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A horizontal load Pis applied to an assembly consisting of two inclined bars, as shown in the figure. The cross-sectional area of bar (1) is 1.45 in.?, and the cross-sectional area of bar (2) is 1.90 in.?. The normal stress in either bar may not exceed 24 ksi. Determine the maximum load P that may be applied to this assembly. Assume dimensions of a = 15.0 ft, b = 7.5 ft, and c = 12.0 ft.

A horizontal load Pis applied to an assembly consisting of two inclined bars, as shown in the figure. The cross-sectional area of bar (1) is 1.45 in.?, and the cross-sectional area of bar (2) is 1.90 in.?. The normal stress in either bar may not exceed 24 ksi. Determine the maximum load P that may be applied to this assembly. Assume dimensions of a = 15.0 ft, b = 7.5 ft, and c = 12.0 ft.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Make a new assumption for the maximum load case, referred to as "Load Case B", in which the force in member (2) will control the capacity of the two-bar assembly. For this assumption, which may or may not be correct, the force in member (2) is F2.B = F2, allow- Enter the value of the force in member 1 for Load Case B. F1B = kips Enter the forces in the members corresponding to the correct maximum load case assumption, Load Case A or Load Case B. F1 = kips F2 = i kips Determine the maximum load P that may be applied to this assembly. Pmax kips
Transcribed Image Text:Make a new assumption for the maximum load case, referred to as "Load Case B", in which the force in member (2) will control the capacity of the two-bar assembly. For this assumption, which may or may not be correct, the force in member (2) is F2.B = F2, allow- Enter the value of the force in member 1 for Load Case B. F1B = kips Enter the forces in the members corresponding to the correct maximum load case assumption, Load Case A or Load Case B. F1 = kips F2 = i kips Determine the maximum load P that may be applied to this assembly. Pmax kips
A horizontal load Pis applied to an assembly consisting of two inclined bars, as shown in the figure. The cross-sectional area of bar (1) is 1.45 in.?, and the cross-sectional area of bar (2) is 1.90 in.². The normal stress in either bar may not exceed 24 ksi. Determine the maximum load P that may be applied to this assembly. Assume dimensions of a = 15.0 ft,b = 7.5 ft, and c = 12.0 ft. (1) a B (2) Determine the allowable force F1 in member (1) and the allowable force F2 in member (2). F1, allow = i kips F2, allow = i kips Find the ratio, (F1 /F2), where F1 is the force in member (1) and F2 is the force in member (2). Both bars are in tension, so these forces are both positive according to the sign conventions. F1/F2 = i Make an assumption for the maximum load case, referred to as "Load Case A", in which the force in member (1) will control the capacity of the two-bar assembly. For this assumption, which may or may not be correct, the force in member (1) is F1A = F1, allow- Enter the value of the force in member 2 for Load Case A. F2.A i kips
Transcribed Image Text:A horizontal load Pis applied to an assembly consisting of two inclined bars, as shown in the figure. The cross-sectional area of bar (1) is 1.45 in.?, and the cross-sectional area of bar (2) is 1.90 in.². The normal stress in either bar may not exceed 24 ksi. Determine the maximum load P that may be applied to this assembly. Assume dimensions of a = 15.0 ft,b = 7.5 ft, and c = 12.0 ft. (1) a B (2) Determine the allowable force F1 in member (1) and the allowable force F2 in member (2). F1, allow = i kips F2, allow = i kips Find the ratio, (F1 /F2), where F1 is the force in member (1) and F2 is the force in member (2). Both bars are in tension, so these forces are both positive according to the sign conventions. F1/F2 = i Make an assumption for the maximum load case, referred to as "Load Case A", in which the force in member (1) will control the capacity of the two-bar assembly. For this assumption, which may or may not be correct, the force in member (1) is F1A = F1, allow- Enter the value of the force in member 2 for Load Case A. F2.A i kips
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