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 aluminum alloy (6061-T6) column allowable stress as follows: σ a = 20.2 − 0.126 ( k L r ) ksi k L r ≤ 66 ....... (3) σ a = 51000 ( k L r ) 2 ksi k L r ≥ 66 (4) The radius of gyration is r and the constant is k . Since the column is fixed at the base and free at the top so, k = 2 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 = 3 in , d 2 = 3.25 in , L = 2 ft , 3 ft , 4 ft and 5 ft . Substitute 3 in for d 1 and 3.25 in for d 2 in equation (6). r = 1 4 ( 3 2 + 3.25 2 ) = 1.106 in When L = 2 ft , k L r = 2 × 2 × 12 1.106 = 43.3996 ≈ 43.40 Since k L r ≤ 66 , so Substitute 43.40 for k L r in equation (3). σ a = 20.2 − ( 0.126 × 43.40 ) ksi = 14 .7316 ksi Substitute 14 .7316 ksi for σ a , 3 in for d 1 , and 3.25 in for d 2 in equation (2). P a = 14 .7316 × 10 3 × π 4 ( 3

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN: 9781337093347

Author: Barry J. Goodno, James M. Gere

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

Videos

Question

Chapter 11, Problem 11.9.23P

To determine

The allowable axial load.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11, Problem 11.9.22P

Chapter 11, Problem 11.9.24P

Chapter 11 Solutions

Mechanics of Materials (MindTap Course List)

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...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The allowable axial load.

Solution Summary: The author explains the allowable axial load and the area of cross-section for the aluminum alloy (6061-T6) column.

Concept explainers

Videos

The allowable axial load.

Want to see the full answer?

Chapter 11 Solutions