For some reason this question keeps getting rejected for b.s. reasons. Please do not reject the question even if I you aren't confident in your answer just poSt it any way Snow Loads :ASCE 7- 16Nominal Snow ForcesRoof slopeHoriz. eave to ridge dist (W) =Roof length parallel to ridge (L) =18.4 deg18.0 ft72.0 ftType of RoofGround Snow LoadHip and gable w/ raftersPg =46.0 psfRisk CategoryImportance FactorThermal FactorII1.0Ct =1.10Exposure FactorCe =1.2Pf = 0.7*Ce*Ct*I'PgUnobstructed Slippery Surface42.5 psf%3DyesSloped-roof FactorBalanced Snow LoadCs =0.8636.5 psfNear ground level surface balanced snow load =46.0 psfRain on Snow Surcharge AngleCode Maximum Rain SurchargeRain on Snow SurchargePs plus rain surchargeMinimum Snow Load0.36 deg5.0 psf0.0 psf36.5 psf0.0 psf%3DPm =NOTE: Alternate spans of continuous beamsshall be loaded with half the design roof snowload so as to produce the greatest possibleeffect - see code for loading diagrams andexceptions for gable roofs..Uniform Roof Design Snow Load =36.5 psfUnbalanced Snow Loads - for Hip & Gable roofs only30.26 degRequired if slope is between 7 on 12 =and 2.38 deg =2.38 deg Unbalanced snow loads must be appliedWindward snow load =Leeward snow load =0.0 psf46.0 psf = 1* PgWindward Snow Drifts 1 - Against walls, parapets, etcUpwind fetchProjection heightSnow densityBalanced snow heightlu =15.0 fth =3.0 ft20.0 pcfhb =1.83 fthd =hc =1.27 ft1.17 ftTherefore, design for drift1.17 ft5.50 ft23.4 psf36.5 psfSurcharge LoadDue to Driftinghc/hb >0.2 = 0.6Drift height (hc)Drift widthSurcharge load:hcW =pd = y*hd =hdpdBalanced Snow LoadBalanced Snow load:hb59.9 psfWindward Snow Drifts 2 - Against walls, parapets, etc24.0 ftLuUpwind fetchProjection heightSnow densityBalanced snow heightlu =h =4.0 ft20.0 pct1.83 ft%3Dhb =Note: If bottom of projection is at least 2 feetabove hb then snow drift is not required.1.42 ft2.17 ftTherefore, design for drift1.42 ft5.68 ft28.4 psf36.5 psf64.9 psfhd =hc =hc/hb >0.2 = 1.2Drift height (hd)Drift widthSurcharge load:W =pd = y*hd =Balanced Snow load: 4. A city of 150,000 people discharges sewage, after removing 90% of the total suspendedsolids (TSS), into a stream with 0 TSS and a flow of 50 m³/min. A newspaper critic stated thatwhat the city is doing is equivalent to discharging untreated sewage from 15,000 people. Isthis criticism fair? Assume that each person produces 2.67 x 104 m3/min of wastewater witha TSS concentration of 200 mg/L.

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1. A city of 150,000 people discharges sewage, after removing 90% of the total suspended solids (TSS), into a stream with 0 TSS and a flow of 50 m3/min. A newspaper critic stated that what the city is doing is equivalent to discharging untreated sewage from 15,000 people. Is this criticism fair? Assume that each person produces 2.67 x 104 m/min of wastewater with a TSS concentration of 200 mg/L.

1. A city of 150,000 people discharges sewage, after removing 90% of the total suspended solids (TSS), into a stream with 0 TSS and a flow of 50 m3/min. A newspaper critic stated that what the city is doing is equivalent to discharging untreated sewage from 15,000 people. Is this criticism fair? Assume that each person produces 2.67 x 104 m/min of wastewater with a TSS concentration of 200 mg/L.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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