypdyX , Y

dR pdx dR pdy

at centroid of volume of pressure distribution

= = = =∫ ∫ ∫ ∫∫ ∫ ∫ ∫

279


5.5 Fluid Statics

Equilibrium of block of liquid

Px, Py = resultants of the pressuresW = weight of the liquid block passing through centroid

of area ABC (constant width b)R = reaction of the surface on the block of liquid

from equilibrium equations

280


5.5 Fluid Statics

Buoyancyreaction force the fluid exertedon the boundary of the cavity

Resultant of the pressure forces over the surface= weight of the fluidResultant must pass through the C.M. of the fluid lump

281


5.5 Fluid Statics

Buoyancy

Replace the fluid lump by an object of the same dimension,the surface forces acting on the body will be the sameas those acting on the fluid lump.

Resultant force (buoyancy) exerted on the surfaceof an object immersed in a fluid = weightof the fluid displaced

Its line of action passes through the C.M. of thedisplaced fluid.

282


5.5 Fluid Statics

Buoyancy

When ρo <ρf and fully immersed,object’s weight < buoyancy force object will riseto the equilibrium position (ρo >ρf) wherethe object’s weight = Σbuoyancy force

F Vgdensity of the displaced fluid

V volume of the displaced fluid

ρρ===

283


5.5 Fluid Statics

P. 5/10

284


5.5 Fluid Statics

P. 5/10Stress in the shell is induced by the compressive forcefrom the surrounding sea water

Write FBD of the vehicle sectioned in hemisphereand display only force in horizontal direction

F

σdA ( ) ( ) yx x

x y

2x

2 2ring

F pdA g hdA ghA

F 0 ghA dA 0

h depth of the centroid of circular area 3000 m

ghA 1030 9.81 3000 0.75 462.4 MPaA 0.75 0.725

ρ ρ

ρ σ

ρ πσπ π

= = =

= − =

= =

× × × ×= = =

× − ×

∫ ∫∑ ∫

x

285


5.5 Fluid Statics

P. 5/11 The cross section of a fresh-water tank with aslanted bottom is shown. A rectangular door1.6 x 0.8 m in the bottom of the tank is hingedat A and is opened against the pressure of thewater by the cable under a tension P as shown.Calculate P.

286


5.5 Fluid Statics

P. 5/11Assumption: door has negligible weight and volume,

i.e., W = 0 and B = 0

( )( )

1

2

A 1 2

pressure force magnitude volume of pressure distributionand divide into two parts -- rectangle and triangleF 1000 9.81 1.2 1.6 0.8

F 1/ 2 1000 9.81 0.8 1.6 0.8

M 0 P 1.6cos30 F 0.8 F 1.6 2 /

=

= × × × ×

= × × × × ×

= − × + × + × × ∑ 3 0

P 12566 N

=

=

AR

F1F2

P 1.2 m

1.2+1.6sin30 m

287


5.5 Fluid Statics

P. 5/12

( )y

pressure referred to atmosphere

F 0 900g A 150 1030g A 250 h

800g A 400 0, h 70.4 mm

= × × + × × − − × × = =

∑

yW

BO

BW

288


5.5 Fluid Statics

P. 5/13 A channel-marker buoy consists of a 2.4 m hollow steelcylinder 300 mm in diameter with a mass of 90 kg andanchored to the bottom with a cable as shown. If h = 0.6 mat high tide, calculate the tension T in the cable. Also findthe value of h when the cable goes slack as the tide drops.

The density of sea water is 1030 kg/m3. Assume the buoyis weighted at its base so that it remains vertical.

If the C.M. of the buoy is in the geometric center of thecylinder, calculate the angleθwhich would be made by thebuoy axis with the vertical when the water surface is 1.5 mabove the lower end of the cylinder. Neglect the diametercompared with the length when locating the center ofbuoyancy.

289


5.5 Fluid Statics

P. 5/13

W

T B

W

T

B

buoy is weighted at its base so it remains vertical

( )

( )

y

2

2

F 0 B Mg T 0

0.3if h 0.6 m, B Vg 1030 1.8g N4

T 402.7 Nwhen the cable goes slack, T 0 B decreases h increases

0.31030 2.4 h g 90g 0, h 1.164 m4

πρ

π

= − − = ×

= = = × ×

∴ =

= ←

×× × − − = =

∑

290


5.5 Fluid Statics

P. 5/13W

T

B

( )2

O

2

0.3B Vg 1030 1.5 / cos g4

M 0 B 0.75tan 90g 1.2sin 0

sin 0, or cos 0.75840, 29.44

πρ θ

θ θ

θ θθ

×= = × ×

= × − × = = =

∴ = ± °

∑

1.5 m

θ

291


5.5 Fluid Statics

P. 5/14 A fresh water channel 3 m wide is blocked atits end by a rectangular barrier, shown insection ABD. Supporting struts BC are spacedevery 0.6 m along the 3 m width. Determinethe compression C in each strut. Neglect theweights of the members.

292


5.5 Fluid Statics

P. 5/14

( )A

R 0.5 1.2 1000 9.81 1.2sin 60 3

M 0 R 0.4 6Csin60 0.6 0, C 2.354 kN

= × × × × ×

= − × + × = = ∑

6C

F

R

Ah = 1.2sin60

293


5.5 Fluid Statics

P. 5/15 The barge crane of rectangular proportions has a 6x15 mcross section over its entire length of 40m. If the maximumpermissible submergence and list in sea water are representedby the position shown, determine the corresponding maximumsafe mass mO that the barge can handle at 10 m extendedposition of the boom. Also find the total displacement m in themetric tons of the unloaded barge. The distribution ofmachinery and ballast places the CG of the barges, minusthe mass mO, at the center of the hull.

294


5.5 Fluid Statics

P. 5/15

( )G O

O

y O

B Vg 1030 15 6 40 / 2 g

M 0 Bcos 2.5 Bsin 1 m g 10cos 23sin 0

m 203 Mg

F 0 B mg m g 0, m 1651 Mg

ρ

θ θ θ θ

= = × × × ×

= × − × − + = =

= − − = =

∑

∑

mOg

mg

B2

3

57.5

295


5.5 Fluid Statics

P. 5/16 The fresh water side of a concrete dam hasthe shape of a vertical parabola with vertex at A.Determine the position b of the base point Bthrough which acts the resultant force of thewater against the dam face C.

296


5.5 Fluid Statics

P. 5/16

( )

( )

2 st 2

y 36362

y 00

362

c0

y ax or x y/a in 1 quadrant contains 27, 36 a 36 / 27

2A xdy xy 648 m3

AX x dA x / 2dy, X 10.125 m

F volume of the pressure distribution1/ 2 36 1000 9.81 36 h

W weight of the fluid bl

=

=

= = ∴ =

= = =

= = =

=

= × × × × ×

=

∫

∫ ∫

D

ock1000 9.81 648 h

M 0 F 18 W 10.125 W b 0, b 28.125 m

= × × ×

= − × − × + × = = ∑

Rx

Ry= W

F

W

x

12 m

D

297


5.5 Fluid Statics

P. 5/17 A flat plate seals a triangular opening in thevertical wall of a tank of liquid of density ρ.The plate is hinged about the upper edge Oof the triangle. Determine the force P requiredto hold the gate in a closed position againstthe pressure of the liquid.

298


5.5 Fluid Statics

P. 5/17

( ) ( ) ( )

( )y a

Oy 0

dR pdA g h y xdy g h y b 1 y/a dy

gbaM 0 Pa ydR 0, P h a/26

ρ ρ

ρ=

=

= = + = + × −

= − + = = + ∑ ∫

y

dyx

x = b(1-y/a)

b

a

O

dR

P

299

ch. 5: distributed forces

Documents