WIND LOADS

IV. Standard Occupancy Structure (Table 103-1)

Basic Wind Speed = 250kPh or 69.44m/s (Figure 207A.5-1A)

Kd = 0.85 (Table 207A.6-1)

Exposure B (207A.7.3)

Kzt = 1 (207A.8.2)

Rigid Structure: G=0.85 (207A.9.4)

Enclosed Building: GCpi = ±0.18 (Table 207A.11-1)

From Table 207A.9-1: α = 7, zg = 365.76

Kh = 2.01(6.35/365.76)2/7 = 0.631

Kz = 2.01(5.6/365.76)2/7 = 0.609

qh = 0.613(0.631)(1)(0.85)(69.44)2 = 1585.361 Pa

qz = 0.613(0.609)(1)(0.85)(69.44)2 = 1530.087 Pa

 External Coefficient Pressure

Walls:
 Windward; Cp = 0.8
 Leeward; L/B = 0.55
Cp = -0.5
 Side; Cp = -0.7

Roofs: (Angle of Inclination = 27.15°)

 Windward: H/L = 1.15

By interpolation:

27.15-25/30-25 = Cp1 - (-0.5)/-0.3- (-0.5) = Cp2 – 0/0.2- 0

Cp1 = -0.414 ; Cp2 = 0.086

 Leeward; Cp = -0.6
Burst Condition: (GCpi = +0.18)

P = qGCp-qi(GCpi)

 Walls
- Windward:
P = 1530.087(0.85)(0.8) - 1585.361(0.18)
P = 755.094 Pa

- Leeward:
P = 1585.361(0.85)(-0.5) - 1585.361(0.18)
P = -959.143 Pa

- Side:
P = 1585.361(0.85)(-0.7) - 1585.361(0.18)
P = -1228.655 Pa

 Roofs
- Windward:
(Cp1) P1 = 1585.361(0.85)(-0.414) - 1585.361(0.18)
P1 = -843.254 Pa; Govern
(Cp2) P2 = 1585.361(0.85)(0.086) – 1585.361(0.18)
P2 = -169.475 Pa

- Leeward:
P = 1585.361(0.85)(-0.6) – 1585.361(0.18)
P = -1093.9 Pa
Suction Condition: (GCpi = -0.18)

P = qGCp-qi(GCpi)

 Walls
- Windward:
P = 1530.087(0.85)(0.8) -
1585.361(-0.18)
P = 1325.824 Pa

- Leeward:
P = 1585.361(0.85)(-0.5) -
1585.361(-0.18)
P = -388.413 Pa

- Side:
P = 1585.361(0.85)(-0.7) - 1585.361(-0.18)
P = -657.925 Pa

 Roofs
- Windward:
(Cp1) P1 = 1585.361(0.85)(-0.414) - 1585.361(-0.18)
P1 = -272.524 Pa
(Cp2) P2 = 1585.361(0.85)(0.086) – 1585.361(-0.18)
P2 = 401.255 Pa; Govern

- Leeward:
P = 1585.361(0.85)(-0.6) – 1585.361(-0.18)
P = -523.169 Pa
 TRUSS

 Angle of Inclination
= 27.15°

 Spacing of Truss
= 2.5m

 Plan Area
5.5m(2.5m) = 13.75m²

 Sloping Area
3.3m(2.5m) = 8.25m²

Loadings on Truss:

Dead Load (Corrugated Asbestos

Cement Roofing) = 0.19 kPa

Assumed Dead Load of Purlin = 0.2 kN/m

Assumed Dead Load of Sag Rod = 0.01 kN/m

Live Load on Roof = 0.75 kPa

Design of Purlins (A36 Steel)

DL = 1.1(0.19) + 0.2 + 0.1 = 0.419 kN/m

LL = 0.75(0.983) = 0.737 kN/m

WL = 0.401(1.1) = 0.4411 kN/m

Wu = 1.2(0.419cos27.15) + 1.6(0.737cos27.15) + 0.5(0.4411) = 1.7172 kN/m

Try C75x6: Sx = 17.04x10³mm³

Mu = 0.106(1.7172)(1.25)² = 0.2844 kN-m
Check for resistance against bending:
Vu = 86/142 (1.82)(1.25) = 1.3 kN
Mb = SxFy
Sx = Mu/Fy
= 17.04x10³mm³ [0.6(250)]
=284400/0.6(250) = 1896mm³
= 2.56 kN-m > 0.2844 kN-m ∴ OK
Design for Sag Rod and Tie Rod (A36 Steel)

DL = 0.19(1.1) + 6(9.81/1000) = 0.268 kN/m

LL = 0.75(1.1) = 0.825 kN/m

WL = 0.401(1.1) = 0.4411 kN/m

Wu = 1.2(0.4411sin27.15) + 1.6(0.825sin27.15) = 0.844 kN/m

Mu = wL²/32

= 0.844(2.5)²/32 = 0.165 kN-m

ΣMBᴬᴮ = -0.165 kN-m

-0.165 = R1(1.25) – 0.844(1.25)(1.25/2)

R1 = 0.4 kN

ΣFv = 0

0.4 + R2 + 0.4 – 0.844(2.5) = 0

R2 = 1.31 kN

For Sag Rod that will support 4 Purlins:

FT = 4(1.31) = 5.24 kN

AREQ of Sag Rod (Fu = 400MPa)

Ab = 5.24x103/0.5625(400) = 23.3 mm2

23.3 = πd2/4

D = 5.4mm < 12mm ∴ use 12mm Dia. Sag Rod

Design for Tie Rod

Tr = FT/cos27.15°

= 5.24/cos27.15°

Tr = 5.9 kN

AREQ of Tie Rod (Fu = 400MPa)

Ab = 5.9x103/0.5625(400) = 26.22 mm2

26.22 = πd2/4

D = 5.78mm < 12mm ∴ use 12mm Dia. Tie Rod

Design for Truss (A36 Steel)

Dead Loads (Upper Panel): Dead Loads (Lower Panel):

Weight of G.I sheet on inclined area - Suspended Steel Channel (0.1 kPa)
(0.19kPa)
= 0.1(2.5)(5.5) = 1.375 kN
= 2[0.19(2.5)(3.3)] = 3.135 kN
- Acoustic Fiber Board (0.05 kPa)
Weight of Purlins (6kg/m)
= 0.05(2.5)(5.5) = 0.687 kN
= 8(6)(9.81/1000)(2.5) = 1.177 kN
- Electrical, Plumbing and other fixtures (0.1
Weight of Truss on Plan Area kPa)

W = 10(L/3 + 5)A = 0.1(2.5)(5.5) = 1.375 kN

= 10(5.5/3 + 5)(5.5)(2.5) = 0.9395kN

WT at Upper Panel = 0.4698 kN Total Dead Load at Lower Panel:

WT at Lower Panel = 0.4698 kN 1.375 + 0.6875 + 1.375 + 0.4698 = 3.073 kN

Total Dead Load at Upper Panel: Live Load: (0.737 kN/m)

3.135 + 1.177 + 0.4698 = 4.7818 kN LL = 0.737(2.5)(8) = 14.74 kN

Wind Load: (0.401 kPa)

WL = 2[0.401(2.5)(3.3)] = 6.617 kN
Dead Load (Upper Panel) = 4.7818 kN PDL(UP) = 4.7818/4 = 1.2 kN
Dead Load (Lower Panel) = 3.9073 kN PDL(LP) = 3.9073/4 = 0.98 kN
Live Load = 14.74 kN PLL = 14.74 /4 = 3.685 kN
Wind Load = 6.617 kN PWL = 6.617cos27.15°/4 = 1.5 kN
Design for Tension Members: (A36 Steel)

Pu = 27.0911 kN

AgREQ = 27.0911x103/0.9(250) = 120.4mm2

AeREQ = 27.0911x103/0.75(400) = 90.3mm2

Try L51 x 51 x 4: (A = 317mm2)

Ag = 2(317) = 634mm2

U = 1 – 13.6/100 =0.864

Ae = 0.864(634) = 547.776mm2

ФPn = 0.75(400)(547.776)

ФPn = 164.33 kN > 27.0911 kN ∴ OK

For Gusset Plate:

Thickness Req = AgREQ/d

TREQ = 120.4/51 = 2.36mm

Try 5mm Thickness Gusset Plate:

Ag = 5(51) = 255mm2

U=1

ФPn = 0.75(400)(255)

ФPn = 76.5 kN > 27.0911 kN ∴ OK

 Block Shear for Gusset Plate:

For Block Shear:

Agv = 100(5)(2) = 1000mm2

Anv = 100(5)(2) = 1000mm2

Agv = 100(3.2)(4) = 1280mm2
Ant = 51(5) = 255mm2
Anv = 100(3.2)(4) = 1280mm2

Ant = 51(3.2)(2) = 326.4mm2

Rn = 0.6(250)(1000) + 1(400)(255)

Rn = 252 kN; Govern

Rn = 0.6(250)(1280) + 1(400)(326.4)

Rn = 322.56 kN; Govern

Rn = 0.6(400)(1000) + 1(400)(255)

Rn = 342 kN
Rn = 0.6(400)(1280) + 1(400)(326.4)

Rn = 437.76 kN
ФPn = 0.75(252)

ФPn = 189 kN > 27.0911 kN ∴ OK

ФPn = 0.75(322.56)

ФPn = 241.92 kN > 27.0911 kN ∴ OK

Design for Compression Members: (A36 Steel)

Pu = 30.4456 kN

rmin = 1650/200 = 8.25mm

KL/r = 0.65(1650)/8.25 = 130

4.71
√ 200000
250
= 133.2 > 130 ∴ Fcr = (0.658Fy/Fe)Fy

2
π (200000)
Fe = 2 = 116.8 MPa
130
Self Weight = 10(5.85/3 + 5)(5.85)(2.5)
Fcr = (0.658250/116.8)250 =102.06 MPa
Self Weight = 1.016 kN
AgREQ = 30.4456x103/102.06 = 298.31mm2

Total Length of Truss Members:

Try 2L 51 x 51 x 5 (A = 466; r = 15.54)
5.85 + 3.3(2) + 1.5 + 0.75(2) + 1.65(2) = 18.75m
AgREQ = 2(466) = 932mm2 > 298.31mm2 ∴ OK

b/t = 51/4.8 = 10.625 < 0.45

√ 200000
250
= 12.73
WT = 18.75(3.63)(9.81/1000)

therefore shape is non-slender WT = = 0.667 kN < 1.016 kN ∴OK

Therefore use 2L 51 x 51 x 5 for both Tension

and Compression members.
KL/r = 0.65(1650)/15.54 = 69.02

4.71
√ 200000
250
= 133.2 > 69.02 ∴Fcr = (0.658Fy/Fe)Fy

2
π (200000)
Fe = = 414.36 MPa
69.022

Fcr = (0.658250/414.36)250 = 194.21 MPa

ФPn = 0.9(194.21)(932) = 162.9 kN > 30.3356 kN ∴ OK

Dead Loads and Live Loads (Second Floor)

Beam Dimensions:

H = 5500/16 = 343.75mm ≈ 400mm

D = 400 - 75 = 325mm

B = 325/1.7 = 191.2 ≈ 200mm

Slab Thickness:

2500/24 = 104.17 ≈ 150mm

- Ext. Wall: a. Dead Loads

150mm CHB, 19.6 kN/m3 Dens. Grout Spacing 400 WBEAMS = 23.6(0.4)(0.2) = 1.9 kN/m

2.11 + 2(0.24) = 2.59 kPa PSLAB = 23.6(0.15) = 3.5 kPa

WEXTWALL = 2.59(2.7) = 7 kN/m

- Floor Finishes:

Terrazzos (25mm) on stone concrete fill = 1.53 kPa Total Weight at Slab:

WSLAB = 3.54 + 0.1 + 0.05 + 1.53 + 0.1 = 5.32 kPa

- Electrical, Plumbing and Other Fixtures = 0.1 kPa

-Ceilings: (Suspended Steel Channel = 0.1 kPa) b. Live Load:

(Acoustic Fiber Board = 0.05 kPa) Dining Rooms and Restaurants = 4.8 kPa
DL at IJ:

WD = WSLAB + WEXTWALL + WBEAM

WD = 5.32(2.5)/2 + 7 +1.9 = 15.55 kN/m

DL at GK, EF:

WD = WSLAB + WBEAM

WD = 5.32(2.5) + 1.9 = 15.2 kN/m

DL at KL:

WD = WSLAB + WBEAM

WD = 5.32(1.5)/2 + 1.9 = 5.89 kN/m

LL at IJ:
DL at CD:
WL = 4.8(2.5/2) = 6 kN/m
WD = WSLAB + WBEAM

WD = 5.32(1.5)/2 + 5.32(2.5)/2 + 1.9 = 12.54 kN/m

LL at GK, EF:

WL = 4.8(2) = 12 kN/m
DL at AB:

WD = WEXTWALL + WBEAM

WD = 7 + 1.9 = 8.9 kN/m LL at KL:

WL = 4.8(1.5/2) = 3.6 kN/m

DL at Longitudinal:

WD = WEXTWALL + WBEAM LL at CD:

WD = 7 + 1.9 = 8.9 kN/m WL = 4.8(1.5/2) + 4.8(2.5/2) = 9.6 kN/m

Load at Windward Wall (1325.824 Pa)

Load at Leeward Wall (-959.143 Pa)

- Windward Wall

W1 = 1325.824(1.25m) = 1657.28 N/m

W2 = 1325.824(2.5m) = 3314.56 N/m

- Leeward Wall

W1 = -959.143(1.25m) = -1200 N/m

W2 = -959.143(2.5m) = -2400 N/m

P1 = 1/2(3314.56 + 2515.5)(2.7/2) = 3935.3 N ≈ 3.94 kN

P2 = 1/2(2515.5 + 828.23)(2.7/2 + 2.9/2) = 4723.22 N ≈ 4.72 kN

P3 = 2400(2.7/2) = 3240 N ≈ 3.24 kN

P4 = 2400(2.7/2 + 2.9/2) = 6720 N ≈ 6.72 kN

Roof Beam Design:

(Use Grade 60 Steel. Fy = 420MPa; Fc’ = 28MPa)

Dead Load at Roof Beam (1.9 kN/m)

Mn = 1.9(4.4)2/8 = 4.6 kN/m

Moment from Portal Lateral Loads (4.85 kN-m)

Mu = 0.9DL + 1WL Try 12mm Bar: (A = 113mm2)

Mu = 0.9(4.6) + 1(4.85) = 9 kN-m N = 214.5/113 = 1.9 ≈ 2pcs

Check if DRB or SRB

S = 200 – 2(40) – 2(10) – 2(2)(10) – 12 = 48mm
Cmax = 3/8 (325) = 121.9mm
D = 400 – 40 – 10 – 12/2 = 344mm
amax = 0.85(121.9) = 103.6mm

800−420
Фmax = 0.65 + 0.25( ¿ = 0.814
1000−420
ФMnmax = 0.814[0.85(28)(103.6)(200)(325 –
103.6/2)]

ФMnmax = 109.67 kN-m > 9 kN-m ∴ Design as SRB

Rn = 9x106/0.9(200)(325)2 = 0.473

ρ = 0.85 (28/420)[1 - 1−
√ 2(0.473)
0.85(28)
] = 0.0011

ρmin = 1.4/420 = 0.0033 ∴ ρ = ρmin

As = 0.0033(200)(325) = 214.5mm2
Check Adequacy:

T=C

2(113)(420) = 0.85(28)(a)(200)

a = 19.94m

C = 19.94/0.85 = 23.46mm
344−23.46
fs = 600( ¿
23.46

fs = 8197.95 > 1000 ∴ Tension Controlled (Φ

= 0.9)

ΦMn = 0.9[2(113)(420)(344 – 19.94/2)]

ΦMn = 28.54 kN-m > 9 kN-m ∴ Adequate

∴ Use 10mm Bar Stirrups Spaced at

160mm
Design for Shear:

Vu = 7.315 – 2.66(0.344) = 6.4 kN

Vc = 0.17(1)√ 28 (200)(344) = 53.6 kN

ΦVc = 0.75(53.6) = 40.2 kN > Vu ∴ Shear

Reinforcement not necessary

ΦVc/2 = 20.1 kN > Vu

Minimum Shear Reinforcement:

2(79)(420)
S1 = = 1011.36mm
0.062 √ 28(200)

2(79)(420)
S2 = =948mm
0.35 (200)

Smax = 325/2 = 162.5mm ≈ 160mm

 ФMnmax = 109.67 kN-m > 77.58 kN-m ∴ Design
as SRB

Design for Beams at Second Floor:

Dead Loads:

WSLAB = 6.65 kN/m

WBEAM = 1.9 kN/m

WEXTWALL 7 kN/m

WTOTAL = 15.55 kN/m

Live Load: (6 kN/m)

Moment from Portal Lateral Loads (18.35 kN-m)

Rn = 77.58x106/0.9(200)(325)2 = 4.08

√
ρ = 0.85 (28/420)[1 - 1−
2(4.08)
0.85(28)
] = 0.0107

ρmin = 1.4/420 = 0.0033 ∴ ρ = ρmin

2
MuDL+LL = 28.26(4.4) /8 = 68.4 kN-m
As = 0.0107(200)(325) = 697.55mm2
Total Mu = 68.4 + 0.5(18.35) = 77.58 kN-m

Try 25mm Bar: (A = 491mm2)

Check if DRB or SRB
N = 697.55/491 = 1.42 ≈ 2pcs
Cmax = 3/8 (325) = 121.9mm
S = 200 – 2(40) – 2(10) – 2(2)(10) – 25
amax = 0.85(121.9) = 103.6mm S = 35mm > 25mm ∴ OK

800−420
Фmax = 0.65 + 0.25( ¿ = 0.814
1000−420
ФMnmax = 0.814[0.85(28)(103.6)(200)(325 –
103.6/2)]
D = 400 - 40 - 10 - 25/2 = 337.5mm

Check Adequacy:

T=C

2(491)(420) = 0.85(28)(a)(200)

a = 86.65mm

C = 86.65/0.85 = 101.94mm
337.5−101.94
fs = 600( ¿
101.94

fs = 1386.46 > 1000 ∴ Tension Controlled (Φ = 0.9)

ΦMn = 0.9[2(491)(420)(337.5 - 86.65/2)]

ΦMn = 109.2 kN-m > 77.58 kN-m ∴ Adequate

Design for Shear and Torsion:

Vu = 77.72 - 28.26(0.3375) = 68.18 kN

Vc = 0.17(1)√ 28 (200)(337.5) = 60.72 kN

ΦVc = 0.75(60.72) = 45.54 kN < Vu ∴ Shear Reinforcement is Necessary

Vs = 68.18/0.75 - 60.72 = 30.2 kN

Acp = 200(400) = 80000mm2

Pcp = 2(200+400) = 1200mm

ΦTth = 0.75[0.083(1)√ 28 (800002/1200)]

ΦTth = 1.76 kN-m < Tu ∴ Torsion Reinforcement is

Necessary

Aoh = 110(310) = 34100mm2

Ao = 0.85(34100) = 28985mm2

Ph = 2(110 + 310) = 840mm

√( )
3 2
68.18 x 10 8.51 x 106 ( 840 )
+[ 2 ¿
]² ¿ = 3.1546
200 ( 337.5 ) 1.7 ( 34100 )

0.75 ¿ ] = 3.3 > 3.1546 ∴ OK

6
At 8.51 x 10 /0.75
= = 0.46603 mm2/mm
s 2(28985)( 420)(cot 45)
3
Av 30.2 x 10
= = 0.22125 mm2/mm
s 420(325)

AH = 0.22125 + 2(0.46603) = 1.1533 mm2/mm

Check for Min. Area:

200
a.) 0.062√ 28 = 0.1562 mm2/mm
420
200
b.) 0.35 = 0.1667 mm2/mm
420

Therefore use AH = 1.1533 mm2/mm

S = 2(79)/1.1533 = 136.9

Smax = Ph/8 = 840/8 = 105 ≈ 100mm

Therefore use S = 100mm

For Longitudinal Reinforcement:

AL = 840(420/420)(0.46603)(cot45) = 391.465 mm 2

Check:

0.42 √ 28(80000) 420

ALmin1 = −0.46603(840)( ) = 31.36mm2
420 420

0.42 √ 28(80000) 0.175 ( 200 ) 420

ALmin2 = −( )(840)( ) = 353.32mm2
420 420 420

Therefore use AL = 391.465mm2

Distribution of Longitdinal Reinforcement:

Layers = 300/300 + 1 = 2

AsLayers = 391.465/2 = 195.73mm2

Try 12mm bar (A = 113mm2) for top and middle layer:

N = 195.73/113 = 1.7 ≈ 2pcs

For Bottom Layer:

As = 2(491) + 195.73 = 1177.73mm2

(Try 28mm bar) N = 1177.73/616 = 1.9 ≈ 2pcs

Design for Slab: (WSLAB = 5.32 kPa)

Slab Thickness:

L/24 = 2500/24 = 104.17 ≈ 150mm

Wu = 1.2[5.32(1)] + 1.6[4.8(1)] = 14.06 kN/m

(Use WuLn2/10 max moment for uniform design of slab)

Mu = 14.06(2)2/10 = 5.624 kN-m

Rn = 5.624x106/0.9(1000)(124)2 = 0.406

√
ρ = 0.85 (28/420)[1 - 1−
2(0.406)
0.85(28)
] = 0.001

ρmin = 1.4/420 = 0.00333 ∴ ρ = ρmin

As = 0.00333(1000)(124) = 412.92mm2

Try 12mmø Bar (A = 113mm2)

113
S= (1000) = 273.66mm ≈ 250mm
412.92

Check Smax:

Smax = 3(150) = 450mm

Therefore use 12mmø Main Bar Spaced at 250mm O.C

For Temp Bars: Try 10mm ø Bars ( A = 79mm2)

0.0018 ( 420 )
AT = (1000 )( 150 ) = 270mm2
420

113
S= (1000) = 292.6mm ≈ 250mm
270

Check Smax:

Smax = 5(150) = 750mm

Therefore use 10mmø Temp. Bar Spaced at 250mm O.C