2nd Moment of Area Tutorial PDF

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2nd Moment of Area The 2nd Moment of Area is an important concept in beam analysis. Essentially this is the geometrical bending stiffness of the cross section. It is a function of both the cross section geometry and the bend axis. It is formally defined as : Ixx = ∫A y2 dA Consider a rectangular cross section b

Then we have Ixx = ∫A y2 dA dy

But dA (the y

x

So :

s shown:

F

d area) is b dy x

Ixx = ∫A y2 b Ixx = [ b y3 / 3 ]-d/2d/2 Ixx = ( b d3 / 24b ) – ( - b d3 / 24 ) = b d 3 / 12 Note that here b is parallel to the bend axis and d perpendicular. Thus by rotating an object through 90o we can change its rigidity about an axis dramatically. Eg. A ruler of approx rectangular cross section, 3mm thick and 30mm broad is : x

y

y

Ixx = (30 x 33) / 12 = 67.5 mm4 Iyy = (3 x 303) / 12 = 6750 mm4

x

ie. our ruler is 100 times more rigid about one axis than the other.

Consider now another typical engineering cross section – a round tube : The value of Ixx for a solid circle can be found as before from first principles using Ixx = ∫A y2 dA. x

x

This gives Ixx = π/4 r4 For a hollow tube we can simply subtract

rinner router

Ixx total = Ixx outer – Ixx inner Ixx total = π/4 ( r outer 4 - r inner 4 )

What about an H section ? We can split it into three rectangles and simply add Ixx for each. Ie.

d2 x

x

d1, d3

Ixx total = Ixx 1 + Ixx 2 + Ixx 3 Ixx total =

b1

b2

b3

(b1 d13 /12) + (b2 d23 /12) + (b3 d33 /12)

What about an I shaped section ? 1

cc1

5mm 2

x

18mm

x 6mm 3

cc3

5mm 20mm

There is a key difference between this and the previous examples of compound shape cross sections. Previously, the centroidal axis of each of the elements in our cross section coincided with the centroid of the whole cross section. In the case of rectangles 1 and 3, their own centroidal axis (cc) is offset from the centroidal axis of the whole shape (xx).

As I is a function of not only the cross sectional geometry but also the reference axis we can not simply add terms based on different axes. Instead we use the parallel axis theorem. This allows us to refer elements back to the main centroidal axis. Ie. Ixx = Icc + A a2 where “A” is the element area and “a” the perpendicular distance between the parallel axes. Thus for our I shaped section : For element 1 : Ixx1 = Icc1 + A1 a12 Ixx1 = (20 x 53) / 12 + (20 x 5) (11.5)2

(The distance between xx & cc is 11.5mm)

Ixx1 = 13433 mm4 For element 2 : Ixx2 = Icc2 + A2 a22 But in this case a2 is zero Ixx2 = (6 x 183) / 12

Ixx2 = 2916 mm4 For element 3 : Ixx3 = Icc3 + A3 a32 Ixx3 = (20 x 53) / 12 + (20 x 5) (11.5)2

(The distance between xx & cc is 11.5mm)

Ixx1 = 13433 mm4 (as element 1) Ixx total = 13433 + 2916 + 13433 = 29782 mm4, or 2.978 E-8 m4

What about a T section ?

Our problem here is we can’t use symmetry to determine the location of the centroidal axis xx.

15mm c

1

c

x

x

c

c

6mm

Instead use : 18mm

2

5mm

So first find location of xx.

ŷ = Σ (Ay) / Σ A where ŷ is the vertical position of axis xx and y the position of the cc axes of the constituent elements.

ŷ = Σ (Ay) / Σ A In our case ŷ = ( A1y1 + A2y2) / (A1 + A2) We will use the base of the T as our datum for measuring y. ŷ = (( 15 x 6 x 21) + (5 x 18 x 9)) / ( (15 x 6) + (5 x 18)) ŷ = 15mm, ie. xx is 15mm from the base of the T. So we can now work out I

For element 1 : Ixx1 = Icc1 + A1 a12 Ixx1 = (15 x 63) / 12 + (15 x 6) (6)2 (The distance “a” between xx & cc1 is 21-15 = 6mm)

Ixx1 = 3510 mm4 For element 2 : Ixx2 = Icc2 + A2 a22 Ixx2 = (5 x 183) / 12 + (5 x 18) (6)2 (The distance “a” between xx & cc2 is 15 – 9 = 6mm – nb : coincident the same as 1)

Ixx2 = 5670 mm4 Itotal = 5670 + 3510 = 9180 mm4 = 9.18e-9 m4...