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ENGR 0135

Chapter 5 –3 Center and centroids

Department of Mechanical Engineering Centers and Centroids

 Center of gravity   Centroid of volume  Centroid of  Centroid of line

Department of Mechanical Engineering 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

Department of Mechanical Engineering Center of Gravity – discrete bodies

 The total weight is  WW i

 The location of the center can be found using the total moments 1  xxWWxM  xW yz G  ii G W  ii 1  yyWWyM  yW zx G  ii G W  ii 1  zzWWzM  zW xy G  ii G W  ii Department of Mechanical Engineering Center of Gravity – continuous bodies  The total weight is   dWW

 The location of the center can be found using the total moments 1 WxM  xdW x  xdW yz G  G W  1 WyM  ydW y  ydW zx G  G W  1 WzM  zdW z  zdW xy G  G W 

Department of Mechanical Engineering 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 1 n 1 n 1 n n xC   ii ymx C   ii zmy C   ii   mmmz i m i m i m i i

 Continuous mass 1 1 1 x  ydmx  zdmy   dmmdmz G m  G m G m 

Department of Mechanical Engineering Example:Example: CenterCenter ofof discretediscrete massmass

 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

Department of Mechanical Engineering CenterCenter ofof massmass –– listlist ofof massmass andand thethe coordinatescoordinates

Labels Mass xi (m) yi (m) zi (m) (kg) 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

Department of Mechanical Engineering CenterCenter ofof discretediscrete massmass –– calculationcalculation ofof thethe centercenter

1st moment of mass

Mass # Mass xi (m) yi(m) zi(m) mix i miy i miz i (kg) 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

xc yc zc

This method applies to discrete weights, lines, etc Department of Mechanical Engineering Centroids of Volumes

 Volumes made of sub vols 1 n 1 n 1 n n xC   ii yVx C   ii zVy C  ii   VVVz i V i V i V i i

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

 Continuous volumes 1 1 1 x  ydVx  zdVy   dVVdVz C V  C V C V 

Department of Mechanical Engineering Centroids of Areas

 Areas made of segments 1 n 1 n 1 n n xC   ii yAx C   ii zAy C  ii   AAAz i A i A i A i i

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

 Continuous areas 1 1 1 x  ydAx  zdAy   dAAdAz C A  C A C A 

Department of Mechanical Engineering Centroids of Lines (xc, yc, zc)

 Lines made of segments 1 n 1 n 1 n n xC   ii yLx C   ii zLy C  ii   LLLz i L i L i L i i

xi, yi, zi = centroids of the line segments Li = length of the segments  Continuous lines

1 1 1 x  ydLx  zdLy   dLLdLz C L  C L C L 

Department of Mechanical Engineering 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

Department of Mechanical Engineering Department of Mechanical Engineering Department of Mechanical Engineering Department of Mechanical Engineering Department of Mechanical Engineering Continuous bodies – crucial tasks

 Choosing the coordinate system

 Determining the differential element for the integration

 Determining the lower and upper limits of the

 Carefully perform the integration (may require integration table)

Department of Mechanical Engineering Example:

dA = differential element = b dy

This is not the only choice of the differential element !! Department of Mechanical Engineering Example:

Many possibilities of differential elements and coordinate system Department of Mechanical Engineering Please read example problems 5-17 and 5-18

 5-17 Centroid of line segments  5-18 Centroid of a

Department of Mechanical Engineering Problem 5-80: Centroid?

Department of Mechanical Engineering CentroidCentroid ofof anan areaarea –– areaarea integrationintegration

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

Department of Mechanical Engineering Centroid of an area – vertical differential element

 The area of the differential element

x  x  bbyyh b1   hdxdA 12   a  a  dx

2

dA h

1

Department of Mechanical Engineering Centroid of an area – vertical differential element

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

x=a

x=0

Department of Mechanical Engineering Centroid of an area – vertical differential element

 Performing the integral to obtain the area

a a a  x  21    2/3   bdAA 1  xbdx  x  0 0  a   a 3 0

 21 2/3   2 ab ab   aba 1    a 3    33

dA

Department of Mechanical Engineering Centroid of an area – Getting the 1st moment of area about y axis - My

 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

 xdAdAsM xy  dA q a  x  x    bxxhdx 1 dx 0 0  a 

Department of Mechanical Engineering Centroid of an area – Getting the 1st moment of area

about y axis (My) and the x coordinate of the centroid

 Performing the integration for the 1st moment of area

a a  x  xdAM bxxhdx 1 dx y    0 0  a 

a 1 2 21 2/5  1 2 21 2/5  2 1  xb    abx    baa 2 a 5 0 2 a 5  10

 Calculating the x coordinate of the centroid

 xdA 2ba 10/ xC  3.0 a dA ab 3/  Department of Mechanical Engineering Centroid of an area – Getting the 1st moment of

area about x axis - Mx

 Mx needs a moment arm parallel to y  The arm is from the x axis to the centroid of the element 1  dAyyM x  2 21

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

1 (y1+y2)/2 y1

x Department of Mechanical Engineering Centroid of an area – Getting the 1st moment of area about x axis (Mx) and the y coordinate of the centroid

 Performing the integration for moment area 1  dAyyM x  2 21 1  x  1  x  x          bb     bbdA   bb dx 2  a  2  a  a  a 2 a  xb  b2  1 x2  ab2 1 dx x    2 0  a  2  a 0 42  Calculating the y coordinate of the centroid

2 Mx ab 4/ yC  75.0 b  dA ab 3/

Department of Mechanical Engineering Problem 5-79: Centroid?

Department of Mechanical Engineering Problem 5-79: Solution

dAv

xc, yc=x, y/2

Department of Mechanical Engineering Problem 5-79: Solution

dAv

xc, yc=x, y/2

Department of Mechanical Engineering Problem 5-79: Solution

dAv

xc, yc=x, y/2

Department of Mechanical Engineering CentroidsCentroids ofof compositecomposite bodiesbodies

 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

Department of Mechanical Engineering CentroidCentroid ofof aa compositecomposite areaarea

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

Department of Mechanical Engineering CentroidsCentroids ofof thethe elementalelemental areasareas

Area 1

120mm

120mm 160/

4r/3 Area 2 Area 3

4r/3

60mm See Table 5-1

60mm Department of Mechanical Engineering CalculationCalculation ofof thethe centroidcentroid relativerelative toto OO

Label Area xi (mm) yi (mm) Aix i (1000 Aiy i (1000 mm3) mm3)

1 57600 120 120 6912 6912

2 -11309 100 80 -1130.9 -904.72

3 -11309 240- 240- -2138.2 -2138.2 160/ 160/ Total 34982 3642.8 3869.1

The centroid 104.1 110.6

Department of Mechanical Engineering CentroidCentroid ofof compositecomposite volumevolume andand lineline

 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

Department of Mechanical Engineering DecompositionDecomposition ofof thethe lineline bodybody

Straight line segments Semicircular arc

Department of Mechanical Engineering How about this?

Department of Mechanical Engineering DistributedDistributed loadsloads onon structuralstructural membersmembers

 Tasks: – Find the resultant

– Find the location of the resultant  Distributed loads: – Continuous distribution  xR R involves some area integral

– Composite of simple distribution – A combination of the two

Department of Mechanical Engineering DistributedDistributed loadload

 The magnitude

L  )( dxxwdRR 0  The location

L )( dxxxw xdR  d   0 R R

Department of Mechanical Engineering CompositeComposite ofof simplesimple distributeddistributed loadload

R1 R R2 3

1 R  300300.2 N 1     60033.91800530033.1 2 d  R2  1800300.6 N  6001800300 1 R  600300.4 N  56.5 m 3 2 Department of Mechanical Engineering ContinuousContinuous distributeddistributed loadload

x  wy sin max 2L

w = y

1 1 L x LL d  )( dxxxw  xw sin dx x  max  )(  wdxxwR sin dx R R 0 2L max 2L 00 2 L L wmax 4 xL 2L x  2Lw  x  2Lw   2 sin  x cos  max  cos max  637.0 Lw R  2L  2L   max  0   2L  0  637.0 DepartmentL of Mechanical Engineering Summary

 Moment about a point O is given by a vector product; ~ ~ ~ o  FrM  The magnitude of the moment is

~ 2 2 2 MM oo Mo ox oy  MMM oz  Moment analysis: – Scalar approach – Vector approach  Moment about a line OB

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

e is the unit vector along OB O is any point on the line OB Department of Mechanical Engineering Summary

1  Couples  xxWWxM  xW yz G  ii G W  ii  Equivalent force-couple system 1 zx G   ii yyWWyM G   yW ii  Finding resultant of general force W system 1 xy G  ii zzWWzM G  zW ii  W   Center of weights and masses 1  Centroids of areas, lines and WxM  xdW x  xdW yz G  G  volumes W 1  Distributed load zx WyM G  ydW yG  ydW  W  1 WzM  zdW z  zdW xy G  G W 

Department of Mechanical Engineering