200 W MARINE VIEW DR ON TRAC LOGISTICS 2025-09-16

SHUTLER CONSULTING ENGINEERS, INC. Job Name : Baywood Industrial Structural Engineers Job No. : 20-33 12503 Bel-Red Road, Suite 100 Engr: : CT Bellevue, Washington 98005 Date: : 10-21-2020 (425) 450-4075 Sheet No.: Description: Panel 37 - 36" leg Page 2 5 Number 6 Bars Results shown are for 36.00 inch wide section. CASE 1 @ Roof, U = 1.4D ( M = P*e --- No P * Deflection. ) Service Load = LOU Service Defl. = N.A. in. Mu = 44.82 in-k Ultimate Defl.= N.A. in. Phi * Mn = 798.81 in-k CASE 1 @ Mid-ht., U = 1.4D ( M = .5*P*e + P * Deflection. ) Service Load = 1.OD Service Defl. = 0.03 in. Mu = 30.48 in-k Ultimate Defl.= 0.19 in, Phi * Mn = 918.75 in-k CASE 2 @ Roof, U = 1.2D + 1.6 (Lr or S or R) + l.OL (M=P*e, No P*Delta) Service Load = 1.0D + 1.0(Lr or S or R) Service Defl, = N.A. in. Mu - 123.78 in-k Ultimate Defl.= N.A. in. Phi * Mn - 831.16 in-k CASE 2 @ Mid-ht, U = 1.2D + 1.6(Lr or S or R) + 1.0L (M=.5*P*e + P*Delta.) Service Load = 1.0➢ + 1.0(Lr or S or R) Service Defl. = 0.08 in. Mu = 87.17 in-k Ultimate Deft.= 0.55 in. Phi * Mn = 933.48 in-k CASE 3 U = 1.2D + 1.6 ( Lr or S or R ) + 0.5W Service load = 1.0D + 0.75L + 0.75*0.6W + 0.75(Lr or S or R) Service Defl. = 0.90 in. Mu = 413.40 in-k i. Ultimate Defl.- 2.59 in. Phi * Mn = 933.48 in-k f. CASE 4 U = 1.2D + 1.0W + 1.0L + 0.5 (Lr or S or R) Service load = 1.OD + O.75L + 0.75*0.6W + 0.75(Lr or S or R) Service Defl. = 0.92 in. Mu = 660.65 in-k Ultimate Defl.= 4.22 in. Phi * Mn = 909.95 in-k CASE 5 U = (1.2 + 0.2 Sds)D + 1.0E + 1.OL + 0.2S Service load = (1.0 + 0.1 Sds)D + 0.75E + 0.525E + 0.75(Lr or S or R) Service Defl. = 1.65 in. Mu = 862.87 in-k Ultimate Defl.= 5.46 in. Phi * Mn - 919.83 in-k II CASE 6 U = 1.2D + 1.0W Service load = 1.0D + 0.6W Service Defl. = 1.21 in. Mu = 627.08 in-k Ultimate Defl.= 4.04 in. Phi * Mn 899.21 in-k CASE 7 U = (1.2 + 0.2 Sds)D + 1.0E Service load = (1 + 0.14Sds)D + 0.7E Service Defl. = 2.18 in. Mu = 847.04 in-k '. Ultimate Defl.= 5.38 in. Phi * Mn = 915.55 in-k CASE 8 U = 0,9D + 1.0W Service load = 0.6D + 0.6W Service Defl. = 1.12 in. Mu = 580.49 in-k Ultimate Defl.= 3.83 in. Phi * Mn = 869.76 in-k == == CASE 9 U = (0.9 - 0.2 Sds)D + 1.0E Service load = (0.6 - 0.14 Sds)D + 0.7E Service Defl. = 2.00 in. Mu - 723.07 in-k Ultimate Defl.- 4.84 in. Phi * Mn = 853.27 in-k _