Question

3. For the following composite area shown below. Shaded regions have material while white regions are empty. Include proper units. a. Find the location of the centroid measured from the shown X and Y axes. b. Calculate the moment of inertia and radius of gyration about the indicated axes fYc centroidal Ixc=bh3/12 and lyc = hb3/12 h Xc b 4Yc centroidal Ixe = 1 r*/4 and Iyo = r4/4 Xc Kx = = TEM Ky = { 6" X- (5 points) Y = (5 points) 899 Ix - (5 points) Iy = (5 points) K, (5 points) 3 о с A Prisc Insert F6 F7 F8 F9 F10 F11 F12 % &

Answer 1

LOCATION OF CENTROIDS

Naming the shaded spaces, we get

AY 2" 6” de 2" lc B D H D E 89 х חד G 3

Now the areas and centroids are as follows. The -ve area means that the area is to be subtracted from the geometry

Part Name	Area (in^2)	Centroid (x,y)	Ax (in^3)	Ay (in^3)
ABCD	10	5,8.05	50	80.5
EFGH	48	5,4	240	192
Circle I	-3.142	5,4	-15.71	-12.56

Thus

$\bar X=\frac{\sum Ax}{\sum A}=\frac{274.29}{54.858}=5$

$\bar Y=\frac{\sum Ay}{\sum A}=\frac{259.94}{54.858}=4.74$

MOMENT OF INERTIA

Let us add the centroidal axes of each part parallel to X axis to the diagram

2" 6" to 2" , C B M- 11" M А H D E N--78 - - N х G F 3"

Before we start, the parallel axis therom says that

$I_a=I_b+At^2$

where

• $I_a$ is the moment of inertia about axis 'a'
• $I_b$ is the moment of inertia about centroidal axis 'b'
• A is the area of the structure.
• t is the perpendicular distance between axis 'a' and 'b'

Now let us proceed to find the moment of inertia about X axis of the individual areas

Part Name	Area (in^2)	$I_{axis}$ (in^4)	t (in)	$I_{X}$ (in^4)
ABCD	10	bh 10 * 13 IM == - 0.833 12 12	8.05	$0.833+10*8.05^2=648.86$
EFGH	48	$I_N=\frac{bh^3}{12}=\frac{6*8^3}{12}=256$	4	$256+48*4^2=1024$
Circle I	-3.142	$I_N=-\frac{\pi r^4}{4}=-\frac{\pi}{4}=-0.785$	4	$-0.785-3.142*16=-51.05$

Thus moment of inertia about X axis is

$I_X=648.86+1024-51.05=1621.81~in^4$

Let us add the centroidal axes of each part parallel to Y axis to the diagram

६ Y 2" 6" 2" C B A H D E 1" 8" 4" X G F 3 W

Now let us proceed to find the moment of inertia about X axis of the individual areas

Part Name	Area (in^2)	$I_{axis}$ (in^4)	t (in)	$I_{Y}$ (in^4)
ABCD	10	$I_W=\frac{hb^3}{12}=\frac{1*10^3}{12}=83.33$	5	$83.33+10*5^2=333.34$
EFGH	48	$I_N=\frac{hb^3}{12}=\frac{8*6^3}{12}=144$	5	$144+48*5^2=1344$
Circle I	-3.142	$I_N=-\frac{\pi r^4}{4}=-\frac{\pi}{4}=-0.785$	5	$-0.785-3.142*5^2=-79.32$

Thus moment of inertia about X axis is

$I_Y=333.34+1344-79.32=1598.02~in^4$

RADIUS OF GYRATION

Using the formulae provided, we get

• $K_x=\sqrt{\frac{I_x}{A}}=\sqrt{\frac{1621.81}{54.858}}=5.437~in$
• $K_y=\sqrt{\frac{I_y}{A}}=\sqrt{\frac{1598.02}{54.858}}=5.397~in$

*A humble request* - If you have any doubt, please use the comment section to communicate. Please be a little patient, it is an honest promise I will reply as soon as possible. This will clarify your doubt, and also help me to get better at answering your next questions. At the same time, If my answer helped you, please consider leaving an upvote. I hope you understand my viewpoint. Thank you :)