solution2_2136

# solution2_2136 - DEPARTMENT OF CIVIL ENGINEERING LAKEHEAD...

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DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD UNIVERSITY Engineering 2136 – Steel Design Solutions for Assignment No.2, Fall 2006 Dr. Y Gong 1. A single-story industrial building in Thunder Bay S s =2.9 kPa, S r =0.4 kPa, q 1/50 =0.39 kPa Importance category=normal (1) Steel joist (j-1) has a tributary width of 2.5 meters. The specified uniformly distributed dead load for joist ( j-1) is: Roof membrane = 0.30 × 2.5 = 0.75 kN/m Insulation and barrier = 0.10 × 2.5 = 0.25 Steel deck = 0.15 × 2.5 = 0.38 Joist = 0.10 × 2.5 = 0.25 Ceiling and ductworks = 0.30 × 2.5 = 0.75 ---------------------------------------------------------------------- Total = 0.95 × 2.5 =2.38 kN/m (2) The specified roof snow load is calculated using equation S = I s [S s C b C w C s C a + S r ] For the roof of this building, C b =0.8, C w =1.0, C s =1.0, C a =1.0, I s =1.0 for ULS and I s =0.9 for SLS S = I s [S s C b C w C s C a + S r ] = I s [2.9 (0.8)(1.0)(1.0)(1.0) + 0.4 ] =I s (2.72) kPa check: S s C b C w C s C a = 2.32 kPa > S r (ok) For ULS, S=2.72 kPa; for SLS, S=2.45 kPa (3) (3.a) Dimension z = ? 10% of least horizontal dimension = 0.1 (37.5) = 3.75 m 40% of height = 0.4 (6) =2.4 m Z = 2.4 m check Z > 1.0 m ( ok ) , and Z 4% (37.5) = 1.5 m (ok) (3.b) Hourly Wind Pressure q (1/50) =0.39 kPa I w =1.0 for ULS, and I w =0.75 for SLS Using P=I w q C e C p C g to calculate external wind pressure. Exposure factor C e =(h/10) 0.2 =(6/10) 0.2 =0.9 where reference height h= 6m, referring to Figure I-9 of Structural Commentary of NBCC-2005 for C p C g values. C e is not less than 0.9 for open terrain. . Area C p C g P=I w q C e C p C g (kPa) r 0.5, -1.7 (1.0)(0.39)(0.9)(0.5) = 0.176, -0.60 s 0.5, -2.5 (1.0)(0.39)(0.9)(0.5) = 0.176, -0.88 c 0.5, -4.2 = 0.176, -1.47 Note: Tributary area =2.5 m 2 for the table The positive wind pressures are compressive wind pressure, and the negative are uplift wind pressure. 1

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DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD UNIVERSITY (3.c) Refer to Figure I-8 of Structural Commentary. For SLS, I w =0.75 Area C p C g P=I w q C e C p C g (kPa) e 1.7, -2.0 (0.75)(0.39)(0.9)(1.7) = 0.45, -0.53 w 1.7, -1.7 (0.75)(0.39)(0.9)(1.7) = 0.45, -0.45 Note: Tributary area = 2 m 2 for the table (3.d) Joist j-1 is a secondary structural member. Use Figure I-9. Joist j-1 is in “r zone”. Tributary area of j-1 is 2.5(12.5) = 31.25 m
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solution2_2136 - DEPARTMENT OF CIVIL ENGINEERING LAKEHEAD...

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