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STRUCTURAL COMPUTATION
OF
PROPOSED TWO-STOREY RESIDENTIAL HOUSE
PHASE 7-A BLOCK 17 LOT 04 ORCHARD & RESIDENTIAL ESTATE, SALITRAN, DASMARIÑAS, CAVITE
Prepared by:
GIL S. BELTRAN Civil Engineer PRC Reg. No. PTR No. Date Issued Place Issued TIN
: 64544 : 8642641 : 1/9/2009 : Dasmariñas, Cavite : 157-467-966
PROPOSED TWO-STOREY RESIDENTIAL HOUSE PHASE 7-A BLOCK 17 LOT 04 ORCHARD & RESIDENTIAL ESTATE, SALITRAN, DASMARIÑAS, CAVITE
Page 1 of 16
Owner: MS. LYDA B. FIX DESIGN OF SLABS A. DESIGN OF TWO-WAY SLAB (S-1) Direct Design Method Design Specifications L1 = 2.30 m column size = 0.20m x 0.40m L2 = 3.35 m beam size = 0.20m x 0.35m Fy = 275 Mpa f'c = 20.7 Mpa Considering the interior slab 1 Relative a values a1 x L2/L1 = 1.46 a2 x L1/L2 = 0.86 2 Slab thickness, t t =
am = 2x(a1 + a2)/4
1
a2 = cc =
1.25 25
=
Ln (800 + 0.73Fy) 36,000 + 5000b [am - 0.12(1 + 1/b)]
where: Ln = L2 - column width Ln = L1 - column width b = (Ln long/Ln short) direction clear span ratio
a1 =
1.16
=
78.05 mm
=
3.35 m
=
2.30 m 1.46
=
It should not be less than tmin = Ln (800 + 0.73Fy) = 36000 + 9000b
68.27 mm
ok
It should not be more than tmax = Ln (800 + 0.73Fy) = 36000
93.13 mm
ok
Try t =
100 mm for all two way slabs
3 Design moments Dead Load = W slab + W ceiling+finish = Live Load
=
4.8 Kpa
Wu = 1.4DL + 1.7LL (consider 1-m strip) = Long Span: 2 Mu = Wu L1 Ln2 /8
=
41.48 KN m;
L1 =
3.35 Kpa
2.3
L2 = f =
3.35 0.9
=
46.09
12.86 Kpa
Mu/f
Mn =
Distribution of Mn between section of positive and negative moment at midspan: +Mms = 0.35 Mn = 16.13 KN m at support: -Ms = 0.65 Mn = -29.96 KN m Percentage of moment at support Column strip: a2 = L1/L2 = 1.25 By interpolation: L1/L2 a2L1/L2 0.5 90 x 0.69 ? 1 75
a2L1/L2 =
0.69
x %M
= =
Code assigns 85% of the column strip moment to the beam Column strip negative Mn = -25.28 KN m Beam negative Mn = -21.49 KN m Slab negative Mn = -3.79 KN m Column strip positive Mn = 13.61 KN m Beam positive Mn = 11.57 KN m Slab positive Mn = 2.04 KN m
Page 2 of 16
0.86
5.60 % 84.40 %
=
-1.90 KN m /half strip
=
1.02 KN m /half strip
Middle strip (takes the remaining moment) Middle strip negative Mn = -4.67 KN m Middle strip positive Mn = 2.52 KN m Short Span: 2 Mu = Wu L2 Ln1 /8
=
28.48 KN m;
Mu/f
Mn =
=
31.64
Distribution of Mn between section of positive and negative moment at midspan: +Mms = 0.35 Mn = 11.07 KN m at support: -Ms = 0.65 Mn = -20.57 KN m Percentage of moment at support Column strip: a1 = 1 L2/L1 = 1.46
a1L2/L1 =
1.46
By interpolation: L2/L1 a1L2/L1 1 75 x x = 1.46 ? %M = 2 45 Code assigns 85% of the column strip moment to the beam Column strip negative Mn = -12.61 KN m Beam negative Mn = -10.72 KN m Slab negative Mn = -1.89 KN m
=
-0.95 KN m /half strip
Column strip positive Beam positive Slab positive
=
0.51 KN m /half strip
Mn Mn Mn
= = =
6.79 KN m 5.77 KN m 1.02 KN m
13.70 % 61.30 %
Middle strip (takes the remaining moment) Middle strip negative Mn = -7.96 KN m Middle strip positive Mn = 4.29 KN m Fy =
275 Mpa
Slab thickness, t (mm) Steel reinforcement f (mm) Middle Strips Strip Width, b (mm) Mn (KN m) Mn (KN m) d (mm) As = [Mn(106)]/[Fy(0.95d)] (mm 2) Min As = 0.002bt (mm2) Spacing, S = Af b /As (mm) Max S = 2t (mm) Column Strips Strip Width, b (mm) Mn (KN m) Mn (KN m) d (mm) As = [Mn(106)]/[Fy(0.95d)] (mm 2) Min As = 0.002bt (mm2) Spacing, S = Af b /As (mm) Max S = 2t (mm) Adopted Spacing, S (mm) Middle Strips (10mm bars) Column Strips (10mm bars) DESIGN OF BEAMS A. FLOOR BEAM, FB1 1 Design loads 2 TA = 10.12 m Beam size: b = 0.20 m
LONG SPAN -M +M 100 10
SHORT SPAN -M +M 100 10
1000 46.09 -4.67 2.52 75 75 238.46 128.40 200 200 329.36 392.70 200 200
1000 31.64 -7.96 4.29 75 50 406.18 328.07 200 200 193.36 239.40 200 200
1000 46.09 -1.90 1.02 75 75 96.78 52.11 200 200 392.70 392.70 200 200
1000 31.64 -0.95 0.51 75 50 48.26 38.98 200 200 392.70 392.70 200 200
200 200
Slab Thickness = Beam span L =
Page 3 of 16
200 200
100 mm 3.35 m
h= d= Dead loads:
Live loads:
0.35 0.34 W slab = W c+f = W beam = W DL =
m m 23.82 7.286 5.52 36.63
LL =
4800 Pa
KN KN KN
=
10.93 KN/m
W LL = 48.58 KN Wu = 1.4DL + 1.7LL
= =
14.50 KN/m 39.96 KN/m
2 Design Moments Assuming fully restrained beam WuL2/24 Max. +Moment = -WuL2/12 Max. -Moment = Max. Shear = WuL/2 3 Reinforcement Design At Support (Negative Moment) Mu = -37.37 KN m Mn = Mu/0.90 = 41.52 KN m r rbal
rmax = 0.75rbal [0.85f'c b/Fy][600+(600+Fy)] =
18.68 KN m -37.37 KN m 66.93 KN
f'c = Fy = AǾ =
20.7 Mpa 275 Mpa 2 201.06 mm
=
rmax = use As1 = a
= = =
=
M1 =
0.75rbal r =
0.3rbal
rbd
=
0.037
=
0.028
=
0.011
As1Fy/(0.85f'cb)
=
As1Fy(d-a/2)
=
No. of bars required (use 16 mm) n = As/AǾ = 3.8 say Reinforcement: 4 pcs - 16mmǾ (Top bar) 2 pcs - 16mmǾ (Bottom bar) At Midspan (Positive Moment) Mu = 18.68 KN/m Mn = Mu/0.90 = 20.76 KN/m r rbal
59.01 mm 63.96 KN m > 37.37 Treat as singly reinforced beam 4
f'c = Fy = AǾ =
rmax = 0.75rbal [0.85f'c b/Fy][600+(600+Fy)] =
pcs
20.7 Mpa 275 Mpa 2 201.06 mm
=
rmax = use As1 = a = M1 =
0.75rbal r =
0.15rbal
rbd As1Fy/(0.85f'cb) As1Fy(d-a/2)
=
0.037
=
0.028
= = = =
No. of bars required (use 16 mm) n = As/AǾ = 1.9 say Reinforcement: 2 pcs - 16mmǾ (Top bar) 4 pcs - 16mmǾ (Bottom bar) 4 Stirrup Vu =
2 755.17 mm
=
66.93 KN Shear capacity of beam Vc = √(f'c)bd/6 fVc/2 Vn =
0.006 2
377.58 mm 29.51 mm 33.51 KN m > 18.68 Treat as singly reinforced beam 2
Vcr = Vu - Wu d = = =
pcs
53.44 KN
51.18 KN 21.75 KN Stirrup is needed 41.12 KN
<
Vcr/f - Vn Vs = = 2 Using 10mmǾ AǾ = 78.54 mm Spacing S = AǾFyd/Vs = 354.53 mm Max S = d/2 = 168.75 mm Use 10mmǾ stirrups spaced @: 3 @ 50mm; 2 @ 100mm; 2 rest @ 150mm
Page 4 of 16
B. FLOOR BEAM, FB2 1 Design loads TA = 5.06 Beam size: b = 0.20 h= 0.30 d= 0.29 Dead loads: W slab = W c+f = W beam = W DL = Live loads:
m2 m m m 11.91 3.643 4.73 20.29
Slab Thickness = Beam span L = LL = KN KN KN
=
6.06 KN/m
W LL = 24.29 KN Wu = 1.4DL + 1.7LL
= =
7.25 KN/m 20.80 KN/m
2 Design Moments Assuming fully restrained beam WuL2/24 Max. +Moment = -WuL2/12 Max. -Moment = Max. Shear = WuL/2 3 Reinforcement Design At Support (Negative Moment) Mu = -19.46 KN m Mn = Mu/0.90 = 21.62 KN m r rbal
rmax = 0.75rbal [0.85f'c b/Fy][600+(600+Fy)] =
= = =
9.73 KN m -19.46 KN m 34.84 KN
f'c = Fy = AǾ =
20.7 Mpa 275 Mpa 2 201.06 mm
=
rmax = use As1 = a
100 mm 3.35 m 4800 Pa
=
M1 =
0.75rbal r =
0.25rbal
rbd
=
0.037
=
0.028
=
0.009
As1Fy/(0.85f'cb)
=
As1Fy(d-a/2)
=
No. of bars required (use 16 mm) n = As/AǾ = 2.7 say Reinforcement: 3 pcs - 16mmǾ (Top bar) 2 pcs - 16mmǾ (Bottom bar) At Midspan (Positive Moment) Mu = 9.73 KN/m Mn = Mu/0.90 = 10.81 KN/m r rbal
41.89 mm 39.30 KN m > 19.46 Treat as singly reinforced beam 3
f'c = Fy = AǾ =
rmax = 0.75rbal [0.85f'c b/Fy][600+(600+Fy)] =
pcs
20.7 Mpa 275 Mpa 2 201.06 mm
=
rmax = use As1 = a = M1 =
0.75rbal r =
0.2rbal
rbd As1Fy/(0.85f'cb) As1Fy(d-a/2)
=
0.037
=
0.028
=
0.007
= = =
No. of bars required (use 16 mm) n = As/AǾ = 2.1 say Reinforcement: 2 pcs - 16mmǾ (Top bar) 3 pcs - 16mmǾ (Bottom bar) 4 Stirrup Vu =
2 536.08 mm
=
34.84 KN Shear capacity of beam Vc = √(f'c)bd/6 fVc/2 Vn =
2
428.86 mm 33.51 mm 31.93 KN m > 9.73 Treat as singly reinforced beam 3
Vcr = Vu - Wu d = = =
Page 5 of 16
pcs
28.86 KN
43.60 KN 18.53 KN Stirrup is needed
<
Vcr/f - Vn Vs = = 15.43 KN 2 Using 10mmǾ AǾ = 78.54 mm Spacing S = AǾFyd/Vs = 805.02 mm Max S = d/2 = 143.75 mm Use 10mmǾ stirrups spaced @: 3 @ 50mm; 2 @ 100mm; rest @ 150mm o.c. DESIGN OF COLUMNS A. EXTERIOR COLUMNS (C1) Trial Section b = 200 mm h = 400 mm Ǿbar = 12 mm n = 6 pcs
Design Specifications f'c = Fy = 1 Axial loads PDL = Pu = PuTotal =
20.7 MPa 275 MPa
Ag Ast Ǿtie
= = =
ǿ β
= =
PLL 18.31 kN (1.4DL + 1.7LL)x1.5 = Pu x no. of sides x no. of floors =
2 80000 mm 2 678.584 mm 10 mm
0.7 0.85
=
24.29 kN 100.39 kN (for one side only) 602.37 kN
2 Check capacity of column ǿ 0.80 [ 0.85f'c (Ag - Ast) + FyAst ] Puall = Puall = since:
886.07 kN Puall >
3 Check steel ratio ρact = Ast/Ag 0.85f'cβ (600) ρmax = Fy (600 + Fy) ρmin = 1.4/Fy 0.005 since: ρmin < 4 Lateral ties s =
Pu
0.008 ρact
16 (Ǿbar)
then:
The column section is safe!
=
0.008
=
0.037
=
0.005 0.037 ρmax then:
<
=
The steel ratio is safe!
192 mm
48 (Ǿtie) s = = 480 mm s = least dimension = 200 mm Thus: Use 6 - 12mm dia bar with 10mm dia lateral ties @ 200mm o.c. B. INTERIOR COLUMNS (C2) Trial Section b = 200 h = 300 Ǿbar = 12 n = 6
Design Specifications f'c = Fy = 1 Axial loads PDL = Pu = PuTotal =
mm mm mm pcs
20.7 MPa 275 MPa
Ag Ast Ǿtie
= = =
ǿ β
= =
PLL 18.31 kN (1.4DL + 1.7LL)x1.5 = Pu x no. of sides x no. of floors =
2 Check capacity of column ǿ 0.80 [ 0.85f'c (Ag - Ast) + FyAst ] Puall = Puall
=
689.01 kN
Page 6 of 16
2 60000 mm 2 678.584 mm 10 mm
0.7 0.85
= 100.39 kN 401.58 kN
24.29 kN (for one side only)
since:
Puall
>
3 Check steel ratio ρact = Ast/Ag 0.85f'cβ (600) ρmax = Fy (600 + Fy) ρmin = 1.4/Fy 0.005 since: ρmin < 4 Lateral ties s =
Pu
0.011 ρact
16 (Ǿbar)
then:
The column section is safe!
=
0.011
=
0.037
=
0.005 0.037 ρmax then:
<
=
The steel ratio is safe!
192 mm
48 (Ǿtie) s = = 480 mm s = least dimension = 200 mm Thus: Use 6 - 16mm dia bar with 10mm dia lateral ties @ 200mm o.c. DESIGN OF FOOTING A. INTERIOR FOOTING (F1) Design Specifications f'c = 20.7 Fy = 275 Ǿbar = 16 σconc = 23.5
MPa MPa mm kN/m3
H h qall
= = =
σsoil
=
600 mm 300 mm 190 kPa 3 15.7 kN/m
colb
=
200 mm
cc
=
100 mm
colh
=
400 mm
ǿ
=
0.85
1 Service loads PT =
PDL + PLL
=
2 Base dimensions qnet qall - [ hσconc + Hσsoil ] = Areqd B2
=
= Trial section: B =
=
173.53 kPa 2 1.47 m
B
=
1.21 m
Aact
=
2 2.25 m
PT / qnet Areqd 1.5 m
255.62 kN
=
3 Soil pressure due to factored loads PU = 1.4DL + 1.7LL qu PU / Aact =
= =
401.58 kN 178.48 kPa
4 Check punching shear d = h - cc colb + d b =
= =
200 mm 400 mm 600 mm 2 2000 mm
c bo
=
colh + d
=
=
=
Vu
=
[2b + 2c] PU - qu x b x c
=
358.74 kN
ǿVc
=
0.33ǿ bo d √f'c
=
510.48 kN SAFE
ǿVc
since:
>
5 Check beam shear x1 (B - colb - 2d)/2 = Vu
=
ǿVc = since:
=
=
0.45 m
qu B x1
=
120.47 kN
0.17ǿ B d √f'c ǿVc >
=
197.23 kN SAFE
6 Design of reinforcement x2 (B - colb)/2 = Mu
→
Vu
2
qu B x2 /2
→
Vu
= =
Page 7 of 16
0.65 m 56.55577 kN m
(assume) (assume)
Mu ω2 ω ρ
ǿf'c Bd2 ω (1-0.59ω) 1.69ω + 0.08576
= = =
ωf'c/Fy
ρmin = use ρ = As =
1.4/Fy
n
Ast/AǾ
Thus:
Use
0 0.052 0.004
= = =
ρ Bd
=
= = =
0.005 0.005 2 1527.27 mm
= 7.59 say 8 pcs 16mm dia bar on both sides of footing.
8
B. EXTERIOR FOOTING (F2) Design Specifications f'c = 20.7 Fy = 275 Ǿbar = 16 σconc = 23.5
MPa MPa mm kN/m3
H h qall
= = =
σsoil
=
500 mm 300 mm 190 kPa 3 15.7 kN/m
colb
=
200 mm
cc
=
100 mm
colh
=
300 mm
ǿ
=
0.85
1 Service loads PT =
PDL + PLL
=
2 Base dimensions qnet qall - [ hσconc + Hσsoil ] = Areqd B2
=
= Trial section: B =
Areqd 1.2 m
= =
B
=
0.99 m
Aact
=
2 1.44 m
3 Soil pressure due to factored loads PU = 1.4DL + 1.7LL qu PU / Aact =
= =
267.72 kN 185.92 kPa
4 Check punching shear d = h - cc colb + d b =
= =
200 mm 400 mm 500 mm 2 1800 mm
c bo
=
colh + d
=
= =
[2b + 2c] PU - qu x b x c
=
Vu
=
230.54 kN
ǿVc
=
0.33ǿ bo d √f'c
=
459.43 kN SAFE
ǿVc
since:
>
Vu
=
ǿVc = since:
→
Vu
5 Check beam shear x1 (B - colb - 2d)/2 =
=
0.3 m
qu B x1
=
66.93 kN
0.17ǿ B d √f'c ǿVc >
=
157.78 kN SAFE
Mu
=
qu B x22/2
Mu ω2 ω ρ
= = =
ǿf'c Bd2 ω (1-0.59ω) 1.69ω + 0.05286 ωf'c/Fy
→
Vu
6 Design of reinforcement x2 (B - colb)/2 =
(assume)
170.41 kN
175.10 kPa 2 0.97 m
PT / qnet
(assume)
=
0.5 m
=
27.88746 kN m =
= =
Page 8 of 16
0 0.032 0.002
ρmin = use ρ = As =
1.4/Fy
n
Ast/AǾ
Thus:
ρ Bd
= Use
7
= = =
0.005 0.005 2 1221.82 mm
= 6.08 say 7 pcs 16mm dia bar on both sides of footing.
Page 9 of 16
OUSE
DASMARIÑAS, CAVITE
N, DASMARIÑAS, CAVITE
Page 10 of 16
(short direction) (long direction) mm
m m
KN m
KN m /half strip
KN m /half strip
Page 11 of 16
KN m
KN m /half strip
KN m /half strip
Page 12 of 16
KN m
KN m
53.44
Page 13 of 16
KN m
KN m
28.86
Page 14 of 16
(for one side only)
The steel ratio is safe!
(for one side only)
Page 15 of 16
The steel ratio is safe!
Page 16 of 16