chapter 3.a(cg centroid)

8/18/2019 Chapter 3.a(CG Centroid)

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Engineering Mechanics :Engineering Mechanics :STATICSSTATICS

Lecture #04By,

Noraniah Kassim

Universiti Tun Hussein Onn Malaysia(UTHM),


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CENTER OF GRAVITY AND CENTROID

Today’s Objective :

Students will:a) Understand the concepts of center of

gravity, center of mass, and centroid.

Learning Topics:

•Applications

• Center of gravity, etc.


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APPLICATIONS

To design the structure forsupporting a water tan, we willneed to now the weights of thetan and water as well as the

locations where the resultantforces representing thesedistri!uted loads are acting.

"ow can we determine theseweights and their locations#


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REVIEW

The net force on the !eam is given !y

$ ↓ F R % ∫ & d' % ∫ & w( x) d % A

"ere A is the area under the loadingcurve w().

Assuming that '* acts at , it will producethe moment a!out point + as

$ M RO % ( ) ('* ) % ∫ & w() d

)

)


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INTRODUCTION

• The earth eerts a gravitational force on each of the particles forming a

!ody. These forces can !e replace !y a single euivalent force eual tothe weight of the !ody and applied at the center of gravity for the

!ody.

• The centroid of an area is analogous to the center of gravity of a !ody.

The concept of the first moment of an area is used to locate thecentroid.

• 0etermination of the area of a surface of revolution and the volume ofa body of revolution are accomplished with the Theorems of Pappus!uldinus.


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CONCEPT OF CG AND CM

The center of gravity (1) is a point whichlocates the resultant weight of a system of

particles or !ody.

'rom the definition of a resultant force, the sum of moments due toindividual particle weight a!out any point is the same as the momentdue to the resultant weight located at 1. 'or the figure a!ove, try taingmoments a!out A and 2.

Also, note that the sum of moments due to the individual particle3sweights a!out point 1 is eual to 4ero.

Similarly, the center of mass is a point which locates the resultantmass of a system of particles or !ody. 1enerally, its location is the

same as that of 1.

••

5m

67

8m

A 28 7

5 71•


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CONCEPT OF CG AND CM (Continued)

• Center of gravity of a plate

∫

∑∑∫

∑∑

=

∆==

∆=

d" y

" y" y M d" x

" x" x M

y

y

• Center of gravity of a wire


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CONCEPT OF CENTROID

The centroid C is a point which defines thegeometric center of an o!9ect.

The centroid coincides with the centerof mass or the center of gravity only ifthe material of the !ody is homogenous(density or specific weight is constantthroughout the !ody).

f an o!9ect has an ais of symmetry, thenthe centroid of o!9ect lies on that ais.

n some cases, the centroid is notlocated on the o!9ect.


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At " ∆=∆ γ

#a" ∆=∆ γ 11

CONCEPT OF CENTROID (Continued)

( ) ( )

x

QdA y A y

y

QdA x A x

dAt x At xd" x" x

x

y

respect tohmoment witfirst

respect tohmoment witfirst

=

==

=

==

=

=

∫

∫

∫ ∫

γ γ

• Centroid of an area

( ) ( )

∫

∫

∫ ∫

=

=

=

=

d# y # y

d# x # x

d#a x #a x

d" x" x

γ γ

• Centroid of a line

; % specified weight (weight per unitvolume)

t % thicness of the plate


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CG / CM FOR A SYSTEM OF PARTICLES

Consider a system of n particles as shown in

the figure. The net or the resultant weight isgiven as >* % ∑>.

Similarly, we can sum moments a!out the and 4=aes to find thecoordinates of 1.

2y replacing the > with a ? in these euations, the coordinatesof the center of mass can !e found.

Summing the moments a!out the y=ais, we get

>* % 8>8 $ @>@ $ .. $ n>nwhere 8 represents coordinate of >8, etc..

BBB

B


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CG / CM / CENTROID OF A BODY (Setion !"#)

A rigid !ody can !e considered as made

up of an infinite num!er of particles."ence, using the same principles as inthe previous slide, we get thecoordinates of 1 !y simply replacing thediscrete summation sign ( ∑ ) !y thecontinuous summation sign ( ∫ ) and >

!y d>.

Similarly, the coordinates of the center of mass and the centroidof volume, area, or length can !e o!tained !y replacing > !y m,

, A, or &, respectively.


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STEPS FOR DETERMING AREA CENTROID

8. Choose an appropriate differential element dA at a general point (,y)."int: 1enerally, if y is easily epressed in terms of (e.g., y % @ $ 8), use a vertical rectangular element. f the converse istrue, then use a hori4ontal rectangular element.

@. Dpress dA in terms of the differentiating element d (or dy).

6. Dpress all the varia!les and integral limits in the formula using

either or y depending on whether the differential element is interms of d or dy, respectively, and integrate.

5. 0etermine coordinates ( , y ) of the centroid of the rectangularelement in terms of the general point (,y).

B B

7ote: Similar steps are used for determining C1, C?, etc.. Thesesteps will !ecome clearer !y doing a few eamples.


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E$AMPLE (Inte%&'tion etod)

Solution

8. Since y is given in terms of , choosedA as a vertical rectangular strip.

•

•,y

, yBB

Given The area as shown.!ind The centroid location ( , y)

"lan 'ollow the steps.

@. dA % y d % (E F @) d

5. % and y % y G @BB


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E$AMPLE (Continued)

6. % ( ∫ A dA ) G ( ∫ A dA )B

H

H

H ∫ ( E F @) d I E (@)G@ F (6) G 6J 5

H ∫ ( E F @

) d I E F (5

) G 5 J5

% ( E ( E ) G @ F K8 G 6 ) G ( E ( 5 ) F ( @L G 5 ) )

% 8#85 m

5% %

5

5

5#MH m ∫ A y dA N H ∫ ( E F @) ( E F @) d

∫ A dA H ∫ ( E F @) d

5

%y % %B


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CONCEPT *UI+

8. The steel plate with nown weight and non=uniform thicness and density is supportedas shown. +f the three parameters (C1, C?,and centroid), which one is needed fordetermining the support reactions# Are allthree parameters located at the same point#

A) (center of gravity, no)2) (center of gravity, yes)C) (centroid, yes)0) (centroid, no)

@. >hen determining the centroid of the area a!ove, which type ofdifferential area element reuires the least computational wor#

A) ertical 2) "ori4ontal

C) Oolar 0) Any one of the a!ove.


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IN CLASS TUTORIAL

Given The area as shown.

!ind The of the centroid.

"lan 'ollow the steps.

Solution

8. Choose dA as a hori4ontalrectangular strip.(8,,y) (@,y)@. dA % (

@ F

8) dy

% ((@ F y) F y@) dy

5. % ( 8 $ @) G @

% H.P (( @ F y) $ y@ )


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IN CLASS TUTORIAL (Continued)

6. % ( ∫ A dA ) G ( ∫ A dA )B

∫ A dA % H∫ ( @ F y F y@) dy

I @ y F y@ G @ F y5 G 5J 8 % 8#8ML m@

8

H

∫ A dA % H∫ H#P ( @ F y $ y@ ) ( @ F y F y@ ) dy

% H#P H∫ ( 6 F 6 y $ y@ F y6 ) dy

% H#P I 6 y F 6 y@ G @ $ y5 G 5 F yP G P J 8

% 8#HML m5

H

8

8B

% 8#HML G 8#8ML % H#E86 m

chapter 3.a(cg centroid)

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