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Material Type: Assignment; Professor: Nichols; Class: ARCHITECTURAL STRUCTURES; Subject: ARCHITECTURE; University: Texas A&M University; Term: Unknown 1989;
Typology: Assignments
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Date: 4/2/09, due 4/9/09 Pass-fail work
Problems: from Onouye Chapters 4 & 10 and as stated. Notes: Problems marked with a * have been altered with respect to the problem stated in the text.
10A) A long span steel joist with a span of 80 feet is required to support a roof. The joists are spaced at 4 ft apart, the dead load is 12 lb/ft 2 , the live load is 28 lb/ft 2 and the live load deflection is limited to L/360 (which is that used to determine the live load limit based on deflection in the Joist catalogue tables). Remembering to estimate a joist weight, use the table provided to select the most economical joist that can be used. (open web joist charts)
Partial answers to check with: 44LH likely
6.5 k 5.5 k
96
10B) If a simply supported 36 ft parallel chord open-web joist has 12 panels at 3 ft for the top chord and the support reactions shown, use the method of sections to determine the member forces in the top chord, bottom chord, and the web for the section indicated in the figure. The joists are 2 ft. on center, the distributed load over the top of the truss is 95 lb/ft 2 and the self weight is 12.2 lb/ft. NOTE: Remember that the tributary width for the end joints is only half what it is for the rest of the top joints. (load tracing and method of sections) Partial answers to check with: top chord = 9.9 k (C) bottom chord = 11.3 k (T) web = 2.6 k (C)
Partial answers to check with: HG = 5.5 k, ED = -7.12 k, EG = 1.77 k.
Partial answers to check with: kL/rx = 57.4, kL/ry = 54.4, Pallowed = 497 k, so...
10C) For the column of problem 10.3.5, assume the roof load is a live load, and the 2nd^ floor framing load is a dead load. Using LRFD design and the tables for the critical unfactored compressive stress, determine if the column section shown is adequate. (LRFD column analysis)
Partial answers to check with:
Problem 10.3.
12 panels @ 3 ft = 36 ft
26 inches 3639.6 lb 3639.6 lb
(method of sections)
( ASD column analysis) Assume A36 steel (F (^) y = 36 ksi, E = 29 x 10^3 ksi)