mechanics of materials - university of pittsburghqiw4/academic/engr0135/chapter5-3.pdf ·...

Department of Mechanical Engineering

ENGR 0135

Chapter 5 –3Center and centroids


Centers and CentroidsCenter of gravityCenter of massCentroid of volumeCentroid of areaCentroid of line


Center of Gravity A point where all of the

weight could be concentrated without changing the external effects of the body

To determine the location of the center, we may consider the weight system as a 3D parallel force system


Center of Gravity – discrete bodies

The total weight is

The location of the center can be found using the total moments

iWW

iiGiiGxy

iiGiiGzx

iiGiiGyz

zWW

zzWWzM

yWW

yyWWyM

xWW

xxWWxM

1

1

1


Center of Gravity – continuous bodies

The total weight is

The location of the center can be found using the total moments

dWW

zdWW

zzdWWzM

ydWW

yydWWyM

xdWW

xxdWWxM

GGxy

GGzx

GGyz

1

1

1


Center of Mass

A point where all of the mass could be concentrated

It is the same as the center of gravity when the body is assumed to have uniform gravitational force

Mass of particles

Continuous mass

n

ii

n

iiiC

n

iiiC

n

iiiC mmmz

mzmy

mymx

mx 111

dmmdmzm

zdmym

ydmxm

x GGG111


Example: Center of discrete massExample: Center of discrete mass

List the masses and the coordinates of their centroids in a table

Compute the first moment of each mass (relative to the planes of the point of interest)

Compute the total mass and total first moment

Compute the center


Center of mass Center of mass –– list of mass and the coordinateslist of mass and the coordinates

Labels Mass (kg)

xi (m) yi (m) zi (m)

A 1 0.3 .24 0.0

B 2 0.15 0.4 0.0

C 1 0.3 0.4 0.27

D 2 0.3 0.0 0.27

E 1 0.0 0.2 0.27


Center of discrete mass Center of discrete mass –– calculation of the centercalculation of the center

Mass # Mass (kg)

xi (m) yi (m) zi (m) mi xi mi yi mi zi

A 1 0.3 0.24 0.0 0.3 0.24 0.0

B 2 0.15 0.4 0.0 0.3 0.8 0.0

C 1 0.3 0.4 0.27 0.3 0.4 0.27

D 2 0.3 0.0 0.27 0.6 0.0 0.54

E 1 0.0 0.2 0.27 0.0 0.2 0.27

total 7 1.5 1.64 1.08

The center 1.5/7 1.64/7 1.08/7

1st moment of mass

xc zcyc

This method applies to discrete weights, lines, areas etc


Centroids of Volumes

Volumes made of sub vols

Continuous volumes

xi , yi , zi = centroids of the sub volumesVi = volumes of the segments

n

ii

n

iiiC

n

iiiC

n

iiiC VVVz

VzVy

VyVx

Vx 111

dVVdVzV

zdVyV

ydVxV

x CCC111


Centroids of AreasAreas made of segments

Continuous areas

n

ii

n

iiiC

n

iiiC

n

iiiC AAAz

AzAy

AyAx

Ax 111

dAAdAzA

zdAyA

ydAxA

x CCC111

xi , yi , zi = centroids of the area segmentsAi = Areas of the segments


Centroids of Lines (xc , yc , zc )

Lines made of segments

Continuous lines

dLLdLzL

zdLyL

ydLxL

x CCC111

n

ii

n

iiiC

n

iiiC

n

iiiC LLLz

LzLy

LyLx

Lx 111

xi , yi , zi = centroids of the line segmentsLi = length of the segments


Tables of special volumetric bodies, areas, and lines

These tables are helpful when the centroid of a composite body (composed of volumes, areas, or lines) is in question

In the following table, the centroids of the body are relative to the given origin O


Continuous bodies – crucial tasks

Choosing the coordinate system

Determining the differential element for the integration

Determining the lower and upper limits of the integral

Carefully perform the integration (may require integration table)


Example:

dA = differential element = b dy

This is not the only choice of the differential element !!


Example:

Many possibilities of differential elements and coordinate system


Please read example problems 5-17 and 5-18

5-17 Centroid of line segments5-18 Centroid of a cone


Problem 5-80: Centroid?


CentroidCentroid of an area of an area –– areaarea integrationintegration

Key components:– The differential element and its definition– The limits of the integration

– The moment arms


CentroidCentroid of an area of an area –– vertical differential elementvertical differential element

The area of the differential element

hdxdA

axb

axbbyyh 112

dA

dx

h

1

2



The limits of the integration– Lower limit x = 0– Upper limit x = a

x=0

x=a



Performing the integral to obtain the area

dA

3321

321

3211

2/3

0 0

2/3

0

ababaa

ab

xa

xbdxaxbdAA

a aa


CentroidCentroid of an area of an area –– Getting the 1Getting the 1stst moment of area about y moment of area about y axis axis -- MMyy

My needs a moment arm parallel to x-axis

The arm is from the y axis to the centroid of the element, here for the element it is x

dAx dx

axbxxhdx

xdAdAsMaq

xy

00

1


CentroidCentroid of an area of an area –– Getting the 1Getting the 1stst moment of area moment of area about y axis (Mabout y axis (Myy )) and the x coordinate of the and the x coordinate of the centroidcentroid

Performing the integration for the 1st moment of area

Calculating the x coordinate of the centroid

dxaxbxxhdxxdAM

aa

y

00

1

101

521

21

521

21 22/52

0

2/52 baaa

abxa

xba

aabba

dA

xdAxC 3.0

3/10/2


CentroidCentroid of an area of an area –– Getting the 1Getting the 1stst moment of moment of area about x axis area about x axis -- MMxx

Mx needs a moment arm parallel to y

The arm is from the x axis to the centroid of the element

dA

(y1+y2)/2

dAyyM x 2121

x

y1

1

2The centroid of the rectangular element is [ x, (y1 +y2 )/2]


CentroidCentroid of an area of an area –– Getting the 1Getting the 1stst moment of area moment of area about x axis (about x axis (MMxx )) and the y coordinate of the and the y coordinate of the centroidcentroid

Performing the integration for moment area

Calculating the y coordinate of the centroid

bab

abdA

My x

C 75.03/4/2

421

21

2

21

21

21

2

0

22

0

2

21

abaxxbdx

axb

dxaxbb

axbbdA

axbb

dAyyM

aa

x


Problem 5-79: Centroid?


Problem 5-79: Solution

xc , yc =x, y/2

dAv


CentroidsCentroids of composite bodiesof composite bodies

Possible elemental bodies:– Basic areas– Basic volumes– Line segments

Similar method to centroid of discrete mass

Pay attention to the centroid of the elemental bodies


CentroidCentroid of a composite areaof a composite area

The composite = A square - a full circle - a quarter circle


CentroidsCentroids of the elemental areasof the elemental areas

120mm

120mm

60mm

60mm

4r/3

4r/3

160/

See Table 5-1

Area 1

Area 2 Area 3


Calculation of the Calculation of the centroidcentroid relative to Orelative to O

Label Area xi (mm) yi (mm) Ai xi (1000 mm3)

Ai yi (1000 mm3)

1 57600 120 120 6912 6912

2 -11309 100 80 -1130.9 -904.72

3 -11309 240- 160/

240- 160/

-2138.2 -2138.2

Total 34982 3642.8 3869.1

The centroid 104.1 110.6


CentroidCentroid of composite volume and lineof composite volume and line

Similar method to composite area can be applied (use volume and length instead of area)

Use Table 5-1 and 5-2 to determine the centroid of the elemental bodies


Straight line segments Semicircular arc

Decomposition of the line bodyDecomposition of the line body


How about this?


Distributed loads on structural membersDistributed loads on structural members

Tasks: – Find the resultant – Find the location of the resultant

Distributed loads:– Continuous distribution

involves some area integral

– Composite of simple distribution – A combination of the two

xR R


Distributed loadDistributed load

The magnitude

The location

L

dxxwdRR0

)(

R

dxxxw

RxdR

d

L

0

)(


Composite of simple distributed loadComposite of simple distributed load

R1 R2R3

NR

NR

NR

600300.421

1800300.6

300300.221

3

2

1

m

d

56.56001800300

60033.91800530033.1


Continuous distributed loadContinuous distributed load

Lxwy

2sinmax

w = y

LwLwLxLw

dxLxwdxxwR

L

L L

maxmax

0

max

0 0max

637.022

cos2

2sin)(

LLxxL

LxL

Rw

dxLxxw

Rdxxxw

Rd

L

L

637.0

2cos2

2sin4

2sin1)(1

02

2max

0max


Summary

Moment about a point O is given by a vector product;

The magnitude of the moment is

Moment analysis:– Scalar approach– Vector approach

Moment about a line OB

FrM o~~~

222~ozoyoxoo MMMMM oM

Mo = r x F Moment about point OMOB = [(r x F) . e] e Moment about line OB

e is the unit vector along OBO is any point on the line OB


Summary

Couples

Equivalent force-couple system

Finding resultant of general force system

Center of weights and masses

Centroids of areas, lines and volumes

Distributed load

iiGiiGxy

iiGiiGzx

iiGiiGyz

zWW

zzWWzM

yWW

yyWWyM

xWW

xxWWxM

1

1

1

zdWW

zzdWWzM

ydWW

yydWWyM

xdWW

xxdWWxM

GGxy

GGzx

GGyz

1

1

1

mechanics of materials - university of pittsburghqiw4/academic/engr0135/chapter5-3.pdf ·...

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