LIST OF SYMBOLS

Mx - Moment in shorter direction

My - Moment in shorter direction
d - Effective depth
D - Overall depth
Ast - Area of Steel
P - Load
Wu (or) Pu - Design load
Mu - Design moment
Asc - Area of concrete
fy - Characteristic strength of steel
fck - Characteristic strength of concrete
B.M - Bending Moment
b - Breadth of beam
D - Overall depth
Vus - Strength of shear reinforcement
L - Clear span
Le - Effective span
N.A - Neutral Axis
MF - Modification factor
Q - Angle of repose of soil
M - Modular of rupture
τc - Permissible shear stress in concrete
τv - Nominal shear stress
1. DESIGN DATA
1.1 SLABS:
The most common type of structural element used to cover floors and roofs of
building are reinforced concrete slabs of different types. One way slabs are
those supported on the two opposite sides so that the loads are carried along
one direction only. Two way slabs are supported on all four sides with such
dimensions such that the loads are carried to the supports along both
directions.
If Ly/Lx < 2, then the slab is designed as two way slab
If Ly/Lx >2, then the slab is designed as one way slab.
Where, Ly = longer span dimension of the slab.
Lx = shorter span dimension of slab.
1.1.1 DESIG_ OF SLAB
Dimensions:
Lx =3.2

Ly =5.5

Span ratio =5.5 /3.2

=1.1<2

ie, Two way slab

Assume,

Overall depth =150-20

D =130mm
Load calculation

Live load =10kN/m2

Light patrician =1kN/m2
D.L =(0.15*25)

=3.75kN/m2
Floor finish =0.75kN/m2
Total load =15.5kN/m2
Factored load =1.5*15.5

= 23.25kN/m2

1) Lx =3.2+(0.23/2)+(0.23/2)

=3.43
2) Ly =3.55+(0.23/2)+(0.23/2)
=3.78

Lx =3.33

Ly =3.78

Bending moment

Mx =αx wlx2
My =αy wlx2
Lx / Ly =1.1

Lx =0.074

Ly =0.061

Mx =0.074*23.25*3.432

=20.24kNm

My =0.061*23.25*3.432

=16.69kNm
Check for depth

Mulim =0.138fckbd2
20.24*103 =0.138*20*1000*d2
D =85.63
D < 130mm

Hence the effective depth selected is sufficient to resist the design

Ultimate moment.

Find the spacing

Use 10mm dia bars

Spacing =(1000*π*102/4)/465.86

=168.59mm

Provide 150 mm c/c spacing

Find Ast provided

Ast provided =1000ast/spacing

=(1000*π*102/4)/150*4

=523.59mm2
For longer span

16069*106 =0.87*415* Ast *130[1-( Ast *415/20*1000*130)]

Ast =378mm2
Using 10mm dia bars

Spacing =(1000*π/4*102)/378
=207mm.

Adopt the spacing 200mm

Checks,
Ast min =0.12% of c/s of Fe 415

=0.12/100[1000*150]

=180mm2
Ast min< Ast for longer and shorter span,

Hence, provide 10mm dia bars @150mmc/c

Ast =465.86mm2[shorter span]

Hence, provide 10mm dia bars @ 200mm c/c

Ast =378mm2[longer span]

Check for shear

Shear stress

τv =Vu/bd

Vu =wl/2

=23.25*3.43/2=39.87Kn

τv =39.87*103/1000*100

τv =0.398N/mm2
Pt =100 Ast pro / bd

=100*523.59/1000*130
To find τc
From IS456:2000,

τc = 0.432

kτc > τv
k*τc=1.3*0.43

=0.5616

0.398 > 0.5616N/mm2

Hence it is safe.

Check for deflection

(L/d) basic =20

Pt =100 Ast pro/bd

=100*523.59/1000*130

=0.4

Fs =0.58*415*523.59/465.86

=270

kc =1

kf =1

kt =1.2

(L/d)max =(L/d)basic*kt*kc*kt
=20*1.3*1*1
=26
(L/d)act =3200/130=24.16

(L/d)act < (L/d)max

Hence safe against deflection.

Check for control

Reinforcement provided is more than, the minimum % of c/s area

Ast =(0.12/100)*1000*150
=180mm2
Spacing of main reinforcement should not be greater than 3d
ie, 3*130 =390mm

Diameter of reinforcement should be less than D/8

150/8 =18.75 Hence cracks will be within safe permissible limits

Torsion reinforcement at corner

Area of torsion steel at each of the corners in 4 layers is computed as
=0.75* Ast along shorter span

=0.75*523.59

=393mm2
Length cover which torsion steel is provided

=1/5*shorter span

=1/5*3200
=640mm

Using 6mm dia bars

Spacing =1000ast/ Ast
=(1000*π*62/4)393

=71.9mm
Provide 6mm bars at 100mm c/c for length and 640mmat all corners in 4
layers

Reinforcement in end strips

Ast =0.12% of c/s

=180mm2
Assume 10mm dia bars

Spacing =(1000*π/4*102)/180

=436 > 300

As per code spacing should not exceed 300mm

Provide 10mm dia bars at 300mm c/c

Ast =(1000*π/4*102)/300

Ast =262mm2

1.2 BEAMS
Beams are defined as structural members subjected to transverse load that
caused bending moment and shear force along the length.
The plane of transverse loads is parallel to the plane of symmetry of the cross
section of the beam and it passes through the shear centre so that the simple
bending of beams occurs. The bending moments and shear force produced by
the transverse loads are called as internal forces.
1.2.1Types of beams

Depending upon the supports and end condition, beams are classified
asbelow.

1. simply supported beams

2. over hanging beams
3. cantilever beam
4. fixed beam

The reinforced concrete beams, in which the steel reinforced is placed only on
tension side, are known as singly reinforced beams, the tension developed due
to bending moment is mainly resisted by steel rein forcementand
compression by concrete. When a singly reinforced beam needs considerable
depth to exist large
Bending moment, then the beam is also reinforced in the compression zone.
The beams having reinforcement in compression and tension zone is called as
doubly reinforced beam.

1.2.2 DESIG_ OF L-BEAMS

Dimensions

c/c of support = 3.2+(0.3/2)+(0.3/2) = 3.5m

Thickness of slabs = 150 mm

fy = 20 N/mm2
fck = 415 N/mm2
Width of beam = 300 mm

Overall depth = 300 mm

Effective cover = 25 mm

Effective depth = 300-25-10=265mm

Effective span

a) c/c of supports = 3.2 +(0.3/2) +(0.32/2) = 3.5 m

b) Clear span + d = 3.2 +0.265 = 3.465m

Hence, l = 3.465 m

Load calculation

Dead load of slab = (3.465/2)*0.15*25 = 6.5 kN/m

Floor finish = 0.75*(3.465/2) = 1.3 kN/m

Self weight of rib = 0.3 *0.15 *25 = 1.125 kN/m

Live load = 4*(3.465/2) = 6.93 kN/m

Total load = 16.855 kN/m

Effective flange width

a)bf = (Lo/12)+bw+3Df = 952.125 mm

c) bf = bw +0.5 times spacing b/w ribs = 1900 m

Ultimate BM and SF
At support,

Mu = 1.5 * wl2/12 = 25.3 kNm

Vu = 1.5 * wl / 2 = 43.8 kN

At centre of span section,

Mu = 1.5 * wl2 / 24 = 12.65kNm

Vu = 1.5 * wl / 2 = 43.8 kN

Torsion moment produced due to dead load of span and live load on it=
working load parameter-rib wt = 16.855-1.125=15.73 kN/m
 Ultimate load on slab = 15.73 *3.465 * 1.5 = 81.8 kN

Total ultimate load = 82/2 =41 kN

Distance of centroid of SF from the centre line of the Beam=(952.125/2)-

150

= 326.06mm

Ultimate tortional moment = 4 * 103 *326.06= 13.37 kNm

Equivalent BM and SF

According to IS456 2000 clause 41.4.2

Mel = Mu +Mt
Mt = Tu*(1+D/b)/1.7 = 15.73 kNm

Mel = 13.37 + 15.73 = 29.09 kNm

Equivalent SF

Ve = Vu + 1.6(Tu/b)

=115.1 kN

Main reinforcement

Mu (lim) = 0.138*fck*bd2 = 58.15 kNm

Mel < Mu (lim)

Hence the section is under reinforced
To find Ast
Mu = 0.87*fy* Ast *d[1-( Ast *fy/bd*fck)]

Ast = 332.83mm2
20mm dia rods are used

Ast pro = 628.32 mm2

Ast min = 0.85*bw*d/ fy = 162.83 mm2
Assume 20mm dia bars, Ast pro = 628 mm2
Provide 2 nos of 20mm dia bars @ side face

Reinforcement Shear reinforcement

τve =Ve/ bw *d = 1.45 N/ mm2

Pt= (100* Ast)/( bw *d) = 0.79

Ref table 19 of IS456 2000

τc=0.56N/ mm2
Hence shear reinforcement are required using 10mm dia 2 legged
stirrups with side cover 25mm top+ bottom cover of 25mm

b1= 300-25-25 = 250mm

d1= 300-25-25 = 250mm

Asv= 157 mm2

σc = Asv *0.87*fy/ (τv – τc)*b = 214.6

Provide 10mm dia 2 legged stirrups @200mm spacing

Check for deflection

(L/d)max = (L/d)basic *kt * kc* kf

(L/d)basic = 20 [for simply supported]

(L/d)max = 20*1*1*1.04 = 20.8

(L/d)actual = 3200/300 = 10.66

(L/d)max > (L/d)actual

Hence the design is safe

1.2.3 DESIG_ OF T- BEAM

Dimensions

Slab thickness =150mm

c/c of support =3.2+(0.3/2)+(0.3/2)

=3.5m

fy =20N/mm2
fck =415N/mm2
Cross sectional dimension

Width of beam =300mm

Overall depth =300mm

Effective cover =25mm

Effective depth =300-2-10

=265mm
Effective span

1. c/c of support =3.2+(0.3/2)+(0.3/2)

=3.5m

2. clear span+depth =3.2+0.265

=3.465m

Load calculation

Dead load of slab =(3.465/2)*0.15*25

=6.5kN/m

Floor finish =0.75*(3.465/2)

=1.3kN/m

Self weight of rib =0.3*0.15*25

=1.125kN/m

Live load =4*(3.465/2)

=6.93kN/m

Light partition =1kN/m

Total load =16.855kN/m
Ultimate moment and shear

Mu =1.5wl2/8

=(1.5*16.855*3.4652)/8

=37.935kN/m

Vu =wl/2

=(16.855*3.465)/2

=43.8kN/m
Effective width of flange

Refer page no 36 clause 23,

1. bf =L0/b+bw+6Df
=(3.465*0.7)/6+300+(6*150)

=1604.25mm

2. c/c of rib =3000-(300/2)-(300/2)

=2700mm

Ie, bf =1604.25mm

Moment capacity of flange

Assume N.A lies with in the flange

Xu(max)=Df , b=bf
Mu(limit) =(0.36*Xu(max))/d *(1-(0.42Xu(max)/d))*(bd2fck)

=0.36*(150/265) *[1-(0.42*150)/26]*(1604.25*2652*20)

=349.98kNm

Mu < Mu(limit)
Hence the section is under reinforced.
Since the section should design as a singly reinforced beam.

Find Ast
Mu =0.87fy Ast d [1-( Ast fy /bf d fck)]

37.935*106=0.87*415* Ast *265*[1-( Ast *415/1604*265*20)]

Ast =404.46mm2
Check for Ast min
Ast min/bw d =0.85/fy

Ast =(0.85*300*265) /415

=162.83mm2
Ast > Ast min
Ast =404.46mm2
N*πd2/4 =404.46

N =2 nos.

Ast provide,

Provide 2 nos of 20mm dia bars

=2*π202/4

=628mm2
And two longer bars of 12mm dia on the compression face .

Shear reinforcement
τv =Vu/bw d

=43.8*103/300*265

=0.55

Pt =100 Ast /bw d

=(100*π/4*202*2)/300*265

=0.79

τc =0.56+((0.62-0.56)/(1-0.75))*(0.79-0.75)
=0.57

τv<τc
Minimum shear reinforcement in the form of stirrups shall be provided.

Design of shear reinforcement

Asv/b Sv =0.4/0.87*fy
Sv =302.47mm

The spacing should not exceed 300mm

Sv =300mm

Provide 8mm dia bars 2 legged stirrups at 300mmc/c.

Check for deflection control

(L/d)max =(L/d)basic*kt*kf*kc
(L/d)basic =20*0.8

=16

fsc =0.58*fy* Ast req/ Ast pro

=0.58*415*404.46/628

fsc =155

kt =1.5,pt=0.78,kf=1,kc=0.8

(L/d)act =3200/265

=12.075

(L/d)max =16*1.5*1*0.8

=19.2
(L/d)act < (L/d)max .
Hence safe.

1.2.3 DESIG_ OF BI AXIALLY LOADED COLUM

Dimension

b = 450mm

D = 600mm

fck =20 N/mm2

fy = 414 N/mm2
Pu = 590.6 kN

Mux= 150 kN

Muy= 106 kN

Reinforcements

Reinforcements are distributed equally on four sides

As a trial adopt percentage of reinforcement in the CS as p =1%
As = pbD/100 = 1*450*600/100

= 2700mm2
Provide 10 bars of 20mm dia on each face
As = 10*π*202/4

= 3141.6mm2
P = (100*3141.6)/(450*600)

= 1.16

p/ fck = 1.16/20 =0.058

d`= 40+10 = 50mm

Pu/ fck bD = (590.6*103)/(20*450*300)

= 0.12

d`/D = 50/600 =0.08

from chart 44 of SP16

Mu/ fck bD2 = 0.09

For moments about minor axis yy

b= 450mm

d`=40+10=50mm

d`/D = 50/450 =0.111

Pu/ fck bD = 0.12

From chart 44 for d`/D=0.15

Mu/ fck bD2 = 0.09

Muy1=0.09*20*600*4502
= 218.7 kNm

Puz= 0.45 fck Ac+0.75 fy As

=0.45*20[(600*450)-314.6]+[0.75*415*3141.6*10-3]

= 3379.55kN

Pu/ Puz = (590.6/3379.55)=0.175

According to IS456 clause 39.6

αn =1.04

(Mux/ Mux1) αn +( Muy/ Muy1) αn = 0.97<1

Hence the section is safe

Design of lateral ties
Dia of lateral ties not less than 6mm and not greater than 16mm
Take 8mm dia ties Spacing should not be greater than 300mm or 16φ=16*20=
320mmHence provide 8mm φ bars @ 300mm c/c

1.2.4. DESIG OF FOOTIG

Dimension

Factored load, Pu = 1200 kN

Size of column = 400 × 300 mm

SBC of soil = 200 kN/m2

fck = 20 N/mm2
fy = 415 N/mm2
Size footing

Load on column = 1200 kN

Weight of footing and backfill at 10% = 120 kN

Area of footing =(1200+120)/(1.5*200)

= 4.4 m2
Adopt 2.5m×2m rectangular footing

Net soil pressure at ultimate load is given by,

qu = 1320/(2.5*2)

= 264 kN/ mm2

One way shear

Critical section is at a distance ‘d’ from the column face
Factored shear force, Vu1 = (0.264*2500)(1050-d)
= 660(1050-d)

Assuming percentage of reinforced cement, Pt = 0.25%

For M20 grade concrete, from table 19 IS456:2000
τc = 0.36 N/mm2 one way shear resistance, Vc1 = 0.36*2500*d
= 900d Equating both,660(1050-d) = 900dHence d = 442.3 mm

Two way shear

Assuming effective depth = 443 mm
Two way shear resistance at a critical section (d/2) from face of
Column, Then, Vu2 = 0.264{(2500*2000)-[(400+d)(300+d)]}

= 0.264{(2500*2000)-[(400+443)(300+443)]}

= 1154.6 kN Two way shear resistance, Vc2 is computed on,

Vc2 = Ks τc [2(400+d)+2(300+d)]dKs
= 1τc
= 0.25√k

= 1.118 N/mm2Hence,

Vc2 = 1565.2d+4.47d2Equating both,1154644=1565.

2d+4.47 d2d= 362 mm

Therefore one way shear is critical

Adopt effective depth = 450mm

Overall depth = 550mmUltimate moment at column face =

0.264*2000*10502/2

= 291.06 kNm291.06*106 = 0.87fyAstd[1-(Astfy/bdfck)]

291.06*106 = 162472.5 Ast – 7.49 Ast2Ast = 1970.4mm2
Pt = 100 Ast/bd

=(100*1970.4)/(1000*450)
= 0.43>0.25

Assuming 60mm dia bars,

Spacing = 1000ast/ Ast

=1000*π*162/(4*1970.4)

=105mm

Hence, provide 16mm dia bars @100mm c/c in both directions.