bmayer@chabotcollege.edu engr-36_lec-232_center_of_gravity.pptx 1 bruce mayer, pe engineering-36:...

BMayer@ChabotCollege.edu • ENGR-36_Lec-232_Center_of_Gravity.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering 36

Chp09: Center

of Gravity

Introduction: Center of Gravity

The earth exerts a gravitational force on each of the particles forming a body. • These forces can be replaced by a

SINGLE equivalent force equal to the weight of the body and applied at the CENTER OF GRAVITY (CG) for the body

The CENTROID of an AREA is analogous to the CG of a body. • The concept of the FIRST MOMENT of

an AREA is used to locate the centroid

Total Mass – General Case

Given a Massive Body in 3D Space

Divide the Body in to Very Small Volumes → dV

Each dVn is located at position (xn,yn,zn)

The DENSITY, ρ, can be a function of POSITION → nnnn zyx ,,

kzjyix nnnˆˆˆr

nnnn zyxdV ,,at

Total Mass – General Case Now since m = ρ•V, then the

incremental mass, dm

nnnnn dVzyxdm ,,kzjyix nnnˆˆˆr

nnnn zyxdV ,,at

Integrate dm over the entire body to obtain the total Mass, M

volume

n dVzyxdmM ,,

Total WEIGHT – General Case Recall that W = Mg Using the the previous

expression for Mkzjyix nnnˆˆˆr

nnnn zyxdV ,,at

Now Define SPECIFIC WEIGHT, γ, as ρ•g →

volume

dVgzyx

dVzyxggdmMgW

volume

dVgzyxW

Uniform Density Case Consider a body with

UNIFORM DENSITY; i.e.

kzjyix nnnˆˆˆr

nnnn zyxdV ,,at

Then M & W→

nmmnnn

zyxzyx

VdVdVW

VdVdVM

volume

Center of Mass Location Use MOMENTS to Locate

Center of Mass/Gravity Recall Defintion of a

MOMENT kzjyix nnnˆˆˆr

nnnn zyxdV ,,at

In the General Center-of-Mass Case• LeverArm ≡ Position Vector, rn, or its components

• Intensity ≡ Incremental Mass, dmn

IntensityleverArm Moment

sIntensitiesIntensitieLeverArmsMR

Center of Mass Location

RM in Component form

kzjyix nnnˆˆˆr

nnnn zyxdV ,,at

Now Define the IncrementalMoment, dΩn

kZjYiXR MMMMˆˆˆ

nnnnn gdVkzjyixd ˆˆˆΩ

LeverArm

Intensity

Center of Mass Location Integrating dΩ to Find Ω for

the entire body

kzjyix nnnˆˆˆr

nnnn zyxdV ,,at

volume

gdVkzjyix

ˆˆˆ

yall z allx all

nnnnnn

dVzkgdVyjgdVxig

ˆˆˆ

ˆˆˆΩ

Center of Mass Location

Now Equate Ω to RM•Mg

kzjyix nnnˆˆˆr

nnnn zyxdV ,,at

Canceling g, and equatingComponents yields, for example, in the X-Dir

LeverArm

Intensity

y all z all xall

ˆˆˆ

nnnnnn

dVzkgdVyjgdVxig

kZjYiXMgMg

nnM dVxMX

Center of Mass Location Divide out the Total Intensity,

M, to Isolate the Overall X-directed Lever Arm, XM

kzjyix nnnˆˆˆr

nnnn zyxdV ,,at

And the Similar expressions for the other CoOrd Directions

x all yall

Center of Gravity of a 2D Body Centroid of an Area

Taking Incremental Plate Areas, Forming the Ωx & Ωy, Along With the Fz=W Yields the Expression for the Equivalent POINT of W application• Note Ω Units = In-lb or N-m

Centroid of a Line

Centroids of Areas & Lines Centroid of an Area

axis x to.t.moment w.r1st

axisy .t.moment w.r1st

dAtxAtx

dLaxLax

For Plate of Uniform Thickness

• Specific Weight• t Plate Thickness• dW = tdA

Wire of Uniform Thickness

• Specific Weight

• a X-Sec Area

• dW = a(dL)

Centroid of a Line

First Moments of Areas & Lines• An area is symmetric with respect to an axis

BB’ if for every point P there exists a point P’ such that PP’ is perpendicular to BB’ and the Area is divided into equal parts by BB’.

• The first moment of an area with respect to a line of symmetry is ZERO.

• If an area possesses a line of SYMMETRY, its centroid LIES on THAT AXIS

• If an area possesses two lines of symmetry, its centroid lies at their INTERSECTION.

• An area is symmetric with respect to a center O if for every element dA at (x,y) there exists an area dA’ of equal area

at (−x, −y). • The centroid of the area coincides

with the center of symmetry, O.

Centroids of Common Area Shapes

Centroids of Common Line Shapes

Recall that for a SMALL Angle, α

sin rx

Composite Plates and Areas• Composite plates

• Composite area

332211 WxWxWxWX

332211 AyAyAyAY

Example: Composite Plate

For the plane area shown, determine the first moments with respect to the x and y axes, and the location of the centroid.

Solution Plan• Divide the area into a triangle,

rectangle, semicircle, and a circular cutout

• Calculate the first moments of each area w/ respect to the axes

• Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout

• Calc the coordinates of the area centroid by dividing the NET first moment by the total area

mm102506

mm107757

x Find the total area and first moments of the

triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout

Find the coordinates of the area centroid by dividing the first moment totals by the total area

Solution

mm1013.828

mm107.757

mm 8.54X

mm1013.828

mm102.506

mm 6.36Y

Centroids by Strip Integration

dyxaxa

dAydydxydAyAy

dAxdydxxdAxAx

el• Double integration to find the first

moment may be avoided by defining dA as a thin rectangle or strip.

Example: Centroid by Integration

Determine by direct integration the location of the centroid of a parabolic spandrel.

Solution Plan• Determine Constant k• Calculate the Total Area• Using either vertical or

horizontal STRIPS, perform a single integration to find the first moments

• Evaluate the centroid coordinates by dividing the Total 1st Moment by Total Area.

Example: Centroid by Integration Solution

• Determine Constant k

• Calculate the Total Area

Strips Vertical Use

2 a when b and ;

bxdxyxdAx

Solution • Calc the 1st Moments, Ωi

Using vertical strips, perform a single integration to find the first moments.

2baabx

2ababy

Evaluate the centroid coordinates• Divide Ωx and Ωy

by the total Area

Finally the Answers

Theorems of Pappus-Guldinus

ydLLyLyA

ydLydLA

Surface of revolution is generated by rotating a plane curve about a fixed axis.

Area of a surface of revolution is equal to the length of the generating curve, L, times the distance traveled by the centroid through the rotation.

Theorems of Pappus-Guldinus

AyydAV

ydAdVV

Body of revolution is generated by rotating a plane area about a fixed axis.

Volume of a body of revolution is equal to the generating area, A, times the distance traveled by the centroid through the rotation.

Example: Pappus-Guldinus

The outside diameter of a pulley is 0.8 m, and the cross section of its rim is as shown. Knowing that the pulley is made of steel and that the density of steel, = 7850 kg/m3, determine the mass and weight of the rim.

Solution Plan• Apply the theorem of Pappus-

Guldinus to evaluate the volumes of revolution for the rectangular rim section and the inner cutout section.

• Multiply by density and acceleration of gravity to get the mass and weight.

Example: P-G

3393633 mmm10mm1065.7mkg1085.7Vm kg 0.60m

2sm 81.9kg 0.60mgW N 589W

Apply Pappus-Guldinus to Sections I & II

Subtract: I-II

WhiteBoard Work

Find theAreal &Lineal

Centroids

Areafor Find

LineOutSidefor Find

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering 36

Appendix 00

bmayer@chabotcollege.edu engr-36_lec-232_center_of_gravity.pptx 1 bruce mayer, pe engineering-36:...

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