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8/13/2019 Chapter 4 Distributed Forces
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VECTOR MECHANICS FOR ENGINEERS:
STATICS
CHAPTER
5Distributed Forces:
1. Centroids and Centers
of Gravity
2. Moment of Inertia of
Areas
4
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Contents
Centroids and Center of Gravity
First Moments of Areas &Lines
Centroids of Common Shapesof Areas
Centroids of Common Shapesof Lines
Composite Plates & Areas
Determination of Centroids byIntegration
Theorems of Pappus-Guldinus
Distributed Loads on Beams
Distributed Force: Centroids and Center of Gravity
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It is expected that students will be able to: Describe the concept of centroids and center of gravity
Calculate the centroids and center of gravity
Calculate the moment of inertia
Distributed Force: Centroids and Centers of Gravity
Topic Outcomes
S
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Introduction
The earth exerts a gravitational force on each of particleforming a rigid body. These large number of small forcescan be replace by a single equivalent force which is theweight of the body, Wand applied at the center of gravityfor the body.
2 main ideas of the chapter:
a. Determination of the center of gravity (centroid ofan area)of plate or wire: The concept of the firstmoment of for a plate or moment at given axisfora wire is used to locate the centroid.
b. Determination of the area of a surface of revolutionor the volume of a body of revolution by Theoremsof Pappus-Guldinus
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Center of Gravity
Center of gravity of a
plate
dWyWyWyM
dWxWxWxM
x
y
:
:
Center of gravity of a
wire
Distributed Force: Centroids and Centers of Gravity
The moment of resultant force W equal to the sum ofthe corresponding moments of elemental force W.From that we can obtain coordinate center of gravity
( , )x y
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Centroids and First Moments of Areas and Lines
dAyAy
dAxAx
dAtxAtx
Centroid of an area
dLyLy
dLxLx
dLaxLax
dWxWx
Centroid of a line
Distributed Force: Centroids and Centers of Gravity
Magnitude of weight, W
plateofareatotal
plateofthickness
eunit volumperweight
A
t
tAW
Coordinate( , ) is known as centroidC of the area Afor a homogenous plateand centroid C of the line Lfor ahomogeneous wire.
x y
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Both integrals are known as the first moment of the area A with
respect to y axis, Qy and first moment of the area A withrespect to x axis, Qx, respectively.
The first moments of the area A can be expressed in area andcoordinates of its centroids
The coordinates of the centeroid of an area can be obtained bydividing the first moment of area by the area
First Moments of Areas and Lines
Distributed Force: Centroids and Centers of Gravity
Qxaxisxrespect toA withareatheofmomentfirst
Qyaxisyrespect toA withareatheofmomentfirst
dAyAy
dAxAx
AyQ
AxQ
x
y
yA
Qx
A
Qxy
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When an area or line is symmetry, its first moment is zeroandits centroid is located on axis BB
An area is said to be symmetric with
respect to an axis BBif for everypoint Pthere exists a point Psuchthat perpendicular PP is divided intotwo equal parts by BB.
First Moments of Areas and Lines
Distributed Force: Centroids and Centers of Gravity
If an area possessestwo lines of symmetry,its centroid lies at their
intersection C.
The centroid of the area coincides with thecenter of symmetry O.
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Centroids of Common Shapes of Areas
Distributed Force: Centroids and Centers of Gravity
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Centroids of Common Shapes of Lines
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Composite Plates and Areas
Composite plates
WyWY
WxWX
Compositehomogenous area
x
y
QAyAY
QAxAX
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Sample Problem 4.1
For the plane area shown,determine the first moments
with respect to thexand yaxes and the location of thecentroid.
SOLUTION:
Divide the area into a triangle,rectangle, and semicircle with acircular cutout.
Compute the coordinates of thearea centroid by dividing the firstmoments by the total area.
Find the total area and firstmoments of the triangle,rectangle, and semicircle.Subtract the area and first
moment of the circular cutout.
Calculate the first moments ofeach area with respect to the
axes.
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Sample Problem 4.1
33
33
mm102.506
mm107.757
y
x
Q
Q Find the total area and first momentsof the triangle, rectangle, andsemicircle. Subtract the area and first
moment of the circular cutout.
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Sample Problem 4.1
23
33
mm1013.828mm107.757
AAxX
mm8.54X
23
33
mm1013.828
mm102.506
A
AyY
mm6.36Y
Compute the coordinates of the
area centroid by dividing thetotal first moments by the totalarea.
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Determination of Centroids by Integration
The coordinate are obtained by expressing the first moments of
the entire area is equal to the sum (integral) of correspondingmoments of elements of area.
ydxy
dAyAyQ
ydxx
dAxAxQ
elx
ely
2
dyxay
dAyAyQ
dyxaxa
dAxAxQ
ely
elx
2
drr
dAyAy
drr
dAxAx
el
el
2
2
2
1sin
3
2
2
1cos
3
2
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Application: Theorems of Pappus-Guldinus
Surface of revolutionis generated byrotating a plane curve, Labout a fixedaxis.
Area of a surface of revolutionis equal to the length of the
generating curve times thedistance traveled by thecentroidthrough the rotation.
LyA 2
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Application: Theorems of Pappus-Guldinus
Body of revolutionis generated byrotating a plane area,Aabout a fixedaxis.
Volume of a body ofrevolutionis equal to the
generating area times thedistance traveled by thecentroidthrough the rotation.
AyV 2
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Di ib d F C id d C f G i
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Sample Problem 4.3
The outside diameter of apulley is 0.1 m, and the crosssection of its rim is as shown.Knowing that the pulley ismade of steel and that thedensity of steel isdetermine the mass andweight of the rim.
33 mkg1085.7
SOLUTION:
Apply the theorem of Pappus-Guldinus to evaluate thevolumes (Volume of a body ofrevolution).
Multiply by density andacceleration to get the massand weight of the rim.
Distributed Force: Centroids and Centers of Gravity
CAUTION!!
You have not been giventhe thickness of the rimto calculate its volume.
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Sample Problem 4.3
SOLUTION:
Apply the theorem of Pappus-Guldinus to evaluate the volumes orfor the rectangular rim sectionandthe inner cutout section. Subtractthe inner section.
3393633 mmm10mm1065.7mkg1085.7Vm kg0.60m
2
sm81.9kg0.60mgWN589W
Multiply by density and accelerationto get the mass and acceleration.
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Di t ib t d F C t id d C t f G it
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Application: Distributed Loads on Beams
A distributed load is represented byplotting the load, wper unit length,x(N/m). The total load, Wis equal to thearea under the load curve,A.
AdAdxwWL
0
AxdAxAOP
dWxWOP
L
0
The total load, Walso can be replace by
a single concentrated loadwhich onlyequivalent considering fee body diagramto the entire beam.
Its magnitude is equal to the area underthe load curve,Aand a line of action
passing through the area centroid, .
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Di t ib t d F C t id d C t f G it
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Sample Problem 4.4
A beam supports a distributedload as shown. Determine the
equivalent concentrated loadand the reactions at thesupports.
SOLUTION:
The magnitude of theconcentrated load is equal to thetotal load or the area under thecurve.
The line of action of theconcentrated load passesthrough the centroid of thearea under the curve.
Determine the support
reactions by summingmoments about the beamends.
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Sample Problem 4.4
SOLUTION:
The magnitude of the concentrated loadis equal to the total load or the areaunder the curve. kN0.18F
The line of action of the concentrated loadpasses through the centroid of the areaunder the curve.
kN18
mkN63 X
m5.3X
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Sample Problem 4.4
Determine the support reactions by
summing moments about the beamends.
0m.53kN18m6:0 yA BM
kN5.10yB
0m.53m6kN18m6:0 yB AM
kN5.7y
A
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Dist ib ted Fo ce Moment of Ine tia of A eas
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Moment of Inertia of Areas
Moment of Inertia by Integration
Polar Moment of Inertia
Radius of Gyration of Area
Parallel Axis Theorem
Moments of Inertia of Composite Area
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Contents
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Consider distributed forces whosemagnitudes are proportional to theelemental areas on which they act andalso vary linearly with the distance of yfrom a given axis.
F
A
R is the resultant of which thenreduced to a couple that provide amoment couple M.The moment called asthe moment of inertia.
inertia)of(momentmomentsecond
axis)on x(centroid0momentfirst
22
dAydAykM
QdAydAykR x
Moment of Inertia of Area
F
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Moment of Inertia of an Area by Integration
Second momentsor moments ofinertiaof an area with respect tothexand yaxes,
dAxIdAyI yx22
Evaluation of the integrals issimplified by choosing dA to be a
thin strip parallel to one of thecoordinate axes.
For a rectangular area,
331
0
22 bhbdyydAyIh
x
The formula for rectangular areasmay also be applied to stripsparallel to the axes,
dxyxdAxdIdxydI yx223
31
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Polar Moment of Inertia
Thepolar moment of inertiais an
important parameter in problemsinvolving torsion of cylindrical shaftsand rotations of slabs.
dArJ2
0
The polar moment of inertia is relatedto the rectangular moments of inertia,
xy II
dAydAxdAyxdArJ
22222
0
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Radius of Gyration of an Area
Consider areaA with moment of inertiaIx
. Imagine that the area isconcentrated in a thin strip parallel tothexaxis with equivalent Ix.
A
IkAkI xxxx
2
kx=radius of gyrationwithrespect to thexaxis
Similarly,
A
JkAkJ
A
IkAkI
OOOO
yyyy
2
2
222yxO kkk
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The radius of gyration of an
areaA with respect to thexaxis is defined asthe distancek
x, where
Ix= k
xA. With similar
definitions for the radius of
gyration ofAwith respectto theyaxis and with
respect toO, we have
kx=
2
IxA
ky=IyA
kO=JOA
y
x
kx
O
A
Radius of Gyration of an Area
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Parallel Axis Theorem
Consider moment of inertia Iof an
areaAwith respect to the axisAA
dAyI2
The axis BBpasses through the area
centroid and is called a centroidalaxis. is moment inertia of the areawith respect to BBis parallel toAA .
dAddAyddAy
dAdydAyI
22
22
2
2AdII parallel axistheorem
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Parallel Axis Theorem
Moment of inertia ITof a circular
area with respect to a tangent tothe circle,
445
224412
r
rrrAdIIT
Moment of inertia of a triangle withrespect to a centroidal axis,
3
361
231
213
1212
2
bh
hbhbhAdII
AdII
AABB
BBAA
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Moments of Inertia of Composite Areas The moment of inertia of a composite areaAabout a given
axis is obtained by adding the moments of inertia of the
component areasA1,A2,A3, ... , with respect to the sameaxis.
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Moments of Inertia of Composite Areas
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Sample Problem 4.7
The strength of a W14x38 rolledsteel beam is increased byattaching a plate to its upperflange.
Determine the moment of inertiaand radius of gyration with respectto an axis which is parallel to the
plate and passes through thecentroid of the section.
SOLUTION:
Determine location of thecentroid of composite sectionwith respect to a coordinatesystem with origin at thecentroid of the beam section.
Apply the parallel axis theoremto determine moments ofinertia of beam section andplate with respect to compositesection centroidal axis.
Calculate the radius of gyrationfrom the moment of inertia ofthe composite section.
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Sample Problem 4.7
SOLUTION:
Determine location of the centroid ofcomposite section with respect to acoordinate system with origin at thecentroid of the beam section.
12.5095.17
0011.20SectionBeam
12.50425.76.75Plate
in,in.,in,Section 32
AyA
AyyA
in.792.2in17.95in12.50 2
3
AAyYAyAY
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Sample Problem 4.7
Apply the parallel axis theorem todetermine moments of inertia of beamsection and plate with respect to compositesection centroidal axis.
4
2343
1212
plate,
4
22sectionbeam,
in2.145
792.2425.775.69
in3.472
792.220.11385
AdII
YAII
xx
xx
Calculate the radius of gyration from themoment of inertia of the composite section.
2
4
in17.95
in5.617
A
Ik xx in.87.5xk
2.1453.472plate,sectionbeam, xxx III
4
in618xI
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Sample Problem 4.8
Determine the moment ofinertia of the shaded area
with respect to thexaxis.
SOLUTION:
Compute the moments of inertiaof the bounding rectangle andhalf-circle with respect to thexaxis.
The moment of inertia of theshaded area is obtained bysubtracting the moment of inertiaof the half-circle from the momentof inertia of the rectangle.
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Sample Problem 4.8SOLUTION: Compute the moments of inertia of the
bounding rectangle and half-circle withrespect to thex axis.
Rectangle: 46
313
31 mm102.138120240 bhIx
Half-circle:moment of inertia with respect toAA,
464814
81 mm1076.2590 rI AA
23
2
2
12
2
1
mm1072.12
90
mm81.8a-120b
mm2.383
904
3
4
rA
ra
moment of inertia with respect tox,
46
362
mm1020.7
1072.121076.25
AaII AAx
moment of inertia with respect tox,
46
2362
mm103.92
8.811072.121020.7
AbII xx
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Sample Problem 4.8
The moment of inertia of the shaded area isobtained by subtracting the moment of inertiaof the half-circle from the moment of inertia ofthe rectangle.
46mm109.45 xI
xI 46mm102.138 46mm103.92
Distributed Force: Moment of Inertia of Areas