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0% found this document useful (0 votes)
144 views6 pages

Struct

Uploaded by

Gerardyne Liu
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Page 1

PROJECT: A PROPOSED WAREHOUSE


LOCATION: MANUEL L. QUEZON ST., ATIMONAN, QUEZON
OWNERS: RICO K. UY married to CARMINA CENDAÑA UY

STRUCTURAL COMPUTATIONS
DESIGN PARAMETERS:

A) DEAD LOADS:
1) STEEL ROOFING SHEETS 0.08 kPa
2) STEEL PURLINS 0.05 kPa
3) STEEL TRUSSES / RAFTERS 0.20 kPa
4) CEILING 0.10 kPa
5) UTILITY LINES 0.01 kPa
6) CHB’s (150-mm thick) 2.82 kPa
7) CHB’s (100-mm thick) 2.50 kPa
8) FLOOR FINISH 0.77 kPa
9) PARTITIONS 1.00 kPa
10) CONCRETE 23.60 kN/m³
11) STEEL 77.30 kN/m³
12) EARTH 17.30 kN/m³

B) LIVE LOADS:
1) ROOF LOAD 0.60 kPa
2) FLOOR LOAD (commercial) 2.40 kPa

C) LATERAL LOADS:
1) SEISMIC LOAD → (Zone 4) conform to NSCP requirements (for 9.0m+ high bldgs)
2) WIND LOAD ( V = 250 kph ) → P net = λK2tPnet1 = (0.4)(3.14) = 1.26 kPa

D) STRENGTH OF MATERIALS:
1) f’c ( class-A concrete ) 20.68 MPa
2) fy ( structural-grade steel ) 228.0 MPa
3) fy ( mild steel’s yield point ) 248.0 MPa
4) f’m ( masonry ) 7.50 MPa
5) q (allowable soil-bearing capacity for stiff sandy clay ) 192.0 kPa
(verify location @ 1.5-m deep)
E) DESIGN METHOD:
1) Ultimate Strength Design

F) LOAD COMBINATION:
1) 1.2DL + 0.60LL + 1.0WL (for components & cladding)
2) 1.2DL + 0.60LL + qnetWL (for main wind-force resisting system)

G) REFERENCES:
1) NSCP, 7th ed. 2015, vol. 1, ASEP
2) Handbook of Structural Steel Shapes and Sections, ASEP
3) Design Formulas, Besavilla

RAFAEL PETER J. R. VILLADIEGO


Civil Engineer
PRC # 54516 PTR # 1230574
Jan.3, 2023 Lucena City
TIN: 176-573-833
Validity : June 21, 2026
Page 2

WIND-LOAD ANALYSIS:

6.00 m
3.00 m P2 P3

9.00 m P1 P4

6.00 m

30.00 m

P1 = 0.80 x 1.57 x 6.00 x 9.01 = 67.90 kN 3.14 x 0.5


P2 = -0.20 x 1.57 x 6.00 x 15.30 = -28.8 kN = 1.57
P3 = 0.50 x 1.57 x 6.00 x 15.30 = 72.0 kN
P4 = 0.50 x 1.57 x 6.00 x 9.01 = 42.44 kN

(72.0-28.8)sin11.31º
Total Horizontal Loads = 67.90 + 42.44 + 8.48 = 118.81 kN
Equal Distribution of Shear = 112.9 ÷ 2 = 59.41 kN
OM @ Base of footing = 59.41 x 0.75 = 44.56 kN-m

Mcol = 44.56 kN-m


Mbeam = (44.56)(6.00 ÷ 2) ÷ 15 = 8.91 kN-m

DESIGN OF PURLIN:
1.2DL 1.2DL 0.60LL 1.0WL tanθ = 3.00 = 0.20
0.096 + 0.06 + 0.36 + 1.26 = 1.77 15.00
kPa θ = 11.31 °
SPACING = 0.60 m
SPAN = 6.000 m cosθ = 0.98 ; sinθ = 0.20

w= 1.77 x 0.60 = 1.063 kN/m

wN = 1.063 x cosθ = 1.043 kN/m ; wT = 1.063 x sinθ = 0.209 kN/m

MN = ,1/16 x wNL² = 1/16 x 1.04 x 6.00 ² = 2.35 kN-m

MT = ,1/90 x wTL² = 1/90 x 0.21 x 6.00 ² = 0.083 kN-m (with sag rods)

TRY LC 150 x 50 x 18 x 1.5 → Sx = 19.87 cm³ → Sy = 4.38 cm³


(from H S S S & S)
fbx = MN = 2.346 x 10 ⁶ N-mm = 118.1 MPa
Sx 19.87 x 10 ³ mm³

fby = MT = 0.083 x 10 ⁶ N-mm = 19.02 MPa


Sy 4.38 x 10 ³ mm³

fbx + fby = 137.1 MPa < Fb = (0.66) Fy = 0.66 x 248 = 163.7 MPa
therefore OK
STEEL PURLIN: LC 150 x 50 x 15 x 1.5 (6" x 2" x 5/8" x #15)
spaced @ 600 mm o.c. with 10-mm Ø sag rods
Page 3

DESIGN OF TRUSS:
qh = 47.3 x 10ˉ⁶kzkztkdV²Iw
qh = (47.3 x 10ˉ⁶)(1.09)(1.0)(0.85)(250)²(1.15) = 3.14 kPa
qnet = (0.50)(3.14) = 1.57 kPa

1200 mm

30.00
no ceiling
wDL = 0.08 + 0.05 + 0.20 + 0.00 + 0.01 = 0.34 x 1.20 = 0.408 kPa

W = 1.2DL + 060LL + (1.0)qnet = 0.41 + 0.36 + 1.57 = 2.34 kPa

PA = 1 x 2.34 x 6.00 x 30.00 x 1 = 210.4 kN


2 SECTION @ SUPPORTS
W= 2.338 x 6.00 = 14.03 kN/m

PT = 210.4 sinθ = 41.27 kN → fa = PT = 41.27 = 1.718 1200


AT 24020 mm

Properties of Angular Section: A= 1269 mm² → Cx = 22.05 mm


2 ∟75 mm x 75 mm x 8.0 mm I= 66.45 cm⁴ t= 8.00 mm
t1 = 8.00 mm

I = Ix + Ad² = 4[66.45 x 10⁴ + (1269)(1200/2 – 22.05)²] + 1/6(8.0)(1200– 16.0)³

I= 1698 + 2213 = 3911 x 10⁶ mm⁴

AT = 1269 4 + 2 8.00 x 1200 -


, 16.00 = 24020 mm²

Mumax = 1/16 wuL² = 1/16 x 14.03 x 30.00 ² = 789.08 kN-m

fb = Mc = (789.68)(106)(1200/2) = 121.05 MPa → Fb = 0.66Fy = (0.66)(248) = 163.68 MPa


I 3911 x 10⁶

Rx = I = 3911 x 10⁶ = 403.5 mm


AT 24020

kL/R = 1.0 x 30000 ÷ 403.5 = 74.34

Fa = 110.38 MPa ( from HSSS&S)


fa + fb ≤ (should be) 1.0 → 0.76 therefore OK
Fa Fb

STEEL TRUSS MEMBERS: TOP & BOTTOM CHORDS: 2∟ - 75 mm x 75 mm x


x 8.0 mm (3" x 3" x 5/16"); DIAGONAL & VERTICAL MEMBERS: 2∟ 75 mm x 75 mm
x 8.0 mm ( 3" x 3" x 5/16")

DESIGN OF ROOF BEAM:


RB DL = 0.08 + 0.05 + 0.20 + 0.00 + 0.01 = 0.34 kPa

19.122 kN/m ROOFINGS BEAM ITSELF

↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓ wu = 0.34 x 15.00 + 0.25 x 0.40 x 23.6 +


6.00 m 4
Page 4

57.4 kN
PARAPET LIVE + WIND LOAD
1.20 x 2.50 x 1.2 + 1.86 x 15.00 x 1.6 = 19.12 kN/m
4

Mumax = 1/12 wuL² = 1/12 x 19.12 x 6.00 ² = 57.37 kN-m

Trial section: # of bars bar diameter

b= 250 mm ; d = 350 mm → 4 , - 16 t= 400 mm

ρ = As/bd = π 4 x 16 ²÷ 4 ÷ 250 ÷ 300 = 0.011 > 1.4/fy = 0.0061

ω = ρfy/f'c = 0.01 x 228 ÷ 20.68 = 0.12

Ru = f’cω (1 – 0.59ω) = 20.68 x 0.12 x 1 , - 0.59 0.118 = 2.274

Mu cap = Φbd2 Ru = 0.85 x 250 x 350 ² 2.27 = 59.20 kN-m


(4 @ top & 2 @ bottom) Mu cap > Mu max, therefore ok

R.C. ROOF BEAM (RB): 250 mm x 400 mm with 4 – 16 mm Ø top bars & 2 - 16 mm Ø
bottom bars @ supports & 4 - 16 mm Ø bottom bars & 2 - 16 mm Ø top bars @ midspan

DESIGN OF CORBEL:
B1 P= 210.42 kN
3.54 kN/m BEAM ITSELF
↓↓↓↓↓↓↓↓↓↓↓ Wu = (0.25)(0.50)(23.6) x 1.2 = 3.54 kN/m
1.00 m
214.0 kN

Mu =1/2 wuL² + PL= 0.50 3.54 x 1.00 ² + 210.4 x 0.50 = 107.0 kN-m

Trial section: # of bars bar diameter

b= 250 mm ; d = 500 mm → 6 , - 16 t= 550 mm

ρ = As/bd = π 6 x 16 ²÷ 4 ÷ 250 ÷ 500 = 0.010 > 1.4/fy = 0.061

ω = ρfy/f'c = 0.01 x 228 ÷ 20.68 = 0.11

Ru = f’cω (1 – 0.59ω) = 20.68 x 0.11 x 1 , - 0.59 0.106 = 2.062

Mu cap = Φbd2 Ru = 0.85 x 250 x 500 ² 2.06 = 109.56 kN-m


(6 @ top & 2 @ bottom) Mu cap > Mu max therefore ok

R.C. CORBEL (B-1): 250 mm x 500 mm with 6 – 16 mm Ø continuous top bars


& 2 - 16 mm Ø continuous bottom bars

DESIGN OF MIDWAY WALL BEAM


WB-1
13.338 kN/m EXTERIOR WALL BEAM ITSELF

↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓ W = [ (ww)(h) + (b)(d)(wc) ] x 1.20


6.00 m
40.01 kN

2.82 x 3.00 + 0.25 x 0.45 x 23.6 x 1.20


Page 5

Wu = 13.34 kN/m LN = 6.00 - 0.00 = 6.00


Mwind
Mmax = 1/12 wuL² = 1/12 x 13.34 x 6.00 ²+ 17.16 = 57.16 kN-m

Mumax = 1/12 wuL² = 1/12 x 13.34 x 6.00 ² = 30.01 kN-m

Trial section: # of bars bar diameter

b= 250 mm ; d = 400 mm → 4 , - 16 t= 450 mm

ρ = As/bd = π 4 x 16 ²÷ 4 ÷ 250 ÷ 400 = 0.008 > 1.4/fy = 0.0061

ω = ρfy/f'c = 0.008 x 228 ÷ 20.68 = 0.09

Ru = f’cω (1 – 0.59ω) = 20.68 x 0.09 x 1 , - 0.59 0.089 = 1.738

Mu cap = Φbd2 Ru = 0.85 x 250 x 400 ² 1.74 = 59.08 kN-m


(4 @ top & 2 @ bottom) Mu cap > Mu max, therefore ok

R.C. WALL BEAM (WB-1): 250 mm x 450 mm with 4 – 16 mm Ø top bars & 2 - 16 mm Ø
bottom bars @ supports & 4 - 16 mm Ø bottom bars & 2 - 16 mm Ø top bars @ midspan

DESIGN OF COLUMN:
C-1 WALL BEAM ROOF BEAM CORBEL
Puact = 40.01 2 + 2 57.37 + 214.0 1 = 408.72 kN

try ρg = 8 x 201 = 0.01 n= 4


400 x 400

MWIND =
44.56
193.10
kN-m
kN-m m == fy
fy = 228
228 = 12.97= 12.97
0.85f’c (0.85)(20.68)
e = Mu ÷ Pu = 0.11
0.85 num1 = 183312 + num2 = 3308800
den1 = 0.86337364 den2 = 2.24787

Puallow = Φ As’ fy + bt fc’ = (0.85) (4)(201)(228) + (400)(400)(20.68)


e + 0.5 3t e + 1.18 110 + 0.50 + 1.18
(3)(400)(110)
d – d’ d2 350– 50 (350)2

Pu allow = 1431646.08 N or 1431.646 kN > 408.72 kN therefore ok

R.C. COLUMN (C-1) 400 x 400 with 8 - 16 mm Ø main bars

DESIGN OF AXIALLY-LOADED FOOTING:


F-1
Pu = 408.7 kN
→ A f = Af
Wf = Wf = 408.7 + 0.06 x 408.7 = 2.256 m²
q 192

LxW= 1.50 x 1.50 m² say 1.75 x 1.75 m²

qnet = P/A = 133.5 kPa → y= L-c = 1.75 - 0.40 = 0.68 m


2 2
Page 6

Mu = (qnet)(y)(y/2)(L) = 133.5 x 0.68 x 0.34 x 1.75 = 53.2 kN-m


MWIND
Mu = 53.21 +
= 44.56 = 97.76 kN-m

Trial section: # of bars bar diameter

b= 1750 mm ; d = 275 mm → 11 , - 16

ρ = As/bd = π 11 x 16 ²÷ 4 ÷ 1750 ÷ 275 = 0.009 > 1.4/fy = 0.0061

ω = ρfy/f'c = 0.01 x 228 ÷ 20.68 = 0.10

Ru = f’cω (1 – 0.59ω) = 20.68 x 0.10 x 1 , - 0.59 0.101 = 1.97

d= Mu = 97.76 x 10 ⁶ = 182.6 mm < 225 mm = d ∴ ok


ΦRub 0.85 x 1.97 x 1,750
L= 1.75 m
t = d + 75 = 300 mm W= 1.75 m

Check for Punching Shear: vpallow = 0.33 f'c = 1.50 MPa


c1 = 0.40 m d= 0.225
c2 = 0.40 m
Vp = qnet [LW - (c1 +d)(c2 + d)] = qnet [LW - d² - (c1+c2)d - c1c2]

Vp = 408.7 ,- 133.5 d²- 106.8 d - 21.35 = 356.6 kN

Ap = 0.85[c1 +d + c2 +d](2)d = 1.36 d + 3.40 d² = 0.478 m²

vpact = Vp/Ap → 356.6 ÷ 0.48 = 0.75 MPa < 1.50 MPa therefore OK

R.C. AXIALLY-LOADED FOOTING (F1): 350 mm x 1750 mm x 1750 mm with


Mu = 30000.00 kN-m
22 – 16 mm Ø rebars bothways (11 @ each way )

DESIGN OF FOOTING TIE BEAM:


FTB Mu = 44.56 kN-m

Trial section: # of bars bar diameter

b= 200 mm ; d = 350 mm → 4 , - 16 t= 350 mm

ρ = As/bd = π 4 x 16 ²÷ 4 ÷ 200 ÷ 300 = 0.013 > 1.4/fy = 0.0061

ω = ρfy/f'c = 0.01 x 228 ÷ 20.68 = 0.15

Ru = f’cω (1 – 0.59ω) = 20.68 x 0.15 x 1 , - 0.59 0.148 = 2.79

Mu cap = Φbd2 Ru = 0.85 x 200 x 350 ² 2.79 = 58.09 kN-m


(4 @ top & 2 @ bottom) Mu cap > Mu max therefore ok

R.C. FOOTING TIE BEAM (FTB): 200 mm x 350 mm with 4 – 16 mm Ø top bars & 2 -
16 mm Ø bottom bars @ supports & 4 - 16 mm Ø bottom bars & 2 - 16 mm Øtop bars
@ midspan

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