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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
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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
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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
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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
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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
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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
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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 =
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