Explanation Given: The width of the rectangular cross-section is 100 mm and the height is 150 mm . The stress is 14 MPa and the modulus of elasticity is 12 GPa . Concept Used: Write the expression for the allowable axial load. P a = F c C p A (1) The allowable load is P a , the column stability factor is C p , and the area is A . Write the expression for the area. A = b × h The area is A , the width is b and the height is h . Substitute b × h for A in equation (1). P a = F c C p ( b × h ) (2) Write the expression for the column stability factor. C p = 1 + ϕ 2 c − ( 1 + ϕ 2 c ) 2 − ( ϕ c ) ....... (3) The sawn number is c and the Euler’s buckling coefficient ratio is ϕ . Write the expression for the Euler’s buckling coefficient ratio. ϕ = K E F c ( L b ) 2 (4) The Buckling coefficient is K and the length is L . Calculation: As per the given problem b = 100 mm , h = 150 mm , E = 12 GPa , F c = 14 MPa, and L = 1.5 m , 2 m and 2.5 m . Refer to property table for the wood block of rectangular cross-section to obtain the sawn number as, c = 0.8 K = 0.3 When L = 1.5 m , Substitute the values in equation (4). ϕ = 0.3 × ( 12 GPa ) ( 1000 MPa 1 GPa ) 14 × ( ( 1.5 m ) ( 1000 mm 1 m ) 100 mm ) 2 = 0.3 × 12000 14 × ( 15 ) 2 = 1.1429 Substitute 1.1429 for ϕ in equation (3). C p = 1 + 1.1429 2 × 0.8 − ( 1 + 1.1429 2 × 0.8 ) 2 − ( 1.1429 0.8 ) = 0.7351 Substitute 0.7351 for C p , 14 MPa for F c , 100 mm for b , and 150 mm for h in equation (2). P a = 0.7351 × 14 × ( 100 × 150 ) = ( 153510 N ) ( 1 kN 1000 N ) = 153.5 kN When L = 2 m , Substitute the values in equation (4). ϕ = 0

9th Edition

ISBN: 9781337093354

Author: Barry J. Goodno, James M. Gere

Publisher: Cengage Learning

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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 …

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Chapter 11, Problem 11.9.30P

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Mechanics of Materials, SI Edition

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The allowable axial load.

Solution Summary: The author explains the allowable axial load, the column stability factor, and the modulus of elasticity.

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