(a) Explanation Given information: E= 200 GPa and s y =290 MPa. Since, the column has a pin support the effective length factor K=1. Also the column will buckle about weak axis of bending, use the smaller radius of gyration r = 51.5 mm. The critical slenderness ratio. ( K L r ) c = 4.71 E σ y ( K L r ) c = 4.71 200 × 1000 290 ( K L r ) c = 123.690 _________ (1) Calculation: To determine the maximum length when the axial load P = 560 kN, use a trial-and-error procedure. Let, the maximum length is 6 m. Therefore, as a trial value, assume L = 6 m. The corresponding slenderness ratio is: L r = 6000 51.5 L r = 116.50 This is less than the critical ratio in equation (1). Now the elastic buckling stress is found by using the equation σ e = π 2 E ( K L r ) 2 σ e = π 2 × 200 × 1000 ( 116.50 ) 2 σ e = 145.29 Μ Ρ a Again, critical stress σ c r = ( 0.658 σ y / σ e ) × σ y σ c r = ( 0.658 290 / 145.29 ) × 290 σ c r = 126.08 Now, the permissible stress is given by the formula, σ a l l o w = σ c r 1.67 σ a l l o w = 126.08 1.67 σ a l l o w = 75.50 Μ Ρ a Now, the permissible load is P a l l o w = σ a l l o w × A P a l l o w = 75 (b) To determine The permissible length L m a x of the column if P=890 KN.

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN: 9781337093347

Author: Barry J. Goodno, James M. Gere

Publisher: Cengage Learning

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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 11, Problem 11.9.16P

(a)

To determine

The permissible length L_m_a_xof the column if P=560 KN.

(b)

To determine

The permissible length L_m_a_xof the column if P=890 KN.

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Chapter 11, Problem 11.9.15P

Chapter 11, Problem 11.9.17P

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The permissible length L m a x of the column if P=560 KN.

Solution Summary: The author calculates the permissible length L m a x of the column if P=560 KN.

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The permissible length L_m_a_xof the column if P=560 KN.

The permissible length L_m_a_xof the column if P=890 KN.

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The permissible length L m a x of the column if P=560 KN.

Solution Summary: The author calculates the permissible length L m a x of the column if P=560 KN.

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The permissible length Lmaxof the column if P=560 KN.

The permissible length Lmaxof the column if P=890 KN.

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The permissible length L_m_a_xof the column if P=560 KN.

The permissible length L_m_a_xof the column if P=890 KN.