ENGI 1313 Mechanics I - Memorial University of …spkenny/Courses/Undergraduate/ENGI1313/...Center...

Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of [email protected]

ENGI 1313 Mechanics I

Lecture 40: Center of Gravity, Center of Mass and Geometric Centroid

2 ENGI 1313 Statics I – Lecture 40© 2007 S. Kenny, Ph.D., P.Eng.

Material Coverage for Final ExamIntroduction (Ch.1: Sections 1.1–1.6)Force Vectors (Ch.2: Sections 2.1–2.9) Particle Equilibrium (Ch.3: Sections 3.1–3.4) Force System Resultants (Ch.4: Sections 4.1–4.10)

Omit Wrench (p.174)Rigid Body Equilibrium (Ch.5: Sections 5.1–5.7)Structural Analysis (Ch.6: Sections 6.1–6.4 & 6.6)Friction (Ch.8: Sections 8.1–8.3)Center of Gravity and Centroid (Ch.9: Sections 9.1–9.3)

Ignore problems involving closed-form integrationSimple shapes such as square, rectangle, triangle and circle


Lecture 40 Objective

to understand the concepts of center of gravity, center of mass, and geometric centroidto be able to determine the location of these points for a system of particles or a body


Center of Gravity

Point locating the equivalent resultant weight of a system of particles or bodyExample: Solid Blocks

Are both final configurations stable?

WR WR

w1

w2

w3

w5

w1

w2

w3

w5


Center of Gravity (cont.)

Resultant Weight

Coordinates

Key Property

41iwW iR K== ∑

w1

w2

w3

w4

WR

x

x1~

41iwx~Wx iiR K== ∑41iwz~Wz iiR K== ∑

zz1~

( ) 41i0x~xwM iGiG K==−= ∑∑

xG

L/2


Center of Gravity (cont.)

Generalized Formulae∑

=

=n

1iiR wW

R

n

1iii

W

wx~

x∑

==

R

n

1iii

W

wy~

y∑

==

R

n

1iii

W

wz~

z∑

==

Moment about y-axis

Moment about x-axis

“Moment”about x-axisor y-axis

z2~

zn~ z1

~


Center of Mass

Point locating the equivalent resultant mass of a system of particles or body

Generally coincides with center of gravity (G)Center of mass coordinates

∑∑=

i

ii

mmx~

x∑

∑=i

ii

mmy~

y∑

∑=i

ii

mmz~

z


Center of Mass (cont.)

Can the Center of Mass be Outside the Body?

Fulcrum / Balance

Center of Mass


Center of Gravity & Mass – Applications

DynamicsInertial termsVehicle roll-over and stability


Geometric Centroid

Point locating the geometric center of an object or bodyHomogeneous body

Body with uniform distribution of density or specific weight• Center of mass and center of gravity coincident• Centroid only dependent on body dimensions and

not weight terms

∑∑=

i

ii

AAx~

x∑

∑=i

ii

AAy~

y∑

∑=i

ii

AAz~

z


Geometric Centroid (cont.)

Common Geometric ShapesSolid structure or frame elements

GC & CM

GC & CM

Median Lines

GC & CM


Composite Body

Find center of gravity or geometric centroid of complex shape based on knowledge of simpler geometric forms


Example 40-01

Determine the location (x, y) of the 7-kg particle so that the three particles, which lie in the x−yplane, have a center of mass located at the origin O.


Example 40-01 (cont.)

Center of Mass

∑∑=

i

ii

mmx~

x

∑∑=

i

ii

mmy~

y

( ) ( )( ) ( )( )kg7kg5kg3

m4kg5m3kg3xkg70x++

−+== m57.1x =⇒

( ) ( )( ) ( )( )kg7kg5kg3

m2kg5m2kg3ykg70y++

++−== m29.2x =⇒


Example 40-02 A rack is made from roll-formed sheet steel and has the cross section shown. Determine the location (x,y) of the centroid of the cross section. The dimensions are indicated at the center thickness of each segment.



Assume Unit ThicknessIgnore bend radiiCenter-to-center distance

Centroid Equations

∑∑=

i

ii

AAx~

x∑

∑=i

ii

AAy~

y∑

∑=i

ii

AAz~

z



Centroid Equations

∑∑=

i

ii

AAx~

x∑

∑=i

ii

AAy~

y

−(15)(7.5) = 112.5−15/2 = 7.5(15mm)(1mm) = 15mm21

Area (mm2)# ( )mmx~ ( )mmy~ ( )3mmAx~ ( )3mmAy~

1x1 = 7.5mm~



Centroid Equations

∑∑=

i

ii

AAx~

x∑

∑=i

ii

AAy~

y

(50)(25) = 1250−50/2 = 25−(50mm)(1mm) = 50mm25


5y5 = 25mm~



Centroid Equations

∑∑=

i

ii

AAx~

x∑

∑=i

ii

AAy~

y

195045050+30/2 = 6515(30mm)(1mm) = 30mm26


6

y6 = 65mm~

x6 = 15mm~



Centroid Equations

∑∑=

i

ii

AAx~

x∑

∑=i

ii

AAy~

y

195045050+30/2 = 6515(30mm)(1mm) = 30mm26

9550mm35737.5mm3−−235mm2Sum3200360080/2 = 4045(80mm)(1mm) = 80mm27

1250−50/2 = 25−(50mm)(1mm) = 50mm2524009008015+30/2 = 30(30mm)(1mm) = 30mm24750112.55015/2 = 7.5(15mm)(1mm) = 15mm23

−562.5−30+15/2 = 37.5(15mm)(1mm) = 15mm22−112.5−15/2 = 7.5(15mm)(1mm) = 15mm21


1 2

3

4

5

6

7



Centroid Equations

2

3

i

ii

mm235mm5.5737

AAx~

x ==∑

∑

2

3

i

ii

mm235mm9550

AAy~

y ==∑

∑

24.4 mm

40.6 mmmm4.24x =

mm6.40y =


Example 40-03

Two blocks of different materials are assembled as shown. The densities of the materials are: ρA = 150 lb/ft3 and ρA = 400 lb/ft3. The center of gravity of this assembly.



Center of Gravity

∑∑=

i

ii

wwx~

x

∑∑=

i

ii

wwy~

y

∑∑=

i

ii

wwz~

z



Center of Gravity

∑∑=

i

ii

wwx~

x∑

∑=i

ii

wwy~

y∑

∑=i

ii

wwz~

z

505016.6733116.67B6.253.12512.52143.125A

56.2553.1329.1719.79 Σ

Weight (lb)#

( )( )( )( )( )

lb125.3ft/in12

in2in6in62/1ft/lb150w 3

3

A ==

( )( )( )( )

lb67.16ft/in12

in2in6in6ft/lb400w 3

3

B ==

( )inx~ ( )iny~ ( )inz~ ( )inlbwx~ ⋅ ( )inlbwy~ ⋅ ( )inlbwz~ ⋅



Center of Gravity

in47.179.1917.29

wwx~

xi

ii ===∑

∑

in68.279.1913.53

wwy~

yi

ii ===∑

∑

in84.279.1925.56

wwz~

zi

ii ===∑

∑


Chapter 9 ProblemsUnderstand principles for simple geometric shapes

Rectangle, square, triangle and circleNo closed form integration knowledge required

ReviewExample 9.9 and 9.10Problems 9-44 to 9-61

OmitExample 9.1 through 9.8Problems 9-1 through 9-43, 9-62, 9-67 to 9-83


References

Hibbeler (2007)http://wps.prenhall.com/esm_hibbeler_engmech_1

ENGI 1313 Mechanics I - Memorial University of …spkenny/Courses/Undergraduate/ENGI1313/...Center...

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