7/29/2019 11 Properties

1/9

Geometric properties Copyright Prof Schierle 2011 1

7/29/2019 11 Properties

2/9

Geometric properties Copyright Prof Schierle 2011 2

Geometric propertiesType of property: Defines:1. Cross section area A Axial stress fa and shear stress fv2. Centroid C Center of mass (Neutral Axis)

3. Moment of Inertia I Bending stress f b and deflection

4. Polar Moment of Inertia J Torsion stress

5. Section Modulus S Max. bending stress f b (S = I/c)

6. Radius of Gyration r Column slenderness r = (I/A)1/2

Todays topics:

Centroid Centroidal Moment of Inertia, etc.Parallel Axis Theorem Moment of Inertia of composite sections

7/29/2019 11 Properties

3/9

Geometric properties Copyright Prof Schierle 2011 3

CentroidCentroid is the center of mass of a body or surface area.

Beam centroid is theNeutral Axisof zero bending stress.Centroid also defines distributed loadcenter of mass, etc.

1 CentroidCof freeform body

2 CentroidCof 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 balancedat the centroid.

Defining the total areaA =dawith lever arms xandyfrom an arbitrary origin to partial areas dawith lever

armsxandyto that origin, yields:

Mx = 0 xA - x da = 0

A =da xda =x da

x= x da / day= y da / da

7/29/2019 11 Properties

4/9

Geometric properties Copyright Prof Schierle 2011 4

Centroid

Beam centroid example

Assume:A1 = 8x2 A1 = 16 in2

A2 = 2 x 2 x 6 A2 = 24 in2

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)

18440

723242

1127161

A Y (in3)Y (in)A (in2)Part

Y = 184 / 40 Y = 4.6

X=8/2=4

7/29/2019 11 Properties

5/9

Geometric properties Copyright Prof Schierle 2011 5

1. T-beam centroid

14828

3632x6 = 12211278x2 = 161

A Y (in3)Y (in)A (in2)Part

Y = 148 / 28 Y = 5.29

2. Facade centroid

48,000,000180,000

12,000,0002002x100x600/2 = 60,0002

36,000,000300200x600 = 120,0001

A Y (ft3)Y (ft)A (ft2)Part

Y = 48,000,000/180.000 Y = 267

3. Plan centroid (eccentricity = seismic torsion

)

13,892284

13,824642x2x(34+20) = 2162

6812x34 = 681

A X (ft3)X (ft)A (ft2)Part

X = 13,892 / 284 X = 48.9

7/29/2019 11 Properties

6/9Geometric properties Copyright Prof Schierle 2011 6

Parallel Axis Theorem

The Parallel Axis Theorem is used to find the

Moment of Inertia for composite sections.

1. Beam for derivation2. T-beam

3. Box beam

Consider the basic Moment of Inertia equation

I =ay2Referring to diagram 1 yields:

Ix=a(y+y)2 =ay2 +2ayy +ay2

Ix=ay2+ 2yay +ay2whereay = 0 since the partial moments above and

below the centroid axis 0-0 cancel out.

Hence:

Ix=ay2+ay2Sinceay2= IoIx=(I0+ay2)The Moment of Inertia of composite beams is the sum of

moment of inertia of each part + the cross section areaof each part times their lever arm to the centroid squared.

(a+b)2 = a2 + 2 ab + b2

a b

a

b

a2

b2

ab

ab

7/29/2019 11 Properties

7/9Geometric properties Copyright Prof Schierle 2011 7

Parallel Axis Theorem examples

T-beam

136Ix=

842x63/12 = 36482122

526x23/12 = 4482121

Ix(in4)I0 = bd

3/12 (in4)Ay2 (in4)Y (in)A (in2)Part

7/29/2019 11 Properties

8/9Geometric properties Copyright Prof Schierle 2011 8

Parallel Axis Theorem examples

T-beam

136Ix=

842x63/12 = 36482122

526x23/12 = 4482121

Ix(in4)I0 = bd

3/12 (in4)Ay2 (in4)Y (in)A (in2)Part

Box-beam(2 MC13x50, A= 2x14.7 = 29.4, I= 2x 314 = 628)

1610Ix=

9822x10x13/12= 29807202

6286280029.41

Ix(in4)I0(in

4)Ay2 (in4)Y (in)A (in2)Part

AISC Table: MC13x50 channel (AISC =American Institute ofSteel Construction)

I =

7/29/2019 11 Properties

9/9Geometric properties Copyright Prof Schierle 2011 9