ColumnC
Axialload = 0 96(3)4 = 11 52kN
kN
kN
2.3
2.4
ColumnA
Axialload = 45(20)25 = 22,500lb
ColumnF
Axialload = 45(40)25 = 45,000lb
ColumnH
Axialload = 45(40)50 = 90,000lb
2.6
SeeSolutionofProblem2.5
BeamCD
Uniformlydistributedload = 662.3 + 120 6 12 (7) = 662.3 + 420 = 1082.3lb/ft
GirderAE
Uniformlydistributedload = 111.3lb/ft
ConcentratedloadatC = 8279 + 420 25 2 = 13,529lb
ConcentratedloadsatAandE = 4529lb
BeamBF
Uniformlydistributedload
Uniformlydistributedload
ConcentratedloadsatBandC
ConcentratedloadsatAandD
Live load = 40 psf
Uniformlydistributedload = 40(12) = 480lb/ft
ConcentratedloadatC = 6000lb
ConcentratedloadsatAandE = 40(6) 25 2 = 3000lb
2.10
Live load = 4.79kPa = 4.79kN/m2 BeamBF
Uniformlydistributedload = 4.79(5) = 23.95kN/m
kN/m
GirderAD
ConcentratedloadsatBandC = 119 75kN
ConcentratedloadsatAandD = 4 79(2 5) 10 2 =
2.11
BeamEF
Uniformlydistributedload = 20(9) = 180lb/ft
GirderAG
ConcentratedloadsatCandE = 1800lb
ConcentratedloadsatAandG = 1800/2 = 900lb
ColumnA
Concentratedload = 2700lb
2.12
V = 95 mp , h = 40 + (15/2) = 47.5 ft, zg = 1200 ft, α = 7.0, Kzt = 1, Kd = 0.85, and Ke = 1. K h =
= 201 475 1200 08 27 / qh = 0.00256(0.8)(1)(0.85)(1)(95)2 = 15.71 psf
G = 0.85
For θ = 45° and h/L = 47.5/30 = 1.58:
Cp = 0.3 for windward side
Cp = - 0.6 for leeward side
Thus, the wind pressures are:
ph = 15.71 (0.85)(0.3) = 4.0 psf for windward side
ph = 15.71 (0.85)(- 0.6) = -8.0 psf for leeward side
2.13
V = 51 m/s, h = 12 + 5 2 = 14.5 m, zg = 365.76 m, α = 7.0, Kzt = 1, Kd = 0.85, and Ke = 1.
Cp = - 0.1 and 0.25 for windward side
Cp = - 0.6 for leeward side
Thus, the wind pressure are: ph = (1084.2)(0.85)(- 0.1) = -92.2 N/m 2 ph = (1084.2)(0.85)(0.25) = 230.4 N/m 2 ph = (1084.2)(0.85)(- 0.6) = -552.9 N/m 2 for leeward side
for windward side
2.14
V = 120 mph, h =+ = 30 11 2 355., ft zg = 900 ft, α = 9.5, Kzt = 1, Kd = 0.85, and Ke = 1. K h =
= 201 355 900 102 29/.5
qh = 0.00256(1.02)(1)(0.85)(1)(120)2 = 32 psf
G = 0.85
Roof slope: θ = tan -1 (11/20) = 28.8° h L == 355 40 089 . .
Cp = - 0.3 and 0.2 for windward side
Cp = - 0.6 for leeward side
Thus, the wind pressures are: p p h h =−
32 0850 38 2 32 0850 25 4 (. )( .) (. )( .) . psf psf for winndward side psffor leeward side ph =− =− 32 0850 6163 (. )( .)
2.15
V = 120 mph, zg = 900 ft, α = 9.5
From the solution of Problem 2.14:
qh = 32 psf and G = 0.85
Leeward wall:
For L/B = 40/30 = 1.33, Cp = - 0.45
Thus, the wind pressure, ph = 32 (0.85) (- 0.45) =−1224.psf
Windward wall: Cp = 0.8
2.16 = 20psf, C e = 1, C t = 1, = 1.2
= 0 7 C e C t I s p g = 0 7(1)(1)(1 2)(20) = 16 8psf
θ = tan 1 (11/20) = 28.8◦ > 15◦
C s = 1
Balancedload = ps = C s pf = 1(16 8) = 16 8psf
θ 16.8 psf
Balanced Snow Load pg pf Is pf Therefore,theminimumvaluesof neednotbeconsidered.
2.17 p g = 1.2kN/m2 , C e = 1, C t = 1, I s = 1.1 pf = 0 7 C e C t I s p g = 0 7(1)(1)(1 1)(1 2) = 0 92kN/m2
θ = tan 1 (5/6) = 39.8◦ > 15◦
Therefore,theminimumvaluesofpf neednotbeconsidered.
C s = 1 θ 30◦ 40◦ = 0.76
BalancedLoad = ps = C s = 0.76(0.92) = 0.7kN/m2 0.7 kN/m2
pf