Question

In: Civil Engineering

A quarter circle of radius 227 mm lies on Quadrant II of a coordinate plane with...

A quarter circle of radius 227 mm lies on Quadrant II of a coordinate plane with reference radius at the origin. A rectangle is attached to the quarter circle at Quadrant I with base at the x-axis of width equal to 45.6 mm and height same as the radius of the quarter circle. Another rectangle with inverse dimension as the previous is attached to the quarter circle at Quadrant III.

a. What is the moment of inertia (m4) of the composite area with respect to the x-axis at point O?

b. What is the moment of inertia (m4) of the composite area with respect to the y-axis at point O?

c. What is the polar moment of inertia (m4) of the composite area?

d.What is the centroidal moment of inertia of the composite area

e. What is the radius of gyration (mm) of the composite area with respect to its

centroid?

Expert Solution

Sol: Information provided in the question:

- A quarter circle of radius (R)= 227 mm ( lies on Quadrant II of a coordinate plane )

- A rectangle is attached to the quarter circle at Quadrant I :

with base at the x-axis of width (b)= 45.6 mm

and height same as the radius of the quarter circle (h)= 227 mm

- Another rectangle with inverse dimension as the previous is attached to the quarter circle at Quadrant III :

Width (b) = 227 mm

Height (h) = 45.6 mm

(a) The moment of inertia (m4) of the composite area with respect to the x-axis at point O :

First moment of inertia for A1 = R⁴/16

= x227⁴/16 = 521354730.9 mm⁴

= 5.2135 x 10^-4 m⁴

Then moment of inertia for A2 = b x h³/12

= 45.6 x 227³/12

= 44448915.4 mm⁴

=4.444 x 10^-5m⁴

Then moment of inertia for A3 = b x h³/12

= 227 x 45.6³/12

= 1793655.936 mm⁴

= 1.7936559 x 10^-6m⁴

Hence The moment of inertia (m4) of the composite area with respect to the x-axis at point O :

= 5.2135 x 10^-4 m⁴+ 4.444 x 10^-5m⁴ +1.7936559 x 10^-6m⁴

= 5.6758 x 10^-4 m⁴

(b) The moment of inertia (m4) of the composite area with respect to the y-axis at point O :

First moment of inertia for A1 = R⁴/16

= x227⁴/16 = 521354730.9 mm⁴

= 5.2135 x 10^-4 m⁴

Then moment of inertia for A2 = h x b³/12

= 227 x 45.6³/12

= 1793655.936 mm⁴

= 1.7936559 x 10^-6m⁴

Then moment of inertia for A3 = h x b³/12

= 45.6 x 227³/12

= 44448915.4 mm⁴

=4.444 x 10^-5m⁴

Hence The moment of inertia (m4) of the composite area with respect to the y-axis at point O :

= 5.2135 x 10^-4 m⁴+ 17936559 x 10^-6m⁴ + 4.444 x 10^-5m⁴

= 5.6758 x 10^-4 m⁴

(c) The polar moment of inertia (m4) of the composite area :

First polar moment of inertia for A1 = R⁴/8

= x227⁴/8 = 1042709462 mm⁴

= 1.04270946 x 10^-3 m⁴

Then polar moment of inertia for A2 = ( b x h³/12 ) + ( h x b³/12 )

= 45.6 x 227³/12 + 227 x 45.6³/12

= 44448915.4 mm⁴ + 1793655.936 mm⁴

= 4.444 x 10^-5m⁴+ 1.7936559 x 10^-6m⁴

= 4.6233655 x 10^-5m⁴

Then polar moment of inertia for A3 = ( b x h³/12 ) + ( h x b³/12 )

= 227 x 45.6³/12 + 45.6 x 227³/12

= 1793655.936 mm⁴ + 44448915.4 mm⁴

= 1.7936559 x 10^-6m⁴+ 4.444 x 10^-5m⁴

= 4.6233655 x 10^-5m⁴

The polar moment of inertia (m4) of the composite area :

= 1.04270946 x 10^-3 m⁴ + 4.6233655 x 10^-5m⁴ + 4.6233655 x 10^-5m⁴

= 1.135176 10^-3 m⁴

(d) The centroidal moment of inertia of the composite area :

For this first we have to find The Centroid of composite Area: (About X axis) with respect to origin

X' = (A1 x X1 + A2 x X2 + A3 x X3 ) / ( A1 + A2 + A3)

= ( R²/4 x (4R/3) + b x h x b/2 + bxhx b/2) / ( R²/4 + b x h + b x h)

= ( .227²/4 x (4x(-0.227)/3) + 0.0456 x 0.227 x 0.0456/2 + 0.227x0.0456x (-0.227)/2) /

( x0.227²/4 + 0.0456 x 0.227 +0.227x0.0456)

=( - 3.8990x10^-3 + 2.36007x10^-4 - 1.17486x10^-3 ) / ( 0.061173 )

X' = - 0.0790852 m

Then we have to find The Centroid of composite Area: (About Y axis) with respect to origin

Y' = (A1 x Y1 + A2 x Y2 + A3 x Y3 ) / ( A1 + A2 + A3)

= ( R²/4 x (4R/3) + b x h x h/2 + bxhx h/2) / ( R²/4 + b x h + b x h)

= ( .227²/4 x (4x(0.227)/3) + 0.0456 x 0.227 x 0.227/2 + 0.227x0.0456x (-0.0456)/2) /

( x0.227²/4 + 0.0456 x 0.227 +0.227x0.0456)

=( + 3.8990x10^-3 + 1.174859x10^-3 - 2.36007x10^-4 ) / ( 0.061173 )

Y' = 0.07908475 m

Now The centroidal moment of inertia of the composite area : (About X axis)

We use Parallel Axis theorem:

The centroidal moment of inertia for A1 = R⁴/16 + (R²/4)x((4R/3) - Y')²

= x0.227⁴/16 + .227²/4 x (4x(0.227)/3 -0.0790847)²

= 5.2135 x 10^-4 m⁴ + 1.2052 x 10^-5 m⁴

= 5.3340 x 10^-4 m⁴

Hence The centroidal moment of inertia of the composite area : (About X axis)

= 5.3340 x 10-4 m4 + 5.67000 x 10-5m4 + 1.092427 x 10-4m4

= 6.99342 10-4 m4

Y axis 45.6 mm A1 A2 227 mm X axis A3 45.6 mm 227 mm

We were unable to transcribe this image

Y axis 45.6 mm A1 A2 227 mm X axis A3 45.6 mm 227 mm

We were unable to transcribe this image

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