Explanation Given information: Consider the strong back BC is made up of an A992 steel hollow circular section. The internal diameter ( d i ) of the strong back is 40 mm . The outer diameter ( d o ) of the strong back is 60 mm . The factor of safety (F.O.S) for buckling is 2. Calculation: Show the forces acting at joint A as shown in Figure 1. Refer Figure 1. The forces in the cable AB and BC is denoted by F A B and F B C . The lifting force is denoted by P . For Equilibrium of forces, Take the summation of all horizontal forces acting at joint A and Equate to zero. Take the forces acting towards right as positive. ∑ F x = 0 F A C cos ( 45 ° ) − F A B cos ( 45 ° ) = 0 F A C ( 1 2 ) − F A B ( 1 2 ) = 0 F A C ( 1 2 ) = F A B ( 1 2 ) F A C = F A B Take the summation of all vertical forces acting at joint A and Equate to zero. Take the forces acting upward as positive. ∑ F y = 0 P − F A C sin ( 45 ° ) − F A B sin ( 45 ° ) = 0 P − F A C ( 1 2 ) − F A B ( 1 2 ) = 0 Consider the value of F A C = F A B = F . P − F ( 1 2 ) − F ( 1 2 ) = 0 P − 2 F ( 1 2 ) = 0 2 F ( 1 2 ) = P F = ( 2 2 ) P F = 0.707 P Show the forces acting in joint B as shown in Figure 2. Refer Figure 2. The forces in the cable AB , BC , and BD is denoted by F A B , F B C , and F B D . For Equilibrium of forces, Take the summation of all horizontal forces acting at joint B and Equate to zero. Take the forces acting towards right as positive. ∑ F x = 0 F A C cos ( 45 ° ) − F B C = 0 F A C ( 1 2 ) − F B C = 0 F B C = F A C ( 1 2 ) Substitute 0.707 P for F A C . F B C = 0.707 P ( 1 2 ) = 0.5 P Calculate the critical load on the strong back using the relation: F B C = P c r F .O .S P c r = ( F .O .S ) F B C Substitute 2 for (F.O.S) and 0.5 P for F B C . P c r = ( 2 ) × 0.5 P = P Calculate the factor of safety (F.O.S) using the relation: P allowable = P c r F .O .S F .O .S = P c r P allowable Substitute 20 kip for P allowable , and 178.9 kip for P c r . F .O .S = 178.9 20 = 8.945 Thus, the factor of safety with respect to buckling about x-x axis is 8.945 _ . Calculate the cross-sectional area ( A ) of the circular hollowtube using the relation: A = π 4 [ ( d o ) 2 − ( d i ) 2 ] (1) Substitute 60 mm for d o and 40 mm for d i in Equation (1): A = π 4 ( 60 2 − 40 2 ) = π 4 ( 3 , 600 − 1 , 600 ) = 1570

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 13.27P

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The maximum value of lifting force P without buckling of strong back.

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Chapter 13.3, Problem 13.26P

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Mechanics of Materials

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