REINFORCED CONCRETE BEAM

FB - 1
Input Parameters :
Sup Mid
Standard Specs Moment Capacity Top Bot Shear Capacity
fc' = 21 Mpa Tension Bar, Dt = 3 3 bw = 250 mm fyt = 230
fy = 275 Mpa Tension Bar 2-L, Nb = 3 3 h = 400 mm Av = 2
b₁ = .85 Compression Bar, Dc = 3 3 Sb = 10 mm Stirrups Bar Ø
Es = 200 Gpa Main Bar dia, D = 16 Cc = 40 mm, Clear Covering
Moment, Mu = 76.0 65.0
CONDITION : Val = 0.65+0.25 (Ԑt - Ԑty/0.005 - Ԑty)
Ø = 0.65 , Ԑt ≤ Ԑty 3-16mmØ S
Val , Ԑty ˂ Ԑt ˂ 0.005 3-16mmØ U
0.90 , Ԑt ≥ 0.005
P
Location of d' & dt; Support Midspan T
As1 = π D² Db /4 = 603.19 603.19 P
2L, As2 = π D² Nb /4 = 603.19 603.19 O
y = (As1 y1 +As2 y 2 )/As = 30.00 30.00 3-16mmØ R
d' = Cc + Sb + (D/2) + y = 88.00 88.00
dt = h - d' = 312.00 312.00
CHECKING : εt ≥ 0.004, Support Midspan
Cmax = 600 d t / fsmin +600 = 133.7 133.71 M
amax = β1 Cmax = 113.7 113.66 3-16mmØ
I
Asmax = 0.85f'c a max b w / fy = 1844.35 1844.35
a = (As fy- As' fy)/0.85 fc' b = 74.34 74.34 D
C= a/β1 = 87.46 87.46 S
fs' = 600 (C-d')/C = -3.69 -3.69
P
fs' = Compression Bars = Not Yield Not Yield 3-16mmØ
600 (dt-C)/C 3-16mmØ
A
fs = = 1540.37 1540.37
fs = Tension Bars = Yield Yield N
BEAM MOMENT CAPACITY; Ø = 0.9 CHECK SHEAR; Ø = 0.75
As fy = 0.85fc' β1 C bw + As' fs'; Where, fs' = (600 C-d'/C) Vc = 0.17 √fc' bw dt = 61 < 75.00
Support Midspan Vsmax = 0.67 √fc' bw dt = 239 KN
C= by quadratic = 87.74 87.74 Vs = (Vu/Ø ) - Vc = 39.24 < 239.49
a= β1 C = 74.58 74.58 Section is Adequate
fs' = 600 (C-d')/C = -1.81 -1.81 Av = 2 (π Sb²/4)= 157.1
fs = 600 (dt-C)/C = 1533.67 1533.67 Vs ≤ 0.33√f'c bw d, d/2
Max S, w/c
Mn = Cc[d t -(a/2)] + Cs (d t -d') = 91.18 91.18 or 600mm otherwise
ever is
SUPPORT, ØMn = 82.06 > 76.00 Pass! Vs > 0.33√f'c bw d,
lesser,mm
MIDSPAN, ØMn = 82.06 > 65.00 Pass! d/4 or 300mm

Therefore use 250x400with 8-16mmØ @ support and 8-16mmØ @ midspan Main Bars (Grade 40);
10mmØ 2 leg-stirrups: Sp. at 4@50, 4@70 and rest 200 O.C
 REINFORCED CONCRETE BEAM
FB - 2
Input Parameters :
Sup Mid
Standard Specs Moment Capacity Top Bot Shear Capacity
fc' = 21 Mpa Tension Bar, Dt = 3 3 bw = 200 mm fyt = 230
fy = 275 Mpa Tension Bar 2-L, Nb = 2 2 h = 400 mm Av = 2
b₁ = .85 Compression Bar, Dc = 3 3 Sb = 10 mm Stirrups Bar Ø
Es = 200 Gpa Main Bar dia, D = 16 Cc = 40 mm, Clear Covering
Moment, Mu = 67.0 62.0
CONDITION : Val = 0.65+0.25 (Ԑt - Ԑty/0.005 - Ԑty)
Ø = 0.65 , Ԑt ≤ Ԑty 3-16mmØ S
Val , Ԑty ˂ Ԑt ˂ 0.005 2-16mmØ U
0.90 , Ԑt ≥ 0.005
P
Location of d' & dt; Support Midspan T
As1 = π D² Db /4 = 603.19 603.19 P
2L, As2 = π D² Nb /4 = 402.12 402.12 O
y = (As1 y1 +As2 y 2 )/As = 24.00 24.00 3-16mmØ R
d' = Cc + Sb + (D/2) + y = 82.00 82.00
dt = h - d' = 318.00 318.00
CHECKING : εt ≥ 0.004, Support Midspan
Cmax = 600 d t / fsmin +600 = 136.3 136.29 M
amax = β1 Cmax = 115.8 115.84 3-16mmØ
I
Asmax = 0.85f'c a max b w / fy = 1503.85 1503.85
a = (As fy- As' fy)/0.85 fc' b = 77.44 77.44 D
C= a/β1 = 91.11 91.11 S
fs' = 600 (C-d')/C = 59.97 59.97
P
fs' = Compression Bars = Not Yield Not Yield 2-16mmØ
600 (dt-C)/C 3-16mmØ
A
fs = = 1494.27 1494.27
fs = Tension Bars = Yield Yield N
BEAM MOMENT CAPACITY; Ø = 0.9 CHECK SHEAR; Ø = 0.75
As fy = 0.85fc' β1 C bw + As' fs'; Where, fs' = (600 C-d'/C) Vc = 0.17 √fc' bw dt = 50 < 75.00
Support Midspan Vsmax = 0.67 √fc' bw dt = 195 KN
C= by quadratic = 85.75 85.75 Vs = (Vu/Ø ) - Vc = 50.45 < 195.27
a= β1 C = 72.88 72.88 Section is Adequate
fs' = 600 (C-d')/C = 26.22 26.22 Av = 2 (π Sb²/4)= 157.1
fs = 600 (dt-C)/C = 1625.15 1625.15 Vs ≤ 0.33√f'c bw d, d/2
Max S, w/c
Mn = Cc[d t -(a/2)] + Cs (d t -d') = 76.99 76.99 or 600mm otherwise
ever is
SUPPORT, ØMn = 69.29 > 67.00 Pass! Vs > 0.33√f'c bw d,
lesser,mm
MIDSPAN, ØMn = 69.29 > 62.00 Pass! d/4 or 300mm

Therefore use 200x400with 7-16mmØ @ support and 7-16mmØ @ midspan Main Bars (Grade 40);
10mmØ 2 leg-stirrups: Sp. at 4@50, 4@70 and rest 200 O.C
 REINFORCED CONCRETE BEAM
CB - 1
Input Parameters :
Sup Mid
Standard Specs Moment Capacity Top Bot Shear Capacity
fc' = 21 Mpa Tension Bar, Dt = 4 2 bw = 200 mm fyt = 230
fy = 275 Mpa Tension Bar 2-L, Nb = 2 1 h = 350 mm Av = 2
b₁ = .85 Compression Bar, Dc = 2 6 Sb = 10 mm Stirrups Bar Ø
Es = 200 Gpa Main Bar dia, D = 16 Cc = 40 mm, Clear Covering
Moment, Mu = 54.0 36.0
CONDITION : Val = 0.65+0.25 (Ԑt - Ԑty/0.005 - Ԑty)
Ø = 0.65 , Ԑt ≤ Ԑty 4-16mmØ S
Val , Ԑty ˂ Ԑt ˂ 0.005 2-16mmØ U
0.90 , Ԑt ≥ 0.005
P
Location of d' & dt; Support Midspan T
As1 = π D² Db /4 = 804.25 402.12 P
2L, As2 = π D² Nb /4 = 402.12 201.06 O
y = (As1 y 1 +As2 y 2 )/As = 20.00 20.00 2-16mmØ R
d' = Cc + Sb + (D/2) + y = 78.00 78.00
dt = h - d' = 272.00 272.00
CHECKING : εt ≥ 0.004, Support Midspan
Cmax = 600 d t / fsmin +600 = 116.6 116.57 M
amax = β1 Cmax = 99.1 99.09 6-16mmØ
I
Asmax = 0.85f'c a max b w / fy = 1286.31 1286.31
a = (As fy- As' fy)/0.85 fc' b = 92.93 46.46 D
C= a/β1 = 109.33 54.66 S
fs' = 600 (C-d')/C = 171.93 -256.15
P
fs' = Compression Bars = Not Yield Not Yield 1-16mmØ
600 (dt-C)/C 2-16mmØ
A
fs = = 892.77 2385.54
fs = Tension Bars = Yield Yield N
BEAM MOMENT CAPACITY; Ø = 0.9 CHECK SHEAR; Ø = 0.75
As fy = 0.85fc' β1 C bw + As' fs'; Where, fs' = (600 C-d'/C) Vc = 0.17 √fc' bw dt = 42 < 75.00
Support Midspan Vsmax = 0.67 √fc' bw dt = 167 KN
C= by quadratic = 95.11 72.54 Vs = (Vu/Ø ) - Vc = 57.62 < 167.03
a= β1 C = 80.85 61.66 Section is Adequate
fs' = 600 (C-d')/C = 107.95 -45.20 Av = 2 (π Sb²/4)= 157.1
fs = 600 (dt-C)/C = 1115.87 1649.92 Vs ≤ 0.33√f'c bw d, d/2
Max S, w/c
Mn = Cc[d t -(a/2)] + Cs (d t -d') = 75.26 42.51 or 600mm otherwise
ever is
SUPPORT, ØMn = 67.73 > 54.00 Pass! Vs > 0.33√f'c bw d,
lesser,mm
MIDSPAN, ØMn = 38.26 > 36.00 Pass! d/4 or 300mm

Therefore use 200x350with 8-16mmØ @ support and 5-16mmØ @ midspan Main Bars (Grade 40);
10mmØ 2 leg-stirrups: Sp. at 4@50, 4@60 and rest 150 O.C
 REINFORCED CONCRETE DESIGN FOR SLAB
S-1 One-way slab

Input Parameters :

f'c = 21 Mpa dᵇ = 10 mm Sup Mid Temp bar

fy = 230 Mpa t= 100 mm Sᵇ = 200 200 300 mm
b₁ = 0.85 Ln = 1.2 m
Es = 200 Gpa L1 = 2.2 m
L2 = 2.2 m

Analysis :

Deadload, D = 5.21 KN/m

Liveload, L = 2.9 KN/m
Wᵤ = 1.2D + 1.6L = 10.89 KN/m
Support, Mᵤ = Wu Ln²/10 = 5.27 KN-m
Midspan, Mᵤ = Wu Ln²/14 = 1.12 KN-m
Check Moment and Shear Capacity b= 1 m
Aᵇ = π dᵇ / 4 = 78.54 mm² Asmin = 200 mm²
Support Midspan
As = Aᵇ b / Sᵇ = 392.70 mm² 392.70 mm²
a = As fy / 0.85 fc' b = 5.06 mm 5.06 mm
C = a / β1 = 5.95 mm 5.95 mm
d = t - (20 + dᵇ/2) = 75.00 mm 75.00 mm
Ԑᵧ =fy / Es= 0.00115
Ԑs=0.003 ((d-c)/c) = 0.0348 > 0.005 Tension Control
Thus use, Ø = 0.90 Support Midspan
Mn=0.85 fc' ab (d-a/2) = 6.55 KNm 6.55 KNm
Support, Ø Mn = 5.89 KNm > 5.27 Pass!
Midspan, Ø Mn = 5.89 KNm > 1.12 Pass!
Ø = 0.75
Vᵤ = 1.15*Wu Ln/2 = 7.52 KN
Ø Vc = Ø 0.17 (√fc' bd) = 43.82 KN Pass!
Check Thickness = Ln / 24 < 100 mm Pass!

Therefore use, 100mm thick slab with 10mmØ Grade (33) rebar spaced @
200mm at support, top bars / 200mm at midspan, bottom bars / 300mm Temp bars O.C
 REINFORCED CONCRETE DESIGN FOR SLAB
S-2 Two-way Slab

Input Parameters :

f'c = 21 Mpa dᵇ = 10 mm², Bar Ø Long Short

fy = 230 Mpa t= 125 mm Bar Ø Sup Mid Sup Mid
b₁ = .85 Top, (-) 200 200
Es = 200 Gpa LA = 3 m, Short Bot, (+) 200 200
ʯᶜ = 23.6 KN/mᶟ LB = 4.5 m, Long

Deadload, D = 4.25 KN/m

Liveload, L = 3.8 KN/m
Analysis :
1.2 D = Du = 5.1 KN/m
1.6 L = Lu = 6.08 KN/m
Wᵤ = 1.2D + 1.6L = 11.2 KN/m
m = LA/LB = 0.65
Moment Coefficients: ACI 318-63 Code Case 2
Ca, neg = 0.077 Cb, neg = 0.01
CaL = 0.053 CbL = 0.01
CaD = 0.032 CbD = 0.006

Moment along Long span Moment along Short span

Ma = Ca Wu LB = 3.17 KNm - Support - Mb = Cb Wu LB = 7.75

Ma = Ca Wu LB = 1.85 KNm - Midspan - Mb = Cb Wu LB = 4.37
Aᵇ =π dᵇ/ 4 = 78.54 mm²
Check Moment and Shear Capacity; b= 1 m strip
Long span Short Span
Sup, As= Aᵇb/ Sᵇ= 392.70 mm² As = Aᵇb/ Sᵇ = 392.70 mm²
Mid, As= Aᵇb/ Sᵇ= 392.70 mm² As = Aᵇb/ Sᵇ = 392.7 mm²
LONG SPAN SHORT SPAN
Top a = As fy/ 0.85 fc' b = 5.06 5.06 mm
Bot a = As fy/ 0.85 fc' b = 5.06 5.06 mm
Top c= a / b₁ = 5.95 5.95 mm
Bot c= a / b₁ = 5.95 5.95 mm
d= t -(20 + dᵇ/2) = 100.00
Ԑᵧ = fy / Es = 0.00115
Ԑs= 0.003[(d-c)/c] = 0.0474 > 0.005 Tension Control
Thus, Use = 0.90
Long Span Short Span
Sup, Mn = 0.85 fc' ab(d-(a/2)) = 8.80 > 3.17 Pass! 8.80 > 7.75 Pass! KNm
Mid, Mn = 0.85 fc' ab(d-(a/2)) = 8.80 > 1.85 Pass! 8.80 > 4.37 Pass! KNm
Minimum Thickness, t = LB / 36 < 125 Pass!

Therefore use, 125mmthick slab with 10mmØ rebar (Grade33)

spaced; along short span top/bot - 200/200: long span 200/200 o.c.
 REINFORCED CONCRETE COLUMN
C-1
Design Parameters:
Main Bars Loc -Z Ast
h= 400 mm d' = 40 mm fs 1 = 3 - 360.00 603.19
b= 250 mm Main bars = 16 mm Ø fs 2 = 0 - 280.00 0
fc' = 21 Mpa Mux = 43 KNm fs 3 = 2 - 200.00 402.12
fy = 275 Mpa Pu = 512 KN fs 4 = 0 - 120.00 0
b₁ = 0.85 Es con = 600 Mpa fs5 = 3 - 40.00 603.19
Es = 200 Gpa Shear parameters
Defining Condition; Nvh = 3 , lateral ties leg
0.65 , Ԑt ≤ Ԑty if fs > fy, use fy Nvb = 3 , lateral ties leg
Ø = Val , Ԑty ˂ Ԑt ˂ 0.005 fyt = 230 Mpa, lateral ties
0.90 , Ԑt ≥ 0.005 Ties, Td = 10 mm Ø
Val = 0.65+0.25 (Ԑt-Ԑty/0.005 - Ԑty)
MAXIMUM AXIAL CAPACITY Ag = 100000 NSCP 422.4.2.2
Pcn = 0.8 [0.85 fc' (Ag-As) + fy As] = 1,758.90 kN
Ø= 0.7 , Pcu = Ø Pcn = 1,143.28 kN
BALANCE CONDITION CAPACITY, Ø = 0.65 COMPRESSION CONTROLLED CAPACITY, Ø = .65
Ԑy = fy / Es = 0.00138 Ԑt = -0.002
Cb = Ԑc d/Ԑc + Ԑy = 246.86 mm Ct = Es con d/(Es con-Es Ԑt) = 216.00 mm
ab = β C = 209.83 mm at = β C = 183.60 mm
from strain diagram As[fy or fs] from strain diagram As[fy or fs]
fs 1 = 600(d-C)/C = 275.00 T1 = 165.88 KN, Tension fs 1 = 600(d-C)/C = 400.00 T1 = 165.88 KN, T
fs 2 = 600(C-Z 2 )/C = 0.00 C2 = 0.00 KN, fs 2 = 600(C-Z 2 )/C = 0.00 C2 = 0.00 KN,
fs 3 = 600(C-Z 3 )/C = 113.89 C3 = 45.80 KN, Compression fs 3 = 600(C-Z 3 )/C = 44.44 C3 = 17.87 KN, C
fs 4 = 600(C-Z 4 )/C = 0.00 C4 = 0.00 KN, fs 4 = 600(C-Z 4 )/C = 0.00 C4 = 0.00 KN,
fs 5 = 600(C-Z 5 )/C = 502.78 C5 = 165.88 KN, Compression fs 5 = 600(C-Z 5 )/C = 488.89 C5 = 165.88 KN, C
Cc = .85 f'c ab = 936.36 KN Cc = .85 f'c ab = 819.315 KN
[∑Fv = 0], Pbn =C2+C3+C4+C5+Cc - T = 982.2 KN [∑Fv = 0], Pbn =C2+C3+C4+C5+Cc - T = 837.2 KN
[∑M T = 0], find e = x-z, z= 160.00 mm [∑M T = 0], find e = x-z, z= 160 mm
P bnX = C 2 X 2+C 3 X 3 +C 4 X 4+C 5 X 5+C c X c PbnX = C 2 X 2 +C 3 X 3+C 4 X 4+C 5 X 5 +C c X c
eb = 144.70 mm ØPb = 638.4 KN et = 102.47 mm ØPt = 544.17 KN
Mb = 142.11 KNm ØMb = 92.37 KNm M t = 55.76 KNm ØMt = 36.25 KNm
ECCENTRICALLY LOADED SECTION CAPACITY, Ø= 0.65 CHECK SHEAR, Øv = 0.75
due to applied load, ex = 83.98 Vc = 0.17[1+(Pu/14Ag)] √fc' b dt = 95.75 89.37 KN
[∑Fv = 0], Pbn =C2+C3+C4+C5+Cc -T Av = π Td² Nv / 4 = 235.6 235.6 mm²
[∑MT = 0], PbnX =C2X2+C3X3+C4X4+C5X5+CcXc Smax, min (16db, 48dties and least col dimension)
C = 311.26 mm a = 264.57 mm 16 db = 256 mm
fs 1 = 600(d-C)/C = 93.95 T1 = 56.67 KN, Tension 48 dties = 480 mm
fs 2 = 600(C-Z 2 )/C = 0.00 C2 = 0.00 KN, Least Column dimension = 250 mm
fs 3 = 600(C-Z 3 )/C = 214.47 C3 = 86.24 KN, Compression
fs 4 = 600(C-Z 4 )/C = 0.00 C4 = 0.00 KN,
fs 5 = 600(C-Z 5 )/C = 522.89 C5 = 165.88 KN, Compression
Cc = .85 f'c ab = 1180.65
Pnx = 1376.10 ØPn = 894.46 KN > 512.00 Pass!
Mnx = 115.57 ØMn = 75.12 KNm > 43.00 Pass!

Therefore use, 400x250mm 8-16mm Ø longitudinal bars (GRADE 40) with 10mm Ø lateral ties space at,
1@50, 8@60mm , and Rest 100mm O.C, BOTH ENDS.
 3 - 16mm Ø b

3 - 16mm Ø
h
FORCE TRANSFER GUIDE FROM STAAD RESULT OR EQUIVALENT SOFTWARE
Mz (Staad) = Mux = 43 < 75.12 Pass! , ex =83.98 mm
My (Staad) = Muy = 32 < 34.41 Pass! , ey =62.5 mm
ØPnx = 894.46 > 512 Pass! > 894.5
ØPny = 550.53 > 512 Pass! < 550.5
, ex = 84.0
Col Height, H = 3 Floor to Beam Bottom , ey = 62.5
Length factor, K = 0.5 Refer (Sheet 2) 34.4
<
Column Slenderness (Unbraced Column, KL/r ≤ 22), 406.2.5
KL/rx = 12.5 < 22 Ok! 12.50
KL/ry = 20 < 22 Ok! 20.00
75.1
Dimension Limits, 418.7.2.1
250 mm 250 Ok!
Ratio Limit 0.4 0.63 Ok!

Main reinforcement Ratio Ast = 1608.5

0.01Ag < Ast > 0.06Ag Ok!
1000
Traverse Reinforcements, 418.7.5.4 #######
No of Leg Req, h = 3 Ok!
No of Leg Req, b = 2 Ok!
 ISOLATED RECTANGULAR FOOTING DESIGN
F-1
Input Parameters:
Concrete Strenght, fc' = 21 Mpa Longitudinal Bar, Ax = 9
16 mmØ
Rebar yield strength, fy = 230 Mpa Traverse Bar, Ay = 9
Net allowable Soil Pressure, qa = 180 kPa Loc X - dir dx = 0 m, 0 from center
Footing Embedment Depth, Df = 1.5 m Loc Y - dir dy = 0 m, 0 from center
Surcharge, qs = 3.5 kPa Depth, Cx = 0.4 m
Soil Weight, ws = 16 kN/m³ Width, Cy = 0.2 m
Footing Thickness, t= 0.3 m Load, Pd = 87 KN
Length, L= 1.5 m Load, Pl = 143 KN
Width, B= 1.5 m Moment X, Mx = 18 KNm
Clear Covering, c= 75 mm Moment Y, My = 17 KNm
Check Soil Bearing Capacity

Applied Load, Pd + Pl = 230.00

Surcharge, qs (LB) = 7.88
Weigth Footing, (23.5-ws) tBL= 5.06
Pn = 242.94 kN
Mx = -18 KNm My = -17 KNm
ex = -0.078 m L/6 = 0.250 m
ey = -0.074 m B/6 = 0.25 m
P 6M < 180.00
qnx(max) = + = 139.97
BL L²B Pass!
P 6M < 180.00
qny(max) = + = 138.19
BL B²L Pass!

Check Thickness ; Two-way Shear Check One-way Shear; must, Vc > Vu

Applied Load, 1.2Pd + 1.6Pl = 333.20 Vux = qu B[(B-C)/2)-d] = 66.63 KN

Surcharge, 1.2 qs (LB) = 9.45 Vuy = qu L[(L-C)/2)-d] = 88.41 KN
Weigth Footing, 1.2(23.5-ws) tLB = 6.08 ØVcx = Ø 0.17 √fc' B d = 183.17 KN Pass!
Pu = 348.73 ØVcy = Ø 0.17 √fc' L d = 183.17 KN Pass!
Mx = -18 My = -17
ex = -0.05 ey = -0.05 Check Flexural Reinforcement; Asmin = 0.002Bt
L/6 = 0.25 L/6 = 0.25
qumaxx = 186.989 qumaxy = 185.211 Kpa Mux = 29.68 KNm Muy = 42.30 KNm

d = 0.209 m Asx = 1809.56 mm² Asy = 1809.56 mm²

Vux = qu (BL-(Cx+d)(Cy+d)) = 310.12 KN Asmx = 900.00 Asmy = 900.00 mm²
Vuy = qu (BL-(Cx+d)(Cy+d)) = 310.12 KN
bo = 2(D1 + d) 2(W1 + d) = 2.036 m Sx = 150.8 Sy = 150.8 mm (Smax=450mm)
Ø Vc 1 = Ø 0.33 √fc' bo d = 482.624 KN Asx fy
ax = = 15.54 mm, ay= 15.54 mm
bc = Long/short side of column = 2 .85 fc B

Ø Vc 2 = Ø( 1 + 2 ).17 √f'c bo d = 497.2 KN Check Tension Contolled Limit Ø= 0.9

βc ab/d = 0.614 > ax/d & ay/d
αs = 40 ax/d = 0.074 ay/d = 0.074

Ø Vc 2 = Ø(2+ αs d ) .083 √f'c bo d = 750.17

βc ax ay
ØMnx = ØAsx fy(d- ) ØMny = ØAsy fy(d- )
2 2

ØMnx = 75.38 KNm, ØMny = 75.38 KNm

Ø Vc = min Ø Vc1 ,Ø Vc2,Ø Vc3 482.6 > 310
Pass! > Mu Pass! > Mu Pass!

Therefore use, 1.5x1.5x0.3m thick Footing with 16mmØ Tension Bar Grade (33) sp. @ 150.75mm and 150mm 3
along BL respectively O.C.
3
 PROPOSED ONE STOREY WITH DECK
STRUCTURAL ANALYSIS

Owner

DESIGNED BY:

ROCKY M. SIORES
Civil Engineer

PRC Reg. No. 0111516

PTR No. 1925073
TIN No. 408-875-292
Issued on: January 08, 2024
Issued at: Puerto Princesa City