COMPOSITE MOMENT OF INERTIA. Built-up Timber I-beams Courtesy Electronic Journal of Polish Agricultural Universities

Transcription

1 COMPOSITE MOMENT OF INERTIA Built-up Timber I-beams Courtesy Electronic Journal of Polish Agricultural Universities

2 Moment of Inertia for Composite Areas Moments of inertia are additive if they reference the same ais. That is: n I = i=1 I i and n I y = i=1 I y i We can use this to our advantage for determination of composite cross sections. For our discussion, a composite cross section is one comprised of mutiple simple geometric shapes. One of the simplest composite shapes is a round or rectangular tube. 2

3 Moment of Inertia for Composite Areas Consider the square tube shown below. c h H b B From the appendi, we know the moment of inertia of a rectangle about each of its centroidal ais is: I = bh3 I y = hb3 3

4 Moment of Inertia for Composite Areas The square tube can be modeled as two concentric rectangles with a common - and y-ais. This allows the moment of inertia of each shape to be added algebraically. Since the interior rectangle is a 'hole', treat this as a negative area and add a negative area and a negative moment of inertia. c b B h H I = BH 3 bh3 I y = HB3 hb3 4

5 Parallel Ais Theorem for Moment of Inertia Since moments of inertia can only be added if they reference the same ais, we must find a way to determine the moments of inertia of composite sections when this is not the case. An eample of this is the concrete T-beam shown. Although it is a simple matter to determine the moment of inertia of each rectangular section that makes up the beam, they will not reference the same ais, thus cannot be added. However, if we found the moment of inertia of each section about some reference ais such as the centroidal ais of the composite, then we could add the individual moments of inertia. 5

6 Parallel Ais Theorem for Moment of Inertia The parallel ais theorem is used to do just that. Consider the following area with a known centroid. y a Let { = a y = b } c da b Then { da = A = da A } = 0 location = 0 of centroid I = A y 2 da = A I = I c b 2 A 6 b 2 da = 2 da A I c 2b da A 0 b 2 da A Area

7 Parallel Ais Theorem for Moment of Inertia y a c da b I = I c b 2 A I y = I yc a 2 A Thus, the area moment of inertia with respect to any ais in its plane is equal to the moment of inertia with respect to the parallel centroidal ais plus the product of the area and the square of the distance between the two ais. 7

8 Eamples for Moment of Inertia Eample 4: Given the moment of inertia of a rectangle about its centroidal ais, apply the parallel ais theorem to find the moment of inertia for a rectangle about its base. I c = b h3 c h distance = h 2 area = b h I = b h3 h 2 2 b h b = b h3 b h3 4 = b h3 3 b h3 = b h3 3 8

9 Eamples for Moment of Inertia Eample 5: Find the centroidal moment of inertia for a T-shaped area. 1) First, locate the centroid of each rectangular area relative to a common base ais, then... 2)...determine the location of the centroid of the composite yc 1 5 = 0 = y i A i A i (by symmetry) = = 608 = 3.4 in

10 Eamples for Moment of Inertia 3) Find centroidal moment of inertia about the -ais I c = I 1 I 2 I 1 = 2 63 where I i = b i h i d 2 i b i h i = in 4 I 2 = = in 4 I c = I 1 I 2 = = in

11 Eamples for Moment of Inertia 4) Find the centroidal moment of inertia about the y-ais. Since the centroidal y-ais for each shape and for the composite is coincident, the moments of inertia are additive. I yc = I y1 I y2 I yi = h 3 i b i I yc = = in4 11

12 Eamples for Moment of Inertia As sections become more comple, it is often easier to perform the calculations breating tables to find centroid and moment of inertia. Part Dimensions Area y A ya Total Part Area I Iy dy d d 2 y(a) d 2 (A) I + d 2 y(a) Iy + d 2 (A) Total

13 Eamples for Moment of Inertia Eample 6: Determine the location of the centroid ('c') of the beam's cross section and the moment of inertia about the centroidal -ais. y 1 Typ. Part Dimensions Area y ya Total in in 3 = y A A = c = 1.25 in 8 4 Part Area Ici dy d 2 y(a) Ici = Ici + d 2 y(a) (4) 3 / = (4) 3 / = (1) 3 / = 5.16 Total 16 in in 4 13

14 Eamples for Moment of Inertia Eample 7: Determine the location of the centroid 'c' the beam's cross section and the moment of inertia about both centroidal ais. y Part Dimensions Area y A ya 4 1 d1 2 d2 c 1 Typ. 8 dy1 dy Total = i A i A = 40 = y i A i A = = 3.33 in = 1.00 in Part Area I d y1 d 2 y1 (A) I + d 2 y1 (A) Iy d 2 d 2 2 (A) Iy + d 2 2 (A) Total