structural computation analysis

8/11/2019 Structural Computation Analysis

1/16

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

Prepared by:

GIL S. BELTRANCivil Engineer

PRC Reg. No. : 64544PTR No. : 8642641Date Issued : 1/9/2009Place Issued : Dasmarias, CaviteTIN : 157-467-966

PROPOSED TWO-STOREY RESIDENTIAL HOUSE

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

STRUCTURAL COMPUTATION

OF

PROPOSED TWO-STOREY RESIDENTIAL HOUSE

Page 1 of 16


2/16

DESIGN OF SLABSA. DESIGN OF TWO-WAY SLAB (S-1)Direct Design MethodDesign Specifications

L1 = 2.30 m column size = 0.20m x 0.40m a1 = 1

L2 = 3.35 m beam size = 0.20m x 0.35m a2 = 1.25Fy = 275 Mpa f'c = 20.7 Mpa cc = 25

Considering the interior slab1 Relative a values

a1 x L2 /L 1 = 1.46 a m = 2x( a 1 + a 2)/4 = 1.16a2 x L1 /L 2 = 0.86

2 Slab thickness, t t = Ln (800 + 0.73Fy) = 78.05 mm

36,000 + 5000 b [a m - 0.12(1 + 1/ b)]

where: Ln = L 2 - column width = 3.35 mLn = L 1 - column width = 2.30 mb = (Ln long/Ln short) direction clear span ratio = 1.46

It should not be less than tmin = Ln (800 + 0.73Fy) = 68.27 mm ok

36000 + 9000 b

It should not be more than tmax = Ln (800 + 0.73Fy) = 93.13 mm ok

36000

Try t = 100 mm for all two way slabs

3 Design momentsDead Load = W slab + W ceiling+finish = 3.35 Kpa L1 = 2.3

Live Load = 4.8 Kpa L2 = 3.35f = 0.9

Wu = 1.4DL + 1.7LL (consider 1-m strip) = 12.86 Kpa

Long Span:Mu = Wu L 1 Ln 22/8 = 41.48 KN m; Mn = Mu/f = 46.09

Distribution of Mn between section of positive and negative momentat midspan: +Mms = 0.35 Mn = 16.13 KN mat support: -Ms = 0.65 Mn = -29.96 KN m

Percentage of moment at supportColumn strip:

a 2 = 1.25 L1/L2 = 0.69 a 2L1/L2 = 0.86By interpolation:L1/L2 a 2L1/L2

0.5 90 x x = 5.60 %0.69 ? % M = 84.40 %

1 75

Code assigns 85% of the column strip moment to the beamColumn strip negative Mn = -25.28 KN mBeam negative Mn = -21.49 KN mSlab negative Mn = -3.79 KN m = -1.90 KN m /halColumn strip positive Mn = 13.61 KN mBeam positive Mn = 11.57 KN mSlab positive Mn = 2.04 KN m = 1.02 KN m /hal

Owner: MS. LYDA B. FIX

Page 2 of 16


3/16

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

Short Span:Mu = Wu L 2 Ln12/8 = 28.48 KN m; Mn = Mu/f = 31.64

Distribution of Mn between section of positive and negative momentat midspan: +Mms = 0.35 Mn = 11.07 KN mat support: -Ms = 0.65 Mn = -20.57 KN m

Percentage of moment at supportColumn strip:

a1 = 1 a1 L2/L1 = 1.46L2/L1 = 1.46

By interpolation:L2/L1 a1 L2/L1

1 75 x x = 13.70 %1.46 ? % M = 61.30 %

2 45Code assigns 85% of the column strip moment to the beamColumn strip negative Mn = -12.61 KN mBeam negative Mn = -10.72 KN mSlab negative Mn = -1.89 KN m = -0.95 KN m /hal

Column strip positive Mn = 6.79 KN mBeam positive Mn = 5.77 KN mSlab positive Mn = 1.02 KN m = 0.51 KN m /hal

Middle strip (takes the remaining moment)

Middle strip negative Mn = -7.96 KN mMiddle strip positive Mn = 4.29 KN m

Fy 275 Mpa-M +M -M +M

Slab thickness, t (mm)Steel reinforcement f (mm)Middle Strips

Strip Width, b (mm)Mn (KN m)Mn (KN m) -4.67 2.52 -7.96 4.29d (mm) 75 75 75 50

As = [Mn(10 6)]/[Fy(0.95d)] (mm 2) 238.46 128.40 406.18 328.07Min As = 0.002bt (mm 2) 200 200 200 200Spacing, S = A f b /As (mm) 329.36 392.70 193.36 239.40Max S = 2t (mm) 200 200 200 200

Column StripsStrip Width, b (mm)Mn (KN m)Mn (KN m) -1.90 1.02 -0.95 0.51d (mm) 75 75 75 50

As = [Mn(10 6)]/[Fy(0.95d)] (mm 2) 96.78 52.11 48.26 38.98Min As = 0.002bt (mm 2) 200 200 200 200Spacing, S = A f b /As (mm) 392.70 392.70 392.70 392.70Max S = 2t (mm) 200 200 200 200

Adopted Spacing, S (mm)Middle Strips (10mm bars)Column Strips (10mm bars)

DESIGN OF BEAMSA. FLOOR BEAM, FB1

1 Design loadsTA = 10.12 m 2 Slab Thickness = 100 mmBeam size: b = 0.20 m Beam span L = 3.35 m

10 10

1000 1000

LONG SPAN SHORT SPAN

100 100

46.09 31.64

200 200

46.09 31.64

1000 1000

200 200

Page 3 of 16


4/16

h = 0.35 m LL = 4800 Pa d = 0.34 m

Dead loads: W slab = 23.82 KNW c+f = 7.286 KNW beam = 5.52W DL = 36.63 KN = 10.93 KN/m

Live loads: W LL = 48.58 KN = 14.50 KN/m

Wu = 1.4DL + 1.7LL = 39.96 KN/m

2 Design Moments Assuming fully restrained beamMax. +Moment = WuL 2/24 = 18.68 KN mMax. -Moment = -WuL 2/12 = -37.37 KN mMax. Shear = WuL/2 = 66.93 KN

3 Reinforcement Design At Support (Negative Moment)Mu = -37.37 KN m f'c = 20.7 MpaMn = Mu/0.90 = 41.52 KN m Fy = 275 Mpa

A = 201.06 mm 2

r = r max = 0.75 r balr bal = [0.85f'c b /Fy][600+(600+Fy)] = 0.037r max = 0.75 r bal = 0.028

use r = 0.3 r bal = 0.011As 1 = r bd = 755.17 mm

a = As 1Fy/(0.85f'cb) = 59.01 mmM1 = As 1Fy(d-a/2) = 63.96 KN m > 37.37

Treat as singly reinforced beamNo. of bars required (use 16 mm )n = As/A = 3.8 say 4 pcsReinforcement:

4 pcs - 16mm (Top bar)2 pcs - 16mm (Bottom bar)

At Midspan (Positive Moment)Mu = 18.68 KN/m f'c = 20.7 MpaMn = Mu/0.90 = 20.76 KN/m Fy = 275 Mpa

A = 201.06 mm 2

r = r max = 0.75 r balr bal = [0.85f'c b /Fy][600+(600+Fy)] = 0.037r max = 0.75 r bal = 0.028

use r = 0.15 r bal = 0.006

As 1 = r bd = 377.58 mma = As1Fy/(0.85f'cb) = 29.51 mmM1 = As1Fy(d-a/2) = 33.51 KN m > 18.68



4 StirrupVu = 66.93 KN Vcr = Vu - Wu d = 53.44 KN

Shear capacity of beamVc = (f'c)bd/6 = 51.18 KNVn = f Vc/2 = 21.75 KN 9.73



4 StirrupVu = 34.84 KN Vcr = Vu - Wu d = 28.86 KN

Shear capacity of beamVc = (f'c)bd/6 = 43.60 KNVn = f Vc/2 = 18.53 KN Pu then: The column section is safe!

3 Check steel ratioact = Ast/Ag = 0.008max = = 0.037

min = 1.4/Fy = 0.0050.005 0.008 0.037

since: min < act < max then: The steel

4 Lateral tiess = 16 ( bar ) = 192 mms = 48 ( tie) = 480 mms = least dimension = 200 mmThus: Use 6 - 12mm dia bar with 10mm dia lateral ties @ 200mm o.c.

B. INTERIOR COLUMNS (C2)Trial Sectionb = 200 mm Ag = 60000 mm 2

h = 300 mm Ast = 678.584 mm 2

bar = 12 mm tie = 10 mmn = 6 pcs

Design Specificationsf'c = 20.7 MPa = 0.7Fy = 275 MPa = 0.85

1 Axial loadsP DL = 18.31 kN P LL = 24.29 kNPu = (1.4DL + 1.7LL)x1.5 = 100.39 kN (for one siPu Total = Pu x no. of sides x no. of floors = 401.58 kN

2 Check capacity of columnPu all = 0.80 [ 0.85f'c (Ag - Ast) + FyAst ]Pu all = 689.01 kN

0.85f'c (600)Fy (600 + Fy)

Page 6 of 16


7/16

since: Puall > Pu then: The column section is safe!

3 Check steel ratioact = Ast/Ag = 0.011max = = 0.037

min = 1.4/Fy = 0.0050.005 0.011 0.037

since: min < act < max then: The steel

4 Lateral tiess = 16 ( bar ) = 192 mms = 48 ( tie) = 480 mms = least dimension = 200 mmThus: Use 6 - 16mm dia bar with 10mm dia lateral ties @ 200mm o.c.

DESIGN OF FOOTINGA. INTERIOR FOOTING (F1)

Design Specificationsf'c = 20.7 MPa H = 600 mmFy = 275 MPa h = 300 mmbar = 16 mm q all = 190 kPa (assume)conc = 23.5 kN/m soil = 15.7 kN/m (assume)col b = 200 mm cc = 100 mmcol h = 400 mm = 0.85

1 Service loadsP T = P DL + P LL = 255.62 kN

2 Base dimensionsqnet = q all - [ h conc + H soil ] = 173.53 kPa

A reqd = P T / q net = 1.47 m

B = A reqd B = 1.21 mTrial section:B = 1.5 m Aact = 2.25 m

3 Soil pressure due to factored loadsPU = 1.4DL + 1.7LL = 401.58 kNqu = P U / A act = 178.48 kPa

4 Check punching shear

d = h - cc = 200 mmb = col b + d = 400 mm

c = col h + d = 600 mmbo = [2b + 2c] = 2000 mm

Vu = P U - q u x b x c = 358.74 kN

Vc = 0.33 bo d f'c = 510.48 kNsince: Vc > Vu SAFE

5 Check beam shear x1 = (B - col b - 2d)/2 = 0.45 m

Vu = qu B x 1 = 120.47 kN

Vc = 0.17 B d f'c = 197.23 kNsince: Vc > Vu SAFE

6 Design of reinforcementx2 = (B - col b)/2 = 0.65 mMu = qu B x 2

2/2 = 56.55577 kN m

0.85f'c (600)Fy (600 + Fy)

Page 7 of 16


8/16

Mu = f'c Bd 2 (1-0.59)2 - 1.69 + = 0 = = 0.052 = f'c/Fy = 0.004

min = 1.4/Fy = 0.005use = = 0.005

As = Bd = 1527.27 mm2

n = Ast/A = 7.59say 8 pcs

Thus: Use 8 16mm dia bar on both sides of footing.

B. EXTERIOR FOOTING (F2)Design Specificationsf'c = 20.7 MPa H = 500 mmFy = 275 MPa h = 300 mmbar = 16 mm q all = 190 kPa (assume)conc = 23.5 kN/m soil = 15.7 kN/m (assume)col b = 200 mm cc = 100 mmcol h = 300 mm = 0.85

1 Service loadsP T = P DL + P LL = 170.41 kN

2 Base dimensionsqnet = q all - [ h conc + H soil ] = 175.10 kPa

A reqd = P T / q net = 0.97 m

B = A reqd B = 0.99 mTrial section:B = 1.2 m Aact = 1.44 m

3 Soil pressure due to factored loadsPU = 1.4DL + 1.7LL = 267.72 kNqu = P U / A act = 185.92 kPa

4 Check punching shear d = h - cc = 200 mmb = col b + d = 400 mm

c = col h + d = 500 mmbo = [2b + 2c] = 1800 mm

Vu = P U - q u x b x c = 230.54 kN

Vc = 0.33 bo d f'c = 459.43 kNsince: Vc > Vu SAFE

5 Check beam shear x1 = (B - col b - 2d)/2 = 0.3 mVu = qu B x 1 = 66.93 kN

Vc = 0.17 B d f'c = 157.78 kNsince: Vc > Vu SAFE

6 Design of reinforcementx2 = (B - col b)/2 = 0.5 mMu = qu B x 2

2/2 = 27.88746 kN m

Mu = f'c Bd 2 (1-0.59)2 - 1.69 + = 0 = = 0.032 = f'c/Fy = 0.002

0.05286

0.08576

Page 8 of 16


9/16

min = 1.4/Fy = 0.005use = = 0.005

As = Bd = 1221.82 mm 2

n = Ast/A = 6.08say 7 pcs

Thus: Use 7 16mm dia bar on both sides of footing.

Page 9 of 16


10/16

S, CAVITE

AS, CAVITE

Page 10 of 16


11/16

(short direction)

(long direction)mm

m

m

KN m

f strip

f strip

Page 11 of 16


12/16

KN m

f strip

f strip

Page 12 of 16


13/16

KN m

KN m

53.44

Page 13 of 16


14/16

KN m

KN m

28.86

Page 14 of 16


15/16

de only)

ratio is safe!

de only)

Page 15 of 16


16/16

ratio is safe!

Page 16 of 16

structural computation analysis

Documents