Structural Computation Analysis

STRUCTURAL COMPUTATION

OF

PROPOSED TWO-STOREY RESIDENTIAL HOUSE

PHA PH ASE 7-A 7-A BLOCK BLOCK 17 17 LOT 04 04 ORCHA ORCHARD RD & RES RESIDE IDENTI NTIA AL ESTA ESTATE, TE, SAL SALITR ITRA AN, DA DASMARI SMARI

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 PHAS PH ASE E 7-A BLOCK BLOCK 17 17 LOT 04 ORC ORCHA HARD RD & RES RESID IDEN ENTIA TIAL L ESTA ESTATE, TE, SAL SALITR ITRAN AN,, DASMA DASMARI RI

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 L2 =

column size =

3.35 m 275 Mpa

Fy =

0.20m x 0.40m

a1 =

1

beam size = 0.20m x 0.35m f'c = 20.7 Mpa

a2 = cc =

1.25 25

Considering the interior slab 1 Relative a values 1.46 a1 x L2 /L1 = 0.86 a2 x L1 /L2 = 2 Slab thickness, t = t

am = 2x(a1 + a2)/4

=

Ln (800 + 0.73Fy) 36,000 + 5000b [am - 0.12(1 + 1/b)]

1.16

78.05 mm

=

where: Ln = L2 - column width

=

3.35 m

Ln = L1 - column width

=

2.30 m 1.46

b = (Ln long/Ln short) direction clear span ratio

=

It should not be less than Ln (800 + 0.73Fy) = tmin = 36000 + 9000b

68.27 mm

ok

It should not be more than Ln (800 + 0.73Fy) = tmax = 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

3.35 Kpa

L1 =

2.3

4.8 Kpa

L2 = =

3.35 0.9

=

46.09

=

f Wu = 1.4DL + 1.7LL (consider 1-m strip) =

Long Span: 2 Mu = Wu L 1 Ln2 /8

=

41.48 KN m;

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 0.69 1

90 ? 75

x

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 /hal

=

1.02 KN m /hal

Middle strip (takes the remaining moment) -4.67 KN m Middle strip negative Mn = Middle strip positive Mn = 2.52 KN m

Short Span: 2 Mu = Wu L 2 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 =

a1L2/L1 =

1.46

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 /hal

Column strip positive Beam positive Slab positive

=

0.51 KN m /hal

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 4.29 KN m Middle strip positive Mn = Fy

LONG SPAN

275 Mpa

-M Slab thickness, t (mm) Steel reinforcement f (mm) Middle Strips Strip Width, b (mm) Mn (KN m) Mn (KN m) d (mm) 6 2 As = [Mn(10 )]/[Fy(0.95d)] (mm ) 2 Min As = 0.002bt (mm ) Spacing, S = Af b /As (mm) Max S = 2t (mm) Column Strips Strip Width, b (mm) Mn (KN m) Mn (KN m) d (mm) 6 2 As = [Mn(10 )]/[Fy(0.95d)] (mm ) 2 Min As = 0.002bt (mm ) Spacing, S = Af b /As (mm) Max S = 2t (mm) Adopted Spacing, S (mm) Middle Strips (10mm bars) Column Strips (10mm bars)

+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

200 200

DESIGN OF BEAMS A. FLOOR BEAM, FB1 1 Design loads TA = Beam size: b =

10.12 m

2

0.20 m

Slab Thickness = Beam span L =

Page 3 of 16

100 mm 3.35 m

h= d= Dead loads:

Live loads:

0.35 m

LL = 0.34 m W slab = 23.82 KN W c+f = 7.286 KN W beam = 5.52 W DL = 36.63 KN =

10.93 KN/m

W LL =

14.50 KN/m

48.58 KN

4800 Pa

=

Wu = 1.4DL + 1.7LL

=

39.96 KN/m

2 Design Moments Assuming fully restrained beam 2 WuL /24 Max. +Moment = 2 -WuL /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

0.75rbal r =

As1 = a

= = =

=

M1 =

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/A = 3.8 say

Reinforcement: 4 pcs - 16mm 2 pcs - 16mm

rbal

4

pcs

f'c = Fy = AǾ =


20.7 Mpa 275 Mpa 2 201.06 mm

=

rmax =

0.75rbal

use

r

As1 = a = M1 =

=

0.15rbal

rbd As1Fy/(0.85f'cb) As1Fy(d-a/2)

=

0.037

=

0.028

=

0.006

= = =


Reinforcement: 2 pcs - 16mm 4 pcs - 16mmǾ 4 Stirrup Vu =

59.01 mm 63.96 KN m > 37.37 Treat as singly reinforced beam

(Top bar) (Bottom bar)

At Midspan (Positive Moment) Mu = 18.68 KN/m Mn = Mu/0.90 = 20.76 KN/m r

755.17 mm

=

66.93 KN Shear capacity of beam Vc = (f'c)bd/6 fVc/2 Vn = Vcr/f - Vn Vs = Using 10mmǾ AǾ = Spacing S = A Fyd/Vs Max S = d/2

377.58 mm 29.51 mm 33.51 KN m > 18.68 Treat as singly reinforced beam

2

pcs

(Top bar) (Bottom bar) Vcr = Vu - Wu d = = = = 2 78.54 mm = =

53.44 KN

51.18 KN 21.75 KN Stirrup is needed 41.12 KN

<

354.53 mm 168.75 mm

Use 10mm stirrups spaced @: 3 @ 50mm; 2 @ 100mm; 2 rest @ 150mm

Page 4 of 16

B. FLOOR BEAM, FB2 1 Design loads TA = Beam size: b =

5.06 m

0.20 m 0.30 m

h= d= Dead loads:

Live loads:

2

Slab Thickness = Beam span L = LL =

0.29 m W slab = 11.91 KN W c+f = 3.643 KN W beam = 4.73 W DL = 20.29 KN

=

6.06 KN/m

W LL =

=

7.25 KN/m

24.29 KN

Wu = 1.4DL + 1.7LL

=

20.80 KN/m

2 Design Moments Assuming fully restrained beam 2 WuL /24 Max. +Moment = 2 -WuL /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


= = =

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

0.75rbal r =

As1 = a

100 mm 3.35 m 4800 Pa

=

M1 =

0.25rbal

rbd

=

0.037

=

0.028

=

0.009

As1Fy/(0.85f'cb)

=

As1Fy(d-a/2)

=


Reinforcement: 3 pcs - 16mm 2 pcs - 16mm

rbal

3

pcs

f'c = Fy = AǾ =


20.7 Mpa 275 Mpa 2 201.06 mm

=

rmax =

0.75rbal

use

r

As1 = a = M1 =

=

0.2rbal

rbd As1Fy/(0.85f'cb) As1Fy(d-a/2)

=

0.037

=

0.028

=

0.007

= = =


Reinforcement: 2 pcs - 16mmǾ 3 pcs - 16mm 4 Stirrup Vu =

41.89 mm 39.30 KN m > 19.46 Treat as singly reinforced beam

(Top bar) (Bottom bar)

At Midspan (Positive Moment) Mu = 9.73 KN/m Mn = Mu/0.90 = 10.81 KN/m r

536.08 mm

=

34.84 KN Shear capacity of beam Vc = √ (f'c)bd/6 fVc/2 Vn =

428.86 mm 33.51 mm 31.93 KN m > 9.73 Treat as singly reinforced beam

3

pcs

(Top bar) (Bottom bar) Vcr = Vu - Wu d = = =

Page 5 of 16

28.86 KN

43.60 KN 18.53 KN Stirrup is needed

<

Vcr/f - Vn Vs = Using 10mmǾ AǾ = Spacing S = A Fyd/Vs Max S = d/2

= 2 78.54 mm = =

15.43 KN 805.02 mm 143.75 mm

Use 10mm stirrups spaced @: 3 @ 50mm; 2 @ 100mm; rest @ 150mm o.c. DESIGN OF COLUMNS A. EXTERIOR COLUMNS (C1) Trial Section b = h = Ǿbar = n =

200 400 12 6

Design Specifications f'c = Fy =

20.7 MPa 275 MPa

1 Axial loads PDL = = Pu PuTotal =

mm mm mm pcs

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 (for one si 100.39 kN

602.37 kN

2 Check capacity of column ǿ 0.80 [ 0.85f'c (Ag - Ast) + FyAst ] Puall =

Puall

=

since:

886.07 kN > Puall

Pu

3 Check steel ratio ρact = Ast/Ag

ρmax ρmin

The column section is safe!

=

0.008

=

0.85f'cβ (600)

=

0.037

=

Fy (600 + Fy) 1.4/Fy

=

0.005

0.005 ρmin

since: 4 Lateral ties s = s s

then:

= =

<

0.008 ρact

<

0.037 ρmax

then:

16 (Ǿbar )

=

192 mm

48 (Ǿtie)

= =

480 mm 200 mm

least dimension

The steel

Thus: Use 6 - 12mm dia bar with 10mm dia lateral ties @ 200mm o.c. B. INTERIOR COLUMNS (C2) Trial Section b = h = Ǿbar = n =

200 300 12 6

Design Specifications f'c = Fy =

20.7 MPa 275 MPa

1 Axial loads PDL = = Pu PuTotal =

mm mm mm pcs

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

=

24.29 kN (for one si 100.39 kN

401.58 kN

Puall

since:

>

Pu

3 Check steel ratio ρact = Ast/Ag

ρmax ρmin

=

0.011

=

0.85f'cβ (600)

=

0.037

=

Fy (600 + Fy) 1.4/Fy

=

0.005

0.005 ρmin

since: 4 Lateral ties s = s s

The column section is safe!

then:

= =

<

0.011 ρact

0.037 ρmax

<

then:

16 (Ǿbar )

=

192 mm

48 (Ǿtie)

= =

480 mm 200 mm

least dimension

The steel

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 = Fy = Ǿbar =

σconc

=

20.7 MPa 275 MPa 16 mm 23.5 kN/m

colb

=

200 mm

cc

=

100 mm

colh

=

400 mm

ǿ

=

0.85

1 Service loads PT =

PDL + PLL

H h qall

= = =

σsoil

=

600 mm 300 mm 190 kPa 15.7 kN/m

2 Base dimensions qnet qall - [ hσconc + Hσsoil ] = Areqd

=

B

= Trial section: B =

255.62 kN

=

=

173.53 kPa 1.47 m

B

=

1.21 m

Aact

=

2.25 m

PT / qnet Areqd

1.5 m

=

3 Soil pressure due to factored loads = 1.4DL + 1.7LL PU 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 2000 mm

c

=

colh + d

=

bo

=

[2b + 2c]

=

Vu

=

PU - qu x b x c

=

358.74 kN

ǿVc

=

0.33ǿ bo d √f'c

=

510.48 kN SAFE

ǿVc

since:

>

→

Vu

5 Check beam shear x1 (B - colb - 2d)/2 =

=

0.45 m

Vu

=

qu B x1

=

120.47 kN

ǿVc

=

0.17ǿ B d f'c

=

197.23 kN SAFE

ǿVc

since:

>

6 Design of reinforcement x2 (B - colb)/2 = Mu

=

qu B

2 x2 /2

→

Vu

= =

Page 7 of 16

0.65 m 56.55577 kN m

(assume) (assume)

= = =

ǿf'c Bd2 ω (1-0.59ω) 1.69ω + 0.08576

1.4/Fy

As

= = =

n

=

Ast/AǾ

Mu

ω2 ω ρ ρmin use

ρ

Thus:

0

= =

ωf'c/Fy

0.052

0.004

= = =

ρ Bd

Use

=

0.005 0.005 2 1527.27 mm

= say

8

7.59

8 pcs

16mm dia bar on both sides of footing.

B. EXTERIOR FOOTING (F2) Design Specifications f'c = Fy = Ǿbar =

σconc

=

20.7 MPa 275 MPa 16 mm 23.5 kN/m

colb

=

200 mm

cc

=

100 mm

colh

=

300 mm

ǿ

=

0.85

1 Service loads PT =

PDL + PLL

H h qall

= = =

σsoil

=

500 mm 300 mm 190 kPa 15.7 kN/m

Areqd

=

B

= Trial section: B =

=

175.10 kPa 0.97 m

B

=

0.99 m

Aact

=

1.44 m

PT / qnet Areqd

1.2 m

=

3 Soil pressure due to factored loads = 1.4DL + 1.7LL PU 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 1800 mm

c

=

colh + d

=

bo

=

[2b + 2c]

=

Vu

=

PU - qu x b x c

=

230.54 kN

ǿVc

=

0.33ǿ bo d √f'c

=

459.43 kN SAFE

ǿVc

since:

>

→

Vu

5 Check beam shear x1 (B - colb - 2d)/2 =

=

0.3 m

Vu

=

qu B x1

=

66.93 kN

ǿVc

=

0.17ǿ B d f'c

=

157.78 kN SAFE

ǿVc

since:

>

2 x2 /2 2

Mu

=

qu B

Mu

= = =

ǿf'c Bd ω (1-0.59ω) 1.69ω + 0.05286

ω ω ρ

2

ωf'c/Fy

→

Vu

6 Design of reinforcement x2 (B - colb)/2 =

(assume)

170.41 kN

=

2 Base dimensions qnet qall - [ hσconc + Hσsoil ] =

(assume)

=

0.5 m

=

27.88746 kN m =

= =

Page 8 of 16

0 0.032

0.002

ρmin

1.4/Fy

As

= = =

n

=

Ast/AǾ

use

ρ

Thus:

ρ Bd

Use

7

= = = = say

0.005 0.005 2 1221.82 mm 6.08

16mm dia bar on both sides of footing.

Page 9 of 16

7 pcs

S, CAVITE

ÑAS, CAVITE

Page 10 of 16

(short direction) (long direction) mm

m m

KN m

f strip f strip

Page 11 of 16

KN m

f strip

f 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

de only)

ratio is safe!

de only)

Page 15 of 16

ratio is safe!

Page 16 of 16

Structural Computation Analysis

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