calculating the center of gravity

Precast Show 2017

NPCA 1

precast.org/education

CALCULATING THE CENTER OF GRAVITY

Claude Goguen, PE, LEED AP

Director of Sustainability and Technical Education


OUTLINE

• Understanding the concepts of gravity

• CoG of 2 dimensional figures

• CoG of 3 dimensional figures

• Examples


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• Moments are used to determine the center of gravity

(CG) of irregular or asymmetrical bodies

• CG of symmetrical bodies lie on their axes of

symmetry

First Moment of Area (Q)

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•Moment of area around an axis is the product of

surface area by the distance of CG from that axis

Moment and Center of Gravity

2b

2ay

x

First moment of area around “x” axis: ))(4( babQx

surface area

Distance from

CG to “x” axis

First moment of area around “y” axis: ))(4( aabQy


Find Qx and Qy

Example

4”

2”

x

y

CG

0.67”

1.33”


Find the CG of the following object

Example #6

3”

2”

1”

1”

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Example

Step 1: Look for an axis of symmetry. CG must lie on

the axis of symmetry

3”

2”

1”

1”

Axis of symmetry


Example

Step 2: Determine the moment of area around a

horizontal axis (for example axis x-x)

3”

2”

1”

1”

x x

5.9)1)(12(

)5.2)(13(

xxQ

in3

5)12()13( A in2

Surface area:


Example

Step 3: Divide Qxx by surface area to determine the

location of CG

3”

2”

1”

1”

x x

9.15/5.9 in

AQxx /

1.9”

Neutral Axis (N.A.)

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Why Locating the CG Is Important?

1- CG must always be below the main hook

(A) Stable:

Hook is above CG

(B) Unstable:

Hook is not above CG

W

Creates unbalanced

moment that tilts the

member

Rigging for Commercial Construction © 1983, Reston Publishing Co.

W



(A) Stable:

Hook is above CG

(B) Unstable:

Hook is not above CG

W

Rigging for Commercial Construction © 1983, Reston Publishing Co.

W

(C) Load will shift until the

CG is below the hook

W


precast.org/educationPicture from: W.E. Rossnagel, Handbook of Rigging © 1964, McGraw-Hill


This may

require using

one or more of

the sling

cables with an

odd length

which is not

readily

availableWhen slings of equal length are used

Load tilts until CG is

directly below hook


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NPCA 6

precast.org/educationPicture from: W.E. Rossnagel, Handbook of Rigging © 1964, McGraw-Hill


Manually

adjusting the

length of the

sling cables

CG

After adjusting sling to

proper length, install a

cable clip at this position


precast.org/educationRigging for Commercial Construction © 1983, Reston Publishing Co.

2- The lifting points must always be above the CG

Lifting points

Unstable: The load can

topple because the

attachments are below CG

Stable: Attachments

are above CG

Load in this

leg is critical


precast.org/educationRigging for Commercial Construction © 1983, Reston Publishing Co.

3- The CG must reside between the sling cables

Stable when sling

angle (A) is much

larger than angle (B)

Unstable when

sling angle (A) is

less than angle (B)


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precast.org/educationW.E. Rossnagel, Handbook of Rigging © 1964, McGraw-Hill

3- The CG must reside between the sling cables

Increasing cable angle will increase stability



Cable Angle

- Tension in the sling cables

depends on their angle as well as

the weight of the load

W=2000 lbs

F F

W

Fy

Fx

45o

FFFx 71.0)45cos(

FFFy 71.0)45sin(

lbsFyFyW 10002

lbsFyF 141471.0/

Tension in the cables:


- Tension in the sling cable depends on its angle as well

as the weight of the load

W.E. Rossnagel, Handbook of Rigging © 1964, McGraw-Hill

- It is recommended to use 45o or larger angles

Cable Angle

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NPCA 8


Remember: CG always lies on the axis of symmetry

a

a/3 2a/3

b

b/3

2b/32b/3b/3

Locating the Center of Gravity


Example 3: Determine the CG of the following object

x

y3’

2’

2’

Qx = (Area)(CG from x axis)

Qy = (Area)(CG from y axis)

Method 2 Using the first moment of area (Q)

Locating the Center of Gravity


Example

x

y3’

2’

2’

Rectangle

Triangle

Area (ft2)

6

2

CG from x

axis (ft)

CG from y

axis (ft)

1 1.5

3.670.67

Qx (ft3) Qy (ft3)

Total 8

6 9

1.33 7.34

7.33 16.34

Qx = (Area)(CG from x axis)

Qy = (Area)(CG from y axis)

(CG from x axis) = Qx/(Area) = 7.33/8 = 0.92 ft

(CG from y axis) = Qy/(Area) = 16.34/8 = 2.04 ft

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Example: Determine the CG of the following object

6’

15’

6’

0.5’

Example

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NPCA 10



1) Plain of symmetry –

so CG must reside on

this plain

2) We can break the

object into two triangles

(sides), and a rectangle

(bottom).

Back side is mostly

empty (hole) so we can

ignore.

6’

15’

6’

0.5’

x

y

z

Example



Rectangle

Triangle

Vol (ft3)

CG from z

axis (ft)

CG from y

axis (ft) Qz (ft4) Qy (ft4)

Total

Triangle

Example

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Rectangle

Triangle

Vol (ft3)

CG from z

axis (ft)

CG from y

axis (ft) Qz (ft4) Qy (ft4)

Total

Triangle

Example

(CG from x axis) = Qx/(Vol) =

(CG from y axis) = Qy/(Vol) =



6’

15’

6’

0.5’

Example


SUMMARY

• Look over plant drawings and make sure you understand every detail

• Are they legible and durable?

• Determine is you can simplify drawings?

• Know your tolerances

• Key is to eliminate confusion and potential for errors

• Training and Education!!!

• ASK QUESTIONS!!!

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AVOID MISCOMMUNICATIONS


Questions?


THANK YOU!

Any more questions?

Contact Claude at

317-571-9500 or at

[email protected]

mailto:[email protected]

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NPCA 13


CALCULATING THE CENTER OF GRAVITY

Claude Goguen, PE, LEED AP

Director of Sustainability and Technical Education

calculating the center of gravity

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