Post on 18-Jul-2016
description
transcript
PROJECT : RESIDENTIAL BUILDING W/ ATTIC LOCATION : Purok SAN Francisco (Puerto Ville), Tiniguiban, Puerto Princesa CityOWNER : Mrs Erlinda M. GayamoSUBJECT : STRUCTURAL ANALYSIS
Prepared by:
JOHN F. QUILLOPECivil EngineerPRC Reg. No : 065611PTR No. : 323024Date Issued: January 25, 2010Place Issued : CTO, PP CityTIN : 146-908-310
1
Design Criteria, Constants and Assumption
a. Design to comply with the latest building ordinances of any municipality of the province of Palawan and its amendments.
b. Design to comply with the National Structural code of the Philippines for Buildings (ASEP 2004)
c. Design to comply with the American Concrete Institute Building Requirements for Reinforced Concrete (ACI 318- ).
DESIGN STRESSES/STRENGTH OF MATERIALS:
Compressive strength of concrete,
fc’ = 20.7 MPa ( 28 days Strength)(for roof beams, Beams, Columns, elev. slab)
for Slab On fill and Footings)
Yield strength of steel,
fy = 375 MPa (16mm dia and up, )
fy = 276 MPa (10mm-12mm dia and below,)
Soil Bearing Capacity
Assumed = 73.4 kPa
Design Loads:
a. Dead Loads
1. Weight of Concrete = 23.60 kN/m3
2. Utilities = 135 kPa
3. Wt. of Ceiling = 50 kPa
4. Floor finish = 1.58 kPa
5. 100 mm. CHB interior wall both face plastered = 1.76 kPa
b. Live Load
1. Bedrooms and corridor = 4.8 kPa
2. Roof Live load = 1.00 kPa
2
Truss Loadings:
Weight of Steel Truss
W = 0. 40 + 0. 04 L
Where:
W = Weight of steel truss in pounds per square foot (lbs/ft2)
L = Span in feet (ft)
Wind Pressure (Zone III (Palawan) from 0 to 40’ high 150 kph
Wind Velocity = 150 kph
a) On vertical plane surfaces of all buildings.b) On inclined surfaces.
Pn = 2 P sinθ1+sin2θ
Where:PN = Normal component on wind pressure / square foot.P = Pressure per square foot on vertical surface.θ = Angle of inclination of surface with the horizontal.
Technical Notations : P = total load acting on the trussLs = tributary area in m2
w = weight of trussV = wind velocityPn = normal wind pressure on a vertical surfaceP = normal wind pressure on a vertical surfaceA = effective width area, (m2)Ag = area of gross section, (mm2)Av = are of ties within a distance, (mm2)Cb = bending coefficient dependent upon momentCs = stiffness factord = depth of the beamDL = dead loadf’c = specified compressive stiffness of concrete, (Mpa.)fy = specified yield strength of steel, (Mpa.)Ld = development length, (mm.)LL = live loadM = momentMFAB = moment (fixed moment)Mn = nominal strength assuming all reinforcement at the section to be stressed to the specified yield strength fy.Ms = moment due to loads causing appreiciable away, or moment of tensile force in reinforcement about centroid of compressive force in masonry, (kN-m.)Mu =factored moment at sectionN = number of bars or wires being developed along the plane of the splitting.φ = strength of reduction factor\Pu = ultimate axial load
3
Qa = allowable soil pressure (Mpa.)qu = ultimate soil pressureR = governing radius of gyration (mm.)Ru = ultimate reactionS = spacing of transverse reinforcement along the longitudinal axis of the structural member, (mm)t = effective thickness of the wall or the column, (mm.)Vc = nominal shear strength provided by concreteVs = shear strengthVu = required shear strength in masonry, (kN.)Wu = factored load per unit length of beam or per unit area of slab (kN/m)β = width of compression of face of member, (mm)βc = constant use to complete Vc in pre-stressed slabπ = constant, equivalent of 3.1416ρ = ratio of non pre-stress tension reinforcement to the concreteρ = factored axial load (kN)Pb = nominal axial load strength at balance strain conditionρg = ratio of toyal reinforcement area to cross-sectional area of columnρn = nominal axial load strength at given electricity
4
Roof Design
Analysis of Loadings on Truss
Joint loads on the top chord
Weight of truss:w=0 . 40+0. 04 Lw=0 . 40+0. 04 (31 . 16))w=1 .6464 psf ( 4 . 89 ) ( 9. 81 )w=78 . 98Pa .
Total weight of truss 78 . 98Pa+500=578. 98 PaP=wLsP=578 . 98(8 )(4 )P=18 ,527 . 33N .
Load per joint = 18 ,527 .234
8
Load per joint = 2 ,315. 92N .
Joint loads due to wind load
Wind velocity = 150KpH.P=0.0000473V 2
P=0.0000473(150 )2
P=1 ,064 . 25Pa .Pn=2 P sinθ
1+sin2θ
θ=2.04
tan−1
Pn=2(1064 . 25)sin 22 °1+sin2 22 °
Pn=699 .4364N .
Load per joint =
699 .435
=138. 88N .
5
Table 1. Bar Stress on the Top chord, Bottom chord, Vertical member and Diagonal member Due To Dead Load.
Top chord Combined Member Computation Stresses
b-1 8.80 x 1,139.08 N. 10,023.904 N.c-3 7.5 x 1,139.08 N. 8,543.1 N.d-5 6.2 x 1,139.08 N. 7,062.30 N.e-7 5.0 x 1,139.08 N. 5,695.4 N.f-8 5.0 x 1,139.08 N. 5,695.4 N.
g-10 6.2 x 1,139.08 N. 7,062.30 N.h-12 7.5 x 1,139.08 N. 8,543.1 N.i-14 8.80 x 1,139.08 N. 10,023.904 N.
Bottom chordk-1 7.8 x 1,139.08 N. 8,884.82 N.k-2 7.8 x 1,139.08 N. 8,884.82 N.k-4 6.7 x 1,139.08 N. 7,631.85 N.k-6 5.6 x 1,139.08 N. 6,378.84 N.k-9 5.6 x 1,139.08 N. 6,378.84 N.k-11 6.7 x 1,139.08 N. 7,631.84 N.k-13 7.8 x 1,139.08 N. 8,884.82 N.k-14 7.8 x 1,139.08 N. 8,884.82 N.
Vertical member1-2 0 x 1,139.08 N. 0 N.3-4 0.5 x 1,139.08 N. 569.54 N.5-6 1.1 x 1,139.08 N. 1,253.00 N.7-8 3.3 x 1,139.08 N. 3,758.96 N.9-10 1.1 x 1,139.08 N. 1,253.00 N.11-12 0.5 x 1,139.08 N. 569.54 N.13-14 0 x 1,139.08 N. 0 N.
Diagonal member
2-3 1.2 x 1,139.08 N. 1,366.90 N.4-5 1.5 x 1,139.08 N. 1,708.62 N.6-7 2.0 x 1,139.08 N. 2,278.16 N.8-9 2.0 x 1,139.08 N. 2,278.16 N.
10-11 1.5 x 1,139.08 N. 1,708.62 N.12-13 1.2 x 1,139.08 N. 1,366.90 N.
Table 6.Bar Stresses on Top chord, Bottom chord, Vertical member and Diagonal member Due to Wind Load.
Top chordMember Computations Stresses
12-e 4.6 x 26.67N. 122.68 N.10-e 3.6 x 26.67N. 96.012 N.8-e 3.0 x 26.67N. 80.01 N.7-d 2.8 x 26.67N. 74.68 N.5-c 3.4 x 26.67N. 90.68 N.3-b 3.95 x 26.67N. 105.35 N.1-a 4.4 x 26.67N. 117.35 N.
Bottom chord
6
14-f 6.9 x 26.67N. 184.02 N.13-f 6.9 x 26.67N. 184.02 N.11-f 3.8 x 26.67N. 101.35 N.9-f 2.8 x 26.67N. 74.68 N.6-f 3.3 x 26.67N. 88.01 N.4-f 4.3 x 26.67N. 114.68 N.2-f 4.9 x 26.67N. 130.68 N.1-f 4.9 x 26.67N. 130.68 N.
Vertical member
14-13 0 x 26.67N. 0 N.12-11 1.5 x 26.67N. 40.0 N.10-9 1.0 x 26.67N. 26.67 N.8-7 2.2 x 26.67N. 58.67 N.6-5 0.9 x 26.67N. 24.0 N.4-3 0.3 x 26.67N. 8.0 N.2-1 0 x 26.67N. 0 N.
Diagonal member
13-12 3.9 x 26.67N. 104.013 N.11-10 1.9 x 26.67N. 50.67 N.9-8 0.9 x 26.67N. 24.00 N.7-6 1.8 x 26.67N. 48.01 N.5-4 1.4 x 26.67N. 37.34 N.3-2 0.6 x 26.67N. 16.0 N.
Design of top chord
Most stress member = 10,023.904 N.Compute for approximate area of the member
Amax .=P69
=10 ,023 .90469
=145.27mm2
Amin .=P103
=10 ,023 . 904103
=97 . 32mm2
Note: from ASEP handbook of structural steel shapes and sections.Compute for minimum radius of gyration.
rmin .=L120
=1 .0(1000)120
rmin .=8 .33mm .
try :∠65 x 65x 4 . 5A=5. 65cm .2Rx=2.03cmRy=2.03cm
Check KL/r and Cc; Assume k = 1.0
7
Klr
=1 .0 (1. 0 )(1000 )20 . 3
Klr
=49. 26
Cc=√2π2Efy
Cc=√2π2(200000 )230
Cc=131. 01
∴Klr
<Cc ( shortcolumn )
Solve for allowable axial stress.
Fs=53
+3(Kl r )8Cc
−(Kl r )
3
8Cc3
Fs=53
+3( 49. 26 )8(131 . 01)
−(49 . 26)3
8(131. 01)3
Fs=1. 80
Fa=[1−(Kl r )
2
2Cc2 ] fyfsFa=[1−
(49 . 26 )2
2 (131 . 01 )2 ]2301 .80
Fa=118. 75MPa.Check for column capacity
P=Fa⋅AP=118 .75(565 . 000)P=67 ,093 . 75>10 ,023 .904 ∴safe
Design of bottom chordMost stress member = 8,884.82 N.Compute for approximate area of the member.
Amax .=P69
=8 ,884 . 8269
=128. 77mm .2
Amin .=P103
=8 ,884 .82103
=86 . 26mm .2
Compute for minimum radius of gyration
rmin=L120
=1. 0(1000 )120
=8 . 33
try :∠65 x 65x 4 . 5A=5. 65 cm .2
Rx=2. 03 cm2
Ry=2. 03 cm .Allowable stress (AISC)
Fs=0 .60 FyFs=0 .60(230 )Fs=138
Safe load for tension member could carry.
8
P=A⋅FsP=565 (138)P=77 ,970>8 ,884 . 82∴ safeLr
=1 . 0(1000 )8. 33
=120<230∴ok
Design of Purlins
Normal pressure Pn = 426.64N.Analyze one typical purlin with S = 1.0m.
WWL=426 .64(1 .0 )=426 .64Nm .
W RL=500 .00 (1 .0)=500.00Nm .
cos26 . 57°=
W RLn
426 .64
sin 26 . 57°=W RLt
426 .64W n=W WL+W RLnW n=426 .64+381. 58
W n=808 .22Nm
WT=W RLt
WT=190 .83Nm
Compute for bending moment
Mn=WnL2
8=
808 .22(0 . 7 )2
8Mn=49 . 50N−m .
Mt=WtL2
8=
190.83(0 .7 )2
8Mt=11.69N−m .
MR=√(Mn )2+(Mt )2
MR=√(49 .50)2+(11.69 )2
MR=50. 86N−m .Compute the section modulus
9
Fb=MSx
Sx=50 .86 (1000)
165[1(10 )3 ]−0 . 165
Sx=0 .31cm3
try :LC 65 x 30 x 15x 1. 2A=1.77cm2
Ix=11.4 cm4
Iy=2.5cm4
Sx=3 .51cm3
Sy=1. 35cm3
Check for bending stress
fbn=MSx
=50 . 86(1000 )3. 51(1000)
=14 . 49Mpa .
fbt=MtSy
=11. 69(1000 )1 .35(1000)
=8 .66Mpa .
fb=14 .49+8 . 66=23.15Mpa .<165Mpa .∴ safeCheck for shearing stress
Vn=WnL2
=808 .22(0.7 )2
=282 . 88Mpa .
Vn=WtL2
=190 . 83(0 . 7)2
=66 . 791Mpa .
fvn=Vndtw
=282 . 8865 (1 . 2)
=3. 63Mpa .
fvt=Vtdtw
=66 .79165(1. 2 )
=0 .86Mpa .
fv=√(3 . 63 )2+(0 . 86 )2
fv=3 . 73Mpa .fv(allow )=0 .40 fy=0 .40 (230)fv(allow )=92Mpa ./¿3 .73Mpa .∴ safe
Design of Welled Connection
Using E 70 XX Electrode
Fu = 70 ksi
Fu = 482 MPa
Fv = 0.30 Fu = (0.30) (482 MPa) = 144.60 MPa
E = 200,000 MPa
For Web Members (Most Stress Member occur @ = 540,064.47 N
(Compression Member)
Try L 100 mm x 100 mm x 11.0 mm for Web Members10
Properties:
A=15.29 cm2 .=1 ,529mm .2
Rx=2.26cm2. .=226mm2 .
Ry=2.26cm. 2=226mm .2
Test: Check kLr and Cc; Assume k = 1.0
kLr
=(1. 0 ) (1 .25 ) (1000 )220
=68 .03
Cc=√(2π 2EFy )=√(2π2 (200000 )
275 )=119 .82
Solve for Allowable Axial Stress
F .S .=53
+3 (kLr )
3
8 (Cc )3−
(kLr )3
8 (Cc )3=5
3+
3 (68. 03 )3
8 (119.82 )3−
(68 .03 )3
8 (119. 82 )3=1. 667
Fa=[1−(kLr )
2
2 (Cc )2 ][ fyFS ]=[1−(68 . 03 )2 (119. 82 ) ][275
1. 71 ]=135. 09
Check for Column Capacity
P=Fa (A )P=( 135.09 N /mm .2) (2 ) (1 ,529 )P=413 ,105.22N .>10 ,023 . 904N∴ safe
For Welded Connection
P = 0.707 F v t L
PT = P1+P2+P3 = 10,023.904N.
t = Thickness of Fillet Weld = 11 mm
To avoid eccentricity, P2 must pass through the centroidal axis thus c2 = 29 mm
c = (29mm ) (2 ) = 58 mm
P2 = (0.707) (144.60) (11) (58 + 42) P2 = 112,455.42 N
∑M P3 = [ 0 ] clockwise direction positive +
P1=(10 , . 023.904 ) (29 .0 )
Say 115 mm (Minimum Length of Full – Welded Connection @ Short Direction)11
∑M P1 = [ 0 ] clockwise direction positive +
P3 = (540,564.87 ) (71.0 )−(112,455.42 ) (71.0 )
100 = 303,602.43 N
P3 = [ 144.60 ] [ (0.707 ) (11.0 ) (b ) ] = 303,602.43 N
b = 303,602.43
[144.60 ] [ (0.707 ) (11.0 ) ] = 269.97 mm
Say 270 mm
(Minimum Length of Full – Welded Connection @ Long Direction)
Design of Base Plate and Anchor Bolt
Allowable Bending Stress in Steel
Fb = 0.75 Fy Fy = 248 MPa A36 Steel Plate
Fb = 0.75 (248) = 186 MPa
t = √ 2 f pn2
Fb
(Thickness of Base Plate) f p=Total Load
Bearing Area of Plate
PT = P1 + P2 = 153, 602.90 N + 75,763.94 N PT = 229,366.84 N
f p=229,366.84 N
(350mm ) (350mm ) = 1.872 MPa Assume k1 = 22.0 mm
n = B2 - k1 =
350mm2 – 22 mm = 153 mm
Thickness of Plate
t = √ 2 f pn2
Fb
= √ 2 (1.872 ) (153 )2
186
t = 26.60 mm = Say 30 mm THK. Base Plate.
For Anchor Bolt
Allowable Stress (AISC)
FS = 0.60 Fy (A36 Steel Bar)
FS = 0.60 Fy = (0.60) (248) = 165 MPa
Safe Load for Tension Member Could Carry
P = (A) (FS)
P = π4
(25mm )2 (6 ) (165 Nmm2 ) = 323,976.76 N > 229,366.84 N
Therefore, safe. Adopt 6-25 mm Ø Anchor Bolt @ 500 mm Length.
12
Slab designSlab-1AWhen: It is one way slab
4 . 05 . 0
=0 .8>0 . 50∴ Two way slab
Minimum thickness
t=perimeter180
t=[5 . 00(2)+4 .0 (2)] 1000180
t=100 .22mmD=100+50mm coveringD=150mm .
Covering (1m-strip)DL=0 .150 (1 . 00)(2 . 4 )(9 . 81)DL=3 .53kN /m .LL=2 . 4kN /m .
Wu=1 . 40DL+1. 70 LLWu=1 . 40(3 .53 )+1. 70(2 . 4 )Wu=4 . 9+4 .08Wu=8 . 98kN /m .
From the table (negative moment at continuous edge), case 4Ca .neg .=0 .071Cb .neg .=0 .029Ms=CswL2
Ms=0 . 071(8 . 98)( 4 .0)2
Ms=10 . 20Kn−m .Mb=0 . 029(8 . 98)( 4 .0)2
Mb=6 . 51kN .m .Coefficient for dead load positive moments in slab
Ca .neg .=0 .039Cb .neg .=0 . 016
Coefficient for live load positive moment in slab
13
Ca .neg .=0 .048Cb .neg .=0 .020
Along short direction+MsDL=0 .039 (4 . 9 )(4 )2
+MsDL=3. 06 kN .m+MsLL=0 . 016(4 .08 )(4 )2
+MsLL=1 .04 kN .m .+MTS=4 .1kN .m .
Along long direction
Negative moment @ discontinuous edges equals 1/3 of positive moment
−Ms=13
(4 . 10)=1. 367kN .m .
−Ms=13
(7 .92 )=2 .64 kN .m .
Along short directionMid spanMu=φ fc' bd2ω (1−0 . 59ω )
Mu=4 .10 kN .m .4 .10 x106=0 .90 (21)(1000)(150 )2ω(1−0 .59ω )ω (1−0 . 59ω)=0 .0096ω2−1 . 69ω+0 . 0163=0
ω=−1 .69±√ (1.69 )2+4 (1)(0 . 0163)2(1)
ω=0 .01
ρ=ωfc 'fy
ρ=0 .01(21 )275
ρ=0 .00076As= ρbdAs=0.00076(1000 )(150)As=114 .55
Use 12mm.φBars1000S
π (12)2
4=114 . 55
S=987 .32mm2
say : 900mm2
Continuous edge
14
+MsDL=0 .048 (4 . 9 )(5 )2
+MsDL=5. 88kN .m .+MsLL=0 . 020(4 . 08 )(5)2
+MsLL=2 .04 kN .m .+Mbt=7 . 92kN .m .
Mu=6 .51 kN .m .Mu=φ fc' bd2ω (1−0 .59ω )6 .51x 106=0 . 90(21)(1000 )(150 )2ω (1−0 .59ω )ω (1−0 . 59ω)=0 .015ω=0 .18
ρ=ωfc 'fy
=0 . 18(21)275
=0 .014
As= ρbdAs=0. 014 (1000)(150 )As=2 ,100mm2
Using 12mmφ1000S
π4
(12)2=2 ,100mm2
S=53. 86mm .say=50mm .O .C .
Discontinuous edgeMoment is only 1/3 at mid spanBent up two of every three bottom bars.
Slab – 2A
SL= 3. 0
4 . 0=0 .75>0. 50∴
Two way slab
Minimum thickness
t=perimeter180
t=[4 .0 (2)+3 . 0(2)](1000)180
t=77 . 78D=77 . 78+50mm(cov ering )D=120mm .
Covering (1m-strip)
DL=3 .53 kN /mLL=2 .4kN /m .Wu=1 .40DL+1.7 LLWu=1 .40(3 .53 )+1.7(2 . 4 )Wu=8 .98 kN /m .
From the table (negative moment at continuous edge) case 4
Ca .neg .=0 .076Cb .neg .=0 .024
15
Ms=CsωL2
Ms=0 . 076(8 . 98 )(3)2
Ms=6 . 14kN−m .Mb=0 . 024(8 .98 )(4 )2
Mb=3 . 45kN−m .
Coefficient for dead-load positive moment in slab
Ca .neg .=0 .043Cb .neg .=0 .013
Coefficient for the live-load positive moment in slab
Ca .neg .=0 .052Cb .neg .=0 . 016
Along short direction
+MsDL=0 .043 (4 .9 )(3)2
+MsDL=1.8963+MsLL=0 .013(4 .08 )(3)2
+MsLL=0 .477+MTS=2 .374 kN−m .
Along long direction
+MsDL=0 .052( 4 .9)( 4 )2
+MsDL=4 . 077 kN−m .+MsLL=0 .016(4 .1 )(4 )2
+MsLL=1 .05 kN−m .+MsbT=5 .127 kN−m .
Negative moment @ discontinuous edge equal 1/3 of positive moment.
−Ms=13
(2. 374 )=0 .79
−Ms=13
(5 . 127 )=1 .704
Along short direction
Mu=φ fcbd2ω (1−0.59ω)Mu=2 .37 kN=m .
2 .374 x 106=0 .90 (21 ) (1000 ) (120 )2ω (1−0 .59ω )ω (1−0 .59ω )=0 . 0087ω=0 . 01
16
ρ=ωfcfy
=0 .01 (21 )275
=0 .00066
As= ρbd=0 .00066 (1000 ) (120 )As=79 .2mm2
Use 12mm.φBars
S=
π (12 )2
4(1000 )
79.2=1427 .99mm.
S=1000mm .
Continuous edge
Mu=6 .14 kN .−m .Mu=φ fcbd2ω (1−0. 59ω)6 . 14 x106=0 . 90 (21 ) (1000 ) (120 )2ω (1−0 . 59ω)ω (1−0 .59ω )=0 .0226ω=0 . 02
ρ=ωfc 'fy
=0 . 02 (21 )275
=0 . 00168
As= ρbd=0 .00168 (1000 ) (120 )As=201 . 6mm .2
Use 12mm.φ Bar
1000S
π (12 )2
4=201 . 6
S=560. 99mm .say :500mm .
Discontinuous edge
Use 3 times the spacing of mid span (moment is only 1/3 if mid span)
Along the long direction
Mid span
17
d=120−50=70mmMu=5 .127kN−m .Mu=φ fcbd2ω (1−0.59ω)5 .127 x 106=0 . 90 (21 ) (1000 ) (70 )2ω (1−0 .59ω )ω (1−0 .59ω )=0 . 0554ω=0 .05
ρ=ωfcfy
=0 . 05 (21 )275
=0.00411
As= ρbd=0 .00411 (1000 ) (70 )As=287 . 7mm .
Use 12mm.φBar
S=
π (12 )2
4(1000 )
287. 7=393 .11mm
say :300mm .
Continuous edge
Mu=3 . 45kN−m .Mu=φ fcbd2ω (1−0.59ω)3 . 45 x106=0 .40 (21 ) (1000 ) (120 )2ω (1−0 . 59ω)ω (1−0 .59ω )=0 . 0127ω=0 . 013
ρ=ωfcfy
=0 . 013 (21 )275
=0 . 001
As= ρbd=0 .001 (1000 ) (120 )As=119.13mm2
Use 12mmφ Bar
S=
π (12 )2
4(1000 )
119.13=949 .36mm.
say : 900mm.
Discontinuous edge
Moment is only 1/3 at mid span, bent up two every three bottom bars.
Beam Design
18
LL=2 .5 kN /m .DL=3 .5kN /m .
Wu=1 .4DL+1 .7 LLWu=1 .4 (3 .5 )+1 .7 (2.5 )Wu=9 .57 kN /m .
Mu=WuL2
8=
9 . 87 (4 )2
8=19. 14kN−m .
R=ωL2
=9 .57 ( 4 )2
=19.14 kN .
ρb=0 .85 fc ' β600fy (600+ fy )
=0 . 85 (21 ) (0 .85 ) (600 )275 (600+275 )
ρb=0 .0378ρmax=0 .75 (0 . 0378 )=0. 02835
assume : ρ=12ρmax=1
2( 0.028 )
ρ=0 .014
ω=ρ fyfc'
=0.014 (275 )21
=0 .18
Mu=φbd 2ω (1−0. 59ω )Mu=φbd 2RR=fc ' ω (1−0 .59ω )R=(21 ) (0 .18 ) [1−0 .59 (0 .18 ) ]R=3.38Mpa .
d=√MuφbRassume : b=200
d=√19 . 14 x106
0 .90 (200 ) (3 .38 )d=177.37mm .
Total depth:
177 . 37+50=227 .37mm .say : d=300mm
As= ρbdAs=0. 014 (200 ) (300 )As=840mm2
Using 16mm. φ RSB
19
n=840π (16 )2
4n=4 .18say : 4 pcs−16mmφ RSB
Vu=WuL2
=Wud
d=L2 (16 )
d=(L12 )=(412 )=0 . 333
Vu=9 ,570 (4 )2 −9 ,570 (0 . 333 )
Vu=17 ,546 .60N .
Shear force that a concrete can carry
Vc=16 √ fc ' bd
Vc=16 √ (21 ) (200 ) (300 )
Vc=187 . 08NφVc=0 . 85 (187. 08 )=159. 0212φVc=159 .02
2=79 .51N<Vu
∴ It needs reinforcement
For nominal shear strength
Vs=Vuφ
−Vc
Vs=17 ,546 .600 .85
−170 .29
Vs=20 ,469 .24N .
Use 10mm.φ Stirrups
Av=2 As
Av=2π4
(10 )2=157mm2
Spacing of stirrups
S=AvfydVs
S=157 (275 ) (300 )20 ,469. 29
S=632. 76mm
20
Check if Vs< 1
3 √ fc ' bd
13 √ (21 ) (200 ) (300 )=374 . 17
Maximum spacing
max .S= d2=300
2=150
Checks on minimum area requirements
Av=bs3 fy
=200 (150 )3 (275 )
=36 .36mm2
Av=36 .36mm2<157mm .2
∴use ,10mm .φ ( steelbars )
Beam -2A
DL=2. 5kN /m .LL=3 .5kN /m .Wu=1 . 40DL+1. 70 LLWu=9 . 57kN /m .
Mu=WuL2
8=
9 . 57 (5 .7 )2
8=38 .87kN /m .
R=wL2 =
9 . 57 (5.7 )2 =27 .27kN .
ρb=0 .85 fc' β600fy (600+ fy )
=0 .85 (21 ) (0 . 85 ) (600 )275 (600+275 )
ρb=0 .038ρmax=0 .028ρ=0 .014
ω=ρ fyfc
ω=0 . 24R=4 . 33MPa.b=300mm .d=350mm.As= ρbdAs=0.014 (300 ) (350 )As=1470mm2
21
No. of 16mm. φ
RSB.
n=1470π (16 )2
4
=7 .31
n=8 pcs .−16mm .φ RSB
Vu=WuL2
−Wud
d=L2 (5 .7
12 )=0 . 48
Vu=9 ,570 (5 .7 )2
−9 ,570 (0 .48 )
Vu=22 ,680.9N .
Shear force that a concrete can carry
Vc=16 √ fc ' bd=1
6 √ (21 ) (300 ) (350 )
Vc=247 . 49N .φVc=0 .85 (247 . 49 )=210 .36N .12φVc=210 .36
2=105N .<Vu
It needs reinforcement
For nominal shear strength
Vs=Vuφ
−Vc
Vs=22 ,680 . 90 .85
−247 .49
Vs=26 ,435 . 92N .
Using 10mm. φ
stirrups
Av=2 As
Av=2π4d2
Av=150mm2
Spacing of stirrups
22
S=AvfydVs
S=157 (275 ) (350 )26 ,435 .92
S=571. 62mm.
Check if Vs< 1
3 √ fcbd
13 √ (21 ) (300 ) (350 )=494 .97 kN .
Maximum spacing
max .S= d2=300
2=150
Check on minimum area requirements
Av=bs3 fy
=300 (150 )3 (275 )
=54 .54
Av=54 . 54mm2<157mm .∴use ,10mm φbar
23
Beam – 4A and 7A
24
Mu=WuL2
8=
8 . 98 (5 )2
8Mu=28 .06 kN ./m .
R=wl2
=8 . 98 (5 )2
=22.45 kN .
ρb=0 .0378ρmax=0 .75 ( ρb )ρmax=0 .028
ρ=12ρmax
ρ=0 .014
ω=ρ fyfc
ω=0 .18Mu=φ fc' bdω (1−0 .59ω )Mu=φbd 2RR=fc ' ω (1−0 .59ω )R=3.38b=200d=300As= ρbdAs=0. 014 (200 ) (300 )As=840mm
n=840π (16 )2
4
=4 .2
n=4 pcs−16mm .φbar .
Vu=WuL2
−Wud
d=L2
=(512 )=0 . 417
Vu=8 . 98 x103 (5 )2
−8 . 98 x103 (0. 417 )
Vu=18 ,705 .34N .
Shear force that a concrete can carry
25
Vc=16 √ fc ' bd=1
6 √21 (200 ) (300 )
Vc=187 . 08NφVc=0 .85 (187.08 )=159.02N .12φVc=159 .02
2=79 .51N .<Vu
It needs reinforcement
For nominal shear strength
Vs=Vuφ
−Vc
Vs=18 ,705 . 340 .85
−187 .08
Vs=21 ,819 . 20N
Using 10 mm. φ
stirrups
Av=2 As
Av=2π4d2
Av=157mm2
Spacing of stirrups
S=AvfydVs
S=157 (275 ) (300 )21,819 . 20
S=593. 63mm .
Check if Vs=< 1
3 √ fc ' bd
13 √ (21 ) (200 ) (300 )=374 . 17
Maximum spacing
max .S= d2=300
2=150
Check on minimum area requirements
26
Av=bs3 fy
=200 (150 )3 (275 )
=36 .36
Av=36 .36mm2<157mm .∴use ,10mm .φbar .
27
Beam -4A
Wu=1 . 40DL+1. 70 LLWu=1 . 40 (2. 1 )+1. 70 (1.5 )Wu=5 . 49kN /m .
Mu=WuL2
8=
5 .49 (4 )2
8=10.98kN /m .
R=wL2 =
5 . 49 (4 )2 =10 . 98kN .
ρb=0 .0378ρ=0 .014ω=0 .18assume : b=150
d=√MuφbRd=√10 .98 x106
0 . 90 (150 ) (3.38 )d=155.12mm.total , depth=155. 12+50=205 .12say :200mmAs= ρbdAs=0. 014 (150 ) (200 )As=420mm2
Using 16 mm. φ
RSB
n=420π (16 )2
4
=2.24
say :3 pcs−16mm .φRSB .
28
Vu=WuL2
−Wud
d=L2 (16 )
d=L12
=412
=0 .333
Vu=9570 ( 4 )2
−9570 (0 .333 )
Vu=15 ,953 .19N .
Shear force that a concrete can carry
Vc=16 √ fc ' bd
Vc=16 √ (21 ) (150 ) (200 )
Vc=132.29N .φVc=0 . 85 (132. 29 )=112.4512φVc=112 .45
2=56. 22N .<Vu
It needs reinforcement
For nominal shear strength
Vs=Vuφ
−Vc
Vs=15 ,953. 190 .85
−132.29
Vs=18 ,636 . 17N .
Using 10 mm. φ
stirrups
Av=2 As
Av=2π4d2
Av=157mm2
Spacing of stirrups
S=AvfydVs
S=157 (275 ) (200 )18 ,636 . 17
S=463 .35mm .29
Check if Vs=< 1
3 √ fc ' bd
13 √ (21 ) (150 ) (200 )=264 .58
Maximum spacing
max .S= d2=200
2=100
Check on minimum area requirements
Av=bs3 fy
=200 (100 )3 (275 )
=24 . 24mm2 .
Av=24 . 24mm2 .
30
Beam – 9A
DL=1. 8kN /m .LL=1 .5kN /m .
Wu=5 . 07kN /m .Mu=10 .14 kN /m .Ru=10 .14kN /m .ρb=0 .0378ρ=0 .014ω=0 .18R=3.38MPa .
d=√MuφbRd=√10 .14 x106
0 . 90 (150 ) (3.38 )d=149mm+50=199 .07mm .say :200mm .
As=ρbdAs=0.014 (150 ) (200 )As=420mm .2
Using 16 mm. φ
RSB
31
n=420π (16 )2
4
=1. 68
say :2−3 pcs−16mm .φ RSB
Vu=WuL2
−Wud
d=0 .333
Vu=5078 (4 )2
−5078 (0 . 333 )
Vu=8 ,451 .69N .
Shear force that a concrete can carry
Vc=16 √ fc ' bd
Vc=16 √ (21 ) (150 ) (200 )
Vc=132.29N .φVc=0 . 85 (132. 29 )=112 .45 N .12φVc=112 . 45
2=56.23N<Vu
It needs reinforcement
For nominal shear strength
Vs=Vuφ
−Vc
Vs=8 ,451 .690 .85
−132.29
Vs=9 ,810. 87 N .
Using 10 mm. φ
bars
Av=3 As
Av=2π4
(d )2
Av=2π4
(10 )2
Av=157mm2
Spacing of stirrups
32
S=AvfydVs
S=157 (275 ) (200 )9 ,810. 87
S=880. 146mm .
Check if Vs< 1
3 √ fcbd
13 √ (21 ) (150 ) (200 )=264 .58
Maximum spacing
max .S= d2=200
2=100
Check on minimum area requirements
Av=bs3 fy
=150 (100 )3 (275 )
=18. 18
Av=18 .18<157mm∴use−10mm .φ
33
Beam – 6A
Mu=WuL2
8=
8 .98 (3 )2
8=10.10kN /m .
Mu=wL2
=8.98 (3 )2
=13 . 47kN .
ρb=0 .0378ρmax .=0.75 ρb
assume : ρ=12ρmax .
ρ=0 .014
ω=ρ fyfc
ω=0 .18R=3.38kN .
d=√MuφbRassume : b=150mm .
d=√10 .10x 106
0 .90 (150 ) (3.38 )d=148.78mm .148 .78+50 (cov ering )=198. 78mm .say :200mm .As= ρbdAs=0. 18 (150 ) (200 )As=420mm .2
34
Using 16 mm. diameters RSB.
n=420π (16 )2
4
=2.08
say :2−5 pcs .−16mm .φRSB .
Vu=WuL2
−Wud
d=L12
=312
=0 .25
Vu=8 ,980 (3 )2
−8 ,980 (0 .25 )
Vu=11 ,225N .
Shear force that a concrete can carry
Vc=16 √ fcbd
Vc=16 √ (21 ) (150 ) (200 )
Vc=170 .78N .φVc=0 .85 (170. 78 )=145 . 16512φVc=145 .165
2=72. 58<Vu
It needs reinforcement.
For nominal shear strength
Vs=Vuφ
−Vc
Vs=11 ,2250 .85
−170 . 78
Vs=13 ,035 . 10
35
Using 10 mm. diameter stirrups
Av=2 As
Av=2π4d2
Av=157mm2
Spacing of stirrups
S=AvfydVs
S=157 (275 ) (200 )13 ,035 . 10
S=662. 44mm.
Check if
13 √ (21 ) (150 ) (200 )=265 .58
Maximum spacing
max .S= d2=200
2=100
Checks on minimum area requirements
Av=bs3 fy
=150 (100 )3 (275 )
=18. 18mm2
Av=18 .18<157mm .∴use :10mm.φ
36
Vs=< 13 √ fc ' bd
Beam @ deck
Beam -7B
37
L=5 . 0m .LL=1 .9 kN /m .DL=3 .33kN /m .Wu=7 .89kN /m .
Mu=WuL2
8=
7 .89 (5 .0 )2
8=24 .66 kN−m .
R=wL2 =
7 .89 (5 )2 =19 . 73kN .
ρb=0 .04ρmax .=0. 03
ρ=12ρmax.
ω=ρ fyfc '
=0. 015 (275 )21
=0 .20
R=3.70b=200mm .
d=√MuφbRd=√24 . 66 x106
0 .90 (200 ) (3 .70 )d=192. 42mm192 .42mm+50 (cov ering )=242.50say :250mm .As=1 ,129. 5mm2
Using 16mm. diameter RSB.
n=1 ,129 .5π (16 )2
4
=5 .62
n=6 pcs .−16mm .φ RSB
Vu=WuL2
−Wud
d=L12
=512
=0 . 42
Vu=7 . 89 (5 )2
−7 . 89 (0 . 42 )
Vu=16 . 41kN .
Shear force that a concrete can carry
38
Vc=16 √ fc ' bd
Vc=16 √ (21 ) (200 ) (250 )
Vc=170 .78N .φVc=0 .85 (170. 78 )=145 . 16512φVc=145 .165
2=72. 58<Vu
It is needs reinforcement
For nominal shear strength
Vs=Vuφ
−Vc
Vs=16 ,411. 20 .85
−170 .72
Vs=19 ,136 . 57N .
Using 10 mm. diameter stirrups
Av=2 As
Av=2π4d2
Av=150mm2
Spacing of stirrups
S=AvfydVs
S=150 (275 ) (250 )19 ,136 . 57
S=538. 89mm .
Check if
13 √ (21 ) (200 ) (250 )=341. 57 N .
Maximum spacing
max .S=2502
=125
Check on minimum area requirements
Av= bs3 fy
=200 (125 )
3 (275 )=30 .30<157mm2
39
Vs=< 13 √ fc ' bd
Used 10mm. diameter RSB
40
Beam -1B, 8B, and 9B
Wu=1 . 4 (DL)+1 .7 (LL )Wu=1 . 4 (3 .33 )+1 . 7 (1. 9 )Wu=7 . 892 kN /m .
Mu=WuL8 =7 .832 (4 )8 =15. 78kN /m .
R=wL2
=7 . 832 (4 )2
=15 .78kN .
ρb=0 .04ρmax .=0. 03
ρ=12ρmax.
ρ=0 .015
ω=ρ fyfc
=0. 015 (275 )21
=0 .20
R=fc ' ω (1−0 .59ω )R=21 (0 . 20 ) [1−0 .59 (0 . 20 ) ]R=3. 70
d=√MuφbRd=√15 . 78 x 106
0 . 90 (200 ) (3 .70 )d=153. 93mm153 . 93+50 (cov ering )=203 . 93mm .say :250mm .b=200mm .As= ρbdAs=( 0. 015 ) (200 ) (250 )As=1 ,129. 5mm2
Using 16mm. diameter RSB
n=1 ,129.5π (16 )2
4
=5 .62 ,−say :6 pcs .−16mm.φ RSb
41
Vu=WuL2
−Wud
d=L12
=(412 )=0 .3
Vu=7 . 89kN /m . ( 4 )2
−7 .89 (0 . 3 )
Vu=13 . 14kN .
Shear force that a concrete can carry
Vc=16 √ fc ' bd
Vc=16 √ (21 ) (200 ) (250 )
Vc=170 .78φVc=0 . 85 (170. 78 )=145 . 16512φVc=145 .165
2=72. 58kN .<Vu
It is needs reinforcement
For nominal shear strength
Vs=Vuφ
−Vc
Vs=13 ,4100 .85
−170 . 78
Vs=15 ,605 . 69
Using 10 mm. diameter stirrups
Av=2 As
Av=2π4d2
Av=150mm2
Spacing of stirrups
S=AvfydVs
S=150 (275 ) (250 )15 ,605 . 69
S=660. 81mm .
Check if
42
Vs=< 13 √ fc ' bd
13 √ (21 ) (200 ) (250 )=341. 57
Maximum spacing
max .S=2502
=125mm .
Check on minimum area requirements
Av=bs3 fy
=200 (125 )3 (275 )
=30 .30mm2
Av=30 .30mm<157mm2
∴use .−10mm .φRSB .
43
Beam -6B
L=3mLL=1 . 9kN /m .DL=3 .33kN /m .Wu=7 . 89kN /m .
Mu=WuL2
8=
7 .89 (3 )2
8=8 .88kN−m .
R=wL2 =
7 .89 (3 )2 =11.48kN .
ρb=0 .04ρmax .=0. 03
ρ=12ρmax
ρ=0 .015ω=0 . 20R=3.70 ;b=200mm.
d=√MuφbRd=√8 . 88 x106
0 . 90 (200 ) (3 .70 )d=115.47+50 (cov ering )=165 . 47say :200mm .As= ρbdAs=0.015 (200 ) (200 )As=600mm2
Using 16 mm. diameters RSB.
44
n=600π (16 )2
4
=2.98 pcs .
say :3 pcs−16mm .φRSB
Vu=WuL2
−Wud
d=L12
=312
=0 .25
Vu=7 ,890 (3 )2
−7 ,890 (0. 25 )
Vu=9 ,862 .5N .
Shear force that a concrete can carry
Vc=16 √ fcbd
Vc=16 √ (21 ) (200 ) (200 )
Vc=152. 75N .φVc=0 . 85 (152. 75 )=129 . 8412φVc=129 .84
2=67. 90<Vu
It needs reinforcement
For nominal shear strength
Vs=Vuφ
−Vc
Vs=9 ,862. 50 .85
−152. 75
Vs=11 ,450 .19 N .
Using 10 mm. diameter stirrups
Av=2 AsAv=150mm2
Spacing of stirrups
S=AvfydVs
S=150 (275 ) (200 )11 ,450 .19
S=720. 51mm.
45
Check if Vs< 1
3 √ fc ' bd
13 √ (21 ) (200 ) (200 )=305 .51<Vs
Maximum spacing
max .S=2002
=100
Check on minimum area requirements
Av=bs3 fy
=200 (100 )3 (275 )
=24 .24
Av=24 .24<(157mm2 )∴use :10mm.φ
46
Beam -2B
47
L=5 . 7m .LL=1 . 9kN /m .DL=3 .33 kN /m .Wu=7 . 89 kN /m .
Mu=WuL8
=7 .89 (5 . 7 )8
=22. 49 kN−m .
R=wL2 =
7 . 89 (5. 7 )2 =22. 49kN .
ρb=0 .04ρmax .=0. 03
ρ=12ρmax.
ρ=0 .015ω=0 . 20R=3. 70assume : b=250
d=√MuφbRd=√32 . 04 x106
0 . 90 (250 ) (3 .70 )d=196. 18+50 (cov ering )=246 . 18mmsay :300mm .As= ρbdAs=0. 015 (200 ) (300 )As=900mm2
Using 16mm diameter RSB
n=900π (16 )2/4
=4 .47
say :5 pcs .−16mm.φ RSB
Vu=WuL2
−Wud
d=L12
=5 .72
=0. 48
Vu=7 .89 (5. 7 )2
−7 . 89 (0. 48 )
Vu=18 .69 kN .
Shear force that a concrete can carry
48
Vc=16 √ fc ' bd
Vc=16 √ (21 ) (250 ) (300 )
Vc=209 .17φVc=0 . 85 (209. 17 )=177 .7912φVc=177 .79
2=88 . 90<Vu
For nominal shear strength
Vs=Vuφ
−Vc
Vs=198 ,699 . 30 .85
−209.17
Vs=21 ,790 . 00N .
Using 10 mm. diameter stirrups
S=AbfydVs
S=150 (275 ) (300 )21,790 . 0
S=567. 92mm .
Check if Vs< 1
3 √ fcbd
13 √21 (250 ) (300 )=418 . 33
Maximum spacing
max .S=3002
=150
Check on minimum area requirements
As=bs3 fy
=300 (150 )3 (275 )
As=54 . 54<157mm .∴use :10mm.φ RSB
49
50
Design of Column
Column -1B, 3B and 4B
Pu=19 ,730N .use :φ=0 . 70
Limits of reinforcement for tied column
ρg=0 .01−0 .08
try : ρg=0 . 03
Ag=Puφ (0 . 80 ) [0 . 85 fc (1−ρg )+ fy ρg ]
Ag=19 ,730N .0 . 56 [ (17. 31 )+(8 .25 ) ]
Ag=1 ,378.37mm .2
h2=1 ,378. 37mm2
h=37 .13mm .try :300mm . x300mm .
Ag=90 ,000mm2
As= ρ gAg=0. 03 (90 ,000 )As=2 ,700mm2
Using 16mm. diameter RSB
n=2700201. 062
=13. 43
say :14 pcs .−16mm .φRSB
ρg=AsAg
=270090 ,000
=0 .03∴ok
Spacing of lateral ties 10 mm. diameter
S=48 (10 )=480mm .S=16 (16 )=256mm .S=300mm. ( least dim ension )
Pu=φ (0. 80 ) [0 . 85 fc ' (Ag−ASt )+fyASt ]Pu=0 . 70 (0 .80 ) [0 . 85 (21 ) (90 ,000−2700 )+275 (2700 ) ]Pu=0 . 56 [ 1,466 . 640+742 ,500 ]Pu=1 ,237 ,118 . 4N .>19 ,730N .∴ safe
51
52
Column -1A
Pu=10 . 14kN .use :φ=0 . 70
Limits of reinforcement for tied column
ρg=0 .01−0 .08try : ρg=0 .03
Ag=Puφ (0 .80 ) [0 . 85 fc ' (1−ρg )+ fy ρg ]
Ag=1 01400 .70 (0.80 ) [0 . 85 (21 ) (1−0. 03 )+ (275 ) (0 . 03 ) ]
Ag=708 . 60mm2
h2=708 . 60mm2
h=26 .62mm .try :350mm . x350mm .Ag=122 ,500mm .As= ρ gAg=0. 03 (122 ,500 )As=3 ,675mm2
Using 25mm. diameter RSB
n=3675490 . 87
=7 .5
say : 8 pcs .
ρg=AsAg
=3675122 ,500
=0 . 03∴ ok
53
Spacing of lateral ties 10mm.diameter
S=48 (10 )=480mm .S=25 (10 )=250mm .S=300mm (least dim ension )
Pu=φ (0. 80 ) [0 . 85 fc ' (Ag−Ast )+ fyAst ]Pu=0 . 70 (0 .80 ) [0 . 85 (21 ) (122 ,500−3675 )+(275 ) (3675 ) ]Pu=1 ,753 ,724 .7>10 ,140N .∴ safe
54
Column -2B
55
Pu=22 .49kN .
try : ρg=0 .03
Ag=Puφ (0 .80 ) [0 . 85 fc ' (1− ρg)+ fy ρg ]
Ag=22 ,4900 .70 (0.80 ) [0 . 85 (21 ) (1−0. 03 )+ (275 ) (0 . 03 ) ]
Ag=1 ,571. 63mm2
h2=1 ,571. 63mm2
h=39. 64
try :300mx300mm .Ag=90 ,000mm2
As= ρ gAg=0. 03 (90 ,000 )As=270mm2
Using 16 mm. diameter RSB
n=2700201. 062
=13. 43
say :14 pcs .−16mm .φRSB
ρd=AsAg
=270090 ,000
=0 . 03∴ok
Spacing of lateral ties 10 mm. diameter
S=48 (10 )=460mm .S=16 (16 )=256mm .S=300mm. (least dim ension )
Pu=φ (0. 80 ) [0 . 85 fc ' (Ag−Ast )+ fyAst ]Pu=0 . 70 (0 .80 ) [0 . 85 (21 ) (90 ,000−2700 )+(275 ) (2700 ) ]Pu=1 ,237 ,118 . 4>22 ,490 . 0N .∴ safe
56
Column -2A
Pu=27 .27kN .
try : ρg=0 .03
Ag=Puφ (0 .80 ) [0 . 85 fc ' (1− ρg)+ fy ρg ]
Ag=27 ,2700 .70 (0.80 ) [0 . 85 (21 ) (1−0. 03 )+ (275 ) (0 . 03 ) ]
Ag=1 ,905 .66mm2
h2=1 ,905 .66mm2
h=43 . 65
try :350mx350mm .Ag=122 ,500mm2
As= ρ gAg=0. 03 (122 ,500 )As=3 ,675mm2
Using 25 mm. diameter RSB
57
n=3675490 . 87
=7 .5
say : 8 pcs .−16mm .φ RSB
ρd=AsAg
=3675122 ,500
=0. 03∴ ok
Spacing of lateral ties 10 mm. diameter
S=48 (10 )=460mm .S=16 (16 )=256mm .S=300mm. ( least dim ension )
Pu=φ (0. 80 ) [0 . 85 fc ' (Ag−Ast )+ fyAst ]Pu=0 . 70 (0 .80 ) [0 . 85 (21 ) (122 ,500−3675 )+(275 ) (3675 ) ]Pu=1 ,753 ,724 N .>27 ,270N .∴ safe
58
Column -3A
Pu=19 . 14kN .
try :0 . 03
Ag=1914014 . 31
Ag=1337 .53h2=1337 . 53h=36 .57
try :350mmx 350mm.Ag=122 ,500mm .As=ρ gAg=0. 03 (122 ,500 )As=3675mm2 .
Using 25 mm. diameter RSB
n=3675490 . 87
=7 .5
say : 8 pcs .−16mm .φ RSB
ρd=AsAg
=3675122 ,500
=0. 03∴ ok
Spacing of lateral ties 10 mm. diameter
S=48 (10 )=460mm .S=16 (16 )=256mm .S=300mm. ( least dim ension )
Pu=φ (0. 80 ) [0 . 85 fc ' (Ag−Ast )+ fyAst ]Pu=0 . 70 (0 .80 ) [0 . 85 (21 ) (122 ,500−3675 )+(275 ) (3675 ) ]Pu=1 ,753 ,724 N .>1 ,940 .0N .∴ safe
59
Column -4A
Pu=38 ,240N .
Ag=38 ,24014 . 13
Ag=2 ,706 .30h2=2 ,706 .30h=52. 02
try :350mmx 350mm.Ag=122 ,500As=ρ gAg=0. 03 (122 ,500 )As=3 ,675mm2
Using 25 mm. diameter RSB
60
n=3675490 . 87
=7 .5
say : 8 pcs .−16mm .φ RSB
ρd=AsAg
=3675122 ,500
=0. 03∴ ok
Spacing of lateral ties 10 mm. diameter
S=48 (10 )=460mm .S=16 (16 )=256mm .S=300mm. ( least dim ension )
FOOTING DESIGN
Footing -1
Pu=29 ,870 N .
qa=PuA
A=29 ,870N .170 (1000 )
A=0 .176m2
61
For square footing
L=√0 .175L=0. 42try :1 .5mx 1 .5mA=1.5 (1.5 )=2 .5m2
b2=√2 .25m2
b=1. 5m .
Net ultimate upward soil pressure
qu=29 ,870N .2. 25m2
=13 ,275 .55N /m .
Mu=13 ,275. 55 (1. 5 ) (0 .58 )(0 .582 )
2
Mu=3 ,349. 42N−m .
Compute W, Ru
ρmin .=1. 40fy
=1 .40275
=0. 0051
ω=ρ fyfc '
=0. 0051 (275 )21
=0.067
Ru=ωfc ' (1−0 .59ω)Ru=0 . 067 (21 ) [1−0 .59 (0 .067 ) ]Ru=1 .35MPa .d=350−50−1.5φd=350−50−1.5 (16 )d=276mm .t=d+1 .5φ+cov eringt=276+1 .5 (16 )+50t=350mm .
Check for bending shearVu=13 ,275 .55 [ (1 .5 )2−(0 .626 )2 ]Vu=24 ,667 .62N .
Actual punching shear
62
Vn=Vuφbod
bo=4 (c+d )bo=4 (350+276 )bo=2504
Vn=24 ,667 . 620 .85 (2.504 ) (0 . 276 )
Vn=43 ,382 . 42Pa .
Actual beam shear
Vn=Vuφbd
=13 ,275.55 (1. 5 ) (0 .350 )0 . 85 (1 .5 ) (0. 226 )
Vn=24 ,187 . 62Pa .≈0.024MPa .
Allowable beam shear ACI code
Vu=16 √ fc '=1
6 √21=0 .76MPa .
Vu=0 . 76Mpa .>Vn0 .024MPa .∴ safe
Compute for required reinforcementAs=ρbdAs=0. 0051 (1500 ) (226 )As=1728 . 9mm2
Using 16mm. diameter RSB
n=1728. 9201. 061
=8 .59
say : 9 pcs .−16mm .φRSB .bothways
63
Footing -2
Pu=49 ,760N .
qa=PuA
A=49 ,760N .170 (1000 )
A=0 .293m2
For square footing
L=√0 .293L=0.42try :1 .2mx 1 .2mA=1. 2 (1 . 2 )=1 .44m2
b2=√1 .44m2
b=1.2m .
Net ultimate upward soil pressure
qu=49 ,760N .1. 44m2
=34 ,555 .56 N /m .
Mu=34 ,555 .56 (1 .2 ) (0 . 45 )(0. 452 )
2
Mu=5 ,038. 20N−m .
Compute W, Ru64
ρmin .=1. 40fy
=1 .40275
=0. 0051
ω=ρ fyfc '
=0. 0051 (275 )21
=0.067
Ru=ωfc ' (1−0 .59ω)Ru=0 . 067 (21 ) [1−0 .59 (0 .067 ) ]Ru=1 .35MPa .d=300−50−1.5φd=300−50−1.5 (16 )d=231mm .t=d+1 .2φ+coveringt=231+1 .2 (16 )+50t=300 .2mm .
Check for bending shearVu=34 ,555 . 56 [ (1. 2 )2− (0 .531 )2 ]Vu=40 ,016 . 69N .
Actual punching shear
Vn=Vuφbod
bo=4 (c+d )bo=4 (300+231 )bo=2324mm .
Vn=40 ,016. 690 . 85 (2. 124 ) (0 .231 )
Vn=87 ,694 . 86 Pa .
Actual beam shear
Vn=Vuφbd
=34 ,555 .56 (1 .2 ) (0 . 300 )0 . 85 (1 . 2 ) (0 .231 )
Vn=5 ,279 .69 Pa .≈0 . 053MPa.
Allowable beam shear ACI code
Vu=16 √ fc '=1
6 √21=0 .76MPa .
Vu=0 . 76Mpa .>Vn0 .053MPa.∴ safe
Compute for required reinforcement
65
As= ρbdAs=0. 0051 (1200 ) (231 )As=1 ,413 . 72mm2
Using 16mm. diameter RSB
n=1 ,413. 72201. 061
=7 .13
say :7 pcs .−16mm .φ RSB .bothways
Footing -3
Pu=38 ,870 N .
qa=PuA
A=38 ,870N .170 (1000 )
A=0 .229m2
For square footing
66
L=√0 .229L=0. 48try :1 .5mx 1 .5mA=1.5 (1.5 )=2 .5m2
b2=√2 .25m2
b=1. 5m .
Net ultimate upward soil pressure
qu=38 ,870N .2.25m2
=17 ,275.56N /m .
Mu=17 ,275. 56 (1 .5 ) (0 .6 )(0 .62 )
2
Mu=1 ,399 . 32N−m .
Compute W, Ru
ρmin .=1. 40fy
=1 .40275
=0. 0051
ω=ρ fyfc '
=0. 0051 (275 )21
=0.067
Ru=ωfc ' (1−0 .59ω)Ru=0 . 067 (21 ) [1−0 .59 (0 .067 ) ]Ru=1 .35MPa .d=300−50−1.5φd=300−50−1.5 (16 )d=226mm .t=d+1 .5φ+cov eringt=226+1 .5 (16 )+50t=300mm .
Check for bending shearVu=17 ,275 .56 [ (1 . 5 )2− (0. 526 )2 ]Vu=34 ,090 .28N .
Actual punching shear
67
Vn=Vuφbod
bo=4 (c+d )bo=4 (300+226 )bo=2 ,104
Vn=34 ,090 . 280 .85 (2.104 ) (0 . 226 )
Vn=84 ,344 .64 Pa .
Actual beam shear
Vn=Vuφbd
=17 ,275 .56 (1 .5 ) (0 . 300 )0 . 85 (1 . 5 ) (0.226 )
Vn=17 ,986 . 01Pa .≈0.084MPa .
Allowable beam shear ACI code
Vu=16 √ fc '=1
6 √21=0 .76MPa .
Vu=0 . 76Mpa .>Vn0 .084MPa .∴ safe
Compute for required reinforcementAs=ρbdAs=0. 0051 (1500 ) (226 )As=1728 . 9mm2
Using 16mm. diameter RSB
n=1728. 9201. 061
=8 .59
say : 9 pcs .−16mm .φRSB .bothways
68
Footing -4
Pu=57 ,970 N .
qa=PuA
A=57 ,970N .170 (1000 )
A=0 .341m2
For square footing
L=√0 . 341L=0. 42try :1 .6mx1. 6mA=1. 6 (1. 6 )=2. 56m2
b2=√2 .56m2
b=1. 6m .
Net ultimate upward soil pressure
qu=57 ,970N .2. 56m2
=22 ,644 .53N /m .
Mu=22 ,644 .53 (1.6 ) (0 .63 )(0 .632 )
2
Mu=2 ,264 .88N−m .
Compute W, Ru69
ρmin .=1. 40fy
=1 .40275
=0. 0051
ω=ρ fyfc '
=0. 0051 (275 )21
=0.067
Ru=ωfc ' (1−0 .59ω)Ru=0 . 067 (21 ) [1−0 .59 (0 .067 ) ]Ru=1 .35MPa .d=350−50−1.5φd=350−50−1.5 (16 )d=276mm .t=d+1 .5φ+cov eringt=276+1 .5 (16 )+50t=350mm .
Check for bending shearVu=22 ,644 .53 [ (1 .6 )2−(0 .626 )2]Vu=43 ,096 . 149N .
Actual punching shear
Vn=Vuφbod
bo=4 (c+d )bo=4 (350+276 )bo=2504
Vn=43 ,096. 1490 .85 (2.504 ) (0 . 276 )
Vn=73 ,362.84 Pa .
Actual beam shear
Vn=Vuφbd
=22 ,644 .53 (1 .6 ) (0 . 350 )0 .85 (1 .6 ) (0 .226 )
Vn=41 ,257 . 602Pa .≈0. 0413MPa .
Allowable beam shear ACI code
Vu=16 √ fc'=1
6 √21=0 .76MPa .
Vu=0 . 76Mpa .>Vn0 . 0413MPa.∴ safe
Compute for required reinforcement
70
As=ρbdAs=0. 0051 (1600 ) (226 )As=1844 . 16mm2
Using 16mm. diameter RSB
n=1844 .16201. 061
=9. 17
say :10 pcs .−16mm.φ RSB .bothways
71
CHAPTER IX
PROJECT PLANNING AND SCHEDULING
Table No. 5 Manpower Schedule
Skilled Non-Skilled Duration
1 st
Quarter 98
2 nd
Quarter 98 days
3 rd
Quarter 98 days
4 th
Quarter 98 days
Foreman 3 4 3 2
Surveyor/Instrumentman 1 0 0 0
Carpenter 9 9 9 0
Tile Settler 0 0 0 5
Painter 0 0 0 9
Steelman 11 11 11 0
Welder 0 9 0 0
Plumber 4 0 0 4
Operator (Equipment) 1 1 1 0
Laborer/Helper 31 31 22 22
Total 60 65 46 38
72
Table No. 6 Schedule of Equipments
Light and Heavy Equipment Duration (Quarterly)
1 st 2 nd 3 rd 4 th
Grader 1 0 0 0
Backhoe 1 0 0 0
Bar Cutter 8 8 8 0
1 Unit bagger concrete mixer 2 2 2 0
1 Unit bagger concrete vibrator 2 2 2 0
Welding Machine 0 8 8 0
Sander 0 4 4 2
Total 14 24 25 2
Table No. 7 Capabilities of Manual Labor per Hour Item Type of work Capability
1. dozer a. clearing 500 sq.m./hr.b. stripping 200 sq.m./hr.c. excavation 25 cu.m./hr.d. quarrying 50 cu.m./hr.
73
CAPABILITIES OF EQUIPMENT
Item Type of work Capability
1. dozer a. clearing 500 sq.m./hr.b. stripping 200 sq.m./hr.c. excavation 25 cu.m./hr.d. quarrying 50 cu.m./hr.e. pushing 3 sq.m./hr.
2. grader a. sub-graving 300 sq.m./hrb. spreading 40 cu.m./hr
3. pay loader a. loading 30 cu.m./hr
4. crane shovel a. loading 35 cuu.m./hr
5. sheep's foot a. static rolling12 passes - 15 cm. lift
roller b. vibrator rolling 4 passes - 15 cm. lift 135 cm./hr.
6. no. 3w road roller a. static rolling 24 cmph 6 passes - 20 cm. lift
7. tractor-drawn roller (1-d) a. vibrator rolling
6 passes-20 cm. lift 240 cmph
8. tandem roller a. static rolling 6 passes-20cm. Lift 24 cmph
9. 5-T dump truck a. hauling common barrow 3.5 cmptb. hauling selected borrow base course 5 cmpt
NOTE: Cmph= cum/hour: cmpt= cum./truck
e. pushing 3 sq.m./hr.
2. grader a. sub-graving 300 sq.m./hrb. spreading 40 cu.m./hr
3. pay loader a. loading 30 cu.m./hr
4. crane shovel a. loading 35 cuu.m./hr
5. sheep's foot a. static rolling
12 passes - 15 cm. lift
roller b. vibrator rolling 4 passes - 15 cm. lift 135 cm./hr.
6. no. 3w road roller a. static rolling 24 cmph
6 passes - 20 cm. lift
7. tractor-drawn roller (1-d) a. vibrator rolling
6 passes-20 cm. lift 240 cmph
8. tandem roller a. static rolling 6 passes-20cm. Lift 24 cmph
9. 5-T dump truck a. hauling common barrow 3.5 cmptb. hauling selected borrow base course 5 cmpt
NOTE: Cmph= cum/hour: cmpt= cum./truck
Table No. 7 Capabilities of Manual Labor per Hour
74
CAPABILITIES OF EQUIPMENT
Item Type of work Capability
1. dozer a. clearing 500 sq.m./hr.b. stripping 200 sq.m./hr.c. excavation 25 cu.m./hr.d. quarrying 50 cu.m./hr.e. pushing 3 sq.m./hr.
2. grader a. sub-graving 300 sq.m./hrb. spreading 40 cu.m./hr
3. pay loader a. loading 30 cu.m./hr
4. crane shovel a. loading 35 cuu.m./hr
5. sheep's foot a. static rolling12 passes - 15 cm. lift
roller b. vibrator rolling 4 passes - 15 cm. lift 135 cm./hr.
6. no. 3w road roller a. static rolling 24 cmph 6 passes - 20 cm. lift
7. tractor-drawn roller (1-d) a. vibrator rolling
6 passes-20 cm. lift 240 cmph
8. tandem roller a. static rolling 6 passes-20cm. Lift 24 cmph
9. 5-T dump truck a. hauling common barrow 3.5 cmptb. hauling selected borrow base course 5 cmpt
NOTE: Cmph= cum/hour: cmpt= cum./truck
Table No. 9 Schedule of Work
Item Description Precedence DurationA Mobilization None 8B Site Clearing A 2C Construction of temporary fences A 7D Storage B,C 6E Layout and Staking out D 3F Excavation/Digging for Foundation/ Footings E 4G Gravel Bedding F 12H Soil poisoning I 2I Fabrication of Footing Rebars H 10J Installation of footing rebars I 6K Fabrication of Footing Tie Beam Rebars I,G 6L Concrete Pouring on Footing D 8M Installation of ground floor column form works K,G 4N Installation of ground floor column rebars M 4O Concrete pouring on Ground Floor Column N 5P Fabrication and installation of beam at second
Floor, Form works.O 8
Q Fabrication and installation of beam at second floor
O 2
R Concrete pouring on second floor beam Q 3S Fabrication and Installation of first floor
formworksR,T 3
T Fabrication and installation of first floor slab rebars
R,S 2
U Concrete pouring on first floor slab S,T 5V Fabrication and installation of second floor
column formworksT
W Fabrication and installation of second floor S 675
column rebarsX Concrete pouring on first floor column A28 10Y Fabrication and installation of formworks (beams
and girders)X,A25 4
Z Fabrication of installation of beams Y 15A1 Concrete pouring on second floor beams and
girders.T 5
A2 Fabrication and installation of second floor.(slab formworks)
Z 5
Table No. 9 Schedule of Work
Item Description Precedence Duration
A3 Fabrication and installation of second (slab rebars)
A1,Z 5
A4 Concrete pouring on second floor slab. A8,A20 15A5 Fabrication and installation of admin column
formworksA4,A31 10
A6 Fabrication and installation of admin column rebars
A4 7
A7 Concrete pouring on admin column A5,A6 5A8 Fabrication and installation of formworks last
deck (beams)A7 8
A9 Fabrication and installation of rebars last deck (beams)
A7 8
A10 Concrete Pouring of beams on last deck A9 8A11 Installation of windows A9,A8 9A12 Installations of glass wall A11,A10 12A13 Installation of doors A12 12A14 Fabrication and installation of slab on fill A13 15A15 Concrete pouring of slab on fill A12 10A16 Installation of CHB wall partition on ground floor A15 7A17 Installation of CHB wall partition A19 6A18 Installation of carpentry works on second floor A14 15A19 Plastering and finishing of CHB wall and
partitionA2 20
A20 Installation of doors and windows glazing A16 5A21 Plumbing at ground floor A19,A18 12A22 Plumbing at second floor A21 10A23 Tile setting at first floor A19 5A24 Tile setting at second floor A23 5A25 Installation of roof drain system 28 4A26 Column design and finishes works A5 5A27 Installation of fixtures A22 10A28 Finishing of floor thru red cement U 5A29 Painting works A23 10A30 Demobilization A29 10
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
CHAPTER X
CONSTRUCTION METHOD
Preliminary Work Plan
106
After the materials are determined and the materials are canvassed, laborers and other needed
skilled worker for the project will be hired to start the project implementation will proceed and the
materials and equipment will be placed in the proper area which will be provided for them.
Preliminary Construction Plan
After the content is awarded and the notice to proceed (NTP) is received, meeting shall be
called up with the management team, the project engineer, the contractor, and the foremen who will
discuss the construction procedure, schedules, and other matters or factor affecting the
implementation of the project.
Construction Method
During the construction, the assigned engineer will supervised and give necessary instructions
to workers. Cleaning will be done first. Batter boards and construction of temporarily facilities are
done. Excavations for foundations will follow. Fabrications of steel bars will be done almost
simultaneously with excavation, layout of steel bars and formworks for column follows after concrete
pouring. After formwork is done in three days, laying of CHB and installation of doors and windows
is done after pouring of beams. Simultaneously with concrete woks for bleachers and stairs, steel
trusses will be fabricated. Construction of septic tank and planning installation will follow. Steel
trusses are installed after poured concrete to of roof beams is sufficiently cured and hardened.
Roofing work is done after installation of steel trusses and pulling. Tile work is done after concrete
pouring for slab on grade is entered. Painting work is done after plastering all walls, beams, columns,
and slabs. Electrical rough-in is immediately preceded. Cleaning out of site will be done before the
project is turned over to the owner.
Construction Work Plan
Division 1. Site Work
107
A. Work Included
1. Site Clearing including removal of bushes, stumps and grading of area
2. Installation of temporary fence made of barbed wire and sawali on wooden
post.
3. Staking out of building, establishment of lines, grades and benchmarks.
4. All excavation work including all necessary shoring bracing, and drainage of
storm water from site.
5. All backfilling, filling and grading, removal of excess material from site.
6. Protection of property work and structures, workmen, and other people from
damage and injury.
7. Temporary facilities and structures shall be constructed prior to execution of the
project to ensure the safety of materials and workers
A. Lines, Grades and Benchmarks
a. Stake out accurately the lines of the building and of the other structures
included in the contract, and establish grades therefore, after which secure
approval by engineer before any excavation work is commenced.
b. Erect basic batter boards and basic reference marks, at such places where
they will not be disturbed during the construction of the foundation.
B. Excavation
a. Structural Excavations – Excavation shall be to the depths indicated bearing
values. Excavation for footings and foundations carried below required depth shall
be filled with concrete, and bottom of such shall be level. All structural
excavations shall extend a sufficient distance from the walls and footings to allow
for proper erection and dismantling of forms, for installation of service and for
inspection. All excavations shall be inspected and approved before pouring any
concrete, laying underground services or placing select fill materials.
b. The Contractor shall control the grading in the vicinity of all excavated areas to
prevent surface drainage running into excavations. Water which accumulates in
108
excavated areas shall be removed by pumping before fill or concrete in placed
therein.
D. Fillings and Backfilling
1. After forms have been removed from footings, piers, foundations, and walls,
etc. and when concrete work is hard enough to resist pressure resulting from
fill,
Backfilling may then be done. Materials excavated may be used for
backfilling. All filling shall be placed in layers not exceeding six (6) inches in
thickness, each layer being thoroughly compacted and rammed by wetting,
tamping, rolling.
E. Placing and Compacting Fill
1. Common Fill – shall be approved site – excavated materials free from roots,
stumps and other perishable or objectionable matter.
2. Select Fill – shall be placed where indicated and shall consist of crushed
gravel crushed rock, or combinations thereof. The material shall be free from
adobe, vegetable matters and shall be thoroughly tamped after placing.
3. Before placing fill materials, the surface upon which it will be placed shall be
cleared of all brush roots, and debris, scattered and thoroughly wetted to
insure good bonding between the grounds.
F. Disposal of Surplus Materials
1. Any excess material remaining after completion of the earthwork shall be
disposed of by hauling and spreading in nearby spoil areas designated by
the owner. Excavated material deposited in spoil areas shall be graded to a
uniform surface.
Division 2. Concrete
II. Concrete and Reinforced Concrete
General
109
1. Unless otherwise specified herein, concrete work shall conform to the
requirements of the ACI Building Code. Full cooperation shall be given other
trades to install embedded items. Provisions shall be made for setting items
not placed in the forms. Before concrete is placed, embedded items shall
have been inspected and tested for concrete aggregates and other materials
shall have been done.
2. Cement for the concrete specified herein, concrete work shall conform to the
requirements of specification for Portland cement (ASTM C-150).
3. Water used in mixing concrete shall be clean and free from other injurious
amounts of oils, acids, alkaline, organic materials or other substances that
may be deleterious to concrete or steel.
4. Fine aggregates shall consist of hard, tough, durable, uncoated particles. The
shape of the particles shall be generally rounded or cubicle and reasonably
free from flat or elongated particles. The stipulated percentages of fines in
the sand shall be obtained either by the processing of natural sand or by the
production of a suitably graded manufactured sand.
5. Coarse aggregates shall consist of crushed gravel or rock, or a Combination
of gravel and rock, coarse aggregates shall consist of hard, touch, durable,
clean and uncoated particles. The sizes of coarse aggregates to be used in the
various parts of the work shall be in accordance with the following:
Size – ¾ “for all concreting work
6. Reinforcing bars shall conform to the requirements of ASTM standard
specification for minimum requirements for the deformed steel bars for
concrete reinforcement (A 305-56).
The main reinforcing bars shall be follows:
No. 4 (½” Q) – 12 mm
No. 5 (5/8” Q) – 16mm
Proportion and Mixing 110
1. Proportion of all materials entering into the concrete shall be as follow:
Cement : Sand : Gravel
Class “A” - 1 2 4
Class “B” - 1 2 ½ 5
2. Class of concrete – concrete shall have a 28 – day cylinder strength of 3,000
psi for all concrete work, unless otherwise indicated in the plans.
3. Mixing – concrete shall be machine mixed, mixing shall begin within 30
minutes after the cement has bear, added to the aggregates.
Forms
4. Generals – Forms shall be used wherever necessary to confine the concrete
and shape it to the required lines, or to insure the concrete of contamination
with materials caving from adjacent excavated surfaces. Forms shall have
sufficient strength to withstand the pressure resulting from placement and
vibration of the concrete, and shall be maintained rigidly in correct position.
Forms shall be sufficiently tight to prevent loss of mortar from the concrete.
Forms for exposed surfaces against which backfill is not be placed shall be
lines with a form grade plywood.
5. Cleaning and oiling of forms – before placing the concrete, the contact
surfaces of the form shall be cleaned of encrustations of mortar, the grout or
other foreign material, and shall be coated with commercial form oil that will
effectively prevent sticking and will not stain the concrete surfaces.
6. Removal of forms – forms shall be removed in a manner, which will prevent
damage to be concrete. Forms shall not be removed without approval. Any
repairs of surface imperfections shall be performed at once and airing shall
be started as soon as the surface is sufficiently hard to permit it without
further damage.
B. Placing Reinforcement
1. General – Steel reinforcement shall be provided as indicated, together with
all necessary wire ties, chains, spacers, supported and other devices
111
necessary to install and secure the reinforcement properly. All reinforcement,
when placed, shall be free from loose, flaky rust and scale, oil grease, clay
and other coating and foreign substances that would reduce or destroy its
bond with concrete reinforcement shall be placed accurately and secured in
place by use of metal or concrete supports, spacers and ties. Such supports
shall be of sufficient strength to maintain the operation. The supports shall be
used in such manner that they will not be exposed or contribute in any way,
to the discoloration or deterioration of the concrete.
C. Conveying and Placing Concrete
1. Conveying – concrete shall be conveyed from mixer to forms as rapidly as
practicable, by methods, which will prevent segregation, or loss of
ingredients. There will be no vertical drop greater than 1.5 meters except
where suitable equipment is provided to prevent segregation and where
specifically authorized.
2. Placing – Concrete shall be worked readily into the corners and angles of the
forms and around all reinforcement and embedded items without permitting
the material to segregate, concrete shall be deposited as close as possible to
its final position in the forms so that flow within the mass does not exceed
two ( 2 ) meters and consequent segregation is reduced to a minimum near
forms or embedded items, or else where as directed, the discharge shall be so
controlled that the concrete may be effectively compacted into horizontal
layers not exceeding 30 centimeters in depth within the maximum lateral
movement specified.
3. Time interval between mixing and placing, concrete shall be placed before
Initial set has occurred and before it has occurred and before it has contained
its water content for more than 45 minutes.
4. Consolidation of concrete – Concrete shall be consolidated with the aid of
mechanical vibrating equipment and supplemented by hand spading and
tamping. Vibrators shall not be inserted into lower coursed that have
112
commenced initial set; and reinforcement embedded in concepts beginning to
set or already set shall not be disturbed by vibrators shall not be used.
5. Placing concrete through reinforcement. In placing concrete through
reinforcement, care shall be taken that no segregation of the coarse aggregate
occurs. On the bottom of beams and slabs, where the congestion of steel near
the forms makes placing difficult, a layer of mortar of the same cement –
sand ratios as used in concrete shall be first deposited to cover the surfaces.
D. Curing
1. General: All concrete shall be moist cured for a period not less than
seven (7) consecutive days by an approved method or combination
applicable to local conditions.
2. Moist Curing – The surface of the concrete shall be kept continuously wet
by covering with burlap plastic or other approved materials thoroughly
saturated with water and keeping the covering wet spraying or intermittent
hosing.
E. Finishing
1. Concrete surfaces shall not be plastered unless otherwise indicated. Exposed
concrete surfaces shall be formed with plywood, and after removal of forms,
the surfaces shall be smooth, true to line and shall present or finished
appearance except for minor defects which can be easily be repaired with
patching with cement mortar, or can be grounded to a smooth surface to
remove all joint marks of the form work.
2. Concrete slabs on fill – The concrete slabs on fill shall be laid on a prepared
foundation consisting of subgrade and granular fill with thickness equal to
the thickness of overlaying slab except as indicated otherwise.
Division III. Structural Steel
Execution
Welding Techniques
113
a. Conform the technique of welding employed, the appearance and
quality of weld made, the methods used in correcting defective work to
the requirements of the standard Code for Welding in building
Construction of the American Welding Society.
a. Make surfaces to be welded free from loose scale, slag, rust grease,
paint and any other foregoing material except that mill scale which
withstands vigorous wire brushing may remain.
b. Prepare edges by cutting, whenever practicable, cut by a mechanically
guided torch.
c. Let gas cut edges which will be subjected to substantial stress or which
are to have weld metal deposited on them be free from gouges. Remove
by grinding any gouges that remain from cutting.
d. In assembling and joining parts of structure or of built-up members,
avoid needless distortion and minimize shrinkage stresses. Where it is
impossible to avoid high residual stresses in the closing welds of a rigid
assembly, make closing welds in compression elements.
Bolts
a. Tighten all bolts to a bolt tension not less than the proof load given in
the applicable ASTM specifications for the type of bolt used.
b. Do tightening with properly calibrated wrenches or by the turn-of-nut
method.
Shop Painting
a. Clean all steelwork specified for painting by hard-ware brushing, or
other methods chosen by the fabricator for cleaning loose mill scale,
loose rust, weld slag, or flux deposit, dirt and other foreign matter after
inspection, approval, and before leaving the shop.
b. Give all steelwork except those to be encased in concrete one coat of
shop paint.
c. Remove oil and grease deposits by solvent.114
d. Do not paint steelwork that is to be encased in concrete. Clean oil or
grease or with solvent cleaners. Remove dirt and foreign materials by
thorough sweeping with fiber brush.
Masonry Works
1. Concrete hollow Blocks shall have a minimum face shell thickness of 1”
(.025). Nominal size shall be 6”x 8”x16” minimum compressive strength
shall be as follows:
All units shall be stored for a period of not less than 28 days (including
curing period) and shall not be delivered to the job site prior to that time
unless the strengths equal or exceed those mentioned in this specification.
2. Wall reinforcement shall be no. 3 (3/8”) or 10 mm steel bars.
3. Sand shall be river sand, well screened, clean, hard, sharp siliceous, free
from loam, silt or other impurities, composed of grains of varying sizes
within the following limits:
4. Cement shall be standard Portland cement, ASTM C – 150 – 68 Type 1.
5. Mortar – Mix mortar from 3 to 5 minutes in such qualities as needed for
immediate use. Re tampering will not be permitted if mortar stiffens because of
pre mature setting. Discard such materials as well as those, which have not been
used within one hour after mixing.
Proportioning – Mix mortar shall be one (1) part Portland cement and two (2)
parts sand by volume but not more than one (1) part Portland cement and three
(3) Parts of sand by volume.
Erection
1. All masonry shall be laid plumb, true to line, with level and accurately spaced
Courses, and with each course breaking joint with the source below. Bond shall
be kept plumb throughout; corners and reveals shall be plumb and true. Units
with greater than 12 percent absorption shall be wet before lying. Work required
115
to be built in with masonry, including anchors, wall plugs and accessories, shall
be built in as the erection progresses.
2. Masonry Units each course shall be solidly bedded n Portland cement mortar. All
units shall be damp when laid units shall be showed into place not laid, in a full
bed of unfurrowed mortar. All horizontal and vertical points shall be completely
filled with mortar when and as laid. Each course shall be bonded at corners and
inter- sections. No cells shall be left open in the face surfaces. All cells shall be
filled up with mortar for exterior walls. Units terminating against beam or slab
soffits shall be wedged tight with mortar. Do not lay cracked, broken or defaced
block.
3. Lintels shall be of concrete and shall be re-enforced as shown in the drawings.
Lintels shall have minimum depth of .20 ( 8”) and shall extend at least .20 (8”) on
each side of opening.
Workmanship and Installation
4. Plastering: Clean and evenly wet surfaces. Apply scratch coat with sufficient
force to form good keys. Cross scratch coat upon attaining its initials set; keep
damp. Apply brown coat after scratch coat has seat at least 24 hours after scratch
coat application. Lightly scratch brown coat; keep moist for 2 days ; allow to dry
out. Do not apply finish until brown coat has seasoned for 7 days. Just before
applying coat, wet brown at again. Float finish coat to true even surface; travel in
manner that will force sand particles down into plaster; with final troweling,
leave surfaces barnished smooth, free from rough areas, trowel marks, cheeks,
other blemishes. Keep finish coat moist for at least 2 days; thereafter protect
against rapid drying until properly, thoroughly cured.
5. Pea Gravel Washout: Before start of work, provide desired pitch for drainage.
Roughen concrete surface with pick or similar tool. Clean off loose particles and
other materials, which may prevent bond, keep surface width for at least 4 hours
before applying. Scratch coat or mortar. Coat not more than ¼” thick. Apply
mixture of pea gravel and Portland cement with pressure to obtain solid adhesion.
116
Trowel pea gravel to hard, smooth, even plane and rod and float to uniform
surface of even texture. When surface is semi-dry evenly spray surfaces with
clean water with spray machine to wash out loose cement to part exposed pea
gravel. Remove and wash down remaining cement paste with soft brush, to leave
pea gravel in its natural’s texture and appearance. Before applying pea gravel
finish, submits samples to owner for approval.
Scaffoldings
Provide all scaffolding required for masonry work, including cleaning down on
completion, remove.
Vitrified Floor Installation
6. Do not start floor tilling occurring in space requiring both floor and wall tile
setting has been completed.
7. Before spreading setting bed, establish border lines center wires in both
directions to permit laying pattern with minimum of cut tiles. Lay floors without
borders from centerline outward. Make adjustment at walls.
8. Clean concrete sub floor and moisten it without soaking. Sprinkle dry cement
over surface. Spread setting bed mortar on concrete and tamp to assure good
bond over the entire area then screed to smooth, level bed. Set average setting
bed thickness at 3/4 “but never less than ½”.
Wall Tile Installation
9. Scratch coat for application as foundation coat shall be at most ½”. While still
plastic, deeply score scratch coat or scratch and cross scratch. Protect scratch coat
and keep reasonably moist within seasoning period. Use mortar for scratch; float
coats, within one hour after mixing. Retampering of partially hardened mortar is
not permitted. Set scratch coat shall be cured for at least 2 days before starting
tile setting.
10. For float coat use one part Portland cement, one part hydrated lime (optional), 3
½ part sand.
117
11. Setting Wall Tiles – Soak wall tile thoroughly in clean water before setting. Set
wall tile by trowelling neat Portland cement skim coat on float coat or apply skim
coat to back of each tile unit. Immediately float tile in place. Make joints straight,
level and perpendicular. Maintain vertical joints plumb.
12. Grouting-Grout joints in wall tile with neat white cement immediately after
suitable area of tile has been set. Tool joints slightly concave, cut off excess
mortar and wipe from face tile. Roughen interstices of depressions. In mortar
joints after grout has been cleaned from surface. Fill to line of cushion tile bases
or covers with mortar. Make joints between wall tile, plumbing and other built in
fixtures with light colored caulking. Immediately after grout has had its initial set,
give tile wall surfaces protective coat of non-corrosive soap.
Division V. Carpentry and Millworks
Treatment of the lumber
a. All concealed lumber shall be sprayed with anti-anay or bukbok liquid.
b. Surface in contact with masonry and concrete coated with creosote or
equivalent.
2. Door Sashes: All door sashes shall be well seasoned, flush type, semi-hollow core
and solid core, tanguile plywood veneers on both sides. Exterior doors shall be of
kiln dried tanguile panel doors.
3. Kind of Lumber:
All unexposed lumber for cabinet and framings shall be tanguile or the same equal
in strength.
All door jambs shall be hard lumber ipil or the same equal in strength.
Workmanship
1. Execute rough carpentry in best, substantial, workmen like manner. Erect
Framing true to line, levels and dimensions, squared, aligned, plumbed, well spliced
and nailed, and adequately braced, properly fitted using mortise and tendon joists.
118
2. Millwork – accurately milled to details, clean cut moldings profiles, lines, scrape,
sand smooth: mortise, tendon, splice, join, block, nail screw, bolt together, as
approved, in manner to allow free play of panels; avoid swelling, shrinkage, ensure
work remaining in place without warping, splitting opening or joist. Do not install
mill work and case until concrete and masonry work have been cured and will not
release moisture harmful to woodwork.
3. Secure work to grounds; otherwise fasten in position to hold correct surfaces, lines
and levels. Make finished work flat, plumb, true.
Division V. Architectural Finishes Schedule
Flooring
1. Toilet floors and laboratory concrete nook shall be vitrified 8” x 8” white or
beige in color, mariwasa brand.
2. Concrete floor shall be plain red cement finished with 16” x 16” nail strip.
3. Corridors shall be black pebble washout finish.
Walling
1. CHB walling shall be plastered and lined with ¼” nail strip.
2. Toilet wall finish shall be of 8” x 8” white glazed tiles.
Ceilings
1. All interior and exterior ceilings shall be of 3/16” x 4’ x8’ cement boards with
¼”x1”x2” galvanized metal furring ceiling joist spaced at 60 cm OC bothways.
2. Outside ceiling eaves shall be with air vents covered with ¼” square screen as
shown in the plan.
Doors
1. Al entrance and exit doors for classroom and office shall be solid panel door
complete with heavy duty lockset and hinges.
2. All toilet doors shall be uPVC plastic doors.
119
Bring float coat flush with screeds or temporary guide strips placed to give true
and even surface at proper distance from the tile finished face.
3. All exterior doors shall be panel doors.
Windows
1. All windows shall be smoke glass jalousie type with rectangular 2”x4”
aluminum jambs secured with 12 mm square steel grills spaced at 6”x6”, framed
with 4.5 mmx25mm x25mm angular bar.
Division VII. Roofing and Tinsmithing Works
1. Roof Sheathing – shall be longspan pre-painted rib type galvamast gauge 26 roofing.
Color shall be upon approval of the owner.
Flashings shall be of gauge 26 plain G.I sheets.
Installation Workmanship
1. Sheathing – layout the roofing sheets in a manner that the side overlap faces away from
the Prevailing wind. Provide not less than 0.30m develop on ends and not less than 1
1/2 “ corrugation on side laps on both sides. Secure the roofing sheets to purlins by
using self driven Tekscrew.
2. Flashing – shall be plain G.I sheet over rib-type roofing of not less than 0.30 overlap
Extend G.I flashing until it covers the top portion of the firewall.
Division VII. Plumbing and Sanitary Works
Installation
1. Install plumbing fixtures free and open to afford easy access for cleaning.
2. Install plumbing fixtures as indicated on drawings, furnishing all brackets, cleats,
Plates and anchors required to support fixtures rigidly in place.
3. Install all fixtures and accessories in locations directed in accordance with
manufacturer’s instructions, minimizing pipe fittings.
120
4. Protect items with approval means to maintain perfect conditions. Remove work
damaged or defective and replace with perfect work without extra cost to owner.
5. All sewer and drainage pipes shall have a minimum slope 1%.
6. Vertical pipes shall be secured strongly by hooks to building framing. Provide suitable
bracket or chairs at the floors from which they start.
Where an end or circuit vent pipe from any fixtures or line of fixtures is connected to a
vent line serving other fixtures, connection shall be at least four (4) feet 1.20m above
floor in which fixtures are located, to prevent use of any vent line as a waste.
Horizontal pipes shall be supported by well-secured straphangers.
7. Connection of water closets to soil pipe shall be made by means of flanged plates and
asbestos packing without use of rubber putty of cement.
8. Make all joints air and water - tight; for jointing pipes, the following shall be used.
a. For PVC sewer pipes, caulk with PVC sealant.
b. Concrete pipes: bell and spigot or tongue and groove use yarning materials
and
Cement mortar.
c. UPVC Pipes-Use Teflon Tape or white lead when tightening threaded joints.
Rough – In
1. Provide correctly located opening of proper sizes where required in walls and floors for
passed of pipes.
2. All times to be embedded in concrete shall be thoroughly clean and free from all rust,
scale and paint.
3. All changes in pipes sizes on soil, wash and drain lines shall be provided with reducing
fittings or recesses reducers. For changes in pipes sizes provide reducing fittings.
4. High corrosive nature ground within site shall be taken into account by plumber.
Protective features shall be installed to prevent corrosion or all water pipes installed
underground.
121
5. Extend piping to all fixtures, outlets an equipment from gate valves installed in the
branch near the riser.
6. All pipes shall be cut accurately to measurements, and worked into place without
springing or forcing.
7. Care shall be taken as not to weaker structural portion of the building.
Division IX. Painting Works
Scope of Work
1. Consists of furnishing all items, articles, materials tools, equipment, labor
scaffoldings, ladders, methods and other incidentals necessary and required for the
satisfactory completion of the work.
2. It covers complete painting and finishing of wood, plasters, concrete, metal or other
surfaces exterior or interior of building.
General Painting and surface finishing shall be interpreted to mean and include Sealers,
primers, fillers, intermediate and finish coats, emulsions, varnish, shellac, stain or
enamels.
1. All paint and accessory materials incorporated in or forming apart thereof shall be
subject to the prior approval and selection for color, tint, finish or shade by the
Architect.
2. In connection with the owner’s determination of color or tint of any particular surface
The depth of any color or tint selected or required shall in no instance be a subject for
an additional cost of the owner.
3. Painting of all surfaces, except as otherwise specified shall be three (3) coat work, one
primer and a finish cost.
1. All materials shall meet the requirements of paint materials under classification
Class “A” as prepared by the institute of Science Manila, use “BOYSEN”
122
2. All paint shall be recommended by the manufacturer for the use intended and shall be
delivered to the jobsite in original containers with seals unbroken and labels intact.
3. Painting materials such as Linseed oil, turpentine, thinner, shellac, lacquer, etc. shall
be pure and of the highest quality obtainable and shall bear the manufacturer’s label
on each container or package.
4. Except for ready mixed materials in original containers, all mixing shall be done in
the jobsite. No materials are to be reduced, changed or mixed except as specified by
manufacturer of said materials.
5. Storage and Protection the resident engineer shall designate a place for the storage of
paint materials whenever it may be necessary to change this designated storage place,
the contractor shall promptly move to the new location. The storage space shall be
adequate protected from damage and paint. Paint shall be covered at all times and
safeguards taken to prevent fire.
6. All surfaces to be painted shall be examined carefully before beginning any work and
see that all work of other trades or subcontractor’s are installed in work manlike
condition to receive paint, stain or particular finish.
7. Before proceeding with any painting or finishing, thoroughly clean, sand and seal if
necessary by removing from all surfaces all dust, dirt, grease, or other foreign
substances which would affect either the satisfactory execution or permanency of the
work. Such cleaning of shall be done after the general cleaning executed under the
separate division of the work.
8. No work shall be done under conditions that are unsuitable for the production of good
results, nor at any time when the plastering is in progress or is being cured, or not dry.
9. Only skilled painters shall be employed in the work. All workmanship shall be
executed in accordance with the best acceptable practices.
10. Finish hardware, lighting fixtures, plates and other similar items shall be removed by
workmen skilled in these trades, or otherwise protected during panting operation and
reposition upon completion of each space.
123
11. Neither paint nor any other finish treatment shall be applied over wet or damp
surfaces. Allow at least two (2) days for drying preceding coat before applying
succeeding coat.
12. Begin work only when resident Engineer has inspected and approved prepared
surface. Otherwise no credit for coat applied shall be given. The contractor shall
assume responsibility to recoat work in question. Notify engineer when particular coat
applied is complete, ready for inspection and approval.
Preparation of Surfaces
1. For bricks, concrete, cement or concrete blocks; cut out scratches, cracks abrasion
in plaster surfaces, openings and adjoining trim as required. Fill flush adjoining
plaster surface. When dry; and smooth and seal before priming coat application.
2. Tint plaster priming coat to approximate shade of final coat. Touch up section spots
in plaster or cement after first coat application, before applying second coat, to
produce even result in finish coat. Secure color schedules for rooms before priming
walls.
3. In cases of presence of high alkali conditions, neutralize surfaces by washing with
zinc Sulphate solution (3 pounds to a gallon of water). Allow to dry thoroughly,
brush free of crystals before priming.
4. Tiny undercoats of paint and enamel to same or approximate coat shade.
5. Sand smoothly woodwork to be finished with enamel or varnish; clean surface
before proceeding with first coat application. Use fine sand paper between coats on
enamel or varnish finish applied to wood to produce even smooth finish.
Varnishing
1. Sand wood surfaces with fine grade sand paper.
2. Do necessary puttying of nail holes, cracks etc. after first coat with putty of color to
match that of finish. Bring putty with adjoining surface in neat, workmanlike
manner.
124
3. Wipe paste wood fillers, applied in open grain wood, when “set”, across wood grain.
Then with grain to secure clean surface.
4. Cover surfaces to be stained with uniform stain coat.
5. Tiny undercoats of paint and enamel to same or approximate coat shade.
6. Sand smoothly woodwork to be finished with enamel or varnish; clean surface
before proceeding with first coat application. Use fine sand paper between coats on
enamel or varnish finish applied to wood to produce even smooth finish.
Wipe dust off with clean cloth dampened with lacquer thinner.
7. Apply wood filler as per manufacturer’s specifications.
8. Apply approved stain in uniform coats until desired shade is achieved.
9. Apply finish coat as per manufacturer’s specifications.
Fire Code Requirements
All interior wooden structures shall be applied with Fire stopper Fire Retardant
solution applied as per manufacturer’s specifications. All other requirement as of the fire code
of the Philippines as far as they relate to this project shall likewise be complied with.
125
BIBLIOGRAPHY
Books
Gillesania, Di.T. 2003, Structural Engineering and Construction. Third Edition. (Used as reference
material in the design computation). Pp. 150-250
Fajardo, M.B. Jr. 2000, Engineering Construction and Management Second Edition. (Used as
reference material in program of works). Pp.67-109
Fajardo, M.B. Jr. 2000, Engineering Project Estimates Second Edition. (Used as reference material
in estimates computation). Pp. All
Besavilla, V.I.Jr., 2007. Fundamentals of Reinforced Concrete.VIB Publisher, #2 Saint John Street,
Don Bosco Village, Punta Princesa Cebu. Pp. 161-245, 412-423
Unpublished books:
Tabinga, DC., 2008. Proposed Design and Construction of a 3- storey Dummy Bridge at the Western
Philippines University Puerto Princesa City. Pp. All
126