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S2015abnSections 1Lecture 8

Elements of Architectural StructuresARCH 614

ELEMENTS OF ARCHITECTURAL STRUCTURES:

FORM, BEHAVIOR, AND DESIGN

ARCH 614

DR. ANNE NICHOLS

SPRING 2015

eight

beam sections -

geometric properties

lecture



Center of Gravity

• location of equivalent weight

• determined with calculus

• sum element weights

∆W1 ∆W4 ∆W2 ∆W3

∑∆W y

x

z

∫= dWW



Center of Gravity

• “average” x & y from moment

∆W1 ∆W4 ∆W2 ∆W3

∑∆W y

x

z

( )W

WWx

xxWxMn

i

iiy

∆∑=⇒=∆=∑ ∑

=1

( )W

WWy

yyWyMn

i

iix

∆∑=⇒=∆=∑ ∑

=1

“bar” means average



Centroid

• “average” x & y of an area

• for a volume of constant thickness

– where is weight/volume

– center of gravity = centroid of area

( )A

Axx

∆∑=

( )A

Ayy

∆∑=

AtW ∆=∆ γ γ



Centroid

• for a line, sum up length

( )L

Lxx

∆∑=

( )L

Lyy

∆∑=

∆L



1st Moment Area

• math concept

• the moment of an area about an axis

AyQx =

( )A

Ayy

∆∑=

x

y

y

x

A (area)

AxQy =



Symmetric Areas

• symmetric about

an axis

• symmetric about

a center point

• mirrored symmetry



Composite Areas

• made up of basic shapes

• areas can be negative

• (centroids can be negative for any area)

-(-)

+⇒



Basic Procedure

1. Draw reference origin (if not given)

2. Divide into basic shapes (+/-)

3. Label shapes

4. Draw table

5. Fill in table

6. Sum necessary columns

7. Calculate x and y

Component Area

Σ

x yAx Ay

yx



Area Centroids

• Figure A.1 – pg 598

b

h

3

b

right triangle only



Moments of Inertia

• 2nd moment area

– math concept

– area x (distance)2

• need for behavior of

– beams

– columns



• about any reference axis

• can be negative

• resistance to bending and buckling

Moment of Inertia

∫=∆∑= dAxAxI iy

22

dx

y

x

elx dx

dA = y⋅dx

∫=∆∑= dAyAyI ix

22

)( 2azIor xx ∑=−



Moment of Inertia

• same area moved away a distance

– larger I

x x xx



Polar Moment of Inertia

• for roundish shapes

• uses polar coordinates (r and θ)

• resistance to twisting

θ pole

o

r

∫= dArJo

2



Radius of Gyration

• measure of inertia with respect to area

A

Ir x

x =



Parallel Axis Theorem

• can find composite I once composite

centroid is known (basic shapes)

axis through centroid

at a distance d away

from the other axis

axis to find moment of

inertia about

y

A

dA

A′

B B′

y′

d

2AdII ∑+∑=

2AdII −=

2

yx AdI +=

2AzII o +=



Basic Procedure

1. Draw reference origin (if not given)

2. Divide into basic shapes (+/-)

3. Label shapes

4. Draw table with A, x, xA, y, yA, I’s, d’s,

and Ad2’s

5. Fill in table and get x and x for composite

6. Sum necessary columns

7. Sum I’s and Ad2’s

yx Ax Ay I

I

yx

)yyd( y −=)xxd( x −=



Area Moments of Inertia

• Figure A.11 – pg. 611: (bars refer to centroid)

– x, y

– x’, y’

– C

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