Principal Moment of Inertia of Angle (L) section

Q: Determine the principal moments of inertia for the angle section oriented as shown below. Geometric Centroid of the Composite Figure F igure Referring to the figure 1, (scaled on 1:1) We split the figure into two rectangles, labeled as (1) and (2); next we find the centroid of the composite composite figure as follows. b denotes dimension along x-axis while subscript number denotes the sub-shape. d denotes dimension along y-axis while subscript number denotes the sub-shape. A denotes area while subscript number denotes the sub-shape.

Given Data: Figure 1

b1  3.5 " d 1  0.5 "  A1  b1d1  3.5 0.5  1.75 .75in 2 b1

 x1 



3.5

2

 y1  5.5 

2 d 1

 1.75 1.75 "  5.75 "

2

Figure 2

b2  0.5 " d 2  6 "  A2  3in 2  x2  3.5   y2 

d 2 2



b2 2 6 2

 3.5 

0.5 2

 3.75 "

3"

Geometric Centroid

 x 

 A1 x1  A2 x2  A1  A2



1.75 1.751.7 1.75  3 3.75 3.75 1.75  3

  x  3.01"  y 

 A1 y1  A2 y2  A1  A2



1.75  5.75  3  3 1.75  3

  y  4.01" Hence Centroid is located at  (̅ ,  )  = (3.01", 1.99") 1.99")

y

y’

4.00" 3.50"

1 1.26"

1.75”

x’ 3.01”

2

3.75”

C

0.50"

Figure 1

x

Moment of Inertia about Centroid Now we find Ix and Iy of the composite figure using parallel axis theorem.

 I x  I x1  I x 2  I x1 

b1d 13 12

 A1 (x  x1 ) 2 

3.5  0.53 12

  1.75  3.01  1.75 

  I x1  0.0364  1.75 1.26  2.81 in 4 2

  I x1  2.81 in 4 b2 d 2 3

 I x 2 

 A2 (| x x 2 |)  2

12

0.5  63

  I x1  9  3  0.74  10.64 in 2

12

 3 | 3.01  3.75 | 2

4

  I x1  10.64 in 4  Now,  I x  I x1  I x 2  2.81 10.64  13.45 in 4

  I x  13.45 in 4  I y  I y1  I y 2  Now, d1b13

 I y1 



 A1  y  y1 

12

0.5  3.53 12

2

 1.75  1.74 2

  I y1  7.08 in 4  I y 2 



d 2b23 12

6  0.53 12

 A2  y  y2 

2

 3  1.012

  I y 2  3.12 in 4  Now,  I y  I y1  I y 2

  I y  7.08  3.12  10.2 in 4   I y  10.2 in 4 Now we find the Product Moment of Inertia

2

Product of Inertia Ixy Rectangle 1  I xy  I xy   Ad x d y

 0  1.75  ( x1  x)  ( y1  y)  0  1 .75   1.75  3.01    5.75  4.01   I xy  3.84 in 4

Rectangle 2  I xy  I xy   Ad x d y

 0  3  ( x2  x)  ( y2  y)  0  3   3.75  3.01    3  4.01   I xy  2.24 in 4

Product of Inertia  I  xy  3.84   2.24 

  I xy  6.08 in 4 Now we have all the prerequisites for Principal Moment of Inertia.

Principal Moment of Inertia We have,

 I x  13.45 in 4  I y  10.2 in 4  I xy  6.08 in 4 Now,

We find the angles of inclinatino of  x, y tan 2  p 

2 I  xy

2   6.08   3.5452 ( I x  I y ) 13.45  10.02 

 2  p  tan 1  3.5452   74.248    p  37.124, 52.876 Now we find the principal moment of inertia as follows

 I max  min

 I x  I y 2

2

 I  I     x y   I  xy2  2 

  I max 

13.45  10.2 2

min

2

2  13.45  10.2       6.08  2  

2

 11.825   1.625   36.9664  11.825  6.293 Which give us,

 I max  18.118 in 4  I min  5.532 in 4 In order to find the correct orientation, we use general equation.

 I x 

 I  x 

 I x  I y 2



I x  I y 2

cos 2  I xy sin 2 

13.45  10.2 13.45  10.2

 cos  2  3 7.124    6.08  sin  2  37.124  2 2  11.825  1.625  0.271474  6.08  0.96244  18.11778  18.118 in 4

Which dictates that the angle of inclination of I max is 37.124° while angle of inclination of I min is -52.876° The oriented figure is shown on next page.

Figure 2