(a) Explanation Calculation: Sketch the annular area as shown in Figure 1. Refer to problem 9.26P. J O = π 2 [ R 2 4 − R 1 4 ] Express the radius of gyration using the relation as follows: k O 2 = J O A (1) Substitute π 2 [ R 2 4 − R 1 4 ] for J O and π [ R 2 2 − R 1 2 ] for A in Equation (1). k O 2 = π 2 [ R 2 4 − R 1 4 ] π [ R 2 2 − R 1 2 ] k O 2 = 1 2 [ R 2 2 + R 1 2 ] (2) Express the thickness of circular section as follows: t = R 2 − R 1 Here, R 1 is inner radius and R 2 is outer radius. Find the mean radius ( R m ) using the relation as follows: ( R m ) = 1 2 ( R 1 + R 2 ) (3) Find the radius ( R 1 ) as shown below: From Equation (3). R 1 = R m − 1 2 t Find the radius ( R 2 ) as shown below: From Equation (3) (b) To determine Find the percentage error introduced by using in place of k O following values of t / R m : 1 , 1 2 , 1 10

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 9.1, Problem 9.26P

(a)

To determine

Prove the polar radius of gyration of the annual area approximately equal to the mean radius kO=Rm_.

(b)

To determine

Find the percentage error introduced by using in place of kO following values of t/Rm:1,12,110

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Chapter 9.1, Problem 9.25P

Chapter 9.1, Problem 9.27P

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(a) Show that the polar radius of gyration kO of the annular area shown is approximately equal to the mean radius Rm= (R1 + R2)/2 2 for small values of th thickness t= R2 - R1. (b) Determine the percentage error introduced by using Rm in place of kO for values of t/ Rm respectively equal to 1, 1/2 and 1/10.Fig. P9.19

(a) Show that the polar radius of gyration rO of the annular area shown is approximately equal to the mean radius Rm = (R1 + R2)/2 for small values of the thickness t = R2 – R1. (b) Determine the percentage error introduced by using Rm in place of rO for the following values of t/Rm: 1, 1/2, and 1/10.

Problem 09.086 - Orientation of the principal axes and the corresponding moments of inertia For the area indicated, determine the orientation of the principal axes at the origin and the corresponding values of the moments of inertia when b= 76 mm and h = 56 mm. b The value of mis The value of 0m2 is The value of Imax is The value of I min is 180 270 1.429 7292 × h x 106 mm4. 106 mm4

Chapter 9 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - 9.1 through 9.4 Determine by direct integration...Ch. 9.1 - Prob. 9.3P Ch. 9.1 - Prob. 9.4P Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - 9.5 through 9.8 Determine by direct integration...Ch. 9.1 - Prob. 9.7P Ch. 9.1 - Prob. 9.8P Ch. 9.1 - 9.9 through 9.11 Determine by direct integration...Ch. 9.1 - Prob. 9.10P

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Prove the polar radius of gyration of the annual area approximately equal to the mean radius k O = R m _ .

Solution Summary: The author demonstrates the polar radius of gyration of the annual area approximately equal to the mean radius.