Explanation Given information: The solid circular rod is made up of 6061-T6 aluminum alloy. The length of the rod is L = 4 m . The both ends of the rod are pinned. The axial load on the rod is P = 15 kN . The factor of safety for buckling ( F .S ) is 2. Calculation: Consider the Young’s modulus of the 6061-T6 aluminum alloy as E = 68.9 × 10 9 N m 2 . Consider the allowable load on the rod ( P allowable ) as 15 kN . Calculate the critical load for the rod using the relation: Factor of safety ( F .S ) = P c r P allowable (1) Here, P c r is the critical buckling load. Substitute 15 kN for P allowable and 2 for ( F . S ) in Equation (1). 15 = P c r 2 P c r = 15 × 2 = 30 kN Show the expression for the moment of inertia of circular cross-section about the as follows: I x = I y = I I = π d 4 64 (2) Here, d is the diameter of the circular cross-section. Show the expression for Euler’s formula as shown below: P cr = π 2 E I ( K L ) 2 (3) Here, P cr is the maximum load on the material just before it begins to buckle, E is the Young’s modulus of the material, I is the least moment of inertia of the material cross-sectional area, L is the length of the material and K is the effective-length factor. Calculate the critical load for the column AB using the relation: The value of effective-length factor for the member AB , K = 0.7 , since one end ofcircular rod is pinned and another end is fixed. Substitute 30 kN for P c r 0.7 for K , 68.9 × 10 9 N m 2 for E , π d 4 64 m 4 for I , and 3 m for L in Equation (3). 30 kN = π 2 ( 68.9 × 10 9 N m 2 ) × π d 4 64 m 4 ( 0.7 × 4 m ) 2 30 kN × 1 , 000 N 1 kN = 33

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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 13.3, Problem 13.35P

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The minimum diameter of the rod.

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Chapter 13.3, Problem 13.34P

Chapter 13.3, Problem 13.36P

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A 6061-T6 aluminum alloy solid circular rod of length 4 m is pinned at both of its ends. If it is subjected to an axial load of 15 kN and F.S. = 2 against buckling, determine the minimum required diameter of the rod to the nearest mm.

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Chapter 13 Solutions

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The minimum diameter of the rod.

Solution Summary: The author calculates the minimum diameter of the rod using the Young's modulus of 6061-T6 aluminum alloy and the factor of safety for buckling.

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