ENGR 0135
Chapter 5 –3 Center and centroids
Department of Mechanical Engineering Centers and Centroids
Center of gravity Center of mass Centroid of volume Centroid of area 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, areas 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 integral
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 cone
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 b1 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 circle - 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