Geometric properties Copyright Prof Schierle 2011 1

Geometric properties Copyright Prof Schierle 2011 1

Geometric properties
Type of property: Defines:
1. Cross section area A Axial stress fa and shear stress fv 2. Centroid C Center of mass (Neutral Axis)
3. Moment of Inertia I Bending stress fb and deflection �
4. Polar Moment of Inertia J Torsion stress �
5. Section Modulus S Max. bending stress fb (S = I/c)
6. Radius of Gyration r Column slenderness r = (I/A) 1/2
Today’s topics:
Centroid Centroidal Moment of Inertia, etc.
Parallel Axis Theorem Moment of Inertia of composite sections
Geometric properties Copyright Prof Schierle 2011 2

Centroid
Centroid is the center of mass of a body or surface area.
Beam centroid is the Neutral Axis of zero bending stress.
Centroid also defines distributed load center of mass, etc.
1 Centroid C of freeform body
2 Centroid C of composite cross section
(with centroid outside cross section area)
Centroid is a point where the moment of all partial areas
is zero, i. e., the area is balanced at the centroid.
Defining the total area A =�da with lever arms x’ and y’
from an arbitrary origin to partial areas da with lever
arms x and y to that origin, yields:
� Mx = 0 � x’A - � x da = 0
A = �da � x’�da = �x da
x’= �x da / �da
y’= �y da / �da
Geometric properties Copyright Prof Schierle 2011 3

X’=8/2=4”
Centroid
Beam centroid example
Assume:
A1 = 8”x2” A1 = 16 in 2
A2 = 2 x 2” x 6” A2 = 24 in 2
Y1 = 6 + 1 Y1 = 7”
Y2 = 6/2 Y2 = 3”
Due to symmetry:
X’ = 8/2 X’ = 4”
Y’= �AY / �A = (A1 Y1+A2 Y2) / (A1+A2)
Part
1
2
�
A (in 2 )
Geometric properties Copyright Prof Schierle 2011 4
16
24
40
Y (in)
7
3
A Y (in 3 )
112
72
184
Y’ = 184 / 40 Y’ = 4.6”

1. T-beam centroid
Part
1
2
�
A (in 2 )
8x2 = 16
2x6 = 12
Geometric properties Copyright Prof Schierle 2011 5
28
Y (in)
7
3
A Y (in 3 )
Y’ = 148 / 28 Y’ = 5.29”
2. Facade centroid
Part
1
2
�
A (ft 2 )
200x600 = 120,000
2x100x600/2 = 60,000
180,000
Y (ft)
300
200
Y’ = 48,000,000/180.000 Y’ = 267’
3. Plan centroid (eccentricity = seismic torsion �)
Part
1
2
�
A (ft 2 )
2x34 = 68
2x2x(34+20) = 216
284
X (ft)
1
64
112
36
148
A Y (ft 3 )
36,000,000
12,000,000
48,000,000
A X (ft 3 )
68
13,824
13,892
X’ = 13,892 / 284 X’ = 48.9’

Parallel Axis Theorem
The Parallel Axis Theorem is used to find the
Moment of Inertia for composite sections.
1. Beam for derivation
2. T-beam
3. Box beam
Consider the basic Moment of Inertia equation
I = �ay2 Referring to diagram 1 yields:
Ix= �a(y+y’) 2 = �ay2 + �2ayy’ + �ay’ 2
(a+ b) 2 = a 2 + 2 ab + b 2
a b
I x= �ay 2 + 2y�ay’ + �ay’ 2
where �ay’ = 0 since the partial moments above and
below the centroid axis 0-0 cancel out.
Hence:
I x= �ay 2 + �ay’ 2
Since �ay’ 2 = I o
I x=�(I 0+ay 2 )
The Moment of Inertia of composite beams is the sum of
moment of inertia of each part + the cross section area
of each part times their lever arm to the centroid squared.
Geometric properties Copyright Prof Schierle 2011 6
a 2
ab
ab
b 2
a
b

Parallel Axis Theorem examples
T-beam
Part
1
2
A (in 2 )
2x6 3 /12 = 36
Geometric properties Copyright Prof Schierle 2011 7
12
12
Y (in)
2
2
Ay 2 (in 4 )
48
48
I 0 = bd 3 /12 (in 4 )
6x2 3 /12 = 4
� I x=
I x (in 4 )
52
84
136

Parallel Axis Theorem examples
T-beam
Part
1
2
A (in 2 )
2x6 3 /12 = 36
Geometric properties Copyright Prof Schierle 2011 8
12
12
Y (in)
2
2
Ay 2 (in 4 )
48
48
I 0 = bd 3 /12 (in 4 )
6x2 3 /12 = 4
� I x=
Box-beam
(2 MC13x50, A= 2x14.7 = 29.4, I= 2x 314 = 628)
Part
1
2
A (in 2 )
29.4
20
Y (in)
0
7
Ay 2 (in 4 )
980
2x10x1 3 /12= 2
AISC Table: MC13x50 channel (AISC = American Institute of Steel Construction)
I =
0
I 0 (in 4 )
628
� I x=
I x (in 4 )
52
84
136
I x (in 4 )
628
982
1610

Geometric properties Copyright Prof Schierle 2011 9