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Problem 10.3

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Problem 10.3 a) Roof shear due to transverse wind load: vroof = (wT L/2)/b = [(375 lb/ft)(100 ft)/2] / 75 ft = 250 lb/ft Wall shear due to transverse wind load: vwall = (wT L/2)/bwall = [(375 lb/ft)(100 ft)/2]/(30 ft + 30 ft) = 312.5 lb/ft b) T1-11 plywood siding is 5/8 in. or 19/32 in. thick. Although the siding may have a thickness of 3/8 in. at grooves, the value for 6d casing nails must be chosen from the line in the table with a minimum net thickness of 5/16 in. at grooves. IBC Table 2306.4.1 (wind): 6d galvanized casing nails spaced 3 in. o.c. at edges, 12 in. o.c. in field; vallow = 1.4(275 lb/ft) = 385 lb/ft > 312.5 lb/ft OK [Alternate solution: 8d galvanized casing nails spaced 4 in. o.c. at edges, 12 in. o.c. in field; vallow = 1.4(240 lb/ft) = 336 lb/ft > 312.5 lb/ft OK] h/bsegment = 12 ft/30 ft = 0.4 < 3.5 OK c) Tgross = vwall(h) = 312.5 lb/ft (12 ft) = 3750 lb Tnet = Tgross – 0.6 (wD)(bsegment/2) = 3750 lb – 0.6(250 lb/ft)(30 ft/2) = 1500 lb Problem 10.10 a) 2nd floor: v2nd = RR/bwall = 2100 lb/8 ft = 263 lb/ft IBC Table 2306.4.1(wind): 15/32 STR I 8d common or galvanized box nails spaced 6 in. o.c. at edges, 12 in. o.c. in field, all edges nailed and blocked; vallow = 1.4(280 lb/ft) = 392 lb/ft > 263lb/ft OK 1st floor: v1st = (RR + R2)/bwall = (2100 lb + 2600 lb)/8 ft = 588 lb/ft IBC Table 2306.4.1(wind): 8d common or galvanized box nails spaced 4 in. o.c. at edges, 12 in. o.c. in field, all edges nailed and blocked; vallow = 1.4(430 lb/ft) = 602 lb/ft > 588 lb/ft OK b) Roof strut: vR = RR/b = 2100 lb/40 ft = 52.5 lb/ft TA = vR (24 ft) = (52.5 lb/ft)(24 ft) = 1260 lb; TB = vR (32 ft) = (52.5 lb/ft)(32 ft) = 1680 lb 2nd floor strut: v2 = R2/b = 2600 lb/40 ft = 65 lb/ft TC = v2 (24 ft) = (65 lb/ft)(24 ft) = 1560 lb; TD = v2 (32 ft) = (65 lb/ft)(32 ft) = 2080 lb c) Load combination p. 2.77: 0.6D + W At 2nd floor: OMgross = RR(h2) = (2100 lb)(9 ft) = 18,900 ft-lb Tgross = OMgross/bwall = 18,900 ft-lb/8 ft = 2360 lb OMnet = OMgross – 0.6(Droof)(bwall)(bwall/2) = 18,900 ft-lb – (0.6)(155 lb/ft)(8 ft)(8 ft/2) = 15,900 ft-lb Tnet = OMnet/bwall = 15,900 ft-lb/8 ft = 1990 lb At foundation: OMgross = RR(h1 + h2) + R2(h1) = (2100 lb)(19 ft) + (2600 lb)(10 ft) = 65,900 ft-lb Tgross = OMgross/bwall = 65,900 ft-lb/8 ft = 8240 lb OMnet = OMgross – 0.6(Droof + Dfloor)(bwall)(bwall/2) = 65,900 ft-lb – (0.6)(155 lb/ft + 250 lb/ft)(8 ft)(8 ft/2) = 58,100 ft-lb Tnet = OMnet/bwall = 58,100 ft-lb/8 ft = 7270 lb d) Load combination p. 2.77: D + 0.75W + 0.75 L + 0.75Lr 2nd floor; D + 0.75(Lr + W) is governing load combination: 155 lb/ft (8 ft) + 0.75[(200 lb/ft)(8 ft) + (2100 lb)(9 ft/8 ft)] = 4210 lb Foundation: D + 0.75(L + Lr + W) is governing load combination:(155+250 lb/ft)(8 ft) + 0.75[(200+500 lb/ft)(8 ft) + 2100 lb(19 ft/8 ft) + 2600 lb(10 ft/8 ft)] = 13,600 lb e) Nbolts ≥ (RR + R2)/Vallow = (2100 lb + 2600 lb)/1500 lb


Problem 10.3

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