Explanation Given: The outside diameter of the pipe is d 2 = 5.60 in and inside diameter of the pipe is d 1 = 4.80 in . Concept Used: Write the expression for the allowable axial load. P a = σ a A (1) The allowable axial load is P a , the allowable stress is σ a , and the area of cross-section is A . Write the expression for the area of cross-section. A = π 4 ( d 2 2 − d 1 2 ) Substitute π 4 ( d 2 2 − d 1 2 ) for A in equation (1). P a = σ a × π 4 ( d 2 2 − d 1 2 ) (2) The equation for the allowable stress as follows: σ a = 30.7 − 0.23 ( k L r ) ksi k L r ≤ 55 ....... (3) σ a = 54000 ( k L r ) 2 ksi k L r ≥ 55 (4) The radius of gyration is r , and the constant is k . Since the column has pinned ends so, k = 1 Write the expression for the radius of gyration. r = I A (5) The area moment of inertia is I . Write the equation of area moment of inertia as follows: I = π 64 ( d 2 4 − d 1 4 ) Substitute π 64 ( d 2 4 − d 1 4 ) for I and π 4 ( d 2 2 − d 1 2 ) for A in equation (5). r = π 64 ( d 2 4 − d 1 4 ) π 4 ( d 2 2 − d 1 2 ) r = 1 16 ( d 2 2 + d 1 2 ) ( d 2 2 − d 1 2 ) ( d 2 2 − d 1 2 ) r = 1 4 ( d 2 2 + d 1 2 ) (6) Calculation: As per the given problem d 1 = 4.80 in , d 2 = 5.60 in , L = 6 ft , 8 ft , 10 ft and 12 ft . Substitute 4.80 in for d 1 and 5.60 in for d 2 in equation (6). r = 1 4 ( 4.80 2 + 5.60 2 ) = 1.84 in When L = 6 ft , k L r = 1 × 6 × 12 1.84 = 39.047 Since k L r ≤ 55 Substitute 39.047 for k L r in equation (3). σ a = 30.7 − 0.23 × 39.047 ksi = 21 .72 ksi Substitute 21 .72 ksi for σ a , 4

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card

9th Edition

ISBN: 9781337594301

Author: Barry J. Goodno, James M. Gere

Publisher: Cengage Learning

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The column is described as a vertical member of the structure that carries an axial compressive load, and buckling is a lateral deformation of the column under the application of load.

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Chapter 11, Problem 11.9.21P

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The tensile 0.2 percent offset yield strength of AISI 1137 cold-drawn steel bars up to 1 inch in diameter from 2 mills and 25 heats is reported as follows: Sy 93 95 101 f 97 99 107 109 111 19 25 38 17 12 10 5 4 103 105 4 2 where Sy is the class midpoint in kpsi and fis the number in each class. Presuming the distribution is normal, determine the yield strength exceeded by 99.0% of the population. The yield strength exceeded by 99.0% of the population is kpsi.

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Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card

Ch. 11 - A rigid bar of length L is supported by a linear...Ch. 11 - The figure shows an idealized structure consisting...Ch. 11 - -2-3. Two rigid bars are connected with a...Ch. 11 - Repeat Problem 11.2-3 assuming that R= 10 kN ·...Ch. 11 - The figure shows an idealized structure consisting...Ch. 11 - An idealized column consists of rigid bar ABCD...Ch. 11 - An idealized column is made up of rigid segments...Ch. 11 - The figure shows an idealized structure consisting...Ch. 11 - The figure shows an idealized structure consisting...Ch. 11 - The figure shows an idealized structure consisting...

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The allowable axial load.

Solution Summary: The author analyzes the allowable axial load and the area of cross-section.

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