STRUCTURAL COMPUTATION

OF

PROPOSED TWO-STOREY RESIDENTIAL HOUSE

PHASE 7-A BLOCK 17 LOT 04 ORCHARD & RESIDENTIAL ESTATE, SALITRAN, DASMARIAS, CAVITE

Prepared by:

GIL S. BELTRAN
Civil Engineer
PRC Reg. No.
PTR No.
Date Issued
Place Issued
TIN

: 64544
: 8642641
: 1/9/2009
: Dasmarias, Cavite
: 157-467-966

Page 1 of 9

PROPOSED TWO-STOREY RESIDENTIAL HOUSE

PHASE 7-A BLOCK 17 LOT 04 ORCHARD & RESIDENTIAL ESTATE, SALITRAN, DASMARIAS, CAVITE
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 =
L2 =
Fy =

3.35 m
275 Mpa

0.20m x 0.40m

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

cc =

Considering the interior slab

1 Relative values
x L2/L1 =
1.46
x L1/L2 =
0.86
2 Slab thickness, t
t
=

m = 2x(1 + 2)/4

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

where: Ln = L2 - column width

Ln = L1 - column width
= (Ln long/Ln short) direction clear span ratio

1.25 (long direction)

25 mm

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 + 9000

68.27 mm

ok

It should not be more than

tmax =
Ln (800 + 0.73Fy) =
36000

93.13 mm

ok

Try t =

1 (short direction)

100 mm for all two way slabs

3 Design moments
Dead Load = W slab + W ceiling+finish =
Live Load

41.48 KN m;

2.3 m

L2 =

4.8 Kpa

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

Long Span:
2
Mu = Wu L1 Ln2 /8

L1 =

3.35 Kpa

3.35 m
0.9

12.86 Kpa

Mu/

Mn =

46.09 KN m

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:
L1/L2 =
2 =
1.25
By interpolation:
L1/L2
2L1/L2
0.5
0.69
1

90
?
75

2L1/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

Page 2 of 9

0.86

5.60 %
84.40 %

-1.90 KN m /half strip

Column strip positive

Beam positive
Slab positive

Mn
Mn
Mn

=
=
=

13.61 KN m
11.57 KN m
2.04 KN m

1.02 KN m /half strip

Mu/

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;

Mn =

31.64 KN m

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:
=
1
L2/L1 =
1.46
By interpolation:
L2/L1

L2/L1 =

1.46

L2/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 (mm)
Middle Strips
Strip Width, b (mm)
Mn (KN m)
Mn (KN m)
d (mm)
As = [Mn(106)]/[Fy(0.95d)] (mm2)
Min As = 0.002bt (mm2)
Spacing, S = Ab /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)] (mm2)
Min As = 0.002bt (mm2)
Spacing, S = Ab /As (mm)
Max S = 2t (mm)
Adopted Spacing, S (mm)
Middle Strips (10mm bars)
Column Strips (10mm bars)

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

Page 3 of 9

200
200

DESIGN OF BEAMS
A. FLOOR BEAM, FB1
1 Design loads
TA =
10.12
Beam size: b =
0.20
h=
0.35
d=
0.34
Dead loads:
W slab =
W c+f =
W beam =
W DL =

m2
m
m
m
23.82
7.2864
5.52

Slab Thickness =
Beam span L =
LL =
KN
KN

36.63 KN

10.93 KN/m

48.576 KN
= 1.4DL + 1.7LL

=
=

14.50 KN/m
39.96 KN/m

W LL =

Live loads:

Wu

2 Design Moments
Assuming fully restrained beam
Max. +Moment =
WuL2/24
Max. -Moment =
-WuL2/12
Max. Shear =
WuL/2
3 Reinforcement Design
At Support (Negative Moment)
Mu =
-37.37 KN m
Mn = Mu/0.90 =
41.52 KN m

bal =
max =

M1 =

=
=
=

18.68 KN m
-37.37 KN m
66.93 KN

f'c =
Fy =
A =

20.7 Mpa
275 Mpa
201.06 mm2

max = 0.75bal
[0.85f'c /Fy][600+(600+Fy)]
0.75bal

0.037

0.028

0.3bal

0.011

=
use
s1bd
a

100 mm
3.35 m
4800 Pa

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
(Top bar)
2 pcs - 16mm
(Bottom bar)
At Midspan (Positive Moment)
Mu =
18.68 KN/m
Mn = Mu/0.90 =
20.76 KN/m

bal =
max =

59.01 mm
63.96 KN m 37.37
KN m
Treat as singly reinforced beam
4

f'c =
Fy =
A =

pcs

20.7 Mpa
275 Mpa
201.06 mm2

max = 0.75bal
[0.85f'c /Fy][600+(600+Fy)]
0.75bal

0.15bal
=
use
s1bd
a
= As1Fy/(0.85f'cb)
M1 =
As1Fy(d-a/2)

0.037

0.028

0.006

=
=
=

No. of bars required (use 16 mm

n = As/A
=
1.9 say
Reinforcement:
2 pcs - 16mm
(Top bar)
4 pcs - 16mm
(Bottom bar)
4 Stirrup
Vu =

755.17 mm2

66.93 KN
Shear capacity of beam

377.58 mm2
29.51 mm
33.51 KN m 18.68
KN m
Treat as singly reinforced beam
2

Vcr = Vu - Wu d =

Page 4 of 9

pcs

53.44 KN

Vc = (f'c)bd/6
Vc/2
Vn =

=
=

51.18 KN
21.75 KN
Stirrup is needed
41.12 KN

<

53.44

Vcr/ - Vn
Vs =
=
Using 10mm
A =
78.54 mm2
Spacing S = AFyd/Vs
=
354.53 mm
Max S = d/2
=
168.75 mm
Use 10mm stirrups spaced @: 3 @ 50mm; 2 @ 100mm; 2 rest @ 150mm
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 =

m2
m
m
m
11.91
3.6432
4.73

Slab Thickness =
Beam span L =
LL =
KN
KN

20.29 KN

6.06 KN/m

24.288 KN
= 1.4DL + 1.7LL

=
=

7.25 KN/m
20.80 KN/m

W LL =

Live loads:

Wu

2 Design Moments
Assuming fully restrained beam
Max. +Moment =
WuL2/24
Max. -Moment =
-WuL2/12
Max. Shear =
WuL/2
3 Reinforcement Design
At Support (Negative Moment)
Mu =
-19.46 KN m
Mn = Mu/0.90 =
21.62 KN m

bal =
max =

M1 =

=
=
=

9.73 KN m
-19.46 KN m
34.84 KN

f'c =
Fy =
A =

20.7 Mpa
275 Mpa
201.06 mm2

max = 0.75bal
[0.85f'c /Fy][600+(600+Fy)]
0.75bal

=
use
s1bd
a

100 mm
3.35 m
4800 Pa

0.25bal

0.037

0.028

0.009

As1Fy/(0.85f'cb)

As1Fy(d-a/2)

No. of bars required (use 16 mm

n = As/A
=
2.7 say
Reinforcement:
3 pcs - 16mm
(Top bar)
2 pcs - 16mm
(Bottom bar)
At Midspan (Positive Moment)
Mu =
9.73 KN/m
Mn = Mu/0.90 =
10.81 KN/m

bal =
max =

536.08 mm2
41.89 mm
39.30 KN m 19.46
KN m
Treat as singly reinforced beam
3

f'c =
Fy =
A =

pcs

20.7 Mpa
275 Mpa
201.06 mm2

max = 0.75bal
[0.85f'c /Fy][600+(600+Fy)]
0.75bal

0.037

0.028

0.2bal

0.007

=
use
s1bd
a
= As1Fy/(0.85f'cb)
M1 =
As1Fy(d-a/2)

=
=
=

No. of bars required (use 16 mm

n = As/A
=
2.1 say

Page 5 of 9

428.86 mm2
33.51 mm
31.93 KN m 9.73
KN m
Treat as singly reinforced beam
3

pcs

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

(Top bar)
(Bottom bar)

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

Vcr = Vu - Wu d =
=
=

28.86 KN

43.60 KN
18.53 KN
Stirrup is needed
15.43 KN

<

28.86

Vcr/ - Vn
Vs =
=
Using 10mm
A =
78.54 mm2
Spacing S = AFyd/Vs
=
805.02 mm
Max S = d/2
=
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
=
200 mm
h
=
400 mm
bar
=
12 mm
n
=
6 pcs

Design Specifications
f'c
=
Fy
=
1 Axial loads
PDL
=
Pu
=
PuTotal =

20.7 MPa
275 MPa

Ag
Ast
tie

=
=

=
=

80000 mm2
678.58401 mm2

PLL
18.31 kN
(1.4DL + 1.7LL)x1.5
=
Pu x no. of sides x no. of floors =

10 mm

0.7
0.85

=
100.39 kN
602.37 kN

24.29 kN
(for one side only)

2 Check capacity of column

Puall
=
0.80 [ 0.85f'c (Ag - Ast) + FyAst ]
Puall
=
886.07 kN
since:

Puall

>

Pu

3 Check steel ratio

act
=
Ast/Ag
max
=
0.85f'c (600)
Fy (600 + Fy)
min
=
1.4/Fy
0.005
min

since:

<

0.008
act

then:

The column section is safe!

0.008

0.037

0.005
0.037
max
then:

<

4 Lateral ties
s

16 (bar)

192 mm

48 (tie)

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 mm
h
=
300 mm
bar
=
12 mm
n
=
6 pcs

Ag
Ast
tie

=
=
=

Page 6 of 9

60000 mm2
678.58401 mm2
10 mm

The steel ratio is safe!

Design Specifications
f'c
=
Fy
=
1 Axial loads
PDL
=
Pu
=
PuTotal =

20.7 MPa
275 MPa

=
=

0.7
0.85

PLL
18.31 kN
(1.4DL + 1.7LL)x1.5
=
Pu x no. of sides x no. of floors =

=
100.39 kN
401.58 kN

24.29 kN
(for one side only)

2 Check capacity of column

Puall
=
0.80 [ 0.85f'c (Ag - Ast) + FyAst ]
Puall
=
689.01 kN
since:

Puall

>

Pu

3 Check steel ratio

act
=
Ast/Ag
max
=
0.85f'c (600)
Fy (600 + Fy)
min
=
1.4/Fy
0.005
min

since:

<

0.011
act

then:

The column section is safe!

0.011

0.037

0.005
0.037
max
then:

<

The steel ratio is safe!

4 Lateral ties
s

16 (bar)

192 mm

48 (tie)

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 MPa
Fy
=
275 MPa
bar
=
16 mm
conc
=
23.5 kN/m3

H
h
qall

=
=

600 mm
300 mm

190 kPa

soil

15.7 kN/m

colb

200 mm

cc

100 mm

colh

400 mm

0.85

1 Service loads
PT
=

PDL + PLL

2 Base dimensions
qnet
qall - [ hconc + Hsoil ]
=

255.62 kN

173.53 kPa

1.47 m2

1.21 m

Aact

2.25 m2

401.58 kN

PU / Aact

178.48 kPa

4 Check punching shear

d
=
h - cc
colb + d
b
=

200 mm

400 mm

600 mm

2000 mm2

Areqd

B2
=
Trial section:
B

PT / qnet
Areqd
1.5 m

3 Soil pressure due to factored loads

PU
=
1.4DL + 1.7LL
qu

c
bo

colh + d

Vu

[2b + 2c]
PU - qu x b x c

Page 7 of 9

358.74 kN

(assume)
3

(assume)

Vc

0.33 bo d f'c
Vc
>

since:

=
Vu

5 Check beam shear

x1
(B - colb - 2d)/2
=
Vu

qu B x1

Vc

0.17 B d f'c
Vc
>

since:

Mu
2

120.47 kN

197.23 kN
SAFE

qu B x /2

0.65 m

56.555771 kN m

f'c Bd (1-0.59)
1.69
+
0.08576
2

=
use =
As
=

1.4/Fy

Ast/A
Use

0
0.052
0.004

=
=
=

Bd

=
=
=

f'c/Fy

min

Thus:

0.45 m

Vu

2
2

=
=
=
=

510.48 kN
SAFE

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

0.005
0.005
1527.27 mm2

=
7.59
say
8 pcs
16mm dia bar on both sides of footing.

B. EXTERIOR FOOTING (F2)

Design Specifications
f'c
=
20.7 MPa
Fy
=
275 MPa
bar
=
16 mm
conc
=
23.5 kN/m3

H
h
qall

=
=

500 mm
300 mm

190 kPa

(assume)

soil

15.7 kN/m3

(assume)

colb

200 mm

cc

100 mm

colh

300 mm

0.85

1 Service loads
PT
=

PDL + PLL

2 Base dimensions
qnet
qall - [ hconc + Hsoil ]
=
Areqd

B2
=
Trial section:
B

175.10 kPa

0.97 m2

0.99 m

Aact

1.44 m2

PT / qnet
Areqd
1.2 m

170.41 kN

3 Soil pressure due to factored loads

PU
=
1.4DL + 1.7LL

267.72 kN

PU / Aact

185.92 kPa

4 Check punching shear

d
=
h - cc
colb + d
b
=

200 mm

400 mm

qu

c
bo

colh + d

500 mm

1800 mm2

Vu

[2b + 2c]
PU - qu x b x c

Vc

since:

0.33 bo d f'c
Vc
>

Vu

230.54 kN

459.43 kN
SAFE

Page 8 of 9

5 Check beam shear

x1
(B - colb - 2d)/2
=
Vu

qu B x1

Vc

0.17 B d f'c
Vc
>

since:

=
=

Mu
2

f'c Bd2 (1-0.59)

1.69
+
0.05286
f'c/Fy

min
=
use =
As
=

1.4/Fy

Ast/A

Thus:

Bd

=
Use

157.78 kN
SAFE

qu B x22/2

=
=
=
=

66.93 kN

=
Vu

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

0.3 m

0.5 m
27.887461 kN m
=

=
=
=
=
=

0
0.032
0.002
0.005
0.005
1221.82 mm2

=
6.08
say
7 pcs
16mm dia bar on both sides of footing.

Page 9 of 9