Machine Elements in Mechanical Design (6th Edition) (What's New in Trades & Technology)
Machine Elements in Mechanical Design (6th Edition) (What's New in Trades & Technology)
6th Edition
ISBN: 9780134441184
Author: Robert L. Mott, Edward M. Vavrek, Jyhwen Wang
Publisher: PEARSON
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Chapter 6, Problem 1P

A column has both ends pinned and has a length of 32 in. It is made of SAE 1040 HR steel and has a circular shape with a diameter of 0.75 in. Determine the critical load.

Expert Solution & Answer
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To determine

The critical load that can be applied to a column of circular cross section whose both ends are pinned.

Answer to Problem 1P

The critical load of the column is given by 4490 lb

Explanation of Solution

Given Information:

Cross section of the column: Circular

Diameter of the column, D = 0.75 in

Length of the column, L = 32 in

End conditions: Both ends pinned

Material = SAE 1040 HR steel

So, Young’s Modulus, E = 30×106 psi

Yield strength, Sy= 42000 psi

Concept used:

For long Column Euler’s Formula used

  Pcr=π2EIminLe2                . (1)

For short Column Euler’s Formula used

  Pcr=Asy[1sySr24π2E]                   . (2)

Here,

Pcr is Critical load

E is Young’s Modulus

Imin is Minimum Moment of Inertia

Le is effective length (Based on the end conditions of Column)

For both ends pinned condition, Le = L

Sy is yield stress

Sr Slenderness ratio = Ler

r is radius of gyration

A cross section area

Calculation:

For solid round section,

r = D/4

= 0.75/4

= 0.188 in

Slenderness ratio,

  Sr=320.188=171

Compute Column constant for the above material

  Cc=2π2ESy=2π2(30×106)42000=119

Sr>Cc, so Column is long, Use Euler’s Formula

Moment of Inertia,

  I=πD464=π(0.754)64=0.01553 mm4

From Equation (1)

  Pcr=π2(30×106)(0.01553)322=4490 lb

Thus, critical load is 4490 lb.

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Machine Elements in Mechanical Design (6th Edition) (What's New in Trades & Technology)

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