Chapter 5 Distributed Forces: Centroids and Center of …cac542/L10.pdf · 1 MEM202 Engineering Mechanics - Statics MEM Chapter 5 Distributed Forces: Centroids and Center of Gravity

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MEM202 Engineering Mechanics - Statics MEM

Chapter 5 Distributed Forces:Centroids and Center of Gravity

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1Fr

2Fr

1x2x

21 FFRrrr

+=

R

x

221121 xFxFMMC +=+=rrr

Simplify

Centroid – An Introduction

RFx i

rr todue Moment todueMoment : gdetermininfor Critirion =

( )xFxRxFM iii ∑∑∑ ===

forces of Sumforces ofmoment of Sum

===∑∑∑

FxF

RM

x iii

3


Centroid – An Introduction

( ) ( )∫∑ ==L

dxxfdRR0

( ) ( ) ( )∫∑∑ =⋅==L

dxxxfdRxdMC0

( ) ( )

( )( )

( )( )

( )xw

xfOxf

dxxf

dxxxfd

dxxfddxxxfC

L

L

LL

of controid

of (area)amount Totalabout ofMoment

0

0

00

=

==⇒

⋅==

∫∫

∫∫

( )∫⋅=⋅=L

dxxfdRdC0

4


5.2 Center of Gravity and Center of MassCenter of Gravity

WdrMdrrr

×=

( )∫∫ ×==VV

WdrMdMrrrrx

y

y

z

xz

V

rr

dV

Wdr

O

WrM G

rrr×=x

y

Gy

z

GxGz

Χ

Grr

Wr

O∫∫

=

=

V

V

dWW

dVV

( )∫ ×=×VG WdrWr

rrrr

WW

dWdW

WW

dWdW

WW

dWdW zzyyxx === :NOTE

5


5.2 Center of Gravity and Center of MassCenter of Gravity

( ) ( ) ( )kWyWxjWxWziWzWyWr xGyGzGxGyGzGG

rrrrr −+−+−=×

( ) ( ) ( ) ( )[ ]∫∫ −+−+−=×V xyzxyzV

kydWxdWjxdWzdWizdWydWWdrrrrrr

( )( )( )( )∫

∫∫

∫−=−

−=−

−=−

⇒×=×

V xyxGyG

V zxzGxG

V yzyGzG

VG

ydWxdWWyWx

xdWzdWWxWz

zdWydWWzWy

WdrWrrrrr

etc. , Recall ∫∫∫ =⎟⎠⎞

⎜⎝⎛=⇒=

Vz

Vz

V zzz ydW

WWdW

WWyydW

WW

dWdW

( ) etc ,∫∫∫ −=−=−V

y

Vz

V yzyGzG zdWWW

ydWWWzdWydWWzWy

∫∫∫ ===VGVGVG zdW

WzydW

WyxdW

Wx 111

∫=VG dWr

Wr rr 1

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5.2 Center of Gravity and Center of MassCenter of Mass

x

y

y

z

xz

V

rr

dm

O

x

y

Gy

z

GxGz

Χ

Grr

m

O∫∫

=

=

V

V

dmm

dVV

∫∫∫

∫∫∫

∫∫∫ ======

V

VVG

V

VVG

V

VVG dm

zdm

m

zdmz

dm

ydm

m

ydmy

dm

xdm

m

xdmx

7


5.3 Centroids of Volumes, Areas, and Lines

x

y

y

z

dA

x

z

x

y

y

z

dV

x

z

x

y

y

z

dL

x

z

V AL

Volume Area Line

C CC

VzdVdVzdVz

VydVdVydVy

VxdVdVxdVx

VVVc

VVVc

VVVc

∫∫∫∫∫∫∫∫∫

==

==

==

AzdAdAzdAz

AydAdAydAy

AxdAdAxdAx

AAAc

AAAc

AAAc

∫∫∫∫∫∫∫∫∫

==

==

==

LzdLdLzdLz

LydLdLydLy

LxdLdLxdLx

LLLc

LLLc

LLLc

∫∫∫∫∫∫∫∫∫

==

==

==

8


5.3 Centroids of Volumes, Areas, and LinesExample: Centroid of A Rectangular Area

b

h

x

ydA

x

ybhdydxdAA

h b

A=⎟

⎠⎞⎜

⎝⎛== ∫ ∫∫ 0 `0

222111 2

0

2

0 0

bbhhbdyb

bhdyxdx

bhxdA

AAM

xhh b

A

yc ===⎟

⎠⎞⎜

⎝⎛=== ∫∫ ∫∫

22111 2

00 0

hbh

bhbydybh

dyydxbh

ydAAA

Myhh b

Ax

c ===⎟⎠⎞⎜

⎝⎛=== ∫∫ ∫∫

b

hy

dA

x

y

( )

2

1

1

0

h

bdyybh

ydAA

y

h

Ac

=

=

=

∫

∫ b

hdA

x

y

( )

2

1

1

0

b

hdxxbh

xdAA

x

b

Ac

=

=

=

∫

∫x

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5.3 Centroids of Volumes, Areas, and LinesExample: Centroid of A Quarter Circle

- Double integral in rectangular coordinates

r

x

y

dA

x

y

dy

dx

22

22

222

xry

yrx

ryx

−=

−=

=+

36222

3

0

32

0

22

00

2

0 0

22

22

rxxrdxxrdxy

dxydyydAM

rrr

xr

r xr

Ax

=⎥⎦

⎤⎢⎣

⎡−=

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛⎥⎦

⎤⎢⎣

⎡=

⎟⎠

⎞⎜⎝

⎛==

∫∫

∫ ∫∫−

−

4 2rdAA

A

π== ∫

ππ 34

4 3

2

3 rr

rA

ydA

AMy Ax

c ==== ∫

π34r

A

xdA

AM

x Ayc === ∫

10



- Single integral using a horizontal strip

( )33

3

0

2322

0

22 ryrdyyryydAMr

r

Ax =⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−=−== ∫∫

ππ 34

4 3

2

3 rr

rA

ydA

AMy Ax

c ==== ∫

r

22 yrx −=

y

dA

x

y

dy

22

22

222

xry

yrx

ryx

−=

−=

=+

( )

3622

222

3

0

32

0

22

2222

22

ryyrdyyrdMM

dyyrdyyryrdAxdM

rr

A yy

y

=⎥⎦

⎤⎢⎣

⎡−=

−==

−=−⎟

⎟⎠

⎞⎜⎜⎝

⎛ −==

∫∫

π34r

AM

x yc ==

11



- Double integral using polar coordinates

( )( )

[ ]

33

cossin

sin

3

0

3

0

2

0

20

2

0

2

0

2

0

2

0

rd

ddd

ddydAM

rr

rr

r

Ax

=⎥⎦

⎤⎢⎣

⎡==

−=⎟⎠⎞⎜

⎝⎛=

==

∫

∫∫ ∫

∫ ∫∫

ρρρ

ρθρρθθρ

ρθρθρ

ππ

π

y

dA

x

y

dy

θ

θd

ρ

ρd

x

ππ 34

4 3

2

3 rr

rA

My xc ===

π34r

AM

x yc ==

θρθρ sincos == yx

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5.4 Centroids of Composite Bodiesy

x

2cx

2cy

1cx

1cy

×× 1C

2C2A1A

∑∑=

++

==

+=

+=

i

cicc

c

yc

C

ccy

AxA

AAxAxA

AM

x

AAA

xAxAM

i

21

21

21

21

21

21

∑∑=

++

==

+=

i

cicc

c

xc

ccx

AyA

AAyAyA

AMy

yAyAM

i

21

21

21

21

21

ExamplePart Ai (in2) xci (in) My (in3) yci (in) Mx (in3)

1 50.0 6.67 333.3 3.33 166.72 100.0 15.0 1,500.0 5.0 500.0Σ 150.0 1,833.3 666.7

in 10

in 10 in 10

2A1A

in 44.4150

7.666in 22.12150

3.833,1======

∑∑

∑∑

AM

yA

Mx x

cy

cin 0.5

in 0.15

in 33.3

in 67.6

1

2

1

1

=

=

=

=

c

c

c

c

y

x

y

x

13


5.4 Centroids of Composite Bodies

−

= +

+ +

Table 5-1, Page 212

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5.6 Distributed Loads on Beams

( )xfw =

O

Find a concentrated force R that is equivalent to the distributed load w.

( )

( )dxxfxdxxwxdRdMM

dxxfwdxdRR

OO ∫∫∫∫∫∫∫

====

===

x dx

wdxdR =

O

cx

R

O

( )( )∫

∫==dxxf

dxxxfR

Mx Oc

cO RxM =

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5.6 Distributed Loads on Beams

O

O

O

m 6 m 51 m 9

N/m 100=w

N 300 N 500,1 N 450

m 42m 3.51

m 4

( )( )

( ) m 4632

32

N 300100621

21

1

1111

1===

====

bx

hbAF

c

( )( )

( ) m 5.1315216

216

N 500,110015

2

2222

2=+=+=

====

bx

hbAF

c

( )( )

( ) m 2493121

3121

N 450100921

21

3

3333

3=+=+=

====

bx

hbAF

c

Example

m 4.31

N ,2502

N 250,2321 =++= FFFR

( )( ) ( )( ) ( )( )m-N 250,32

244505.13500,14300321 321

=++=

++= cccO xFxFxFM

m 3.14== RMx Oc

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5.7 Forces on Submerged Surfaces

psdVpdAdR ==

psps VdVpdAdRR ==== ∫∫∫

pscpsx VxxdVxpdAxdRRdps

==== ∫∫∫

pscpsy VyydVypdAydRRdps

==== ∫∫∫

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Documents

Chapter 5 Distributed Forces: Centroids and Center of …cac542/L10.pdf · 1 MEM202 Engineering Mechanics - Statics MEM Chapter 5 Distributed Forces: Centroids and Center of Gravity