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425. In the Fink truss shown in Fig. P-425, the web members BC and EF are perpendicular to the inclined members at their midpoints. Use the method of sec- tions to determine the force in members DF, DE, and CE.

425. In the Fink truss shown in Fig. P-425, the web members BC and EF are perpendicular to the inclined members at their midpoints. Use the method of sec- tions to determine the force in members DF, DE, and CE.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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I cannot arrive with the same answer on some members with the book . Please give me a detailed solution on 425 for me to check where did I get wrong. Thank you!

Art. 4-4] Method of Sections the method of sections, first obtain the value of BE by inspection. Ans. BF-2500 lb C 425. In the Fink truss shown in Fig. P-425, the web members BC and EF are perpendicular to the inclined members at their midpoints. Use the method of sec- tions to determine the force in members DF, DE, and CE. Ans. DF 5.82 kips C; DE -2.00 kips T; CE - 4 kips T 426. Show thst the method of joints cannot determine the forces in all bars of the Fan Fink truss in Fig. P-426. Then A use the method of sections to compute the force in bars FH, GH, and EK. 1100 lb C. GH = 520 lb 2400 lb
Transcribed Image Text:Art. 4-4] Method of Sections the method of sections, first obtain the value of BE by inspection. Ans. BF-2500 lb C 425. In the Fink truss shown in Fig. P-425, the web members BC and EF are perpendicular to the inclined members at their midpoints. Use the method of sec- tions to determine the force in members DF, DE, and CE. Ans. DF 5.82 kips C; DE -2.00 kips T; CE - 4 kips T 426. Show thst the method of joints cannot determine the forces in all bars of the Fan Fink truss in Fig. P-426. Then A use the method of sections to compute the force in bars FH, GH, and EK. 1100 lb C. GH = 520 lb 2400 lb
Thên une the method of sections to compute the force in bars FH, GH, and EK. Ans. FH 1100 lb C; GH = 520 lb T;EK 693 lb T 2400 Ib 1200 lb FIG. P-424. 24 2K DY 2 B. to to 1K, 10 12.5 C. 10 -40- FIG. P-425. 200 lb T人T- 200 lb 200 lb H. 200 lb 200 lb 200 lb
Transcribed Image Text:Thên une the method of sections to compute the force in bars FH, GH, and EK. Ans. FH 1100 lb C; GH = 520 lb T;EK 693 lb T 2400 Ib 1200 lb FIG. P-424. 24 2K DY 2 B. to to 1K, 10 12.5 C. 10 -40- FIG. P-425. 200 lb T人T- 200 lb 200 lb H. 200 lb 200 lb 200 lb
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