CE 71 - RC&BM Monograph

7/27/2019 CE 71 - RC&BM Monograph

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CE1401 Design of Reinforced Concrete & Brick Masonry Structures

UNIT IRETAINING WALL

Retaining wall Retains Earth when level difference exists between two surfaces

A) Gravity wall (h


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E) "uttress wall ransverse stem su**ort *rovided on front side+

,) "ridge abutment Additional hori-ontal restraint from bridge dec.+

!tability /verturning and !liding Avoided by *roviding sufficient base width0

Earth *ressure and stability re1uirements2

ressure4 ZCP e=

5here4 6 7 de*th4 e7 8nit weight

#a7

Sin

Sin

+

1

19

Sin

SinCp

+=

1

1

Always4 #* $ #a0 Eg2 :f ; 7 34 #a 7 =>3 and #* 7 30

:n slo*ed bac.fill4

#a7

Cos

CosCosCos

CosCosCos

+

22

22

9

Sin

SinCp

+=11

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=0 Effect of surcharge on level bac.fill2

a7 a=? a@4 where4a=7 #a05s0h 7 #a0e0hs0h B h>@ above heel+a@7 #a0 e0h

@>@ h>3 above heel+

Cote 2 ur*ose of retaining wall is to retain earth and not water0 herefore4 submergedcondition should be avoided by *roviding and maintaining *ro*er drainage facilitiesincluding *rovision of wee* holes+0

@0 Effect of water in the bac.fill2

!tability re1uirements ,/! againstSliding

OvertuningD =0

i) /verturning2

,/!overturning7Mo

Mr9.0 D =0

a) ,or slo*ing bac.fill4 o7 (a0#os )0

!h7 #a0 e 0

"

#!0#os

r 7 5(F w) ? (a0!in )0F

b) ,or level bac.fill with surcharge4 o7 (a=(h>@) ? (a@(h>3)0

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r 7 5(F w) as 7 o+

ii) !liding2 ,riction between base slab and su**orting soil+, 7 H0R where4 R 7 5+

R I$ Resultant soil *ressure at footing baseH I$ #oefficient of static friction 03J !ilt B 0% Rough roc.+

,/!sliding7CosP

F

a

9.0 D =0

5hen ais very high4 shear .ey *roKection can be *rovided below footing base roduces*assive resistance *s4 which is generally neglected4 otherwise+0

!liding is reduced by *roviding shear .ey li.e a *lug4 anchors inside+

2$%&2

1

2

2 hhCP eps =

s.I$ ,lexural reinforcement from stem is extended straight into shear .ey near thetoe0

Cote2 ,or economical design4 soil *ressure resultant(R) must be in line with front face ofwall0

reliminary *ro*ortioning of cantilever retaining wall2

=0 he thic.ness of base slab is h>=@ or &L of the height of wall ? surcharge0@0 he base thic.ness of stem should be greater than the thic.ness of base slab30 he to* thic.ness of stem should not be less than =Jmm00 #lear cover for stem is Jmm and base slab is MJmmJ0 inimum length of base slab is given by

R

a hCL

!

'in

=

where4 NR7 #oefficient de*ending on the *ressure distributionNR 7 0J for rectangular *ressure distribution B 0%M for tra*e-oidal *r0dist0

%0 inimum length of heel slab is given by

!

hC

X a

=

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Cotes2=0 he critical section for moment is at front face of stem0@0 he critical section for shear is at OdP from face of stem030 he stem4 heel and toe slabs are designed as cantilever slabs for the resultant

*ressure0

0 em*erature and shrin.age reinforcement is *rovided as 0=@L of cross sectionalong the transverse direction to the main reinforcement and front face of the stem0

=) 'etermine suitable dimensions of a cantilever retaining wall4 which is re1uired to su**orta ban. of earth 0m high above ground level on the toe side of the wall0 #onsider thebac.fill surface to be inclined at an angle of =J owith the hori-ontal0 Assume good soil forfoundation at a de*th of =0@Jm below ground level with !"# of =%.C>m @0 ,urther assumethe bac.fill to com*rise of granular soil with unit weight of =%.C>m3and an angle of shearingresistance of 3o0 Assume the coefficient of friction between soil and concrete to be 0J0

Given2 h 7 0 ? =0@JmQ 7 =Jo

; 7 3o

e 7 =%.C>m3

1a7 =%.C>m@

H 7 0J

inimum de*th of foundation (Ran.inePs)4

dm7

2

sin1

sin1

+

e

aq

7

2

1

1"

1"0

7 =0==mEarth *ressure coefficient4

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#a7

Cos

CosCosCos

CosCosCos

+

22

22

7 03M3

Sin

SinCp

+=

1

1

7 30

reliminary *ro*ortioning2hic.ness of footing base slab 7 0&h 7 0& x J0@J 7 0@mrovide a base thic.ness of @mm for base slab0

Assume stem thic.ness of Jmm at base of stem ta*ering to =Jmm at to* of wall0

,or economical *ro*ortioning of length OFP4 assume vertical reaction R at the footing base tobe in line with front face of the stem0

!

hC

X a

= 7 %4.02(.(&

).0+ 7 @0m where 0m is assumed as

height above wall+Assuming a triangular base *ressure distribution4F 7 =0J 7 30mreliminary *ro*ortions are shown in figure0

,or the assumed *ro*ortions4 the retaining wall is chec.ed for stability against overturningand sliding0

!tability against overturning2

,orce

:'

,orce (.C) 'istance from heel

(m)

oment

(.Cm)5= =%(=0&J)(J0@J I 0@) 7 =30 0@J =3@03

5@ =%(=0&J)(0Jx0J3%) 7 M0 @tan=Jo70J3%+ 0%=M 0

53 @J(0=J)(J0@JI0@) 7 =&0= =0@J 30&

5 (@JI=%)(0&3)(0Jx03) 7 %0J =0MJ ==0

5J @J(3)(0@) 7 3=0J =0J M0@

a!inQ

@J0

Total W = 232.9 Mw = 230.6

a 7 Active *ressure exerted by retained earth on wall both wall and earth move in samedirection+* 7 assive *ressure exerted by wall on retained earth both move in o**osite direction+

#a I$ same for dry and submerged condition4 since ; for granular soil does not changesignificantly0

,orce due to active *ressure4

a7 #a0 e0hP

@

>@5here4 hP 7 h ? tanQ7 J@J ? @tan=Jo7 JM&%mm

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a7 03M3(=%)(J0M&%)@>@ 7 0.C *er m length of wall+

,/! 7 0 x !tabilising force or moment D =0'estabilising force or moment

herefore4 ,/!(overturning)7

Mo

Mr9.0D =0

/verturning moment4 o7 (a#osQ)hP>3 7 %0J(J0M&%>3) 7 =&%0=.Cm

o find the distance of resultant vertical force from heel4

'istance of resultant vertical force from heel4w7 w>5 7 @30%>@3@0 7 0m

!tabilising moment (about toe)4r7 5(F w) ? a!inQ(F) 7 @3@0(3 0) ? MM0%

7 %&0=.Cm *er m length of wall+

,/!(overturning)7Mo

Mr9.07

1.1*"

%".))1.4"*&9.0 +x7 @0@% $ =0

!oil *ressure at footing base2

Resultant vertical reaction4 R 7 5 7 @[email protected] *er m length of wall+

'istance of R from heel4 FR7 (w? o) > R 7 (@30% ? =&%0=)>@3@0 7 =0M&m

Eccentricity4 e 7 FR F>@ 7 =0M& 3>@ 7 0@&m < F>% I$0Jm+

Sence the resultant lies within the middle third of the base4 which is desirable0

aximum *ressure as base4

qq!NL

e

L

Rq


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0$*.2%()*.01&

9.22"1 2'in >==

= !N

L

e

L

Rq Co tension develo*s+ !afe+0

!tability against sliding2

!liding force4 a#osQ 7 %0J.C

Resisting force4 , 7 HR 7 0J x @3@0 7 ==%0.C :gnoring *assive *ressure on toe side+

,/!(!liding)7CosP

F

a

9.07 40.10*(.1

(.9"

4.11"9.0@ 7 3 x =% x (@0=@ 0J@)>@ 7 M0==.C0

,/! (!liding) 7 4.1)*.1

(.9"

%44.)44.11"&9.0>=

+!A,E+

'esign of toe slab2

Assuming a clear cover of MJmm and =%mm T used4

d 7 @ MJ & 7 33Mmm

Uu7 !Nx 42.9"%).01&2

9.*1112(.1 =

+

Uuis design shear at OdP from face of stem+

u7 =0JV(&=0 x =@>@) ? (==@ I &=0) x =>@ x =@x @>3W

7 M%0& .Cm>m length

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Cominal shear stress4 2

$2*".0)1000

1042.9"N

x

x

"d

#uv ===

8sing @ concrete4

,or a =$ 0@C>mm@4 *t(re1uired) 7 0@L age =M&4 !I=%+

X 7 22

"

2$").0

)1000

104*.)"N

x

x

"d

Mu == age &4 !I=%+

,or *t7 0@L4 Ast7 0@>= x = x 33M 7 %M mm@> m

!*acing 7")4

4$1"1000 2xx7 @&mm

rovide =%mm Y @mm c>c at bottom of toe slab

Fd 7"d

%

"d

s &

4

*).0%.1"&

4= 7=% x M 7 MJ@mm4 beyond face of stem0

!ince length available is =m4 no curtailment is sorted0

'esign of heel slab2

Uu7 !Nx 0".12*%).0((.1&2

".12*(4.*2(.1 =

+

u7 =0JV(&@0J x =0JJ@>@) ? (=@&0% I &@0J) x =>@ x =0JJ@x @>3W

7 @30% .Cm>m length

Cominal shear stress4 2

$*.0)1000

100".12*N

x

x

"d

#uv ===

8sing @ concrete4

,or a =$ 03C>mm@4 *t(re1uired) 7 03L age =M&4 !I=%+

X 7 22

"

2$*.1

)1000

109".20N

x

x

"d

Mu == age &4 !I=%+

,or *t7 0J%JL4 Ast7 0J%J>= x = x 33M 7 =0J mm@> m

!*acing 70(.1904

4$1"1000 2xx7 =J0%=mm

rovide =%mm Y =mm c>c at bottom of toe slab

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Fd7"d

%

"d

s &

4

*).0%.1"&

4= 7=% x M 7 MJ@mm4 beyond face of stem0

!ince length available is =0JJm4 no curtailment is sorted0

'esign of vertical stem2

Seight of cantilever above base 7 J@J @ 7 &3mm

Assume a clear cover of Jmm and =%mmT bar4

dat base 7 J J =%>@ 7 3@mm

u7 =0J(#a0e0h3>%) 7 =0J(=>3)(=% x 0@3>%) 7 [email protected]

X 7 2

"d

Mu7

2

"

92.41000

1024.1(0

x

x7 =C>mm@0

*t7 03L4 Ast7 0@J>= x = x 3@ 7 =@mm@0

!*acing 7 = x @=>=@ 7 =%mm

rovide =%mm Y =%mm c>c in the stem4 extending into the shear .ey u*to MT 7

MJ@mm0

#hec. for shear2 at OdP from base+

Uu(stem) 7 =0J(#a0e06@>@) 7 =0J(=>3 x =% x (0&3 03@)@>@) 7 J30&3.C

$u

v Nx

x

"d

# mm

@for *t7 03L))

Sence4 !A,E0

#urtailment of bars2

#urtailments of bars in stem are done in two stages2

At =>3rdand @>3rdheight of the stem above base0

em*erature and shrin.age reinforcement2

Ast 7 0=@>= x = x J 7 Jmm@

,or : =>3rdheight4 *rovide @>3rdof bar near front face (ex*osed to weather) and =>3rdnear

rear face0

,or :: =>3rdheight4 *rovide =>@ the above0

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,or ::: =>3rdheight4 *rovide =>3rdof : case0

rovide nominal bars of =mm Y 3mm c>c vertically near front face0

'E!:GC /, #/8CER,/R REA:C:CG 5AFF

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reliminary *ro*ortioning of counterfort retaining wall2

=0 hic.ness of heel slab and stem 7 JL of Seight of wall

@0 hic.ness of toe slab buttress not *rovided+ 7 &L of Seight of wall

30 hic.ness of counterfort 7 %L of height of wall

0 :n no case thic.ness of any com*onent be less than 3mm0

J0 !*acing of counterforts 7 =>3rdto =>@ of Seight of wall

%0 Each *anel of stem and heel slab are designed as two way slab with one edge free

(one way continuous slab)0

M0 he toe slab is designed as

a0 #antilever slab when buttress is not *rovided

b0 /ne way continuous slab4 when buttress is *rovided

&0 #ounterfort is a triangular sha*ed structure designed similar to a I"eam as vertical

cantilever with varying de*th (stem acts as flange)0 he main reinforcement is along

the slo*ing side0 !tirru*s are *rovided in the counterfort to secure them firmly with

the stem0 Additional ties are *rovided to securely tie the counterfort to the heel slab0

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=) 'esign a suitable counterfort retaining wall to su**ort a leveled bac.fill of height M0Jm

above ground level on the toe side0 Assume good soil for the foundation at a de*th of =0Jm

below ground level0 he !"# of soil is =M.C>m@with unit weight as =%.C>m30 he angle

of internal friction is [ 7 3o

0 he coefficient of friction between the soil and concrete is 0J0

8se @J concrete and ,e=J steel0

inimum de*th of foundation 7 P

(.11*1.10sin1

0sin1

1"

1)0

sin1

sin122

3 B #*7 3+

Fmin7 =0J x 3 7 0Jm

hic.ness of counterfort 7 %L of 7 0Jm

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!tability conditions2

Earth *ressure calculations2

,orce:'

,orce (.C) 'istance from heel(m)

oment(.Cm)

5= =%(M0J?=0JI0J)(@0J) 7 3 (3 I 0J)>@ 7 =0@J @J5@ @J(0J)( 0J) 7 =%0@J 0J>@ ? @0J 7 @0MJ @@0=&

53 @J(0J)(3) 7 3M0J =0J J%0@J

5 @J(=0J)(0M@) 7 @M =0J>@ ? 3 7 30MJ ==0@JTotal W = 510.75 Mw = 874.69

w 7 &M0%>J=0MJ 7 =0M=3m

,/! (overturning)7 0r>o

5here4 o 7 a0h>3 7 #a0e0h3>% 7 033x=%x3>% 7 %M03J.Cm0

r 7 (F w)05 7 J=0MJ(0J =0M=3) 7 =@30%.Cm0

,/! (overturning)7 =0& $ =0

Sence4 section is safe against overturning0

!liding2

,/! (sliding)7 0(HR)>a#osQ

, 7 HR 7 0J x J=0MJ 7 @JJ03MJ.C

a7 #a0e0h@>@ 7 @=J0M&

"ase *ressure calculation2

qq!Nx

L

e

L

Rq >=+=

+= 2'a+ $9).22%

(.4

).0"1&

(.4

)(.(10"1 where 1a7 =M.C>m@+ 8n!afe+0

0$02).%(.4

).0"1&

(.4

)(.(10"1 2'in >==

= !N

x

L

e

L


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5here4 FR7 ( ? o)>R4 e 7 FR F>@4 where4

FR7 (&M0%&& ? %M03J@)>J=0MJ 7 @0&m

B e 7 FR F>@ 7 @0& (0J>@) 7 0M3 < F>% (0MJm)

!ince the maximum earth *ressure is greater than !"# of soil4 the length of base slab has

to be increased *referably along the toe side0 :ncrease the toe slab by 0Jm in length0

]5 7 J=0MJ ? 0J x @J x 0M@ 7 J=0MJ.C

Additional load due to increase in toe slab by 0Jm is4

oment 7 0J>@ ? 0J 7 0MJm

] 7 &M0% ? @0MJ 7 =M03&.Cm0

FR7 (o ? ) > R 7 (=M03& ? %M03J@)>J=0MJ 7 30==m

e 7 FR F>@ 7 30== (J>@) 7 0J==m < F>% (0&33m)

qq!Nx

L

e

L

Rq ==

= !Nx

L

e

L


,/! (!liding)7 0(HR)>a7 0(0J x J=0MJ)>@=J0M& 7 =0& < =00

Sence the section is not safe against sliding0 !hear .ey is *rovided to resist sliding0

Assume shear .ey of si-e 3 x 3mm0

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*s7 #*0e0(h@@ h=@)>@ 7 =%0J.C>m

,/! (sliding)7 0(HR ? *s)>a7 =0MM $ =0 where4 h= 7 =0@m4 h@ 7 =0@ ? 03 ? =03 7 @0&&m+

Sence4 section is safe in sliding with shear .ey 3 x 3mm0

'esign of oe !lab2

Effective cover 7 MJ ? @>@ 7 &Jmm

oe slab is designed similar to cantilever slab with maximum moment at front face of the

stem and maximum shear atOdP from front face of stem0

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d 7 M@ &J 7 %3Jmm0

7 &03&x@@>@ ? ^ x @ x 0 x @>3 x @ 7 =%0M% ? =@@0M% 7 @@M03J.Cm0

!, at 0%3Jm4 7 0>@ x 0%3J 7 =J0%%.C

Area of tra*e-oid 7 ^0h0(a ? b) 7 ^(@ 0%3J)(=303@ ? J0&) 7 =J0.C

,actored !, 7 @3=0%%.C9 ,actored oment 7 [email protected]

X 7 u>bd@Ast7 =JJ=0@Jmm

@!*acing 7 =ast>Ast=%mm Y=@Jmmc>c0

ransverse reinforcement2 7 0=@L of c>s7 0=@>= x = x M@ 7 &%mm@

rovide =mm Y=mm c>c0

'esign of heel slab2

he heel slab is designed as an one way continuous slab with moment wl @>=@ at the su**ort

and wl@>=% at the mids*an0 he maximum shear at the su**ort is w(l>@ d)0

he maximum *ressure at the heel slab is considered for the design0

oment at the su**ort4 su* 7 wl@>=@ 7 =%0@ x @0J@>=@ 7 JJ0%&&.Cm0

oment at the mids*an4 mid7 wl@>=% 7 =0M%.Cm

he maximum *ressure acting on the heel slab is ta.en as OwP for which the A stre1uired at

mids*an and su**ort are found0

,actored su*7 &30J3.Cm Ast 7 JM0Mmm@

,actored mid7 %@0%.Cm Ast 7 @J0mm@

8sing =%mm [ bar4 !*acing 7 =ast > Ast 7 ==0@mm rovide =%mm Y ==mm c>c

At mids*an4 s*acing 7 =J%0M@mm rovide =%mm Y =Jmm c>c

ransverse reinforcement 7 0=@L of c>s 7 0=@>= x = x J 7 %mm@

,or &mm bar4 !*acing 7 &30MMJmm rovide &mm Y&mm c>c0

#hec. for shear2

aximum shear 7 w (l>@ d) 7 =M (@0J>@ 0=J) 7 &03J.C

,actored shear force 7 =30=MJ.C

Zv 7 033C>mm@4 Zc 7 0@C>mm@4 Zcmax 7 30=C>mm@

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'e*th has to be increased0

'esign of stem2

he stem is also designed as one way continuous slab with su**ort moment wl @>=@ and

mids*an moment wl@

>=%0 ,or the negative moment at the su**ort4 reinforcement is

*rovided at the rear side and for *ositive moment at mids*an4 reinforcement is *rovided at

front face of the stem0

he maximum moment varies from a base intensity of Xa0e0h7=>3x=%x(0J)7 J033.C>m

su*7 wl@>=@ 7 =0J x J033 x 30J@>=@ 7 M=.Cm

mid7 wl@>=% 7 =0J x J033 x 30J@>=% 7 J30@%.Cm

Effective de*th 7 J (J ? @>@) 7 mm

Astat su**ort 7 =J&mm@4 ,or =%mm [4 !*acing 7 =mm0 rovide =%mm Y =mm c>c

Astat mids*an 7M=&mm@4 ,or =%mm [4 !*acing 7 @&mm0 rovide =%mmY@&mm c>c

ax0 !, 7 w (l>@ d) 7 %[email protected] ,actored !, 7 0.C

ransverse reinforcement 7 0=@L of c>s &mm Y &mmc>c

Zv7 0=&&C>mm@4 Zc7 0%JC>mm@4 Zcmax7 30=C>mm@ !afe in !hear0

'esign of #ounterfort2

he counterfort is designed as a cantilever beam whose de*th is e1ual to the length of the

heel slab at the base and reduces to the thic.ness of the stem at the to*0 aximum

moment at the base of counterfort4 max7 Xa0e0h3>% x Fe

5here4 Fe

c>c distance from counterfort

max7 [email protected] ,actored max 7 @&&0MJ.Cm

Ast7 @MJJ0Jmm@4 Assume @Jmm[ bar4 Co0 of bars re1uired 7 @MJJ0J>=0J 7 J0%= _ %

he main reinforcement is *rovided along the slanting face of the counterfort0

#urtailment of reinforcement2

Cot all the % bars need to be ta.en to the free end0 hree bars are ta.en straight to the

entire s*an of the beam0 /ne bar is cut at a distance of4

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2

2

(.*

11 h

n

n=

4 where n is the total number of bars and h= is the distance from to*0

5hen n 7 %4 h= 7 M0MJm from bottom+

he second *art is cut at a distance of4

2

2

(.*

22 h

n

n=

4 h@ 7 %0m from bottom+

he third *art is cut at a distance of4

2

2

(.*

h

n

n=

4 h3 7 %0=m from bottom+

Uertical ties and hori-ontal ties are *rovided to connect the counterfort with the vertical

stem and the heel slab0

'esign of hori-ontal ties2

#losed stirru*s are *rovided to the vertical stem and the counterfort0 #onsidering =m stri*4

the tension resisted by reinforcement is given by lateral *ressure on the wall multi*lied by

contributing area0

7 #a0e0h x h4 where4 Ast7 &%T*).0

7 =>3 x =% x ( 0J) x 30J 7 =%0&.C

,actored force4 7 =0J x =%0&.C

Ast7 %%%0M@mm@0 ,or =mm [4 !*acing 7 ==mm0

rovide =mmY==mm c>c closed stirru*s as hori-ontal ties0

'esign of vertical ties2

he vertical stirru* connects the counterfort and the heel slab0 #onsidering =m stri*4 the

tensile force is the *roduct of the average downward *ressure and the s*acing between the

counterforts0 7 Avg(30J% B =M) x Fe 7 @%%0 .C

,actored 7 30M .C

Ast 7 ==M0=Jmm

@

0 ,or =mm [4 !*acing 7 M03mm0

rovide =mm Y Mmm c>c0

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UNIT II

DESIGN ! WATER TAN"IS3370 #$a%t I & I'(

#ylindrical Rectangular

,urther4 water tan.s are classified based on their *ositions as4

=0 Resting on ground

a0 ,lexible or free base

b0 Singed

c0 ,ixed

@0 Elevated or overhead

30 8nderground

#om*onents of water tan.2

=0 !ide 5alls Rectangular or cylindrical+

@0 "ase slab

30 #over slab or dome

4. !taging /verhead+ #olumns4 "eams4 "racings

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y*es of `oints2

i) ,lexible base

ii) Singed base

iii) ,ixed base

5ater *ressure distribution2

=0 /n side walls4 it acts linearly varying from at to* to OwhP at bottiom0

@0 /n base slab4 uniform *ressure acts with intensity OwhP

ermissible stresses2he method of design ado*ted for design of water tan. is wor.ing stress method0

he *ermissible stress values in concrete and steel are given in ables @= and @@ of :!J%I

@4 as follows2

,or @ cbc7 M C>mm@4 t7 J C>mm

@

,or ,e@J steel4 st 7 =3 = C>mm@

,or ,e=J steel4 st 7 @3 @ C>mm@

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,or li1uid retaining structures4 the stress values are further reduced and given in :!33M

art :: Reinforced concrete structures+0

#oncrete Grade 'irect ension "ending ension

=J =0= =0J@ =0@ =0M

@J =03 =0&

3 =0J @0

3J =0% @0@

=0M @0

As *er clause " @0=0=4 the tensile stress is given by4

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7st$ ''

Ft

+ where4 m modular ratio 7 @& > 3 cbc7 Es>Ec

cbc is the *ermissible com*ressive stress0

ermissible stress in steel is given in :!33M art :: as4

ensile stress2 ,e@J

==J C>mm

@

4 ,e=J

=J C>mm

@

"ending stress2

< @@Jmm $ @@Jmm

/n ,ace /n face away /n ,ace /n face away

,e@J ==J ==J ==J =@J

,e=J =J =J = =MJ

Reinforcement re1uirements2 As *er :!33M+

=0 inimum Astis 03L for =mm section and 0@L for Jmm section0

@0 :f thic.ness exceeds @mm4 the reinforcement is *rovided in double layers0

A minimum cover of @Jmm is *rovided along the li1uid retained face and the

cover is increased by =@mm (3Mmm) if the wall is subKected to aggressive soil

or li1uid0

S// EC!:/C AC' "EC':CG /EC! :C #F:C'R:#AF ACX 5AFF!

#F:C'R:#AF ACX 5:S ,FE:"FE "A!E2

_________________________________________________________________________________________________________________________

KLNCIT 2(


26/117


he wall is designed for hoo* stress4 for which circumferential hori-ontal steel is *rovided0

inimum reinforcement is *rovided along the vertical direction

=) 'esign a circular water tan. with flexible base for a ca*acity of la.h litres with the tan.

having a de*th of m4 including a free base of @mm0 8se @ concrete and ,e=J steel0

Area of water tan. 7 Uolume > Seight7 > 7 =m@ As Uol0 7 Fa.h Fitres 7 m3+

0d@> 7 = m@d 7 ==0@&m

rovide a diameter of ==0Jm

he height of water to be retained 7 30&m

he wall is subKected to hoo* tension acting along the circumferential direction0 he hoo*

tension *er metre height is given by4

Soo* tension 7 2

.. (h7 (0&= x 30& x ==0J)>@ 7 @=03 .C

ermissible stress in tension as *er :!33M for ,e=J steel is =J C>mm@0

Astre1uired 7 @=03 x =3> =J 7 =@&0 mm@

!*acing 7 =0ast > Ast4 =%mm Y =mm c>c

hic.ness of tan. is ado*ted based on the tensile stress concrete can ta.e0

ct7st$ ''

Ft

+

Ac 7 =t4 , 7 @=03.C4 m 7 @& > 3 cbc7 @& > (3 x M) 7 =3033 \ =3

ct7 =0@ 7 %9.142*.1&1000

4.214

xt+

t 7 =%mm

inimum thic.ness as *er em*irical formula is4

tmin7 (3h ? J)mm where h m7 3 x 30& ? J 7 =%mm

rovide a thic.ness of =Mmm0

inimum reinforcement is *rovided as vertical steel0

inimum Ast is 03L for =mm section and 0@L for Jmm section0

herefore4 for =Mmm thic.ness4 Ast re1uired is 0@&L of c>s

Ast7 (0@&>=) x = x =M 7 M%mm@

rovide &mm Y =mm c>c

#urtailment of reinforcement2

_________________________________________________________________________________________________________________________

KLNCIT 2"


27/117


At @m height4

Soo* tension 72

.. (h7 (0&= x =0& x ==0J) > @ 7 ==0J33J .C

Ast 7 ==0J3J x =3 > =J 7 %M%0& mm@

rovide =mm Y ==mm c>c (or) =%mm Y @mm c>c

'esign of base slab2

!ince the base slab rests directly on the ground4 a nominal thic.ness of =Jmm is *rovided

and a minimum reinforcement of 03L of c>s is *rovided along both ways and along both

the faces0

03L of c>s 03>= x =J x = 7 J mm@4 Re1uired &mm Y== mm c>c

rovide &mm Y@@ mm c>c on both faces0

"elow the base slab4 a layer of lean concrete mix @ is *rovided for MJmm thic.ness with

a layer of tar felt0

'esign of dome2 I Roof covering for cylindrical water tan.s

A dome acts as a roof covering for cylindrical water tan.s0 he dome slab is cast between

the ring beams *rovided at the to* edge of the side walls of the water tan.0 he dome is

designed for meridonial thrust and hoo* force+0

_________________________________________________________________________________________________________________________

KLNCIT 2)


28/117


@) :n the above *roblem4 design a s*herical dome having a central rise of one fifth the

diameter0

Seight 7 =>J x ==0J 7 @03m

Radius of curvature of the dome R+

R@7 (R @03)@? J0MJ@

R 7 &033m

#osQ 7 %0 > &033 7 0M@

he dome is subKected to meridonial thrust and hoo* force4 for which the *ermissible stress

should be within *ermissible com*ressive strength of concrete0

c 7 J C>mm

@

Assume thic.ness of dome as =mm

eridonial thrust4 7Cos

)R

+14 where OwP is the loading of dome slab4 self weight of slab

and any live load acting on it0

!elf weight of slab 7 0= x @J 7 @0J .C>m@

Five load 7 @.C>m@

5 7 0J .C>m@

7)4.01

.*(.4

+

x7 @=0%.C

eridonial thrust 7 > (c>s area) 7 @=0% x =3> (= x =) 7 0@= C>mm@< J C>mm@

he hoo* stress develo*ed in dome slab is given by4

Soo* stress 7

+

CosCos

t

)R

1

17

+

)4.01

1)4.0

100

.*(.4 x7 0% C>mm@< J C>mm@

he *rovided =mm section is sufficient0 inimum reinforcement of 03L of c>s is

*rovided both ways0

03L of c>s 7 03>= x = x = 7 3mm@ &mm Y =%mm c>c

_________________________________________________________________________________________________________________________

KLNCIT 2*


29/117


'esign of ring beam2

he hori-ontal com*onent of meridonial thrust acts on the ring beam0 he hori-ontal thrust4

Ast 7 SF #om*> s where4 SF #om*7 0#osQ x '>@ 7 @=0% x =3x 0M x J0MJ 7 =03.C

Ast7 =03 x =3 > =J 7 %& mm@

rovide (@ =%mm[ ? @ =@mm[) (%@%mm@)

!i-e of ring beam is obtained based on tensile stress relation4

ct7st$

t

''

F

%1& +

Sere ring beam is not subKected to water load0 !o4 *ermissible tensile stress for @

concrete is @0& C>mm@(Annex "I30==) of :!J%I@0

ct7 @0& C>mm@

@0& 7"0*.1 x'

F

$

t

+ Ast7 %& mm@

, 7 #osQ x '>@ 7 @=0% x 0M x J0MJ 7 =03= x =3C

Ac7 @J3=mm@0

rovide =J x =J mm si-e of ring beam with (@ =%mm[ ? @ =@mm[) bars0 :f any tensile

stress is develo*ed4 the steel will ta.e the tensile stress0

_________________________________________________________________________________________________________________________

KLNCIT 29


30/117


'E!:GC /, #F:C'R:#AF 5AER ACX2 5ith fixed base+

,or cylindrical tan.s fixed at the base4 bending moment and hoo* tension are develo*ed4

whose values are based on nonIdimensional *arameter4(t

*20 he vertical reinforcement is

*rovided for the bending moment develo*ed and the transverse reinforcement is *rovided

for hoo* tension develo*ed0

he coefficients of the nonIdimensional *arameter(t

*2are given in ables (S) and =

(")0

aximum hoo* tension 7 #oefficient x 50S0'>@

"ending moment 7 #oefficient x w0S3

5here4 R Radius of tan.4 ' 'iameter of tan.

3) :n the above *roblem4 design a water tan. for fixed base condition0 ermissible stresses

are4 for ,e=J4 st7 =JC>mm@4 as *er :!33M art :: and for @4 ct7 =0@ C>mm

@0

S 7 m4 ' 7 ==0Jm

t 7 3h ? J 7 =Mmm

(t

*2

7 1).0(.11

42

x 7 &0=&

,rom able of :!33M art :U4 the coefficient for maximum hoo* tension is ta.en and from

able = of :!33M art :U4 the coefficient of bending moment is ta.en0

(t

*2&4 Soo* ension #oeff0 0JMJ Y 0%S

(t

*2=4 Soo* ension #oeff0 0%& Y 0%S

(t

*2&0=&4 Soo* ension #oeff0 0JM& Y 0%S

herefore4 7 #oeff0 wSR 7 0JM& x 0&= x x ==0J>@ 7 =30= .C

(t

*2&4 " #oeff0 I0=%

(t

*2=4 " #oeff0 I0=@@

(t

*2

&0=&4 " #oeff00=3&

ax0 " 7 I0=3& x 0&= x 37 I03= .Cm

he vertical reinforcement is *rovided for the above moment and transverse steel is

*rovided for hoo* tension0

_________________________________________________________________________________________________________________________

KLNCIT 0


31/117


,or hoo* tension as4

Ast7 S>=J 7 =30= x =3> =J 7 &%033 mm30

rovide =@mm Y =3mm c>c along the transverse direction0

,or bending moment4

Ast7 d+

M

st ..

5here4 for @4 K 7 0&4 R 7 =0=%4 cbc7 M C>mm@

Ast7129*4.01(0

0.9

xx7 JJJ mm@

d 7 ' [S [">@ I ##7 =M =@ &>@ @J 7 =@mm

d 7"R

M

.7

10001".1

102.9 "

x

x7 &&0@3mm < d =@mm+

in0 de*th re1uired 7 &&0@3mm

03L of c>s (vertical steel) 7 0@&>= x = x =M 7 M% mm

@

rovide &mm Y =mm c>c

he maximum of Ast for " at AstVmaximumWis *rovided as vertical reinforcement JJJmm@+0

rovide &mm Y mm c>c

rovide =mm Y =3mm c>c as vertical reinforcement0 :f the reinforcement is *rovided as

double layers on both the faces4 the s*acing is doubled0

rovide =mm Y @%mm c>c as vertical reinforcement along the two faces with transverse

hoo. reinforcement at =@mm Y @%mm c>c along both the faces0


rovide a nominal thic.ness of =Jmm for base slab with 03L distribution steel &mm Y

@mm c>c4 along both faces4 both ways+0

Also *rovide a haunch of si-e =%mm at the Kunction of the wall and the base slab0

_________________________________________________________________________________________________________________________

KLNCIT 1

,aunc#1(0''

*'' - 200'' c$c

12'' - 2"0'' c$c

10mm @ 260mmc$c


32/117


'E!:GC /, RE#ACG8FAR 5AER ACX

F

F

""

F>" $ @F>" < @

5hen F > " < @4 cantilever moment generates at the base and maximum bending action

ta.es *lace along the continuous edges of the side walls in both short and long direction0

5hen F > " $ @4 only cantilever action ta.es *lace in long wall0 5hereas4 a short wall is

designed for hori-ontal bending and cantilever action0

#ylindricalSoo* ension ransverse reinforcement +

5hen F>" < @4

i) #antilever moment at base 7 w0S0h@> %

where4 h 7 =m or S>4 whichever greater

S

h

ii) ,or hori-ontal bending of the walls4 the maximum moment is found from4

a) ,ixed end moments4

F@>=@ B "@>=@

b) ositive moment at mids*an4

f7 F@>& B f7 "

@>&

5here4 f is the final moment at the su**orts (Kunction between long wall and short

wall)0 he maximum value is ta.en for design0 Sere4

water *ressure given as * 7 w(S h)

,or the maximum moment4 the area of steel along the longer direction and shorter direction

are found from the relation4

AstF7st

L

st

L P

d+

xPM

+

..9 Ast"7

st

,

st

, P

d+

xPM

+

..

_________________________________________________________________________________________________________________________

KLNCIT 2


33/117


5here4 FB "are tension in long and short walls

* ressure exerted by water

ension in the walls is given by4 F7 * x ">@ B "7 *

=0 'esign a rectangular tan. of si-e m x %m with height 3m0 he tan. rests on firmground0 8se @ concrete and ,e=J steel0 a.e design constants K 7 0&J3 B R 7 =03@0

ressure exerted by water4 * 7 w (S h)

5here4 h 7 =m or S> =m or 0MJm 7 =m greater+

7 0&= (3 =) 7 =0%@ .C>m@

o find the final moment at the Kunction of long wall and short wall based on the fixed end

moment and distribution factor4 the moment distribution is done0

`oint A2

'0,0 72211

11

$$

$

LILI

LI

+0

he stiffness along the long wall and short wall are the same ( 21 II = )4 since uniform

thic.ness of wall is ado*ted along long wall and short wall0

A" A'

'0,02211

11

$$

$

LILI

LI

+ 02211

22

$$

$

LILI

LI

+ 0

4$1"$1

"$1

+0

4$1"$1

4$1

+0

!hort wall is stiffer than the long wall0

,A"7 *F@> =@ 7 =0%@ x %@> =@ 7 J&0&% .Cm (3*)

,A'7 *"@> =@ 7 =0%@ x @> =@ 7 @%0=% .Cm (=033*)

`oint A4

A" A'

,E 3* I=033*

', 0 0%

" (I=0%M x 0) (I=0%M x 0%)

7 I0%M* 7 I=0@*

#/ I I

f ? @033* I @033*

f 7 @033* 7 @033 x =0%@ 7 [email protected] at mids*an4

_________________________________________________________________________________________________________________________

KLNCIT


34/117


Fong wall2

&MpL

*

2

7 @0JM .Cm4

!hort wall2

&Mp,

*

2

7 I%0& .Cm4

he reinforcement is *rovided for maximum moment generated0

herefore4 maximum moment generated in the water tan. is

[email protected]

AstF7st

L

st

L P

d+

xPM

+

..

9 Ast"7st

,

st

, P

d+

xPM

+

..

F7 * x " > @ 7 =0%@ x >@ 7 30@ .C

"7 * x F > @ 7 =0%@ x %>@ 7 J&0&% .C

dre17"R

M

.7

10002.1

10)2.4( "

x

x7 =&%mm

rovide d 7 =mm4 ' 7 @@mm

rovide effective de*th of =mm and effective cover of

3mm

x 7 ' >@ Eff0 #over

7 @@ > @ 37 &mm

Area of steel re1uired for ension in the wall

along the longer side and shorter side4

AstF71(0

1024.9

190*(.01(0

%10*024.9&10)2.4(" x

xx

xxx+

7 @=3mm@

Ast"71(0

10*".(*

1*0*(.01(0

%10*0*".(*&10)2.4(" x

xx

xxx+

7 @Mmm@

rovide @mm [ bar4 s*acing re1uired along

hori-ontal direction is =Jmm

rovide @mm Y 3mm c>c along both faces in

the hori-ontal direction4 along short wall and long

wall0

_________________________________________________________________________________________________________________________

KLNCIT 4

=J


35/117


he vertical cantilever moment for a height of OhP is 7 w0S0h@> % F > " < @+

7 0&= x 3 x =@> % 7 0J .Cm

Ast7d+

M

st ..7 @=0M%mm@

in0 Ast 7 03L of c>s

7 03>= x = x @@

7 %% mm@

!*acing is *rovided for the maximum of the above two %%mm@+

Re1uired4 =mm Y ==mm c>c

rovide4 =mm Y @@mm c>c as vertical reinforcement along both the faces0 ,or the base

slab4 *rovide a nominal thic.ness of =Jmm and minimum Astof 03L of c>s0

Ast7 03>= x = x =J 7 J mm@

!*acing of =mm bars re1uired 7 =Mmm

rovide =mm Y 3 mm c>c along both faces4 both ways0

rovide MJmm lean mix with a layer of tar felt which acts as a water bar4 *rovided between

the tan. and lean mix concrete0

@) 'esign a water tan. of si-e m x m with height 3m0 8se @ concrete and ,e=J

steel0 he design constants are K 7 0&J3 and R 7 =03@0

!ince F>" $ @4 the tan. behaves such that the long wall acts as a cantilever member with

moment w0S3>% and short wall is subKected to both cantilever moment and hori-ontal

bending moment0

Fong wall2

#antilever moment at base 7 w0S3>%

5here4 h 7 =m or 3>m 7 =m 7 0&= x 33> % 7 0=J .Cm

!hort wall2

#antilever moment 7 w0S0h@>=@ 7 0&= x 3 x =@>@ 7 =0M=J .Cm

Sori-ontal bending moment 7 *"@>=% 7 =0%@ x @> =% 7 =0%@.Cm

(5here4 * 7 w (S h) 7 0&= (3 =) 7 =0%@.C>m)

aximum of the three moments 7 0=J.Cm

AstF7st

L

st

L P

d+

xPM

+

..

4 where4 F7 * x ">@ 7 =0%@ x >@ 7 30@ .C and "7 * 7 =0%@ .C

_________________________________________________________________________________________________________________________

KLNCIT (


36/117


dre17"R

M

.7

10002.1

1014(.44 "

x

x7 =&@0&Mmm

rovide d 7 =mm4 ' 7 @@mm

rovide effective de*th of =mm and effective cover of

3mm0x 7 ' >@ Eff0 #over

7 @@ > @ 3 7 &mm

Area of steel re1uired for ension in the wall along the longer

side4

AstF71(0

1024.9

190*(.01(0

%10*024.9&1014(.44 " x

xx

xxx+

7

=&03Jmm@

rovide @mm [ bar4 s*acing re1uired along hori-ontal direction is =%mm0 rovide @mm

Y 3mm c>c along both faces in the hori-ontal direction4 along short wall and long wall0

Along the shorter side4

Along vertical direction4

Ast"71(0

10"2.19

1*0*(.01(0

%10*0"2.19&10)2.14" x

xx

xxx+

7 %0Jmm@

Re1uired I =mm Y @@mm c>c on both faces0

Along hori-ontal direction4

Ast"71(0

10"2.19

1*0*(.01(0

%10*0"2.19&10"2.19" x

xx

xxx+

7 &M0Jmm@

Re1uired I =mm Y=%mm c>c on both faces0

'istribution steel2

=mm 03L

Jmm 0@L

@@mm 0@%L

in Ast7 0@%L of c>s 7 0@% > = x = x @@ 7 JM@mm@

in0 Re1uired I =mm Y @%mm c>c4 which is less than the above two values0

_________________________________________________________________________________________________________________________

KLNCIT "


37/117


"ase slab2

!ince base slab is resting directly on the ground4 nominal thic.ness of =Jmm is *rovided

and minimum reinforcement of 03L of crossIsection is *rovided both ways along both

faces0

Astre1uired 7 03L of c>s 7 %Jmm@

0rovide &mm Y @mm c>c0

"elow the base slab4 a layer of lean concrete mix is *rovided with MJmm thic. tar felt layer

between them0

'E!:GC /, 8C'ERGR/8C' 5AER ACX

'esign of underground water tan. is similar to that of tan.s resting on grounds (for

rectangular water tan.s based on F>" ratio)4 where additional moment if any due to the

earth *ressure on the side walls need to be considered0 :f the soil is submerged4 *ressure

exerted by water is also considered0 hus the side walls are chec.ed for the two critical

conditions4

i) Co earth *ressure with *ressure from water inside4 * 7 w(S h)

ii) Earth *ressure exerted on wall under tan. em*ty condition4 * 7S>30(Iw)?wS

he tan. has to be chec.ed for u*lift water *ressure for which frictional resistance should

be sufficient0

_________________________________________________________________________________________________________________________

KLNCIT )

@@


38/117


'esign an underground water tan. of si-e =@m x Jm with height m0 he density of soil is

=%.C>m3and coefficient of friction between soil and concrete is 0=J0 he soil is saturated0

Sere4 F > " 7 =@ > J 7 @0 $ @

he tan. walls are chec.ed for two critical conditions4

i) Co earth *ressure with *ressure from water inside4 * 7 w(S h)7 = (I=) 7 3.C>m@

ii) Earth *ressure exerted on wall under tan. em*ty condition4 * 7S>30(Iw)?wS

7 &.C>m@

5here4 h 7 =m or (S> 7 =m)

herefore4 the maximum *ressure is used in finding out bending moment on the wall0

Fong wall4

#antilever moment 7"

*- 7 = x 3>% 7 =%0%M.Cm

!hort wall4

#antilever moment 72

2*h- 7 @.Cm

Sori-ontal bending moment 71"

2p,7

1"

(02x

7 %0&MJ .Cm B

7

1"

(4* 2x7 MJ .Cm

Fong wall2

AstF7st

L

st

L P

d+

xPM

+

..

4 where4 F7 * x ">@ 7 & x @0J 7 =@ .C

"7 * 7 & .C

dre17"R

M

.7

10002.1

10").10" "

x

x7 @&0@Mmm

rovide d 7 @mm

rovide4 ' 7 @ ? 3 7 3@mm

x 7 3@ > @ 3 7 =3mm

Ast(UF) 7 @J03J ? & 7 3@J03J mm@

Ast(SF)7 0@3ML of c>s

(,or =mm 03L

,or Jmm 0@L

,or 3@mm 0@3ML)

Ast(SF)7 0@3M x = x 3@ 7 MJ&0 mm@

rovide =mm Y @mm c>c

_________________________________________________________________________________________________________________________

KLNCIT *


39/117


!hort wall2

Ast(UF) 7 3M0& ? 3@ 7 %0& mm@< Astmin

rovide Astmin4 =mm Y @mm c>c

Ast(SF) 7 @=M30 mm@

rovide =%mm Y =&mm c>c

#hec. for u*lift2

he tan. is chec.ed for u*lift *ressure0 5hen the u*lift *ressure exceeds the downward

load due to the self weight of the tan. and the frictional resistance re1uired0 he base slab

is *roKected all around thereby increasing the downward load0 he *ressure of submerged

earth and water at the bottom of the base slab for =m length of the wall is found0 he

frictional resistance of the tan. is found by multi*lying the coefficient of friction between soil

and concrete with the *ressure exerted0

Assume the thic.ness of the base slab as mm and *rovide a *roKection of

3mm all around the water tan.0

'ownward load due to self weight of tan.2

Fong wall 7 @ x =@0% x 03@ x x @J 7 &&03% .C

!hort wall 7 @ x J x 03@ x x @J 7 3@ .C

"ase slab 7 (=@0% ? 0%)0(J0% ? 0%) x 0 x @J 7 &@%0=M%.C

otal 7 =JJ0=3% .C

5eight of earth retained over *roKection4

Fong wall 7 @ x (=@0J ? 0%) x x 03 x =% 7 J&0@ .C

!hort wall 7 @ x J0% x x 03 x =% 7 @=%0JM% .C

7 M@0% .C

otal load 7 2680.132 )N

,rictional force re1uired 7 3%3J0=M @%&0=3@ 7 JJ .C

_________________________________________________________________________________________________________________________

KLNCIT 9


40/117


ressure exerted by water at a de*th of 0m4

* 7 S > 30( w) ? w0S

7

4.4(=% =) ? = x 0 7 J@0& .C>m@

#onsidering =m length of wall4 the force exerted is 7 ^ x S x *

7 ^ x 0 x J@0& 7 ==%0=%.C>m

he frictional resistance *er metre length of wall 7 0, 7 0=J x ==%0=% 7 =M0@ .C>m

,or the entire *erimeter of the wall4 the frictional resistance offered4

7 @ x (=@0% ? J0%) x =M0@ 7 %3M .C < JJ .C

Sence the downward load due to self weight and cantilever *roKection of bse slab is

insufficient against the u*lift force0 he length of *roKection has to be increased0 :ncrease

the base *roKection to 0Mm4 all around0

'ownward load due to self weight of tan.2

Fong wall 7 @ x =@0% x 03@ x x @J 7 &&03% .C

!hort wall 7 @ x J x 03@ x x @J 7 3@ .C

"ase slab 7 (=@0% ? =0)0(J0% ? =0) x 0 x @J 7 &&0=%.C

otal 7 @==%0M& .C

5eight of earth retained over *roKection4

Fong wall 7 @ x (=@0J ? =0) x x 03 x =% 7 =@JM0& .C

!hort wall 7 @ x J0% x x 0M x =% 7 JJ03 .C

7 =M%303@ .C

otal load 7 3880.1 )N

,rictional force re1uired 7 303 3&& 7 %03 .C < %3M .C

(,rictional force offered by walls)

* 7 J@0& .C>m@

0

he water tan. is safe against u*lift0

he base is design as a continuous slab4 su**orted between two short walls4 for the self

weight and weight of side walls0

_________________________________________________________________________________________________________________________

KLNCIT 40


41/117



Foading on the base slab includes the self weight of base slab and weight of side walls0

#onsidering =m stri*4

5eight of base slab 7 M0 x 0 x @J 7 M0 .C>m

5eight of walls 7 @ x x 03@ x @J 7 % .C>motal 7 =30 .C>m

Reaction at the su**ort 7 =30 x M0 > @ 7 M30 .C

"00 at the su**ort 7 w0l@>@ 7 =30 x 0&%@> @ 7 0M .Cm

"00 at centre 7 =30 x 30J@@ > @ M30 (@0%%) 7 I@J0&.Cm sagging moment+

Astre1uired 72

"

2.911*"(*(.01(0

10*.42(

..

xx

x

d+

M

st

==

' 7 mm4 d 7 3J 7 3%Jmm+

rovide @Jmm in two layers both faces+0

rovide @Jmm Y =mm c>c on both faces0

UNIT III

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*IELD LINE T+ER*

ield lines y*ical crac. *atterns generated when ultimate moment is reached

#haracteristics of yield lines are4

i) ield lines are straight

ii) ield lines end at su**orting edges of slabiii) ield lines *asses through intersection of axis of rotation of adKacent slab elements

iv) Axis of rotation lies along lines of su**orts and *asses over columns

Assum*tions2he following are the assum*tions of the yield line analysis of reinforced concrete slabs0

=0 he steel reinforcement is fully yielded along the yield lines at colla*se0 Rotation

following yield is at constant moment0

@0 he slab deforms *lastically at colla*se and is se*arated into segments by the yield

lines0 he individual segments of the slab behave elastically0

30 he elastic deformations are neglected and *lastic deformations are only considered0

he entire deformations4 therefore4 ta.e *lace only along the yield lines0 he individualsegments of the slab remain *lane even in the colla*se condition0

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0 he bending and twisting moments are uniformly distributed along the yield lines0 he

maximum values of the moments de*end on the ca*acities of the section based on the

amount of reinforcement *rovided in the section0

J0 he yield lines are straight lines as they are the lines of intersection between two *lanes0

Rules of yield lines:

The two terms, positive and negative yield lines, are used in the analysis to

designate the yield lines for positive bending moments having tension at the

bottom and negative bending moments having tension at the top of the slab,

respectively.

The following are the guidelines for predicting the yield lines and axes of

rotation:1. Yield lines between two intersecting planes are straight lines.

2. ositive yield line will be at the mid!span of one!way simply supported slabs.

". #egative yield lines will occur at the supports in addition to the positive yield

lines at the mid!span of one!way continuous slabs.

$. Yield lines will occur under point loads and they will be radiating outward

from the point of application of the point loads.

%. Yield line between two slab segments should pass through the point of

intersection of the axes of rotation of the ad&acent slab segments.

'. Yield lines should end at the boundary of the slab or at another yield line.

(. Yield lines represent the axes of rotation.

). *upported edges of the slab will also act as axes of rotation. +owever, the

fixed supports provide constant resistance to rotation having negative yield

lines at the supported edges. n the other hand, axes of rotation at the simply

supported edges will not provide any resistance to rotation of the segment.

-. xis of rotation will pass over any column support, if provided, whose

orientation will depend on other considerations.

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/pper and 0ower ound Theorems

ccording to the general theory of structural plasticity, the collapse load

of a structure lies in between the upper bound and lower bound of the true

collapse load. Therefore, the solution employing the theory of plasticity should

ensure that lower and upper bounds converge to the uniue and correct values

of the collapse load.

The statements of the two theorems applied to slabs are given below:

(A) Lower bound theorem:The lower bound of the true collapse load is that

external load for which a distribution of moments can be found satisfying the

reuirements of euilibrium and boundary conditions so that the moments at

any location do not exceed the yield moment.

(B) Upper bound theorem:The upper bound of the true collapse load is that

external load for which the internal wor3 done by the slab for a small increment

of displacement, assuming that moment at every plastic hinge is eual to the

yield moment and satisfying the boundary conditions, is eual to the external

wor3 done by that external load for the same amount of small increment of

displacement.

Thus, the collapse load satisfying the lower bound theorem is always lower than

or eual to the true collapse load. n the other hand, the collapse load

satisfying the upper bound theorem is always higher than or eual to the true

collapse load.

The yield line analysis is an upper bound method in which the predicted failure

load of a slab for given moment of resistance 4capacity5 may be higher than the

true value. Thus, the solution of the upper bound method 4yield line analysis5

may result into unsafe design if the lowest mechanism could not be chosen.

+owever, it has been observed that the prediction of the most probable true

mechanism in slab is not difficult. Thus, the solution is safe and adeuate in

most of the cases. +owever, it is always desirable to employ a lower boundmethod, which is totally safe from the design point of view.

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ield moments2

m 7 u7 0&M x fyx Ast(d 0@xu)

mN0a0b 7 m0cosN0c0d0 mN 7

"ad$.. m0cosN mN7 m0cos

@N

where4 m yield moment *er m length along the yield line

/ne mesh reinforcement

ii)

mN7 m0cos@N ? Hm0cos@0( N)

7 m0cos@N ? Hm0cos@ ( N)

7 m0cos@N ? Hm0sin@N

,or isotro*ically reinforced slab4

(H 7 =) B mN7 m

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5hen reinforcement is same hori-ontally and vertically4 H 7 = B mN7 m

ethods of yield line analysis2

=0 Uirtual wor. method@0 E1uilibrium method (E1uilibrium of individual elements of slab along yield line)

Uirtual wor. method A**lied load causing virtual dis*lacement is e1ual to internal wor.

done or energy dissi*ated in rotation along the yield lines0

:) :sotro*ically reinforced s1uare slab sim*ly su**orted udl

External wor. done 7 w0

5here4 w load4 virtual dis*lacement

:nternal wor. done 7 0Q 7 m0F0Q

m 8ltimate moment > unit length

F Fength of yield line

otal moment *roduced along all the yield lines

/**osite of isotro*ically reinforced is orthotro*ically reinforced+

#entre *oint is the *lace where the first ultimate moment is reached and the crac.

originates at this *oint as4

F> 2an Q 7 = > (F> 2 ) 7 2 >F

=

(0Q)ac7 (m0F0Q)ac7 m0 2 F0L

227 m

L

22since defl0 :s on two sides+

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5or. done by ObdP is same as OacP

otal internal wor. done 7 0Q 7 &m

,or a virtual dis*lacement of = at centre (i0e) #G of each triangular deflects =>3

50 7 (=>@ x F x F>@ x w) x =>3+ x 7

2-L4 where4 w udl on slab

0Q 7 50 &m 7

2-l

m 724

2-l

::) :sotro*ically reinforced s1uare slab fixed on all edges udl


5here4 w load4 virtual dis*lacement

:nternal wor. done 7 0Q 7 m0F0Q 7 m0 2 F0L

227 m

5or. done by ObdP is same as OacP

otal internal wor. done 7 0Q 7 &m

:nternal wor. done by negative yield line (ab4 bc4 cd4 de) for4

7 = > F>@ 7 @>F4 0Q 7 w x F x @>F+ 7 &motal internal wor. done 7 =%m

External wor. done4

50 7 (=>@ x F x F>@ x w) x =>3+ x 7 wF@> 3

0Q 7 50

=%m 7 wF@> 3 m 7 wF@> &

moment *er metre length along the yield line

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:::) /rthotro*ically reinforced ('iff0 rein0 bothways) rectangular slab sim*ly su**orted udl

,or element A4

x7L

2B y7

x7 mF

(x0 x? y0 y+A7

2

,or element '4

:nternal wor. done4

x7 4 y7 =>(F)4 y7 N0F0Hm

0Q 7 @ x

+

2


m 7 2222

&/24

.

+L-

:U) /rthotro*ically reinforced rectangular slab fixed along long edges I sim*ly su**orted

along short edges udl

m 7

22

tan

24

xuL-4where4 tan

+

24(.1

2

2

H #oefficient of orthotro*hy Ratio between reinforcement

*rovided along shorter direction and longer direction+U) /rthotro*ically reinforced rectangular slab all four edges fixed udl

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m 7

22 tan

4*

%uL-

=0 'esign a s1uare slab fixed along all four edges4 which is of side Jm0 he slab has to

su**ort a service load of .C>m@

0 8se @ concrete and ,e=J steel0As *er :! J%I@4

l>d 7 (0& x 3J) 7 @& J > @& 7 d

d 7 =M&0%mm 7 =&mm

rovide ' 7 @mm

Foading on slab2

!elf weight 7 0@ x @J 7 J .C>m@

Five load 7 .C>m@

,loor finish 7 = .C>m@

otal 7 = .C>m@

,actored load (wu) 7 =0J x = 7 =J .C>m@

"y yield line theory44*

2-L= !N*12(.)=

i'iting 'o'ent Mu3i' 0.1*.fck..d2 0.1* + 20 + 1000 + 1*02 *9.424 + 10"5''

Mu 6 Mu3i'

7 2"d

Mu 0.241

8st 122''2

ro:ide *'' - 00'' c$c

1% ;n t#e ao:e lx7 =

,our edges are discontinuous

Nx7 Ny7 0J%

x7 Nx0w0lx@7 0J% x =J x J@7 @= .Cm

y7 Ny0w0lx@7 0J% x =J x J@7 @= .Cm

Re1uired s*acing of &mm bar is =mm0

rovide &mm Y =mm c>c0

@0 'esign a s1uare slab of si-e Jm4 sim*ly su**orted along its four edges and subKected to

a live load of .C>m@0

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24

2-L=

As *er :! J%I@4

l>d 7 (0& x 3J) 7 @& J > @& 7 d

d 7 =M&0%mm 7 =&mmrovide ' 7 @mm

Foading on slab2


Five load 7 .C>m@


otal 7 = .C>m@

,actored load (wu) 7 =0J x = 7 =J .C>m@

"y yield line theory424

2-L= !N"2(.1(=

i'iting 'o'ent Mu3i' 0.1*.fck..d2 0.1* + 20 + 1000 + 1*02 *9.424 + 10"5''

Mu 6 Mu3i'

8st 24).(9''2

ro:ide *'' - 200'' c$c

30 'esign a rectangular slab of si-e m x %m sim*ly su**orted along all its edges4

subKected to a live load of .C>m@0 he coefficient of orthotro*hy is 0M0 8se @ and

,e=J0

,or four edges sim*ly su**orted condition4

m 7 2222

&/24

.

+

L-

Assume s*an>de*th 7 @& Eff0de*th 7 >@& 7 =@0&%mm

rovide d 7 =Jmm4 ' 7 =MmmFoading on slab2

!elf weight 7 0=M x @J 7 0@J .C>m@

Five load 7 .C>m@

,loor finish 7 0MJ .C>m@

otal 7 .C>m@

,actored load (wu) 7 =0J x 7 =30J .C>m@

m 7 2222

&/24. +L- 7 &0&@ x =0%@ 7 =0=3 .Cm

Fimiting moment4 ulim7 0=3&0fc.0b0d@7 %@0= .Cm

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u < ulim0 he section is underIreinforced0

X 7 u>b0d@Ast7 @MMmm@

inimum Ast7 0=@L of c>s 7 @mm@

Astalong longer s*an 7 H x Ast7 0M x @3M 7 =mm@< @ mm@

rovide &mm Y @mm c>c#hec. for shear (factored)2

Uu72

. xu l-7 =30J x > @ 7 @M .C

X0Zv7"d

#u7

1(01000

102)

x

x7 0=& C>mm@

Zcfor =Ast>bd 7 (= x @MM)>(= x =J) 7 0=&JC>mm@+

0=J 0@&

0@J 03%

0=&J 03&

As *er #l0@0 of :!J%I@4

X =03 for d 7 =Jmm

X = for d 7 3mm

X ((=0=@% ? 0=33) 7 =0@%) for d 7 =Mmm

X0 Zc 7 =0@% x 03& 7 03&& C>mm@$ Zv(0=& C>mm

@)

!ection is safe in shear0

0 :n the above *roblem4 design the slab if all the su**orts are fixed0

,or four edges sim*ly su**orted condition4

m 7 tan

/4*

. 22

%L-4

where4 tan [ 7

24(.1

2

2

+

H 7 0M4 N 7 0%%4 tan[ 7 0&4 [ 7 3&0%Jo

Assume s*an>de*th 7 @& Eff0de*th 7 >@& 7 =@0&%mm

rovide d 7 =Jmm4 ' 7 =Mmm

Foading on slab2

!elf weight 7 0=M x @J 7 0@J .C>m@

Five load 7 .C>m@

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,loor finish 7 0MJ .C>m@

otal 7 .C>m@


m 7 tan

/

4*

. 22

%L-7

).0

*01.0/

4*

"(.1 22x7 0@& .Cm

Fimiting moment4 ulim7 0=3&0fc.0b0d@7 %@0= .Cm

u< ulim0 he section is underIreinforced0

X 7 u>b0d@Ast7 =MJmm@


Astalong longer s*an 7 H x Ast7 0M x =MJ 7 =@@0Jmm@< @ mm@

rovide &mm Y @mm c>c bothways in0 Ast+

#hec. for shear (factored)2 VCo changeW

Uu72

. xu l-7 =30J x > @ 7 @M .C

X0Zv7"d

#u7

1(01000

102)

x

x7 0=& C>mm@


0=J 0@&

0@J 03%

0=&J 03&

As *er #l0@0 of :!J%I@4

X =03 for d 7 =Jmm

X = for d 7 3mm

X ((=0=@% ? 0=33) 7 =0@%) for d 7 =Mmm

X0 Zc 7 =0@% x 03& 7 03&& C>mm@$ Zv(0=& C>mm

@)


J0 !olve the above *roblem if two long edges are fixed0

m 7

22

tan

24

xuL-

rovide &mm Y @mm bothways0

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riangular slab2

:) :sotro*ically reinforced E1uilateral triangular slab sim*ly su**orted along all edges

udl

7)2

2L-u

::) :sotro*ically reinforced E1uilateral triangular slab sim*ly su**orted along two

adKacent edges udl

72"

2

2

SinL-u

:::) :sotro*ically reinforced Right angled triangular slab sim*ly su**orted along all edges

udl

7"

2L-u

:U) :sotro*ically reinforced #ircular slab sim*ly su**orted along edges udl

,ailure ta.es *lace by formation of infinite number of *ositive yield lines running radially

from centre to circumference4 forming a flat cone at colla*se0

7"

2r-u

%0 'esign an e1uilateral triangular slab of side Jm4 isotro*ically reinforced and is sim*ly

su**orted along its edges0 he slab is subKected to a su*erim*osed load of 3.C>m@0 8se

@ concrete and ,e=J steel0

Assume s*an>de*th 7 @& Eff0de*th 7 J>@& 7 =M&0JMmm

rovide d 7 =&mm4 ' 7 @mm

Foading on slab2


Five load 7 3 .C>m@


otal 7 .C>m@


m 7

)2

2L-u 7 0%&& .Cm

Fimiting moment4 ulim7 0=3&0fc.0b0d@7 &0@ .Cm

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u < ulim0 he section is underIreinforced0

X 7 u>b0d@Ast7 M@0MJmm@


Ast< min Ast @ mm@+

rovide &mm Y @mm c>c#hec. for shear (factored)2

Uu72

. xu l-7 =30J x J > @ 7 330MJ .C

X0Zv7"d

#u7

1(01000

102)

x

x7 0=&MJ C>mm@


0=3 0@&

As *er #l0@0 of :!J%I@4

X =03 for d 7 =Jmm

X = for d 7 3mm

X ((=0 ? 0@) 7 =0@) for d 7 =&mm

X0 Zc 7 =0@ x 0@& 7 03M@ C>mm@$ Zv(0=& C>mm

@)


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UNIT I'

MIS,ELLANEUS STRU,TURES #STAIR,ASE- !LAT SLA- R, WALL(

'E!:GC /, !A:R#A!E

GE/ER:#AF E! /, !A:R#A!E2

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I "ased on loading and su**ort conditions2

o !*anning along transverse direction

#antilever staircase

!lab su**orted between stringer beams

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o !*anning along longitudinal direction

ost commonly ado*ted+

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Foading on staircase2

'ead load2

I !elf weight of slab

I !elf weight of ste*I read finish 0% = .C>m@+

Five load2

,or overcrowding J .C>m@

Co overcrowding 3 .C>m@

,or inde*endent cantilever state4 the following live load condition is also chec.ed2

!A:R#A!E !ACC:CG RAC!UER!EF

=) A straight staircase is made of structurally inde*endent tread slab cantilevered from a

R# wall0 Given4 the riser is =Jmm and tread is 3mm with width of flight =0Jm0 'esign a

ty*ical cantilever tread slab0 A**ly live load for overcrowding0 8se @ concrete and

,e@J steel0

Foading on the staircase (03m width)

'ead load2

!elf weight of tread slab 7 @J x 0=J x 03 7 =0=@J .C>m

,loor finish (0% .C>m@) 7 0% x 03 7 0=& .C>m

otal 7 =03J .C>m

'ead load moment4 '7 =03J x =0J@

> @ 7 =0%& .Cm

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Five load2 aximum of4

i) /vercrowding J .C>m@

F0F0 7 J x 03 7 =0J .C>m

F7 wl@

>@ 7 =0J x =0J@

> @ 7 =0% .Cm

ii)

F7 =03 x =0J 7 =0J .Cm

he maximum of the above two values is F7 =0J .Cm

otal moment 7 =0%& ? =0J 7 30@ .Cm

,actored moment 7 J0=3 .Cm

Effective de*th 7 =J (@ ? =>@) 7 =@J mm

!lab cover =Jmm to @ mm+

=

"&

'&d'&M

$!

st%

st%u..".0

.*).042.0...*).0 /R+ X 7

2"d

Mu a.e *t

from !=%

J0=3 x =%7 %J0@J x =30Ast 0=0Ast@

Ast7 @@ mm@

Co0 of bars 7st

st

a

'7

4$10.

2022

7 3

3 nos0 of =mm [ are *rovided at the to*0

'istribution steel2

! 0=JL of c>s 0=J>= x 3 x =J 7 %M0Jmm@

!*acing 7 = xst

st

'

a7 M0%M mm

rovide &mm Y 3 mm c>c

#hec. for shear2

Zv7 Uu> b0d

!hear force due to dead load 7 w0l 7 =03J x =0J 7 =0J& .C

!hear force due to live load4

i) =0J x =0J 7 @0@J .C_________________________________________________________________________________________________________________________

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ii) =03 .C

Uu7 @0@J ? =0J& 7 0@& .C

,actored Uu 7 %03=@ .C

As *er #l00@0=0=4 :!J% @4 Zcvalue is modified when thic.ness is less than 3mm

8*to a de*th of =Jmm4 X 7 =035hen de*th is $ 3mm4 X 7 =

'evelo*ment length 7"d

s

4

.7 J30=@J mm

rovide a develo*ment length of Jmm

roviding a bent4 length re1uired 7 J & 7 3Mmm

Zv 7d"

#u

.7 0=%& C>mm@

,or *t7d"

's

.

1007 0JL4

Zc7 =03 x 0@& 7 0%%C>mm@

Zv < Zc

Sence4 safe0

@) 'esign a waist slab ty*e staircase com*rising of a straight flight of ste*s su**orted

between two stringer beams along the two sides0 Assume an effective s*an =0Jm4 a riser of

=Jmm and tread of @Mmm0 Assume a live load of 3.C>m@0 8se @ concrete and

,e@J steel0 Assume mild ex*osure condition0

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he staircase is s*anning along the transverse direction0

he main reinforcement should be *rovided along the transverse direction and distribution

steel is *rovided at the to*0

:nclined length of one ste* 7 22 TR + 7 3&0&M mm _ 3mm

he loading on the slab is found for an inclined width of 3mm4 which is later converted for

=m length0

Assume l>d 7 3=J>d 7 3 d 7 Jmm

Assume d 7 %mm

' 7 % ? cover ? "ar dia0>@ 7 % ? @ ? =>@ 7 &Jmm

Foading on slab over each tread width2

!elf weight of slab 7 03 x @J x 0&J 7 0%JM .C>m

!elf weight of ste* 7 ^ x 0=J x 0@M x @J 7 0J%@J .C>m

read finish 7 0@M x 0% 7 0=%@ .C>m

Five load (3.C>m@) 7 0@M x 3 7 0&= .C>m

otal 7 @0=3J .C>m

he load @0=3J .C>m acts vertically downwards0 he load acting along the inclined slab is

the cosine value of the above0

@0=3J x cosQ 7 @0=3J (@M>3) 7 =0&%% .C>m

he distributed load for =m ste* along the inclined slab is =0&% x =>03 7 %0@ .C>m

,actored load 7 03 .C>m

7 wl@>& 7 03 x =0J@> & 7 @0J .Cm

=100020".0

2(0*).042.0"0..2(0.*).010(4.2 "

xx

x'x'x stst

Ast7 @J0M@ mm@>m

rovide &mm4 s*acing re1uired is @mm c>c

!*acing < 3mm and 3d 7 =&mm+

rovide &mm Y =&mm c>c

'istribution steel2

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! 0=JL of c>s

7 0=J>= x = x % 7 mm@

rovide %mm [4 s*acing re1uired 7 3=0=Jmm < (Jd 7 3mm)

rovide %mm Y 3mm c>c

!A:R#A!E !ACC:CG F/CG:8':CAFF

=) 'esign the staircase slab shown in figure0 he stairs are sim*ly su**orted on beams

*rovided at the first riser and at the edge of u**er landing0 Assume a floor finish of

0&.C>m@and a live load of J .C>m@0 8se @ concrete and ,e=J steel0 Assume mild

ex*osure condition0

F 7 3 ? =0J ? (0=J ? 0=J) 7 0&m

Assume F>d as @4

0&>d 7 @ d 7 @mm

' 7 @ ? @ ? =>@ 7 @%Jmm

Foading on going slab2

Fength of the inclination of one ste* is 22 TR + 4

5here4 R 7 =Jmm4 7 3mm

F 7 33J0=mm!elf weight of waist slab 7 @J x (0@%J x (033J>03) 7 M0 .C>m@

!elf weight of ste* 7 @J x x 0=J 7 =0&MJ .C>m@

,loor finish 7 0& .C>m@

Five load 7 J .C>m@

otal 7 =J .C>m@

Foading on going slab2

!elf weight of slab 7 @J x 0@%J 7 %0%@J .C>m@

,loor finish 7 0& .C>m@

Five load 7 J .C>m@

otal 7 =@0@J .C>m@

#onsidering =m stri*4

he staircase slab is ideali-ed as given below2

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RAx 0& (=J x 30J x 30MJ) (=@0@J x 0%MJ x =03J) 7

RA7 3J0% .C4 R"7 330== .C

o 7 RA> udl 7 3J0%>=J 7 @03M m

oment at @03Mm is4

3J0% x @03M (=J x @03M x @03M>@) 7 @0@3 .Cm

,actored moment 7 %303J .Cm

=

"&

'&d'&M

$!

st%

st%u..".0

.*).042.0...*).0 /R+ X 7

2"d

Mu a.e *t

from !=%

%303J x =%7 &%0%@J x =30Ast M0%0Ast@

Ast7 M&J0= mm@

rovide =mm Y =mm c>c

'istribution steel2

0=@L of c>s 0=@>= x = x @%J 7 3=&mm@

rovide &mm Y =Jmm c>c

#hec. for shear2

Zv 7 Uu>b0d

aximum shear force 7 3J0% (=J0& x 0@)+ x =0J 7 &0= .C

Zv72401000

101.4*

.

x

x

d"

#u = 7 0@ C>mm@

*t7 =0Ast>b0d 7 03@ML

,or 0@JL 03%

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,or 0JL 0& ,or 033L 003&

Zc 7 03& C>mm@

he Zc value is modified based on #l00@0=0= for ' 7 @%Jmm

,or ' 7 =Jmm4 =03

,or ' 7 3mm4 = ,or ' 7 @%Jmm4 =0MZc modified 7 =0M x 03& 7 0@% C>mm

@

Zv < Zc

Sence4 safe0

@) 'esign a dog legged staircase having a waist slab for an office building for the following

data2

i) Seight between floor 7 30@m

ii) Riser 7 =%mm

iii) read 7 @Mmm

iv) 5idth of flight is e1ual to the landing width 7 =0@Jm

FF 7 J .C>m@4 ,, 7 0% C>mm@

Assume the stairs to be su**orted on @3mm thic. masonry walls at the outer edges of the

landing *arallel to the risers0 8se @ concrete and ,e=J steel0

Cote 2 "ased on riser4 number of ste*s is found0 "ased on tread4 length of staircase is found0

Co0 of ste*s 7 30@>0=% 7 @

= numbers of ste*s are used for first flight and other = to the second flight0

Foading on going2

!elf weight of waist slab 7 @J x 0@&3 x (03=3&J>0@M) 7 &0@@ .C>m@

!elf weight of ste* 7 @J x x 0=% 7 @ .C>m@

read finish 7 0% .C>m@

Five load 7 J .C>m@

otal 7 =J0&@ .C>m@

Foading on landing slab2

!elf weight of slab 7 @J x 0@&3 7 M0MJ .C>m@

l>d 7 @ J0=%>d 7 @

d 7 @J&mm

' 7 @J& ? @ ? =>@ 7 @&3mm

Fength of inclination of one ste*4

R 7 =%mm4 7 @Mmm4 F 7 3=30&Jmm

!elf weight of slab 7 @J x 0@&3 7 M0MJ .C>m@

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,, 7 0% .C>m@

FF 7 J .C>m@

otal 7 =@0%MJ .C>m@

RAx J0=% (=@0%MJ x =03%J x 0MMJ) (=J0&@ x @03 x @0J&) (=@0%MJ x =03%J x 0%&MJ) 7

RA 7 3%0J .C

R" 7 3%0J= .C

aximum moment at centre 7 3%0J x @0J& (=@0%MJ x =03%J x (0%&@J x =0@=J) (=J0&@ x=0@=J@>@) 7 0%% .Cm

,actored moment 7 M0 .Cm

=

"&

'&d'&M

$!

st%

st%u..".0

.*).042.0...*).0

b 7 =mm4 d 7 @J&mm

M0 x =%7 30=J x =3Ast M0 Ast@

Ast7 &%&0 mm@

rovide =@mm [ Y =3mm c>c

'istribution steel2

0=JL of c>s 7 0=J>= x = x @&3 7 @0J mm@

&mm Y ==mm c>c

#hec. for shear2

Zv 7 Uu>b0d

aximum shear force 7 3%0J (=@0%MJ x 0@J&)+ x =0J 7 0& .CZv 7 0=3 C>mm

@

Zc2

t 7 =0Ast>b0d 7 033%L

,or *t 7 0@JL 03%

,or *t 7 0JL 0&

,or *t 7 033%L 0

Zc7 0 C>mm@

odification factor4

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,or ' 7 =Jmm4 =03

,or ' 7 3mm4 =

,or ' 7 @&3mm4 =03

Zc modified7 =03 x 0 7 0=@ C>mm@

Zv < Zc

!afe in shear0

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REA' R:!ER !A:R#A!E

Actual analysisheory of *lates

l>d limited to @J

!u**ort conditions2

i) ransverse direction !tair is s*anning along transverse directionii) Fongitudinal direction

iii) !tair slab s*anning longitudinally and landing slab su**orted transversely

:n read Riser stair s*an by de*th ratio is ta.en as @J and the loading on the folded slab

com*rising the tread and riser is ideali-ed as a sim*ly su**orted slab with loading on

landing slab and going similar to a waist li.e slab0 he loading on folded slab includes4

i) self weight of tread riser slab

ii) floor finish

iii% live load J .C>m@(overcrowded)4 3 .C>m@(Co overcrowding)

Cote2

,or staircase s*anning longitudinally where the landing is su**orted along the transverse

direction only0 5hile finding the effective length along the longitudinal direction only half the

length of the landing slab is considered0 here is no change in the loading of going slab0

"ut the loading on landing slab is half (waist ty*e and treadIriser ty*e)0 he landing slab is

designed se*arately for the full load on landing *lus half the loading from going slab0

=) 'esign a dog legged staircase having a treadIriser slab for an office building for the

following data2

i) Seight between floor 7 30@m

ii) Riser 7 =%mm

iii) read 7 @Mmm

iv) 5idth of flight is e1ual to the landing width 7 =0@Jm

FF 7 J .C>m@4 ,, 7 0% C>mm@

Assume the stairs to be su**orted on @3mm thic. masonry walls su**orted only on two

edges *er*endicular to the risers0 8se @ concrete and ,e=J steel0

he length of the landing slab is halved while finding the effective length along the

longitudinal direction since the staircase is su**orted only on the landing slab along the

transverse direction0

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Effective length 7 @03 ? =0@J 7 30%&m

Assume l>d 7 @

CE 71 - RC&BM Monograph

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