Steel Connections Part 1_Part1.pdf - Staff.zu.edu.eg

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• For member 3: • F3 = +23.371t, 2┘└11011010, then n3 = 23.371 / 3.456 = 6.76 8, arranged in a double gauges as a‐t/ > 6. • 5‐ For member 4: • F4 = +41.569t, 2┘└13013012, then n4 = 41.569 / 3.456 = 12.1 14, arranged in a double gauges as a‐t/ > 6. • As the number of bolts in each gauge line exceeds 6 bolts then, there is a need to reduce the capacity of the bolt by a reduction factor , therefore R L = 0.9853.456 = 3.4t , and the new number of bolts is n = 41.569/3.4 = 12.21 14 and there is no further reduction.
Analysis of eccentric shear connections
Page 1 and 2: Steel Connections Non‐pre‐tensi
Page 3 and 4: Objectives Through the following 3
Page 6 and 7: Desseldorf airport Germany
Page 8 and 9: Kolen banhof Germany
Page 10 and 11: 2‐ According to type of transferr
Page 12 and 13: 4‐ According to whether the conne
Page 14 and 15: • A major part of labour costs ha
Page 16 and 17: Our Course shall handles:‐ • Bo
Page 18 and 19: Bearing and friction type bolted co
Page 20 and 21: Bolts Grades According to ECP Bolt
Page 22 and 23: Common Bolt Sizes (AISC) • A325 a
Page 24 and 25: Parts of the Bolt Assembly (AISC) G
Page 26 and 27: Gallery of column and beam splices
Page 29: Bolted Joint Failure Modes (AISC) B
Page 34 and 35: • Bolts in a connection is to be
Page 36 and 37: Allowable stresses for Bearing type
Page 38 and 39: • 3‐ Bolts subjected to tension
Page 40 and 41: General Requirements Regarding Bolt
Page 42 and 43: • Gusset plates should be checked
Page 44 and 45: Examples Example 5.1 • Given: The
Page 48 and 49: Example for an eccentric connection
Page 50 and 51: Example
Page 52: Example 5.4

Steel Connections Part 1_Part1.pdf - Staff.zu.edu.eg

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