# 5. Design of Intermediate Purlins

July 16, 2017 | Author: Niño Ediliz B. Garcia | Category: Structural Load, Motion (Physics), Structural Engineering, Mechanical Engineering, Building Engineering

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Design of intermediate purlins...

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STEP 5. DESIGN OF INTERMEDIATE PURLINS and SAG RODS 1. Specifications a.) Length of purlins L = 6.00 m b.) Slope of roof θ = tan-1(2.00/7.50) = 14.93° c.) Inclined length of roof Inclined length = (7.52+22)1/2 = 7.762 m d.) Purlin spacing Assuming 11 purlins will be used S = 7.762/11 = 0.70 m e.) No. of lines of sag rod = 3 lines f.) Roof Slope = (2/7.5)*100 = 15% f.) Tributary area of purlin = 0.70(6.00) = 4.20 m2

WL

DL , LL

x y

2. Load computations a.) Dead Loads Weight of G.I. Sheet = 108(0.70) = 75.6 N/m Purlin self-weight + accessories = 100 N/m Total dead load = 75.6 + 100 = 175.6 N/m b.) Live Loads From Table 205 – 3 of NSCP 2010, Total LL = 1000 Pa Total live load = 1000(0.70) = 700 N/m c.) Wind Loads Wn

Maximum Pressure = -1807 Pa Total wind load = -1807(0.70) = -1264.9 N/m

Wt LOAD SUMMARY Dead load Wdn = 175.6cos(14.93°) = 169.67 N/m Wdt = 175.6sin(14.93°) = 45.25 N/m Live load Wln = 700cos(14.93°) = 676.36 N/m Wlt = 700sin(14.93°) = 180.36 N/m Wind load Wwn = -1264.9 N/m Wwt = 0

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y

FACTORED DESIGN LOADS Load Case 1: DL + LL Wn = 169.67 + 676.36 = 846.04 N/m Wt = 45.25 + 180/36 = 225.61 N/m Load Case 2: 0.75(DL + WL) Wn = 0.75(169.67 – 1264.9) = -821.42 N/m Wt = 0.75(45.25 + 0) = 33.93 N/m Among the two combinations Load Case 2 governs Therefore use, Wn = 846.04 N/m and Wt = 225.61 N/m 3. Select trial section a.) Moment at midspan Mx = WnL2/8 = 846.04(6.00)2/8 = 3807.159 N-m My = WtL2/90 = 225.61(6.00)2/90 = 90.244 N-m b.) Allowable bending stress Fbx = 0.6Fy = 0.6(250) = 150 MPa Fby = 0.75Fy = 0.75(250) = 187.5 MPa c.) Required section modules Sx = Mx/Fbx = 3807.159/150 = 25.38x103 mm3 Try C 130x10 Ix = 3.09x106 mm4 Sx = 48.6x103 mm3 Sy = 6.14x103 mm3 d.) Actual bending stress fbx = Mx/Sx = 3807.159/48.6 = 78.35 MPa fby = My/Sy = 90.244/6.14 = 14.70 MPa e.) Check for biaxial bending criteria (fbx/Fbx) + (fby/Fby) < 1.0 (78.34/150) + (14.70/187.5) = 0.601< 1.0 OK! f.) Check for deflection Wmax = 846.04 N/m E = 200 GPa Allowable deflection δall = 6000/180 = 33.33 mm Actual deflection δactual = 5WL4/384EI = 5(846.06)(6)4109/384(200)(3.09)109 δactual = 23.10 mm < 33.33 mm OK! g.) Actual weight of purlin 9.9 x 9.81 = 97.12 < 100 N/m , OK! Therefore USE C 130 x 10 intermediate purlins spaced at 0.70 m O.C. with 2 lines of sag rods

Tie Rod Critical Sag Rod

DESIGN OF SAG RODS MAXIMUM UNIFORM LOAD IS DUE TO LOAD CASE 1 CHECK FOR LOAD CASE 1 Wt = 225.61 N/m

1.) Total uniform load acting on purlin 45.25 + 180.36 = 225.61 N/m 2.) Max. Tension in Sag Rod to support one purlin 2.0m R1 = 4w/5 = 4(225.61)/5 = 180.488 R2 = 11w/5 = 11(225.61)/5 = 496.342 N 3.) Axial load on critical sag rod, Critical sag rod is carrying, n = 11 spans ΣR1 = 180.488(11) = 1985.368 ΣR2 = 496.342(11) = 5459.762 N 4.) Allowable Axial Stress, Ft = 250(0.60) = 150 MPa 5.) Required Diameter of Sag Rod, A = P/Ft = 5459.762 / 150 = 36.398 m² πdr²/4 = ΣR2 / Ft πdr²/4 = 36.398 mm² dr = 6.807 mm, say 10 mm Therefore use 10 mm dia. Sag rods.

R1

2.0 m

2.0 m R2

R2

R1