Explanation Given information: The outer diameter of pipe is 1 in. The thickness of pipe is 3 16 in . The diameter of solid rod is 1 2 in . Calculation: Calculate the inner diameter of the pipe ( d i ) as follows: d i = d o − 2 t (1) Here, d o is outer diameter and t is the thickness of the pipe. Substitute 1 in. for d o and 3 16 in . for t in Equation (1). d i = 1 − 2 ( 3 16 ) = 1 − 0.375 = 0.625 in . Calculate the cross section area ( A hallow pipe ) of hallow pipe: ( A hallow pipe ) = π 4 ( d 0 2 − d i 2 ) (2) Substitute 1 in. for d 0 and 0.625 in. for d i in Equation (2). ( A hallow pipe ) = π 4 ( 1 2 − 0.625 2 ) = π 4 ( 0.609 ) = 0.478 in 2 Calculate the cross section area ( A solid pipe ) of solid pipe: ( A solid pipe ) = π 4 ( r 2 ) (3) Here, r is the radius of solid pipe. Substitute 0.5 in. for r in Equation (3). ( A solid pipe ) = π 4 ( 0.5 2 ) = 0.1963 in 2 Find the value of ( y 1 ) . Consider the origin of the coordinate system is at centroid of the solid pipe that is centre of solid pipe. y 1 = 0 Find the value of ( y 2 ) centroid of the hallow pipe from origin. y 2 = 1 2 2 + 5.75 + 1 2 = 6.5 in . Calculate the location of centroid ( C ) combined pipe. y ¯ = a 1 y 1 + a 2 y 2 a 1 + a 2 (4) Here, a 1 is the area of solid pipe, a 2 is area of hallow pipe, y 1 is centroid of solid pipe, and y 2 is centroid of hallow pipe. Substitute 0.1963 in 2 for a 1 , 0 for y 1 , 0.478 in 2 for a 2 , and 6.5 in . for y 2 in Equation (4). y ¯ = ( 0.1963 × 0 ) + ( 0.478 × 6.5 ) 0.1963 + 0.478 = 0 + 3.107 0.6743 = 4.6077 in . Calculate the moment of inertia ( I 1 ) hollow pipe section about centroid C as follows: I 1 = π 64 ( d o 4 − d i 4 ) + A 2 ( y 2 − y ¯ ) 2 (5) Substitute 1 in

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 6.4, Problem 6.70P

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