CARMEL B. SABADO CE-162 PROF. GERONIDES P. ANCOGBSCE-5 2nd Excel Program SEPT. 6, 2009*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program. =)

****====DESIGN OF ONE-WAY SLAB====****

4.6 m 4.6 m

GIVEN:

27.5 Mpa

414 Mpa

23.5 1) L / 20 ; simply supported

Slab live load = 4.8 kPa 2) L / 24 ; one end continuous

clear span,L = 4.6 m 3) L / 28 ; two end continuous

main bar diameter = 10 mm Ø 4) L / 10 ; cantilever

temp. bar diameter = 8 mm Ø

TYPE OF SLAB SECTION: 3 h = L / 28 ; TWO END CONTINUOUS

h = L

= 164.2857 mm28

say h = 165 mm

COMPUTE LOADS:DEAD LOAD(DL)

= 3.8775 Kpa

assume floor finish = 25 mm

= 0.5875 Kpa

total DL = 4.465 KPa

LIVE LOAD(LL)

service load = 4.8 Kpa

wu = 1.2DL + 1.6LL = 13.038 Kpa

COMPUTE MOMENTS:

coefficients, c L(m) w(kPa) (kN.m/m)0.111111111 4.6 13.038 30.653786670.071428571 4.6 13.038 19.706005710.041666667 4.6 13.038 11.49517

COMPUTE FOR ρ:

Concrete,fc' =

Steel,fy =

Unit weight of concrete,wc = kN/m3

self weight = wch

assume floor finish = ywc

M = cwL2

= 0.0033816

= 0.0031667

= 0.00316669

0.02129829

main bar reinforcement = 78.5398163

temp bar reinforcement = 50.2654825

COMPUTE THE REQUIRED As AND SPACING OF BAR REINFORCEMENTSinterior support at mid-span exterior support

= 30.653787 19.706006 11.49517 (kN-m/m)

d = h - (20+Ø/2) = = 120 120 120

= 2.3652613 1.5205251 0.886972994 ρ = = 0.0060358 0.0038007 0.002184714 use ρ = 0.0060358 0.0038007 0.003166693

= 724.29727 456.08248 380.0032059 s = 1000Ao/As = 108.43589 172.20529 206.6819835

TEMPERATURE BAR SPACING REQUIREMENT:

fy < 414MPa = 0.002

fy = 414MPa = 0.0018

fy > 414MPa = 0.0018(414)/fy

MINIMUM RHO FOR TEMPERATURE BARS:

= 200/fy

= 0.48309179

= 0.0018

= 216

temp bar spacing = 215

SLAB DETAILS FOR BENDING OF REINFORCEMENTSfixed bar spacing at mid-span, s = 150

= 456.0824839

523.5987756design is ok!!!=)

at exterior support

= 349.0658504

= 380.0032059design is not safe!!!=(

provide extra bars,n = 1

= 427.6056667design is ok!!!=)

at interior support

= 698.1317008

= 724.2972694design is not safe!!!=(

provide extra bars, n = 2

= 855.2113335design is ok!!!=)

1.4/ fy

√¯fc'/ 4fy

use ρmin

Mu =

(mm2)

Ru = Mu/(.9bd2)

As = ρbd (mm2)

ρtemp.

ρtemp.

ρtemp.

ρmin

ρtemp.

As,temp =ρtempbd (mm2)

(mm2)

As(required)

As(actual) = 1000Ao/s(fixed)

Ap = As(actual)2/3

As(required)

Ap = Ap + n(Ao)

Ap = As(actual)4/3

As(required)

Ap = Ap + n(Ao)

ρmax=. 75[ .85 fc ' β1600fy(600+ fy ) ]=

Slab Details Scheme 1: Bending of reinforcements

10 mm Ø bent 2/3 spaced @150 mm

1 -extra top bar is needed 2 -extra top bars are needed

165mm

1 8 10 mm Ø bent 2/3 spaced @-extra top bar is needed mmØ temp bars @ 150 mm

215 mm

2-extra top bars are needed

Slab Details Scheme 2: Cutting of reinforcements

10 mmØ 10 mmØ 10 8 mmØ 190 oc 380 oc 190 oc 215 oc

10 mmØ 8 mmØ 10 mmØ

190 mm spacing 215 mm spacing 190 mm spacing 10mmØ 95

mm spacing

*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program. =)

CARMEL B. SABADO CE-162 PROF. GERONIDES P. ANCOGBSCE-5 2nd Excel Program SEPT. 6, 2009*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program. =)

****====DESIGN OF TWO-WAY SLAB====****

7.7 m

6.2 m

Estimate slab thickness based on code minimum thickness requirement.GIVEN: Trial depth, h:

20.7 MPa = 154.444 mm

414 MPa say 160 mm

23.6bar size = 12Live load = 6.65 kPa

6.2 m

7.7 m

Slab load:DL:

3.776 kPa

0.59 kPa

TOTAL = 4.366 kPaLL: 6.65 kPa

17.417 kPaSlab aspect ratio, m:

= 0.805194805

Positve moments:Case 4: w L

0.039 4.366 6.2 6.545

0.048 6.65 6.2 12.270

27.486

0.016 4.366 7.7 4.142

0.02 6.65 7.7 7.886

17.587

fc' =

fy =

wc = kN/m³mmφ

LA =

LB =

slab: wch =

finish, assume 25mm cement finish 0.025wc =

Ultimate load: wu = 1.4DL + 1.7LL =

coeff, c M=cwL²

Ma, pos DL

Ma, pos LL

Mau, pos = 1.2DL + 1.6LL

Mb, pos DL

Mb, posLL

Mbu, pos = 1.2DL + 1.6LL

h=2 ( A+B )180

m=LALB

Negative moments:

At continuous edge: L

0.071 17.4174 6.2 47.536 kN-m/m

0.029 17.4174 7.7 29.948 kN-m/m

9.162 kN-m/m

5.862 kN-m/m

Design of middle strip in the short direction:h d

Mu = Mmax = = 47.536 Kn-mTrial d = h - (20 + φ/2) = = 134 mm

= 2.941527733

= 0.002747419

= 0.016031805

= 0.007826 ok, use this rho!!!=)

= 113.0973355 mm2

= 1048.632079 mm2

= 107.8522561 mm oc

At mid-span:Mu = 27.486 Kn-mRu = 1.700853684p = 0.004328795 ok, use this rho!!!=)As = 580.0584963 mm2s = 194.9757417 mm oc

At discontinuous end:Mu = 9.162 Kn-mRu = 0.566951228p = 0.001392252 use pmin!!!=(As = 368.1540976 mm2s = 307.2010776 mm oc

Design od middle strip in the long direction:h d

Mu = 29.948 Kn-mdL = h - (20 + 1.5φ) = 122 mm

Ru = 2.235633472p = 0.005795189 ok, use this rho!!!=)Ao = 113.0973355 mm2As = 776.5553675 mm2

coeff, c wu M=cwL²

Ma, neg

Mb,neg

At discontinuous edge: M = 1/3(Mpos)

Ma, neg

Mb,neg

As = ρbd

Ru=M u /φ

bd2

ρmin=[ 1. 4f y,√ f c '4 f y ]min

ρmax=. 75[ . 85 f c ' β1

f y

. 003 Es

. 003 Es+ f y ]ρ= 1ω [1−√1−

2ωRuf y ]=. 85 f c '

f y [1−√1−2Ru

. 85 f c ' ]Ao=

π4Db2

s=1000 AoAs

s = 145.6397576 mm oc

At mid-span:Mu = 17.587 Kn-mRu = 1.312895778p = 0.00329931 ok, use this rho!!!=)As = 402.5158091 mm2s = 280.976133 mm oc

At discontinuous end:Mu = 5.862 Kn-mRu = 0.437631926p = 0.001070566 use pmin!!!=(As = 335.185074 mm2s = 337.4175771 mm oc

Check for shear:Total load on panel, Wt = LALBwu = 831.5067 Kn

= 0.71

= 38.3357 Kn/mShear concrete:

d = 134 mm

= 91.44948 shear is okay!!=)

SLAB DETAILS FOR BENDING OF REINFORCEMENTS

Short Direction:fixed bar spacing at mid-span, s = 150 mm oc

As(actual) = 1000Ao/s(fixed) = 753.9822 >As required at mid-span, ok As(required) = 580.0585

at exterior support Ap = As(actual)2/3 = 502.6548 >As required at exterior support,ok As(required) = 368.1541

at interior support Ap = As(actual)4/3 = 1005.31 <As required,provide extra bars As(required) = 1048.632 provide extra bars, n = 2

Ap = Ap + n(Ao) = 1231.504 >As required at interior support,ok

Long Direction:fixed bar spacing at mid-span, s = 220 mm oc

As(actual) = 1000Ao/s(fixed) = 514.0788 >As required at mid-span, ok As(required) = 402.5158

at exterior support Ap = As(actual)2/3 = 342.7192 >As required at exterior support,ok As(required) = 335.1851

at interior support Ap = As(actual)4/3 = 685.4384 <As required,provide extra bars As(required) = 776.5554 provide extra bars, n = 2 Ap = Ap + n(Ao) = 911.6331 >As required at interior support,ok

Shear per m of long beam, Case 4: CA

v A=CAWT

2 LB

V uc=φvc√ f c '

6bd

colu

mn

stri

p

provide 2 extra top bars

mid

dle

stri

pco

lum

n st

rip

provide 2 extra top bars

column strip middle strip column strip column strip

Slab thickness = 160 mm

Slab Details Scheme 2: Cutting of Reinforcements

colu

mn

stri

pco

lum

n st

rip

column strip middle strip column strip column strip

LB/4

12mmφ @150 oc bent 2/3

LB/2

12mmφ @ 220 oc bent 2/3

LB/4

LB/4

LA/4 LA/2 LA/4 LA/4

12φ top bars @ 245

LB/4

12φ

top

bars

@ 2

45

12φ

bot

bar

s @

175

12φ

top

bars

@ 1

05

12φ bot bars @ 250

LB/2

LB/4

12φ top bars @ 150

LB/4

LA/4 LA/2 LA/4 LA/4

Slab thickness = 160 mm

*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program. =)

>As required at exterior support,ok

>As required at interior support,ok

>As required at exterior support,ok

>As required at interior support,ok