Chapter 4 Distributed Forces

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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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


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


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


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


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


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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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b d d d f


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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.


CAUTION!!

You have not been giventhe thickness of the rimto calculate its volume.

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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 4.3

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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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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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

Distributed Force: Moment of Inertia of Areas

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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


AF ky

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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

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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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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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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


Chapter 4 Distributed Forces

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