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UNIVERSITY OF CALIFORNIA, BERKELEYSpring 2016Department of Civil and Environmental Engineering Instructor: J. P. Moehle CE 120 - Structural Engineering Solutions of Assignment No. 5 Due: February 26, 2016Solution: !Dead Load from ASCE-7-10 (Table C-3-1): Component Load(psf) Four ply felt and gravel 5.5psf Mechanical Duct 4psf ¾” Subflooring (plywood) 3psf !Live loads 20psf !Material Density: Douglas fir 34pcf Fig: Schematic Drawing of the Roof. (a)Loads Supported by Plywood: ¾in plywood carries weight from everything above it and it’s self weight. Table 1: Loads carried by plywood Dead load Live load Four ply felt and gravel =5.5psf Roof Live load =20psf ¾” Plywood =3psf Total =8.5psf[ = 7.9psf]=20psf (b)Loads Supported by Joist: Joist supports loads that are on the plywood (from part a), mechanical duct and its self-weight.•Self-weight of joist: Actual dimensions of 2X10 Douglas Fir roof joist are 1.5”X 9.25”Therefore self-weight= (34pcf)*(1.5/12 ft)*(9.25/12 ft) = 3.27plf[OR: ¾” Plywood =0.4psf/(1/8")*(3/4")=2.4psf]
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Civil Engineering, Department of Civil Engineering, University of California, Construction terminology, J. P. Moehle