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Problem II. The fabricated steel members are being assembled into a truss on-site in the municipality of Alfonso, Cavite. The assembly started at the elevated ground at both sides and is to meet at the midspan. Nearing to the end of construction, it was found that the last member, CD, was fabricated too short by A mm (see the given table in MS teams). The engineer then came up with an idea of applying a force P at joints C and D as shown to close the gap. A in mm = 4.090

Problem II. The fabricated steel members are being assembled into a truss on-site in the municipality of Alfonso, Cavite. The assembly started at the elevated ground at both sides and is to meet at the midspan. Nearing to the end of construction, it was found that the last member, CD, was fabricated too short by A mm (see the given table in MS teams). The engineer then came up with an idea of applying a force P at joints C and D as shown to close the gap. A in mm = 4.090

Construction Materials, Methods and Techniques (MindTap Course List)
Construction Materials, Methods and Techniques (MindTap Course List)
4th Edition
ISBN:9781305086272
Author:William P. Spence, Eva Kultermann
Publisher:William P. Spence, Eva Kultermann
Chapter16: Nonferrous Metals
Section: Chapter Questions
Problem 16RQ
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Problem II. The fabricated steel members are being assembled into a truss on-site in the municipality of Alfonso, Cavite. The assembly started at the elevated ground at both sides and is to meet at the midspan. Nearing to the end of construction, it was found that the last member, CD, was fabricated too short by A mm (see the given table in MS teams). The engineer then came up with an idea of applying a force P at joints C and D as shown to close the gap. A in mm = 4.090 P. C' 3 m A B 3 m 3 m 3 m Use E = 200 GPa and cross-sectional area of members BC, DE, CD each 4500 mm² while AC and DF being 6363.961 mm² Determine the necessary force, P, to close the gap in kN. b) If force P will then be removed after being assembled, calculate the force in CD in kN. a)
Transcribed Image Text:Problem II. The fabricated steel members are being assembled into a truss on-site in the municipality of Alfonso, Cavite. The assembly started at the elevated ground at both sides and is to meet at the midspan. Nearing to the end of construction, it was found that the last member, CD, was fabricated too short by A mm (see the given table in MS teams). The engineer then came up with an idea of applying a force P at joints C and D as shown to close the gap. A in mm = 4.090 P. C' 3 m A B 3 m 3 m 3 m Use E = 200 GPa and cross-sectional area of members BC, DE, CD each 4500 mm² while AC and DF being 6363.961 mm² Determine the necessary force, P, to close the gap in kN. b) If force P will then be removed after being assembled, calculate the force in CD in kN. a)
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