[email protected] ENGR-36_Lec-232_Center_of_Gravity.pptx 1 Bruce Mayer, PE Engineering-36:...

[email protected] • ENGR-36_Lec-232_Center_of_Gravity.pptx1

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

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

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

nnnn

body

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

nnnn

volume

nnnn

volume

nnnn

body

n

dVgzyx

dVgzyx

dVzyxggdmMgW

,,

,,

,,

volume

nnnn

volume

nnnn

dVzyx

dVgzyxW

,,

,,



Uniform Density Case Consider a body with

UNIFORM DENSITY; i.e.

kzjyix nnnˆˆˆr

nnnn zyxdV ,,at

Then M & W→

nmmnnn

nmmnnn

zyxzyx

zyxzyx

,,,,

,,,,

VdVdVW

VdVdVM

volume

n

volume

n

volume

n

volume

n



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

nnnn

body

n

gdVkzjyix

d

ˆˆˆ

ΩΩ

yall z allx all

nnnnnn

xyx

dVzkgdVyjgdVxig

kji

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

ˆˆˆ

nnnnnn

MMMM

dVzkgdVyjgdVxig

kZjYiXMgMg

R

x all

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

M

dVx

Xnn

M

x all

M

dVz

ZM

dVy

Ynn

M

nn

M

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

dWy

WyWy

dWx

WxWx

y

x

Centroid of a Line



Centroids of Areas & Lines Centroid of an Area

axis x to.t.moment w.r1st

axisy .t.moment w.r1st

y

x

dAyAy

dAxAx

dAtxAtx

dWxWx

dLyLy

dLxLx

dLaxLax

dWxWx

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

kk

kk

WyWY

WxWX

• Composite area

kk

kk

AyAY

AxAX

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




33

33

mm102506

mm107757

.

.

y

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

23

33

mm1013.828

mm107.757

A

AxX

mm 8.54X

23

33

mm1013.828

mm102.506

A

AyY

mm 6.36Y



Centroids by Strip Integration

ydxy

dAyAy

ydxx

dAxAx

el

el

2

dyxay

dAyAy

dyxaxa

dAxAx

el

el

2

drr

dAyAy

drr

dAxAx

el

el

2

2

2

1sin

3

2

2

1cos

3

2

dAydydxydAyAy

dAxdydxxdAxAx

el

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

3

3

Strips Vertical Use

0

3

20

22

ab

x

a

bdxx

a

bdxy

dAA

aa

2121

22

22

2 a when b and ;

yb

axorx

a

by

a

bkakb

xyxky




44

2

0

4

20

32

0

22

bax

a

bdxx

a

b

dxxa

bxdxyxdAx

aa

a

elx

Solution • Calc the 1st Moments, Ωi

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

1052

2

1

2

2

0

5

4

2

0

22

2

abx

a

b

dxxa

bdxy

ydAy

a

a

ely




43

2baabx

Ax x

ax

4

3

103

2ababy

Ay y

by

10

3

Evaluate the centroid coordinates• Divide Ωx and Ωy

by the total Area

Finally the Answers



Theorems of Pappus-Guldinus

ydLLyLyA

ydLydLA

yas2

22

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

22

2

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

AA YX

Areafor Find

&

LL YX

LineOutSidefor Find

&



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix 00

sinhT

µs

T

µx

dx

dy

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