aect360 lecture 4

8/14/2019 AECT360 Lecture 4

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Lecture 4 - Page 1 of 9

Lecture 4 Section Properties (symmetric shapes)

1. Area

Area has units of length squared typically in 2, ft2, mm 2, etc.

a) Rectangle:

Area = base x height

b) Circle:

Area = 24

Dia

Circumference = D

Hollow circle area = A outer - A inner

c) Triangle:

Area = (Base)(Height)

d) Trapezoid:

Area = (Height 1 + Height 2)(Base)

e) Hollow shapes:

Area = A outer - A inner

Base

Height 1Height 2

Base

Height

Dia

Base

Height


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2. Location of Centroid

The centroid of a shape is the location of the center of its weight . It is thepoint on a shape where you could balance it in equilibrium. It is the locationof the intersection of the neutral axes in the primary X and Y axes. The

location of the centroid is essential for determining other cross-sectionalproperties such as moment of inertia.

a) Rectangle:

y = (Height)

x = (Base)

b) Circle:

y = (Dia)

c) Triangle:

y =31 (Height)

x =31

(Base)

X axis

x

Height

Base

y

y

x

y

Dia

Neutral axes Y a

x i s


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3. Moment of Inert ia

The moment of inertia, or I is a measure of the stiffness of the cross-sectionalshape. It is the property which explains why, for example, a 2x10 wood joist isoriented vertically rather than flat to carry more load. It is sometimes referred to

the second moment of area. The moment of inertia has units of length raised tothe 4 th power typically in 4 of mm 4. The moment of inertia is typically measuredabout the strong axis (usually the X axis) and the weak axis (usually the Y axis).

a) Rectangle:

12

3bh I x = where b = base and h = height

12

3hb I y =

b) Circle:

64

4 D I I y x

== where D = Diameter

c) Triangle:

36

3bh I

x =

d) Symmetric Hollow Shapes:

I = I outer - I inner

Height h

Base b

Y a

x i s

x

y

Neutral axes

X axis

x

Height h

Base b

y


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4. Section Modulus

The section modulus or S is also a measure of the stiffness of a cross-sectional shape. It is sometimes more convenient to use this rather than momentof inertia for calculations such as bending stresses. It has units of length raisedto the 3 rd power typically in 3 or mm 3. Do NOT get it confused with volume!!Similar to moment of inertia, the section modulus is typically measured about thestrong axis (usually the X axis) and the weak axis (usually the Y axis).

General formula for determining section modulus:

y

I S x x = where I x = moment of inertia about x axis

y = distance to neutral axis

x

I S y y = where I y = moment of inertia about y axis

x = distance to neutral axis

a) Rectangle:

62

12 2

3

bh

h

bh

S x =

=

similar S y =

b) Other shapes:

Section modulus shall be determined as dividing moment of inertia bydistance to neutral axis as mentioned above.

Base b

Height h

Neutral axes

X axis


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5. Radius of Gyration

The radius of gyration is used most often in column design as a measure of across-sectional shapes ability to resist buckling . It has units of length typically inches or mm.

General formula for determining radius of gyration:

A

I r x x = where I x = moment of inertia about x axis

A = cross-sectional area

A

I r y y = where I y = moment of inertia about y axis

A = cross-sectional area


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Example 1:

GIVEN: A wood 2x8 (actual 1 x 7)REQUIRED: Determine

a) Area

b) Location of centroidc) Moment of inertia about strong and weak axesd) Section modulus about strong and weak axese) Radius of gyration about strong and weak axes.

a) Area:

A = bh= (1.5)(7.25)= 10.88 in 2

b) Location of centroid:y = (Height)

= (7.25)= 3.625

x = (Base)= (1.5)= 0.75

c) Moment of inertia:

12

3bh I x = where b = base and h = height

= (1.5)(7.25) 3 12

= 47.63 in 4 Strong axis

12

3hb I y =

= (7.25)(1.5) 3 12

= 2.04 in 4 Weak axis

b = 1

h = 7


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d) Section modulus:

62

12 2

3

bhh

bhS x ==

6)"25.7)("5.1(

2"25.7

12)"25.7)("5.1( 2

3

== xS

= 13.14 in 3 Strong axis

62

12 2

3

hbb

hbS y ==

6)"5.1)("25.7(

2"5.1

12)"5.1)("25.7( 23

== yS

= 2.72 in 3 Weak axis

e) Radius of gyration:

A

I r x x =

2

4

88.1063.47

inin

r x =

= 2.09 in. Strong axis

A

I r

y y

=

2

4

88.1004.2

ininr y =

= 0.43 in. Weak axis


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Example 2:

GIVEN: A 6 O.D. hollow steel pipe with wall thicknessREQUIRED: Determine

a) Area

b) Location of centroidc) Moment of inertia about strong and weak axesd) Section modulus about strong and weak axese) Radius of gyration about strong and weak axes.

a) Area:

A tot = A out A in

=inner outer

Dia Dia 22

44

=inner outer

22 )"

21

5(4

)"6(4

= 28.27 in 2 23.76 in 2

A tot = 4.51 in 2

b) Location of Centroid:

Centroid location y = ..21

DO

= )"6(21

Centroid location y = 3 from bottom of pipe

6

y


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c) Moment of Inertia about Strong & Weak Axis:

inout tot I I I =

6464

44inout D D

= where D = Diameter

64

)"21

5(

64)"6(

44

=

= 63.62 in 4 44.92 in 4

I tot = 18.70 in 4

d) Section Modulus about Strong & Weak Axis:

y

I S tot =

"370.18 4in=

S = 6.23 in 3

e) Radius of Gyration about Strong & Weak Axis:

tot

tot x A

I r =

2

4

51.470.18

in

in=

r = 2.04 in

aect360 lecture 4

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