Explanation Given information: Both the ends of the columns are fixed. EI is constant. Calculation: Sketch the Free Body Diagram of the deflected of the column as shown in Figure 1. Refer to Figure 1: Take moment about the section as shown below. ∑ M = 0 M ( x ) − M ′ + P v = 0 M = M ′ − P v (1) Apply the slope and elastic curve as shown below. E I d 2 v d x 2 = M ( x ) (2) Here, EI is the flexural rigidity, d 2 v d x 2 is the second derivatives of the function v , and M ( x ) is the moment at the section x . Substitute M ′ − P v for M ( x ) in Equation (2). E I d 2 v d x 2 = M ′ − P v d 2 v d x 2 = M ′ E I − P v E I d 2 v d x 2 + P v E I = M ′ E I The general solution for the differential equation. v = C 1 sin P E I x + C 2 cos P E I x + M ′ P (3) Differentiate both sides of the Equation (3). d v d x = C 1 cos P E I x × ( P E I ) + C 2 ( − sin P E I x ) × ( P E I ) = C 1 P E I cos P E I x − C 2 P E I sin P E I x (4) Apply the boundary conditions as shown below. i. The value of v = 0 at x = 0 . ii. The value of d v d x = 0 at x = 0 . Apply boundary condition (i) in Equation (3). Substitute 0 for v and 0 for x in Equation (3). 0 = C 1 sin P E I ( 0 ) + C 2 cos P E I ( 0 ) + M ′ P C 2 = − M ′ P Apply boundary condition (ii) in Equation (4). Substitute 0 for d v d x , − M ′ P for C 2 , and 0 for x in Equation (4)

Statics And Mechanics Of Materials, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (5th Edition)

5th Edition

ISBN: 9780134380704

Author: Russell C. Hibbeler

Publisher: PEARSON

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Buckling of columns

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 17.3, Problem 44P

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Show that the critical load on the column Pcr=4π2EIL2.

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Chapter 17.3, Problem 43P

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Chapter 17 Solutions

Statics And Mechanics Of Materials, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (5th Edition)

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Show that the critical load on the column P cr = 4 π 2 E I L 2 .

Solution Summary: The author illustrates how the critical load on the column is proved. Both the ends of the columns are fixed, and EI is constant.

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Show that the critical load on the column Pcr=4π2EIL2.

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Chapter 17 Solutions