simply supported slab

7/27/2019 Simply Supported Slab

1/4

Project

Office Block

Job Ref.

Example 5.24

Section

Two-W ay spanning Slab

Sheet no./rev.

1

Calc. by

R.M.I

Date

11/4/2013

Chk'd by

R.M.I

Date

11/5/2013

A pp 'd by

I.R

Date

11/10/2013

RC SLAB DESIGN (BS8110:PART1:1997)

TEDD S calculation version 1.0.0

T W O W AY SPANNING SL AB DEF I N IT IO N S I M PL Y SUPPO RTED

; Overal l depth of slab ;h = 275 m m

Outer sagging steel

; Cover to outer tension reinforceme nt resist ing sagg ing ;c sa g= 20 m m

; Trial bar diameter;D tryx= 12 m m

Depth to ou ter tension steel (resist ing sagging)

d x= h - c sa g- D tryx/2 = 249 m m

Inner sagging steel

; Trial bar diameter;D tryy= 12 m m

Depth to inner tension steel (resist ing sagg ing)

d y= h - c sa g- D tryx- D tryy/2 = 237 m m

Materials

; Charac teristic strength of reinforceme nt ;fy= 460 N /mm 2

; Charac teristic strength of concrete ;fcu = 40 N /mm 2

N o m i n a l 1 m w i d t h

N o m i n a l 1 m w i d t h

d x

d y

T w o - w a y s p a n n in g s l a b

h

h

A s x

A s x

A s y

A s y

( s im p l e )

L o n g e r S p a n

S h o r te r S p a n

M AXI M UM DESI G N M O M ENT S

; Length of shorter side of slab ; lx= 5.000m

; Length of longer side of slab ; ly= 7.500m

; Design u lt ima te load per unit area ; n s= 15.0 kN/m 2

Moment coeff ic ients

sx = (l y/ lx)4/ (8 (1+(ly/ lx)4)) = 0.104

sy = (l y/ lx)2/ (8 (1+(ly/ lx)4)) = 0.046

Maximum moments per unit width - simply supported slabs

m sx = sx n s lx2= 39.1 kNm/m


2/4

Project

Office Block

Job Ref.

Example 5.24

Section


Sheet no./rev.

2

Calc. by

R.M.I

Date

11/4/2013

Chk'd by

R.M.I

Date

11/5/2013

A pp 'd by

I.R

Date

11/10/2013

m sy = sy n s lx2= 17.4 kNm/m

CO NCRET E SL AB DESIG N SAG G ING O UT ER L AYER O F ST EEL (CL 3 .5 .4 )

; Des ign sagging mom ent (per m width of s lab) ;m sx = 39.1 kNm/m

; Moment Redistr ibution Factor; bx = 1.0

Area of reinforcement required

;; K x= abs(m sx ) / ( d x2 fcu ) = 0.016

K'x= m in (0.156 , (0.402 ( bx - 0.4)) - (0.18 ( bx - 0.4)2) ) = 0.156

O u t e r c o m p r e s s i o n s t e e l n o t r e q u i r ed t o r e s i s t s a g g i n g

S l a b r e q u i r i n g o u t e r te n s i o n s t e e l o n l y - b a r s ( s a g g i n g )

;; zx= m in (( 0.95 d x),(dx(0.5+0.25-K x/0.9)))) = 237 m m

Neutral axis depth ;x x= (d x- zx) / 0.45 = 28 m m

Are a of te nsi on s te el r equired

;; ; A sx_req= abs(m sx ) / (1/m s fy z x) = 378 m m 2/m

Tension steel

; ;P r o v i d e 1 2 d i a b a r s @ 2 5 0 c e n t r e s ;o u t e r t e n s i o n s t e e l r e s is t i n g s a g g i n g

A sx_prov= A sx = 452 m m 2/m

A r e a o f o u t e r t e n s io n s t e e l p r o v i d e d s u f f i c ie n t t o r e s i s t s a g g i n g

Con crete Slab D esign - Sag ging - Inner layer of steel (cl. 3.5.4)

; Des ign sagging mom ent (per m w idth of s lab);m sy = 17.4 kNm/m

; Moment Redistr ibution Factor; by = 1.0

Area of reinforcement required

;; K y= abs(m sy ) / ( d y2

fcu ) = 0.008K'y= m in (0.156 , (0.402 ( by - 0.4)) - (0.18 ( by - 0.4)2) ) = 0.156

In n e r c o m p r e s s i o n

s t e e l n o t r e q u i r e d t o

r e s i s t s a g g i n g

S l a b r e q u i r i n g i n n e r t en s i o n s t e e l o n l y - b a r s ( s a g g i n g )

;; zy= m in (( 0.95 d y),(d y (0.5+0.25-K y/0.9)))) = 225 m m

Neutral axis depth ;x y= (d y- zy) / 0.45 = 26 m m

Are a of te nsi on s te el r equired

;; ; A sy_req= abs(m sy) / (1/ms fy z y) = 176 m m 2/m

Tension steel

; ;P r o v i d e 1 2 d i a b a r s @ 3 0 0 c e n t r e s ;i n n e r t en s i o n s t e e l r e s i s t i n g s a g g i n g

A sy_prov= A sy = 377 m m 2/m

A r e a o f i n n e r t e n s i o n s t e e l p r o v i d e d s u f f i c i e n t t o r es i s t s a g g i n g

C h e c k m i n a n d m a x a r e as o f s t e e l re s i s t in g s a g g i n g

;Total area of concrete ;A c= h = 275000 m m 2/m

; Min imum % re inforcement;k = 0.13 %

A st_min = k A c= 358 m m 2/m

A st_max= 4 % A c= 11000 m m 2/m


3/4

Project

Office Block

Job Ref.

Example 5.24

Section


Sheet no./rev.

3

Calc. by

R.M.I

Date

11/4/2013

Chk'd by

R.M.I

Date

11/5/2013

A pp 'd by

I.R

Date

11/10/2013

Steel defined:

; Outer steel resist ing sagging ;A sx_prov= 452 m m 2/m

A r e a o f o u t e r s t e e l p r o v i d e d ( s a g g i n g ) OK

; Inner steel resist ing sag ging ;A sy_prov= 377 m m 2/m

A r e a o f i n n e r s t e e l p r o v i d e d ( s a g g i n g ) OK

SHEAR RE SISTANCE OF CONCR ETE SLABS (CL 3 .5 .5 )

Outer tension steel resist ing sagging mom ents

; Depth to tension steel from comp ression face ;d x= 249 m m

; Are a of te nsi on re in fo rc em ent pro vided (p er m wid th of sl ab) ;A sx_prov= 452 m m 2/m

; Design ult imate shear force (per m width of slab) ;V x= 38 kN/m


Applied shear stress

vx= V x/ d x= 0.15 N /mm 2

Check shear stress to clause 3.5.5.2

vallowable= m in ((0.8 N1/ 2/mm) (fcu ) , 5 N/mm 2) = 5.00 N /mm 2

S h e a r s t r e s s - OK

Shear stresses to clause 3.5.5.3

Design shear stress

fcu_ratio = if (fcu > 40 N/mm 2, 40/25 , f cu / (25 N/mm 2)) = 1.600

vcx = 0.79 N/mm 2 min(3,100 A sx_prov/ d x)1/ 3 max(0.67,(400 mm / d x)1/ 4) / 1.25 fcu_ratio 1/ 3

vcx = 0.47 N /mm 2

Applie d sh ear s tress

vx= 0.15 N /mm 2

N o s h e a r r e i n fo r c e m e n t r e q u i r e d

SHEAR RE SISTANCE OF CONCR ETE SLABS (CL 3 .5 .5 )

Inner tension steel resist ing sagging m oments

; Depth to tension steel from comp ression face ;d y= 237 m m

; Are a of te nsi on re in fo rc em ent pro vided (p er m wid th of sl ab) ;A sy_prov= 377 m m 2/m

; Design ult imate shear force (per m width of slab) ;V y= 19 kN/m


Applied shear stress

vy= V y/ d y= 0.08 N /mm 2

Check shear stress to clause 3.5.5.2

vallowable= m in ((0.8 N1/ 2/mm) (fcu ) , 5 N/mm 2) = 5.00 N /mm 2

S h e a r s t r e s s - OK

Shear stresses to clause 3.5.5.3

Design shear stress

fcu_ratio = if (fcu > 40 N/mm 2, 40/25 , f cu / (25 N/mm 2)) = 1.600


4/4

Project

Office Block

Job Ref.

Example 5.24

Section


Sheet no./rev.

4

Calc. by

R.M.I

Date

11/4/2013

Chk'd by

R.M.I

Date

11/5/2013

A pp 'd by

I.R

Date

11/10/2013

vcy = 0.79 N/mm 2 min(3,100 A sy_prov/ d y)1/ 3 m ax(0.67,(400 mm) / d y)1/ 4/ 1.25 fcu_ratio 1/ 3

vcy = 0.46 N /mm 2

Applie d sh ear s tress

vy= 0.08 N /mm 2

N o s h e a r r e i n fo r c e m e n t r e q u i r e d

CONCRE TE SLAB DEFLECTION CHECK (CL 3.5 .7 )

; Slab span length ;l x= 5.000 m

; Design ult imate moment in shorter span per m width ;m sx = 39 kNm/m

; Depth to outer tension steel;d x= 249 m m

Tension steel

; Are a of oute r te nsi on re in fo rc em ent pro vided ;A sx_prov= 452 m m 2/m

; Are a of te nsi on re in fo rc em ent re quired ;A sx_req= 378 m m 2/m

; Moment Redistr ibution Factor; bx = 1.00

Modif ication Factors

;Basic spa n / effective depth rat io (Table 3.9) ;ratio span_depth = 20

The m odif ication factor for spans in excess of 10m (ref. c l 3.4.6.4) has not been included.

;fs= 2 fy A sx_req/ (3 A sx_prov bx ) = 256.3N /mm 2

factortens= m in ( 2 , 0 .55 + ( 477 N /mm 2- fs) / ( 120 ( 0 .9 N/mm 2+ m sx / d x2))) = 1.751

Calculate Maximum Span

This is a simpl i f ied approa ch and further attention should be given w here special circumstance s exist. Refer to clauses

3.4.6.4 and 3.4.6.7.

Maximum span ;lma x= ratio span_depth factortens d x= 8.72 mCheck the actual beam span

Actu al sp an/d epth ra tio ;l x/ d x= 20.08

Span depth l imit;ratio span_depth factortens= 35.02

S p a n / D e p t h r a t io c h e c k s a t i s f i e d

CHECK O F NOMINAL COVER (SAGGING) (BS8110:PT 1, TABLE 3 .4 )

; Slab thickness ;h = 275 m m

; Effective depth to bottom ou ter tension reinforcemen t ;d x= 249.0 m m

; Diameter of tension reinforcement;D x= 12 m m

; Diam eter of l inks ;L diax= 0 m m

Cover to outer tension reinforcement

c tenx= h - d x- D x/ 2 = 20.0 m m

Nom inal cover to l inks steel

c nom x= c tenx- L diax= 20.0 m m

Permissable minimum nominal cover to al l reinforcement (Table 3.4)

; cmi n= 20 m m

C o v e r o v e r s t e e l r es i s t in g s a g g i n g O K

simply supported slab

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