Explanation Given information: The linkage is made with two A992 steel rods. The load on the linkage is 6 kip . The cross-section of the rod is circular. The ends of the rods are pinned. The factor of safety ( F .S ) against buckling is 1.8. Calculation: Consider the Young’s modulus ( E ) of the steel as 29 × 10 3 ksi . Calculate the length of the rod AB using the relation: Consider the length of the rod AB as L A B . L A B cos ( 45 ° ) = 12 L A B ( 1 2 ) = 12 L A B = 12 2 = 16.971 ft Calculate the length of the rod BC using the relation: Consider the length of the rod BC as L B C . L B C cos ( 30 ° ) = 12 L B C ( 3 2 ) = 12 L B C = 12 × 2 3 = 13.856 ft Show the free body diagram of forces acting at joint B as shown in Figure 1. Refer Figure 1. For Equilibrium of forces, Take summation of horizontal forces and Equate to zero. ∑ H = 0 Take the forces acting towards the right as positive. F A B sin ( 45 ° ) − F B C sin ( 30 ° ) = 0 F A B ( 1 2 ) − F B C ( 1 2 ) = 0 F A B ( 1 2 ) = F B C ( 1 2 ) F A B = F B C × ( 2 2 ) (1) Take summation of vertical forces and Equate to zero. ∑ V = 0 Take the forces acting towards upward as positive. F A B cos ( 45 ° ) + F B C cos ( 30 ° ) − 6 = 0 F A B ( 1 2 ) + F B C ( 3 2 ) = 6 (2) Substitute F B C × ( 2 2 ) for F A B in Equation (2). F A B cos ( 45 ° ) + F B C cos ( 30 ° ) − 6 = 0 F B C × ( 2 2 ) ( 1 2 ) + F B C ( 3 2 ) = 6 F B C × ( 1 2 ) + F B C ( 3 2 ) = 6 F B C ( 1 + 3 2 ) = 6 F B C = 6 ( 2 1 + 3 ) = 4.392 kip Substitute 4.392 kip for F B C in Equation (1). F A B = 4.392 × ( 2 2 ) = 4.392 × ( 0.707 ) = 3.106 kip Thus, the value of forces F B C is 4.392 kip and F A B is 3.106 kip . Consider the rod AB . Consider the diameter of the rod AB as d A B . Show the moment of inertia ( I ) of the rod AB as follows: I = π d A B 4 64 The ends of the rod AB are pinned, thus the value of effective length factor ( K ) is 1. The length ( L A B ) of the rod AB is 16.971 ft . Calculate the critical load for the rod AB using the relation: P c r = F .S ( F A B ) Substitute 1.8 for F.S and 3.106 kip for F A B . P c r = 1.8 × 3.106 kip = 5.5908 kip Calculate the diameter of the rod AB using Euler’s formula. Show the expression for Euler’s formula as shown below: P cr = π 2 E I ( K L A B ) 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. Substitute 5.5908 kip for P c r , 29 × 10 3 ksi for E , π d A B 4 64 for I , 1 for K , and 16.971 ft for L A B in Equation (3). 5.5908 = π 2 × 29 × 10 3 ( π d A B 4 64 ) ( 1 × 16.971 ft × 12 in . 1 ft ) 2 = π 2 × 29 × 10 3 × π × d A B 4 41 , 474

9th Edition

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 13.3, Problem 13.22P

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The diameter of the rod to the nearest 18 in. to support a load of 6 kip .

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Chapter 13.3, Problem 13.21P

Chapter 13.3, Problem 13.23P

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Mechanics of Materials

Ch. 13.3 - A 50-in long steel rod has a diameter of 1 in....Ch. 13.3 - A 12-ft wooden rectangular column has the...Ch. 13.3 - The A992 steel column can be considered pinned at...Ch. 13.3 - A steel pipe is fixed supported at its ends. If it...Ch. 13.3 - Determine the maximum force P that can be...Ch. 13.3 - The A992 steel rod BC has a diameter of 50 mm and...Ch. 13.3 - 13-1. Determine the critical buckling load for the...Ch. 13.3 - The column consists of a rigid member that is...Ch. 13.3 - The aircraft link is made from an A992 steel rod....Ch. 13.3 - Rigid bars AB and BC are pin connected at B. If...

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