Explanation Given information: The load applied on the wooden column is P = 1.5 kip . The eccentricity ( e ) of loading is 3 in . The one end of the column end is pinned and another end is fixed. The width of cross-sectional area of the column is b = 1.5 in . The depth of cross-sectional area of the column is d = 12 in . The length of the column, L = 6 ft . Show the expression of NFPA Equation of Sec 13.6 as follows: ( σ a ) allow = [ 540 ( K L d ) 2 ] ksi, 26 < K L r < 50 Here, ( σ a ) allow is the allowable stress and ( K L r ) is the slenderness ratio. Show the expression for Equation 13-30 as follows: σ max = P A + M c I Here, σ max is the maximum compressive stress, P is the load, A is the area, M is the moment, c is the distance between the centroid of member and outer fiber, and I is the moment of inertia. Calculation: Calculate the cross-sectional area ( A ) of the rectangular column using the relation: A = b d (1) Substitute 1.5 in . for b and 12 in . for d in Equation (1): A = 1.5 × 12 = 18 in . 2 Calculate the moment of inertia using the relation: I = d b 3 12 (2) Substitute 1.5 in . for b and 12 in . for d in Equation (2): I = 1.5 × 12 3 12 = 216 in . 4 Calculate the value of ( K L d ) as follows: Theone end of the wood column are pinned and another end is fixed, thus the value of K = 0.7 . Substitute 0.7 for K , 6 ft for L, and 1.5 in . for d . ( K L d ) = ( 0.7 × 6 ft × 12 in 1 ft 1

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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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 13.7, Problem 13.124P

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To check: whether the wood column can adequately support eccentric load of 1.5 kip .

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Chapter 13.7, Problem 13.123P

Chapter 13.7, Problem 13.125P

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Chapter 13 Solutions

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To check: whether the wood column can adequately support eccentric load of 1 .5 kip .

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