STRUCTURAL ANALYSIS AND DESIGN

DATA AND SPECIFICATIONS : Owner : MS. LEA MAYLA B. JACKSON Name of Project : Proposed ONE STOREY RESL. BLDG

Location of Building : DUNKIN DRIVE ST., CARIG-ALIMANNAO, TUGUEGARAO CITY Unit Wt. Of concrete : 24.00 KN / m3. Exterior Walls : 2.11 KN / m2. Slab Thickness ( t ) : 0.125 m.

Live Loads ( LL ) : 4.80 KN / m2. Live Loads on stairs : 2.40 KN / m2. Super imposed DL : 1.45 KN / m2. height of CHB : 2.80 m. Stair : 4.80 KN / m2.

Design Specifications / Criteria : fc' = 20.70 Mpa n = 10.00 fs = 124.00 Mpa E= 17,300.00 Mpa fy = 280.00 Mpa Sp= 170.00 Mpa

Fy= 0.70 Mpa

0.17(fc') - for beam stress = 0.33(fc') - for punching shear

10.12(fc')/D for Bond Shear

Seismic load provisions - Allowable 10 - 20 % of DL and LL

Prepared by :

RICARDO V. ALLAMCivil Engineer

PRC No. : 53492PRTR No. : 264358D-Issued: : April 1, 2008P-Issued: : Tuguegarao City

۷allow. =

μallow. =

DESIGN OF BEAM : B1

length(L) d (m,) b (m,) Size of BEAM - 6.00 0.50 0.30

DEAD LOAD DL :

Wt. OF Beam = b x d x (24) = 3.60 KN/m Wt. OF Slab = t x L x 24 x (2/3) 9.00 KN/m CHB WALL = 2.103 x h = 5.90 KN/m S.I. Dead load = 1.45 x L x 2/3 5.80 KN/m dead load ( B ) =

========= = 24.30 KN/m

LIVE LOAD LL : 14.40 KN/m=========

Total Load 38.70 KN/mUSING " USD " :

wu = (1.4DL+1.7LL) 58.49552 KN/m Add seismic load provision :

wu = 1.15 x wu = 67.269848 KN/m

Use Mu = wu x L^2 for most critical seciton12

201.8095 KN. M

CHECK FOR d:

Mu Ǿ bd fy [ 1 - 0.59 x fy x ]

0.013fy

2.02E+08 = bd x 2.9362788

d = 201.8095442.9362788

= 473.64 add 40 mm steel cover

= 513.642852924 mm. 500.00

try = ρ min = 0.18 fc' =

mm SAFE !!!

THEREFORE ADOPT BEAM SECTION : 500.00 mm X 300.00 mm

CHECK FOR RSB :

As = bd : use d = 460.00 mm

= 1,794.00 mm^2

Using 16.00 mm Ǿ RSB

n = 8.92260313216 say 6 pcs. 16 mm Ǿ RSB

@ SUPPORT @ MIDSPAN

n.a. d = 500 mm.

b= 300 mm

V = 1.15 w x L = 232.0809756 KN2

vu = 1.978524941176 Mpa

vu allow. = 0.17 fc' = 0.773453295293 Mpa

Since vu allow. < vu actual therefore provide stirrups of 10 mm Ǿ RSB spaced 5 @ 0.05m.;4 at 0.10 m. : 3 at 0.15 m and rest at 0.20 m. on center.

CHECKED FOR BENDING STRESS :

10.12 fc' = 2.877701231018 MpaD

μ act. = V / (o x j x d) = 1.945186009129 Mpa

Adopt 250 x 400 mm Beam Section w/ 6 - 16mmØ RSB & 2 - 12mmØ RSB provided stirrups of

V / Φ bd =

μallow. =

therefore μ actual < μ allow. SAFE ! ! !

10mmØ RSB spaced 5 @ 50mm, 4 @ 100mm, 3 @ 150mm and rest at 200mm O.C.

DESIGN OF COLUMNS :

C 1: Design Load P = 235.85 KN

Add 15 % for Seismic load provision :

Pu= 271.22 KN

Using = 0.013

Ag =

= 23056.2542039 mm^2? ?

CHECK ACTUAL COLUMN SIZE : USE 250 X 250

Ag = 62500 mm^2 23056.2542 mm^2 Required

Therefore Very S A F E ! ! !

CHECK FOR STEEL REQUIREMENT :

812.5 mm2

Using 16 4.041034 4 pcs.

Check Area of Steel :

Ast = 812.5 mm^2 812.5 mm^2 = As req'd

Therefore Very S A F E ! ! !

therefore adopt : 250 x 250 column with 4 16

rest at 0.20 m. on center.

check Spacing of stirrups :

256 mm.

480 mm. 3. Least dimension = 250 mm.

CHECK FOR Pu allow. :

Pu / [ø (0.80) { 0.85 x fc' ( 1 - ρg ) + (fy x ρg) }]

As = ρg x Ag =

mm ø RSB : n =

mm^2 øRSB with 10 mm ø RSB as stirrups spaced at 7 @ 0.05 m.; 3 @ 0.10 m.; 3 @ 0.15 m. and

1 16 ø =

2. 48 x tie ø =

Pu =

= 607.82519 KN 235.85 KNtherefore S A F E ! ! !

Φ (0.80) { 0.85 x Fc' x ( Ag - Ast ) + fy x Ast }

DESIGN OF FOOTING:

Use Pu = 235.85 KN

Using Sp = 170.00 Mpa

A req'd = Pu / Sp = 1.387327058824 m2

Using 1.17784848721 1.2

APPROX. USE 1.2 X 1.2

Sp net = Pu / A = 163.7816666667 KN/ m2

CHECK FOR BENDING SHEAR :

use = 1.4 / fy = 0.005

w = fy / fc' = 0.067632850242 X = 0.475

01.2 0.25

Mu = Sp x Ab x X/2

= 22.171943125 KN.m.1.2 m.

DESIGN DEPTH d OF FOOTING :

Mu = 1451.66608696 d^2

d = 123.585782678 mm. + 100 mm steel cover = 223.5858

therefore d 300 mm.

STEEL REQUIREMENT :

As = min. L . d min = 0.0035

ρmin =

Mu = Φ fc' . Bd^2 x (ω) ( 1 - 0.59ω)

= 939.06028725 mm^2

TRY = 16

Ao = 201.0624 mm^2

N = As / Ao = 4.870491783892

S A Y 5 PCS. - 16

THEREFORE USE 5 PCS. - 16 mm ø RSB MATTINGBOTHWAYS FOR 1.2 X 1.2 X 300FOOTING .

P = 235.85 Kn.

300 mm.

L = 1.20 m. w = 1.20 m.

1.20

1.20 0.25

mm ø RSB

mm ø RSB

P

0.25

DESIGN OF FLOOR SLAB

fc' = 20.70 Mpa. DCE fs = 124.00 Mpa.

Consider typical Slab :

6.00 DCE CE m = S / L = 1.5

CE CASE NO. 3 - TWO EDGE DISCONTTINUES : 4.00

SOLVE FOR THE SLAB THICKNESS :

t = perimeter / 180 = 111.1111111111 mm. say 111.11111111 mm.

total depth = t + steel cover(25mm) = 136.1111111 mm.

say St = 136.111111111 mm.

DEAD LOAD :Slab = 24 x d = 3.266666666667 KN/m2Ceiling = 0.24 KN/m2Floor fin.= 0.72 KN/m2super imposed dead load= 1.45 KN/m2

total = 5.676666666667 KN/m2

Live load = 2.39 KN/m2

* Consider 1.0 m. Strip : Total w = 1.4 DL + 1.7 LL = 12.01033333 KN/ m.

m = 1

FROM TABLE ( Case 3 ) ::: Along short span / long span

+ M @ midspan = cw s^2 = 14.124152 KN.m. - M @ DCE = 7.2062 KN.m. - M @ CE = 10.665176 KN.m.

Consider Max Moment to check for thickness of SLAB t =? : * * * M = 14.124152 KN.m.

b = 1000 ( 1 m. strip ) d = M / (R x b) = n = 10 ; k = 0.42 ; j = 0.86

R = 1/2 x fc' x k x j = 3.73842 Mpa d = 61.4663105186 mm. - add 25 mm. Steel cover

d = 86.4663105186 mm. d = 136.11111 mm. OK ! ! !

Therefore adopth Slab Thickness St = 136.1111111111 mm.

use effective d = 111.1111111111 mm. @ Midspan use M= 14.124152 Mpa.

As = M / (fs x j x d) = 1192.023331 mm2

Using 12 113.0976 mm2

Spacing S = Ast x 1000 / As = 94.87867987 mm.

CHECK FOR MAXIMUM SLAB SPACING S = :S = 3 x t = 408.3333333333 mm.

THEREFORE ADOPT ; S = 94.87867986694 mm.

* * @ DCE max M = 7.2062 Mpa

As = M / (fs x j x d) = 608.1751688 mm2

S = Ast*1000/ As = 185.9622125 mm.

Adopt S = 185.9622125392 mm. 408.33333 mm. O K !!!

* * @ CE max M = 10.665176 Mpa

As = M / (fs x j x d) = 900.0992498 mm^2

S = Ast*1000/ As = 125.6501436 mm. 408.33333333 mm.

Adopt S = 125.6501436076 mm. L/4 L/2 L/4

mm ø RSB Ast =

SUMMARY :

185.9622 mm. on center DCE

95 mm.

12 mm ø RSB @

10 mm ø RSB @ 125mm

65.28 32.64 5.04 2.16 5.90 4.80 2.11 117.92