Two Wayslabs Typical

8/18/2019 Two Wayslabs Typical

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By: JURONGCONSULTANTS(INDIA) PVT.LTD Project: Proposed Residentail buildin

Date: Client: Doc. No.:

Chd. Subject: Design o S!"b(S#)

Date: Job No.- Office Dept: h. ! of ! Re".

TWO WAY SLAB DESIGN

Panal size = 5.60 4.20

= 30

= 500

EDGE COND. 1 INTERIOR PANEL

= 1.34

= 125 mm

Dead load of the slab = D*25 = 3.13

Floor finishes = 2.00

Live load = 2.00

. = 7.13

0.0486

0.0372

0.0320

0.0240

he Fa!tored "oments are Moment At 8 tor 10 tor 12 tor

in KNm mm c/c mm c/c mm c/c

= #.16 216 232 363 523

= 7.01 164 306 478 68#

= 6.03 153 328 512 737

= 4.52 114 440 687 ##0

= = 47 mm

= 125 mm

= 101 mm

= #3 mm

$rovide 125 mm h% tor & 200 mm !'! in (horter dire!tion )

tor & 200 mm !'! in loner dire!tion

Check for Deflection :

$er!entae of ension renfor!ement & mid s+an = 0.25

Area of c/s of steel require = 1#0

Area of c/s of steel !ro"ie

,asi! s+an'eff de+th ratio for !ontin-o-s slabs = 26

"odifi!ation fa!tor = 2.00

oeffi!ient for ma/ ratio of s+an'eff de+th = 52

ffe!tive De+th re-ired = 81 101 mm

SA!E

l". S#an $ L y% s&. S#an$ L

x %

f ck N'((2

f y N'((

2

L # /L

$

% ss-me &'

Kn/m2

Kn/m2

Kn/m2

otal load ( Kn/m2

α $%)"e* =

α $%+"e =

α #%)"e =

α #%+"e* =

in m2

M$%)"e = α $%)"e* ,(,l $2,1-.

M$%+"e* = α $%+"e* ,(,l $2,1-.

M#%)"e* = α #%)"e* ,(,l $2,1-.

M#%+"e* = α #%+"e* ,(,l $2,1-.

req

RT% %M u,10 */%0-18,f

c3 ,4**

& !ro"

$ !ro"

# !ro"

f s '0-.8,f

#, N/mm2


2/10


3/10

s

o. R#


4/10


5/10

By: JURONGCONSULTANTS(INDIA) PVT.LTD Project: Proposed R


Chd. Subject: Design o

Date: Job No.- Office Dept: h. ! of !

TWO WAY SLAB DESIGN


= 30

= 500


= 1.08

= 125 mmDead load of the slab = D*25 = 3.13


Live load = 2.00

. = 7.13

0.0360

0.0272

0.0320

0.0240



= 10.40 247 204 318 458

= 7.86 185 272 425 613

= #.25 23# 210 32# 473

= 6.#4 177 284 443 638

= = 71 mm

= 125 mm

= 101 mm

= #3 mm





Area of c/s of steel require = 160




x %

f ck N'((2

f y N'((2

L # /L

$

% ss-me &' Kn/m2

Kn/m2

Kn/m2

otal load ( Kn/m2

α $%)"e* =

α $%+"e =

α #%)"e =

α #%+"e* =

in m2

M$%)"e = α $%)"e* ,(,l $2,1-.

M$%+"e* = α $%+"e* ,(,l $2,1-.

M#%)"e* = α #%)"e* ,(,l $2,1-.

M#%+"e* = α #%+"e* ,(,l $2,1-.

req

RT% %M u,10 */%0-18,f

c3 ,4**

& !ro"

$ !ro"#

!ro"

f s '0-.8,f

#, N/mm2


6/10




SA!E


7/10





TWO WAY SLAB DESIGN


= 30

= 500

EDGE COND. 4 T5O A&6A7ENT E&E &I7ONT-

= 1.20



Live load = 2.00

. = 7.13

0.0600

0.0450

0.0470

0.0350



= 11.31 270 186 2#1 420

= 8.48 200 252 3#3 566

= 8.86 228 220 344 4#5

= 6.60 168 2## 467 672

= = 74 mm

= 125 mm

= 101 mm

= #3 mm

$rovide 125 mm h% tor & 1)5 mm !'! in (horter dire!tion )




Area of c/s of steel require = 202




x %

f ck N'((2

f y N'((2

L # /L

$

% ss-me &' Kn/m2

Kn/m2

Kn/m2

otal load ( Kn/m2

α $%)"e* =

α $%+"e =

α #%)"e =

α #%+"e* =

in m2

M$%)"e = α $%)"e* ,(,l $2,1-.

M$%+"e* = α $%+"e* ,(,l $2,1-.

M#%)"e* = α #%)"e* ,(,l $2,1-.

M#%+"e* = α #%+"e* ,(,l $2,1-.

req

RT% %M u,10 */%0-18,f

c3 ,4**

& !ro"

$ !ro"#

!ro"

f s '0-.8,f

#, N/mm2


8/10



ffe!tive De+th re-ired = #2 101 mm

SA!E


9/10





TWO WAY SLAB DESIGN


= 30

= 500


= 1.63



Live load = 2.00

. = 7.13

0.0566

0.0431

0.0320

0.0240



= #.6# 22# 21# 343 4#3

= 7.37 173 2#1 455 655

= 5.47 13# 362 566 815

= 4.10 103 486 75# 10#4

= = 68 mm

= 125 mm

= 101 mm

= #3 mm





Area of c/s of steel require = 1##




x %

f ck N'((2

f y N'((2

L # /L

$

% ss-me &' Kn/m2

Kn/m2

Kn/m2

otal load ( Kn/m2

α $%)"e* =

α $%+"e =

α #%)"e =

α #%+"e* =

in m2

M$%)"e = α $%)"e* ,(,l $2,1-.

M$%+"e* = α $%+"e* ,(,l $2,1-.

M#%)"e* = α #%)"e* ,(,l $2,1-.

M#%+"e* = α #%+"e* ,(,l $2,1-.

req

RT% %M u,10 */%0-18,f

c3 ,4**

& !ro"

$ !ro"#

!ro"

f s '0-.8,f

#, N/mm2


10/10

"odifi!ation fa!tor = 1.#0

oeffi!ient for ma/ ratio of s+an'eff de+th = 4#


SA!E

Two Wayslabs Typical

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